DIRECT STIFFNESS METHOD FOR ANALYSIS OF SKELETAL STRUCTURES

http://web.iitd.ac.in/~sbhalla/cvl756.html

Dr. Suresh Bhalla Professor Department of Civil Engineering Indian Institute of Technology Delhi 1 The audio of this lecture is based on classroom recording from CVL 756 class for batch 2019-20. Use earphone/ amplifier for best audio experience.

2 SLOPE DEFLECTION METHOD

Applied moment M θ Carry over moment 4EI  2EI L 6EI 6EI L 2 L2 L

12EI 3 6EI L 2 L  6EI 12EI L2 L3 3 Generalized slope deflection equations

2EI 3(  ) θ  2 1  M1 2 M1  21  2   2 L  L  θ1  2EI  3( 2 1) 1 M2 M 2  1  2 2   L  L 

F F1 2 M  M F  F  1 2 1 2 L

2EI 6(  ) 2EI  6( 2 1)  2 1  F  3  3  F2   31  3 2  1 2  1 2  2   L  L  L  L 

4 PUT IN MATRIX FORM

 12EI 6EI 12EI 6EI    3 2 3 2  F L L L L   1   6EI 4EI 6EI 2EI  1        M1 2 2 1 • Symmetric     L L L L    F  12EI 6EI 12EI 6EI   2     2 • K +ive    3 2 3 2   ii M L L L L   2   6EI 2EI 6EI 4EI  2      L2 L L2 L   Force Stiffness Displacement vector matrix vector (symmetric)

5 F1  K11 K12 . K1n 1  1 2 3 F  K K . K    2    21 22 2n  2   .   . . . .  .       Fn  Kn1 Kn1 . Knn n  n = Degrees of freedom (DKI)

th Fi = Force along i degree of freedom th i = Displacement along i degree of freedom th j col. of [K] : th Kij =Force generated along i Forces generated along the various degrees of degrees of freedom under a unit freedom under a unit displacement along the jth displacement along the jth degree of degree of freedom (j = 1), with all other freedom with all other degrees of degrees of freedom locked (x = 0, where x j ) freedom locked.

Matrix Stiffness Approach (MSA) : The elements of [K] are obtained by first principles using the definition of kij from the deformation pattern of the structure and force-deformation relations of the members 6 Matrix Stiffness Approach (MSA) : Human judgment needed 1 2 3

The direct stiffness approach (DSA), on the other hand, enables computation of the overall stiffness of any complicated structure using computer program, based on finite element formulation, without human visualization of the

overall structure. 7 SIGNIFICANCE OF DIRECT STIFFNESS METHOD

ALL COMMERCIAL STRUCTURAL ENGINEERING ANALYSIS PACKAGES ARE BASED ON THE DIRECT STIFFNESS APPROACH.

UNDERSTANDING AND IMPLEMENTING THE CONCEPTS WILL HELP YOU IN: 1. MAKING YOUR OWN CUSTOMIZED RESULT ORIENTED SOFTWARE WITHOUT SPENDING ANY PENNY. 2. USING THE EXISTING SOFTWARE IN ERROR FREE MANNER, WITH UNDERSTANDING, RATHER THAN AS A “BLACK BOX” APPROACH. 8 DIRECT STIFFNESS METHOD: ASSUMPTIONS

1. Restricted to frame and truss structures (skeletal structures) only. Members assumed as line elements (passing through neutral axis) with lumped sectional properties. At first, we restrict analysis to prismatic members only.

2. Hooke’s law of elasticity holds.

3. Small deflections => no change in overall geometry of structure.

9 DIRECT STIFFNESS METHOD: ASSUMPTIONS

4. Plane sections remain plane after bending.

5. In bending mode, very small slope d 2 y / dx2 d 2 y Curvature = 3/ 2  1 dy / dx 2  dx2

6. If displacement takes place normal to member, no change in length of the member. Change in length of an element due to flexural deformation (curvature effects) is also negligible.

10 DIRECT STIFFNESS METHOD: ASSUMPTIONS

7. Principle of superposition holds good. • Loads can be superimposed • Boundary conditions can be superimposed • Displacements can be superimposed • BMD, SFD can be superimposed.

IMPORTANT: All assumptions of slope deflection method are repeated except one….. We have discarded the assumption regarding inextensibility of the members…… Unlike manual approach, digital computers will not have no problem in tackling additional degrees of freedom. 11 DIRECT STIFFNESS METHOD FOR COMPUTER APPLICATIONS

• Each individual member is treated as structure (called element).

