Introduction: Structural Analysis

INTRODUCTION: STRUCTURAL ANALYSIS

! Deflected Shape of Structures

! Method of Consistent Deformations

! Maxwell’s Theorem of Reciprocal Displacement

1 Deflection Diagrams and the Elastic Curve P

fixed support ∆ = 0 θ = 0

-M

roller or rocker θ support ∆ = 0 inflection point

-M 2 P θ pined support ∆ = 0

-M

3 inflection point • Fixed-connected joint P

fixed-connected inflection point joint

Moment diagram

4 • Pined-connected joint pined-connected P joint

Moment diagram

5 P

inflection point

Moment diagram

6 P1

C A B D

M +M x -M

inflection point

7 P1

+M x -M

inflection point

8 Method of Consistent Deformations Beam 1 DOF P

Ax = 0 B A C =

A RB y P

B A C ∆´B +

f BB x RB B A C

∆´ + f R = ∆ = 0 B BB B B 1 9 Beam 2 DOF Compatibility Equations. w ∆´1 + f11R1 + f12R2 = ∆1 = 0 3

∆´2 + f21R1 + f22R2 = ∆2 = 0 4 1 2 5 = 0 w ∆´ f f R 1 11 12 1 ∆1 0 + = ∆´ f12 f ∆ 2 22 R2 2 ∆´1 ∆´2 +

f11 f21 ×R1

1 +

f12 f22 ×R2 1

10 Beam 3 DOF Compatibility Equations.

P1 w P2 θ´ + f M + f R + f R = θ = 0 1 6 1 11 1 12 2 13 3 1 = ∆´2 + f21M1 + f22R2 + f23R3 = ∆2 = 0 4 2 3 5 ∆´ + f M + f R + f R = ∆ = 0 P1 w P2 3 31 1 32 2 33 3 3 0 θ´ f f f M 1 11 12 13 1 θ1 0 ∆´2 θ´ ∆´3 + 1 ∆´2 f21 f22 f23 R f f + 2 = ∆2 0 1 11 21 f31 ∆´3 f31 f32 f33 R3 ∆3 ×M1 + f f12 22 f32

×R2

1 + f33 f13 f23

×R3 1 11 Compatibility Equation for n span. Equilibrium Equations

∆´1 + f11R1 + f12R2 + f13R3 = ∆ 1 ∆´ + f R + f R + f R = ∆ 2 21 1 22 2 23 3 2 Fixed-end force matrix

∆´n + fn1R1 + fn2R2 + fnnRn = ∆n [Q] = [K][D] + [Qf] If ∆1 = ∆2=……= ∆n = 0 ;

Stiffness matrix ∆ ´1 f11 f12 f13 R1 0 Solve for displacement [D]; ∆´2 f21 f22 f23 R + 2 = 0

∆´ f n 1 f n 2 f -1 f n n n R n 0 [D] = [K] [Q] - [Q ] or [∆´] + [f][R] = 0 1 k = ij f [R] = - [f]-1[∆´] ij

[fij] = Flexibility matrix dependent on 1 1 1 , , EI EA GJ 12 Maxwell’s Theorem of reciprocal displacements

1=i 2=j 1 EI A B f = f f11 = fii 21 ji dx f = m M ij ∫ i j EI

Mi = mi dx = m m ∫ i j EI

dx 1 f = m M A B ji ∫ j i EI

f22 = fjj f = f dx 12 ij = m m ∫ j i EI

M = m j j fij = f ji

13 Example 1

Determine the reaction at all supports and the displacement at C.

50 kN

B A C 6 m 6 m

14 SOLUTION

• Principle of superposition 50 kN

MA B A C 6 m 6 m RA RB =

50 kN

∆´B

x R fBB B

-----(1) 1 kN Compatibility equation : ∆'B + f BB RB = 0 15 • Use conjugate beam in obtaining ∆´B and fBB 50 kN A B 300 kN•m Real beam

∆´B 50 kN 6 m 6 m 9000/EI 6 + (2/3)6 = 10 m Conjugate beam

900/EI

/EI 900/EI 300 ∆´B = M´B = -9000/EI ,

12 kN•m fBB Real beam

1 kN 1 kN 72/EI 12 /EI (2/3)12 = 8 m 576/EI Conjugate beam f = M´´ = 576/EI, BB B 72/EI 16 • Substitute ∆´B and fBB in Eq. (1)

9000 576 + ↑: − + ( )R = 0 EI EI B

RB = +15.63 kN, (same direction as 1 kN)

50 kN 300 kN•m

∆´B 50 kN +

f 12 kN•m BB x RB = 15.63 kN

1 kN = 1 kN 50 kN B C 34.37 kN•m A 34.37 kN 15.63 kN 17 Use conjugate beam in obtaining the displacement 50 kN 113 kN•m 6 mC 6 m Real Beam ∆ B A C 15.6 kN 34.4 kN 93.6

M (kN•m) x (m) 3.28 m 6 m 12 m

-113 93.6/EI

Conjugate Beam

-113/EI 223/(EI) 281/(EI) M´ 281 223 776 C M ' = (2) − (6) = − C EI EI EI

776 V´ ∆C = M 'C = − , ↓ C 2 m 4 m 223/(EI) EI 18 Example 2

Determine the reaction at all supports and the displacement at C. Take E = 200 GPa and I = 5(106) mm4

10 kN 3EI 2EI A B C 4 m 2 m 2 m

19 10 kN 3EI 2EI A B C 4 m 2 m 2 m =

10 kN

∆´B +

fBB x RB 1 kN

Compatibility equation:

∆´B + fBBRB = ∆B = 0

20 • Use conjugate beam in obtaining ∆´ B 10 kN 40 kN•m 3EI 2EI A B Real Beam C 4 m 2 m 2 m 10 kN

10 V (kN) 10 + x (m)

M x (m) (kN•m) -

177.7/EI Conjugate Beam

40/3EI = 13.33/EI 26.66/EI ∆´ = M´ = 177.7/EI B B 21 • Use conjugate beam for fBB 3EI 2EI 8 kN•m A B Real Beam C 1 kN 1 kN 4 m 2 m 2 m

V (kN) x (m) - -1 -1

8 4 + M x (m) (kN•m)

4/(3EI)=1.33EI 2.67 8 4/(2EI)=2EI = EI 3EI 60.44/EI

12/EI Conjugate Beam fBB = M´B = 60.44/EI 22 10 kN 3EI 2EI A B C 4 m 2 m 2 m =

10 kN

∆´B +

fBB x RB 1 kN Compatibility. equation:

∆´B + fBBRB = ∆B = 0 −177.7 60.44 + ↑: − + ( )R = 0 EI EI B

RB = +2.941 kN, (same direction as 1 kN) 23 • The quantitative shear and moment diagram and the qualitative deflected curve 10 kN 16.48 kN•m 3EI 2EI A B ∆ C C 7.06 kN 4 m 2 m 2 m 2.94 kN

7.06

+ V (kN) - x (m) -2.94 -2.94

11.76 M 2.33 m (kN•m) + x (m) - 1.67 m

-16.48 24 • Use the conjugate beam for find ∆C 10 kN 3EI 16.48 kN•m C 2EI A B Real beam ∆C 7.059 kN 4 m 2 m 2 m 2.941 kN 11.76 2EI 11.76 2.335 m 3EI Conjugate beam

1.665 m 16.48 3EI 3.263 EI 3.263 6.413 −18.85 M´ = (0.555) − (3.222) = ,↓ C EI EI EI

6.413 (1.665)/3=0.555 m EI −18.85 ∆C = M'C = = −18.85 mm,↓ 1.665+(2/3)(2.335) = 3.222 m (200×5) 25 Example 3

Draw the quantitative Shear and moment diagram and the qualitative deflected curve for the beam shown below.EI is constant. Neglect the effects of axial load.

5 kN/m

AB 4 m 4 m

26 SOLUTION

• Principle of superposition 5 kN/m

AB θ θ A 4 m 4 m B = 5 kN/m

θ´A θ´B

1 kN•m + × M A αAA αBA

+ 1 kN•m × M B α AB α Compatibility equations: BB

θ A = 0 =θ 'A + f AAM A + f AB M B − − − (1) θ = 0 = θ ' + f M + f M − − − (2) B B BA A BB B 27 • Use formula provided in obtaining θ´A, θ´B, αAA, αBA, αBB, αAB 5 kN/m

θ´A θ´B 4 m 4 m

3wL3 3(5)(8)3 60 θ ' = = = A 128EI 128EI EI 7wL3 7(5)(8)3 46.67 θ 'B = = = 1 kN•m 384EI 384EI EI 1 kN•m

α α αAB AA BA α 8 m 8 m BB

M o L 1(8) 2.667 M o L 1(8) 2.667 α = = = α BB = = = AA 3EI 3EI EI 3EI 3EI EI M L 1(8) 1.333 M L 1(8) 1.333 α = o = = α = o = = BA 6EI 6EI EI AB 6EI 6EI EI

Note: Maxwell’s theorem of reciprocal displacement, αAB = αBA 28 • Use conjugate beam for αAA, αBA, αBB, αAB

Real Beam 1 kN•m 1 kN•m Real Beam

α α α AA BA AB α (1/8) BB (1/8) (1/8) (1/8) 8 m 8 m

4/EI 4/EI 1/EI 1/EI

Conjugate Beam Conjugate Beam 2.67/EI 1.33/EI 1.33/EI 2.67/EI − 2.667 −1.333 α = V ' = α = V ' = AA A EI AB A EI

1.333 2.667 α = V ' = α = V ' = BA B EI BB B EI

29 5 kN/m

ABCompatibility equation θ θ A 4 m 4 m B 60 2.667 1.333 + + ( )M A + ( )M B = 0 = EI EI EI 5 kN/m 46.67 1.333 2.667 + ( )M + ( )M = 0 + EI EI A EI B 60 46.67 θ ' = θ 'B = A EI EI Solve simultaneous equations,

1 kN•m + M = -18.33 kN•m, + × M A A 1.333 MB = -8.335 kN•m, + 2.667 α BA = V 'B = α AA = EI EI

+ 1 kN•m

× M B 1.333 2.667 α AB = α = EI BB EI 30 MA = -18.33 kN•m,

MB = -8.335 kN•m,

5 kN/m

18.33 kN•m AB8.335 kN•m 4 m 4 m RA RB

18.33− 20(2) + R (8) −8.355 = 0 R = 3.753 kN, + ΣMA = 0: B B 3.753 R = 16.25 kN, + ΣFy = 0: RA + RB − 20 = 0 a

31 • Quantitative shear and bending diagram and qualitative deflected curve

5 kN/m

18.33 kN•m AB8.36 kN•m 16.25 kN 4 m 4 m 3.75 kN

16.25 V diagram 3.25 m -3.75

M 8.08 diagram 6.67 -8.36 -18.33

Deflected Curve 32 Example 4

Determine the reactions at the supports for the beam shown and draw the quantitative shear and moment diagram and the qualitative deflected curve. EI is constant.

2 kN/m

C A 4 m B 4 m

33 • Principle of superposition 2 kN/m

C A B 4 m 4 m Compatibility equations:

2 kN/m ∆'B + f 'BB RB + f 'CB RC = 0 − − − (1) ∆' + f ' R + f ' R = 0 − − − (2) C C BC B CC C ' A ∆ B 4 m B 4 m ∆'C

f ' f 'BB CB × RB C A 1 kN 4 m B 4 m

f 'CC f 'BC × RC C A 4 m B 4 m 1 kN 34 • Solve equation 2 kN/m

C Compatibility equations: A 4 m B 4 m 64 21.33 53.33 − + RB + RC = 0 − −(1) 2 kN/m EI EI EI B 149.33 53.33 170.67 C − + R + R = 0 − −(2) EI EI B EI C A 64 149.33 ∆' = − ∆'C = − B EI EI RB = 3.71 kN,↑ 21.33 53.33 f 'CB = RC = −0.29 kN,↓ f 'BB = EI EI × RB C A 1 kN

53.33 170.67 f ' = f 'CC = BC EI EI × RC C A 1 kN 35 • Diagram

M = 8(2) + 0.29(8) − 3.71(4) A 2 kN/m = 3.48 kN • m C A 4 m B 4 m

Ay = 8 + 0.29 − 3.71 = 4.58 kN 3.71 kN 0.29 kN

V (kN) 4.58 0.29 x (m) 2.29 m -3.42

M (kN•m) 1.76 x (m) -1.16 -3.48 Deflected shape x (m)

Point of inflection 36 Example 5

Draw the quantitative Shear and moment diagram and the qualitative deflected curve for the beam shown below. (a) The support at B does not settle (b) The support at B settles 5 mm. Take E = 200 GPa, I = 60(106) mm4.

16 kN

BC A

2 m 2 m 4 m

37 SOLUTION

• Principle of superposition 16 kN

BC A

∆B = 5 mm 2 m 2 m 4 m

16 kN =

∆´B +

fBB

× RB 1 kN

Compatibility equation : + ∆ B = 0 = ∆'B + f BB RB -----(1) : no settlement

+ ∆ = −0.005m = ∆' + f R -----(2) : with settlement B B BB B 38 • Use conjugate beam method in obtaining ∆´B 16 kN Real AB Real C beam beam ∆´ 12 kN B 4 kN 2 m 2 m 4 m

24 ∆´B M´ 16 diagram 16 32 24 24 72 EI EI EI EI EI Conjugate M´´B beam 40 V´´B 4 2 EI 4 2 40 4 56 2 m 4 m 3 3 3 3 EI EI 32 4 40 + ΣM = 0: − M '' + ( ) − (4) = 0 B B EI 3 EI 117.33 ∆' = M '' = − , ↓ B B EI 39 • Use conjugate beam method in obtaining fBB

fBB Real A C beam B 1 kN 0.5 kN 0.5 kN 4 m 4 m 4 2 4 m´ 3 3 m´´ diagram B -2 v´´B 4 4 − Conjugate EI EI beam + ΣMB = 0: 4 4 4 4 − m''B − ( ) + (4) = 0 4 − 2 4 4 EI 3 EI EI − − EI EI EI EI 10.67 f = m'' = ,↑ BB B EI

fBB

40 117.33 • Substitute ∆´B and fBB in Eq. (1) ∆' = − , ↓ B EI 10.67 f = ,↑ 117.33 10.67 BB EI + ↑ ; 0 = − + R , R = 11.0 kN, EI EI B B

16 kN

fBB

+ xRB = 11.0 kN 1 kN ∆´B 12 kN 4 kN 0.5 kN 0.5 kN

16 kN =

BCno settlement A

R 11.0 kN A = 6.5 kN RC = 1.5 kN

41 117.33 • Substitute ∆´B and fBB in Eq. (2) ∆' = − , ↓ B EI 10.67 117.33 10.67 f = ,↑ + ↑ ; − 0.005m = − + R BB EI EI EI B

(−0.005m)EI = −117.33 +10.67RB

(−0.005)(200×60) = −117.33+10.67RB

RB = 5.37 kN, 16 kN

fBB

+ xRB = 5.37 ∆´ 1 kN B 4 kN 12 kN 0.5 kN 0.5 kN

16 kN =

BCwith 5 mm settlement A

5.37 kN RA = 9.31 kN R = 1.32 kN C 42 • Quantitative shear and bending diagram and qualitative deflected curve

16 kN 16 kN

BC BC A A

∆B = 5 mm 6.5 kN 11 kN 1.5 kN 9.31 kN 5.37 kN 1.32 kN 2 m 2 m 4 m 2 m 2 m 4 m

6.5 V 1.5 9.31 diagram V diagram -9.5 -1.32 -6.69 M 13 18.62 diagram + M 5.24 - diagram + -6

Deflected Deflected Curve Curve ∆ = 5 mm B 43 Example 6

Calculate supports reactions and draw the bending moment diagrams for (a)

D2 = 0 and (b) D2 = 2 mm.

1 2 w 2 kN/m 3

4 m 6 m

44 Compatibility equation:

∆´2 + f22R2 = ∆2

1 2 2 kN/m 3 w Use formulations provided : wx ∆' = − (x3 − 2Lx2 + L3 ) 2 24EI 4 m 6 m (2)(4) 3 2 3 = = − (4 − 2×10× 4 +10 ) 24EI 2 kN/m 248 = − = −6.2mm, ↓ (200)(200) ∆´ 2 Pbx f = (L2 − b2 − x2 ) 22 6LEI + 1×6× 4 = − (102 − 62 − 42 ) f22 6×10EI 19.2 ×R2 = = +0.48 mm, ↑ (200)(200) 1

45 For ∆2 = 0 For ∆2 = 2 mm Compatibility equation: Compatibility.equation:

-6.2 + 0.48R2 = 0 -6.2 + 0.48R2 = -2 R2 = 12.92 kN R2 = 8.75 kN

1 2 w 2 kN/m 3 1 2 w 2 kN/m 3

12.92 kN 2.25 kN4 m 6 m 4.83 kN 4.75 kN 4 m8.75 kN 6 m 6.5 kN

7.17 4.75 5.5 2.25 2.415 m 3.25 m V + V + + (kN) x (m) - x (m) - - (kN) - 2.375 m-3.25 1.125 m -5.75 -4.83 -6.5 5.85 10.56 1.27 5.64 M + + x (m) + (kN•m) - M + 3 (kN•m) x (m) -7 46 APPENDIX Basic Beams: Single span

1 w

wL/2 wL/2 L/3 L/3 5wL4 384EI L

wL/2 V

- wL/2

wL2/8

47 • Find ∆1 by Castigliano’s 2 1 3 P w

L/2 L/2 wL P wL P + x x + 2 2 1 L 2 2 2

w M1 w M2

wL P x1 V + 1 x2 wL P 2 2 V2 + 2 2

+ ΣΜ = 0; wx 2 wL P M + 1 − ( + )x = 0 1 2 2 2 1 wx 2 wL P M = − 1 + ( + )x = M 1 2 2 2 1 2 48 wx 2 wL P M = − 1 + ( + )x = M 1 2 2 2 1 2

L / 2 ∂M dx L / 2 ∂M dx ∆ = ∆ = ( 1 )M 1 + ( 2 )M 2 1 max ∫ 1 ∫ 2 0 ∂P EI 0 ∂P EI

L / 2 ∂M dx = 2 ( 1 )M 1 ∫ 1 0 ∂P EI 0 2 L / 2 x − wx 2 wL P = ( 1 )( 1 + ( + )x )dx ∫ 1 1 EI 0 2 2 2 2

5wL4 = 384EI

49 Single span 1 P

P/2 P/2 PL3 L3 δ = = f P → f = 11 48EI 11 11 48EI

L /2 L /2 L P/2 V + - - P/2

PL/4

50 Double span wl 4 ∆ = 4 max 185EI 1 wl ∆max = w 185EI

0.375 wl 1.25 wl 0.375 wl

l = L/2 l = L/2 L = 2l 0.625 wl 0.375 wl + + - - -0.625 wl -0.375 wl

0.070 wl2 0.070 wl2 ++ - -0.125 wl2 51 Triple span ∆ = wl4 1 2 max w 145 EI

0.4wl 0.4wl 1.1wl 1.1wl l = L/3 l = L/3 l = L/3

0.6wl 0.4wl 0.5wl V + + + - - - - 0.4wl - 0.6wl -0.5wl

0.08 wl2 0.08 wl2 0.025 wl2 M + + + - - -0.1 wl2 -0.1 wl2 52 INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS

! Comparison Between Indeterminate and Determinate

! Influence line for Statically Indeterminate Beams

! Qualitative Influence Lines for Frames

1 Comparison between Indeterminate and Determinate

Indeterminate Determinate

1 D B E C 1 D B E C R A RA A A

2 Indeterminate Determinate

1 D B E C 1 D B E C A RA A RA

1 D B E C 1 D B E C V A VD A D

1 1 D B E C D B E C M A ME A E

3 Influence Lines for Reaction

Redundant R1 applied

1 4 2 j 3 1 Compatibility equation:

= f1 j + f11R1 = ∆1 = 0

1 R1 = − f1 j ( ) f11 ∆´1 = f1j 1 f R = ( j1 ) f 1 + jj f11

f11

×R1 1

fj1 4 Redundant R2 applied

1 4 2 j 3 1 Compatibility equation.

∆'2 + f22 R2 = ∆2 = 0 =

f2 j + f22 R2 = ∆2 = 0

1 1 R2 = − f2 j ( ) f22 f 2j + fjj f j2 R2 = ( ) f22 f22 fj2

×R2 1

5 Influence Lines for Shear

1 4 2 j 3

1 VE = ( ) f j4 f44 1

f44

fj4 1

6 Influence Lines for Bending Moment

1 4 2 j 3

11 1 M E = ( ) f j4 α 44

α44 11

fj4

7 Using Equilibrium Condition for Shear and Bending Moment • Influence line of Reaction

1 4 2 j 3

f41 f11 f j1 f11 =1 R1 = ( ) f11 R1 f11

1 f j1 f f 11 f 22 42 =1 f j2 f22 f22 f22 f j2 R2 R = ( ) 2 f 1 22

f 33 =1 f33 f R R = ( j3 ) 3 3 f f j3 33 f43 1 f33 8 f33 • Influence line of Shear • Unit load to the left of 4 x 1 1 1 1 4 2 j 3 M4

V4 R1

R1 R2 R3 + ↑ ΣFy = 0; R1 −1−V4 = 0

V4 = R1 - 1

R 1 • Unit load to the right of 4 1

1 M4 V V4 = R1 R 4 4 1 1 V4 + ↑ ΣFy = 0; R1 −V4 = 0 1

V4 = R1 9 1 V4 = R1 - 1 • Influence line of Bending moment

4 l • Unit load to the left of x 1 1 1 4 2 j 3 1 M l 4 V4 R1

R1 R2 R3 + Σ M4 = 0: M4 - 1 (l-x) - l R1 = 0 l1 l2 M4 = - l + x + l R1

R1 • Unit load to right of 4 1 1 M4 l V M = - l + x + l R R 4 4 4 1 1 1 1 1 + Σ M4 = 0: M4 - l R1 = 0 M4

M4 = l R1 10 M4 = lR1 Qualitative Influence Lines for Frames

I 1

Influence Line of VI

11 Maximum positive shear Maximum negative shear 1 1

Influence Line of MI

Maximum positive moment Maximum negative moment 12 Influence Line for MOF A D G MA

Ay Dy Gy 15 m 15 m

1.0

A y 13 A D H G

15 m 15 m

MA 1

14 G

AEBCD

1.0

MG 15 G

AEBCD

1 VG

1 VF

1 VH

16 Example 1

Draw the influence line for - the vertical reaction at A and B - shear at C - bending moment at A and C EI is constant . Plot numerical values every 2 m.

ABCD

2 m 2 m 2 m

17 • Influence line of RB

ABCD

2 m 2 m 2 m

f DB f f AB CB f BB f BB =1 f f BB BB f BB

18 • Find fxB by conjugate beam

ABCD 6 kN•m Real Beam

1 1 2 m 2 m 2 m

6 18 x EI EI 72 EI Conjugate Beam

2 EI x x EI 2EI x x3 72 18x 72 + − = M´x 6EI EI EI EI x 2x V´x 3 3 18 EI 19 x x3 72 18x ABCD f = M ' = + − xB x 6EI EI EI

2 m 2 m 2 m Point x (m) fxB fxB / fBB 72 B 0 1 EI 72/EI 37.33/EI 37.33 0 10.67/EI f D 2 0.518 BB EI fxB 1 10.67 C 4 0.148 EI 37.33 /72 = 0.518 10.67 /72 = 0.148 f 72 A 6 0 0 BB = =1 0 f BB 72

1 Influence line of RB 20 Influence line of RA

ABCD

2 m 2 m 2 m

1 kN fCA f DA f AA f f AA AA =1 f AA

21 • Find fxA by conjugate beam 1 kN A CD B 6 kN•m Real Beam

1 kN 2 m 2 m 2 m x 72 M ' = Conjugate Beam A EI 18 B' = y EI 18 6 EI EI x 2x 18x x3 − = M´ 3 3 EI 6EI x V´x x 18 B' = x x2 y EI EI 2EI

22 x 18x x3 ABCD f = M ' = − xA x EI 6EI

2 m 2 m 2 m Point x (m) fxA fxA / fAA

B 0 0 0

72/EI 61.33 /EI 1 kN 34.67 34.67 /EI D 2 0.482 fAA EI fxA 61.33 C 4 0.852 EI 72 /72 = 1.0 72 61.33 /72=0.852 A 6 EI 1.0 1 kN f 34.67 /72 = 0.482 AA =1 f AA Influence line of RA 23 Alternate Method: Use equilibrium conditions for the influence line of RA x 1

MA ABCD

+ ↑ ΣFy = 0; R R A 2 m 2 m 2 m B RA + RB −1 = 0

RB =1− RA R B 1 .518 0.148 RB 1

RA = 1- RB 1.0 0.852 1 kN 0.482

24 Using equilibrium conditions for the influence line of VC 1 x • Unit load to the left of C M MA ABCD C

VC R RA 2 m 2 m 2 m RB B

+ ↑ ΣFy = 0;

1 + RB +VC = 0 0.518 0.148 VC = - RB RB • Unit load to the left of C 1 1 x MC

0.852 V R 0.482 C B 1 + ↑ ΣFy = 0; VC +V −1+ R = 0 -0.148 C B

VC = 1 - RB 25 Using equilibrium conditions for the influence line of MA 1 x

MA ABCD

+ ΣM A = 0;

− M A −1(6 − x) + 6RB = 0 RA 2 m 2 m 2 m RB M A = −6 + x + 6RB

1 0.518 0.148 RB 1

MA -1.112 -0.892 1 26 Using equilibrium conditions for the influence line of MC 1 x • Unit load to the left of C

MA ABCD MC V 4 m C RB RA 2 m 2 m 2 m RB + ΣM C = 0;

− M C + 4RB = 0 1 0.518 MC = 4RB 0.148 RB • Unit load to the left of C 1 1 x

MC 1 4 m 0.592 V R C C B 0.074 + ΣM C = 0; MC − M C −1(4 − x) + 4RB = 0

MC = -4 + x + 4RB 27 Example 2

Draw the influence line and plot numerical values every 2 m for - the vertical reaction at supports A, B and C - Shear at G and E - Bending moment at G and E EI is constant.

ABG DE F C

2@2=4 m 4@2 = 8 m

28 Influence line of RA

ABG DE F C

2@2=4 m 4@2 = 8 m

f AA =1 f AA

1 f DA f f EA FA f AA f f AA AA

29 • Find fxA by conjugate beam

ABDE F C Real beam 1 1.5 0.5 4 m2 m 6 m

4/EI

Conjugate beam 4/EI

0 5.33 10.67 EI EI 4/EI 64/EI 0 64 f = M ' = 10.67 AA A EI 18.67/EI EI 30 4/EI x x 64/EI 2 1 Conjugate beam

4 m 8 m 18.67 5.33 EI 2 EI x1 x1 4EI 2EI x1 x 3 5.33x 1 − 1 = M´ 12EI EI x1 V´ x1 5.33 x1 2x1 2 EI x2 3 3 2EI x 64 2 x2 EI EI x 3 64 18.67 M´ = 2 + − x2 6EI EI EI V´ x2 18.67 2x2 x2 EI 3 3 31 x3 5.33x x x f = M ' = − , for B to C 2 1 xA x1 12EI EI x 3 18.67x 64 ABG DE F Cf = M ' = 2 − 2 + , for A to B xA x2 6EI EI EI

Point f 4 m2 m 6 m x (m) xA fxA / fAA C 0 64 0 0 28 EI 10 EI F − 2 EI -0.1562 fAA fxA 16 1 − −14 −16 −10 E 4 EI -0.25 EI EI EI 14 D 6 − -0.2188 EI f AA =1 Influence line of RA B 8 0 0 f AA 28 0.438 G 10 0.4375 EI 1 -0.219 -0.25 -0.156 64 A 1 12 EI 32 Using equilibrium conditions for the influence line of RB

1 x + ΣM C = 0; ABG DE F C −8RB + x −12RA = 0 x 12 RB = − RA RA RB RC 8 8 4 m2 m 6 m Point x (m) RA RB C 0 0 0 1 F -0.1562 0.485 0.438 2 -0.25 0.875 RA E 4 1 -0.219 -0.25 -0.156 D 6 -0.2188 1.078 1 B 8 0 1.078 0.875 0.4375 0.5939 0.59 1 0.485 G 10 A 1 0 RB 12 1 33 Using equilibrium conditions for the influence line of RC

1 x + ΣM B = 0; ABG DE F C 4RA −1(x −8) −8RC = 0 x RC = 0.5RA − +1 RA RB RC 8

4 m2 m 6 m Point x (m) RA RC C 0 0 1 1 F 2 -0.1562 0.6719 0.438 E 4 -0.25 0.375 RA 1 -0.219 -0.25 -0.156 D 6 -0.2188 0.1406 B 8 0 0 1 0.672 G 10 0.4375 -0.0312 0.1410.375 A 1 0 RC 12 -0.0312 1 34 • Check ΣFy = 0

1 x ABG DE F C + ↑ ΣFy = 0;

RA + RB + RC =1 RA RB RC R R R ΣR 4 m2 m 6 m Point A B C 1 C 0 0 1 1 0.438 F -0.1562 0.485 0.6719 1 RA 1 -0.219 -0.25 -0.156 E -0.25 0.875 0.375 1

1.08 D -0.2188 1.078 0.1406 1 1 0.875 0.59 0.49 B 0 1 0 1 RB G 0.4375 0.5939 -0.0312 1 1 1 0.672 A 1 0 0 1 0.1410.375 R C 35 -0.0312 1 Using equilibrium conditions for the influence line of VG

1 • Unit load to the left of G x x 1 ABG DE F C

A MG V R G 4 m2 m 6 m A

+ ↑ ΣFy = 0; RA −1−VG = 0 1

0.438 VG = RA - 1

RA -0.219 -0.25 -0.156 1 • Unit load to the right of G

0.438 A MG V V R G 1 G A -0.562 -0.219 -0.25 -0.156 VG = RA 36 Using equilibrium conditions for the influence line of VE

1 • Unit load to the left of E x

ABG DE F CME V E RC

+ ↑ ΣFy = 0; RC +VE = 0 4 m2 m 6 m

VE = - RC 1 0.672 • Unit load to the right of E 0.1410.375 R C 1 x -0.0312 1

ME 0.625 V R 0.328 E C 0.0312 1 + ↑ ΣFy = 0; VE −1+ RC = 0 VE -0.141 -0.375 VE = 1 - RC 37 Using equilibrium conditions for the influence line of MG

1 • Unit load to the left of G x x 1 ABG DE F C

A MG 2 m VG RA 4 m2 m 6 m

+ ΣM G = 0;

M G +1(2 − x) − 2RA = 0 1 M = -2 + x + 2R 0.438 G A

RA • Unit load to the right of G 1 -0.219 -0.25 -0.156 A MG 2 m 0.876 VG 1 RA

+ ΣM G = 0; MG M − 2R = 0 -0.312 G A -0.438 -0.5 MG = 2RA 38 Using equilibrium conditions for the influence line of ME 1 • Unit load to the left of E x ABG DE F C 4 m ME V E RC 4 m2 m 6 m + ΣME = 0; M = 4R 1 E C 0.672 0.375 • Unit load to the right of E 0.141 RC 1 x -0.0312 1 ME 1.5 V 4 m R 1 E C 0.564 0.688 + ΣM E = 0; M E − M E −1(4 − x) + 4RC = 0 -0.125 ME = - 4 + x+ 4RC 39 Example 3

For the beam shown (a) Draw quantitative influence lines for the reaction at supports A and B, and bending moment at B. (b) Determine all the reactions at supports, and also draw its quantitative shear, bending moment diagrams, and qualitative deflected curve for - Only 10 kN downward at 6 m from A - Both 10 kN downward at 6 m from A and 20 kN downward at 4 m from A

20 kN 10 kN 2EI 3EI A B C 4 m 2 m 2 m

40 Influence line of RA 2EI 3EI A B C 4 m 2 m 2 m

fAA fDA f CA fEA

f AA /fAA fDA /f f AA CA /fAA f EA /fAA

41 1 • Find fxA by conjugate beam 8 kN•m 2EI 3EI B A Real Beam 1 C 4 m 2 m 2 m 1 kN

1 1 + V (kN) x (m)

8 4 + M x (m) (kN•m)

2.67 2EI 1.33EI EI

60.44/EI

42 12/EI fAA = M´A = 60.44/EI Conjugate Beam • Quantitative influence line of RA 2.67 2EI 1.33EI EI

60.44/EI A C B 12/EI Conjugate Beam

60.44/EI 37.11/EI 17.77/EI 4.88/EI 0 fxA

1 0.614 0.294 0.081 0 RA = fxA/fAA

43 Using equilibrium conditions for the influence line of RB and MB

x 1 MB A B

RA 4 m 2 m 2 m RB

1 0.614 0.294 0.081 0 RA

0.706 0.919 1 0.386 0 RB = 1 - RA

MB = 8RA - (8-x)(1) 0 0 -1.352 -1.088 -1.648 44 Using equilibrium conditions for the influence line of VB

x 1 A B

RA 4 m 2 m 2 m VB = RA - 1

1 0.614 0.294 0.081 0 RA

0 VB = RA -1 -0.386 -0.706 -0.919 -1

45 Using equilibrium conditions for the influence line of VC and MC

x 1 C MB A B

RA 4 m 2 m 2 m RB 1 0.614 0.294 0.081 0 RA

VC = RA - 1 VC = RA

0.294 1 0.081 0 VC

-0.386 -0.706

MC = 4RA - (4-x)(1) MC = 4RA 1.176 0.456 0.324 MC 46 The quantitative shear and bending moment diagram and qualitative deflected curve 10 kN MB= 13.53 kN•m A B

10(0.081)=0.81 kN 4 m 2 m 2 m RB=9.19 kN

1 0.614 0.294 0.081 0 RA

V (kN) 0.81 0.81

-9.19

4.86 MA (kN•m) + -

-13.53 47 The quantitative shear and bending moment diagram and qualitative deflected curve 20 kN 10 kN MB=46.48 kN•m A B

20(.294) +1(0.081) 4 m 2 m 2 m R =23.31 kN = 6.69 kN B

1 0.614 0.294 0.081 0 RA

V (kN) 6.69 6.69

- -13.31 -23.31 -23.31

M kN•m) 26.76 A ( 0.14 + -

-46.48 48 APPENDIX •Muller-Breslau for the influence line of reaction, shear and moment •Influence lines for MDOF beams

Example 1

Draw the influence line for - the vertical reaction at B

ABCD

2 m 2 m 2 m

49 Influence line of RA

ABCD

2 m 2 m 2 m

1 kN fCA f DA f AA f f AA =1 AA f AA

50 • Find fxA by conjugate beam 1 kN A CD B 6 kN•m Real Beam

1 kN 2 m 2 m 2 m x 72 M ' = Conjugate Beam A EI 18 B' = y EI 18 6 /EI EI x 2x 18x x3 − = M´ 3 3 EI 6EI x V´x x 18 B' = x x2 y EI EI 2EI

51 x 18x x3 ABCD f = M ' = − xA x EI 6EI

2 m 2 m 2 m Point x (m) fxA fxA / fAA

B 0 0 0

72/EI 1 kN 61.33 /EI 34.67 34.67 /EI D 2 0.482 fAA EI fxA 61.33 C 4 EI 0.852

72/72 = 1.0 72 61.33 /72=0.852 A 6 EI 1.0 1 kN f 34.67 /72 = 0.482 AA =1 f AA Influence line of RA 52 • Influence line of RB

ABCD

2 m 2 m 2 m

fDB f fCB f f AB BB BB =1 f BB f BB f BB

53 • Find fxB by conjugate beam

ABCD 6 kN•m Real Beam

1 1 2 m 2 m 2 m

6 18 x EI EI 72 EI Conjugate Beam 18 EI x2 x 2EI EI x x3 72 18x 72 + − = M´x 6EI EI EI EI x 2x V´x 3 3 18 EI 54 x x3 72 18x ABCD f = M ' = + − xB x 6EI EI EI

2 m 2 m 2 m Point x (m) fxB fxB / fBB

B 0 72 1 EI 72/EI 37.33/EI 37.33 0 10.67/EI f D 2 0.518 BB EI fxB 1 C 4 10.67 0.148 EI 37.33 /72 = 0.518 72 /72 = 1 10.67 /72 = 0.148 f A 6 0 0 BB = 1 0 fBB

1 Influence line of RB 55 Example 2

For the beam shown (a) Draw the influence line for the shear at D for the beam (b) Draw the influence line for the bending moment at D for the beam EI is constant.Plot numerical values every 2 m.

ABDEC

2 m 2 m 2 m 2 m

56 The influence line for the shear at D

1 kN ABDEC D 1 kN

2 m 2 m 2 m 2 m

f DD =1 f DD 1 kN

VD D f ED

f DD 1 kN

57 • Using conjugate beam for find fxD

1 kN ABDEC

1 kN 2 m 2 m 2 m 2 m

2 kN•m

1 kN 1 kN

1 kN 2 kN•m

2 k 1 kN

58 1 kN 2 kN 1 kN 1 kN ABDE C Real beam

1 kN 2kN 1 kN 1 kN 2 m 2 m 2 m 2 m

V( kN) 1 x (m)

-1 4 M (kN •m) x (m)

4/EI

Conjugate beam 59 M´D • Determine M´D at D 4/EI

A C Conjugate beam M´D DEB 2 m 2 m 2 m 2 m

8/EI 4/EI 8 m 3 0

16 8 3EI 3EI 8/EI 8 m 3 4/EI

16 128/3EI 3EI 40 3EI 60 4/EI

Conjugate beam A C 40 DEB 128/3EI 8 3EI 3EI 2 m 2 m 2 m 2 m

2/EI 2/EI 2 2 40 76 M´ = ( )( ) − ( )(2) = − DL EI 3 3EI 3EI 40 2 V´DL 3EI 3 2/EI

128/3EI 2 2 128 40 52 M´ = ( )( ) + − ( )(2) = DR EI 3 3EI 3EI 3EI 40 2 2/EI V´DR 2/EI 3EI 3 4 2 2 8 − = ( )( ) − ( )(2) = M´E EI EI 3 3EI 2 8 V´ E 3 3EI 61 4/EI

Conjugate beam A C DEB 40 128/3EI 8 3EI 3EI 2 m 2 m 2 m 2 m

2 8 V ´ 3EI 3EI x (m) θ 16 − 40 34 3EI − − 3EI 3EI 52 3EI 128/3EI = M´D = fDD x (m) ∆ f = M ´ xD 4 − 76 − EI 3EI 52 0.406 = 128 Influence line of VD = fxD/fDD 4/(128/3) = -0.094 62 76/128 = -0.594 The influence line for the bending moment at D

αDD

ABDEC

1 kN •m 1 kN •m

2 m 2 m 2 m 2 m

f DD

α DD

f ED

α DD

63 • Using conjugate beam for find fxD

ABDEC

1 kN •m 1 kN •m 2 m 2 m 2 m 2 m

1 kN•m

0.5 kN 0.5 kN

0.5 kN 1 kN•m

1 k 0.5 kN

0.5 kN 1 kN 0.5 kN 64 1 kN •m 1 kN •m ABDE C Real beam

0.5 kN 1 kN 0.5 kN 2 m 2 m 2 m 2 m

V (kN) 0.5 x (m)

1 2 M (kN •m) x (m)

2/EI

Conjugate beam

65 2/EI

Conjugate beam

2 m 2 m 2 m 2 m

4/EI 2/EI 8 m 3 0

8 4 3EI 4/EI 3EI 8 m 3 2/EI

8 4 32 3EI 3EI 3EI 66 2/EI

Conjugate beam

4 32 4 EI 3EI 3EI 2 m 2 m 2 m 2 m

1/EI 1/EI 1 2 4 26 M´ = ( )( ) + ( )(2) = D EI 3 EI 3EI

4 2 V´D EI 3 1/EI 1/EI 2 1 2 4 − = ( )( ) − ( )(2) =M´E EI EI 3 3EI 2 4 V´ E 3 3EI 67 2/EI

Conjugate beam

4 4 32 3EI EI 3EI 2 m 2 m 2 m 2 m 5 4 EI 1 4 EI 3EI 3EI αDD = 32/3EI θ V ´ x (m) 8 17 − − 3EI 3EI 26/3EI fxD = M ´ x (m) ∆ -2/EI 26 = 0.813 32 θDL = 5/(32/3) = 0.469 rad. θ = -17/32 = -0.531 rad. f xD DR Influence line of MD α xD θD = 0.469 + 0.531 = 1 rad -2 /(32/3) = -0.188 68 Example 3

Draw the influence line for the reactions at supports for the beam shown in the figure below. EI is constant.

