Beam Analysis Using the Stiffness Method

Home , Deflection (engineering), Stiffness

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

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

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