• Stiffness matrix of each individual element is obtained.

• Total stiffness matrix of the entire structure is then computationally obtained by superimposing the matrices of elements, without human intervention.

• Hence, analysis can be broken down into small steps and programmed, in a finite element procedure.

12 GENERATION OF ELEMENT STIFFNESS MATRIX 2D STRUCTURES x’ 5 4  EA   L   12EI  Local coordinate  f1  d1   0 3  system  f   L d   2  6EI 4EI  2  y’ 6  0    2    f 3  L  d 3    L   f  EA EA  d y  4   0 0  4    L L   2 f 5   d 5 3    12EI 6EI 12EI    f  0  3  2 0 3 d  1 Global coordinate  6   L L L  6  system  6EI 2EI 6EI 4EI   0 0   x  L2 L L2 L  z axis normal to plane of board towards viewer

In short form, {f}= [k] L {d}

[k]L = Element stiffness matrix with respect to local coordinate system. 13 TRANSFORMATION OF COORDINATE SYSTEM

At a joint, members of different orientations may meet.

The forces and displacements at member ends cannot be easily related.

To consider equilibrium of the joint and compatibility of member displacements, the member end forces and displacements must be transformed to a common coordinate system.

14  V = A vector (force or displacement) y

y’ l1 m1 Vx  cos sin Vx  Vy      l2 m2   V Vy  sin cos V V y     y  x’ (Local) V    x l x (Global) 1 m1 Direction V x cosines l m2   2 V =Vx î + Vy ĵ V = Vx’ î’ + Vy ’ĵ’

= (Vcosφ)î + (Vsinφ)ĵ = Vcos(φ-θ) î’ + Vsin(φ-θ) ĵ’ = [Vcosφcosθ + Vsinφsinθ] î’ + [Vsinφcosθ -Vcosφsinθ] ĵ’

= [Vxcosθ + Vysinθ] î’ + [-Vxsinθ +Vycosθ] ĵ’

V ’ = V cosθ + V sinθ Hence x x y 15 Vy’ = -Vxsinθ + Vycosθ MEMBER FORCES IN GLOBAL & LOCAL COORDINATES

F5 Local coordinate system f5 f4 F4 y’ x’ F6 f6 F2 f2  y

F 1 F3 Member end forces f3 f1 x in global coordinates

Global coordinate system Member end forces in local coordinates

 f1   cos sin 0F1   f4   cos sin 0F4            f2   sin cos 0 F2 f   sin cos 0 F       5    5   f3   0 0 1F3   f6   0 0 1F6  16  f1   cos sin 0F1  F1         f2   sin cos 0 F2  F 1      R  2   f3   0 0 1F3  F3  F f f  1  -1  1  T  1  F    f   f   2  R  2  = R  2  F3   f3   f3 

R-1 = RT R is an orthogonal matrix. R = JOINT TRANSFORMATION MATRIX

 f4  F4   f   F  2 Similarly  5  R  5   f  F   6   6  17 Combining equations 1 and 2

 f1  F1  0 0 0  f  F   2  0 0 0  2  R 000 0  f3  F3     0 0 0   f4 F4   0 0 0    f  F   5  0 00 0 R  5   f6  F6  R 0 T= {f}=[T]{F} 0 R

[T] = MEMBER TRANSFORMATION MATRIX

Similarly for displacements {d}=[T]{D}

18 HOW WILL PROGRAM OBTAIN THE NECCESARY INFORMATION FOR COORDINATE TRANSFORMATION?? User should provide the coordinates of all joints…. x  x cos  2 1 L (x2,, y2) Local coordinate y  y system sin  2 1 y’ L 

y 2 2 L  (x2  x1)  (y2  y1)

(x1,, y1) Global coordinate system x  cos sin 0   R   sin cos 0  0 0 1 R 0 [T] = T-1 TT 10/1/20200 R = 19 19 HOW IS TRANSFORMATION UTILIZED??