ADBC EF G

5 m5 m 5 m 5 m 5 m 5 m

69 Influence line for RD

ADBC EF G

5 m5 m 5 m5 m5 m 5 m

f f CD DD fED fBD fFD fXD 1

f fCD DD fED fBD =1 fFD f DD fDD f DD f DD fDD fXD/fDD = Influence line for RD 1

70 • Use the consistency deformation method

A G 3@5 =15 m 3@5 =15 m 1 =

∆´G

1 + fGG x RG

∆´G + fGGRG = 0 ------(1) 1

- Use conjugate beam for find ∆´G and fGG 15 30 A Real beam G A Real beam G 15 m 15 m 30 m 1 1 1 450 1 112.5 15 EI 15 2812.5 9000 EI ∆'C = M 'C = EI f 'GG = M ''G = EI EI EI Conjugate beam Conjugate beam 450 71 15 + (2/3)(15) 112.5/EI 20 m EI Substitute ∆´G and fGG in (1) : 2812.5 9000 + R = 0 EI EI G

RG = −0.3125 kN,↓

5.625 A G

0.6875 1 0.3125 = 15

1 1 + 30 x RG = -0.3125 kN

1 1 72 • Use the conjugate beam for find fXD f f 5.625 CD DD fED fBD fFD A G Real beam 3@5 =15 m 3@5 =15 m 0.6875 23.01 1 0.3125 5.625 EI x EI 6.818 m 2 A G Conjugate beam 8.182 m 28.13 2 − 4.688 35.16 0.6875x1 x1 EI 15.98 EI 2EI 5.625x − 0.6875x 2 EI 1 1 EI x EI 2 5.625 V´2 (5.625-0.6875x )/EI M´ G EI 1 2 A 2 2 5.625x1 − 0.6875x1 x1 0.3125x2 x2 x M´1= ( ) − ( ) 0.3125x2 28.13 1 EI 2 2 3 V´ 2 2 EI 1 .6875x 2x 0.3125x2 + 1 ( 1 ) 28.13 + x2 = M ' x2 2EI 2EI 3 EI

x1 = 5 m -----> fBD = M´1= 56/EI x2 = 5 m -----> fFD = M´2= 134.1/EI

x1 = 10 m -----> fCD = M´1= 166.7/EI x2 = 10 m -----> fED = M´2= 229.1/EI 73 x1 = 15 m -----> fDD = M´1= 246.1/EI x2 = 15 m -----> fDD = M´2= 246.1/EI • Influence Line for R D 246.1 166.7 229.2 56 EI 134.1 EI EI EI EI fXD 1

166.7 246.1 56 229.2 134.1 246.1 246.1 246.1 246.1 246.1 fXD/fDD 1

1.0 0.228 0.677 0.931 0.545

Influence Line for RD

74 Influence line for RG

ADBC EF G

5 m5 m 5 m5 m5 m 5 m

f fFG GG fEG fXG f f BG CG 1

fGG f FG =1 fEG fGG fGG fGG fXG/fGG f f BG CG 1 fGG fGG

75 • Use consistency deformations

fXG 1 3@5 =15 m 3@5 =15 m = ∆´D

+ 1 fDD X RD 1 ∆´D + fDDRD = 0 ------(2)

- Use conjugate beam for find ∆´D and fDD 30 15 A Real beam G A Real beam G 30 m 15 m 15 m 1 450 1 1 1 30 112.5 EI 9000 15 EI EI EI EI Conjugate beam Conjugate beam 76 20 m 450/EI 15 + (2/3)(15) 112.5 9000 112.5 1.5 EI 15 EI M´ EI EI EI M´´ 15 m V´ 450/EI V´´ 112.5 15 9000 450 2812.5 112.5 2 1125 ∆' = M '= ( ) + − (15) = f = M ''= ( ×15) = D EI 3 EI EI EI DD EI 3 EI

2812.5 1125 Substitute ∆´ and f in (2) : + R = 0, R = −2.5kN = 2.5kN,↓ D DD EI EI D D 7.5

2.5 1.5 1 30 =

1 1 + 15 x RD = -2.5 kN 77 1 1 • Use the conjugate beam for find f XG f fFG GG 7.5 fEG Real beam f f BG CG 2.5 1 1.5 3@5 =15 m 3@5 =15 m 112.5 75 EI EI 15 fGG = M´G = 1968.56/EI EI A G Conjugate beam 10 m − 7.5 168.75/EI EI 15 + (10/3) = 18.33 m 18.75 EI 25 + (2/3)(5) = 28.33 m 5 m 18.75 2 62.5 A M´ = f = − ( ×5) = − 1 BG EI 3 EI − 7.5 V´1 18.75 EI EI 18.75 6.67 m EI 18.75 125.06 A M´ = f = − (6.67) = − − 7.5 2 CG EI EI −18.75 V´ EI 2 78 EI x2

2 fGG = M´G = 1968.56/EI x 2 x 1968.56 168.75 x 3 ( 3 ) + − x =M´ G 2 3 EI EI 3 x V´2 168.75/EI x = 5 m -----> fFG = M´= 1145.64/EI x = 10 m -----> f = M´ = 447.73/EI 1968.56 EG 1145.64 447.73 EI EI EI fXG −125 − 62.5 1 EI EI 1145.64 1968.56 447.73 1968.56 1968.56 1968.56 fXG/fGG − 62.5 −125 1 1968.56 1968.56 1.0 0.227 0.582 Influence line for RG -0.032 -0.064 79 Using equilibrium condition for the influence line for Ay x 1 ADBC EF G MA 5 m5 m 5 m5 m5 m 5 m

Ay RD RG

+ ↑ ΣFy = 0 : RA =1− RD − RC

1 1

Unit load

1.0 0.228 0.678 0.929 0.542

Influence Line for RD 1.0 0.227 0.582 Influence line for RG -0.032 -0.064 1.0 0.804 0.386 Influence line for Ay 80 -0.156 -0.124 Using equilibrium condition for the influence line for MA x 1 ADBC EF G MA 5 m5 m 5 m5 m5 m 5 m

Ay RD RG

+ ΣM A = 0 : 1x −15RD − 30RC

20 25 30 5 10 15 1x

1.0 0.228 0.678 0.929 0.542

RD x 15 1.0 0.227 0.582 RG x 30 -0.032 -0.064 2.54 1.75 Influence line for MA 81 -0.745 -0.59 DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS

! General Case

! Stiffness Coefficients

! Stiffness Coefficients Derivation

! Fixed-End Moments

! Pin-Supported End Span

! Typical Problems

! Analysis of Beams

! Analysis of Frames: No Sidesway

! Analysis of Frames: Sidesway

1 Slope – Deflection Equations

i P j k w Cj

settlement = ∆j

i P j M w ij Mji

θi

θ ψ j

2 Degrees of Freedom

M θΑ

A B 1 DOF: θΑ L

P θΑ B 2 DOF: θ , θ A C Α Β θΒ

3 Stiffness

k kAA 1 BA

A B L

4EI k = AA L 2EI k = BA L

4 k

kAB BB A 1 B L

4EI k = BB L 2EI k = AB L

5 Fixed-End Forces Fixed-End Moments: Loads P

PL L/2 L/2 PL 8 8 L

P P 2 2 w wL2 wL2 12 12 L wL wL 2 2

6 General Case i P j k w Cj

settlement = ∆j

i P j M w ij Mji

θi

ψ θj

7 P i w j Mij Mji θi L settlement = ∆ θ j ψ j θ θ 4EI 2EI 2EI 4EI i + θ j = M M = + θ L L ij ji L i L j θj

θi + F (M ij) F ∆ (M ji)∆

settlement = ∆ + j P w F (MF ) (M ij)Load ji Load

θ θ 4EI 2EI F F 2EI 4EI F F M = ( ) + ( )θ + (M ij ) + (M ij ) , M = ( ) + ( )θ + (M ji ) + (M ji ) ij L i L j ∆ Load ji L i L j ∆ Load 8 Equilibrium Equations

i P j k w Cj

C Mji j Mjk

Mji Mjk

+ ΣM j = 0 : − M ji − M jk + C j = 0

9 Stiffness Coefficients

Mij i j Mji L θj

θi

4EI kii = 2EI L k ji = ×θi L 1 + 2EI kij = 4EI L k = ×θ j jj L 1

10 Matrix Formulation

θ 4EI 2EI F M = ( ) + ( )θ + (M ij ) ij L i L j θ 2EI 4EI F M = ( ) + ( )θ + (M ji ) ji L i L j

θ F M ij  (4EI / L) (2EI / L)  iI  M ij    =   +  F  M (2EI / L) (4EI / L) θ  ji     j  M ji 

kii kij  []k =   k ji k jj 

Stiffness Matrix

11 P i w j Mij Mji θi [M ] = [K][θ ]+[FEM ] L θ ([M ]−[FEM ]) = [K][θ ] ψ j ∆j [θ ] = [K]−1[M ]−[FEM ]

Mij Mji

θj

θi Fixed-end moment + Stiffness matrix matrix F (M ij) F ∆ (M ji)∆ [D] = [K]-1([Q] - [FEM]) +

Displacement Force matrix F P F (M ij)Load w (M ji)Load matrix

12 Stiffness Coefficients Derivation: Fixed-End Support M Mi θi j Real beam i j L

M i + M j M i + M j L L L/3 M j L M j 2EI EI Conjugate beam

M i EI M i L 2EI θι From(1)and (2); M i L L M j L 2L + ΣM 'i = 0 : − ( )( ) + ( )( ) = 0 2EI 3 2EI 3 4EIθ M i = ( ) i M i = 2M j − − − (1) L 2EI M i L M j L M j = ( )θi + ↑ ΣFy = 0 : θ i − ( ) + ( ) = 0 − − − (2) L 2EI 2EI 13 Stiffness Coefficients Derivation: Pinned-End Support

Mi θi Real beam θ i j j L

M i M i L L 2L 3 Conjugate beam

M i EI M i L 2EI θi θj

M L 2L M L M L + ΣM ' = 0 : ( i )( ) −θ L = 0 + ↑ ΣF = 0 : ( i ) − ( i ) +θ = 0 j 2EI 3 i y 3EI 2EI j M L i − M i L θi = ( ) θ = ( ) 3EI j 6EI

M i L 3EI θi =1 = ( ) → M i = 3EI L 14 Fixed end moment : Point Load P Real beam Conjugate beam A B

ABL M M M EI EI

M EI ML M 2EI

M ML EI P 2EI PL2 PL PL2 16EI 4EI 16EI

ML ML 2PL2 PL + ↑ ΣF = 0 : − − + = 0, M = y 2EI 2EI 16EI 8 15 P

PL PL 8 L 8 P P P/2 2 2

P/2 PL/8

-PL/8 -PL/8

- -PL/8 -PL/16

- -PL/16 -PL/8 − PL − PL PL PL PL/4 + + = + 16 16 4 8 16 Uniform load

w Real beam Conjugate beam A B ABL M M M EI EI

M EI ML M 2EI

M ML EI 2EI 2 wL3 wL wL3 w 24EI 8EI 24EI

ML ML 2wL3 wL2 + ↑ ΣF = 0 : − − + = 0, M = y 2EI 2EI 24EI 12 17 Settlements M M Mi = Mj Real beam j Conjugate beam EI L A B

M + M ∆ ∆ i j M L M i + M j M EI L

M ML EI ML 2EI 2EI M M EI

∆

ML L ML 2L + ΣM = 0 : − ∆ − ( )( ) + ( )( ) = 0, B 2EI 3 2EI 3 6EI∆ M = L2 18 Pin-Supported End Span: Simple Case P w B A L θ θ 2EI 4EI 4EI 2EI A + θ B + θ L L L A L B θ A θ AB+ B P w (FEM) (FEM)AB BA

M AB = 0 = (4EI / L)θ A + (2EI / L)θ B + (FEM ) AB − − − (1)

M BA = 0 = (2EI / L)θ A + (4EI / L)θ B + (FEM )BA − − − (2)

2(2) − (1) : 2M BA = (6EI / L)θ B + 2(FEM )BA − (FEM )BA (FEM ) M = (3EI / L)θ + (FEM ) − BA BA B BA 2 19 Pin-Supported End Span: With End Couple and Settlement P w M A B A L ∆ θ θ 2EI 4EI 4EI 2EI + θ + θ A B L A L B L L θ A θ ABP B w F (MF ) (M AB)load BA load

ABF (MF ) B (M BA) ∆ AB ∆ A

θ 4EI 2EI M = M = + θ + (M F ) + (M F ) − − − (1) AB A L A L B AB load AB ∆ θ θ 2EI 4EI M = + θ + (M F ) + (M F ) − − − (2) BA L A L B BA load BA ∆

2(2) − (1) 3EI F 1 F 1 F M A E liminate A by : M BA = θ B +[(M BA )load − (M AB )load ]+ (M BA )∆ + 2 L 2 2 2 20 Fixed-End Moments Fixed-End Moments: Loads P

PL L/2 L/2 PL 8 8

PL 1 PL 3PL L/2 L/2 + ( )[−(− )] = 8 2 8 16

2 wL2 wL 12 12

wL2 1 wL2 wL2 + ( )[−(− )] = 12 2 12 8 21 C Typical Problem B P P1 w 2

A C B L1 L2

2 wL 2 PL P PL w wL 12 8 8 12 L L θ 0 4EI 2EI P1L1 M AB = A + θ B + 0 + L1 L1 8 θ 0 2EI 4EI P1L1 M BA = A + θ B + 0 − L1 L1 8 θ 0 2 4EI 2EI P2 L2 wL2 M BC = B + θC + 0 + + L2 L2 8 12 θ 0 2 2EI 4EI − P2 L2 wL2 M CB = B + θC + 0 + − L2 L2 8 12 22 C B P P1 w 2

A C B L1 L2

C MBA B MBC

θ 2EI 4EI P1L1 M BA = A + θ B + 0 − L1 L1 8 θ 2 4EI 2EI P2L2 wL2 M BC = B + θC + 0 + + L2 L2 8 12

+ ΣM B = 0 : CB − M BA − M BC = 0 → Solve for θ B

23 C B P P1 w 2 M M BA AB C A MCB M B BC L1 L2

Substitute θB in MAB, MBA, MBC, MCB

θ 0 4EI 2EI P1L1 M AB = A + θ B + 0 + L1 L1 8 θ 0 2EI 4EI P1L1 M BA = A + θ B + 0 − L1 L1 8 θ 0 2 4EI 2EI P2 L2 wL2 M BC = B + θC + 0 + + L2 L2 8 12 θ 0 2 2EI 4EI − P2 L2 wL2 M CB = B + θC + 0 + − L2 L2 8 12

24 C B P P1 w 2 MBA MAB MCB A MBC C Ay B L1 L2 Cy

By = ByL + ByR

B C P P 1 M 2 B BA MCB MAB A MBC

A ByR C y ByL y L1 L2

25 Example of Beams

26 Example 1

Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

10 kN 6 kN/m

A C

4 m 4 m B 6 m

27 10 kN 6 kN/m

A C

4 m 4 m B 6 m PL P wL2 wL2 8 PL w 8 30 20 FEM M BA MBC [M] = [K][Q] + [FEM]

B θ 0 4EI 2EI (10)(8) + ΣM = 0 : − M − M = 0 M = + θ + B BA BC AB 8 A 8 B 8 θ 0 4EI 4EI (6)(62 ) 2EI 4EI (10)(8) ( + θ ) −10 + = 0 M = + θ − 8 6 B 30 BA 8 A 8 B 8 2.4 θ 0 θ = 4EI 2EI (6)(62 ) B EI M BC = B + θC + 6 6 30 Substitute θB in the moment equations:

θ 0 2 2EI 4EI (6)(6) MAB = 10.6 kN•m, MBC = 8.8 kN•m M CB = B + θC − 6 6 20 M = - 8.8 kN•m, M = -10 kN•m BA CB 28 10 kN 6 kN/m 8.8 kN•m 10.6 kN•m A C 10 kN•m 8.8 kN•m 4 m 4 m B 6 m

MAB = 10.6 kN•m, MBC = 8.8 kN•m M = - 8.8 kN•m, BA MCB= -10 kN•m 2 m

10 kN 18 kN 6 kN/m A B 8.8 kN•m B 10.6 kN•m 10 kN•m 8.8 kN•m A B y = 5.23 kNyL = 4.78 kN ByR = 5.8 kN Cy = 12.2 kN

29 10 kN 6 kN/m

10.6 kN•m 10 kN•m A C B 5.23 kN 4 m 4 m 6 m 12.2 kN

4.78 + 5.8 = 10.58 kN

5.8 5.23 V (kN) + + x (m) - - - 4.78

10.3 -12.2 M (kN•m) + x (m) - - - -10.6 -8.8 -10 2.4 θ = B EI Deflected shape x (m) 30 Example 2

Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

10 kN 6 kN/m

A C

4 m 4 m B 6 m

31 10 kN 6 kN/m

A C

4 m 4 m B 6 m PL P wL2 wL2 8 PL w 8 30 20 FEM [M] = [K][Q] + [FEM] 10 θ 0 4EI 2EI (10)(8) M = + θ + − − − (1) AB 8 A 8 B 8 θ 10 2EI 4EI (10)(8) M = + θ − − − − (2) BA 8 A 8 B 8 θ 4EI 2EI 0 (6)(62 ) M = + θ + − − − (3) BC 6 B 6 C 30 θ 2EI 4EI 0 (6)(6)2 M = + θ − − − − (4) CB 6 B 6 C 20 6EI 2(2) − (1) : 2M = θ − 30 BA 8 B 3EI M = θ −15 − − − (5) BA 8 B 32 M BA MBC

B 2 4EI (6)(6 ) Substitute θA and θB in (5), (3) and (4): M = θ + − − − (3) BC 6 B 30 M = - 12.19 kN•m 2EI (6)(6)2 BA M CB = θ B − − − − (4) 6 20 MBC = 12.19 kN•m 3EI M BA = θ B −15 − − − (5) M = - 8.30 kN•m 8 CB

+ ΣM B = 0 : − M BA − M BC = 0

3EI 4EI (6)(62 ) ( + θ ) −15 + = 0 − − − (6) 8 6 B 30 7.488 θ B = θ EI 4EI 2EI Substitute B in (1) : 0 = A + B −10 8θ 8 − 23.74 θ θ A = EI 33 10 kN 6 kN/m 12.19 kN•m A C 8.30 kN•m 12.19 kN•m 4 m 4 m B 6 m

MBA = - 12.19 kN•m, MBC = 12.19 kN•m, MCB = - 8.30 kN•m

2 m

10 kN 18 kN 6 kN/m B A B 12.19 kN•m C 8.30 kN•m 12.19 kN•m A B y = 3.48 kNyL = 6.52 kN ByR = 6.65 kN Cy = 11.35 kN

34 10 kN 6 kN/m

A C

3.48 kN B 4 m 4 m 6 m 11.35 kN

6.52 + 6.65 = 13.17 kN

6.65 V (kN) 3.48 x (m)

- 6.52 -11.35 14 M (kN•m) x (m) -8.3 -12.2 7.49 θ = B EI Deflected shape x (m) − 23.74 θ = A EI 35 Example 3

Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

10 kN 4 kN/m

A C 2EI 3EI B 4 m 4 m 6 m

36 10 kN 4 kN/m

A C 2EI 3EI 2 (10)(8)/8 (10)(8)/8 B (4)(62)/12 (4)(6 )/12 4 m 4 m 6 m

0 θ 10 4(2EI) 2(2EI) (10)(8) M = + θ + − − − (1) AB 8 A 8 B 8 10 θ 2(2EI) 4(2EI) (10)(8) M BA = A + θ B − − − − (2) 8 8 15 8 2(2) − (1) 3(2EI) (3/ 2)(10)(8) : M = θ − − − − (2a) 2 BA 8 B 8 12 4(3EI) (4)(62 ) M = θ + − − − (3) BC 6 B 12

37 10 kN 4 kN/m

A C 2EI 3EI 2 (3/2)(10)(8)/8B (4)(62)/12 (4)(6 )/12 4 m 4 m 6 m

15 3(2EI) (3/ 2)(10)(8) M = θ − − − − (2a) BA 8 B 8 12 4(3EI) (4)(62 ) M = θ + − − − (3) BC 6 B 12

− M BA − M BC = 0 : 2.75EIθ B = −12 +15 = 3

θ B =1.091/ EI

3(2EI) 1.091 M = ( ) −15 = −14.18 kN • m BA 8 EI 4(3EI) 1.091 M = ( ) −12 =14.18 kN • m BC 6 EI 2(3EI) M CB = θ B −12 = −10.91 kN • m 6 38 10 kN 4 kN/m

A C 10.91 2EI 14.18 14.18 3EI B 4 m 4 m 6 m

MBA = - 14.18 kN•m, MBC = 14.18 kN•m, MCB = -10.91 kN•m

10 kN 24 kN 4 kN/m A B 14.18 kN•m 140.18 kN•m 10.91 kN•m C A B y = 3.23 kN yL = 6.73 kN ByR = 12.55 kN Cy = 11.46 kN

39 10 kN 4 kN/m

A C 10.91 kN•m 2EI 3EI 3.23 kN B 4 m 4 m 6 m 11.46 kN

6.77 + 12.55 = 19.32 kN 12.55 V (kN) 3.23 2.86 + + x (m) - - -6.73 -11.46 12.91 M 5.53 (kN•m) + + x (m) - - -10.91 -14.18 θB = 1.091/EI Deflected shape x (m)

40 Example 4

Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

10 kN 12 kN•m 4 kN/m

A C 2EI 3EI B 4 m 4 m 6 m

41 10 kN 12 kN•m 4 kN/m

A C 2EI 3EI 1.5PL/8 = 15 B wL2/12 = 12 wL2/12 = 12 4 m 4 m 6 m

3(2EI) 0 θ -3.273/EI M = θ −15 − − − (1) BA 8 B 4(2EI) 2(3EI) (10)(8) M AB = A + θ B + 4(3EI) 8 8 8 M BC = θ B +12 − − − (2) 7.21 6 θ = − A EI 2(3EI) M CB = θ B −12 − − − (3) 3.273 6 M BA = 0.75EI (− ) −15 = −17.45 kN • m 12 kN•m EI M M 3.273 BA BA M = 2EI (− ) +12 = 5.45 kN • m BC EI M 3.273 B BC MBC M = EI(− ) −12 = −15.27 kN • m CB EI Joint B : − M BA − M BC −12 = 0

− (0.75EI −15) − (2EIθ B +12) −12 = 0 3.273 θ = − B EI 42 3.273 M = 0.75EI(− ) −15 = −17.45 kN • m BA EI 3.273 M = 2EI (− ) +12 = 5.45 kN • m BC EI 3.273 M = EI (− ) −12 = −15.27 kN • m CB EI

10 kN 24 kN 4 kN/m A B 5.45 kN•m 17.45 kN•m 15.27 kN•m C 2.82 kN7.18 kN 10.36 kN 13.64 kN

10 kN 12 kN•m 4 kN/m

A C 15.27 kN•m B 2.82 kN 17.54 kN 13.64 kN 43 10 kN 12 kN•m 4 kN/m

A C 15.27 kN•m 2EI 3EI B 7.21 2.82 kN 13.64 kN θ = − 17.54 kN A 4 m 4 m 6 m EI 3.273 θ B = − 10.36 EI V (kN) 2.82 + + 3.41 m x (m) - - -7.18 -13.64 11.28 M 7.98 (kN•m) + + x (m) - -5.45 - -17.45 -15.27

Deflected shape x (m) − 7.21 3.273 θ = θ B = A EI EI 44 Example 5

Draw the quantitative shear, bending moment diagrams, and qualitative deflected curve for the beam shown. Support B settles 10 mm, and EI is constant. Take E = 200 GPa, I = 200x106 mm4.

10 kN 12 kN•m 6 kN/m B A C 2EI 3EI 10 mm 4 m 4 m 6 m

45 10 kN 12 kN•m 6 kN/m B A C 2EI 3EI 10 mm 4 m 4 m 6 m 6EI∆ 6EI∆ A L2 L2 6EI∆ 6EI∆ 2 L2 ∆ L ∆

[FEM]∆ C B wL2 B wL2 PL P PL w 8 8 30 30

[FEM]load -12 θ 4(2EI) 2(2EI) 6(2EI)(0.01) (10)(8) M AB = A + θ B + + − − − (1) 8 θ 8 82 8 2(2EI) 4(2EI) 6(2EI)(0.01) (10)(8) M = + θ + − − − − (2) BA 8 A 8 B 82 8 θ 0 4(3EI) 2(3EI) 6(3EI)(0.01) (6)(62 ) M BC = B + θC − 2 + − − − (3) 6 θ 6 0 6 30 2(3EI) 4(3EI) 6(3EI)(0.01) (6)(6)2 M = + θ − − − − − (4) CB 6 B 6 C 62 30 46 10 kN 12 kN•m 6 kN/m B A C 2EI 3EI 10 mm 4 m 4 m 6 m θ 4(2EI) 2(2EI) 6(2EI)(0.01) (10)(8) M AB = A + θ B + + − − − (1) 8 θ 8 82 8 2(2EI) 4(2EI) 6(2EI)(0.01) (10)(8) M = + θ + − − − − (2) BA 8 A 8 B 82 8 Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN• m2 : θ 4(2EI) 2(2EI) M AB = A + θ B + 75 +10 − − − (1) 8 θ 8 2(2EI) 4(2EI) M = + θ + 75−10 − − − (2) BA 8 A 8 B 16.5 2(2) − (1) 3(2EI) : M = θ + 75 − (75/ 2) −10 − (10 / 2) −12 / 2 − − − (2a) 2 BA 8 B

47 10 kN 12 kN•m 6 kN/m B A C 2EI 3EI 10 mm 4 m 4 m 6 m

MBA = (3/4)(2EI)θB + 16.5

MBC = (4/6)(3EI)θB - 192.8 (3/4 + 2)EI + 16.5 - 192.8 = 0 + ΣMB = 0: - MBA - MBC = 0 θB

MBA MBC θB = 64.109/ EI

Substitute θB in (1): θA = -129.06/EI B

Substitute θA and θB in (5), (3), (4):

MBA = 64.58 kN•m,

MBC = -64.58 kN•m

MCB = -146.69 kN•m 48 10 kN 12 kN•m 6 kN/m B 146.69 kN•m A C 64.58 kN•m 64.58 kN•m

4 m 4 m 6 m

MBA = 64.58 kN•m,

MBC = -64.58 kN•m

MCB = -146.69 kN•m 2 m

18 kN 12 kN•m 10 kN 6 kN/m 64.58 kN•m 64.58 kN•m A B 146.69 kN•m B C

= 11.57 kN = -1.57 kN B C = 47.21 kN Ay ByL yR = -29.21 kN y

49 10 kN 12 kN•m 6 kN/m

A C 146.69 kN•m 2EI 3EI 11.57 kN B 47.21 kN 4 m 4 m 6 m θA = -129.06/EI 1.57 + 29.21 = 30.78 kN θB = 64.109/ EI 11.57 V (kN) 1.57 + x (m) - -29.21 -47.21 58.29 64.58 12 M + (kN•m) x (m) -

-146.69 Deflected shape x (m) 10 mm θ = 64.109/ EI 50 θA = -129.06/EI B Example 6

For the beam shown, support A settles 10 mm downward, use the slope-deflection method to (a)Determine all the slopes at supports (b)Determine all the reactions at supports (c)Draw its quantitative shear, bending moment diagrams, and qualitative deflected shape. (3 points) Take E= 200 GPa, I = 50(106) mm4.

12 kN•m 6 kN/m

B A 2EI C 1.5EI 10 mm 3 m 3 m

51 12 kN•m 6 kN/m

B A 2EI C 1.5EI 10 mm 3 m 3 m 6 kN/m 6(32 ) = 4.5 4.5 12 F C M w A 6(1.5× 200×50)(0.01) C 32 0.01 m =100 kN • m F 100 kN • m M ∆ A 4(2EI) M CB = θC − − − (1) 3 θ 4(1.5EI) 2(1.5EI) M CA = C + θ A + 4.5 +100 − − − (2) 3 θ 3 12 2(1.5EI ) 4(1.5EI) M = + θ − 4.5 +100 − − − (3) AC 3 C 3 A 2(2) − (2) 3(1.5EI) 3(4.5) 100 12 : M CA = θC + + + − − − (2a) 2 3 2 2 2 52 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3 m 3 m

4(2EI) M = θ − − − (1) CB 3 C 3(1.5EI) 3(4.5) 100 12 M = θ + + + − − − (2a) CA 3 C 2 2 2 • Equilibrium equation: M MCB CA M CB + M CA = 0 (8+ 4.5)EI 3(4.5) 100 12 θ + + + = 0 C 3 C 2 2 2 −15.06 θ = = −0.0015 rad C EI Substitute θ in eq.(3) 2(1.5EI ) −15.06 4(1.5EI) C 12 = ( ) + θ − 4.5 +100 − − − (3) 3 EI 3 A − 34.22 θ A = = −0.0034 rad EI 53 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3 m 3 m

−15.06 − 34.22 θ = = −0.0015 rad θ = = −0.0034 rad C EI A EI 2(2EI ) 2(2EI) −15.06 M = θ = ( ) = −20.08 kN • m BC 3 C 3 EI 4(2EI) 4(2EI) −15.06 M = θ = ( ) = −40.16 kN • m CB 3 C 3 EI

20.08 kN•m 40.16 kN•m B C 40.16 + 20.08 20.08 kN = 20.08 kN 3 18 kN 12 kN•m 6 kN/m 40.16 kN•m C A 8.39 kN 26.39 kN 54 12 kN•m 6 kN/m θ = −0.0015 rad B C A C 1.5EI 10 mm θ A = −0.0034 rad 2EI 3 m 3 m 20.08 kN•m 40.16 kN•m B C 20.08 kN 20.08 kN 12 kN•m 6 kN/m 40.16 kN•m C A 26.39 kN 8.39 kN V (kN) 26.39 8.39 + x (m) -

M (kN•m) -20.08 20.08 12 x (m)

-40.16 Deflected shape θ = 0.0015 rad C x (m)

θ A = 0.0034 rad 55 Example 7

12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3 m 3 m

56 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3 m 3 m

C EI∆C C

B ∆C ∆C 6(1.5EI)∆ 4EI∆ C C 2 = EI∆C 3 6 kN/m A 3 6(2EI)∆C 4EI∆C = 2 2 6(3 ) 3 3 4.5 = 4.5 C A 12 100 C 2(2EI) 4EI 6(1.5× 200×50)(0.01) M BC = θC − ∆C − − − (1) 0.01 m 3 3 32 4(2EI) 4EI A =100 kN • m M CB = θC − ∆C − − − (2) 3 θ 3 4(1.5EI) 2(1.5EI) M CA = C + θ A + EI∆C + 4.5 +100 − −(3) 3 θ 3 12 2(1.5EI) 4(1.5EI) M = + θ + EI∆ − 4.5 +100 − −(4) AC 3 C 3 A C 2(3) − (4) 3(1.5EI) EI 3(4.5) 100 12 : M = θ + ∆ + + + − − − (3a) 57 2 CA 3 C 2 C 2 2 2 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3 m 3 m 18 kN • Equilibrium equation: 12 kN•m 6 kN/m M MBC M CA B C CB C A Ay M + M M +12 +18(1.5) M + 39 By (C ) = −( BC CB ) (C ) = CA = CA y CB 3 y CA 3 3

MCB MCA ΣM C = 0 : M CB + M CA = 0 − − − (1*) ΣC = 0 : (C ) + (C ) = 0 − − − (2*) C y y CB y CA (C ) y CB (Cy)CA

Substitute in (1*) 4.167EIθC − 0.8333EI∆C = −62.15 − − − (5)

Substitute in (2*) − 2.5EIθC + 3.167EI∆C = −101.75 − − − (6)

From (5) and (6) θ C = −25.51/ EI = −0.00255 rad ∆C = −52.27 / EI = −5.227 mm 58 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3 m 3 m • Solve equation − 25.51 θ = = −0.00255 rad Substitute θC and ∆C in (1), (2) and (3a) C EI M = 35.68 kN • m − 52.27 BC ∆C = = −5.227 mm EI M CB =1.67 kN • m

Substitute θC and ∆C in (4) M CA = −1.67 kN • m − 2.86 θ = = −0.000286 rad A EI 12 kN•m 6 kN/m B 35.68 kN•m A C 2EI 1.5EI 18(4.5) −12 − 35.68 B =18 − 5.55 A = y 3 m 3 m y 6 =12.45 kN = 5.55 kN 59 12 kN•m 6 kN/m 35.68 kN•m B A 2EI C 1.5EI 10 mm 12.45 kN 3 m 3 m 5.55 kN

12.45 V (kN) 0.925 m θC = −0.00255 rad + x (m) ∆ = −5.227 mm C -5.55

θ A = −0.000286 rad M (kN•m) 14.57 12 1.67 + x (m) -

-35.68 Deflected shape ∆C = 5.227 mm x (m) θ = 0.00255 rad C θ = 0.000286 rad A 60 Example of Frame: No Sidesway

61 Example 6

For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports (b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.

10 kN 12 kN/m

C B 2EI 3 m 3EI 40 kN

3 m A

1 m 6 m

62 10 kN 36/2 = 18 12 kN/m • Equilibrium equations C 10 B 2EI 36 2 3 m PL/8 = 30 (wL /12 ) =36 MBC 3EI 40 kN MBA 3 m 10 − M − M = 0 − − − (1*) A PL/8 = 30 BA BC

1 m 6 m Substitute (2) and (3) in (1*)

10 − 3EIθ + 30 − 54 = 0 • Slope-Deflection Equations B −14 − 4.667 2(3EI ) θ B = = M = θ + 30 − − − (1) (3EI) EI AB 6 B − 4.667 4(3EI) Substituteθ = in (1) to (3) M = θ − 30 − − − (2) B EI BA 6 B M = 25.33 kN • m 3(2EI) AB M BC = θ B + 36 +18 − − − (3) M = −39.33 kN • m 6 BA M BC = 49.33 kN • m 63 10 kN 12 kN/m 20.58 C 10 -39.3 B 49.33 2EI -49.33 3 m 39.33 3EI 40 kN MAB = 25.33 kN•m 27.7

3 m MBA = -39.33 kN•m A 25.33 -25.33 MBC = 49.33 kN•m

1 m 6 m Bending moment diagram 12 kN/m θ B C B 49.33 θB = -4.667/EI θB B 39.33 27.78 kN

40 kN

A 17.67 kN Deflected curve 25.33 64 Example 7

Draw the quantitative shear, bending moment diagrams and qualitative deflected curve for the frame shown. E = 200 GPa.

25 kN 5 kN/m

B E 240(106) mm4 C 180(106)

5 m 6 4 120(106) mm4 60(10 ) mm

A D 3 m 3 m 4 m

65 25 kN 5 kN/m PL/8 = 18.75 18.75 B E 240(106) mm4 C 180(106) 6.667+ 3.333 (wL2/12 ) = 6.667 5 m 6 4 120(106) mm4 60(10 ) mm

A D 3 m 4 m θ 3 m 0 4(2EI) 2(2EI) M AB = A + θ B 5 5 M + M = 0 θ 0 BA BC 2(2EI) 4(2EI) θ M = + θ 8 16 8 BA 5 A 5 B ( + )EI + ( )EIθ = −18.75 − − − (1) θ 5 6 B 6 C 4(4EI) 2(4EI) M = + θ +18.75 BC B C θ M + M + M = 0 6 θ 6 CB CD CE 2(4EI) 4(4EI) 8 16 3 9 M CB = B + θC −18.75 ( )EI B + ( + + )EIθC = 8.75 − − − (2) 6 6 6 6 5 θ4 3(EI) M = θ − 5.29 2.86 CD 5 C From (1) and (2) : = θ = B EI C EI 3(3EI) M = θ +10 66 CE 4 C Substitute θB = -1.11/EI, θc = -20.59/EI below

θ 0 4(2EI) 2(2EI) M = kN•m M = + θ AB −4.23 AB 5 A 5 B θ 0 2(2EI) 4(2EI) MBA = −8.46 kN•m M BA = A + θ B 5 θ 5 4(4EI) 2(4EI) MBC = 8.46 kN•m M BC = B + θC +18.75 6 θ 6 2(4EI) 4(4EI) M = + θ −18.75 MCB = −18.18 kN•m CB 6 B 6 C 3(EI) M = 1.72 kN•m M = θ CD CD 5 C 3(3EI) M = θ +10 MCE = 16.44 kN•m CE 4 C

67 MAB = -4.23 kN•m, MBA = -8.46 kN•m, MBC = 8.46 kN•m, MCB = -18.18 kN•m, MCD = 1.72 kN•m, MCE = 16.44 kN•m 20 kN 25 kN 16.44 kN•m 3 m 3 m 2.2 kN 2.54 kN B C 2.54 kN2.54-0.34 CE =2.2 kN 8.46 kN•m 18.18 kN•m (25(3)+8.46-18.18)/6 14.12 kN (20(2)+16.44)/4 5.89 kN = 10.88 kN = 14.11 kN

10.88 kN 14.12+14.11=28.23 kN 8.46 kN•m 1.72 kN•m B B (1.72)/5 = 0.34 kN (8.46 + 4.23)/5 = 2.54 kN 5 m 5 m

A 2.54 kN A 0.34 kN 4.23 kN•m 10.88 kN 28.23 kN 68 24.18 14.11 1.29 m 10.88 2.33 m 1.18 m + + 0.78 m 3.46 + 1.72 - + -2.54 2.82 m -8.46 - - - -5.89 -8.46 - 1.18 m - -14.12 -16.44 + -18.18 1.67m -2.54 0.34 + 4.23 Moment diagram

Shear diagram 1.29 m 2.33 m 0.78 m

θB = −5.29/EI 1.18 m

θC = 2.86/EI 1.67m

Deflected curve 69 Example of Frames: Sidesway

70 Example 8

Determine the moments at each joint of the frame and draw the quantitative bending moment diagrams and qualitative deflected curve . The joints at A and D are fixed and joint C is assumed pin-connected. EI is constant for each member

3m B C 1 m 10 kN

3 m

A D

71 • Overview • Unknowns B C 1 m 10 kN θB and ∆ • Boundary Conditions 3 m θ = θ = 0 MDC A D MAB A D D Ax x • Equilibrium Conditions 3m - Joint B A Dy y B

M MBA BC

ΣM B = 0 : M BA + M BC = 0 − − − (1*)

- Entire Frame

+ → ΣFx = 0 : 10 − Ax − Dx = 0 − − − (2*) 72 ∆ ∆ MAB

(0.375EI∆)∆ B C 1 m B C 10 kN 10 kN (0.375EI∆)∆ (5.625)load 4 m 4 m 3 m (1/2)(0.375EI∆)∆ (1.875)load (0.375EI∆)∆ A D A D Ax Dx (0.375EI∆)∆ 3m

• Slope-Deflection Equations MBA MDC 5.625 0.375EI∆ 2(EI) 10(3)(12 ) 6EI∆ M = θ + + − − − (1) + ΣM B = 0 : AB 4 B 42 42 (M AB + M BA ) 5.625 0.375EI∆ Ax = 4(EI) 10(32 )(1) 6EI∆ 4 M BA = θ B − 2 + 2 − − − (2) 4 4 4 Ax = 0.375EIθ B + 0.1875EI∆ +1.563− − − (5) 3(EI) M BC = θ B − − − (3) 3 + ΣM C = 0 : 1 M M DC = 0.375EI∆ − 0.375EI∆ = 0.1875EI∆ − − − (4) DC Dx = = 0.0468EI∆ − − − (6) 2 4 73 Equilibrium Conditions: • Solve equation

M BA + M BC = 0 − − − (1*) Substitute (2) and (3) in (1*)

10 − Ax − Dx = 0 − − − (2*) 2EI θB + 0.375EI ∆ = 5.625 ----(7) Slope-Deflection Equations: Substitute (5) and (6) in (2*) 2(EI) M AB = θ B + 5.625 + 0.375EI∆ − − − (1) − 0.375EIθ − 0.235EI∆ = −8.437 − − − (8) 4 B 4(EI) M = θ − 5.625 + 0.375EI∆ − − − (2) From (7) and (8) can solve; BA 4 B 3(EI) − 5.6 44.8 M = θ − − − (3) θ B = ∆ = BC 3 B EI EI − 5.6 44.8 M DC = 0.1875EI∆ − − − (4) Substituteθ = and ∆ = in (1)to (6) B EI EI

Horizontal reaction at supports: MAB = 15.88 kN•m MBA = 5.6 kN•m Ax = 0.375EIθ B + 0.1875EI∆ +1.563− − − (5) MBC = -5.6 kN•m Dx = 0.0468EI∆ − − − (6) MDC = 8.42 kN•m Ax = 7.9 kN D = 2.1 kN x 74 B C 1 m 10 kN M = 15.88 kN•m, M = 5.6 kN•m, 5.6 AB BA MBC = -5.6 kN•m, MDC = 8.42 kN•m, Ax = 7.9 kN, Dx = 2.1 kN, 3 m 15.88 8.42 A 7.9 kN D 2.1 kN 3m

5.6 ∆ = 44.8/EI ∆ = 44.8/EI 5.6 C C B B θ = -5.6/EI 7.8 B

A A 15.88 D 8.42 D

Bending moment diagram Deflected curve 75 Example 9

From the frame shown use the slope-deflection method to: (a) Determine the end moments of each member and reactions at supports (b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.