{f} = [k]L {d}

[T] {F} = [k]L[T]{D}

T {F} = [ T ] [k]L [T]{D}

[K] Stiffness matrix of member in global G coordinates

T [K]G =[ T ] [k]L [ T ]

20 SPECIAL CASE: TRUSS STRUCTURES

x’ 4 3 EA EA f   d  1   0  0 1    L L   Local coordinate system  f 2   0 0 0 0d 2        y’ f EA EA d   3   0 0 3   f   L L d  y  4   0 0 0 0 4  2  

1 Global coordinate system x  cos sin  R    No member end moments or  sin cos  rotations

21 MEMBER FORCES IN GLOBAL & LOCAL COORDINATES

F Local coordinate 5 system f5 f4 F4 y’ x’

F6 f6

F2 f2  y

F 1 F3 f f3 1 x Member end forces in global coordinates Global coordinate Member end forces system in local coordinates

 f1   cos sin 0F1   f4   cos sin 0F4            f2   sin cos 0 F2 f   sin cos 0 F       5    5   f3   0 0 1F3   f6   0 0 1F6  22 GENERATION OF TOTAL STRUCTURAL STIFFNESS MATRIX

1 2 3 4

6 7 5 8

i+1 i

n n - 1

We shall first derive formulations for simple 2D case: (1)Supports are fixed (2) All joints are rigid with no internal hinges. (3) Joints can be sequentially numbered as above

We shall introduce complications into analysis one by one. 23 NUMBERING SCHEME Joints numbered sequentially, restrained joints numbered in end

P2 , u2 Joint loads and joint displacements

 P1   k11 k12 . k1(3n) u1  P1, u1      2 3 4 P . . . . u 1  2     2   .   . . . .  .       k k . k 6 7 X 3n   (3n)1 (3n)2 (3n)(3n) un  P3, u3 5 8 Total structural stiffness matrix should relate global structural loads to global structural displacements P3 i - 1 c P 3 i - 2 b i+1 i X3 n - 1

a X3n - 2 n n - 1 P3 i

Similar pattern for numbering X3 n of displacements Degrees of freedom shall include those at supports also 24 5 6 4

c 1 3 3 i - 1 2 5 2 3 b 3 i - 2 Joint i 5 1 4 3 j - 1 6 4 6 3 j - 2 Joint 3 i a j=i+1 1 3 3 j 2 Member degrees of freedom: From element point of view (1..6) Structural degrees of freedom: From global (overall structures) point of view (1..3n) Each member degree of freedom in global coordinates (1,2,…,6) corresponds to a particular structural degree of freedom (1,2,….,3n).

a b c COMPATIBILITY u3i2  D4  D1  D1 25 EQUILIBRIUM CONDITIONS P2 , u2

P1, u1 1 2 3 4

6 7 P3, u3 5 8

P3 i - 1 c P 3 i - 2 b i+1 i

a n n - 1 P3 i

Let a unit displacement be applied along d.o.f (3i-2) and all other d.o.f. =0

26 5 6 4

c 1 3 3 i - 1 2 5 2 3 b 3 i - 2 Joint i 5 1 4

6 4 6 Compatibility Joint 3 i a a b c j=i+1 1 3 u3i2  D4  D1  D1 2

Let a unit displacement be applied along d.o.f (3i-2) and all other d.o.f. =0 By joint equilibrium,

P3i-2 = Sum of member end forces of a, b, c in X-direction a b c P3i-2 = F 4 +F1 +F 1

Recall: kmn = Force induced along d.o.f.’m’ due to unit displacement along d.o.f ’n’, all other displacements maintained zero. K , = ka + kb + kc 3i-2 3i-2 44 11 11 27 a b c K3i-2, 3i-2 = k 44 + k 11 +k 11

Elements of member Element of total structural stiffness matrices in global stiffness matrix coordinates

• An element of [K]TS can be obtained by summing the elements of member stiffness matrices (in global coordinates) of corresponding d.o.f from members that frame into that joint.

• In order to carryout smoothly, we follow Code Number Approach.

28 Each member degree of freedom in global coordinates (1,2,…,6) corresponds to a particular structural degree of freedom (1,2,….,3n). This information can be stored in the association matrix of the member.

K34 of [K]L will correspond to K3i, 3(i+1)-2 of [K]TS (i.e. will transfer to that location)

For member ‘b’, the association matrix is

3 4

3i-2 3i-1 3i 3(i+1)-2 3(i+1)-1 3(i+1)

2 5 3i-1 3(i+1)-1 3 b 3i 1 4 b 3i-2 3(i+1)-2

6 3(i+1)

29 EXAMPLE

1 2 3   4 5 6     7 8 9      

30 KTS GENERATION: MATRIX STIFFNESS VS DIRECT STIFFNESS APPROACH 1 2 3   4 5 6     7 8 9      

KTS (9,9) =??

31 KTS GENERATION: MATRIX STIFFNESS VS DIRECT STIFFNESS APPROACH 1 2 3   4 5 6     7 8 9      

KTS (6,9) =??

32 HOW TO GENERATE THE TOTAL STRUCTURAL STIFFNESS MATRIX

. All joints of the structure should be numbered sequentially, starting from the unstrained joints. . Restrained joints should be numbered in the end. . Initialize the total structural stiffness matrix to ‘0’. . Consider each member; compute its member stiffness matrix in global coordinates. . Then send its elements into appropriate location of the global stiffness matrix of the entire structure, one at a time. . Repeat this process for each member; keep adding its elements to the appropriate elements of the total structural stiffness matrix. . Finally, the total stiffness matrix of the structure will result.

33 STEPWISE PROCEDURE FOR PROGRAMMING

1.Label all elements (or members) 1…..m.

2.Label all joints 1….n, first unrestrained, then the restrained ones. D.O.F associated with ith node: 3i-2, 3i-1, 3i. Hence, all d.o.f are also numbered.

3. Compute the size of structural stiffness matrix & initialize it to 0.

34 STEPWISE PROCEDURE FOR PROGRAMMING (Contd…)

4. Repeat for each element (from i=1 to m)

• Compute [k]L • Compute [R] from end coordinates. T • Compute [k]G = [T] [k]L[T] • Establish association matrix (from node number of the two nodes of member.

• Transfer each element of [kG] to appropriate location of [K]TS (kTS)ij = Σ(kG)mn Extends over all members meeting at a joint. m: Corresponds to ith dof and n to the jth dof.

Need to do this process 36 times for each member, no discount from symmetry….Why? 35 STEPWISE PROCEDURE FOR PROGRAMMING (Contd…) 5. Obtain nodal loads P • Direct nodal loads • Equivalent nodal loads (from member loads such as distributed loads)

 +

Fixed ended forces (to be Equivalent joint added in the loads end)

T f Equivalent joint load contributed by a member = [T ] { f } (Global coordinates) 36 STEPWISE PROCEDURE FOR PROGRAMMING (Contd…)

 P  K PP K PX uP  6. Set up equations       X  K XP K XX uX 

ux : prescribed displacements : corresponding to X (unknown reactions) up : un-prescribed displacements: corresponding to P (known loads)

Let {ux} =0 (no support movement )

{P}= [kpp] {up} -1  {up}= [kpp] {P} and hence {X}= [kxp] {up} Support If not zero {P}= [kpp] {up} + [kpx] {ux} settlement -1 {up}= [kpp] {P}-[kpx] {ux}

Reactions {X}=[kxp]{up} + [kxx]{ux} 37 STEPWISE PROCEDURE FOR PROGRAMMING (Contd…) 7. Member end forces (go back to member level) For each member (from 1 to m), do- a) Using association matrix, get nodal displacements{D}(global coordinates) Eg, for this member D =u 1 13 2 3 D2=u14 1 D3=u15 D4=u4 D5=u5 4 5 6 D6=u6

(b)[d]= [T][D] Member end forces (c){f}= [k]L{d} (d) Correction for fixed ended action 7 8 9

 +

L f {f}=[k ][d]+ [f]

38 EFFICIENT STORAGE SCHEME FOR KPP [Kpp] is a huge matrix. For example, for n= 100 unrestrained joints, it will be 300x300 in size. i.e. 9x104 elements.

However, the fact is that it is symmetric and banded (why??) R (i,j) i <= j

[K]pp = [B]=

All elements outside the band are zero Therefore, we only need to generate elements which are within the band. Generally, this is achieved by storing right half band in a rotated rectangular matrix.

An element (i,j) in the original matrix will go to: Row = i Column = c=(j-i+1) in the banded matrix

Diagonal element kii : ki1

39 Correlation between half banded and full [K] Element (i, r) (i, c+i-1) Diagonal (i, 1) (i, i)

HOW TO COMPUTE ‘R’ , THE HALF BAND WIDTH  Depends on structure size and also how we do numbering of joints  For each member find – x = (max d.o.f. – min d.o.f) +1  The max value of x will be equal to “R”, the half band width. We need to store [k]pp, [k]px, [k]xx BASIS OF THIS FORMULATION ??