B C pin 2 EI 10 kN EI 2.5 EI 4 m 2 m

A D 4 m 3m

76 • Overview C´ B ∆ 2EI ∆ CD • Unknowns ∆BC B´ C ∆ θ and ∆ 10 kN B EI 2.5EI 4 m • Boundary Conditions 2 m MDC MAB A D A x D x θA = θD = 0 Ay D 4 m 3m y • Equilibrium Conditions - Joint B B

M MBA BC

ΣM B = 0 : M BA + M BC = 0 − − − (1*)

- Entire Frame

+ → ΣFx = 0 : 10 − Ax − Dx = 0 − − − (2*) 77 • Slope-Deflection Equation C´ B ∆ CD = ∆ / cos 36.87° = 1.25 ∆ ∆ ∆ 5 BC C´ ∆ B ´ 2EI C ∆CD = ∆ tan 36.87° = 0.75 ∆ 10 kN 36.87° ∆BC EI 2.5EI 4 m 36.87° 2 m C ∆ PL/8 = 5 D A 4 m 3m 2(EI) M = θ + 0.375EI∆ + 5 − − − (1) AB 4 B 0.375EI∆ (6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆ 4(EI) B C´ M BA = θ B + 0.375EI∆ − 5 − − − (2) 4 ∆ = 1.25 ∆ ∆´ C CD 3(2EI) B´ M = θ − 0.2813EI∆ − − − (3) BC 4 B

M DC = 0.375EI∆ − − − (4) A D 0.75EI∆ (1/2) 0.75EI∆ 6EI∆/(4) 2 = 0.375EI∆ C´ (6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆ 0.5625EI∆

B ∆BC= 0.75 ∆ B´ C (1/2) 0.5625EI∆ 78 • Horizontal reactions B 2 EI pin C 10 kN EI 2.5 EI 4 m 2 m

A D 4 m 3m

M BC B C MBA M B BC MBC 4 10 kN 4 C

A Ax = (MBA+ MAB-20)/4 -----(5) M AB Dx= (MDC-(3/4)MBC)/4 ---(6) D MDC

MBC/4 79 Equilibrium Conditions: • Solve equations

M BA + M BC = 0 − − − (1*) Substitute (2) and (3) in (1*)

10 − Ax − Dx = 0 − − − (2*) 2.5EIθ B + 0.0938EI∆ − 5 = 0 − − − (7)

Slope-Deflection Equation: Substitute (5) and (6) in (2*) 2(EI) 6EI∆ M = θ + 5 + − − − (1) 0.0938EIθ B + 0.334EI∆ −5 = 0 − − − (8) AB 4 B 42 4(EI) 6EI∆ From (7) and (8) can solve; M BA = θ B − 5 + 2 − − − (2) 4 4 1.45 −14.56 3(2EI) 3(2EI)(0.75∆) θ B = ∆ = M = θ − − − − (3) EI EI BC 4 B 42 1.45 −14.56 3(2.5EI)(1.25∆) Substituteθ B = and ∆ = in (1)to (6) M DC = − − − (4) EI EI 52

Horizontal reactions at supports: MAB = 15.88 kN•m MBA = 5.6 kN•m (M BA + M AB − 20) Ax = − − − (5) M = -5.6 kN•m 4 BC M = 8.42 kN•m 3 DC M − M A = 7.9 kN DC 4 BC x Dx = − − − (6) D = 2.1 kN 4 x 80 MAB = 11.19 kN•m B 1.91 2 EI pin M = 1.91 kN•m 1.91 BA C MBC = -1.91 kN•m 10 kN M = 5.46 kN•m EI 2.5 EI 4 m DC 5.46 = 8.28 kN•m 2 m 11.19 kN•m Ax 8.27 kN 1.73 Dx = 1.72 kN•m A D 0.478 kN 4 m 3m 0.478 kN ∆ ∆ 1.91 θ =1.45/EI B B B 1.91 C C

5.35 θB=1.45/EI 5.46 11.19 A D A D

Bending-moment diagram Deflected shape

81 Example 10

From the frame shown use the moment distribution method to: (a) Determine all the reactions at supports, and also (b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve.

20 kN/m 3m B pin C 3EI

3 m 2EI 4EI 4m

A D

82 • Overview ∆ ∆ • Unknowns B θ and ∆ C B 3EI • Boundary Conditions

3 m 20 kN/m 2EI 4EI θ = θ = 0 4m A D • Equilibrium Conditions

A - Joint B D [FEM]load B 3m M MBA BC

ΣM B = 0 : M BA + M BC = 0 − − − (1*)

- Entire Frame

+ → ΣFx = 0 : 60 − Ax − Dx = 0 − − − (2*) 83 • Slope-Deflection Equation 3m ∆ ∆ B B C 2 C 6(2EI∆)/(3) 3EI = 1.333EI∆ wL2/12 = 15 1.5EI∆

3 m 20 kN/m 2EI 4EI 2 4m wL /12 = 15 6(2EI∆)/(3) 2 = 1.333EI∆ 6(4EI∆)/(4) 2 A A = 1.5EI∆ D (1/2)(1.5EI∆) [FEM]load [FEM] θ0 ∆ D 4(2EI) 2(2EI) M = + θ +15 +1.333EI∆ = 1.333EIθ +15 +1.333EI∆ ------(1) AB 3 A 3 B B θ0 2(2EI) 4(2EI) ------(2) M = + θ −15 +1.333EI∆ = 2.667EIθ B −15 +1.333EI∆ BA 3 A 3 B 3(3EI) M = θ = 3EIθ B ------(3) BC 3 B 0 3(4EI) M = θ + 0.75EI∆ = 0.75EI∆ ------(4) DC 4 D 84 • Horizontal reactions

MAB C B + ΣM B = 0 : 1.5 m M + M + 60(1.5) A = BA AB 60 kN 4 m x 3

Ax =1.333EIθ B + 0.889EI∆ + 30 − − − (5) 1.5 m A

Ax D Dx + ΣMC = 0: MDC M M D = DC = 0.188EI∆ − − − (6) BA x 4

85 Equilibrium Conditions • Solve equation

M BA + M BC = 0 − − − (1*) Substitute (2) and (3) in (1*)

60 − Ax − Dx = 0 − − − (2*) 5.667EIθ B +1.333EI∆ =15 − − − (7)

Equation of moment Substitute (5) and (6) in (2*)

−1.333EIθ B −1.077EI∆ = −30 − − − (8) M AB =1.333EIθ B +15 +1.333EI∆ − − − (1) From (7) and (8), solve equations; M BA = 2.667EIθ B −15 +1.333EI∆ − − − (2) − 5.51 34.67 M = 3EIθ − − − (3) θ B = ∆ = BC B EI EI M = 0.75EI∆ − − − (4) − 5.51 34.67 DC Substituteθ = and ∆ = in (1)to (6) B EI EI

Horizontal reaction at support M AB = 53.87 kN • m

M BA =16.52 kN • m

Ax =1.333EIθ B + 0.889EI∆ + 30 − − − (5) M BC = −16.52 kN • m

M DC = 26.0 kN • m Dx = 0.188EI∆ − − − (6) Ax = 53.48 kN

Dx = 6.52 kN 86 B 3m C M = 53.87 kN • m 16.52 kN•m AB M BA =16.52 kN • m

3 m M BC = −16.52 kN • m 20 kN/m 4m 53.87 kN•m M DC = 26.0 kN • m A = 53.48 kN 53.48 kN x

A 26 kN•m Dx = 6.52 kN D 5.55 kN 6.52 kN 5.55 kN 16.52 ∆ ∆ 16.52 B C B C

53.87 A A 26. Deflected shape Moment diagram 87 D D DISPLACEMENT MEDTHOD OF ANALYSIS: MOMENT DISTRIBUTION

! Member Stiffness Factor (K)

! Distribution Factor (DF)

! Carry-Over Factor

! Distribution of Couple at Node

! Moment Distribution for Beams

! General Beams

! Symmetric Beams

! Moment Distribution for Frames: No Sidesway

! Moment Distribution for Frames: Sidesway

1 Member Stiffness Factor (K) & Carry-Over Factor (COF) C P B w

EI ACB D L1 /2 L1/2 L2 L3

Internal members and far-end member fixed at end support: COF = 0.5 2EI 4EI k = k = DC 4EI CC L L K = L 1 CD COF = 0.5

K(BC) = 4EI/L2, K(CD) = 4EI/L3 2 C P B w

EI ACB D L1 /2 L1/2 L2 L3

Far-end member pinned or roller end support: COF = 0 3EI 0 3EI k AA = K = L L 1 CD

K(AB) = 3EI/L1, K(BC) = 4EI/L2, K(CD) = 4EI/L3

3 Joint Stiffness Factor (K) C P B w

EI ACB D L1 /2 L1/2 L2 L3

K(AB) = 3EI/L1 K(BC) = 4EI/L2, K(CD) = 4EI/L3

Kjoint = KT = ΣKmember

4 Distribution Factor (DF) C P B w

EI ACB D L1 /2 L1/2 L2 L3

K DF = ΣK

Notes: - far-end pined (DF = 1) - far-end fixed (DF = 0)

A KAB/(K(AB) + K(BC) ) BCDKBC/(K(AB) + K(BC) ) K(CD)/(K(BC) + K(CD) )

DF 1 DFBA) DFBC DFCB DFCD 0

K(BC)/(K(BC) + K(CD) ) 5 Distribution of Couple at Node

(EI) (EI) (EI) AC1 BD2 3 L1 /2 L1/2 L2 L3

A BCD

DF1 DFBA DFBC DFCA DFCD 0

CB CO=0 C (DF ) CO=0.5 B BC CB( DFBC)

CB(DFBC) B CB( DFBC)

6 50 kN•m

2EI 3EI 3EI ACB D 4 m 4 m 8 m 8 m

L1= L2 = L3

A BCD DF 1 0.333 0.667 0.5 0.5 0

50 kN•m CO=0 CO=0.5 50(.667) 50(.667) 16.67

50(.333) 50(.333) B

7 Distribution of Fixed-End Moments

P w

(EI)1 M (EI)2 (EI)13 ACF B D

L1 /2 L1/2 L2 L3

L1= L2 = 8 m, L3 = 10 m A BCD

DF1 DFBA DFBC DFCB DFCD 0

MF MF B

MF 0 MF(DFBC) MF( DFBC) 0.5

B 0 M (DF ) M ( DF ) F BC F BC 8 P w EI

2 2 wL2 /12=16 wL /12=25 AC1.5PL1/8=30B 3 D

L1 /2 L1/2 L2 L3

L1= L2 = 8 m, L3 = 10 m A BCD DF 1 0.4 0.6 0.5 0.5 0 14

30 30 B 16 16

14 0 5.6 5.6 0.5 4.2

0 B 8.4 8.4 9 Moment Distribution for Beams

10 Example 1

The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ). Use the moment distribution method to: (a) Determine all the reactions at supports, and also (b) Draw its quantitative shear and bending moment diagrams,and qualitative deflected shape.

20 kN 3 kN/m

2EI 3EI ACB 4 m 4 m 8 m

11 20 kN 3 kN/m

2EI 3EI 30 16 16 ACB CO 0.5 0 0.5 0 4 m 4 m 8 m

K1 = 3(2EI)/8 K2 = 4(3EI)/8 K1/(K1+ K2) K2/(K1+ K2)

DF1 0.333 0.667 0

[FEM]load -30 16 -16 Dist. 4.662 9.338 CO 4.669

Σ -25.3425.34 -11.33

20 kN 24 kN A B B C 25.34 kN•m 11.33 kN•m

6.83 kN 13.17 kN 13.75 kN 10.25 kN 12 Note : Using the Slope 20 kN 3 kN/m Deflection

2EI 3EI 20+10 16 16 ACB 4 m 4 m 8 m

3(2EI ) M = θ − 30 − − − (1) BA 8 B 4(3EI) M = θ +16 − − − (2) BC 8 B

MBA MBC + ΣMB = 0: -MBA - MBC = 0

B (0.75 + 1.5)EIθB - 30 + 16 = 0

θB = 6.22/EI

MBA = -25.33 kN•m,

MBC = 25.33 kN•m 2(3EI) M CB = θ B −16 = −11.33 kN • m 8 13 20 kN 3 kN/m

A C 11.33 B 6.83 kN 13.17 + 13.75 = 26.92 kN 10.25 kN 4 m 4 m 8 m 13.75 V (kN) 6.83 + + x (m) 4.58 m - - -10.25 -13.17 27.32 6.13 M + + (kN•m) - x (m) - -11.33 -25.33 Deflected shape x (m) 14 Example 2

From the beam shown use the moment distribution method to: (a) Determine all the reactions at supports, and also (b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.

10 kN 50 kN•m 5 kN/m

2EI 3EI 3EI ACB D 4 m 4 m 8 m 8 m

15 10 kN 50 kN•m 5 kN/m

2EI 15 26.67 26.67 40 3EI AC0.5 B 0.5 0.5 D 4 m 4 m 8 m 8 m

K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8

K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3) DF1 0.333 0.6670.571 0.428 1 Joint couple -16.65 -33.35

50 kN•m CO=0 CO=0.5 50(.667) 50(.667)

50(.333) 50(.333) B

16 10 kN 50 kN•m 5 kN/m

2EI 15 26.67 26.67 40 3EI AC0.5 B 0.5 0.5 D 4 m 4 m 8 m 8 m

K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8

K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3) DF1 0.333 0.6670.571 0.429 1 Joint couple -16.65 -33.35 CO -16.675 FEM -15 26.667 -26.667 40 Dist. -3.885 -7.782 1.905 1.437 CO 0.953 -3.891 Dist. -0.317 -0.636 2.218 1.673 CO 1.109 -0.318 Dist. -0.369 -0.740 0.181 0.137 Σ -36.22 -13.78 -43.28 43.25

17 10 kN 50 kN•m 5 kN/m

2EI 36.22 13.78 43.25 43.25 3EI ACB D 4 m 4 m 8 m 8 m

10 kN 40 kN 36.22 kN•m A B 43.25 kN•m

CD A B y = 0.47 kNyL = 9.53 kN CyR = 25.41 kNDy = 14.59 kN

40 kN 13.78 kN•m 43.25 kN•m

ByR= 12.87 kNCyL = 27.13 kN 18 10 kN 50 kN•m 5 kN/m

A D 2EI B C 3EI 0.47 kN 14.59 kN 9.53+12.87=22.4 kN 27.13+25.41=52.54 kN 4 m 4 m 8 m 8 m

12.87 25.41 2.92 m 0.47 V (kN) x (m) 2.57 m -9.53 -14.59 30.32 -27.13 21.29 13.78 1.88 M(kN•m) x (m)

-36.22 -43.25

Deflected x (m) shape 19 Example 3

2EI 40 kN 50 kN•m 10 kN/m 40 kN

A D EI B C 3 m 3 m 9 m 3 m

20 120 kN•m 40 kN 2EI 40 kN 50 kN•m 10 kN/m

A D 30 30 101.25 EI C 0.5 B 0.5 3 m 3 m 9 m 3 m

K1 = 4(2EI)/6 K2 = 3(EI)/9

K1/(K1+ K2) K2/(K1+ K2)

DF 00.2010.80

Joint couple 40 10 -120 CO 20 -60 FEM 30-30 101.25 Dist.

Dist. -9 -2.25 CO -4.5

Σ 45.5 1 49 -120

21 2EI 40 kN 50 kN•m 10 kN/m 40 kN 1 A D 120 45.5 49 EI 120 B C 3 m 3 m 9 m 3 m 40 kN 40 kN 45.5 kN•m 1 kN•m A B 120 kN•m

Ay = 27.75 kNByL = 12.25 kN CyR = 40 kN 90 kN 49 kN•m 120 kN•m

ByR = 37.11 kNCyL = 52.89 kN

22 2EI 40 kN 50 kN•m 10 kN/m 40 kN

A D EI C 27.75 kN B 12.25+37.11 = 49.36 kN 52.89+40 = 92.89 kN 3 m 3 m 9 m 3 m

37.11 40 27.75 + + + V (kN) - x (m) -12.25 3.71 m - 37.75 -52.89 19.84 M(kN•m) + 1 + x (m) - - - -45.5 -49

-120 Deflected x (m) shape 23 Example 4

The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ) settles 10 mm. Use the moment distribution method to: (a) Determine all the reactions at supports, and also (b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.

20 kN 15 kN•m 12 kN•m 3 kN/m

2EI 3EI ACB 4 m 4 m 8 m

24 20 kN 15 kN•m 12 kN•m 3 kN/m B AC 2EI 30 16 3EI 16 0.5 10 mm 4 m 4 m 8 m 0.5 6(3EI)∆ A 6(2EI)∆ 6(2EI)∆ − = 9.375 6(3EI)∆ 2 = 28.125 L2 2L2 = 28.125 L L2 ∆ ∆ B [FEM]∆ B 6(2EI)∆ C 2 L K1 = 3(2EI)/8 K2 = 4(3EI)/8 K1/(K1+ K2) K2/(K1+ K2)

DF1 0.333 0.667 0 Joint couple -12 510 CO -6 5

[FEM]load -30 16 -16

[FEM]∆ 9.375 -28.125 -28.125 Dist. 12.90 25.85 CO 12.92

Σ -12-8.72 23.72 -26.20 25 15 kN•m Note : Using the 20 kN 18.75-9.375 12 kN•m 3 kN/m slope deflection (9.375 ) ∆ B (28.125)∆ AC 2EI 3EI 28.125 (20+10)load (16)load 16 4 m 4 m10 mm 8 m 15 kN•m -12 θ MBA 4(2EI) 2(2EI) MBC M AB = A + θ B + 20 −18.75 − − − (1) 8 θ 8 2(2EI ) 4(2EI) B M = + θ − 20 +18.75 − − − (2) BA 8 A 8 B + ΣMB = 0: - MBA - MBC + 15 = 0 (2) − (1) 3(2EI) : M BA = θ B − 30 + 9.375 −12 / 2 − − − (2a) 2 8 (0.75 + 1.5)EIθB - 38.75 - 15 = 0 4(3EI) M = θ +16 − 28.125 − − − (3) θB = 23.9/EI BC 8 B M = -8.7 kN•m, 2(3EI) BA M CB = θ B −16 − 28.125 − − − (4) 8 MBC = 23.72 kN•m 2(3EI) M = θ −16 − 28.125 CB 8 B = −26.2 kN • m 26 20 kN 15 kN•m 12 kN•m 3 kN/m

2EI 8.725 23.7253 EI 26.205 ACB 4 m 4 m 8 m

20 kN 12 kN•m 24 kN 8.725 kN•m 26.205 kN•m A B B C

23.725 kN•m B C Ay = 7.41 kNByL = 12.59 kN yR = 11.69 kN y = 12.31 kN

27 20 kN 15 kN•m 12 kN•m 3 kN/m

A C 2EI B 3EI 26.205 7.41 kN 12.59+11.69 = 24.28 kN 12.31 kN 4 m 4 m 8 m

11.69 V (kN) 7.41 + + x (m) 3.9 m - - -12.31 -12.59 41.64 12 + M (kN•m) x (m) -8.725 - -0.93 - -23.725 -26.205

Deflected 10 mm θB = 23.9/EI shape x (m) 28 Example 5

For the beam shown, support A settles 10 mm downward, use the moment distribution method to (a)Determine all the reactions at supports (b)Draw its quantitative shear, bending moment diagrams, and qualitative deflected shape. Take E= 200 GPa, I = 50(106) mm4.

12 kN•m 6 kN/m B A C 2EI 1.5EI 10 mm 3 m 3 m

29 12 kN•m 4.5+(4.5/2) 6 kN/m = 6.75 B A 4.5 10 mm 2EI C 1.5EI 3 m0.5 3 m 0.5

6(1.5× 200×50)(0.01) 100-(100/2) = 50 0.01 m 2 C F 3 M ∆ A =100 kN • m

K1 = 4(2EI)/3 K2 = 3(1.5EI)/3

K1/(K1+ K2) K2/(K1+ K2)

DF00.36 0.64 1 Joint couple 12 CO 6

[FEM]load 6.75

[FEM]∆ 50 Dist. -40.16 -22.59 CO -20.08

Σ -20.08-40.16 40.16 12 30 12 kN•m 6 kN/m B A 2EI C 1.5EI 3 m 3 m

ΣΜ -20.08-10.16 40.16 12

20.08 kN•m 40.16 kN•m B C 40.16 + 20.08 20.08 kN = 20.08 kN 3 18 kN 12 kN•m 6 kN/m 40.16 kN•m C A 26.39 kN 8.39 kN 31 12 kN•m 6 kN/m B A 2EI C 1.5EI 10 mm 3 m 3 m 20.08 kN•m 40.16 kN•m B C 20.08 kN 20.08 kN 12 kN•m 6 kN/m 40.16 kN•m C A 26.39 kN 8.39 kN V (kN) 26.39 8.39 + x (m) -

M (kN•m) -20.08 20.08 12 x (m)

Deflected shape -40.16 x (m) 32 Example 6

12 kN•m 6 kN/m B A C 2EI 1.5EI 10 mm 3 m 3 m

33 12 kN•m • Overview 6 kN/m B A C 10 mm 2EI 1.5EI 3 m 3 m

12 kN•m 6 kN/m B A 10 mm C R

∆ B A ×C

R´

R + R'C = 0 − − − (1*) 34 12 kN•m • Artificial joint applied 4.5+(4.5/2) 6 kN/m = 6.75 B A 4.5 10 mm 2EI C 1.5EI 3 m0.5 3 m 0.5

6(1.5× 200×50)(0.01) 100-(100/2) = 50 0.01 m 2 C F 3 M ∆ A =100 kN • m

K1 = 4(2EI)/3 K2 = 3(1.5EI)/3

K1/(K1+ K2) K2/(K1+ K2)

DF00.36 0.64 1 Joint couple 12 CO 6

[FEM]load 6.75

[FEM]∆ 50 Dist. -40.16 -22.59 CO -20.08

Σ -20.08-40.16 40.16 12 35 12 kN•m 6 kN/m B A 2EI C 1.5EI 3 m 3 m

ΣΜ -20.08-40.16 40.16 12

20.08 kN•m 40.16 kN•m B C 40.16 + 20.08 18 kN 20.08 kN = 20.08 kN 12 kN•m 3 6 kN/m 40.16 kN•m C A 26.39 kN 8.39 kN 40.16 kN•m 40.16 kN•m

20.08 C 26.39 kN + ↑ ΣFy = 0 : − 20.08 − 26.39 + R = 0 R R = 46.47 kN 36 • Artificial joint removed ∆ B A C 2EI 0.5 1.5EI 0.5 3 mR´ 3 m C C 75 B ∆ ∆ 6(1.5EI)( ) 100 EI = 75 75-(75/2) A 32 6(2EI)∆ 75 C =100 → ∆ = = 37.5 32 EI DF00.36 0.64 1

[FEM]∆ -100 -100 +37.5 Dist. 40 22.5 CO 20 Σ -80-60 60

80 kN•m 60 kN•m 60 kN•m B C C A 46.67 kN 46.67 kN 20 kN 20 kN C 46.67 20 kN + ↑ ΣFy = 0 : R'= 66.67 kN 37 R´ • Solve equation Substitute R = 46.47 kN and R'= 66.67 kN in (1*)

46.47 + 66.67C = 0 C = −0.6970 12 kN•m 6 kN/m B 20.08 kN•m A 10 mm 20.08 kN C R = 46.47 kN 8.39 kN

∆ B 80 kN•m A ×C = −0.6970 46.67 kN 20 kN R´ = 66.67 kN 12 kN•m 6 kN/m 35.68 kN•m B A 2EI C 1.5EI 12.45 kN 5.55 kN 38 12 kN•m 6 kN/m 35.68 kN•m B A 2EI C 1.5EI 10 mm 12.45 kN 3 m 3 m 5.55 kN

12.45 V (kN) 0.925 m + x (m) -5.55

M (kN•m) 14.57 12 1.67 + x (m) -

-35.68 Deflected shape x (m)

39 Symmetric Beam • Symmetric Beam and Loading P P

θ θ

ACBD L´ L L´ real beam L L V´B V´C M L 2 2 + ΣMC´ = 0: −VB' (L) + (L)( ) = 0 EI 2 B´ C´ ML M V =θ = M B' 2EI L EI EI 2EI conjugate beam M = θ L The stiffness factor for the center span is, therefore,

2EI K = L 40 • Symmetric Beam with Antisymmetric Loading

P θ A D B θ C P L´ L L´ real beam 1 M L 1 M L 2L ( )( ) M + ΣM = 0: −V (L) + ( )( )( ) = 0 C´ B' 2 EI 2 3 2 EI 2 EI 2 V´B L ML B´ V =θ = 3 B' 6EI C´ 6EI 1 M L V´C M = θ ( )( ) L M 2 EI 2 EI The stiffness factor for the center span is, therefore, conjugate beam 6EI K = L 41 Example 5a

Determine all the reactions at supports for the beam below. EI is constant.

15 kN/m

A D B C 4 m 6 m 4 m

42 15 kN/m

A D wL2/15 = 16 B wL2/12 = 45 C wL2/15 = 16 4 m 6 m 4 m 3EI 3EI 2EI 2EI K = = , K = = ( AB) L 4 (BC) L 6

K( AB) K( AB) (3EI / 4) (DF) AB = =1,(DF)BA = = = 0.692, K( AB) K( AB) + K(BC) (3EI / 4) + (2EI / 6)

K(BC) (2EI / 6) (DF)BC = = = 0.308 K( AB) + K(BC) (3EI / 4) + (2EI / 6) DF 1.0 0.692 0.308

[FEM]load 0-16+45 Dist. -20.07 -8.93 ΣΜ -36.07 +36.07 30 kN 30 kN 8 90 kN 8 36.07 kN•m 36.07 kN•m 3 3 A B B C C D 4 m 3 m 3 m 4 m 0.98 kN 29.02 kN 45 kN 45 kN 29.02 kN 0.98 kN 43 30 kN 30 kN 8 90 kN 8 36.07 kN•m 36.07 kN•m 3 3 A B B C C D 4 m 3 m 3 m 4 m 0.98 kN 29.02 kN 45 kN 45 kN 29.02 kN 0.98 kN

15 kN/m

A D B C 0.98 kN 74.02 kN 74.02 kN 0.98 kN 4 m 6 m 4 m 45 29.02 V 0.98 (kN) x (m) -0.98 -29.02 -45 31.42 M + (kN•m) x (m) - - -36.07 -36.07 Deflected shape 44 Example 5b

Determine all the reactions at supports for the beam below. EI is constant.

15 kN/m

A C D B 15 kN/m 4 m 3 m3 m 4 m

45 Fixed End Moment 15 kN/m A C D wL2/15 = 16 B 11wL2/192 5wL2/192 = 30.938 = 14.063

5wL2/192 11wL2/192 = 14.063 = 30.938 wL2/15 = 16 A C D B 15 kN/m

15 kN/m 16.875 16 A C D 16 B 16.875 15 kN/m

46 15 kN/m 16.875 16 A C D 16 B 16.875 15 kN/m 4 m 3 m3 m 4 m

3EI 3EI 6EI 6EI K = = = 0.75EI, K = = = EI ( AB) L 4 (BC) L 6 0.75 1 (DF) =1, (DF) = = 0.429, (DF) = = 0.571 AB BA 0.75 +1 BC 0.75 +1 DF 1.0 0.429 0.571

[FEM]load 0 -16 16.875 Dist. -0.375 -0.50 ΣΜ -16.37516.375 8 30 kN 8 45 kN 3 16.375 kN•m 16.375 kN•m 3 4 m A B B C C D 4 m 5.91 kN 24.09 kN 27.96 kN kN kN 5.91 27.96 45 kN 24.09 kN 30 kN 47 8 30 kN 8 45 kN 3 16.375 kN•m 16.375 kN•m 3 4 m A B B C C D 4 m 5.91 kN 24.09 kN 27.96 kN kN kN 5.91 27.96 45 kN 24.09 kN 30 kN

15 kN/m

A C D B 15 kN/m 5.91 kN 52.05 kN 52.05 kN 5.91 kN 27.96 27.96 V 5.91 5.91 (kN) x (m) -24.09 -24.09 M 16.375 (kN•m) x (m) -16.375 Deflected shape 48 Moment Distribution Frames: No Sidesway

49 Example 6

From the frame shown use the moment distribution method to: (a) Determine all the reactions at supports (b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape. 40 kN•m 48 kN 8 kN/m B C A 2.5EI 2.5EI

4 m 3EI

5 m D 5 m

50 40 kN•m 48 kN 8 kN/m B C A 2.5EI 45 25 2.5EI 0.5 0.5 4 m 3EI 0.5 KAB = KBC = 3(2.5EI)/5 = 1.5 EI

KBD = 4(3EI)/4 = 3EI 5 m D 5 m

ABDC

Member ABBA BC BD DB CB

DF100.25 0.25 0.5 1 Joint load-10 -10 -20 CO -10 FEM-45 25 Dist.55 10 CO 5

Σ 0-5-5020 -10 0 51 40 kN•m 48 kN 8 kN/m B C A 20 2.5EI 50 2.5EI 14 kN 10 16 kN 3EI 5 Member AB BA BC BD DB CB D Σ 0-5-5020 -10 0

40 kN•m 58 kN 34 24 3.75 50 20 48 kN 3.75 40 kN A B 0 50 kN•m 58 3.75 10 BC 58 20 3.75 14 kN 34 kN 10 kN•m 24 kN 16 kN 3.75 kN

5 3.75 kN 58 52 40 kN•m 48 kN 8 kN/m B C A 2.5EI 20 2.5EI 14 kN 50 16 kN 3EI 10 4 m 5 D 5 m 5 m

58 kN

35 16 10

50 Moment diagram Deflected shape

53 Moment Distribution for Frames: Sidesway

54 Single Frames

P P ∆ ∆ C C R´ BCB R B

x C1

A D A D A D

Artificial joint applied Artificial joint removed (no sidesway) (sidesway)

0 = R + C1R´

55 P3 Multistory Frames ∆2 P2

P1 ∆1

P3 ∆´´ ∆´´ P 2 R2´´ R R ´ 2 ∆´ ∆´ 2 P4 x C1 x C2 P1 R1´ R1 R1´´

0 =R2 + C1R2´ + C2R2´´

0 =R + C R ´ + C R ´´ 1 1 1 2 2 56 Example 7

5 m

A D

57 • Overview 16 kN 16 kN 1 m 4 m C C R´ BCB R B

5 m 5 m = + x C1

A D A D A D

artificial joint applied artificial joint removed (no sidesway) ( sidesway)

R + C1R´ = 0 ------(1)

58 • Artificial joint applied (no sidesway)

Fixed end moment:

16 kN b = 1 m a = 4 m C B R Pa2b/L2 Pb2a/L2 = 10.24 = 2.56

A D

Equilibrium condition :

+ ΣFx = 0: Ax + Dx + R = 0

59 16 kN 1 m 4 m C A BC D B R 0.5 0.50 0.50 10.24 2.56 DF00.50 0.50 0 FEM 10.24 -2.56 5 m 5 m 0.5 0.5 Dist. -5.12 -5.12 1.28 1.28 CO -2.56 0.64-2.56 0.64 A D Dist. -0.32 -0.32 1.28 1.28 CO -0.16 0.64-0.16 0.64 Dist. -0.32 -0.32 0.08 0.08 5.78 kN•m 2.72 kN•m CO -0.16 0.04-0.16 0.04 Dist. -0.02 -0.02 0.08 0.08 B C Σ -2.88 -5.78 5.78-2.72 2.72 1.32

5 m 5 m Equilibrium condition :

+ ΣF = 0: 1.73 - 0.81 + R = 0 A D x Ax = 1.73 kN Dx = 0.81 kN R = - 0.92 kN

2.88 kN•m 1.32 kN•m 60 • Artificial joint removed ( sidesway)

Fixed end moment: Since both B and C happen to be displaced the same amount ∆, and AB and DC have the same E, I, and L so we will assume fixed-end moment to be 100 kN•m.

∆ ∆ B 5m C 100 kN•m R´ 100 kN•m

5 m 5 m 100 kN•m 100 kN•m A D

Equilibrium condition :

+ ΣFx = 0: Ax + Dx + R´ = 0

61 ∆ ∆ B 5m C R´ A BC D 0.5 100 kN•m 100 kN•m DF00.500.50 0.50 0.50 0 FEM 100 100 100 100 5 m 5 m 0.5 0.5 Dist. -50-50 -50 -50 100 kN•m 100 kN•m CO -25.0 -25.0-25.0 -25.0 A D Dist. 12.512.5 12.5 12.5 CO 6.5 6.5 6.5 6.5 Dist. -3.125-3.125 -3.125 -3.125 60 kN•m 60 kN•m CO -1.56 -1.56-1.56 -1.56 Dist. 0.780.78 0.78 0.78 CO 0.39 0.39 0.39 0.39 B C Dist. -0.195-0.195 -0.195 -0.195 Σ 80 60 -60-60 60 80 5 m 5 m

Equilibrium condition: + ΣF = 0: A D x Ax = 28 kN Dx = 28 kN -28 - 28 + R´ = 0

80 kN•m 80 kN•m R´ = 56 kN 62 Substitute R = -0.92 and R´= 56 in (1) :

R + C1R´ = 0

-0.92 + C1(56) = 0 0.92 C = 1 56

16 kN 16 kN BC C R´ 1 m 4 m C R B B 5.78 2.72 60 60 4.79 3.71 2.72 60 5.78 60 4.79 3.71 + x C1 = 5 m 5 m 2.88 80 1.32 80 1.57 2.63 1.73 kN 0.81 28 28 1.27 kN A D A D A D 1.27

63 8.22 16 kN 3.71 1 m 4 m C B 4.79 4.79 3.71 4.79 3.71 4.79 3.71 5 m 5 m

1.57 2.63

1.27 kN 1.27 kN 1.57 2.63 Bending moment A D 2.99 kN 13.01 kN diagram (kN•m) ∆ ∆

Deflected shape 64 Example 8

20 kN/m 3m B pin C 3EI

3 m 2EI 4EI 4m

A D

65 • Overview

20 kN/m B 3m, 3EI B C B C C R R´

3 m , 2EI = 4m , 4EI + x C1

A A A D D D

artificial joint applied artificial joint removed (no sidesway) (sidesway)

R + C1R´ = 0 ------(1)

66 • Artificial joint applied (no sidesway) 20 kN/m A BC D B 3m , 3EI C R DF01.000.471 0.529 1.00 0 0.5 15 FEM 15.00 -15.00 Dist. 7.065 7.935 0.5 0.5 CO 3.533 3 m , 2EI 4m, 4EI 15 Σ 18.53 -7.94 7.94

A D 7.94 kN•m

KBA = 4(2EI)/3 = 2.667EI B + C ΣFx = 0: KBC = 3(3EI)/3 = 3EI 60 - 33.53 - 0 + R = 0 60 K = 3(4EI)/4 = 3EI 3 m 4 m CD R = - 26.47 kN

A D Ax = 33.53 kN Dx = 0

18.53 kN•m 0 67 • Artificial joint removed ( sidesway)

• Fixed end moment ∆ ∆ ∆ ∆ B B 3m, 3EI C R´ 3m, 3EI C R´ 100 kN•m 2 6(2EI∆)/(3) 2 6(4EI)∆/(4) 100 kN•m 3 m, 2EI 100 kN•m 4m, 4EI 3 m, 2EI 4m, 4EI 6(2EI∆)/(3) 2 100 kN•m 3(4EI∆)/(4) 2 3(4EI)(75/EI)/(4) 2 A A = 56.25 kN•m

D D

Assign a value of (FEM)AB = (FEM)BA = 100 kN•m

6(2EI)∆ =100 32

∆AB = 75/EI 68 ∆ ∆ A BC D B 3m C R´ 0.5 DF01.000.471 0.529 1.00 0 100 FEM 100 100 56.25 0.5 Dist. -47.1 -52.9 0.5 0 3 m 4m CO -28.55 100 Σ 76.45 52.9 -52.9 56.25 56.25 A

D 52.9. kN•m + C ΣFx = 0: B -43.12 - 14.06 + R´ = 0 3 m 4 m R´ = 57.18 kN A A = 43.12kN x D 14.06 kN

76.45 kN•m 56.25 kN•m 69 Substitute R = -26.37 and R´= 57.18 in (1) :

R + C1R´ = 0

-26.47 + C1(57.18) = 0 26.47 C = 1 57.18 20 kN/m B C B 52.9 kN•m C B 3m R C 7.94 kN•m R´ 16.55 kN•m 7.94 kN•m 52.9 = 3 m + x C1 4m 18.53 kN•m 76.45 kN•m 53.92 kN•m

33.53 kN 43.12 kN 53.49 kN 56.25 A 0 A A 26.04 kN•m 5.52 kN 0 6.51 kN D D 14.06 D 5.52 kN 70 20 kN/m 16.55 B 3m 16.55 C B C 16.55 kN•m

3 m 53.92 kN•m 4m 53.92 53.49 kN A 26.04 kN•m A 5.52 kN 26.04 6.51 kN Moment diagram D ∆ ∆ D 5.52 kN B C

A Deflected shape D 71 Example 8

10 kN B C

4 m

A D 4 m 3m

72 • Overview 10 kN B C

4 m

A D 4 m 3m

R + C R´ = 0 ------(1) = 1

10 kN B C B C R R´

x C + 1

A D A D artificial joint applied artificial joint removed (no sidesway) (sidesway) 73 • Artificial joint applied (no sidesway)

10 kN B C R

0 0 A D

+ Equilibrium condition : ΣFx = 0: 10 + R = 0 R = - 10 kN

74 • Artificial joint removed (sidesway)

• Fixed end moment 6EI∆ /(4) 2 B BC 2 R´ 6EI∆AB/(4) 2 C 3EI∆CD/(5) 100 kN•m 100 kN•m 6EI∆ /(5) 2 4 m 2 CD 6EI∆AB/(4) A D 4 m 3m 6EI∆ Assign a value of (FEM)AB = (FEM)BA = 100 kN•m : AB =100 42

∆AB = 266.667/EI ∆ C´∆ B ∆ CD BC R´ ∆ ∆CD = ∆ / cos 36.87° = 1.25 ∆ = 1.25(266.667/EI) B´ C = 333.334/EI 36.87°4 m C´ ∆CD ∆BC = ∆ tan 36.87° = 0.75 ∆ 36.87° = 0.75(266.667/EI) A D C ∆AB = ∆ = 200/EI 4 m 3m 75 ∆BC= 200/EI, ∆CD = 333.334/EI

6EI∆ /(4) 2 = 6(200)/42 = 75 kN•m B BC R´ 100 kN•m 2 2 C 3EI∆CD/(5) = 3(333.334)/5 = 40 kN•m

100 kN•m

A D

Equilibrium condition :

+ ΣFx = 0: Ax + Dx + R´ = 0

76 75 A BCD B 75 R´ DF00.50 0.50 0.625 0.375 1 C 100 0.5 100 100 -75 -75 40 40 FEM 4 m Dist. -12.5 -12.5 21.875 13.125 0.5 0.5 100 CO -6.25 10.938 -6.25 A D Dist. -5.469 -5.469 3.906 2.344 4 m 3m CO -2.735 1.953 -2.735 Dist. -0.977 -0.977 1.709 1.026 K = 4EI/4 = EI, K = 4EI/4 = EI, BA BC Σ 91.02 81.05 -81.05 -56.48 56.48 KCD = 3EI/5 = 0.6EI 34.38 kN 34.38 kN 81.05 81.05 B C 56.48 B 56.48 C 34.38 kN 34.38 kN