Horizontal numbering

[k]pp = 45x45 Half band width = 18

Vertical numbering

[k]pp = 45x45 Half band width = 12

CONCLUSION?? 40 INVERSE OF [K]PP: CHOLESKY’S ALGORITHM

To solve – [P] = [kpp] [up] (if [ux]=0 or [P*] = [kpp][up] (if [ux] is not equal to 0)

K u = P {let us relax notation}

Since K is symmetric K = VT V Where V is upper triangular matrix.

 K1 K2 K3 K4  V1 0 0 0  V1 V2 V3 V4   K K K K  V V 0 0   0 V V V   5 6 7 8    2 3   5 6 7   K9 K10 K11 K12  V4 V5 V6 0   0 0 V8 V9        K13 K14 K15 K16  V7 V8 V9 V10   0 0 0 V10  [K] [VT] [V] 41 CHOLESKY’S ALGORITHM (contd….) Ku = P

T V Vu =P VT w = P where (w = Vu)

w1 =P1/V1 V1 0 0 0  w1  P1    w2 = (P2-V2w1)/V3 V V 0 0  w P   2 3   2   2    = V V 0 0  w3 P  4 5   3   w   Similarly we can find w , V7 V8 V9 V10   4  P4  3 w4 ,……wN

Further, Vu= w

V1 V2 V3 V4  u1  w1   0 V V V  u  w   5 6 7   2  =  2  On similar lines, we can  0 0 V8 V9  u3  w3        find uN,uN-1 ,……u1  0 0 0 V10  u4  w4 

42 HOW TO OBTAIN [V] THE UPPER TRIANGULAR MATRIX

V11 = K11

V1i =k1i/V11 i1 V = [(K - V2 )] i>1 ii ii m1 mi i1 V = (k - V V )/V j>i ij ij m1 mi mj ii Vij = 0 for i>j

V & VT both will be band matrices, with same bandwidth as [k]pp. We can over write ‘V’ on [k]pp . Hence, no need to create a new matrix.

43 INTERACTIVE EXERCISE FORM ALL MATRICES FOR THE STRUCTURE w = 12 kN/m, L = 6m W

L L

44 HOW TO ANALYSE 3D STRUCTURES

Option1: A space frame can be broken down into plane frames.

z y

x

Member forces to be superimposed for columns 45 OPTION 2: MATRIX FORMULATIONS FOR 3D STRUCTURES

Alternatively, the 2D formulations can be extended into 3D

Additional term (dof 4, 10) GJ  T    d -d  L  10 4

7 10

Local Rotations 1 x’ 4 Displacements Displacements z Rotations y

Effect of slab in lateral load distribution Global x ignored….. 46  EA   L   12EI   0 z   L3   12EI y  0 0  L3   GJ   0 0 0   L   6EI y 4EI y  0 0  0  L2 L   6EI 4EI   0 z 0 0 0 z   L2 L  EA EA  0 0 0 0 0   L L    12EI z 6EI z 12EI z  0  3 0 0 0  2 0 3   L L L  [K] 12EI 6EI 12EI L  0 0  y 0 y 0 0 0 y   L3 L2 L3   GL GJ   0 0 0  0 0 0 0 0   L L  6EI 2EI 6EI 4EI  0 0  y 0 y 0 0 0 y 0 y   L2 L L2 L   6EI 2EI 6EI 4EI   0 z 0 0 0 z 0  z 0 0 0 z   L2 L L2 L 

47 x’ → along centroidal axis of the member. z’→ towards viewer. y’→ can be ascertained by right hand system rule (y’ and z’ should be along the principal axes of cross section) kˆ  iˆxˆj

z y

COORDINATE TRANSFORMATION x

x  cos sin x l1 m1 x For 2D           y  sin cos y l2 m2 y

l1,m1 : direction cosines of x’ axis w.r.t global system. l2,m2 : direction cosines of y’ axis w.r.t. global system.

48 Direction cosines x' l m n x 1 1 1 l1, m1, n1 : x’ y'  l m n y    2 2 2   l2, m2, n2 : y’ l m n : z’ z' l3 m3 n3 z 3, 3, 3

R R 0 0 0 0 R 0 0 0 0 R 0 12 x 12 0 0 0 R

49 2 2 2 1/2 L = [(x2-x1) + (y2-y1) + (z2-z1) ] l1 = (x2-x1)/L m1 = (y2-y1)/L n1 = (z2-z1)/L Unit vector along x’ =l î +m ĵ+n k 1 1 1  iˆ' We must specify the direction of y-axis

1. Unit vector along y’ = l î+m ĵ+n k 2 2 2  ˆj' OR

2. Two points along y’ : (x3,y3,z3) & (x4,y4,z4)so that we can find: l2, m2, n2

Since x’y’z’ from right handed coordinate system, unit vector along z’

kˆ'  iˆ'xˆj' 50 z y

x

51 ANALYSIS STEPS

• Fix member dimensions tentatively. • Perform analysis. • Check for adequacy of member sizes at key locations. • Revise dimensions if necessary

52 TEMPERATURE VARIATION

L L

L

Let there be uniform temperature change ∆T throughout the member Determinate vs indeterminate structures, any difference??

53 2 5 8 7 1 4 L 6 L 9 3 L

11 14 17 10 13 16

12 15 18

Identify and mark the degrees of freedom L Unconstrained L+∆L length

54 2 5 8 7 1 4 ∆T L 6 L 9 3 L

11 14 17 10 13 16

12 15 18 Convert thermal effect into fixed ended forces

f F ∆T

 EA  F f   L L  LT F f  EAT  L  55 APPLY OPPOSITE OF THE FEF ON THE STRUCTURE F f  EAT

Solve the matrix equation as before  0  and obtain displacements and    0  member forces  0     EAT  Final member end forces can be P   0    obtained by superimposing the fixed  0   EAT  ended condition with above solution    0     0  F { f }  {K L}{d}{ f }

What would happen in case of 56 temperature fall?? EXTENSION TO LACK OF FIT Longer member by construction flaw

L L

L

Longer member : Analogous to temperature rise Smaller member : Analogous to temperature fall

57 HINGED/ GUIDED SUPPORTS

L L

L

58 Let us first consider all supports to be fully rigid as treated so far…… 8 2 5 7 1 4 L 9 6 KTS 3

L 11 14 P KPP 10 13 6x1 6x6 12 15 X KXX 9x1 9x9

59 Let us now altogether remove the right support…… 8 2 5 7 1 4 L 9 6 3

L 11 14 P KPP 10 13 9x1 9x9 12 15 X KXX 6x1 6x6

P7 = P8 = P9 = 0 60 Let us now introduce the hinge

9 2 5 7 1 4 L 8 6 3

L 11 14 P KPP 10 13 8x1 8x8 12 15 X KXX 7x1 7x7

P7 = P8 = 0 (if no load acting at those points) 61 u7 ≠ 0 and u8 ≠ 0 WHAT HAPPENS WHEN THERE IS A LOAD ACTING ALONG THE RELEASE?

Mo

P KPP KPX uP 8x1 = 8x8 8x1 X KXX ux 7x1KXP 7x7 7x1

How would P change ? After solution, corresponding member end moment = M 62 P8 = Mo o UDL IN ADJOINING SPAN

Convert the UDL into equivalent joint loads

Any change in KTS ? YES/ NO Any change in P, X? YES/ NO

Upon solving, P KPP KPX uP 8x1 8x8 8x1 we will get the = values of u7 and X KXX u u8 x P3  ?? KXP wL 7x1 7x7 7x1 P2  P5   2 P  ?? EXERCISE: Form the matrices P and X 6 63 WHEN EQUIVALENT JOINT LOAD HAS COMPONENT ALONG THE RELEASE Convert UDL into equivalent joint loads wL 2

wL2 12 wL wL 2 2 P = -wL/2 W 5 2 2 2 wL wL P6 = -wL /12 12 2 12 L P8 = +wL /12 F In end,{ f }  {K L}{d}{ f } X9 also gets 64 Final moment at right end of member = 0 additional term ALTERNATE APPROACH Convert UDL into equivalent joint loads, but considering the far end to be hinged. 3wL 5wL 8 8

wL2 5wL 3wL 8 8 8 W P = -5wL/8 wL2 5 P = -wL2/8 8 L 6

F P8 = 0 In end, { f }  {K L}{d}{ f } Caution: Displacement correction needed in end. df to be added to 65 the displacement from the output INTERNAL HINGE

h

L

h

Hinge in beam only (not in column)