39.91 kN + ΣF = 0: A 43.02 kN x D 91.02 -43.02 - 39.91 + R´ = 0 34.38 kN R´ = 82.93 kN 34.38 kN 77 Substitute R = -10 kN and R´= 82.93 kN in (1) : -10 + C1(82.93) = 0

R + C1R´ = 0 ------(1) C1 = 10/82.93

81.05 10 kN B C B C R R´ 56.48 81.05 x C = 10/82.93 + 56.48 1 0 0 91.02 39.91 kN A 0 D A 43.02 kN D = 0 34.38 kN 0 34.38 kN 9.77 10 kN B C 9.77 6.81 6.81 4 m

10.98 4.81 kN A 5.19 kN D 4.15 kN 4.15 kN 4 m 3m 78 9.77 10 kN B C 9.77 6.81 6.81 4 m 10.98 A 5.19 kN 4.81 kN D 4.15 kN 4 m 3m 4.15 kN

9.77 B B 9.77 C C

6.81

10.98 A D A D

Bending moment diagram Deflected shape (kN•m)

79 Example 9

A D 2 m 3 m 2m

80 40 kN • Overview 20 kN B C 3EI 3 m 4EI 4EI 4 m

A D 2 m 3 m 2m

R + C1R´ = 0 ------(1)

40 kN = 20 kN B C B C R R´

+ x C artificial joint applied artificial joint removed 1 (no sidesway) (no sidesway) A A D D 2 m 3 m 2m 2 m 3 m 2m 81 • Artificial joint applied (no sidesway)

Fixed end moments:

40 kN PL/8 = 15 B C 20 kN R 15+(15/2) = 22.5 kN•m

A D 2 m 3 m 2m

Equilibrium condition :

+ ΣFx = 0: Ax + Dx + R = 0

82 40 kN A BC D 20 kN B C DF01.000.60 0.40 1.00 0 R 22.5 FEM 22.5 15 0.5 Dist. -13.5 -9.0 0.5 0.5 CO -6.75 Σ -6.75 -13.5 13.5 A D 2 m 3 m 2m KBA = 4(4EI)/3.6 = 4.444EI, KBC = 3(3EI)/3 = 3EI,

24.5 kN 15.5 kN 13.5 kN•m C + B ΣFx = 0: 40 kN 23.08 + 20 -7.75 + R´ = 0 13.5 B C R´ = - 35.33 kN A 23.08 kN 7.75 kN 24.5 kN 15.5 kN D 6.75 kN•m 0 24.5 kN 15.5 kN 83 • Artificial joint removed (sidesway)

Fixed end moments: 3 m B C R´ m 6 3EI 4 0 . 6 4 . 7 3 2 m

4EI 4EI A D 3(3EI)∆ /(3) 2 2 BC 6(3EI)∆BC/(3) 100 kN•m 2 R´ 6(4EI)∆AB/(3.606) BC 2 6(4EI)∆CD/(4.47) 100 kN•m 6(4EI)∆ /(3.61) 2 AB 2 3(4EI)∆CD/(4.472) A D

Assign a value of (FEM) = (FEM) = 100 kN•m : 6(4EI)∆ AB AB BA =100, ∆AB = 54.18/EI 3.612 84 CC´ ∆ = ∆ cos 33.69° = 45.08/EI BB´ AB B C C´ R´ C´

o CD 9 B´ ∆ .6 3 ∆ tan 26.57 = 22.54/EI 3 C 26.57° B 33.69o

26.57° ∆ AB = ∆ tan 33.69 = 30.05/EI 5 4.3 A /EI D B´

∆BC = B'C'= 22.54 / EI + 30.05/ EI = 52.59 / EI

∆CD = ∆/cos 26.57°= 50.4/EI

2 2 3(3EI)∆BC/(3) = 3(3EI)(52.59/EI) /(3) = 52.59 kN•m R´ 100 kN•m BC

100 kN•m 2 3(4EI)∆CD/(4.472) A = 3(4EI)(50.4/EI)/(4.472) 2 D = 30.24 kN•m 85 3EI A BC D 52.59 B C DF01.000.60 0.40 1.00 0 R´ 100 0.5 FEM 100 100-52.59 30.24 3 m Dist. -28.45 -18.96 100 0.5 0.5 4 m CO -14.223 4EI 4EI Σ 85.78 71.55 -71.55 30.24 A D 30.24 2 m 3 m 2m 23.85 kN 23.85 kN 71.55 kN•m + ΣF = 0: C x B -68.34 - 19.49 + R´ = 0 71.55 B C R´ = 87.83 kN

A 68.34 kN 23.85 23.85 19.49 kN D 85.78kN•m 30.24 kN•m 23.85 kN 23.85 kN 86 Substitute R = -35.33 and R´= 87.83 in (1) : -35.33 + C1(87.83) = 0 40 kN C1 = 35.33/87.83 20 kN B C B C 35.33 kN 90.59 kN 13.5 kN•m 71.55 kN•m

+ x C1 6.75 kN•m 85.78 kN•m 23.08 kN 68.34 kN 0 30.24 kN•m A 24.5 kN A 23.85 kN 7.75 kN 19.485 kN D D = 15.5 kN 23.85 kN 40 kN 20 kN B C 15.28 kN•m

27.76 kN•m 4.41 kN 12.16 kN•m A 14.91 kN 15.59 kN D 25.09 kN 87 40 kN 20 kN B C 15.28 kN•m

27.76 kN•m 4.41 kN 12.16 kN•m A 14.91 kN 15.59 kN D 37.65 25.09 kN

15.28 C C B B

A A 27.76 12.16 Deflected shape Bending moment diagram D D

88 TRUSSES ANALYSIS

! Fundamentals of the Stiffness Method

! Member Local Stiffness Matrix

! Displacement and Force Transformation Matrices

! Member Global Stiffness Matrix

! Application of the Stiffness Method for Truss Analysis

! Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors

! Space-Truss Analysis

1 2-Dimension Trusses

2 Fundamentals of the Stiffness Method

4 2 3 1 2 1 1 (x , y ) (x2, y2) 1 1 • Node and Member Identification 2 3 6 8 5 3 4 7 • Global and Member Coordinates (x , y ) 3 3 (x4, y4)

• Degrees of Freedom

•Known degrees of freedom D3, D4, D5, D6, D7 and D8 • Unknown degrees of freedom D1 and D2

3 Member Local Stiffness Matrix

x´ y´ q´ AE AE j q'i = d'i − d' j j L L

i AE AE q´ q' = − d' + d' i j L i L j x´ y´ x dd´d q'i  AE  1 −1 d'i  = 1 AE/L ii i d´ i   =   q' L −1 1  d' AE/L  j     j 

= 1 d´ j x´ [q´] = [k´][d´] ------(1) y´ AE/L x d´ j j AE  1 −1 [k'] =   L −1 1  AE/L 4 Displacement and Force Transformation Matrices

y y´ x´

j θy m (xj,yj) x θx i

(xi,yi)

λ x − x x − x = cosθ = j i = j i x x L 2 2 (x j − xi ) + (y j − yi ) λ y − y y − y = cosθ = j i = j i y y L 2 2 (x j − xi ) + (y j − yi )

5 • Displacement Transformation Matrices y y´ x´ djy d´j d j jx θ m m j y diy x θx i dix d´i i Local Global

θ λx λy

d'i = dix cos x + diy cosθ y

d' j = d jx cosθ x + d jy cosθ y λ λ λ λ  d ix    d '   0 0  d iy  0 0  i = x y λ   [T ] = x y λ     d   0 0  d ' j   0 0 x λ y  jx  x λ y    d jy  [d´] = [T][d] ------(2) 6 • Force Transformation Matrices

qjy y´ x´ y y q´j qjx m j θy j θy m qiy θ x x θ i x x qix q´i i Global Local

λ x λ q = q' cosθ where ix i x q  λ ix λ x 0  λy  0  q = q' cosθ q    λ x iy i y iy y 0  q 'i      =   0   λ   T  y  q = q' cosθ q jx  0 x  q ' j [T ] = jx j x   λ      0 x  q 0 λ y   q = q' cosθ  jy    jy j y  0 λ y  [q] = [T]T[q´] ------(3) 7 Member Global Stiffness Matrix [q] = [T]T[q´] ------(3) Substitute ( [q´] = [k´][d´] + [q´F] ) into Eq. 3, yields the result, [ q ] = [ T ]T ([k´][d´] + [q´F] ) = [ T ]T [ k´ ][T][d] + [ T ]T [q´F] = [k][d] + [qF] [ k ] = [ T ]T[ k´ ][T] [ k ] [qF] = [ T ]T [q´F]

λx 0 λ 0 AE 1 -1 λx λy 0 0 [ k ] = y L 0 λx -1 1 0 0 λx λy 0 λy U VUV

U λxλx λxλy −λxλx −λxλy AE V λ λ λ λ −λ λ −λ λ [ k ] = y x y y y x y y L U −λxλx −λxλy λxλx λxλy

V −λyλx −λyλy λyλx λyλy

8 Application of the Stiffness Method for Truss Analysis

Equilibrium Equation:

[Qa] = [K][D] + [QF] Partitioned Form:

Joint Load Unknown Displacement

Q D F k K11 K12 u Q k = + Q D F u K21 K22 k Q u

Reaction Boundary Condition

F [Qk] = [K11][Du] + [K12][Dk] + [Q ] -1 F [Du] =[Ku] (([Qk] - [Q ]) - [K12][Dk])

9 Member Forces y´ x´ y q´j j θy m

θx x

q´i i

F q'i  AE  1 −1 d'i  q' i  = +        F  q' j L −1 1 d' j λ q' j        λ   d ix     0 0  d iy x y λ     d   0 0 x λ y  jx   d λ  jy  λ Dix    Diy F q'i  AE  1 −1  x y 0 0  q' i  = λ   +      D   F  q' j  L −1 1   0 0 x λy  jx q' j    D jy  10 λ λ λ λ Dix  λ   λ Diy F q'i  AE  x y − x − y  q' i  = λ   +    D   F  q' j  L  − x − y x λy  jx q' j    D jy 

Dxi D q´ = AE −λ −λ λ λ yi + q ´ F j L x y x y j Dxj

Dyj y´ x´ y q´j j θy m

θx x

q´i i Member Forces

11 Member Forces

Dxi D q = AE −λ −λ λ λ yi + q ´F m L x y x y j Dxj

y´ x´ Dyj y qm j θy m

θx x i Member Forces

12 Example 1

For the truss shown, use the stiffness method to: (a) Determine the deflections of the loaded joint. (b) Determine the end forces of each member and reactions at supports. Assume EA to be the same for each member.

5 m 3 m 50 kN 5 m m 3 m 5 80 kN

4 m 4 m

13 6 5 m 3 5 3 m (-4,3) 2 2 50 kN 5 m 1 m 3 m 5 1 1 4 3 8 80 kN (0,0) (-4,-3) 7 3 4 4 m 4 m 2 (4,-3)

(x − x )iˆ (y − y ) ˆj ˆ j i j i λij = + L L Member λx λy

#1 -4/5 = -0.8 -3/5 = -0.6 cosθx = λx cosθy = λy #2 -4/5 = -0.8 3/5 = 0.6 U V U V i i j j #3 4/5 = 0.8 -3/5 = -0.6

Ui λxλx λxλy −λxλx −λxλy AE V λ λ λ λ −λ λ −λ λ [ k ] = i y x y y y x y y m L Uj −λxλx −λxλy λxλx λxλy

Vj −λ λ −λ λ λ λ λ λ y x y y y x y y 14 6 2 2 3 5 Member λx λy λx λx λy λy 2 2 #1 -0.8 -0.6 0.64 0.48 0.36 1 #2 - 0.8 0.6 0.64 -0.48 0.36 1 1 4 3 8 #3 0.8 -0.6 0.64 -0.48 0.36 7 3 4 2 1 234 1 256 1 0.64 0.48 -0.64 -0.48 1 0.64 -0.48 -0.64 0.48 AE 2 0.48 0.36 -0.48 -0.36 AE 2 -0.48 0.36 0.48 -0.36 [ k ]1 = [ k ]2 = 5 3 -0.64 -0.48 0.64 0.48 5 5 -0.64 0.48 0.64 -0.48 4 -0.48 -0.36 0.48 0.36 6 0.48 -0.36 -0.48 0.36

1 278 12 1 0.64 -0.48 -0.64 0.48 AE 2 -0.48 0.36 0.48 -0.36 1 1.92 -0.48 [ k ] = 3 5 AE 7 -0.64 0.48 0.64 -0.48 [K] = 5 2 -0.48 1.08 8 0.48 -0.36 -0.48 0.36 15 6 5 m Global 5 3 m 2 50 kN 2 5 m 1 m 3 m 5 1 3 80 kN 4 8 3 7 4 m 4 m

1 2 1 Q1 = -50 1 1.92 -0.48 D1 0 AE + = 0 2 Q2 = -80 5 2 -0.48 1.08 D2

D1 -250.65/AE = D2 -481.77/AE

16 6 Local 3 5 λ Dxi  λ D  2 2 AE λ  yi  F [q'F ]m = []− x − y x λx + []q' L Dxj  1   1 1 Dyj  4 3 8 D = -250.65/AE 3 7 1 4 D = -481.77/AE ] = AE 0.8 0.6 -0.8 -0.6 2 2 [q´F 1 5 D3= 0.0 Member λ λ x y D = = -97.9 kN (C) 4 0.0 #1 -0.8 -0.6 #2 -0.8 0.6 D1= -250.65/AE #3 0.8 -0.6 D = AE 0.8 -0.6 -0.8 0.6 2 -481.77/AE [q´F]2 = 5 D = 0.0 17.7 kN 5 D = = +17.7 kN (T) 6 0.0 36.87o 50 kN D1= -250.65/AE D = -481.77/AE 17.7 kN [q´ ] = AE -0.8 +0.6 +0.8 -0.6 2 97.9 kN 80 kN F 3 5 D7= 0.0 D = = -17.7 kN (C) 8 0.0 17 6 3 5 17.7 kN 2 2 36.89o 1 50 kN 1 1 3 4 8 17.7 kN 3 7 97.9 kN 80 kN 4 Check : 2 + 17.7 + 17.7 +50cos 36.89 Member λ λ ΣFx ´ = 0: x y - 97.9cos73.78 - 80cos53.11 = 0, O.K #1 -0.8 -0.6 #2 -0.8 0.6 17.7(0.6)=10.62 kN #3 0.8 -0.6 17.7(0.8)=14.16 kN

50 kN 97.9(0.6)=58.74 kN 17.7(0.6)=10.62 kN

80 kN 17.7(0.8)=14.16 kN 97.9(0.8)=78.32 kN 18 Example 2

For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the deflections of the loaded joint. The support B settles downward 2.5 mm. Temperature in member BD increase 20 oC. Take α = 12x10-6 /oC, AE = 8(103) kN.

8 kN D 4 kN A

o C 3 m 0 +2 C B 4 m ∆B = 2.5 mm

19 8 kN 4 2 D 3 1 4 kN 2 1 1 A (-4,0) (0,0) 2 o C 3 3 m 0 +2 6 8 5 C 3 4 7 B 4 m (-4,-3) (0,-3) ∆B = 2.5 mm Member λ λ (x − x )iˆ (y − y ) ˆj x y λˆ = j i + j i ij L L #1 -4/4 = -1 0 #2 -4/5 = -0.8 -3/5 = -0.6 cosθx = λx cosθy = λy #3 0 -3/3 = -1

Ui Vi Uj Vj

Ui λxλx λxλy −λxλx −λxλy AE V λ λ λ λ −λ λ −λ λ [ k ] = i y x y y y x y y m L Uj −λxλx −λxλy λxλx λxλy V −λ λ −λ λ λ λ λ λ j y x y y y x y y 20 4 2 3 1 1 Member λ λ λ 2/L λ λ /L λ 2/L 2 1 x y x x y y (0,0) (-4,0) #1 -1 0 0.25 0 0 2 3 #2 - 0.8 -0.6 0.128 0.096 0.072 6 8 5 #3 0 -1 0 0 0.333 3 4 7 (-4,-3) (0,-3) 1 234 1 2 5 6 1 0.25 0 -0.25 0 1 0.128 0.096 -0.128 -0.096 2 0 0 0 0 3 3 2 0.096 0.072 -0.096 -0.072 [k]1 = 8x10 [k]2 = 8x10 3 -0.25 0 0.25 0 5 -0.128 -0.096 0.128 0.096 4 0 0 0 0 6 -0.096 -0.072 0.096 0.072

1 278 1 2 1 0 0 0 0 1 0.378 0.096 3 2 0 0.333 0 -0.333 3 [k]3 = 8x10 [K] = 8x10 7 0 0 0 0 2 0.096 0.405 8 0 -0.333 0 0.333 21 Member 2: [q] = [k] [d] + [qF] 4 m 2 1.92 kN 1.152 kN 3 1 2 1 1 (-4,0) (0,0) 2 2 1.536 kN 2 o o C 3 +20 C +20oC 20 1.536 kN 6 + 8 3 5 4 7 1.92 kΝ =α(∆T1)AE 1.152 kN (-4,-3) (0,-3) = (12x10-6)(20)(8x103) ∆B = 2.5 mm 1 256 q 1 1 1 0.128 0.096 -0.128 -0.096 d1 -1.536 q2 2 0.096 0.072 -0.096 -0.072 d -1.152 2 = 8x103 2 + q5 5 -0.128 -0.096 0.128 0.096 d5 = 0 1.536 5 q -3 6 6 -0.096 -0.072 0.096 0.072 d6 = -2.5x10 1.152 6 1 2 56 1 q1 1 0.128 0.096 d1 -0.128 -0.096 0 5 -1.536 = 8x103 3 + 8x10 -3 + q2 2 0.096 0.072 d2 -0.096 -0.072 -2.5x10 6 -1.152 2

1 q1 1 0.128 0.096 d1 1.92 -1.536 = 8x103 + + q 2 0.096 0.072 d 1.44 -1.152 2 2 222 8 kN 4 2 D 3 1 4 kN 2 1 1 A (-4,0) (0,0) 2 o C 3 3 m 0 +2 6 8 5 C 3 4 B 7 4 m (-4,-3) (0,-3) ∆B = 2.5 mm

[Q] = [K][D] + [QF]

Global: 1 2

Q1 = -4 1 0.378 0.096 D1 1.92 -1.536 = 8x103 + + Q2 = -8 2 0.096 0.405 D2 1.44 -1.152

-3 D1 -0.8514x10 m = -3 D2 -2.356x10 m 23 4 2 Local λ Dxi  3 1 1 2 1 λ D  AE λ  yi  F [q'F ]m = []− x − y x λx + []q' L Dxj  2 3   Dyj  6 8 5 D = -0.8514x10-3 3 4 7 1 D = -2.356x10-3 3 1.0 0.0 -1.0 0.0 2 [q´F]1 = 8x10 4 D3= 0.0 Member λx λy D = 0.0 = -1.70 kN (C) 4 #1 -1 0 -0.8514x10-3 #2 - 0.8 -0.6 D1= D = -2.356x10-3 #3 0-1[q´ ] = 8x103 0.8 0.6 -0.8 -0.6 2 + -1.92 F 2 0.0 5 D5= 1.92 kN D = -0.0025 = -2.87 kN (C) 6

D =-0.8514x10-3 2 1 +20oC D =-2.356x10-3 [q´ ] = 8x103 0.0 1.0 0.0 -1.0 2 F 3 0.0 3 D7= 1.92 kN D = 0.0 = -6.28 kN (C) 8 24 4 2 3 1 Member cosθ cosθ [q´] 2 1 1 x y m #1 -1 0 -1.70 2 3 6 #2 - 0.8 -0.6 -2.87 8 5 3 4 7 #3 0 -1 -6.28

8 kN 1 8 kN 4 kN 1.70 kN 1.70 kN 4 kN 2 3

2.87(0.8) = 2.30 kN 2.87 kN 6.28 kN

2.87(0.6) = 1.72 kN 6.28 kN

25 Example 3

For the truss shown, use the stiffness method to: (a) Determine the end forces of each member and reactions at supports. (b) Determine the displacement of the loaded joint. Take AE = 8(103) kN. 8 kN 4 kN D ∆

m 4 mm =- m 3 3 m + = D ∆ A A C B 4 m 4 m

26 8 kN 2 4 kN (0,0) D 1 1 ∆

m 4 mm =- m 2 3 m 3 1 + 3 = D ∆ A 4 6 A 8 C 2 3 4 3 5 5 4 B 7 (0,-3) (4,-3) 4 m 4 m (-4,-3)

(x − x )iˆ (y − y ) ˆj λˆ = j i + j i Member λx λy ij L L #1 -4/5 =-0.8 -3/5 = -0.6 cosθ = λ cosθ = λ x x y y #2 0 -3/3 = -1 U V U V i i j j #3 4/5 = 0.8 -3/5 = -0.6 U λ λ λ λ −λ λ −λ λ i x x x y x x x y #4 4/4 = 1 0 AE V λ λ λ λ −λ λ −λ λ [ k ] = i y x y y y x y y m L #5 4/4 = 1 0 Uj −λxλx −λxλy λxλx λxλy

Vj −λyλx −λyλy λyλx λyλy 27 2 2 2 (0,0) Member λx λy λx /L λxλy/L λy /L 1 1 #1 -0.8 -0.6 0.1280.096 0.072 2 1 3 m #2 0 -1 00 0.333 5 3 m m 4 5 6 #3 0.8 -0.6 0.128-0.096 0.072 8 3 3 5 2 4 5 4 7 4 m(0,-3) 4 m (-4,-3) (4,-3)

1 234 1 0.128 0.096 -0.128 -0.096

3 2 0.096 0.072 -0.096 -0.072 [k]1 = 8x10 3 -0.128 -0.096 0.128 0.096 1 27 8 4 -0.096 -0.072 0.096 0.072 1 0.128 -0.096 -0.128 0.096 2 -0.096 0.072 0.096 -0.072 1 256 3 [k]3 = 8x10 7 -0.128 0.096 0.128 -0.096 1 0 0 0 0 8 0.096 -0.072 -0.096 0.072 3 2 0 0.333 0 -0.333 [k]2 = 8x10 5 0 0 0 0 6 0 -0.333 0 0.333 28 2 2 2 (0,0) Member λx λy λx /L λxλy/L λy /L 1 1 #4 1 0 0.250 0 2 1 3 m #5 1 0 0.250 0 5 3 m m 4 5 6 8 3 3 5 2 4 5 4 7 4 m(0,-3) 4 m (-4,-3) (4,-3) 3 456 5 678 3 0.25 0 -0.25 0 5 0.25 0 -0.25 0

3 4 0 0 0 0 3 6 0 0 0 0 [k]4= 8x10 [k]5= 8x10 5 -0.25 0 0.25 0 7 -0.25 0 0.25 0 6 0 0 0 0 8 0 0 0 0

Global Stiffness Matrix 1 257 1 2 [K] = 8x103 5 7 29 Global Stiffness Matrix 1 234 3 456 1 0.128 0.096 -0.128 -0.096 3 0.25 0 -0.25 0 4 0 0 0 0 3 2 0.096 0.072 -0.096 -0.072 3 [k]1 = 8x10 [k]4= 8x10 3 -0.128 -0.096 0.128 0.096 5 -0.25 0 0.25 0 4 -0.096 -0.072 0.096 0.072 6 0 0 0 0 1 256 5 678 1 0 0 0 0 5 0.25 0 -0.25 0

3 2 0 0.333 0 -0.333 3 6 0 0 0 0 [k]2 = 8x10 [k]5= 8x10 5 0 0 0 0 7 -0.25 0 0.25 0 6 0 -0.333 0 0.333 8 0 0 0 0

1 27 8 1 257 1 0.128 -0.096 -0.128 0.096 1 0.256 0.0 0.0 -0.128 2 -0.096 0.072 0.096 -0.072 3 2 0.0 0.477 0.0 0.096 [k]3 = 8x10 [K] = 8x103 7 -0.128 0.096 0.128 -0.096 5 0.0 0.0 0.50 -0.25 8 0.096 -0.072 -0.096 0.072 7 -0.128 0.096 -0.25 0.378 30 8 kN Global Fixed end forces 2 4 kN D 1 ∆

m 4 mm =- m 2 3 m 3 1 + 3 = D ∆ A 4 6 A 8 C 3 4 5 5 B 7 4 m 4 m

2.88 kN ∆AE/L = 4.8 kN ∆AE/L = 10.67 kN -3.84 1 ∆ 3.84 kN mm =-4 -2.88 + 10.67 = 7.79 2 m 2 1 m Fixed End 5 3.84 kN 3 0.0 + = 7 D 0.0 ∆ A 10.67 kN 4.8 kN 2.88 kN

31 8 kN 2 4 kN D 1 ∆

m mm =-4 m 2 3 m 3 1 + 3 = D ∆ A 4 6 A 8 C 3 4 5 5 B 7 4 m 4 m

Global: [Q] = [K][D] + [QF]

1 257

Q1 = 4 -3.84 1 0.256 0.0 0.0 -0.128 D1 Q2 = -8 2 0.0 0.477 0.0 0.096 D 7.79 8x103 2 + Q = 0 = 5 5 0.0 0.0 0.50 -0.25 D5 0.0 Q = 0 7 7 -0.128 0.096 -0.25 0.378 D7 0.0

-3 D1 6.4426x10 m D -3 2 = -5.1902x10 m -3 D5 2.6144x10 m D -3 7 5.2288x10 m 32 Member forces 2 λ Dxi  λ   AE λ Dyi 1 [q' ] = []− − λ   + []q'F F m x y x x D  2 L xj 1 3   Dyj  4 6 8 3 5 Member λ λ 4 5 7 ix iy #1 -0.8 -0.6 #2 0-1 -3 D1 6.4426x10 m D -3 2 = -5.1902x10 m D -3 1 D5 2.6144x10 m D D -3 3 0.8 0.6 -0.8 -0.6 2 7 5.2288x10 m [q´F]1 = 8x10 + -4.8 5 0 = -1.54 kN (C) 0 4.8 kN 10.67 kN D1 m m 2 D 1 3 3 0.0 1.0 0.0 -1.0 2 + 10.67 + [q´F]2 = 8x10 = D D 3 5 ∆ A 10.67 kN 0 4.8 kN = -3.17 kN (C) 33 2 λ Dxi  λ   AE λ Dyi 1 [q' ] = []− − λ   + []q'F F m x y x x D  2 L xj 1 3   Dyj  4 6 8 D 3 4 5 5 1 7 D 3 -0.8 0.6 0.8 -0.6 2 [q´F]3 = 8x10 5 D7 -3 = -6.54 kN (C) D1 6.4426x10 m 0 D -5.1902x10-3 2 = m -3 D5 2.6144x10 m 0 D -3 7 5.2288x10 m 0 3 -1.0 0.0 1.0 0.0 [q´F]4 = 8x10 4 D5 = 5.23 kN (T) 0 Member λx λy

#3 0.8 -0.6 D5 0 #4 1 0 3 -1.0 0.0 1.0 0.0 [q´F]5 = 8x10 #5 10 4 D7 = 5.23 kN (T) 0 34 2 Member λx λy [q´] 1 #1 -0.8 -0.6 -1.54 2 #2 0 -1 -3.17 1 3 4 6 #3 0.8 -0.6 -6.54 8 3 5 #4 1 0 5.23 4 5 7 #5 1 0 5.23

8 kN 4 kN 8 kN 4 kN

6.54 kN 1.54 kN 3.17 kN

3.17 kN 4 kN 1.54 kN 6.54 kN 0.92 kN 5.23 kN 5.23 kN 3.17 kN 3.92 kN

35 Special Trusses (Inclined roller supports)

36 Transformation Matrices 2

1 [ q* ] = [ T ]T[ q´ ] 3 1 4 * q 3 * 6 λix 0 4 8 q* 5 4 λ 0 q´i 2 5 7 = iy * q1 q´ 3 0 λjx j λ = cos θ q2 0 λ y jx j jy [T]T 2 λjy = sin θj θj 1 x y * j 1 λix λiy 0 0 4* [ T ] = [[ T ]T]T = λ = cos θ 0 0 λjx λjy θ ix i i i λ = sin θ 3* iy i x * q´j

1 j

i q´i 37 [ k ] = [ T ]T[ k´ ][T]

λix 0 λ 0 AE 1 -1 λix λiy 0 0 [ k ] = iy m L 0 λjx -1 1 0 0 λjx λjy 0 λjy

Ui Vi Uj Vj

Ui λixλix λixλiy −λixλjx −λixλjy AE V λ λ λ λ −λ λ −λ λ [ k ] = i iy ix iy iy iy jx iy jy m L Uj −λjxλix −λjxλiy λjxλjx λjxλjy

Vj −λjyλix −λjyλiy λjyλjx λjyλjy

38 Example 5

For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. AE is constant.

30 kN

3 m

45o

4 m

39 2 Member 1: 1 6 4* 3* 1 o θ = 0, 45 3 i 5 3 m 2 θ = -45o = 135o, λix = cos 0 = 1, [q*] ij 6 4* λ = cos (-45o) = 0.707, 3* λiy = sin 0 = 0 ix 1 o λiy = sin(- 45 ) = -0.707 5 45o q´ q´ 4 m i 1 j i j

[q*] = [T*]T[q´] + [T*]T[q´F]

q5 1 0 q´ q6 0 0 i = q´ q3* 0 0.707 j q 0 -0.707 4* [T*]T

40 2 Member 1: 1 6 4* 3* 1 o θ = 0 ; 45 3 i 5 3 m 2 θ = -45o = 135o, λix = cos 0 = 1, [q*] ij 6 4* λ = cos (-45o) = 0.707, 3* λiy = sin 0 = 0 ix 1 o λiy = sin(- 45 ) = -0.707 5 45o q´ q´ 4 m i 1 j i j T AE  1 −1 [k*] = [T ]  []T L −1 1 

1 0 0 0 AE 1 -1 1 0 0 0 [k*] = 1 0 0.707 4 -1 1 0 0 0.707 -0.707 0 -0.707 5 6 3* 4* 5 0.25 0 -0.1768 0.1768 6 0 0 0 0 [k*] = AE 1 3* -0.1768 0 0.125 -0.125 4* 0.1768 0 -0.125 0.125 41 2 Member 2: q´ 1 2 i o o 1 θi = -90 = 270 , i 3 m 3 o o 2 90 λix = cos(-90 ) = 0, 6 4* o 2 λ = sin(-90 ) = -1 2 3* iy 1 4* 5 45o 3* 4 m o o o o j 90 +45 θj = -135 = 215 , q´ =135o o j 1 −1 λix = cos (-135 ) = -0.707, T AE   o [k*] = [T ]  []T λiy = sin(- 135 ) = -0.707 L −1 1 

0 0 -1 0 AE 1 -1 0 -1 0 0 [k*] = 2 0 -0.707 3 -1 1 0 0 -0.707 -0.707 0 -0.707 1 24*3* 1 0 0 0 0 2 0 0.3333 -0.2357 -0.2357 [k*] = AE 2 3* 0 -0.2357 0.1667 0.1667 4* 0 -0.2357 0.1667 0.1667 42 2 2 Member 3: 36.87o 1 1 j q´j

3 m 3 2 3 3 6 4* 6 3* o 1 36.87 i 5 45o 5 o q´ 4 m θi = θj = 36.87 ; i o λix = λjx = cos (36.87 ) = 0.8, o T AE  1 −1 λiy = λjy = sin(36.87 ) = 0.6 [k] = [T ]  []T L −1 1 

0.8 0 0.6 0 AE 1 -1 0.8 0.6 0 0 [k] = 3 0 0.8 5 -1 1 0 0 0.8 0.6 0 0.6 5 621 5 0.128 0.096 -0.128 -0.096 6 0.096 0.072 -0.096 -0.072 [k] = AE 3 1 -0.128 -0.096 0.128 0.096 2 -0.096 -0.072 0.096 0.072 43 2 Global Stiffness: 1

5 6 3* 4* 3 m 3 2 6 4* 5 0.25 0 -0.1768 0.1768 3* 1 6 0 0 0 0 [k*]1 = AE 5 45o 3* -0.1768 0 0.125 -0.125 4 m 4* 0.1768 0 -0.125 0.125

1 24*3* 1 0 0 0 0 2 0 0.3333 -0.2357 -0.2357 [k*] = AE 2 3* 0 -0.2357 0.1667 0.1667 1 2 3* 4* 0 -0.2357 0.1667 0.1667 1 0.128 0.096 0 2 0.096 0.4053 -0.2357 [K] = AE 5 621 3* 0 -0.2357 0.2917 5 0.128 0.096 -0.128 -0.096 6 0.096 0.072 -0.096 -0.072 [k] = AE 3 1 -0.128 -0.096 0.128 0.096 2 -0.096 -0.072 0.096 0.072 44 2 Global : 30 kN 1

3 m 3 m 3 2 6 4* 3* 1 45o 5 45o 4 m 4 m

[Q] = [K][D] + [QF]

1 2 3* Q1 = 30 1 0.128 0.096 0 D1 Q2 = 0 = AE 2 0.096 0.4053 -0.2357 D2 Q3*= 0 3* 0 -0.2357 0.2917 D3*

D 352.5 1 1 D = -157.5 2 AE D3* -127.3 45 2 Member Forces : 1 λ Dxi  λ   AE λ Dyi [q' ] = []− − λ   + []q'F 3 F m x y x x 3 m 2 L Dxj    6 4* D 3*  yj  1 5 45o 0 4 m -1 0 0.707 -0.707 0 [q´F]1 = AE 4 D3* D 352.5 = -22.50 kN, (C) 1 1 0 D = -157.5 2 AE D3* -127.3 D1 D 0 1 -0.707 -0.707 2 [q´F]2 = AE 3 D3* Member λix λiy λjx λjy = -22.50 kN, (C) 0 #1 1 0 0.707 -0.707 0 #2 0 -1 -0.707 -0.707 -0.8 -0.6 0.8 0.6 0 #3 0.8 0.6 0.8 0.6 [q´F]3 = AE 5 D1 = 37.50 kN, (T) D 246 2 Reactions : 30 kN 1

3 m 3 m 3 2 6 4* 3* 1 45o 5 45o 4 m 4 m

Member [q´]1 [q´]2 [q´]3 Member Force (kN) -22.50 -22.50 37.50

22.50 kN 37.50 kN 45o 36.87o 45o 7.50 kN 22.50 kN 31.82 kN 22.50 kN 47 Example 6

For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. AE is constant.

30 kN

3 m

45o

4 m 4 m

48 Member 1: 2 8 1 4 7 6 4* 3* 1 o 2 θ = 0o, 45 3 5 i 5 3 m o θ = -45o λix = cos 0 = 1, [q*] ij 6 4* 3* o λ = cos (-45o) = 0.707, λiy = sin 0 = 0 ix 1 λ = sin(- 45o) = -0.707 5 iy 4 m 4 m q´ q´ i 1 j i j

[q*] = [T*]T[q´] + [T*]T[q´F]

q5 1 0 q´ q6 0 0 i = q´ q3* 0 0.707 j q 0 -0.707 4* [T*]T

49 Member 1: 2 8 1 4 7 6 4* 3* 1 o 2 θ = 0o, 45 3 5 i 5 3 m o θ = -45o = 315o, λix = cos 0 = 1, [q*] ij 6 4* 3* o λ = cos (-45o) = 0.707, λiy = sin 0 = 0 ix 1 λ = sin(- 45o) = -0.707 5 iy 4 m 4 m q´ q´ i 1 j i j T AE  1 −1 [k*] = [T ]  []T L −1 1 

1 0 0 0 AE 1 -1 1 0 0 0 [k*] = 1 0 0.707 4 -1 1 0 0 0.707 -0.707 0 -0.707 5 6 3* 4* 5 0.25 0 -0.1768 0.1768 6 0 0 0 0 [k*] = AE 1 3* -0.1768 0 0.125 -0.125 4* 0.1768 0 -0.125 0.125 50 Member 2: 2 8 1 4 7 q´ 2 i o 2 θi = -90 , i 5 1 3 m 3 o o 90 λix = cos(-90 ) = 0, 6 4* 3* o 2 λiy = sin(-90 ) = -1 2 1 5 4* 4 m 4 m 3* 90o+45o o o j θj = -135 = 215 , q´ =135o o j 1 −1 λix = cos (-135 ) = -0.707, T AE   o [k*] = [T ]  []T λiy = sin(- 135 ) = -0.707 L −1 1 

0 0 -1 0 AE 1 -1 0 -1 0 0 [k*] = 2 0 -0.707 3 -1 1 0 0 -0.707 -0.707 0 -0.707 1 24*3* 1 0 0 0 0 2 0 0.3333 -0.2357 -0.2357 [k*] = AE 2 3* 0 -0.2357 0.1667 0.1667 4* 0 -0.2357 0.1667 0.1667 51 2 8 2 Member 3: o 1 4 7 36.87 1 j q´j 2 3 m 3 5 3 3 6 4* 3* 6 o 1 36.87 i 5 5 4 m 4 m o q´ θi = θj = 36.87 ; i o λix = λjx = cos (36.87 ) = 0.8, o T AE  1 −1 λiy = λjy = sin(36.87 ) = 0.6 [k*] = [T ]  []T L −1 1 

0.8 0 0.6 0 AE 1 -1 0.8 0.6 0 0 [k] = 3 0 0.8 5 -1 1 0 0 0.8 0.6 0 0.6 5 621 5 0.128 0.096 -0.128 -0.096 6 0.096 0.072 -0.096 -0.072 [k] = AE 3 1 -0.128 -0.096 0.128 0.096 2 -0.096 -0.072 0.096 0.072 52 Member 4: 2 8 1 4 7 2 8

2 4 3 m 3 5 1 7 [q] 6 4* 3* o o 1 θi = θij = 0 ; λix = λjx = cos 0 = 1, o 5 λiy = λjy = sin 0 = 0 4 m 4 m q´ q´ i 1 j i j T AE  1 −1 [k*] = [T ]  []T L −1 1 

1 0 0 0 AE 1 -1 1 0 0 0 [k] = 1 0 1 4 -1 1 0 0 1 0 0 0 1 2 7 8 1 0.25 0 -0.25 0 2 0 0 0 0 [k] = AE 4 7 -0.25 0 0.25 0 8 0 0 0 0 53 2 8 8 Member 5: o 1 4 7 36.87 , j 0.8 q´ 7 = j o ; o ) .6 2 .87 .87 0 5 o 36 36 o ) = 3 m 3 8.13 5 = s ( 7 5 θj co 6.8 6 4* 3* 4* 3* = (3 λjx sin = 1 λjy i 5 4 m 4 m o q´ θi = - 8.13 ; i o λix = cos (- 8.13 ) = 0.9899, o T AE  1 −1 λiy = sin(- 8.13 ) = -0.1414 [k*] = [T ]  []T L −1 1 

0.9899 0 -0.1414 0 AE 1 -1 0.9899 -0.1414 0 0 [k*] = 5 0 0.8 5 -1 1 0 0 0.8 0.6 0 0.6 3* 4*7 8 3* 0.196 -0.028 -0.1584 -0.1188 4* -0.028 0.004 0.02263 0.01697 [k*] = AE 5 7 -0.1584 0.02263 0.128 0.096 8 -0.1188 0.01697 0.096 0.072 54 Global Stiffness: 2 8 5 621 1 4 7 5 0.128 0.096 -0.128 -0.096 6 0.096 0.072 -0.096 -0.072 2 [k*] = AE 3 m 3 5 3 1 -0.128 -0.096 0.128 0.096 6 4* 3* 2 -0.096 -0.072 0.096 0.072 1 5 4 m 4 m 1 2 7 8 1 0.25 0 -0.25 0 5 6 3* 4* 2 0 0 0 0 [k] = AE 5 0.25 0 -0.1768 0.1768 4 7 -0.25 0 0.25 0 6 0 0 0 0 8 0 0 0 0 [k*]1 = AE 3* -0.1768 0 0.125 -0.125 3* 4*7 8 4* 0.1768 0 -0.125 0.125 3* 0.196 -0.028 -0.1584 -0.1188 4* -0.028 0.004 0.02263 0.01697 [k*] = AE 5 7 -0.1584 0.02263 0.128 0.096 1 24*3* 8 -0.1188 0.01697 0.096 0.072 1 0 0 0 0 12 3* 2 0 0.3333 -0.2357 -0.2357 1 0.378 0.096 0 [k*]2 = AE 3* 0 -0.2357 0.1667 0.1667 2 0.096 0.4053 -0.2357 [K] = AE 4* 0 -0.2357 0.1667 0.1667 3* 0 -0.2357 0.4877 55 Global : 2 8 30 kN 1 4 7

2 3 m 3 5 6 4* 3* 1 5 4 m 4 m

[Q] = [K][D] + [QF]

1 2 3* Q1 = 30 1 0.378 0.096 0 D1 Q2 = 0 = AE 2 0.096 0.4053 -0.2357 D2 Q3*= 0 3* 0 -0.2357 0.4877 D3*

D 86.612 1 1 D = -28.535 2 AE D3* -13.791 56 Member Forces : 2 8 1 4 7 λ Dxi  λ   AE λ Dyi 2 [q' ] = []− − λ   + []q'F 3 5 F m x y x x 3 m L Dxj  6 4* 3*   Dyj  1 5 0 4 m 4 m -1 0 0.707 -0.707 0 [q´F]1 = AE 4 D3* D 86.612 1 1 = -2.44 kN, (C) 0 D = -28.535 2 AE D3* -13.791 D1 D 0 1 -0.707 -0.707 2 [q´F]2 = AE D Member λix λiy λjx λjy 3 3* = -6.26 kN, (C) 0 #1 1 0 0.707 -0.707 #2 0 -1 -0.707 -0.707 0 #3 0.8 0.6 0.8 0.6 -0.8 -0.6 0.8 0.6 0 [q´F]3 = AE 5 D1 = 10.43 kN, (T) D 2 57 Member Forces : 2 8 1 4 7 λ Dxi  λ   AE λ Dyi 2 [q' ] = []− − λ   + []q'F 3 5 F m x y x x 3 m L Dxj  6 4* 3*   Dyj  1 5 D 4 m 4 m 1 D -1010 2 [q´F]4 = AE 4 0 D 86.612 1 1 = -21.65 kN, (C) 0 D = -28.535 2 AE D3* -13.791 D3* -0.9899 0.141 0.8 0.6 0 [q´F]5 = AE 0 Member λix λiy λjx λjy 5 = 2.73 kN, (T) 0 #4 10 1 0 #5 0.9899 -0.141 0.8 0.6

58 Reactions : 2 8 30 kN 1 4 7

2 3 m 3 5 6 4* 3* 1 5 4 m 4 m

Member [q´]1 [q´]2 [q´]3 [q´]4 [q´]5 Member Force (kN) -2.44 -6.26 10.43 -21.65 2.73

1.64 kN 21.65 kN 19.47 kN 36.87o

10.43 kN 6.26 kN 2.73 kN 45o 36.87o 45o 5.90 kN 81.87o 2.44 kN 2.44 kN 6.54 kN 6.26 kN 59 Example 7

For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. Take AE = 8(103) kN. 8 kN 4 kN D Ui Vi Uj Vj

Ui λixλix λixλiy −λixλjx −λixλjy 3 m AE V λ λ λ λ −λ λ −λ λ [ k *] = i iy ix iy iy iy jx iy jy m L A Uj −λjxλix −λjxλiy λjxλjx λjxλjy C B Vj −λjyλix −λjyλiy λjyλjx λjyλjy 36.87o 4 m 4 m

60 8 kN 2 4 kN D 1 3 1 4 3 m 4* 6 8 A 4 m 4 m 5 C 2 5 7 B 3* 36.87o Member 1: o λjx = cos 36.87 = 0.8, λ = sin 36.87o = 0.6 jy * * T * T F y [q ] = [T ] [q´] + [T ] [q´ ] 2 36.87o 1 x q3* 0.28 0 y * j q´ 1 j q q´i * 4* 0.96 0 4 j = q´ 1 q1 0 0.8 j o i 73.74 q 0 0.6 2 [T*]T 3* i x * q´i

o λix = cos 73.74 = 0.28, λ = sin 73.74o = 0.96 iy 61 8 kN 2 4 kN D 1 3 1 4 3 m 4* 6 8 A 4 m 4 m 5 C 2 5 7 B 3* 36.87o T AE  1 −1 [k] = [T ]  []T L −1 1 

0.28 0 0.96 0 8x103 1 -1 0.28 0.96 0 0 [k] = 1 0 0.8 5 -1 1 0 0 0.8 0.6 0 0.6 3* 4* 1 2 3* 0.01568 0.05376 -0.0448 -0.0336 4* 0.05376 0.18432 -0.1536 -0.1152 [k] = 8x103 1 1 -0.0448 -0.1536 0.128 0.096 2 -0.0336 -0.1152 0.096 0.072 62 8 kN 2 4 kN D 1 3 1 4 3 m 4* 6 8 A 4 m 4 m 5 C 2 5 7 B 3* 36.87o [q*] = [T*]T[q´] + [T*]T[q´F] Member 2:

y* y q3* 0.8 0 4* 6 q´ q4* 0.6 0 i 2 5 = q´ x q5 0 1 j i 36.87 j o * λ = cos 0 = 1, q 3 x * jx 6 0 0 * T λ = sin 0o = 0 [T ] o jy 1 -1 λix = cos 36.87 = 0.8, [k] = [TT] AE [T] o λiy = sin 36.87 = 0.6 L -1 1 3* 4* 5 6 q´ q´ i 2 j 3* 0.16 0.12 -0.2 0 i j 4* 0.12 0.09 -0.15 0 [k] = 8x103 2 5 -0.2 -0.15 0.25 0 6 0 0 0 0 63 8 kN 2 4 kN D 1 3 1 4 3 m 4* 6 8 A 4 m 4 m 5 C 2 5 7 B 3* 36.87o λ = cos 270o = 0, Member 3: x Member 4: λ = cos 323.13o = 0.8, o y x y λy = sin 270 = -1 2 o 2 λy = sin 323.13 = -0.6 270o 1 x 1 x 323.13o 4 3 8 6 7 5

1 256 1 27 8 1 0 0 0 0 1 0.128 -0.096 -0.128 0.096 3 8x103 2 -0.096 0.072 0.096 -0.072 [k]3 = 8x10 2 0 0.333 0 -0.333 [k]4 = 5 0 0 0 0 7 -0.128 0.096 0.128 -0.096 6 0 -0.333 0 0.333 8 0.096 -0.072 -0.096 0.07264 8 kN 2 4 kN D 1 3 1 4 3 m 4* 6 8 A 4 m 4 m 5 C 2 5 7 B 3* 36.87o y Member 5: 6 8 5 5 7 x o λx = cos 0 = 1, o λy = sin 0 = 0 5 678 5 0.25 0 -0.25 0 6 0 0 0 0 [k] = 8x103 5 7 -0.25 0 0.25 0 8 0 0 0 0

65 * * 2 3 4 1 2 1 3* 0.01568 0.05376 -0.0448 -0.0336 3 4* 0.05376 0.18432 -0.1536 -0.1152 1 4 [k] = 8x103 1 1 -0.0448 -0.1536 0.128 0.096 4* 6 8 2 -0.0336 -0.1152 0.096 0.072 5 2 5 7 3* 3* 4* 5 6 1 256 3* 0.16 0.12 -0.2 0 1 0 0 0 0 4* 0.12 0.09 -0.15 0 [k] = 8x103 2 0 0.333 0 -0.333 [k] = 8x103 3 2 5 -0.2 -0.15 0.25 0 5 0 0 0 0 6 0 0 0 0 6 0 -0.333 0 0.333 1 27 8 5 678 1 0.128 -0.096 -0.128 0.096 5 0.25 0 -0.25 0 [k] 8x103 2 -0.096 0.072 0.096 -0.072 6 0 0 0 0 4 = [k] = 8x103 7 -0.128 0.096 0.128 -0.096 5 7 -0.25 0 0.25 0 8 0.096 -0.072 -0.096 0.072 8 0 0 0 0 1 23* 5 1 0.256 0.0 -0.0448 0 2 0.0 0.474 -0.0336 0 [K] = 8x103 3* -0.0448 -0.0336 0.17568 -0.2 5 00-0.2 0.5 66 8 kN 2 4 kN D 1 3 1 4 3 m * 6 4 8 A 4 m 4 m 5 C 2 5 7 B 3* 36.87o

Global: [Q] = [K][D] + [QF]

1 23* 5 Q1 = 4 1 0.256 0.0 -0.0448 0 D1 Q2 = -8 2 0.0 0.474 -0.0336 0 D2 Q = 0 = 8x103 3* * 3 -0.0448 -0.0336 0.17568 -0.2 D3* Q = 0 5 5 00-0.2 0.5 D5

-3 D1 1.988x10 m D -3 2 = -2.0824x10 m -4 D3* 1.996x10 m D -5 5 7.984x10 m 67 8 kN Member forces

4 kN λ Dxi  D λ D  AE λ  yi  F [q'F ]m = []− x − y x λx + []q' L Dxj  3 m   Dyj  A 4 m 4 m C Member λ λ λ λ B ix iy jx jy

o #1 0.28 0.96 0.8 0.6 36.87 2 #2 0.8 0.6 1 0 1

3 D3* 1 4 0 [q´ ] = 8x103 -0.28 -0.96 0.8 0.6 * 6 F 1 4 8 D 5 5 1 2 5 7 = 0.46 kN, (T) 3* D2

D3* -3 0 D 1.988x10 m 3 -0.8 -0.6 1 0 1 [q´F]2 = 8x10 D -3 D 2 = -2.0824x10 m 4 5 -4 D3* 1.996x10 m = -0.16 kN, (C) 0 D -5 5 7.984x10 m 68 2

1 λ Dxi  λ D  3 AE λ  yi  F 1 4 [q'F ]m = []− x − y x λx + []q' L Dxj    4* 6 8 Dyj  2 5 5 * 7 3 D1 D 3 01 0-1 2 [q´F]3 = 8x10 -3 D D1 1.988x10 m 3 5 D -2.0824x10-3 = -5.55 kN 0 2 = m * -4 D 3 1.996x10 m -5 D5 7.984x10 m D1 D 3 -0.8 0.6 0.8 - 0.6 2 [q´F]4 = 8x10 0 Member λix λiy λjx λjy 5 = -4.54 kN 0 #3 0-10 -1 #4 0.8 -0.6 0.8 -0.6 D5 #5 10 1 0 0 3 -1 0 1 0 [q´F]5 = 8x10 4 0 = -0.16 kN 0 69 8 kN 2 4 kN D 1 3 1 3 m 4 * 6 4 8 A 4 m 4 m 5 C 2 5 7 B 3* 36.87o Member [q]1 [q]2 [q]3 [q]4 [q]5 Member Force (kN) 0.46 -0.16 -5.55 -4.54 -0.16

4.54 kN y* 0.46 kN 5.55 kN 36.87o 36.87o 3.79 kN 0.16 kN 0.16 kN 36.87o 5.55 kN 0.36 kN 2.72 kN

x * 70 Space-Truss Analysis

71 Member Local Stiffness [k´]:

[q´] = [k´][d´] + [q´F] = [k´][T][d] + [q´F]

F q'i  EA  1 −1 d'i  q'i  = +   q'    d'  F  j  L −1 1   j  q' j 

dix  d   iy    EA  1 −1 diz F =   []T   + []q' L −1 1  d jx  d   jy  λ d jz  λ λ where,

 x y z λ0 0 0  []T =  λ   0 0 0 x y λz 

72 λ

Memberλ Global Stiffness [km]: λ T [km]= [T] [k´] [T] λ λ  x 0  λ λ λ  0   y λ  z 0  EA  1 −1  x y z λ0 0 0  []km =   λ    0 0 0 λ   0 x  L −1 1   x y z   0   y   0 z 

73 Global equilibrium matrix:

[Q] = [K][D] + [QF]

Joint Load Unknown Displacement Fixed End Forces Q D QF I KI,I KI,II u I =+ Q D QF II KII,I KII,II k II Reaction Support Boundary Condition

74 q´j

j d qix λxλx λxλy λxλz −λxλx −λxλy −λxλz ix d qiy λyλx λyλy λyλz −λyλx−λyλy −λyλz iy m d qiz EA λzλx λzλy λzλz −λzλx −λzλy −λzλz iz = d qjx L jx −λxλx −λxλy −λxλz λxλx λxλy λxλz i q djy q jy −λyλx−λyλy −λyλz λyλx λyλy λyλz q´i jy q djz jz −λzλx −λzλy −λzλz λzλx λzλy λzλz qjx j

qjz q iy m F q'i  EA  1 −1 d'i  q'i  q   =   +  F  ix q' L −1 1  d' q' i  j     j   j  q iz λ dix  λ d  λ  iy  λ EA λ diz  []q' = []− − − λ + []q'F j m x y z x y z   L d jx  d   jy  d jz    75 Example 6

For the truss shown, use the stiffness method to: (b) Determine the end forces of each member. (a) Determine the deflections of the loaded joint. 2 Take E = 200 GPa, A = 1000 mmz . 60 kN 80 kN

10 m y

4 m O 4 m 3 m 3 m x

76 z

60 kN 3 80 kN 1 2 (0, 0, 10) 1

10 m 1 4 2 y 3 2 4 m 5 O (-4, 3, 0) 3 4 m 3 m (-4, -3, 0) 3 m x (4, 3, 0) 4 λm = λxi + λyj + λzk (4, -3, 0)

λ1 = (-4/11.18)i + (3/11.18)j + (-10/11.18)k Member λ λ λ = -0.3578 i + 0.2683 j - 0.8944 k x y z

λ2 = (+4/11.18)i + (3/11.18)j + (-10/11.18)k #1 -0.3578 +0.2683 -0.8944 = +0.3578 i + 0.2683 j - 0.8944 k #2 +0.3578 +0.2683 -0.8944 λ3 = (+4/11.18)i + (-3/11.18)j + (-10/11.18)k #3 +0.3578 -0.2683 -0.8944 = +0.3578 i - 0.2683 j - 0.8944 k #4 -0.3578 -0.2683 -0.8944 λ4 = (-4/11.18)i + (-3/11.18)j + (-10/11.18)k = -0.3578 i - 0.2683 j - 0.8944 k 77 Member Stiffness Matrix [k]6x6 Member λx λy λz

#1 -0.3578 +0.2683 -0.8944 [k11]3x3 [k12]3x3

#2 +0.3578 +0.2683 -0.8944 [k]m = [k ] [k ] #3 +0.3578 -0.2683 -0.8944 21 3x3 22 3x3 #4 -0.3578 -0.2683 -0.8944 1 23 1 +0.128 -0.096 -0.320 AE 1 23 [k ] = 2 -0.096 +0.072 +0.240 11 3 L 1 +0.128 -0.096 +0.320 3 -0.320 +0.240 +0.80 AE [k ] = 2 -0.096 +0.072 -0.240 11 1 L 1 23 3 +0.320 -0.240 +0.80 1 +0.128 +0.096 +0.320 AE [k11]4 = 2 +0.096 +0.072 +0.240 1 23 L 3 +0.320 +0.240 +0.80 1 +0.128 +0.096 -0.320 AE 1 23 [k11]2 = 2 +0.096 +0.072 -0.240 L 1 0.512 0.0 0.0 3 -0.320 -0.240 +0.80 AE [K ] = 2 0.0 0.288 0.0 I,I L 3 0.0 0.0 3.2 78 z

60 kN 3 80 kN 1 2 (0, 0, 10) 1

10 m 1 4 2 y 3 2 4 m 5 O (-4, 3, 0) 3 4 m 3 m (-4, -3, 0) 3 m x (4, 3, 0) 4 [Q] = [K][D] + [QF] (4, -3, 0) 1 23

60 1 0.512 0.0 0.0 D1 0.0 AE -80 2 0.0 0.288 0.0 D + 0.0 = L 2 0.0 3 0.0 0.0 3.2 D3 0.0

79 Global equilibrium matrix:

[Q] = [K][D] + [QF]

Joint Load Unknown Displacement Fixed End Forces Q D QF I KI,I KI,II u I =+ Q D QF II KII,I KII,II k II Reaction Support Boundary Condition

(AE/L) = (1x10-3)(200x106)/(11.18) = 17.89x103 kN

1 23

60 1 0.512 0.0 0.0 D1 0.0 AE -80 2 0.0 0.288 0.0 D + 0.0 = L 2 0.0 3 0.0 0.0 3.2 D3 0.0

D1 +117.2 6.551 mm L = = D2 AE -277.8 -15.53 mm D 0.0 0.0 mm 3 80 z 60 kN 80 kN Member λx λy λz #1 -0.3578 +0.2683 -0.8944 10 m #2 +0.3578 +0.2683 -0.8944 y #3 +0.3578 -0.2683 -0.8944 #4 -0.3578 -0.2683 -0.8944 4 m O D 4 m 3 m 1 x D2 3 m dxi D3 Member forces: dyi dzi [q´ ] = AE −λ −λ −λ λ λ λ + q´F j m L x y z x y z dxj

dyj

dzj [ 0 ] 117.2 AE L [q´ ] = +0.3578 -0.2683 +0.8944 j 1 L AE -277.8 0.0 = +116.5 kN (T) 81 z 60 kN 80 kN Member λx λy λz #1 -0.3578 +0.2683 -0.8944 10 m #2 +0.3578 +0.2683 -0.8944 y #3 +0.3578 -0.2683 -0.8944 #4 -0.3578 -0.2683 -0.8944 4 m O 4 m 3 m 3 m x

117.2 AE L [q´ ] = -0.3578 -0.2683 +0.8944 = +32.61 kN (T) j 2 L AE -277.8 0.0 117.2 AE L [q´ ] = -0.3578 +0.2683 +0.8944 = -116.5 kN (T) j 3 L AE -277.8 0.0 117.2 AE L [q´ ] = +0.3578 +0.2683 +0.8944 = -32.61 kN (T) j 4 L AE -277.8 0.0 82 Member λx λy λz [q´j]m #1 -0.3578 +0.2683 -0.8944 116.5 #2 +0.3578 +0.2683 -0.8944 32.6 #3 +0.3578 -0.2683 -0.8944 -116.5 #4 -0.3578 -0.2683 -0.8944 -32.6

60 kN 80 kN

1 4 2 3

32.6 kN 116.5 kN 32.6 kN R5z = (-32.6)(-0.8944) 116.5 kN = 29.16 kN

5 R5y = (-32.6)(-0.2683) = 8.75 kN

R5x = (-32.6)(-0.3578) = 11.66 kN 83 BEAM ANALYSIS USING THE STIFFNESS METHOD

! Development: The Slope-Deflection Equations

! Stiffness Matrix

! General Procedures

! Internal Hinges

! Temperature Effects

! Force & Displacement Transformation

! Skew Roller Support

1 Slope – Deflection Equations

i P j k w Cj

settlement = ∆j

i P j M w ij Mji

θi

θ ψ j

2 • Degrees of Freedom

M θΑ

A B 1 DOF: θΑ L

P θΑ B θ θ A C 2 DOF: Α , Β θΒ

3 • Stiffness Definition

k kAA 1 BA

A B L

4EI k = AA L 2EI k = BA L

4 k

kAB BB A 1 B L

4EI k = BB L 2EI k = AB L

5 • Fixed-End Forces Fixed-End Forces: Loads P

PL L/2 L/2 PL 8 8 L

P P 2 2 w wL2 wL2 12 12 L wL wL 2 2

6 • General Case i P j k w Cj

settlement = ∆j

i P j M w ij Mji

θi

ψ θj

7 P i w j Mij Mji θi L settlement = ∆ θ j ψ j θ θ 4EI 2EI 2EI 4EI i + θ j = M M = + θ L L ij ji L i L j θj

θi + F (M ij) F ∆ (M ji)∆

settlement = ∆ + j P w F (MF ) (M ij)Load ji Load

θ θ 4EI 2EI F F 2EI 4EI F F M = ( ) + ( )θ + (M ij ) + (M ij ) , M = ( ) + ( )θ + (M ji ) + (M ji ) ij L i L j ∆ Load ji L i L j ∆ Load 8 • Equilibrium Equations

i P j k w Cj

C Mji j Mjk

Mji Mjk

+ ΣM j = 0 : − M ji − M jk + C j = 0

9 • Stiffness Coefficients

Mij i j Mji L θj

θi

4EI kii = 2EI L k ji = ×θi L 1 + 2EI kij = 4EI L k = ×θ j jj L 1

10 • Matrix Formulation

θ 4EI 2EI F M = ( ) + ( )θ + (M ij ) ij L i L j θ 2EI 4EI F M = ( ) + ( )θ + (M ji ) ji L i L j

θ F M ij  (4EI / L) (2EI / L)  iI  M ij    =   +  F  M (2EI / L) (4EI / L) θ  ji     j  M ji 

kii kij  []k =   k ji k jj 

Stiffness Matrix

11 P i w j Mij Mji θi [M ] = [K][θ ]+[FEM ] L θ ([M ]−[FEM ]) = [K][θ ] ψ j ∆j [θ ] = [K]−1[M ]−[FEM ]

Mij Mji

θj

θi Fixed-end moment + Stiffness matrix matrix F (M ij) F ∆ (M ji)∆ [D] = [K]-1([Q] - [FEM]) +

Displacement Force matrix F P F (M ij)Load w (M ji)Load matrix

12 • Stiffness Coefficients Derivation M Mi θi j Real beam i j L

M i + M j M i + M j L L L/3 M j L M j 2EI EI Conjugate beam

M i EI M i L 2EI θι From(1)and (2); M i L L M j L 2L + ΣM 'i = 0 : − ( )( ) + ( )( ) = 0 2EI 3 2EI 3 4EIθ M i = ( ) i M i = 2M j − − − (1) L 2EI M i L M j L M j = ( )θi + ↑ ΣFy = 0 : θ i − ( ) + ( ) = 0 − − − (2) L 2EI 2EI 13 • Derivation of Fixed-End Moment Point load P Real beam Conjugate beam A B

ABL M M M EI EI

M EI ML M 2EI

M ML EI 2EI P PL2 PL PL2 16EI 4EI 16EI

ML ML 2PL2 PL + ↑ ΣF = 0 : − − + = 0, M = y 2EI 2EI 16EI 8 14 P

PL PL 8 L 8 P P P/2 2 2

P/2 PL/8

-PL/8 -PL/8

- -PL/8 -PL/16

- -PL/16 -PL/8 − PL − PL PL PL PL/4 + + = + 16 16 4 8 15 Uniform load

w Real beam Conjugate beam A B ABL M M M EI EI

M EI ML M 2EI

M ML EI 2EI 2 wL3 wL wL3 w 24EI 8EI 24EI

ML ML 2wL3 wL2 + ↑ ΣF = 0 : − − + = 0, M = y 2EI 2EI 24EI 12 16 Settlements M M Mi = Mj Real beam j Conjugate beam EI L A B

M + M ∆ ∆ i j M L M i + M j M EI L

M ML EI ML 2EI 2EI M M EI

∆

ML L ML 2L + ΣM = 0 : − ∆ − ( )( ) + ( )( ) = 0, B 2EI 3 2EI 3 6EI∆ M = L2 17 C • Typical Problem B P P1 w 2

A C B L1 L2

2 wL 2 PL P PL w wL 12 8 8 12 L L θ 0 4EI 2EI P1L1 M AB = A + θ B + 0 + L1 L1 8 θ 0 2EI 4EI P1L1 M BA = A + θ B + 0 − L1 L1 8 θ 0 2 4EI 2EI P2 L2 wL2 M BC = B + θC + 0 + + L2 L2 8 12 θ 0 2 2EI 4EI − P2 L2 wL2 M CB = B + θC + 0 + − L2 L2 8 12 18 C B P P1 w 2

A C B L1 L2

C MBA B MBC

θ 2EI 4EI P1L1 M BA = A + θ B + 0 − L1 L1 8 θ 2 4EI 2EI P2L2 wL2 M BC = B + θC + 0 + + L2 L2 8 12

+ ΣM B = 0 : CB − M BA − M BC = 0 → Solve for θ B

19 C B P P1 w 2 M M BA AB C A MCB M B BC L1 L2

Substitute θB in MAB, MBA, MBC, MCB

20 C B P P1 w 2 MBA MAB MCB A MBC C Ay B L1 L2 Cy

By = ByL + ByR

B C P P 1 M 2 B BA MCB MAB A MBC

A ByR C y ByL y L1 L2

21 Stiffness Matrix

• Node and Member Identification

• Global and Member Coordinates

• Degrees of Freedom

•Known degrees of freedom D4, D5, D6, D7, D8 and D9 • Unknown degrees of freedom D1, D2 and D3

6 9 5 3 8 2EI 2 EI 2 1 4 1 21 3 7

22 Beam-Member Stiffness Matrix i j 1 4 3 6 E, I, A, L 2 5

k41 k14 AE/L = k k11 = AE/L AE/L AE/L 44

d1 = 1 d4 = 1

12 3 45 6 1 AE/L - AE/L

2 0 0 [k] = 3 0 0 4 -AE/L AE/L 5 0 0

6 0 0 23 i j 1 4 3 6 6EI/L2 = k E, I, A, L 32 2 2 5 6EI/L = k65

2 2 k62 = 6EI/L 6EI/L = k35 d2 = 1 d5 = 1

3 3 12EI/L = k52 12EI/L = k25 k = 12EI/L3 3 22 12EI/L = k55

12 3 45 6 1 AE/L 0 - AE/L 0

2 0 12EI/L3 0 - 12EI/L3 [k] = 3 0 6EI/L2 0 - 6EI/L2 4 -AE/L0 AE/L 0 5 0 -12EI/L3 0 12EI/L3

6 0 6EI/L2 0 - 6EI/L2 24 i j 1 4 3 6 E, I, A, L 2 5 k33 = 4EI/L = k 2EI/L = k63 = k 4EI/L 66 d3 = 1 2EI/L 36

d6 = 1 k = 2 2 2 2 23 6EI/L 6EI/L = k53 k26 = 6EI/L 6EI/L = k56 12 3 45 6 1 AE/L 0 0 - AE/L 0 0

2 0 12EI/L3 6EI/L2 0 - 12EI/L3 6EI/L2 [k] = 3 0 6EI/L2 4EI/L 0 - 6EI/L2 2EI/L 4 -AE/L0 0 AE/L 0 0 5 0 -12EI/L3 -6EI/L2 0 12EI/L3 -6EI/L2

6 0 6EI/L2 2EI/L 0 - 6EI/L2 4EI/L 25 • Member Equilibrium Equations i j Fxi F M xj i Mj E, I, A, L

F = yi Fyj AE/L AE/L AE/L AE/L x δi x δj 1 1 + + 6EI/L2 6EI/L2 6EI/L2 6EI/L2 1 1 x ∆i x ∆j 3 3 3 12EI/L 12EI/L + 12EI/L + 12EI/L3

4EI/L 1 2EI/L 2EI/L 4EI/L

x θi x θj 1 + 6EI/L2 6EI/L2 6EI/L2 2 FF 6EI/L F yi FFF FFF yj FFF FF xi xj M ii FF M jj 26 F Fxi = (AE / L)δ i + (0)∆i (0)θi + (−AE / L)δ j + (0)∆j + (0)θ j + Fxi 3 2 3 2 F Fyi = (0)δ i + (12EI / L )∆i (6EI / L )θi (0)δ j (−12EI / L )∆j (6EI / L )θ j Fyi 2 2 F Mxi = (0)δ i (6EI / L )∆i (4EI / L)θi (0)δ j (−6EI / L )∆j (2EI / L)θ j Mi F Fxj = (−AE / L)δ i (0)∆i (0)θi (AE / L)δ j (0)∆j (0)θ j Fxi 3 2 2 F Fyj = (0)δ i (−12EI / L )∆i (−6EI / L )θi (0)δ j (0)∆j (−6EI / L )θ j Fyj 2 2 F Mj = (0)δ i (6EI / L )∆i (2EI / L)θi (0)δ j (−6EI / L )∆j (4EI / L)θ j Mj

F Fxi   AE/L 0 0 − AE/L 0 0 δ i  Fxi     3 2 3 2    F  F 0 12EI/L 6EI/L 0 −12EI/L 6EI/L ∆ F  yj    i   yi    2 2    F  Mi  0 6EI/L 4EI/L 0 − 6EI/L 2EI/L  θ i Mi   =    +   F δ F  xj  − AE/L 0 0 AE/L 0 0  j  Fxj  F   0 −12EI/L3 − 6EI/L2 0 12EI/L3 − 6EI/L2 ∆  F F   yj    j   yi  M 2 2 θ F  j   0 6EI/L 2EI/L 0 − 6EI/L 4EI/L  j  M j 

Stiffness matrix Fixed-end force matrix

[q] = [k][d] + [qF]

End-force matrix Displacement matrix 27 6x6 Stiffness Matrix

δi ∆i θi δj ∆j θj

Ni  AE/L 0 0 − AE/L 0 0  V  3 2 3 2  i  0 12EI/L 6EI/L 0 −12EI/L 6EI/L  2 2 Mi  0 6EI/L 4EI/L 0 − 6EI/L 2EI/L  []k 6×6 =   Nj − AE/L 0 0 AE/L 0 0   3 2 3 2  Vj 0 −12EI/L − 6EI/L 0 12EI/L − 6EI/L  2 2  Mj  0 6EI/L 2EI/L 0 − 6EI/L 4EI/L 

4x4 Stiffness Matrix

∆i θi ∆j θj 3 2 3 2 Vi  12EI/L 6EI/L −12EI/L 6EI/L    M 6EI/L2 4EI/L − 6EI/L2 2EI/L []k = i   4×4  3 2 3 2  Vj −12EI/L − 6EI/L 12EI/L − 6EI/L  2 2  Mj  6EI/L 2EI/L − 6EI/L 4EI/L 

28 2x2 Stiffness Matrix

θi θj

Mi 4EI / L 2EI / L []k 2×2 =   Mj 2EI / L 4EI / L

Comment: - When use 4x4 stiffness matrix, specify settlement. - When use 2x2 stiffness matrix, fixed-end forces must be included.

29 General Procedures: Application of the Stiffness Method for Beam Analysis

P P2y w M M1 2

2 4 6

Global 1 3 1 2 2 1 3 5

2´ 4´ 2´ 4´

Local 1 2 1´ 3´ 1´ 3´

30 P P2y w M M1 2

2 4 6

Global 1 3 1 2 2 1 3 5 2 4 4 6

Member 1 2 1 3 3 5 2´ 4´ 2´ 4´

Local 1 2 1´ 3´ 1´ 3´

31 P P2y w M M1 2

2 4 6

Global 1 3 1 2 2 1 3 5 2´ 4´ 2´ 4´

Local 1 2 1´ 3´ 1´ 3´

P w F F (q´ ) (q´ 1)2 2 1 (q´F ) (q´F ) 4 2 [FEF] 1 4 1 2 F (q´F ) (q´F ) (q´F ) (q´ 1 )1 3 1 2 2 3 2 32 2 4 6

Global 1 3 1 2 2 1 3 5 2´ 4´

Local 1 1´ 3´

[q] = [T]T[q´] 1´2´ 3´ 4´

q1  1 1 0 0 0 q1'      q 2 0 1 0 0 q  2  =    2'  q3  3 0 0 1 0 q3'        q 0 0 0 1 q  4  1 4    4' 

[k] = [T]T[q´] [T]

33 2 4 6

Global 1 3 1 2 2 1 3 5 2´ 4´

Local 2 1´ 3´

[q] = [T]T[q´] 1´2´ 3´ 4´

q3  3 1 0 0 0 q1'      q 4 0 1 0 0 q  4  =    2'  q5  5 0 0 1 0 q3'        q 0 0 0 1 q  6  2 6    4' 

[k] = [T]T[q´] [T]

34 2 4 6

1 3 1 2 2 1 3 5 Stiffness Matrix:

1234 1 123456 2 1 [k]1 = 3 2 Member 1 4 3 2 [K] = 4 Node 3456 5 3 Member 2 6 4 [k]2 = 5 6

35 2 4 6

1 3 1 2 2 Joint Load 1 3 5 Du=Dunknown Q F k 2341 56 Q A M1z F Q2  2   D2  Q2   -P       F  Q 2y 3 D Q  3   KAA KAB   3   3  M3z  F  Q4  4   D4  Q   =     +  4  Q D 0 F  1  1    1  Q1  0 F Q    D  Q   5  5  5   5   KBA KBB  0 F Q6  6   D6  Q6 

Reaction F D = D Q B Qu k known F [][Qk = K AA Du + ][][Q A ]

−1 F [][Du = K AA + ]( Qk − [] []QA )

Member Force :[]q = k [][d]+ [q F ] 36 Global: 2 4 6

1 3 1 2 2 1 P 3 5 w (qF ) (qF ) 2 1 4 2 (q´F ) (qF ) 6 2 [FEF] 1 4 1 2 F (qF ) (qF ) (q´F ) (q 1 )1 3 1 3 2 5 2

123456 0  F  Q1  1   D1  Q1   Q M  Member 1  D  Q F  2 1 2    2   2  -P  F F F  Q3 2y 3   D3  Q3 = (q 3 )1 + (q 3 )2   =  2    +  F F F  Q M Node D Q = (q ) + (q )  4 2 4    4   4 4 1 4 2  Q    D 0  Q F  5 5 5  5     Member 2   0 F Q6  6   D6   Q6  234

Q = M F  2 1  2   D  Q    2 2 Q = −P      F   3 2 y  = 3 D + Q    3   3   Q = M  F  4 2  4  Node 2  D  Q     4   4  37 2 4 6

1 1 3 2 2 1 3 5

234 D  2    F  2  Q2 = M1  Q2         F  D3 = 3 Q = −P − Q      3 2y   3    F D4 4  2   Q = M  Q     Node   4 2   4 

38 2 4 6

1 3 1 2 2 1 3 5 P F q 2 F 1 q 4 F qF q 1 [FEF] 3

Member 1: 1234 1 F q 1     d 1 = 0  q 1          q 2 k d = D q F  2  =  1   2 2  +  2         F  q 3 3 d 3 = D3 q 3        F  q 4  4   d 4 = D4  q 4 

39 2 4 6

1 3 1 2 2 1 3 5 w F q 6 F 2 q 4 F F q 3 q 5 [FEM]

Member 2: 1234 1 F q3    d3 = D3  q3         F  q 2 k d = D q  4  =  1   4 4  +  4     F  q5  3  d5 = 0  q5        F  q6  4    d6 = 0  q6 

40 Example 1

For the beam shown, use the stiffness method to: (a) Determine the deflection and rotation at B. (b) Determine all the reactions at supports. (c) Draw the quantitative shear and bending moment diagrams.

10 kN 1 kN/m

C A B 1.5 m 9 m 3 m

41 10 kN 1 kN/m

C A B 1.5 m 9 m 3 m

3 2 1

Global 1 2

3 2 2 1 Members 1 2 10 kN 1 kN/m 1.5 m 1.5 m [FEF] 9 m 1 2 wL2/12 = 6.75 wL2/12 = 6.75 PL/8 = 3.75 PL/8 = 3.75

42 3 2 1

1 2

9 m 3 m

Stiffness Matrix: θi θj

Mi 4EI / L 2EI / L []k 2×2 =   Mj 2EI / L 4EI / L

3 2 21 4/9 2/9 3 4/3 2/3 2 [k]1= EI [k]2 = EI 2/9 4/9 2 2/3 4/3 1

21 (4/9)+(4/3) 2/3 2 [K] = EI 2/3 4/3 1 43 3 2 10 kN 1 1 kN/m

A C 6.75 6.75 3.75 B 9 m 1.5 m 1.5 m

Equilibrium equations: MCB = 0 MBA + MBC = 0 Global Equilibrium: [Q] = [K][D] + [QF]

21 2 2 MBA + MBC = 0 (4/9)+(4/3) 2/3 θB -6.75 + 3.75 = -3 = EI + 1 1 MCB = 0 2/3 4/3 θC -3.75

θ 0.779/EI B = θC 2.423/EI 44 3 2

1 1 kN/m [FEF] 9 m 1 wL2/12 = 6.75 wL2/12 = 6.75

Substitute θB and θC in the member matrix,

F Member 1 : [q]1 = [k]1[d]1 + [q ]1

3 2 0 3 MAB 4/9 2/9 θA 6.75 6.92 = EI = 0.779/EI + = 2 MBA 2/9 4/9 θB -6.75 -6.40

1 kN/m 6.92 kN•m 6.40 kN•m 9 m 1 4.56 kN 4.44 kN

45 10 kN 2 1 1.5 m 1.5 m

2 2 PL/8 = 3.75 PL/8 = 3.75 [FEF]

Substitute θB and θC in the member matrix,

F Member 2 : [q]2 = [k]2[d]2 + [q ]2

21 2 MBC 4/3 2/3 θB = 0.779/EI 3.75 6.40 = EI + = 1 MCB 2/3 4/3 θC = 2.423/EI -3.75 0

10 kN

6.40 kN•m 2 7.13 kN 2.87 kN 46 10 kN 1 kN/m 6.92 6.40 6.40 4.56 kN 4.44 kN 2.87 kN 7.13 kN 10 kN 1 kN/m θB = +0.779/EI θ = +2.423/EI 6.92 kN•m C A C 4.56 kN B 11.57 kN 2.87 kN 9 m 1.5 m 1.5 m 7.13 4.56

V (kN) x (m) 4.56 m -4.44 -2.87 3.48 4.32 M (kN•m) x (m)

-6.92 -6.40 47 Example 2

For the beam shown, use the stiffness method to: (a) Determine the deflection and rotation at B. (b) Determine all the reactions at supports.

20 kN 9kN/m 40 kN•m

2EI EI ACB 4 m 4 m

48 2 4 6

1 2EI 3 EI 5 2 1 1 2 3

Use 4x4 stiffness matrix, ∆ θ ∆ θ i i j j 34 Vi  12EI/L3 6EI/L2 −12EI/L3 6EI/L2  M   3 0.5625 -0.375 i 6EI/L2 4EI/L − 6EI/L2 2EI/L []k =   [K] = EI V 3 2 3 2 j −12EI/L − 6EI/L 12EI/L − 6EI/L  4 -0.375 3 M  2 2  j  6EI/L 2EI/L − 6EI/L 4EI/L 

[k]1 [k]2 1234 3456 1 0.375EI 0.75EI - 0.375EI0.75EI 3 0.1875EI 0.375EI - 0.1875EI 0.375EI 2 0.75EI 2EI -0.75EI EI 4 0.375EI EI -0.375EI 0.5EI 3 -0.375EI -0.75EI 0.375EI -0.75EI 5 -0.1875EI -0.375EI 0.1875EI-0.375EI 4 0.75EI EI -0.75EI 2EI 6 0.375EI 0.5EI -0.375EI EI 49 20 kN 9kN/m 40 kN•m

12 kN•m 12 kN•m 18 kN 18 kN 4 m 4 m 2 4 6

1 3 2EI EI 5 2 1 1 2 3 [Q] = [K][D] + [QF] Global: 34 0.5625 D 3 Q3 = -20 3 -0.375 3 18 3 = EI 4 D + 4 Q4 = 40 -0.375 3 4 -12 4

D3 -61.09/EI = D4 9.697/EI 50 2 4 9 kN/m 1 3 2EI 12 kN•m 12 kN•m A B 1 1 2 1 F 18 kN [q ]1 18 kN Member 1:

F [q]1 = [k]1[d]1 + [q ]1 1234

3 q1 1 12(2EI)/4 0.75EI - 0.375EI0.75EI d1 = 0 18 48.18 q2 2 0.75EI 2EI -0.75EI EI d2 = 0 12 67.51 d = -61.09/EI 18 -12.18 q3L 3 -0.375EI -0.75EI 0.375EI -0.75EI 3 q4L 4 0.75EI EI -0.75EI 2EI d4 = 9.697/EI -12 53.21

9 kN/m 53.21 kN•m

A B 1 67.51 kN•m 12.18 kN 48.18 kN [q]1 51 4 6

3 EI 5 2 2 3

Member 2:

F [q]2 = [k]2[d]2 + [q ]2

3456 q3R 3 0.1875EI 0.375EI - 0.1875EI 0.375EI d3 =-61.09/EI 0 -7.818 d = 9.697/EI q4R 4 0.375EI EI -0.375EI 0.5EI 4 0 -13.21 q5 5 -0.1875EI -0.375EI 0.1875EI-0.375EI d5 = 0 0 7.818 q6 6 0.375EI 0.5EI -0.375EI EI d6 = 0 0 -18.06

13.21 kN•m 18.06 kN•m B C 2 7.82 kN 7.82 kN [q]2 52 9 kN/m 53.21 kN•m 13.21 kN•m 18.06 kN•m A B B C 1 67.51 kN•m 2 12.18 kN 7.82 kN 48.18 kN [q]1 7.82 kN [q]2 20 kN 9kN/m 40 kN•m D3 = ∆B = -61.09/EI 67.51 kN•m 18.06 kN•m

D = = +9.697/EI 48.18 kN 4 m 4 θB 4 m 7.818 kN 48.18

V (kN) + 12.18 x (m) - -7.818 -7.818 53.21 M (kN•m) + 13.21 x (m) - - -18.08 -67.51 53 Example 3

For the beam shown, use the stiffness method to: (a) Determine the deflection and rotation at B. (b) Determine all the reactions at supports.