66 INTERNAL HINGE 9 12

8 11 10 13

2 1 6 3 1 2 5

3 4 Internal hinge 7 15 5 18 4 is tackled by duplication of 14 the degree of 17 16 19 freedom

DOF (7) : Common for Members 2, 3, 5 DOF (3) : Common for Members 1, 4 67 DOF (4) : Member 2 ONLY INTERNAL HINGE Terms corresponding to DOF (4) will get contribution from member 2 only. Take care of the DOF in code number approach format

Association matrices:

1 14 2     15 4 16 A[2]    A[4]    P KPP KPX uP 5  1  7x1 7x7 7x1 6  2  =     X KXX u 7  3  x   12x1KXP 12x12 12x1 68 INTERNAL HINGE 12WITH UDL 9 P = 0 8 11 1 10 13 P2 = -wL/2 2 6 1 3 P3 = 0 1 2 2 P4 = -wL /12 7 5 P5 = 0 3 4 Influence of 15 5 18 P = -wL/2 4 UDL absorbed 6 by (4) only 19 2 14 16 17 P7 = +wL /12

wL wL W wL2 2 2 wL2 12 L 12 69 INTERNAL HINGE 12WITH UDL 9 P = 0 8 11 1 10 13 P2 = -wL/2 2 6 1 3 P3 = 0 1 2 2 P4 = -wL /12 7 5 P5 = 0 3 4 Influence of 15 5 18 P = -wL/2 4 UDL absorbed 6 by (4) only 2 14 17 P7 = +wL /12

16 19 ALTERNATE APPROACH ONE wL SIDE HINGED…… 2 wL W wL2 2 2 wL2 12 L 12 70 INTERNAL HINGE 12WITH UDL 9 P = 0 8 11 1 10 13 P2 = -3wL/8 2 6 1 3 P3 = 0 1 2 P4 = -0 7 5 Influence of P5 = 0 3 4 UDL absorbed 15 4 5 18 P = -5wL/8 by (4) only 6 19 2 14 F17 P7 = +wL /8 { f }  {K L}{d}{ f } ALTERNATE 2 W 6 APPROACH ONE 1 SIDE HINGED…… 4 7 L 5 71 HINGE THROUGH COLUMN AND BEAM

3 5

4

Independent DOF for all members meeting at the joint, rest of the procedure same. 72 INTERNAL HINGE : ALTERNATE

APPROACH- TO MODIFY [K]L

2 5 1 4 3 6 Permanent hinge This boundary condition is not to be altered while deriving the member stiffness matrix Hinge not to be fixed while deriving the member stiffness matrix 73 INTERNAL HINGE : ALTERNATE

APPROACH- TO MODIFY [K]L EA L 3EI 0 L3 [K] = 0 0 0 L EA EA  0 0 L L 3EI 3EI 0  0 0 L3 L3 3EI 3EI 3EI 0 0 0  L2 L2 L

3rd row and 3rd column: All terms zero 74 Let us derive second column ALTERNATE APPROACH: INTERNAL HINGE

h

L

h

Use the code number approach, but here there is no duplication of the DOF as in the earlier approach. 75 8 11 7 10 9 12

2 1 5 3 1 2 4

3 4 KPP KPX 6 14 4 5 17 6x6 13 16 15 KXX 18 KXP 2 12x12 Automatically, member 2 will not make any contribution in the third row or column of KTS. Members 1 and 4 will make contribution as before The displacement corresponding to DOF 3 remains unknown for member 2. Corresponding displacement of the column can be obtained 76 ALTERNATE APPROACH: HINGE THROUGH BEAM AND COLUMNS BOTH

h

L

h

77 ALTERNATE APPROACH: HINGE THROUGH BEAM AND COLUMNS BOTH 8 11 7 10 9 12

2 1 6 3 1 2

5 4 3 14 4 5 17 18 13 16 15

No duplication of DOF as before…… 78 IMPLICATIONS

All three members 1, 2 and 3 have modified [K]L

After KTS is formed, we will find the third row and third column to be zero. WHY????

The diagonal element of KTS (3,3) shall be ZERO. This would imply the matrix to be singular, |KTS | =0, hence, we will encounter run time error. To circumvent this situation, eliminate the DOF (3). Renumber the DOFs and skip numbering this DOF.