20 kN 10 kN 9kN/m 40 kN•m

2EI EI ACB 4 m 2 m 2 m

54 20 kN 10 kN 9kN/m 40 kN•m A C 2EI B EI 4 m 2 m 2 m 2 4 6

1 3 5

Global 2 1 1 2 3 10 kN 9 kN/m 5 kN•m 12 kN•m 5 kN•m 12 kN•m [FEF] 1 2 18 kN 18 kN 5 kN 5 kN ∆i θi ∆j θj Vi  12EI/L3 6EI/L2 −12EI/L3 6EI/L2  M   i 6EI/L2 4EI/L − 6EI/L2 2EI/L []k =   4×4 V 3 2 3 2 j −12EI/L − 6EI/L 12EI/L − 6EI/L  M  2 2  j 6EI/L 2EI/L − 6EI/L 4EI/L   55 2 4 6

1 3 5

2 1 1 2 3 4 m 2 m 2 m

12 34 1 12(2EI)/43 0.75EI - 0.375EI 0.75EI 2 0.75EI 2EI -0.75EI EI 346 [k]1 = 3 -0.375EI -0.75EI 0.375EI -0.75EI 3 0.5625 -0.375 0.375 4 0.75EI EI -0.75EI 2EI/L [K] = EI 4 -0.375 3 0.5 3456 6 0.375 0.5 1 3 0.1875EI 0.375EI - 0.1875EI 0.375EI 4 0.375EI EI -0.375EI 0.5EI [k] = 2 5 -0.1875EI -0.375EI 0.1875EI -0.375EI 6 0.375EI 0.5EI -0.375EI EI 56 20 kN 10 kN 9kN/m 40 kN•m

A C 2 B 4 6

1 3 5

1 2 1 2 10 kN 3 9 kN/m 5 kN•m 12 kN•m 5 kN•m 12 kN•m [FEF] 1 2 18 kN 18 kN 5 kN 5 kN Global: [Q] = [K][D] + [QF] 346

VBL+VBR = -20 3 0.5625 -0.375 0.375 ∆B 18 + 5 = 23 + MBA+MBC = 40 = EI 4 -0.375 3 0.5 θB -12 + 5 = -7 M = 0 6 0.375 0.5 1 CB θC -5

∆B -116.593/EI

θB = -7.667/EI 52.556/EI 57 θC 2 4 9 kN/m 1 3 12 kN•m 12 kN•m 1 1 1 2 18 kN 18 kN [FEF]

F Member 1: [q]1 = [k]1[d]1 + [q ]1

12 34

VA 1 0.375EI 0.75EI - 0.375EI 0.75EI 0 18 55.97 M 2 0.75EI 2EI -0.75EI EI 0 12 91.78 AB = + = VBL 3 -0.375EI -0.75EI 0.375EI -0.75EI ∆B =-116.593/EI 18 -19.97

MBA 4 0.75EI EI -0.75EI 2EI/L θB =-7.667/EI -12 60.11

9kN/m

91.78 kN•m 60.11 kN•m 4 m 1

55.97 kN 19.97 kN 58 4 6 10 kN 5 kN•m 3 5 5 kN•m 3 2 2 2 5 kN [FEM] 5 kN F Member 2: [q]1 = [k]1[d]1 + [q ]1

3456

VBR 3 0.1875EI 0.375EI - 0.1875EI 0.375EI ∆B =-116.593/EI 5 -0.0278

MBC 4 0.375EI EI -0.375EI 0.5EI θB =-7.667/EI 5 -20.11 = + = VC 5 -0.1875EI -0.375EI 0.1875EI -0.375EI 0 5 10.03

MCB 6 0.375EI 0.5EI -0.375EI EI θC =52.556/EI -5 0

10 kN

20.11 kN•m 2 m 2 m 2 0.0278 kN 10.03 kN 59 9kN/m 20.11 kN•m 10 kN 91.78 kN•m 2 m 2 m 4 m 1 2 60.11 kN•m 0.0278 kN 10.03 kN 55.97 kN 19.97 kN 20 kN 10 kN 9kN/m 40 kN•m θC = 52.556/EI 91.78 kN•m

55.97 kN 10.03 kN θB = -7.667/EI ∆B = -116.593/EI 55.97 4 m 2 m 2 m 19.97 V (kN) + x (m) - -0.0278 60.11 -10.03 -10.03 20.11 M + (kN•m) x (m) -

-91.78 60 Example 4

For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative shear diagram, bending moment diagram and qualitative deflected shape. Take I = 200(106) mm4 and E = 200 GPa and support B settlement 10 mm.

40 kN 6 kN/m B A C 2EI EI ∆B = -10 mm 8 m 4 m 4 m

61 1 2 3

2EI EI

1 2

θi θj

Use 2x2 stiffness matrix: Mi 4EI / L 2EI / L []k 2×2 =   Mj 2EI / L 4EI / L

12 23

1 8 4 2 4 2 EI EI [k]1 = 8 [k]2 = 8 2 4 8 3 2 4

2 12 2 EI [K] = 8 3 2 4 62 40 kN 1 2 3 6 kN/m B C

A 2EI EI 1 2 ∆B = -10 mm 8 m 4 m 4 m 40 kN 6 kN/m [FEM]load

1 2 wL2/12 = 32 kN•m 32 kN•m PL/8 = 40 kN•m 40 kN•m 2 37.5 kN•m 1 75 kN•m 10 mm [FEM]∆ 10 mm 6(EI)∆/L2 = 37.5 kN•m 6(2EI)∆/L2 = 75 kN•m Global: [Q] = [K][D] +23 [QF]

Q = 0 2 12 2 D -32 + 40 + 75 -37.5 = 45.5 2 EI 2 = + 8 Q3 = 0 3 2 4 D3 -40 - 37.5 = -77.5

D2 -61.27/EI rad = D 185.64/EI rad 3 63 1 2 6 kN/m [FEM]load 2EI 2 1 32 kN•m 1 wL /12 = 32 kN•m

1 75 kN•m [FEM]∆ 10 mm 6(2EI)∆/L2 = 75 kN•m

F Member 1: [q]1 = [k]1[d]1 + [q ]1

12 q 1 8 4 d = 0 32 + 75 = 107 76.37 kN•m 1 EI 1 = 8 + = q2 2 4 8 d2 = -61.27/EI -32 + 75 = 43 -18.27 kN•m

6 kN/m 76.37 kN•m 18.27 kN•m

8 m 1 31.26 kN 16.74 kN 64 2 3 40 kN

[FEM]load 2 2 PL/8 = 40 kN•m 40 kN•m 2 [FEM] 37.5 kN•m ∆ 10 mm 6(EI)D/L2 = 37.5 kN•m

F Member 2: [q]2 = [k]2[d]2 + [q ]2

23 q2 2 4 2 d2 = -61.27/EI 40 - 37.5 = 2.5 18.27 kN•m EI = + = 8 q3 3 2 4 d3 = 185.64/EI -40 - 37.5 = -77.5 0 kN•m

40 kN 18.27 kN•m

2 22.28 kN 17.72 kN 65 2 4 6 2EI EI

1 2 1 3 5 Alternate method: Use 4x4 stiffness matrix

[k]1 [k]2 12 3 4 34 5 6 1 3 12(2)/82 1.5 -0.375 1.5 12/82 0.75 -0.1875 0.75 EI 2 4 = 1.5 8 -1.5 4 EI 0.75 4 -0.75 2 8 = 3 -0.375 -1.5 0.375 -1.5 8 5 -0.1875 -0.75 0.1875 -0.75 4 1.5 4 -1.5 8 6 0.75 2 -0.75 4 145623 1 0.375 1.5 -0.375 1.5 0 0 2 1.5 8 -1.5 4 0 0 EI 3 -0.375 -1.5 0.5625 -0.75 -0.1875 0.75 [K] = 8 4 1.5 4 -0.75 12 -0.75 2 5 0 0 -0.1875 -0.75 0.1875 -0.75 6 0 0 0.75 2 -0.75 4 66 40 kN 2 6 kN/m 4 6 B C

A 2EI EI 1 2 ∆B = -10 mm 1 3 5 8 m 4 m 4 m 40 kN 6 kN/m 40 kN•m 32 kN•m 40 kN•m 32 kN•m [FEF] 1 2 24 kN 24 kN 20 kN 20 kN Global: [Q] = [K][D] + [QF] 46 5 Q = 0 4 12 2 D 4 EI 4 200× 200 4 -0.75 D = -0.01 8 = + ( ) 5 + Q6= 0 8 6 2 4 D6 8 6 -0.75 -40 46

Q4 = 0 EI 4 12 2 D4 (200x200/8)(-0.75)(-0.01) = 37.5 8 = + + Q6= 0 8 6 2 4 D6 (200x200/8)(0.75)(-0.01) = -37.5 -40

-3 D4 -61.27/EI = -1.532x10 rad = -3 D6 185.64/EI = 4.641x10 rad 67 2 4 6 kN/m 32 kN•m 32 kN•m 1 ∆B = -10 mm 1 [FEF]load 3 1 24 kN 24 kN F Member 1: [q]1 = [k]1[d]1 + [q ]1

12 3 4 q 1 2 1 12(2)/8 1.5 -0.375 1.5 d1 = 0 24 q 2 2 (200x200) 1.5 8 -1.5 4 d2 = 0 32 q = 8 3 + 3 -0.375 -1.5 0.375 -1.5 d3 = -0.01 24 q4 4 -3 1.5 4 -1.5 8 d4 = -1.532x10 -32 q1 31.26 kN 6 kN/m 76.37 kN•m 18.27 kN•m q2 76.37 kN•m q = 3 16.74 kN 8 m 1 q 4 -18.27 kN•m 31.26 kN 16.74 kN

68 6 40 kN 4 40 kN•m 40 kN•m -10 mm = ∆ [FEF] B 2 2 3 5 20 kN 20 kN Member 2: 345 6 q 3 2 3 12/8 0.75 -0.1875 0.75 d3 = -0.01 20 4 -3 q4 0.75 4 -0.75 2 d = -1.532x10 40 = (200x200) 4 q 8 5 + 5 -0.1875 -0.75 0.1875 -0.75 d5 = 0 20 q6 6 -3 0.75 2 -0.75 4 d6 = 4.641x10 -40

40 kN q3 22.28 kN 18.27 kN•m q4 18.27 kN•m = 2 q5 17.72 kN 17.72 kN q6 0 kN•m 22.28 kN 69 40 kN 6 kN/m 76.37 kN•m B C

A 2EI EI ∆B = -10 mm 31.26 kN 16.74 + 22.28 kN 17.72 kN 8 m 4 m 4 m

31.26 V (kN) 22.28 + + - x (m) 5.21 m -16.74 - -17.72 70.85

M 5.06 + (kN•m) - x (m) - -18.27

-76.37 ∆ = -10 mm D = θ = 4.641x10-3 rad Deflected B 6 C Curve

-3 d4 = θB = -1.532x10 rad 70 Example 5

4 kN 0.6 kN/m 20 kN•m B A C 2EI EI ∆C = -10 mm 8 m 4 m 4 m

71 2 4 6 2EI EI

1 2 1 -10 mm 3 5 •Member stiffness matrix [k]4x4

[k]1 [k]2 12 3 4 34 5 6 1 3 12(2)/82 1.5 -0.375 1.5 12/82 0.75 -0.1875 0.75 EI 2 4 = 1.5 8 -1.5 4 EI 0.75 4 -0.75 2 8 3 = 5 -0.375 -1.5 0.375 -1.5 8 -0.1875 -0.75 0.1875 -0.75 4 1.5 4 -1.5 8 6 0.75 2 -0.75 4

• Global: [Q] = [K][D] + [QF] 346 5 F Q3 3 0.5625 -0.75 0.75 D3 3 -0.1875 Q 3 EI 200× 200 F Q4 = 4 -0.75 12 2 D4 + ( )4 -0.75 D5 = -0.01 + Q 4 8 8 F Q6 6 0.75 2 4 D6 6 -0.75 Q 6

72 4 kN 2 0.6 kN/m 20 kN•m 4 6 B 2EI EI C A EI 1 2 2EI 10 mm 1 5 8 m 4 m 4 m 3 4 kN 0.6 kN/m 4 kN•m 3.2 kN•m 4 kN•m 3.2 kN•m [FEF] 1 2 2.4 kN 2.4 kN 2 kN 2 kN Global: [Q] = [K][D] + [QF] 346

Q3 = 0 3 0.5625 -0.75 0.75 D3 9.375 2.4+2 = 4.4 EI Q4 = 0 = 4 -0.75 12 2 D4 + 37.5 + -3.2+4 = 0.8 8 Q6 = 20 6 0.75 2 4 D6 37.5 -4.0

-3 D3 -377.30/EI = -9.433x10 m -3 D4 = -61.53/EI = -1.538x10 rad -3 D6 +74.50/EI = +1.863x10 rad 73 2 4 0.6 kN/m 3.2 kN•m 1 3.2 kN•m 1 [FEF]load 3 8 m 1 2.4 kN 2.4 kN F Member 1: [q]1 = [k]1[d]1 + [q ]1

12 3 4 q 1 2 1 12(2)/8 1.5 -0.375 1.5 d1 = 0 2.4 2 q2 200× 200 1.5 8 -1.5 4 d = 0 3.2 = 2 q 8 3 -3 + 3 -0.375 -1.5 0.375 -1.5 d3 = -9.433x10 2.4 q4 4 -3 1.5 4 -1.5 8 d4 = -1.538x10 -3.2 q1 8.55 kN 0.6 kN/m 43.19 kN•m 6.03 kN•m q2 43.19 kN•m q = 3 -3.75 kN 8 m 1 q 4 6.03 kN•m 8.55 kN 3.75 kN

74 6 4 kN 4 4 kN•m 4 kN•m [FEF] 2 10 mm 8 m 2 3 5 2 kN 2 kN Member 2: 345 6 q 3 2 -3 3 12/8 0.75 -0.1875 0.75 d3 = -9.433x10 2 4 -3 q4 200× 200 0.75 4 -0.75 2 d = -1.538x10 4 = 4 q 8 5 + 5 -0.1875 -0.75 0.1875 -0.75 d5 = -0.01 2 q6 6 -3 0.75 2 -0.75 4 d6 = 1.863x10 -4

4 kN q3 3.75 kN 6.0 kN•m q4 -6.0 kN•m 20 kN•m = 2 q5 0.25 kN 0.25 kN q6 20.0 kN•m 3.75 kN 75 4 kN 0.6 kN/m 20 kN•m B C A 2EI EI10 mm 8 m 4 m 4 m 0.6 kN/m 4 kN 43.19 kN•m 6.03 kN•m 20 kN•m 6.0 kN•m 8 m 1 2 8.55 kN 3.75 kN 3.75 kN 0.25 kN 8.55 3.75 V (kN) + -0.25 x (m) 21 20 6 M + (kN•m) - x (m)

-43.19 Deflected D3 = -9.433 mm D5 = -10 mm Curve -3 D = +1.863x10-3 rad D4 = -1.538x10 rad 6 76 Internal Hinges Example 6

For the beam shown, use the stiffness method to: (a) Determine all the reactions at supports. (b) Draw the quantitative shear and bending moment diagrams and qualitative deflected shape. E = 200 GPa, I = 50x10-6 m4.

30 kN 9kN/m Hinge

2EI EI AC B 4 m 2 m 2 m

77 3 4 2 2EI EI C A 2 1 1 B 4 m 2 m 2 m

θi θj Use 2x2 stiffness matrix, Mi 4EI / L 2EI / L []k 2×2 =   Mj 2EI / L 4EI / L

31 24

3 2EI EI 2 1EI 0.5EI [k]1 = [k]2 = 1 EI 2EI 4 0.5EI 1EI 12 1 2.0 0.0 [K] = EI 2 0.0 1.0 78 30 kN 9kN/m Hinge

3 4 2 A C 1 2 1 B 30 kN 12 kN•m 9kN/m 12 kN•m 15 kN•m 15 kN•m B C [FEF] A 1 B 2

Global matrix: 12

0.0 1 2.0 0.0 D1 -12 = EI + 0.0 2 0.0 1.0 D2 15

D1 0.0006 rad = D2 -0.0015 rad 79 3 12 kN•m 9kN/m 1 12 kN•m A [FEF] 1 A 1 B B

Member 1: 31

q3 3 2EI EI d3 = 0.0 12 18 = + = = 0.0006 q1 1 EI 2EI d1 -12 0.0

9 kN/m

A B 18 kN•m 1 22.5 kN 13.5 kN

80 4 15 kN•m 30 kN 15 kN•m B B C C 2 2 2

Member 2: 24

q2 2 1EI 0.5EI d2 = -0.0015 15 0.0 = + = q4 4 0.5EI 1EI d4 = 0.0 -15 -22.5

30 kN B C 22.5 kN•m 9.37 kN 20.63 kN

81 30 kN 9 kN/m B C 13.5 kN A B 9.37 kN 22.5 kN•m 18 kN•m 22.5 kN 13.5 kN 22.87 kN 9.37 kN 20.63 kN 30 kN 9kN/m Hinge

2EI EI θBR= -0.0015 rad ACθBL= 0.0006 rad B 4 m 2 m 2 m

22.5 9.37 V (kN) + x (m) 2.5 m -13.5 -20.63 18.75 10.13 M + + (kN•m) - - - x (m) -18 -22.5 82 Example 7

20 kN 9kN/m

EI 2EI Hinge ACB 4 m 4 m

83 5 7 1 4 2EI3 EI 6 1 2 AC2 B

45 1 2 4 0.375EI 0.75EI - 0.375EI 0.75EI 5 0.75EI 2EI -0.75EI EI [k]1 = 1 -0.375EI -0.75EI 0.375EI -0.75EI 123 2 0.75EI EI -0.75EI 2EI/L 1 0.5625 -0.75 0.375

[K] = EI 2 -0.75 2.0 0 3 0.375 0.0 1.0 13 6 7 1 0.1875EI 0.375EI - 0.1875EI 0.375EI 3 0.375EI EI -0.375EI 0.5EI [k]2 = 6 -0.1875EI -0.375EI 0.1875EI -0.375EI 7 0.375EI 0.5EI -0.375EI EI

84 9kN/m 20 kN B A C 5 7 1 3 4 2EI EI 6 1 2 AC2 B 9kN/m

12 kN•m 12 kN•m A 1 B 18 kN [FEM] 18 kN Global: 1 2 3

Q1 = -20 1 0.5625 -0.75 0.375 D1 18

Q2 = 0.0 = EI 2 -0.75 2.0 0 D2 + -12

Q3 = 0.0 3 0.375 0.0 1.0 D3 0.0

D1 -0.02382 m

D2 = -0.008333 rad D 0.008933 rad 3 85 5 9kN/m 1 4 2EI 2 12 kN•m 12 kN•m 1 B 1 A AB18 kN [FEF] 18 kN Member 1: 45 1 2

q4 4 0.375EI 0.75EI - 0.375EI 0.75EI d4 = 0.0 18 44.83 q5 5 0.75EI 2EI -0.75EI EI d 12 107.32 = 5 = 0.0 + = q1 1 -0.375EI -0.75EI 0.375EI -0.75EI d1 = -0.02382 18 -8.83

q2 2 0.75EI EI -0.75EI 2EI/L d2 = -0.00833 -12 0.0

107.32 kN•m 9kN/m AB 1 8.83 kN 44.83 kN

86 7 1 EI 6 2 B 3 C

Member 2: 13 6 7 q 1 1 0.1875EI 0.375EI - 0.1875EI 0.375EI d1 = -0.02382 0 -11.16 q 3 3 0.375EI EI -0.375EI 0.5EI d3 = 0.008933 0 0.0 q =6 + = 6 -0.1875EI -0.375EI 0.1875EI -0.375EI d6= 0.0 0 11.16 q7 7 = 0.0 0.375EI 0.5EI -0.375EI EI d7 0 -44.66

B C 44.66 kN•m 2 11.16 kN 11.16 kN

87 20 kN 9kN/m

θ = -0.008333 rad θ = 0.008933 rad ACBL B BR 4 m 4 m

107.32 kN•m 9kN/m B C AB 44.66 kN•m 1 2 44.83 kN 8.83 kN 11.16 kN 11.16 kN 44.83

V (kN) + 8.83 x (m) - -11.16 M -11.16 (kN•m) x (m) - - -44.66

-107.32 88 Example 8

89 3 4 2 2EI EI A C 1 2 1 B 4 m 2 m 2 m

θ θ Use 2x2 stiffness matrix: i j Mi 4EI / L 2EI / L []k 2×2 =   Mj 2EI / L 4EI / L

31 24

3 2EI EI 2 1EI 0.5EI [k]1 = [k]2 = 1 EI 2EI 4 0.5EI 1EI 12 1 2.0 0.0 [K] = EI 2 0.0 1.0 90 30 kN 9kN/m Hinge AC 2EI 40 kN•m EI 3 B 4 2 A C 1 2 1 B 30 kN 12 kN•m 9kN/m 12 kN•m 15 kN•m 15 kN•m B C [FEM] A 1 B 2

Global matrix: 12

Q1 = 40 1 2.0 0.0 D1 -12 = EI + Q2 = 0.0 2 0.0 1.0 D2 15

D1 0.0026 rad = D2 -0.0015 rad 91 3 12 kN•m 9kN/m 1 12 kN•m A [FEF] 1 A 1 B B

Member 1:

q3 3 2EI EI d3 = 0.0 12 38 = + = = 0.0026 q1 1 EI 2EI d1 -12 40

9 kN/m

A B 40 kN•m 38 kN•m 37.5 kN 1.5 kN

92 4 15 kN•m 30 kN 15 kN•m B B C C 2 2 2

Member 2:

q2 2 1EI 0.5EI d2 = -0.0015 15 0.0 = + = q4 4 0.5EI 1EI d4 = 0.0 -15 -22.5

30 kN B C 22.5 kN•m 9.37 kN 20.63 kN

93 1.5 kN 30 kN 9 kN/m B C A B 40 kN•m 9.37 kN 22.5 kN•m 38 kN•m 37.5 kN 1.5 kN 7.87 kN 9.37 kN 20.63 kN 30 kN 9kN/m 40 kN•m Hinge AC

B θBR=- 0.0015 rad θBL= 0.0026 rad 4 m 2 m 2 m 37.5

1.5 9.37 V (kN) + x (m)

-20.63

M 40 18.75 (kN•m) + + x (m) - - - -22.5 -38 94 Example 9

EI 2EI 40 kN•m Hinge ACB 4 m 4 m

95 20 kN 9kN/m

EI 2EI 40 kN•m Hinge ACB 4 m 4 m

5 7 1 4 3 6 AC 1 2 2 B

9 kN/m 12 kN•m 12 kN•m [FEM] 1 ∆ θ ∆ θ 18 kN 18 kN i i j j Vi  12EI/L3 6EI/L2 −12EI/L3 6EI/L2  M   i 6EI/L2 4EI/L − 6EI/L2 2EI/L []k =   4×4 V 3 2 3 2 j −12EI/L − 6EI/L 12EI/L − 6EI/L  M  2 2  j  6EI/L 2EI/L − 6EI/L 4EI/L  96 5 7 1 4 2EI3 EI 6 AC 1 2 2 B

45 1 2 4 0.375EI 0.75EI - 0.375EI 0.75EI [k] = 5 0.75EI 2EI -0.75EI EI 1 1 -0.375EI -0.75EI 0.375EI -0.75EI 123 2 0.75EI EI -0.75EI 2EI 1 0.5625 -0.75 0.375 [K] = EI 2 -0.75 2.0 0 3 0.375 0.0 1.0 1 3 6 7 1 0.1875EI 0.375EI - 0.1875EI 0.375EI 3 0.375EI EI -0.375EI 0.5EI [k]2 = 6 -0.1875EI -0.375EI 0.1875EI -0.375EI 7 0.375EI 0.5EI -0.375EI EI 97 9kN/m 20 kN

EI 5 40 kN•m 7 1 4 2EI3 EI 6 AC 1 2 2 B 9 kN/m 12 kN•m 12 kN•m [FEF] 1 18 kN 18 kN Global: 123

Q1 = -20 1 0.5625 -0.75 0.375 D1 18 Q = 40 2 -0.75 2.0 0 D 2 = EI 2 + -12

Q3 = 0.0 3 0.375 0.0 1.0 D3 0.0

D1 -0.01316 m

D2 = -0.002333 rad

D3 0.0049333 rad 98 5 1 9 kN/m 4 12 kN•m 2EI 2 12 kN•m 1 [FEF] 1 18 kN 18 kN

F Member 1: [q]1 = [k]1[d]1 + [q ]1

45 1 2 q 4 4 0.375EI 0.75EI - 0.375EI 0.75EI d4 = 0.0 18 49.85 q 5 5 0.75EI 2EI -0.75EI EI d = 0.0 12 87.37 = 5 + = q1 1 -0.375EI -0.75EI 0.375EI -0.75EI d1 = -0.01316 18 -13.85 q2 2 0.75EI EI -0.75EI 2EI d2 = -0.002333 -12 40

87.37 kN•m 2EI 40 kN•m 1 49.85 kN 13.85 kN

[q]1 99 7 1 EI 6 2 C B 3

F Member 2: [q]2 = [k]2[d]2 + [q ]2

13 6 7 q 1 1 0.1875EI 0.375EI - 0.1875EI 0.375EI d1 = -0.01316 0 -6.18 q 3 3 0.375EI EI -0.375EI 0.5EI d3 = 0.004933 0 0.0 q =6 + = 6 -0.1875EI -0.375EI 0.1875EI -0.375EI d6= 0.0 0 6.18 q7 7 0.375EI 0.5EI -0.375EI EI d7= 0.0 0 -24.69

24.69 kN•m 2 6.18 kN 6.18 kN [q]2

100 20 kN 9kN/m 40 kN•m

2EI EI ACθBL= -0.002333 rad θBR = 0.004933 rad 4 m B 4 m

A B B C 40 kN•m 24.69 kN•m 87.37 kN•m 49.85 kN 13.85 kN 6.18 kN 6.18 kN

49.85

V (kN) + 13.85 x (m) - -6.18 -6.18

M 40 (kN•m) + - x (m) - -24.69

-87.37 101 Temperature Effects

• Fixed-End Forces (FEF) - Axial - Bending

• Curvature

102 T −T ∆T tan β = b t = ( ) • Thermal Fixed-End Forces (FEF) d d Room temp = TR Tt TR Tt β F F F T > T FB c A m R t NA Tt +Tb d Tm = cb 2 F T > T F M A b t M B Tl A B Tb - Tt

σaxial σaxial TR Tm

AB(∆T)axial = Tm −TR

∆T (∆T ) = y( ) σ y y Tt d y y β

AB Tb

(∆T )bending = Tb −Tt 103 - Axial

σaxial σaxial Tt TR Tm

F Tm> TR F FA FB

ABTb > Tt (∆T)axial = Tm −TR

F F = σ dA A ∫ axial A = Eε dA ∫ axial

= Eα (∆T) dA ∫ axial

= Eα (∆T) dA axial ∫

= EAα (∆T )axial

F (Faxial ) A = α (Tm −TR )AE

104 - Bending ∆T (∆T ) = y( ) σ y y Tt d y y β

M F M F A ABB Tb (∆T) = T −T M F = yσ dA bending b t A ∫ y A = ∫ yEε dA

= yEα(∆T ) dA ∫ y ∆T = yEαy( ) dA ∫ d ∆T = Eα( ) y 2dA d ∫ T −T (F F ) = α( l u )EI bending A d 105 • Elastic Curve: Bending O´ dθ

ρ (dx) = ρ (dθ ) 1 dθ y (dθ ) = ( ) ρ dx ds = dx y ds´ dθ y dx dx

Before After deformation deformation

106 • Bending Temperature Tl > Tu Tt

Tb O dx

dθ ∆T = Tb - Tt T +T T T T = t b T dθ t t ∆T m 2 u ∆T y y c β = y d M M t d cb T > T l u Tb Tb θ ∆T (d )y = αy( )dx d θ ∆T (d ) = α( )dx d dθ 1 ∆T M ( ) = = α( ) = dx ρ d EI

107 Example 10

o T2 = 40 C 182 mm 2EI EI T = 25oC AC1 B 4 m 4 m

108 o T2 = 40 C 182 mm 2EI EI T = 25oC AC1 B 4 m 4 m 2 3 1

1 2

Mean temperature = (40+25)/2 = 32.5 Room temp = 32.5oC 19.78 kN•m a(∆T /d)EI = 19.78 kN•m +7.5 oC FEM -7.5 oC

∆T 40 − 25 F F =α ( )(2EI) = (12×10−6 )( )(2× 200×50) =19.78 kN • m bending d 0.182

109 2 3 1 19.78 kN•m 19.78 kN•m 182 mm A C 1 2EI 2 EI B 4 m 4 m

F Element 1: [q] = [k][d] + [q ] 21 2 M2 2 1 q2 -1.978 = EI + (10-3) EI M1 1 1 2 q1 1.978

Element 2: 13 1 M1 1 0.5 q1 0 = EI + M3 3 0.5 1 q3 0 0 -3 [M1] = 3EIθ1 + (1.978x10 )EI

-3 θ1 = -0.659x10 rad 110 2 3 1 19.78 kN•m 19.78 kN•m 182 mm A C 1 2EI 2 EI B 4 m 4 m Element 1: 21 2 = 0 M2 2 1 q2 -19.78 -26.37 kN•m EI = -3 + = M1 1 1 2 q1 = -0.659x10 19.78 6.59 kN•m

Element 2: 13

1 -3 M1 1 0.5 q1 = -0.659x10 0 -6.59 kN•m EI =+= -3.30 kN•m M3 3 0.5 1 q3 = 0 0

26.37 kN•m 6.59 kN•m 6.59 kN•m 3.3 kN•m ABB C

4.95 kN 2.47 kN 2.47 kN 4.95 kN 111 26.37 kN•m +7.5 oC

ACo B -7.5 C 2.47 kN 3.3 kN•m 4.95 kN 2.47 kN 4 m 4 m

V (kN) x (m) - - -2.47 -4.95

26.37 6.59 M + (kN•m) - x (m) -3.30

-3 Deflected θB = -0.659x10 rad curve x (m)

112 Example 11

o T2 = 40 C 182 mm 2EI, 2AE EI, AE T = 25oC AC1 B 4 m 4 m

113 6 9 T = 40oC 5 2 2 3 8 A 1 2 EI 4 2EI 1 7 T = 25oC 1 B C 4 m 4 m Element 1:

(2AE) 2(20×10−3 m2 )(200×106 kN / m2 ) = = 2(106 ) kN / m L (4 m)

4(2EI) 4× 2(200×106 kN / m2 )(50×10−6 m4 ) = = 20(103 ) kN • m L (4 m)

2(2EI) 2× 2(200×106 kN / m2 )(50×10−6 m4 ) = =10(103 ) kN • m L (4 m)

6(2EI) 6× 2(200×106 kN / m2 )(50×10−6 m4 ) = = 7.5(103 ) kN L2 (4 m)2

12(2EI) 12× 2(200×106 kN / m2 )(50×10−6 m4 ) = = 3.75(103 ) kN / m L3 (4 m)3 114 o T2 = 40 C 182 mm 2EI, 2AE EI, AE T = 25oC AC1 B 4 m 4 m

Fixed-end forces due to temperatures

∆T 40 − 25 F F =α ( )(2EI) = (12×10−6 )( )(2× 200×50) =19.78 kN • m bending d 0.182

o Mean temperature(Tm) = (40+25)/2 = 32.5 C , o TR = 28 C

F −6 2 6 2 Faxial =α (∆T )AE = (12×10 )(32.5 − 28)(2× 20×10 − 3 m )(200×10 kN / m ) = 432 kN

19.78 kN•m 19.78 kN•m +12 oC 432 kN 432 kN A -3 oC B

115 6 9 T = 40oC 5 2 2 3 8 182 mm 2 1 2EI, 2AE 1 EI, AE 7 4 T = 25oC AC1 B 4 m 4 m Element 1: FEM 19.78 kN•m 19.78 kN•m +12 oC 432 kN 432 kN A -3 oC B [q] = [k][d] + [qF]

4 5 6 1 2 3

6 6 q4 4 2x10 0.00 0.00 - 2x10 0.00 0.00 0 432

q5 5 0.00 3750 7500 0.00 -3750 7500 0 0.00

3 3 q6 6 0.00 7500 20x10 0.00 -7500 10x10 0 -19.78

= 6 6 + q1 1 -2x10 0.00 0.00 2x10 0.00 0.00 d1 -432

q2 2 0.00 -3750 -7500 0.00 3750 -7500 d2 0.00 q 3 3 0.00 7500 10x103 0.00 -7500 20x103 d 19.78 3 116 6 9 T = 40oC 5 2 2 3 8 182 mm 2 1 2EI, 2AE 1 EI, AE 7 4 T = 25oC AC1 B 4 m 4 m Element 2:

[q] = [k][d] + [qF]

12378 9

6 6 q1 1 1x10 0.00 0.00 - 1x10 0.00 0.00 d1 0

q2 2 0.00 1875 3750 0.00 -1875 3750 d2 0 3 3 q3 3 0.00 3750 10x10 0.00 -3750 5x10 d3 0

= 6 6 + q7 7 -1x10 0.00 0.00 1x10 0.00 0.00 0 0

q8 8 0.00 -1875 -3750 0.00 1875 -3750 0 0 q 9 9 0.00 3750 5x103 0.00 -3750 10x103 0 0 117 6 9 T = 40oC 5 2 2 3 8 182 mm 2 1 2EI, 2AE 1 EI, AE 7 4 T = 25oC AC1 B 4 m 4 m Global: 1 2 3

6 Q1 = 0.0 1 3x10 0.0 0.0 D1 -432

Q2 = 0.0 = 2 0.0 5625 -3750 D2 + 0.0 3 Q3 = 0.0 3 0.0 -3750 30x10 D3 19.78

D1 0.000144 m

D2 = -0.0004795 m -6 D3 -719.3x10 rad

118 Element 1: 4 5 6 1 2 3

6 6 q4 4 2x10 0.00 0.00 - 2x10 0.00 0.00 0 432 q5 5 0.00 3750 7500 0.00 -3750 7500 0 0.00

3 3 q6 6 0.00 7500 20x10 0.00 -7500 10x10 0 -19.78 = + 6 6 -6 q1 1 -2x10 0.00 0.00 2x10 0.00 0.00 d1= 144x10 -432 -6 q2 2 0.00 -3750 -7500 0.00 3750 -7500 d2= -479.5x10 0.00 q 3 3 3 -6 3 0.00 7500 10x10 0.00 -7500 20x10 d3= -719.3x10 19.78 6 q 144.0 kN 4 5 2 3 q5 -3.60 kN A 4 B 1 q6 -23.38 kN•m = q1 -144.0 kN 23.38 kN•m 9 kN•m q2 3.60 kN q 144 kN 144 kN 3 9.00 kN•m A B 3.60 kN 3.60 kN 119 Element 2: 12378 9

6 6 -6 q1 1 1x10 0.00 0.00 - 1x10 0.00 0.00 d1 = 144x10 0 -6 q2 2 0.00 1875 3750 0.00 -1875 3750 d2 = -479.5x10 0

3 3 -6 q3 3 0.00 37500 10x10 0.00 -3750 5x10 d3 = -719.3x10 0

= 6 6 + q7 7 -1x10 0.00 0.00 1x10 0.00 0.00 0 0 q8 8 0.00 -1875 -3750 0.00 1875 -3750 0 0 q 9 9 0.00 3750 5x103 0.00 -3750 10x103 0 0 3 q 144 kN 1 2 8 9 q2 -3.6 kN B 1 C 7 q3 -9 kN•m = q7 -144 kN 9 kN•m 5.39 kN•m q8 3.6 kN q 144 kN 144 kN 9 -5.39 kN•m B C 3.60 kN 3.60 kN 120 Isolate axial part from the system o T2 = 40 C R A RC 2EI EI o ACT1 = 25 C B 4 m 4 m

RA RA = RC +

Compatibility equation: dC/A = 0

R (4) R (4) A + A +12×10−6 (32.5 − 28)(4) = 0 2AE AE

RA = 144 kN

RC = -144 kN

121 o T2 = 40 C

2EI EI o ACT1 = 25 C B 4 m 4 m

23.38 kN•m 9 kN•m 9 kN•m 5.39 kN•m 144 kN 144 kN 144 kN 144 kN A B B C 3.60 kN 3.60 kN 3.60 kN 3.60 kN V (kN) x (m) - -3.6 23.38 8.98 M + (kN•m) - x (m) -5.39 D2 = -0.0004795 mm -6 Deflected D3 = -719.3x10 rad curve x (m) 122 Skew Roller Support

- Force Transformation - Displacement Transformation - Stiffness Matrix

123 • Displacement and Force Transformation Matrices

x´ y´ 5´ 6´ 4´ j

2´ m

3´ 1´ i

y 6

5 4 j 3 θy m θx 2 1 i x 124 Force Transformation λ 6´ x´ x λy q = q cosθ − q cosθ 4 4' θ x 5' y θy 5´ 4´ y´ q5 = q4' cos y − q5' cosθ x j θx θ q6 = q6' 2´ y θx x j − xi q 4  λx − λy 0 q4'  λx = 3´ 1´ L q  = λ λ 0 q  i y − y  5   y x   5'  j i       λy = q 6   0 0 1 q6'  L

q 1  λx − λy 0 0 0 0q 1'  y     6 q λ λ 0 0 0 0 q  2   y x  2'  5 q 3   0 0 1 0 0 0q 3'    =    4 q 0 0 0 λ − λ 0 q j  4   x y  4'  q   0 0 0 λ λ 0q  3 m  5   y x  5'  q 6   0 0 0 0 0 1q 6'  2 1 i x []q = T []T [q '] 125 Displacement Transformation

λx λ y 6 x´ y d '4 = d 4 cos θ x + d 5 cos θ y 5 5´ s θx co d' = −d cos θ + d cos θ d 4 5 4 y 5 x 4 j 6´ θx4´ d4 d '6 = d 6 j θy s θy θy co d 4 d 4'   λx λy 0 d 4  θx d  = − λ λ 0 d   5'   y x   5        x d 6'   0 0 1 d 6  y θx os d   λ λ 0 0 0 0d  c 1' x y 1 θx d 5     d − λ λ 0 0 0 0 d d  2'   y x  2  5      θy θy d 3' 0 0 1 0 0 0 d 3 os   =    c d 5 d 4'   0 0 0 λx λy 0d 4       θ d 5' 0 0 0 − λy λx 0 d 5 y      θ x d 6'   0 0 0 0 0 1d 6 

x [d']= [T ][d] 126 []q = [T ][T q ' ]

= [T ][][]T ( k' d' + [q'F ])

= [T ][T k' ][]d' + [T ]T [q'F ]

[]q = T []T [k'][T ][d + ]T T [[q'F ]]= [][]k d + [q F ]

Therefore, []k = [T ][T k' T ][]

[q F ]= [T ]T []q 'F

[]q = T [][]T q '

[d'][][]= T d

[k][][][]= T T k' T

127 Stiffness matrix 6 * 3´ 6´ 2 * 5 * 5´ 3 * 1 2´ 4´ 1 * θi 4 * θj 1´ i j i j

λix = cos θi λjx = cos θj

λiy = sin θi λjy = sin θj

[q*] = [T]T[q´]

145623 1* q 1*  λix − λiy 0 0 0 0 q 1'    q  2* λ λ 0 0 0 0 q   2*   iy ix   2' 

q 3*  3*  0 0 1 0 0 0 q 3'    =     4* 0 0 0 0 q 4*   λ jx − λ jy  q 4'  q  5*  0 0 0 λ λ 0 q   5*   jy jx   5'  q 6*  6*  0 0 0 0 0 1 q 6'  [ T ]T 128 145623

1  λix λiy 0 0 0 0 2 − λ λ 0 0 0 0  iy ix  3  0 0 1 0 0 0 []T =   4  0 0 0 λ jx λ jy 0 5  0 0 0 − λ λ 0  jy jx  6  0 0 0 0 0 1

123456 1  AE/L 0 0 − AE/L 0 0   3 2 3 2  2  0 12EI/L 6EI/L 0 −12EI/L 6EI/L  3  0 6EI/L2 4EI/L 0 − 6EI/L2 2EI/L  []k ' =   4 − AE/L 0 0 AE/L 0 0  5  0 −12EI/L3 − 6EI/L2 0 12EI/L3 − 6EI/L2   2 2  6  0 6EI/L 2EI/L 0 − 6EI/L 4EI/L  129 [ k ] = [ T ]T[ k´ ][T] =

Ui Vi Mi Uj Vj Mj

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - - - U ( λix + λiy ) ( )λixλiy λiy -( λixλjx + λiyλjy) -( λixλjy - λiyλjx) λiy i L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 λ λ V ( )λixλiy ( λiy + λix ) λix -( λiyλjx - λixλjy) -( λiyλjy + ix jx) λix i L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 4EI 6EI 6EI 2EI M - λiy λix λjy - λjx i L2 L2 L L2 L2 L

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - U -( λixλjx + λiy λjy) -( λiyλjx - λixλjy) λjy ( λjx + λjy ) ( )λjxλjy λjy j L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 - V - ( λixλjy- λiyλjx ) -( λiyλjy + λix λjx ) - λjx ( )λjxλjy ( λjy + λjx ) λjx j L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 2EI 6EI 6EI 4EI M - λiy λix λjy - λjx j L2 L2 L L2 L2 L

130 Example 12

For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative bending moment diagrams and qualitative deflected shape. Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members. Include axial deformation in the stiffness matrix.

40 kN

4 m 4 m 22.02 o

131 40 kN

4 m 4 m 22.02o

6 3´ 6´

5 2 * 2´ 5´ Global 1 3 * Local 1 x´ 1 * i 4 j 22.02 o i 1´ j 4´ * λ = cos 0o = 1, x ix o λ = cos 90o = 0 λjx = cos 22.02 = 0.9271, iy o λjy = cos 67.98 = 0.3749

40 kN 40 kN•m 40 kN•m [ q´F ] [FEF] 20sin22.02=7.5 20cos22.02=18.54 20 kN 20 kN 132 • Transformation matrix 6 3´ 6´

5 2 * 2´ λ 5´ Global 1 3 * Local 1 x´ λ 1 * λ 4 i 1´ λ j 4´ o * λix = cos 0 = 1, x  ix − iy 0 0 0 0 o λ = cos 22.02o = 0.9271,   λiy = cos 90 = 0 jx o iy ix 0 0 0 0 λjy = cos 67.98 = 0.3749   T  0 0 1 0 0 0 []T =  λ  0 0 0 −λ 0  λ jx jy   0 0 0 λ 0 Member 1: [ q ] = [ T ]T[q´]  jy jx   0 0 0 0 0 1 1´ 2´ 3´ 4´ 5´ 6´

q 4  4 1 0 0 0 0 0 q 1'      q 0 1 0 0 0 0 q  5  5    2' 

q 6  6 0 0 1 0 0 0 q 3'    =   *   q 1*  1 0 0 0 0.9271 − 0.3749 0 q 4'  q  2* 0 0 0 0.3749 0.9271 0 q   2*     5'  * q 3*  3 0 0 0 0 0 1 q 6'  133 3´ 6´ • Local stiffness matrix 2´ 5´ Local 1

i 1´ j 4´

δi ∆i θi δj ∆j θj

1´ 2´ 3´ 4´ 5´ 6´ 1´ 150.0 0.000 0.000 -150.0 0.000 0.000 2´ 0.000 0.9375 3.750 0.000 -0.9375 3.750 3´ 0.000 3.750 20.00 0.000 -3.750 10.00 [k´] 103 1 = 4´ -150.0 0.000 0.000 150.0 0.000 0.000 5´ 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 6´ 0.000 3.750 10.00 0.000 -3.750 20.00 134 6 3´ 6´

5 2 * 2´ 5´ Global 1 3 * Local 1 x´ 1 * 4 i 1´ j 4´ Stiffness matrix [k´]: x* 1´ 2´ 3´ 4´ 5´ 6´ 1´ 150.0 0.000 0.000 -150.0 0.000 0.000 2´ 0.000 0.9375 3.750 0.000 -0.9375 3.750 3´ 0.000 3.750 20.00 0.000 -3.750 10.00 [k´] 103 1 = 4´ -150.0 0.000 0.000 150.0 0.000 0.000 5´ 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 6´ 0.000 3.750 10.00 0.000 -3.750 20.00 Stiffness matrix [k*]: [ k* ] = [ T ]T[ k´ ][T] 4156 * 2* 3* 4 150.0 0.000 0.000 -139.0 -56.25 0.000 5 0.000 0.9375 3.750 0.351 -0.869 3.750 6 0.000 3.750 20.00 1.406 -3.750 10.00 [k*] 103 1 = 1* -139.0 0.351 1.406 129.0 51.82 1.406 2* -56.25 -0.869 -3.750 51.82 21.90 -3.476 3* 0.000 3.750 10.00 1.406 -3.476 20.00 135 6

40 kN 5 2 * 3 * 1 x´ 4 1 * 4 m 4 m 22.02 o x* 40 kN 40 kN•m 40 kN•m [ q´F ] 20sin22.02=7.5 Global Equilibrium: 20cos22.02=18.54 20 kN 20 kN

[Q] = [K][D] + [QF]

1* 3*

* Q1 = 0.0 1 129 1.406 D1* -7.5 3 = 10 * + Q3 = 0.0 3 1.406 20.0 D3* -40

D 36.37x10-6 m 1* = D3* 0.002 rad 136 6

40 kN 5 2 * 1 3 * x´ 4 1 * 40 kN 4 m 4 m 22.02 o 40 kN•m x* 40 kN•m

7.5 Member Force : [q] = [k*][D] + [qF] 20 kN 18.54

4156 * 2* 3* q4 4 150.0 0.000 0.000 -139.0 -56.25 0.000 0 0 -5.06 q5 5 0.000 0.9375 3.750 0.351 -0.869 3.750 0 20 27.5 q6 6 0.000 3.750 20.00 1.406 -3.750 10.00 0 40.0 60 3 = 10 -6 q1* 1* -139.0 0.351 1.406 129.0 51.82 1.406 36.4x10 + -7.54 = 0 q2* 2* -56.25 -0.869 -3.750 51.82 21.90 -3.476 0 18.54 13.48 q3* 2x10-3 3* 0.000 3.750 10.00 1.406 -3.476 20.00 -40.0 0 60 kN•m 40 kN

5.06 kN 27.5 kN 13.48 kN 137 40 kN 40 kN 60.05 kN•m

5.05 kN 4 m 4 m 22.02 o 27.51 kN 13.47 kN

50.04 + -

-60.05

Bending moment diagram (kN•m)

0.002 rad

Deflected shape 138 Example 13

40 kN 6 kN/m

EI, AE 2EI, 2AE 22.02o 8 m 4 m 4 m

139 40 kN 6 kN/m

EI, AE 2EI, 2AE 22.02 o AE (0.006 m2 )(200×106 kN / m2 ) 8 m 4 m 4 m = L (8 m) =150×103 kN • m 6 3 4EI 4(200×106 kN / m2 )(0.0002 m4 ) 5 2 8 * = 9 * L (8 m) 2 7 * = 20×103 kN • m 4 1 1 Global 2EI =10×103 kN • m L 3´ 6´ 3´ 6 2 4 2´ 5´ 6´ 6EI 6(200×10 kN / m )(0.0002 m ) 2´ 5´ 2 = 2 L (8 m) 1´ 1 4´ 1´ 2 4´ = 3.75×103 kN

Local 12EI 12(200×106 kN / m2 )(0.0002 m4 ) = L2 (8 m)3 3 = 0.9375×10 kN / m 140 6 3

5 2 8 * 9 * 2 7 * 4 1 1 Global Local : Member 1 6 3 3´ 6´ 5 2 2´ 5´ [ q ] [ q´]

4 1 1 1´ 1 4´ [q] = [q´] Thus, [k] = [k´] 412356 4 150.0 0.000 0.000 -150.0 0.000 0.000 5 0.000 0.9375 3.750 0.000 -0.9375 3.750 6 0.000 3.750 20.00 0.000 -3.750 10.00 ´ 2x103 [k]1 = [k ]1 = 1 -150.0 0.000 0.000 150.0 0.000 0.000 2 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 3 0.000 3.750 10.00 0.000 -3.750 20.00

141 6 3

5 2 8 * 9 * 1 1 2 7 * 3 4 2 3´ 6´ * 8 * [ q ] 9 * 2´ [ q´ ] 5´ x´ 7 * 1 2 1´ 2 4´ o * λix = cos 0 = 1, x o λ = cos 22.02o = 0.9271, λiy = cos 90 = 0 jx o λjy = cos 67.98 = 0.3749 Member 2: Use transformation matrix, [q*] = [T]T[q´]

1´ 2´ 3´ 4´ 5´ 6´

q1 1 1 0 0 0 0 0 q1´

q2 2 0 1 0 0 0 0 q2´

q3 3 001 0 0 0 q3´ = * q7* 7 0 0 0 0.9271 -0.3749 0 q4´ * q8* 8 0 0 0 0.3749 0.9271 0 q5´ q9* * 9 0 0 0 001 q6´ 142 3 3´ 6´ 8 * 2 [ q* ] 9 * 2´ [ q´ ] 5´ x´ 7 * 1 2 1´ 2 4´ x* Stiffness matrix [k´]: 1´ 2´ 3´ 4´ 5´ 6´ 1´ 150.0 0.000 0.000 -150.0 0.000 0.000 2´ 0.000 0.9375 3.750 0.000 -0.9375 3.750 3´ 0.000 3.750 20.00 0.000 -3.750 10.00 3 [k´]2 = 10 4´ -150.0 0.000 0.000 150.0 0.000 0.000 5´ 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 6´ 0.000 3.750 10.00 0.000 -3.750 20.00 Stiffness matrix [k*]: [k*] = [T]T[k´][T] 1723 * 8* 9* 1 150.0 0.000 0.000 -139.0 -56.25 0.000 2 0.000 0.9375 3.750 0.351 -0.869 3.750 3 0.000 3.750 20.00 1.406 -3.477 10.00 * 3 [k ]2 = 10 7* -139.0 0.351 1.406 129.0 51.82 1.406 8* -56.25 -0.869 -3.477 51.82 21.90 -3.476 9* 0.000 3.750 10.00 1.406 -3.476 20.00 143 6 3

5 2 8 * 9 * 2 7 * 4 1 1 412356 4 150.0 0.000 0.000 -150.0 0.000 0.000 5 0.000 0.9375 3.750 0.000 -0.9375 3.750 6 0.000 3.750 20.00 0.000 -3.750 10.00 3 [k]1= 2x10 1 -150.0 0.000 0.000 150.0 0.000 0.000 2 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 3 0.000 3.750 10.00 0.000 -3.750 20.00

1723 * 8* 9* 1 150.0 0.000 0.000 -139.0 -56.25 0.000 2 0.000 0.9375 3.750 0.351 -0.869 3.750 3 0.000 3.750 20.00 1.406 -3.477 10.00 * 3 [k ]2 = 10 7* -139.0 0.351 1.406 129.0 51.82 1.406 8* -56.25 -0.869 -3.477 51.82 21.90 -3.476 9* 0.000 3.750 10.00 1.406 -3.476 20.00 144 6 3 40 kN * 6 kN/m 9 5 2 8 *

2 4 1 1 7 * 40 kN 6 kN/m 40 kN•m 32 kN•m 40 kN•m 32 kN•m 1 7.5 24 kN 24 kN 20 kN 18.54

Global: 13 7* 9* 0 1 450 0 -139 0 D1 0 0 3 0 60 1.406 10 D3 -32 + 40 = 8 = 103 * 0 7 -139 1.406 1.406 + 129 D7* -7.5 0 9* 0 10 1.406 20 D9* -40

-6 D1 18.15x10 m -6 D3 -509.84x10 rad = -6 D7* 58.73x10 m

D9* 0.00225 rad 145 6 3 6 kN/m 5 2 32 kN•m 32 kN•m 4 1 1 1 24 kN 24 kN Member 1: [ q ] = [k ][d] + [qF] 412356 q4 4 150.0 0.000 0.000 -150.0 0.000 0.000 0 0 q5 5 0.000 0.9375 3.750 0.000 -0.9375 3.750 0 24 q6 6 0.000 3.750 20.00 0.000 -3.750 10.00 0 32 = 2x103 -6 + q1 1 -150.0 0.000 0.000 150.0 0.000 0.000 d1=18.15x10 0 q2 2 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 0 24 q3 3 0.000 3.750 10.00 0.000 -3.750 20.00 d3=-509.84 -32 x10-6 q4 -5.45 kN q 20.18 kN 6 kN/m 5 21.80 kN•m 52.39 kN•m q 21.80 kN•m 6 = 5.45 kN 5.45 kN 5.45 kN q1 27.82 kN q2 20.18 kN 27.82 kN q3 -52.39 kN•m 146 40 kN 2 3 8 * 40 kN•m 9 * 40 kN•m [ q´F ] x´ 7 * 20sin22.02=7.5 1 2 2 * 20cos22.02=18.54 20 kN x 20 kN Member 2: [ q ] = [k ][d] + [qF] 1723 * 8* 9* -6 q1 1 150.0 0.000 0.000 -139.0 -56.25 0.000 18.15x10 0 q2 2 0.000 0.9375 3.750 0.351 -0.869 3.750 0 20 -6 q3 3 0.000 3.750 20.00 1.406 -3.750 10.00 -509.84x10 40 = 103 -6 + q7* 7* -139.0 0.351 1.406 129.0 51.82 1.406 58.73x10 -7.5 q8* 8* -56.25 -0.869 -3.750 51.82 21.90 -3.476 0 18.54 q9* 9* 0.000 3.750 10.00 1.406 -3.476 20.00 0.00225 -40

q1 -5.45 kN 40 kN 52.39 kN•m q2 26.55 kN

q3 52.39 kN•m = 5.45 kN q7* 0kN q 14.51 kN 14.51 kN 8* 26.55 kN q 9* 0 kN•m 147 40 kN 6 kN/m 52.39 kN•m 21.80 kN•m 52.39 kN•m

5.45 kN 5.45 kN 5.45 kN 14.51 kN 27.82 kN 20.18 kN 26.55 kN 14.51cos 22.02o =13.45 kN 40 kN 21.80 kN•m 6 kN/m

5.45 kN 22.02o 20.18 kN 54.37 kN 8 m 4 m 4 m 14.51 kN

20.18 26.55 V (kN) + + x (m) 3.36 m - - -27.82 -13.45 53.81 M 12.14 + - x (m) (kN•m) - -21.8 -52.39 148 FRAME ANALYSIS USING THE STIFFNESS METHOD

! Simple Frames

! Frame-Member Stiffness Matrix

! Displacement and Force Transformation Matrices

! Frame-Member Global Stiffness Matrix

! Special Frames

! Frame-Member Global Stiffness Matrix

1 Simple Frames

2 Frame-Member Stiffness Matrix 1 AE/L ´ = d 1 y´ 6EI/L2 6´ x´ AE/L 5´ ´ 1 4 2 ´ = 6EI/L d 2 j 3 3´ 12EI/L m 2´ 12EI/L3 1´ 2EI/L i 1 ´ = 4EI/L d 3 2 [k´] 6EI/L 6EI/L2 AE/L 1´ 2´ 3´ 4´ 5´ 6´ 1 ´ = d 4 1´ AE/L 0-000 AE/L 6EI/L2

2´ 0 12EI/L3 6EI/L2 0 - 12EI/L3 6EI/L2 AE/L 1 2 ´ = 6EI/L d 5 3´ 0 6EI/L2 4EI/L 0 - 6EI/L2 2EI/L 12EI/L3 3 4´ -AE/L 0 0 AE/L 0 0 12EI/L 4EI/L 5´ 0 -12EI/L3 -6EI/L2 0 12EI/L3 -6EI/L2 1 ´ = 2EI/L d 6 6´ 0 6EI/L2 2EI/L 0 - 6EI/L2 4EI/L 6EI/L2 6EI/L2 3 Displacement and Force Transformation Matrices

x´ y´ 5´ 6´ 4´ j

2´ m

3´ 1´ i

y 6

5 4 j 3 θy m θx 2 1 i x 4 Force Transformation λ x λy 6´ x´ q4 = q4´ cos θx - q5´ cos θy q = q cos θ + q cos θ θy 5 4´ y 5´ x 5´ 4´ y´ q6 = q6´ j θx θy 2´ q 4  λx − λy 0 q 4'  θx       x − x q = λ λ 0 q λ = j i  5   y x   5'  x       3´ 1´ L q 6   0 0 1 q 6'  i y − y λ = j i y L q 1  λx − λy 0 0 0 0 q 1'      q λ λ 0 0 0 0 q y  2   y x   2'  6 q 3   0 0 1 0 0 0 q 3'    =     q 0 0 0 λ − λ 0 q 5  4   x y   4'        4 q 5 0 0 0 λy λx 0 q 5' j       q 6   0 0 0 0 0 1 q 6'  3 m

2 T 1 []q = T [][q '] i x 5 [q] = [T]T[q´] = [T]T ( [k´][d´] + [q´F] ) = [T]T [k´][d´] + [T]T [q´F] [q] = [T]T [k´][T][d] + [T]T [q´F] = [k][d] + [qF]

Therefore, [k] = [T]T [k´][T]

[qF] = [T]T [q´F]

[q] = [T]T[q´]

[d´] = [T][d]

[k] = [T]T [k´][T]

6 Frame Member Global Stiffness Matrix

[q] = [T]T[q´]= [T]T ( [k´][d´] + [q´F] ) = [T]T[k´][d´] + [T]T[q´F] = [T]T [k´][T][d] + [T]T [q´F]

[k] [qF] [ k ] = [ T ]T[ k´ ][T] =

Ui Vi Mi Uj Vj Mj

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - - - U ( λix + λiy ) ( )λixλiy λiy -( λixλjx + λiyλjy) -( λixλjy - λiyλjx) λiy i L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 λ λ V ( )λixλiy ( λiy + λix ) λix -( λiyλjx - λixλjy) -( λiyλjy + ix jx) λix i L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 4EI 6EI 6EI 2EI M - λiy λix λjy - λjx i L2 L2 L L2 L2 L

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - U -( λixλjx + λiy λjy) -( λiyλjx - λixλjy) λjy ( λjx + λjy ) ( )λjxλjy λjy j L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 - V - ( λixλjy- λiyλjx ) -( λiyλjy + λix λjx ) - λjx ( )λjxλjy ( λjy + λjx ) λjx j L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 2EI 6EI 6EI 4EI M - λiy λix λjy - λjx j L2 L2 L L2 L2 L

7 Example 1

For the frame shown, use the stiffness method to: (a) Determine the deflection and rotation at B. (b) Determine all the reactions at supports. (c) Draw the quantitative shear and bending moment diagrams. E = 200 GPa, I = 60(106) mm4, A = 600 mm2

6 m A 5 kN B

6 m

8 6 m 5 kN A kN B (600 ×10 −6 m2 )(200 ×10 6 ) AE 2 = m = 20000 kN/m L 6m 6 m kN 12(200 ×10 6 )(60 ×10 −6 m 4 ) 12EI 2 = m = 666.667 kN/m C L3 (6m) 3 kN Global : 6(200 ×10 6 )(60 ×10 −6 m 4 ) 6EI m2 2 = 2 = 2000 kN 6 5 2 3 L (6m) A 1 kN 4 1 4(200 ×10 6 )(60 ×10 −6 m 4 ) B 4EI 2 = m = 8000 kN• m L 6m 2 kN 2(200 ×10 6 )(60 ×10 −6 m 4 ) 2EI 2 = m = 4000 kN• m 8 9 C L 6m 7

9 Using Transformation Matrix: Global : Local :

2´ 3´ 5´ 6´ 6 5 2 3 4´ A 1 1 1 A 1´ B 4 B 1´ 3´ 2

2 2´ 8 9 C 7 6´ 5´ • Member Stiffness Matrix 4´ δi ∆i θi δj ∆j θj

Ni  AE/L 0 0 − AE/L 0 0  V  3 2 3 2  i  0 12EI/L 6EI/L 0 −12EI/L 6EI/L  2 2 Mi  0 6EI/L 4EI/L 0 − 6EI/L 2EI/L  []k' =   AE/L 0 0 AE/L 0 0 Nj −  V  0 −12EI/L3 − 6EI/L2 0 12EI/L3 − 6EI/L2  j    0 6EI/L2 2EI/L 0 − 6EI/L2 4EI/L  Mj   10 Stiffness Matrix: Member 1 Local : 3´ 6´ 6 5 2 3 2´ 5´ A 1 4´ 1 1´ 1 4 B A B 2 Global: [q] = [q´]

8 9 -> [k] = [k´] C 1 1 7

465123 4 20000 0 0 -20000 0 0 6 0 666.667 2000 0 -666.667 2000 5 0 2000 8000 0 -2000 4000 [k]1 = 1 -20000 0 0 20000 0 0 2 0 -666.667 -2000 0 666.667 -2000 3 0 2000 4000 0 -2000 8000 11 Stiffness Matrix: Member 2

6 5 2 3 1´ 3´ A 1 λ = cos (-90o) = 0 1 ix 4 B λ = sin (-90o) = -1 2 2´ iy 90o 2 Global: Local: 6´ o 9 λjx = cos (-90 ) = 0 8 5´ o C λjy = sin (-90 ) = -1 7 T 4´ [q]2 = [ T ] [ q´]2 1´ 2´ 3´ 4´ 5´ 6´

q1 1 0 1 0 0 0 0 q1´ q2 2 -1 0 0 0 0 0 q2´ q 3 0 0 1 0 0 0 q 3 = 3´ q7 7 0 0 0 0 1 0 q4´

q8 8 0 0 0 -1 0 0 q5´ q9 9 0 0 0 0 0 1 q6´ [T]T 12 1´ 2´ 3´ 4´ 5´ 6´ 1´ 20000 0 0 -20000 0 0 2´ 0 666.667 2000 0 -666.667 2000 3´ 0 2000 8000 0 -2000 4000 [k´] = 2 4´ -20000 0 0 20000 0 0 5´ 0 -666.667 -2000 0 666.667 -2000 6´ 0 2000 4000 0 -2000 8000

T [k]2 = [ T ] [ k´ ]2[ T ] 123789 1 666.6670 2000 -666.667 0 2000 2 0 20000 0 0 -20000 0 3 2000 0 8000 -2000 0 4000 [k] = 2 7 -666.667 0 -2000 666.667 0 -2000 8 0 -20000 0 0 20000 0 9 2000 0 4000 -2000 0 8000 13 Global Stiffness Matrix: [k]1 465123 6 5 2 3 4 20000 0 0 -20000 0 0 A 1 1 4 B 6 0 666.667 2000 0 -666.667 2000

2 5 0 2000 8000 0 -2000 4000 Global: 1 -20000 0 0 20000 0 0 8 9 2 0 -666.667 -2000 0 666.667 -2000 C 7 3 0 2000 4000 0 -2000 8000

[k]2 [K] 123789 45 1 2 3 1 666.6670 2000 666.667 0 2000 4 20000 0 -20000 0 0 2 0 20000 0 0 -20000 0 5 0 8000 0 -2000 4000 3 2000 0 8000 -2000 0 4000 1 -20000 0 20666.667 0 2000 7 -666.667 0 -2000 666.667 0 -2000 2 0 -2000 0 20666.667 -2000 8 0 -20000 0 0 20000 0 3 0 4000 2000 -2000 16000 9 2000 0 4000 2000 0 8000 14 6 m 6 5 2 3 5 kN A A 1 1 1 B 4 B 2 2 6 m

8 9 C [Q] = [K][D] + [QF] C Global: 7 45123 4 20000 0 -20000 0 0 D Q4 = 0 4 0 0 8000 0 -2000 4000 D Q5 = 0 5 5 0 -20000 0 20666.667 0 2000 D 0 Q1 = 5 = 1 1 + 0 -2000 0 20666.667 -2000 D Q2 = 0 2 2 0 0 4000 D Q3 = 0 3 2000 -2000 16000 3 0

D4 0.01316 m -4 D5 9.199(10 ) rad

D1 = 0.01316 m -5 D2 -9.355(10 ) m -3 D3 -1.887(10 ) rad 15 6 m Member 1 5 kN A 5 3 1 6 2 B 1 4 1 2 A B 6 m 1.87 kN 1.87 kN 11.22 kN•m C A 1 B

F [q]1 = [k]1[d]1 + [q ]1 465123 q D = 0.01316 0 4 4 20000 0 0 -20000 0 0 4

q6 6 0 666.667 2000 0 -666.667 2000 D6 = 0 -1.87

-4 q5 5 0 2000 8000 0 -2000 4000 D5 = 9.199(10 ) 0

q1 1 -20000 0 0 20000 0 0 D1 = 0.01316 0

-5 q2 2 0 -666.667 -2000 0 666.667 -2000 D2 = -9.355(10 ) 1.87

-3 q3 3 0 2000 4000 0 -2000 8000 D3 = -1.887(10 ) -11.22 16 6 m Member 2 5 kN A 1 3 11.22 kN•m B 2 1.87 kN

2 1 5 kN 2 6 m 2

8 9 18.77 kN•m C 7 5 kN F 1.87 kN [q]2 = [k]2[d]2 + [q ]2 123789 1 q1 666.6670 2000 -666.667 0 2000 D1 = 0.01316 5

-5 q2 2 0 20000 0 0 -20000 0 D2 = -9.355(10 ) -1.87

-3 q3 3 2000 0 8000 -2000 0 4000 D3 = -1.887(10 ) 11.22

q7 7 -666.667 0 -2000 666.667 0 -2000 D7 = 0 -5

q8 8 0 -20000 0 0 20000 0 D8 = 0 1.87

q9 9 2000 0 4000 -2000 0 8000 D9 = 0 18.77 17 1.87 kN 1.87 kN 11.22 kN•m 5 kN A A 1 B B 1.87 kN 1.87 kN 11.22 kN•m 6 m 5 kN 2 18.77 kN•m C 18.77 kN•m 5 kN 6 m 5 kN 1.87 kN

1.87 kN

11.22 B B A - A - 5 -1.87 -11.22

Bending moment Shear diagram + diagram + C C 5 18.77 18 D4=13.16 mm D1=13.16 mm D =0.00092 rad D =-0.00189 rad 5 3 11.22 B A A B - -11.22 D3=-0.00189 rad

Deflected shape Bending moment diagram + C C 18.77

6 5 2 3 A 1 D 0.01316 m 1 4 4 B -4 D5 9.199(10 ) rad 2 D 0.01316 m 1 = Global : -5 D2 -9.355(10 ) m -3 8 9 D3 -1.887(10 ) rad C 7 19 Example 2

For the beam shown, use the stiffness method to: (a) Determine the deflection and rotation at B (b) Determine all the reactions at supports (c) Draw the quantitative shear and bending moment diagrams. E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.

3 kN/m

B C

4.5 m A

6 m 6 m

20 3 kN/m 2 3 8 9 Global 1 2 7 4.5 m 6 5 1

4 6 m 6 m

2 3 8 9 2 3 Members 1 2 7 1

6 1 5

4 3 kN/m 9 kN•m 9 kN•m [FEM] 2 9 kN 9 kN

21 3 kN/m AE (600×10−6 m2 )(200×106 kN / m2 ) = m L 7.5 m 5 4.5 m 7. =16000 kN / m

12EI 12(200×106 kN / m2 )(60×10−6 m4 ) = 6 m 6 m L3 (7.5 m)3 = 341.33 kN / m

Member 1: 6EI 6(200×106 kN / m2 )(60×10−6 m4 ) = L2 (7.5 m)2 2 3 =1280 kN

θ 1 4EI 4(200×106 kN / m2 )(60×10−6 m4 ) y = 6 1 L 7.5 m θx 5 = 6400 kN • m 4 2EI 2(200×106 kN / m2 )(60×10−6 m4 ) = λx = cos θx = 6/7.5 = 0.8 L 7.5 m = 3200 kN • m λy = cos θy = 4.5/7.5 = 0.6 22 Member m: M V j λx = cos θx j

λy = cos θy θy Uj M m T i [ km ] = [ T ] [ k´ ][T ] = θx Vi

U V M U V M i i i j Ui j j

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - - - U ( λix + λiy ) ( )λixλiy λiy -( λixλjx + λiyλjy) -( λixλjy - λiyλjx) λiy i L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 λ λ V ( )λixλiy ( λiy + λix ) λix -( λiyλjx - λixλjy) -( λiyλjy + ix jx) λix i L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 4EI 6EI 6EI 2EI M - λiy λix λjy - λjx i L2 L2 L L2 L2 L

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - U -( λixλjx + λiy λjy) -( λiyλjx - λixλjy) λjy ( λjx + λjy ) ( )λjxλjy λjy j L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 - V - ( λixλjy- λiyλjx ) -( λiyλjy + λix λjx ) - λjx ( )λjxλjy ( λjy + λjx ) λjx j L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 2EI 6EI 6EI 4EI M - λiy λix λjy - λjx j L2 L2 L L2 L2 L

23 3 kN/m Member 1: 2 3

m θ 1 5 y 4.5 m 7. 6 1 θ 5 x

6 m 6 m 4

λx = cos θx = 6/7.5 = 0.8

λy = cos θy = 4.5/7.5 = 0.6

4 5 6 1 2 3 4 10362.879 7516.162 -768 -10362.879 -7516.162 -768 5 7516.162 5978.451 1024 -7516.162 -5978.451 1024 6 -768 1024 6400 768 -1024 3200 [k ] = 1 1 -10362.879 -7516.162 768 10362.879 7516.162 768 2 -7516.162 -5978.451 -1024 7516.162 5978.451 -1024 3 -768 1024 3200 768 -1024 6400 24 3 kN/m AE (600×10−6 m2 )(200×106 kN / m2 ) = m L 6 m 5 4.5 m 7. = 20000 kN / m

12EI 12(200×106 kN / m2 )(60×10−6 m4 ) = 6 m 6 m L3 (6 m)3 = 666.667 kN / m

Member 2: 6EI 6(200×106 kN / m2 )(60×10−6 m4 ) = L2 (6 m)2 3 9 2 8 = 2000 kN 2 1 7 4EI 4(200×106 kN / m2 )(60×10−6 m4 ) = o o L 6 m λx = cos 0 = 1.0, λy = cos 90 = 0 = 8000 kN • m

2EI 2(200×106 kN / m2 )(60×10−6 m4 ) = L 6 m = 4000 kN • m

25 1 27398 2 3 8 9 1 AE/L 0-0 AE/L 0 0 2 1 7 3 2 3 2 2 0 12EI/L 6EI/L 0 - 12EI/L 6EI/L 2 2 3 0 6EI/L 4EI/L 0 - 6EI/L 2EI/L [k2] = 7 -AE/L 0 0 AE/L 0 0 3 2 3 2 8 0 -12EI/L -6EI/L 0 12EI/L -6EI/L 2 2 9 0 6EI/L 2EI/L 0 - 6EI/L 4EI/L

1 27398

1 20000 00 - 20000 0 0 2 0 666.667 2000 0 - 666.667 2000 3 0 2000 8000 0 - 2000 4000 [k2] = 7 -20000 0 0 20000 0 0 8 0 -666.667 -2000 0 666.667 -2000

9 0 2000 4000 0 - 2000 8000 26 4 5 6 1 2 3 4 10362.879 7516.162 -768 -10362.879 -7516.162 -768 5 7516.162 5978.451 1024 -7516.162 -5978.451 1024 6 -768 1024 6400 768 -1024 3200 [k ] = 1 1 -10362.879 -7516.162 768 10362.879 7516.162 768 2 -7516.162 -5978.451 -1024 7516.162 5978.451 -1024 3 -768 1024 3200 768 -1024 6400

1 27398

1 20000 00 - 20000 0 0 2 0 666.667 2000 0 - 666.667 2000 3 0 2000 8000 0 - 2000 4000 [k2] = 7 -20000 0 0 20000 0 0 8 0 -666.667 -2000 0 666.667 -2000

9 0 2000 4000 0 - 2000 8000 27 3 kN/m 2 3 8 9 9 kN•m m 1 2 7 5 4.5 m 7. 9 kN 9 kN 1 5 6

6 m 6 m 4

Global: 1 2 3 0 Q 1 30362.9 7516.16 768 D 0 10 1 Q 2 7516.16 6645.12 976 D 9 20 = 2 + Q3 3 768 976 14400 D3 9

-4 D1 4.575(10 ) m -3 D2 = -1.794(10 ) m -4 D3 -5.278(10 ) rad 28 Member 1: 1.19 kN•m 11.37 kN 2 3 6.75 kN 1.19 kN•m 9.15 kN θy 1 1 0.50 kN•m 0.09 kN 6 1 1 θx 0.50 kN•m 5 6.75 kN

4 9.15 kN 0.09 kN 11.37 kN

λx = cos θx = 6/7.5 = 0.8

λy = cos θy = 4.5/7.5 = 0.6 465123 q 0 0 9.15 4 4 q 0 0 6.75 5 5 q6 0 0 0.50 = 6 k + 1 -4 = q1 D1 = 4.575(10 ) 0 -9.15 1 -3 q2 D2 = -1.794(10 ) 0 -6.75 2 q3 -4 D3 = -5.278(10 ) 0 -1.19 3 29 Member 2:

2 3 8 9 1.19 kN•m 3 kN/m 14.70 kN•m 1 2 7 9.15 kN 2 9.15 kN 3 kN/m 6.75 kN 11.25 kN 9 kN•m 9 kN•m 2 9 kN [FEM] 9 kN

132789 -4 q1 1 D1 = 4.575(10 ) 0 9.15 q -3 2 2 D2 = -1.794(10 ) 9 6.75 q -4 9 1.19 3 3 k D3 = -5.278(10 ) = 2 + q7 7 0 0 = -9.15

q8 8 0 9 11.25 q 9 9 0 -9 -14.70

30 1.19 kN•m 3 kN/m 14.70 kN•m 6.75 kN 1.19 kN•m 9.15 kN 9.15 kN 2 9.15 kN 0.50 kN•m 6.75 kN 11.25 kN 1 6.75 kN

9.15 kN

3 kN/m 14.70 kN•m

9.15 kN 0.50 kN•m 11.25 kN All Reactions 9.15 kN

6.75 kN

31 11.37 kN 1.19 kN•m 3 kN/m 14.70 kN•m 1.19 kN•m 9.15 kN 2 9.15 kN 0.09 kN 6.75 kN 11.25 kN 1 6.75 0.50 kN•m 0.09 kN + 11.37 kN - -4 -0.09 D1 4.575(10 ) m -3 D2 = -1.794(10 ) m -11.25 Shear diagram -4 D3 -5.278(10 ) rad -0.09 (kN)

D1 = 0.46 mm

D2 = -1.79 mm -1.19 -4 D3 =-5.278(10 ) rad -14.70 Deflected shape Bending-moment diagram 0.5 (kN•m) 32 Example 3

20 kN•m 10 kN 15 kN C B /m 4.5 m kN 3

A 6 m3 m 3 m

33 20 kN•m 10 kN 3 9 Global 2 8 15 kN C 1 2 7 B θy = 53.13 /m 1 N 6 4.5 m k θ =36.87 3 5 x λx = cos θx = 6/7.5 = 0.8 4 A 6 m3 m 3 m λy = cos θy = 4.5/7.5 = 0.6

2 3 8 9 2 3 Members 1 2 7 1 6 1 5 14.06 kN•m 10 kN 4 7.5 kN•m 7.5 kN•m 6.75 kN 2 /m 11.25 kN [FEM] kN 3 1 9 kN 5 kN 5 kN wL2/12 = 14.06 kN•m 11.25(0.6) = 6.75 kN 11.25(0.8) = 9 kN wL/2 = 11.25 kN 34 20 kN•m 10 kN 2 3 8 9 15 kN C B 1 2 7 /m N k 3 1 5 6

A 4 14.06 kN•m 10 kN Global: 7.5 kN•m 7.5 kN•m 6.75 kN 2 /m 11.25 kN kN 3 1 9 kN 5 kN 5 kN [FEM] 14.06 kN•m 11.25(0.6) = 6.75 kN

9 kN 11.25 kN

1 2 3

Q1 = 15 1 30362.9 7516.16 768 D1 -6.75

Q2 = 0 = 2 7516.16 6645.2 976 D2 + 9 + 5 Q = 20 3 768 976 14400 D -14.06 + 7.5 3 3 35 2 3 8 9

1 2 7

1 5 6

-3 D1 1.751(10 ) m -3 D2 = -4.388(10 ) m -3 D3 2.049(10 ) rad

36 Member 1: 14.06 kN•m 2 3 [FEM] 6.75 kN θy 1 /m 11.25 kN 1 kN 6 θ 3 1 9 kN 5 x 14.06 kN•m 11.25(0.6) = 6.75 kN 4 9 kN o 11.25 kN λx = cos 36.87 = 0.8, o λy = cos 53.13 = 0.6

465123 q 0 -6.75 6.51 4 4 q 0 9 24.17 5 5 q 0 14.06 26.46 6 6 = k1 + -3 = q1 1 D1 = 1.751(10 ) -6.75 -20.01 -3 q2 2 D2 = -4.388(10 ) 9 -6.17 q 3 D = 2.049(10-3) -14.06 4.89 3 3 37 2 3

o q4 6.51 θy = 53.13 1 1 q5 24.17 6 o θx = 36.87 5 q6 26.46 = 4 q1 -20.01

q2 -6.17 q 3 4.89

4.89 kN•m 19.71 kN 6.17 kN 4.89 kN•m 20.01 kN /m N 7.07 kN /m k N 3 k 1 3 m m 5 .5 26.46 kN•m 7. 26.46 kN•m 7 6.51 kN 15.43 kN 19.71 kN 24.17 kN 38 Member 2:

2 3 8 9 15.12 kN•m 10 kN 8.08 kN•m 1 2 7 10 kN 35.02 kN 2 35.02 kN 7.5 kN•m 7.5 kN•m 6.17 kN 3.83 kN 2 5 kN 5 kN [FEM]

132789 -3 q1 1 D1 = 1.751(10 ) 0 35.02 q -3 2 2 D2 = -4.388(10 ) 5 6.17 q -3 7.5 15.12 3 3 k D3 = 2.049(10 ) = 2 + q7 7 0 0 = -35.02

q8 8 0 5 3.83 q 9 9 0 -7.5 -8.08

39 6.17 kN 4.89 kN•m 15.12 kN•m 10 kN 8.08 kN•m 20.01 kN 35.02 kN 2 35.02 kN /m N k 6.17 kN 3.83 kN 3 m 5 26.46 kN•m 7. 6.51 kN 24.17 kN

20 kN•m 10 kN 8.08 kN•m 15 kN C B 35.02 kN /m kN 3.83 kN 3 26.46 kN•m

A 6.51 kN 24.17 kN 6 m3 m 3 m

40 4.89 kN•m 19.71 kN 15.12 kN•m 10 kN 8.08 kN•m

35.02 kN 2 35.02 kN /m N k 7.07 kN 3 6.17 kN 3.83 kN 1m 5 26.46 kN•m 7. 6.17 15.43 kN 19.71 kN -7.07 -3.83 -3 D1 1.751(10 ) m Shear diagram D -4.388(10-3) m 15.43 2 = (kN) -3 D3 2.049(10 ) rad

D1 =1.75 mm 4.89 D2 = -4.39 mm

-3 -15.12 -8.08 D3 = 2.05(10 ) rad Deflected shape Bending-moment diagram (kN•m)

-26.46 41 Special Frames

42 Stiffness matrix

6 * 3´ 6´ 2 * 5 * 5´ 3 * 1 2´ 4´ 1´ 1 * θi 4 * θj i j i j

λix = cos θi λjx = cos θj

λiy = sin θi λjy = sin θj

[ q* ] = [ T ]T[ q´ ]

1´ 2´ 3´ 4´ 5´ 6´

q 1*  1*λix − λiy 0 0 0 0 q 1'      q λ λ 0 0 0 0 q  2*  2* iy ix   2' 

q 3*  3* 0 0 1 0 0 0 q 3'    =     q 4*  4* 0 0 0 λ jx − λ jy 0 q 4'  q  5* 0 0 0 λ λ 0 q   5*   jy jx   5'  q 6*  6* 0 0 0 0 0 1 q 6'  [ T ]T 43 1* 2* 3* 4* 5* 6*

1´  λix λiy 0 0 0 0 − λ λ 0 0 0 0 2´  iy ix  3´  0 0 1 0 0 0 []T =   4´  0 0 0 λ jx λ jy 0 5´  0 0 0 − λ λ 0  jy jx  6´  0 0 0 0 0 1

• Member Stiffness Matrix

1´ 2´ 3´ 4´ 5´ 6´ 1´  AE/L 0 0 − AE/L 0 0   3 2 3 2  2´  0 12EI/L 6EI/L 0 −12EI/L 6EI/L  3´  0 6EI/L2 4EI/L 0 − 6EI/L2 2EI/L  []k' =   4´ − AE/L 0 0 AE/L 0 0  5´  0 −12EI/L3 − 6EI/L2 0 12EI/L3 − 6EI/L2   2 2  6´  0 6EI/L 2EI/L 0 − 6EI/L 4EI/L  44 [ k ] = [ T ]T[ k´ ][T] =

Ui Vi Mi Uj Vj Mj

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - - - U ( λix + λiy ) ( )λixλiy λiy -( λixλjx + λiyλjy) -( λixλjy - λiyλjx) λiy i L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 λ λ V ( )λixλiy ( λiy + λix ) λix -( λiyλjx - λixλjy) -( λiyλjy + ix jx) λix i L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 4EI 6EI 6EI 2EI M - λiy λix λjy - λjx i L2 L2 L L2 L2 L

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI 2 2 - U -( λixλjx + λiy λjy) -( λiyλjx - λixλjy) λjy ( λjx + λjy ) ( )λjxλjy λjy j L L3 L L3 L2 L L3 L L3 L2

AE 12EI AE 12EI 6EI AE 12EI AE 12EI 6EI - 2 2 - V - ( λixλjy- λiyλjx ) -( λiyλjy + λix λjx ) - λjx ( )λjxλjy ( λjy + λjx ) λjx j L L3 L L3 L2 L L3 L L3 L2

6EI 6EI 2EI 6EI 6EI 4EI M - λiy λix λjy - λjx j L2 L2 L L2 L2 L

45 Example 4

For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative bending moment diagrams and qualitative deflected shape. Take I = 200(106) mm4 , A = 6(103) mm2, and E = 200 GPa for all members. Include axial deformation in the stiffness matrix.

40 kN 20 kN 200 kN•m 6 k N/m 3 m 22.02 o

4 m 4 m 7.416 m

46 40 kN 20 kN 200 kN•m 6 k N/m 3 m 22.02 o

4 m 4 m 7.416 m

3 * 6 2 * 5 Global 1 4 1 * 9 22.02 o 2 8 7

3´ 6´ 5´ 3´ 2´ 2´ Local 4´ 1´ 1 1´ 6´ 5´ 2

4´

47 δi ∆i θi δj ∆j θj

Ni  AE/L 0 0 − AE/L 0 0  V  3 2 3 2  i  0 12EI/L 6EI/L 0 −12EI/L 6EI/L  2 2 Mi  0 6EI/L 4EI/L 0 − 6EI/L 2EI/L  []k' =   Nj − AE/L 0 0 AE/L 0 0  V  0 −12EI/L3 − 6EI/L2 0 12EI/L3 − 6EI/L2  j    0 6EI/L2 2EI/L 0 − 6EI/L2 4EI/L  Mj  

AE (0.006 m2 )(200×106 kN / m2 ) = =150×103 kN / m L 8 m 4EI 4(200×106 kN / m2 )(0.0002 m4 ) = = 20×103 kN • m L 8 m 2EI =10×103 kN • m L 6EI 6(200×106 kN / m2 )(0.0002 m4 ) = = 3.75×103 kN L2 (8 m)2 12EI 12(200×106 kN / m2 )(0.0002 m4 ) = = 0.9375×103 kN / m L3 (8 m)3 48 3´ 6´ 5´ 3´ 2´ 8 m 2´ 4´ 1´ 1 1´ 8 m 6´ 5´ 2 Local 4´

δi ∆i θi δj ∆j θj

Ni 150000 00 - 150000 0 0

Vi 0 937.5 3750 0 - 937.5 3750

Mi 0 3750 20000 0 - 3750 10000 [ k´]1 = [ k´]2 = Nj -150000 0 0 150000 0 0

Vj 0 -937.5 -3750 0 937.5 -3750

Mj 0 3750 10000 0 - 3750 20000

49 Member 1: 3 * 6 3´ 6´ 2 * 5 5´ 2´ Local Global 1 θ =22.02o 4 1 4´ 1 * i o 1´ θj = 0 40 kN o o λix = cos (22.02 ) = 0.927, λjx = cos (0 ) = 1, o o 40 kN•m 40 kN•m λiy = sin (22.02 ) = 0.375 λjy = sin (0 ) = 0

1 [ q* ] = [ T ]T[ q´ ] 20 kN [FEM] 20 kN

1´ 2´ 3´ 4´ 5´ 6´

* 1* 0.927 - 0.375 0 0 0 0 q 1 q1´ * 2* 0.375 0.927 0 0 0 0 q 2 q2´ q* 3* 001 0 0 0 q 3 = 3´ 4* 0 0 0 1 0 0 q4 q4´ 5* 0 0 0 0 1 0 q5 q5´ q6 6* 0 0 0 001 q6´ [ T ] T 1 50 3 * 6 3´ 6´ 2 * 5 5´ 2´ Local Global 1 θ =22.02o 4 1 4´ 1 * i o 1´ θj = 0

* T [ k ]1 = [ T ]1 [ k´ ]1[ T ]1

1* 2* 3* 4 5 6 1* 129.046 51.811 -1.406 -139.058 0.351 -1.406 2* 51.811 21.892 3.476 -56.240 -0.869 3.476 3* -1.406 3.476 20.00 0 -3.75 10.00 [ k* ] = 103 1 4 -139.058 -56.240 0.00 150 0 0 5 0.351 -0.869 -3.75 0 0.938 -3.75 6 -1.406 3.476 10.00 0 -3.75 20

51 40 kN 3 * 6 2 * 5 40 kN•m 40 kN•m Global 1 θ =22.02o 4 1 1 * i o θj = 0 20 kN [FEM] 20 kN

[ qF* ] = [ T ]T[ qF´]

0 -7.50 1* 40 kN 20 * 18.54 2 40 kN•m 40 kN•m 40 * F* T 40 3 [ q ] = [ T ]1 = 7.5 1 0 0 4 18.54 20 kN 20 20 5 -40 -40 6 52 3´ Member 2 2´ 6 λ = λ = cos (-22.02o) = 0.927, 1 5 ix jx ´ λ = λ = sin (-22.02o) = -0.375 6´ 4 iy jy 5´ 22.02o 2 9 [ q´] 2 8 4 32 6 ´ 7 kN kN/ [ q ] 22.02o •m m 24 kN 2 [ q´F] 24 kN 3 2 k [ q ] = [ T ]T[ q´ ] N•m

1´ 2´ 3´ 4´ 5´ 6´ 4 0.927 0.375 0 0 0 0 q4 q1´ 5 -0.375 0.927 0 0 0 0 q5 q2´ q 6 001 0 0 0 q 6 = 3´ 7 0 0 0 0.927 0.375 0 q7 q4´ 8 0 0 0 -0.375 0.927 0 q8 q5´ q9 9 0 0 0 001 q6´ [ T ] T 2 53 6 3´ 5 2´ 4 22.02o 1´ 9 6´ 2 5´ 8 2 7 [ q ] 22.02o [ q´ ] 4´

T [ k ]2 = [ T ]2 [ k´]2[ T ]2 4 5 6 7 8 9 4 129.046 -51.811 1.406 -129.046 51.811 1.406 5 -51.811 21.892 3.476 51.811 -21.892 3.476 6 1.406 3.476 20 -1.406 -3.476 10 [ k ] = 103 2 7 -129.046 51.811 -1.406 129.046 -51.811 -1.406 8 51.811 -21.892 -3.476 -51.811 21.892 -3.476 9 1.4056 3.476 10 -1.406 -3.476 20

54 6 5 3 6 4 2 k kN 22.02o N•m /m 9 24 2 8 kN 2 7 [ q´F ] [ q ] 22.02o 24 kN 3 2 k N•m

[ qF* ] = [ T ]T[ qF´ ]

0 32 kN•m 8.998 4 [ qF ] 24 5 6 k 22.249 8.998kN N/m 32 6 F T 32 22.249 kN 2 32 kN•m [ q ] = [ T ]2 = 0 8.998 7 8.998 kN

24 22.249 8 22.249 kN -32 -32 9 55 3 * 6 Global Stiffness: 2 * 5

1 4 1 * 9 2 8 7 1* 2* 3* 4 5 6 1* 129.046 51.811 -1.406 -139.058 0.351 -1.406 2* 51.811 21.892 3.476 -56.240 -0.869 3.476 3* -1.406 3.476 20.00 0 -3.75 10.00 [ k* ] = 103 1 4 -139.058 -56.240 0.00 150 0 0 5 0.351 -0.869 -3.75 0 0.938 -3.75 6 -1.406 3.476 10.00 0 -3.75 20 4 5 6 7 8 9 4 129.046 -51.811 1.406 -129.046 51.811 1.406 5 -51.811 21.892 3.476 51.811 -21.892 3.476 6 1.406 3.476 20 -1.406 -3.476 10 [ k ] = 103 2 7 -129.046 51.811 -1.406 129.046 -51.811 -1.406 8 51.811 -21.892 -3.476 -51.811 21.892 -3.476

9 1.4056 3.476 10 -1.406 -3.476 20 56 40 kN Global: 20 kN 3 * 6 200 kN•m 2 * 5 6 kN /m 1 4 1 * 9 2 8 40 kN 7 40 kN•m 32 kN•m 40 kN•m [ qF ] 6 8.998 kN kN 7.5 1 /m 18.54 kN [ q*F ] 20 kN 22.249 kN 2 32 kN•m 8.998 kN [ Q ] = [ K ][ D ] + [ QF ] 22.249 kN 1* 3* 4 5 6

Q1* = 0.0 1* 129.046 -1.406 -139.058 0.351 -1.406 D1* -7.5

Q3* = 0.0 3* -1.406 20 0 -3.75 10 D3* 40

Q 3 D 8.998 4 = 0.0 = 10 4 -139.058 0 279.046 -51.811 1.406 4 +

Q5 = -20 5 0.352 -3.75 -51.811 22.829 -0.274 D5 20 +22.249

Q6 = -200 6 -1.406 10 1.406 -0.274 40 D6 -40 + 32 57 40 kN Global: 20 kN 3 * 6 200 kN•m 2 * 5 6 kN /m 1 4 1 * 9 2 8 7

D1* -0.0205 m

D3* -0.0112 rad D -0.0191 m 4 =

D5 -0.0476 m

D6 -0.0024 rad

58 3 * 6 40 kN•m 40 kN 2 * 5 40 kN•m [ q ] 1 4 1 1 * 7.5 18.54 [ qF* ] 20 kN Member 1: [ q*] = [ k*][ d*] + [ qF*] 1* 2* 3* 4 5 6 1* 129.046 51.811 -1.406 -139.058 0.351 -1.406 -7.50 q1* D1*=-0.0205 q 2* 51.811 3.476 -56.240 -0.869 3.476 18.54 2* 21.892 D2*= 0.0 0 q3* 3* -1.406 3.476 20.00 -3.75 10.00 D =-0.0112 40 = 103 3* + 4 -139.058 -56.240 0.00 150 0 0 q4 D4= -0.0191 0 q 5 0.351 -0.869 -3.75 0 0.938 -3.75 5 D5= -0.0476 20 q6 6 -1.406 3.476 10.00 0 -3.75 20 D6=-0.0024 -40 q1* 0 40 kN q2* 22.63 kN 7.87 kN•m q3* 0 = 8.49 kN q4 -8.49 kN 22 .63 19.02 kN q5 19.02 kN kN q6 7.87 kN•m 59 6 32 kN•m 5 [ qF ] 6 4 8.998 kN kN/ 9 m 2 8 22.249 kN 2 32 kN•m [ q ] 7 8.998 kN Member 2 : [ q ] = [ k ][ d ] + [ qF ] 22.249 kN 4 5 6 7 8 9 4 129.046 -51.811 1.406 -129.046 51.811 1.406 D = -0.0191 q4 4 8.998 5 -51.811 21.892 3.476 51.811 -21.892 3.476 D = -0.0476 q5 5 22.249 q 6 1.406 3.476 20 -1.406 -3.476 10 D6=-0.0024 32 6 =103 7 -129.046 51.811 -1.406 129.046 -51.811 -1.406 D = 0 + q7 7 8.998 8 51.811 -21.892 -3.476 -51.811 21.892 -3.476 D = 0 q8 8 22.249 q 9 9 1.4056 3.476 10 -1.406 -3.476 20 D9 = 0 -32 q4 8.49 kN 207.87 kN•m q5 -39.02 kN 6 k 8.49 kN N/m q6 -207.87 kN•m = 248.04 kN•m q7 9.51 kN 39.02 kN q8 83.51 kN 9.51 kN 83.51 kN 60 q9 -248.04 kN•m 207.87 kN•m 40 kN 22 7.87 kN•m .5 k 8.49 kN N 6 8.49 kN 8.49 kN kN /m 248.04 kN•m 2 20.98 kN 33 kN 2.6 19.02 kN 3 kN 39.02 kN 9.51 kN 207.87 22 81 kN .5 k 83.56 N 83.51 kN 7.87 + +

D1* -0.0205 m Bending-moment diagram (kN•m) D3* -0.0112 rad - D -0.0191 m 4 =

D5 -0.0476 m -248.04 D1*=- 20.5 mm D -0.0024 rad D4=-19.1 mm 6

D5=-47.6 mm

D3*=-0.0112 rad D = 6 -0.0 024 rad Deflected shape 61 Example 5

For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative bending moment diagrams and qualitative deflected shape. Take I = 400(106) mm4 , A = 60(103) mm2, and E = 200 GPa for all members.

20 kN/m 80 kN•m 50 kN AB 3 m

C 20o 4 m 2 m

62 20 kN/m 80 kN•m 50 kN AB 3 m

C 76.31o 20o 4 m 2 m

2 2 3 5 6 2´ 3´ 5´ 6´ ´ 1 1 3 ´ 1 4 1´ 4´ 1 ´ 2 2 9* Global Local 5 8* 7* ´ 6 ´ 4 ´

63 20 kN/m 80 kN•m 50 kN AB 2 3 5 6 3 m 1 1 4

C 20o 4 m 2 m

Member 1: 2EI 2(200 ×10 6 kN/m 2 )(400 ×10 −6 m 4 ) = L 4 m AE (60 ×10 −3 m2 )(200 ×10 6 kN/m 2 ) = = 40 ×10 3 kN• m L 4 m 3 6EI 6(200 ×10 6 kN/m 2 )(400 ×10 −6 m 4 ) = 3000 ×10 kN/m = L2 (4 m) 2 4EI 4(200 ×10 6 kN/m 2 )(400 ×10 −6 m 4 ) = 30 ×10 3 kN = L 4 m 12EI 12(200 ×10 6 kN/m 2 )(400 ×10 −6 m 4 ) 3 = 80 ×10 kN• m 3 = 3 L (4 m) = 15 ×10 3 kN/m 64 20 kN/m 80 kN•m 50 kN AB 2 3 5 6 3 m 1 1 4 56.31o

o C 20 76.31o 20 kN/m 4 m 2 m 26.67 kN•m 26.67 kN•m 1 40 kN 40 kN Member 1: [ q ] = [ k ][ d ] + [ qF ]

1 2 3 456 q1 1 3000 0 0 -3000 0 0 d1 0 q2 2 0 15 30 0 -15 30 d2 40 q3 3 0 30 80 0 -30 40 d3 26.67 = 103 + q4 4 -3000 0 0 3000 0 0 d4 0 q5 5 0 -15 -30 0 15 -30 d5 40 q6 6 0 30 40 0 -30 80 d6 -26.67 65 2 ´ 5 6 Member 2: 3 i 4 ´ 1 6 2 −6 4 ´ 6EI 6(200 ×10 kN/m )(400 ×10 m ) 2 = 2 L2 (3.61 m) 2 9* 5 ´ = 36.83 ×10 3 kN * * 8 7 6 ´ 12EI 12(200 ×10 6 kN/m 2 )(400 ×10 −6 m 4 ) j 4 = ´ L3 (3.61 m) 3 −3 2 6 2 AE (60 ×10 m )(200 ×10 kN/m ) 3 = = 20.41 ×10 kN/m L 3.61 m 3 = 3324 ×10 kN/m 2EI 2(200 ×10 6 kN/m 2 )(400 ×10 −6 m 4 ) = 4EI 4(200 ×10 6 kN/m 2 )(400 ×10 −6 m 4 ) L 3.61 m = L 3.61 m = 44.32 ×10 3 kN• m = 88.64 ×10 3 kN• m 1´ 2´ 3´ 4´ 5´ 6´ 1´ 3324 0 0 -3324 0 0 2´ 0 20.41 36.83 0 -20.41 36.83 3´ 0 36.83 88.64 0 -36.83 44.32 [ k´ ] 103 2 = 4´ -3324 0 0 3324 0 0 5´ 0 -20.41 -36.83 0 20.41 -36.83 6´ 0 36.83 44.32 0 -36.83 88.64 66 6 2 5 ´ 4 i 3 56.31o i ´ o 1 λix = cos (-56.31 ) = 0.55, 2 ´ o λ = sin (-56.31 ) = -0.83 * 2 iy 9 * * 5 o 8 7 ´ λjx = cos (-76.31 ) = 0.24, o λ = sin (-76.31o) = -0.97 j 76.31 6 jy j ´ 4 [ q* ] = [ T ]T[ q´ ] ´ 1´ 2´ 3´ 4´ 5´ 6´

q4 4 0.55 0.83 0 0 0 0 q1´ q5 5 -0.83 0.55 0 0 0 0 q2´ q 6 0 0 1 0 0 0 q 6 = 3´ q7* 7* 0 0 0 0.24 0.97 0 q4´ q8* 8* 0 0 0 -0.97 0.24 0 q5´ q9* 9* 0 0 0 0 0 1 q6´ 47*8*9*56 4 1036.923 -1524.780 30.646 -452.884 1787.474 30.646 5 -1524.780 2307.582 20.431 643.585 -2689.923 20.431 6 30.646 20.431 88.643 -35.786 -8.717 44.321 * T 3 [ k ]2 = [ T ] [ k´]2[ T ] = 10 7* -452.884 643.585 -35.786 205.452 -759.668 -35.786 8* 1787.474 -2689.923 -8.717 -759.668 3139.053 -8.717 67 9* 30.646 20.431 44.321 -35.786 -8.717 88.643 2 3 6 1 5 1 4 2 9* 8* 7*

1 2 3 456 1 3000 0 0 -3000 0 0 2 0 15 30 0 -15 30 3 0 30 80 0 -30 40 3 [k]1 = 10 4 -3000 0 0 3000 0 0 5 0 -15 -30 0 15 -30 6 0 30 40 0 -30 80 47*8*9*56 4 1036.923 -1524.780 30.646 -452.884 1787.474 30.646 5 -1524.780 2307.582 20.431 643.585 -2689.923 20.431 6 30.646 20.431 88.643 -35.786 -8.717 44.321 * 3 [k ]2 = 10 7* -452.884 643.585 -35.786 205.452 -759.668 -35.786 8* 1787.474 -2689.923 -8.717 -759.668 3139.053 -8.717 9* 30.646 20.431 44.321 -35.786 -8.717 88.643 68 20 kN/m 80 kN•m 2 3 6 1 5 50 kN AB 1 4 3 m 2 26.67 kN•m 20 kN/m 9* 8* 7* 1 C 26.67 kN•m 20o 40 kN 40 kN 4 m 2 m Global: 47*9*56 D 0 Q4 = -50 4 4036.923 -1524.780 30.646 -452.884 30.646 4 -1524.780 232.582 -9.569 643.585 -9.569 D 40 Q5 = 0 5 5 3 30.646 -9.569 168.643 -35.786 44.321 D -26.67 Q6 = -80 = 10 6 6 + 7 -452.884 643.585 -35.786 205.452 -35.786 D 0 Q7* = 0 7* 9 30.646 20.431 44.321 -35.786 88.643 D 0 Q9* = 0 9*

-5 D4 -2.199x10 m -4 D5 -3.095x10 m -4 D6 = -2.840x10 rad -3 D7* 0.979x10 m -4 D9* 6.161x10 rad 69 Member 1: 1 2 3 456

q1 1 3000 0 0 -3000 0 0 0 0

q2 2 0 15 30 0 -15 30 0 40

q3 3 0 30 80 0 -30 40 0 26.67 = 103 + q4 4 -3000 0 0 3000 0 0 D4 0

q5 5 0 -15 -30 0 15 -30 D5 40 q6 6 0 30 40 0 -30 80 D6 -26.67

2 3 5 6 1 q1 65.97 kN 1 4 q2 36.12 kN

q3 24.59 kN•m -65.97 kN 24.59 kN•m 20 kN/m 40.11 kN•m q4 = q 43.88 kN 65.97 kN 5 65.97 kN 1 q6 -40.11 kN•m 36.12 kN 43.88 kN

70 Member 2: 47*8*9*56 q4 4 1036.923 -1508.14 30.57 -455.21 1769.34 30.57 D4 q5 5 -1508.14 2296.15 20.26 651.27 -2678.93 20.26 D5 q6 6 30.57 20.26 88.64 -35.73 -8.84 44.32 D6 = 103 q7* 7* -455.21 651.27 -35.73 210.67 -769.1 -35.73 D*7 q8* 8* 1769.34 -2678.93 -8.84 -769.1 3128.82 -8.84 0 q9* 9* 30.57 20.26 44.32 -35.73 -8.84 88.64 D*9

q 15.97 kN 39.89 kN•m 4 5 6 43.88 kN q -43.88 kN 5 i 4 15.97 kN q -39.89 kN•m 6 2 2 q = 0kN * 7* 9 46.69 kN * * q8* 8 7 q9* 0kN•m j 46.69 kN

71 2 3 6 20 kN/m 1 5 24.59 kN•m 40.11 kN•m 1 65.97 kN 4 1 2 65.97 kN * 9 36.12 kN 43.88 kN 8* 7* 43.88 kN 39.89 kN•m 15.97 kN

-5 2 D4 -2.199x10 m -4 D5 -3.095x10 m -4 D6 = -2.840x10 rad -3 D7* 0.979x10 m -4 46.69 kN D9* 6.161x10 rad

-5 39.89 D4=-2.2x10 m + - -4 - + D5=-3.1x10 m -24.59 -4 D6 = -2.84x10 rad -40.11 -3 D7*=0.979x10 m

Bending-moment Deflected shape diagram (kN•m) -4 D9*=6.161x10 rad 72 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES

! Trusses

! Vertical Loads on Building Frames

! Lateral Loads on Building Frames: Portal Method

! Lateral Loads on Building Frames: Cantilever Method Problems

1 Trusses

P1 P2 a a

Fa V = R1 Fb

F2 a a R1 (a) R2 R1 (b)

Method 1 : If the diagonals are intentionally designed to be long and slender, it is reasonable to assume that the panel shear is resisted entirely by the tension diagonal, whereas the compressive diagonal is assumed to be a zero-force member.

Method 2 : If the diagonal members are intended to be constructed from large rolled sections such as angles or channels, we will assume that the tension and compression diagonals each carry half the panel shear.

2 Example 1

Determine (approximately) the forces in the members of the truss shown in Figure. The diagonals are to be designed to support both tensile and compressive forces, and therefore each is assumed to carry half the panel shear. The support reactions have been computed.

F E D

3 m C A B 4 m 4 m 10 kN 20 kN

3 20 kN F E D o + ΣMA = 0: FFE(3) - 8.33cos36.87 (3) = 0

3 m FFE = 6.67 kN (C) C A B o 4 m 4 m 10 kN + ΣMF = 0: FAB(3) - 8.33cos36.87 (3) = 0 10 kN 20 kN FAB = 6.67 kN (T) 20 kN o F θ = 36.87 FFE θ FAF 3 m FFB= F V = 10 kN 8.33 F = F AE θ θ A 6.67 kN A FAB 10 kN 10 kN o + ΣFy = 0: 20 - 10 - 2Fsin(36.87 ) = 0

F = 8.33 kN o + ΣFy = 0: FAF - 10 - 8.33sin(36.87 ) = 0

FFB = 8.33 kN (T) FAF = 15 kN (T) F = 8.33 kN (C) AE 4 θ = 36.87o D θ = 36.87o FED D θ 6.67 kN F = FDB θ F = F V = 10 kN EC 3 m 8.33 kN θ C F FBC DC 10 kN

o o + ΣFy = 0: 10 - 2Fsin(36.87 ) = 0 + ΣFy = 0: FDC - 8.33sin(36.87 ) = 0

F = 8.33 kN FDC = 5 kN (C) F = 8.33 kN (T) DB θ = 36.87o E FEC = 8.33 kN (C) 6.67 kN 6.67 kN θθ

o 8.33 kN 8.33 kN + ΣMC = 0: FED(3) - 8.33cos36.87 (3) = 0 FEB FED = 6.67 kN (C) + ΣFy = 0:

+ ΣM = 0: F (3) - 8.33cos36.87o(3) = 0 o D BC FEB = 2(8.33sin36.87 ) = 10 kN (T) F = 6.67 kN (T) BC 5 Example 2

Determine (approximately) the forces in the members of the truss shown in Figure. The diagonals are slender and therefore will not support a compressive force. The support reactions have been computed.

10 kN 20 kN 20 kN 20 kN 10 kN

J I H G F

4 m

ABC4 m4 m4 m4DE m 40 kN 40 kN

6 10 kN FJA 0 J θ FJI A F 0 45o JB 4 m V = 30 kN 40 kN FAI = 0 F A AB + ΣFy = 0: FJA = 40 kN (C) 40 kN 10 kN 20 kN

FAI = 0 J I FIH o + ΣFy = 0: 40 - 10 - F cos 45 = 0 F JB 45o IC 4 m V = 10 kN F = 0 FJB = 42.43 kN (T) BH FBC AB4 m o + ΣMA = 0: FJI(4) - 42.43sin 45 (4) = 0 40 kN F = 30 kN (C) JI FBH = 0

o + ΣFy = 0: 40 - 10 - 20 - F cos 45 = 0 + ΣMJ = 0: FAB(4) = 0 IC

F = 0 FIC = 14.14 kN (T) AB 7 10 kN 20 kN FBI 42.3 kN J I o 0 F 45 IH o o F 45 45 45o IC 0 30 kN 4 m V = 10 kN B FBH = 0 FBC AB o 4 m + ΣFy = 0: FBI = 42.3 sin 45 = 30 kN (T) 40 kN

+ ΣMB = 0: F 14.14 kN CH 14.14 kN o FIH(4) - 14.14sin 45 (4) + 10(4) - 40(4) = 0 45o 45o FJH = 40 kN (C) 30 kN 30 kN C

+ ΣMI = 0: FBC(4) - 40(4) + 10(4) = 0 + ΣFy = 0: FBC = 30 kN (T) o FBI = 2(14.1442.3 sin 45 ) = 20 kN (C)

8 Vertical Loads on Building Frames

typical building frame

9 • Assumptions for Approximate Analysis

columnw column

girder AB L (a) w w assumed points of ABpoint of zero 0.1L zero moment 0.1L 0.21L moment 0.21L L L (b) approximate case w (d) Point of Point of zero A B zero w moment moment L 0.1L 0.1L Simply supported 0.8L (c) model (e) 10 Example 3

Determine (approximately) the moment at the joints E and C caused by members EF and CD of the building bent in the figure.

1 kN/m

E F

1 kN/m

C D

A B

6 m

11 1 kN/m

4.8 kN 1 kN/m

4.8 m 4.8 m 2.4 kN 2.4 kN 0.1L=0.6 m 0.6 m

0.6 kN 2.4 kN 2.4 kN 0.6 kN

0.6(0.3) + 2.4(0.6) = 1.62 kN•m 1.62 kN•m

3 kN 0.6 m 0.6 m 3 kN

12 Portal Frames and Trusses ∆∆ • Frames: Pin-Supported P P assumed h h hinge

l l (a) (b) Ph/2 L/2 L/2 Ph/2 Ph/2 P P/2 P/2 Ph/2 Ph/l Ph/l h h

P/2 P/2 (d) Ph/l (c) Ph/l 13 • Frames : Fixed-Supported ∆∆ P P assumed hinges h h

l l (a) L/2 (b) L/2 P P/2 P/2 h/2 Ph/4 h/2 Ph/2l Ph/2l Ph/4 Ph/4 P/2 P/2

Ph/4 Ph/2l Ph/2l Ph/2l Ph/2l P/2 P/2 h/2 Ph/4 Ph/4 P/2 P/2 (d) Ph/4 (c) Ph/4 Ph/2l Ph/2l 14 • Frames : Partial Fixity P P

θθassumed h hinges

h/3 h/3

l (a) (b) • Trusses

P P

∆∆ assumed h hinges h/2 P/2 P/2

l l (a) (b) 15 Example 4

Determine by approximate methods the forces acting in the members of the Warren portal shown in the figure.

2 m 4 m 2 m

40 kN C DEF 2 m

B HG 4 m 7 m

A I

8 m

16 2 m 4 m 2 m

40 kN C DEF 2 m + ΣMJ = 0: B HG N(8) - 40(5.5) = 0 3.5 m JK N = 27.5 kN 40/2 = 20 kN = V 20 kN = V NN N N V = 20 kN V = 20 kN + ΣMA = 0: 3.5 m M - 20(3.5) = 0 A I 20 kN = V V = 20 kN MMM = 70 kN•m NN

17 2 m 2 m

C F FEF F 40 kN CD D E

2 m F FEG 2 m 45o BD 45o F FGH B BH G 3.5 m 3.5 m J K 20 kN 20 kN = V 27.5 kN 27.5 kN

o o + ΣFy = 0: -27.5 + FBDcos 45 = 0 + ΣFy = 0: 27.5 - FEGcos 45 = 0

FBD = 38.9 kN (T) FEG = 38.9 kN (C)

+ ΣMB = 0: FCD(2) - 40(2) - 20(3.5) = 0 + ΣMG = 0: FEF(2) - 20(3.5) = 0

FCD = 75 kN (C) FEF = 35 kN (T)

+ ΣMD = 0: FBH(2) + 27.5(2) - 20(5.5) = 0 + ΣME = 0: FGH(2) + 27.5(2) - 20(5.5) = 0 F = 27.5 kN (T) F = 27.5 kN (C) BH GH 18 y y

38.9 kN FHE

D F 45o 45o 75 kN DE x 27.5 kN x 45o 45o H 27.5 kN

38.9 kN FDH

o o o o + ΣFy = 0: FDHsin 45 - 38.9sin 45 = 0 + ΣFy = 0: FHEsin 45 - 38.9sin 45 = 0

FDH = 38.9 kN (C) FHE = 38.9 kN (T)

+ 75 - 2(38.9 cos 45o) - F = 0 ΣFx = 0: DE

FDE = 20 kN (C)

19 Lateral Loads on Building Frames: Portal Method

= inflection point (a)

V V V V (b) 20 Example 5

Determine by approximate methods the forces acting in the members of the Warren portal shown in the figure.

5 kN B D FG

3 m

A C E H

4 m 4 m 4 m

21 5 kN B MND FG O

3 m I JKL

A C E H

4 m 4 m 4 m

5 kN BDMN FG O

1.5 m VVI 2VJKL 2V

Iy Jy Ky Ly

+ ΣFx = 0: 5 - 6V = 0

V = 0.833 kN 22 2 m 5 kN B 2 mD 2 m N 4.167 kN 2.501kN M 4.167 kN 1.5 m 1.5 m 0.625 kN 0.625 kN 0.625kN 0.833 kN I 1.666 kN J

Iy = 0.625kN Jy = 0

2.501 kN NO2 m F 2 m 0.835 kN O 2 m G 0.835 kN 0.625kN 1.5 m 0.625 kN 1.5 m 0.625kN 1.666 kN K 0.833 kN L

Ky = 0 0.625 kN = Ly 0.625 kN 0.625 kN 1.666 kN 1.666 kN 0.833kN I J K L 0.833 kN 1.5 m 1.5 m 1.5 m 1.5 m A C E H 0.833 kN 1.666kN 1.666kN 0.833 kN 0.625 kN 1.25 kN•m 2.50 kN•m 2.5 kN•m 0.625 kN 1.25 kN•m 23 Example 6

Determine (approximately) the reactions at the base of the columns of the frame shown in Fig. 7-14a. Use the portal method of analysis.

GH I 20 kN

5 m DEF 30 kN

6 m

ABC

8 m 8 m

24 GHR S I 20 kN

OPQ5 m DEFM N 30 kN

JKL6 m

ABC

8 m 8 m

25 GI 20 kN 2.5 m V 2V V Oy Py Py

+ ΣFx = 0: 20 - 4V = 0 V = 5 kN

GH I 20 kN

5 m DEF 30 kN 3 m

V´ 2V´ V´ Jy Ky Ly

+ ΣFx = 0: 20 + 30 - 4V´ = 0 V´ = 12.5 kN 26 Ry = 3.125 kN 3.125 kN Sy = 3.125 kN G 4 m R R 4 m H S 20 kN Rx = 15 kN Sx = 5 kN 15 kN 4 m 2.5 m 2.5 m 10 kN 5 kN Oy = 3.125 Py = 0 kN 3.125 kN

5 kN 10 kN M = 12.5 kN O y P N = 12.5 kN 2.5 m 2.5 m y M 22.5 kN M 30 kN Nx = 7.5 kN 4 m 4 m 3 m 4 m Mx = 22.5 kN N 12.5 kN 3 m 12.5 kN J K 25 kN K = 0 kN Jy = 15.625 kN y 15.625kN 25 kN J 12.5 kN K 3 m 3 m A B Ax = 12.5 kN Bx = 25 kN A = 15.625 kN MA = 37.5 kN•m B = 0 MB = 75 kN•m x x 27 Lateral Loads on Building Frames: Cantilever Method P

beam building frame (a) (b) In summary, using the cantilever method, the following assumptions apply to a fixed-supported frame. 1. A hinge is place at the center of each girder, since this is assumed to be point of zero moment. 2. A hinge is placed at the center of each column, since this is assumed to be a point of zero moment. 3. The axial stress in a column is proportional to its distance from the centroid of the cross-sectional areas of the columns at a given floor level. Since stress equals force per area, then in the special case of the columns having equal cross-sectional areas, the force in a column is also proportional to its distance from the centroid of the column areas. 28 Example 7

Determine (approximately) the reactions at the base of the columns of the frame shown. The columns are assumed to have equal crossectional areas. Use the cantilever method of analysis.

C 30 kN D

4 m B 15 kN E

4 m AF

6 m

29 C x 30 kN D I 4 m H K B J 15 kN E 6 m

4 m G L

AF~ xA 0(A) + 6(A) x = ∑ = = 3 A A + A 6 m ∑

30 3 m 3 m C + ΣMM = 0: -30(2) + 3Hy + 3Ky = 0 30 kN D 2 m I The unknowns can be related by proportional triangles, Kx that is H M x H K y = y or H = K 3 3 y y Hy Ky

H y = K y =10kN

C 30 kN D + ΣM = 0: -30(6) - 15(2) + 3G + 3L = 0 I N y y 4 m H K The unknowns can be related by proportional triangles, that is B J 15 kN E G G 3 m 3 m y = y or G = L 2 m 3 3 y y Lx G N x Gy = Ly = 35kN

Gy Ly 31 Iy = 10 kN 10 kN C 3 m I 3 m D 30 kN Ix = 15 kN 15 kN 2 m 2 m Kx = 15 kN Hx = 15 kN 10 kN 10 kN 10 kN 10 kN 15 kN 15 kN H Jy = 25 kN K 2 m 2 m J J 3 m 15 kN 7.5 kN 2 m 3 m Jx = 7.5 kN 2 m 25 kN G L Lx = 22.5 kN Gx = 22.5 kN 35 kN 35 kN 35 kN 35 kN 22.5 kN G 22.5 kN L 2 m 2 m A F Ax = 22.5 kN Fx = 22.5 kN A = 35 kN MA = 45 kN•m F = 35 kN MF = 45 kN•m x y 32 Example 8

Show how to determine (approximately) the reactions at the base of the columns of the frame shown. The columns have the crossectional areas show. Use the cantilever of analysis.

PQ R 35 kN 6000 mm2 5000mm2 4000 mm2 6000 mm2 4 mLMN O IJ K 45 kN

6 mEFG H 6000 mm2 5000mm2 4000 mm2 6000 mm2 ABC D

6 m 4 m 8 m

33 PQ R 35 kN

4 mLMN O IJ K 45 kN

6 mEFG H

ABC D

6 m 4 m 8 m

2 6000 mm2 5000mm 4000 mm2 6000 mm2

6 m 4 m 8 m x

~ xA 6000(0) + 5000(6) + 4000(10) + 6000(18) x = ∑ = = 8.48m ∑ A 6000 + 5000 + 4000 + 6000 34 PQ R 35 kN

2 m Mx Nx Ox

Lx My Ly Ny Oy

2.48 m 1.52 m 8.48 m 9.52 m

+ ΣMσNA = 0: -35(2) + Ly(8.48) + My(2.48) + Ny(1.52) + Oy(9.52) = 0 -----(1) σ Since any column stress σ is proportional to its distance from the neutral axis σ σ 2.48 M 2.48 L M = σ L ; = σ ; y = ( y ) − − − − − (2) 2.48 8.48 M 8.48 L 5000(10−6 ) 8.48 6000(10−6 ) σ σ 1.52 N 1.52 L N = σ L ; = σ ; y = ( y ) − − − − − (3) 1.52 8.48 N 8.48 L 4000(10−6 ) 8.48 6000(10−6 ) σ 9.52 O 9.52 L O = L ; = σ ; y = ( y ) − − − − − (4) 9.52 8.48 O 8.48 L 6000(10−6 ) 8.48 6000(10−6 )

Solving Eqs. (1) - (4) yields Ly = 3.508 kN My = 0.855 kN N = 0.419 kN O = 3.938 kN y y 35 35 kN

4 m LMN O IJ K 45 kN 3 m Fx Gx Hx

Ex Fy Gy Ey Hy 2.48 m 1.52 m 8.48 m 9.52 m

+ ΣMσ = 0: -45(3) - 35(7) + E (8.48) + F (2.48) + G (1.52) + H (9.52) = 0 -----(5) NA σ y y y y Since any column stress σ is proportional to its distance from the neutral axis ; σ σ 2.48 F 2.48 E F = σ E ; = σ ; y = ( y ) − − − − − (6) 2.48 8.48 F 8.48 E 5000(10−6 ) 8.48 6000(10−6 ) σ σ 1.52 G 1.52 E G = σ E ; = σ ; y = ( y ) − − − − − (7) 1.52 8.48 G 8.48 E 4000(10−6 ) 8.48 6000(10−6 ) σ 9.52 H 9.52 E H = E ; = σ ; y = ( y ) − − − − − (8) 9.52 8.48 H 8.48 E 6000(10−6 ) 8.48 6000(10−6 ) Solving Eqs. (1) - (4) yields E = 19.044 kN F = 4.641 kN G = 2.276 kN H = 21.38 kN y y y y 36 P = 3.508 kN 3 m y 35 kN P = 29.738 kN 2 m x

Lx= 5.262 kN

3.508 kN

3.508 kN 5.262 kN Iy= 15.536 kN 2 m I 45 kN Ix= 114.702 kN 3 m 3 m

Ex= 64.44 kN 19.044 kN

19.044 kN E 64.44 kN 3 m One can continue to analyze the Ax = 64.44 kN other segments in sequence, i.e., M = 193.32 kN•m PQM, then MJFI, then FB, and so on. Ax = 19.044 kN A 37