79 ALTERNATE APPROACH: HINGE THROUGH BEAM AND COLUMNS BOTH 7 10 6 9 8 11

2 1 4 3 1 2

3 5

13 4 5 16 17 12 15 14

Need to skip the DOF corresponding to rotation 80 (displacement output will be devoid of the values of these) PRACTICE EXERCISE <1>

L

L

81 PRACTICE EXERCISE <1>

L

L

82 PRACTICE EXERCISE <1>

L

L

83 INTERNAL HINGE : ALTERNATE

APPROACH- TO MODIFY [K]L

2 5 1 4 3 6 Permanent hinge This boundary condition is not to be altered while deriving the member stiffness matrix Hinge not to be fixed while deriving the member stiffness matrix 84 PRACTICE EXERCISE <2> 500 kNm 500 kN L L

L

85 PRACTICE EXERCISE <3> h

L h

86 PRACTICE EXERCISE <4> 200 kN 15 kN/m 15 kN/m = = 4m 4m 4m

87 PRACTICE EXERCISE <4> 200 kN 15 kN/m 15 kN/m = = 4m 4m 4m

88 PRACTICE EXERCISE <5>

L

L

K

89 P KPP up 6x1 6x6 6x1 = X KXX ux 6x1 6x6 6x1

 X Alternate approach: u  11 11 k Add one more member (a link element)

90 TREATMENT OF NON-PRISMATIC MEMEBERS Case I: Determinate Structures

P



For a determinate structure, both member end forces as well as deflections can be directly calculated….. 91 SOLUTION BY PRINICIPLE OF VIRTUAL WORK 1 P 

BMD for the M (x) actual load

BMD for unit (virtual) load at m(x) the point of displacement Int. Virtual Work = Ext. Virtual Work M (x)m(x) To take care of the non-prismatic 1.  dx nature of the member  EI Conclusion: For determinate structures, both member end forces and deflections can 92 be easily computed by incorporating the variation of EI TREATMENT OF NON-PRISMATIC MEMEBERS Case II: Indeterminate Structures

How to determine KL?? Any issue? 93 TREATMENT OF NON-PRISMATIC MEMEBERS Case II: Indeterminate Structures 2 5 3 1 4

6 In order to derive stiffness matrix, as per first principles, we need to apply unit displacement along a particular DOF keeping all other displacements zero Basic slope-deflection formulations no longer valid. Use of classical force method too tedious.. M Θ = 1 ??? Being indeterminate, principle of 94 Virtual Work cannot be applied TREATMENT OF NON-PRISMATIC MEMEBERS Case II: Indeterminate Structures

How to determine KL?? Solution? We have to use indirect approach, employing flexibility method 95 WE WILL UTILIZE THE FLEXIBILITY METHOD

Fij = Displacement along the line of action of the ith force when we apply unit force along the line of action of the jth force….such that… (no force acting along the lines of action of other designated forces=> no restraint)

Unlike the stiffness approach (which emphasizes on locking remaining displacements), this process creates a determinate structure… How this is done?....see the next step. 96 USE OF FLEXIBILITY METHOD

F63 F33 1

Apply unit force along “3”, no force to be applied along other force lines.

Hence, no other force is generated, except reactions. The structure is determinate, so that we may easily apply the principle of virtual work

97 APPLY PRINCIPLE OF VIRTUAL WORK

F63 Real system 1 F33

1 M (x) Virtual 1 system 1

Int. Virtual Work = Ext. Virtual Work 1 m1(x)

L L 2 M ( x ) m ( x ) [m 1 ( x )] 1 .F  dx F33  dx 33  EI  EI 0 0 98 APPLY PRINCIPLE OF VIRTUAL WORK

F63 Real system 1 F33

1 M (x)  m1 (x) 1 Virtual system 2

1 m2 (x) L L M (x)m(x) m1(x)m2 (x) 1.F63  dx F  dx  EI 63  0 0 EI 99 SIMILARLY….

F66 Real 1 system F36

L 2 [m2 (x)] F36  F66  dx L  EI m (x)m (x) 0 1 2 dx  EI 0 100 BUT THIS IS NOT WHAT WE FINALLY WANT

Can we use superposition??

101 WE USE THE PRINCIPLE OF SUPERPOSITION TO GET FINAL SOLUTION F 66 1 F F63 F 1 33 36 A B Combine (A) and (B) in following fashion:

 F63  A   B  F66 

102 1

 F36 F63   F63  F33      F66   F66 

Hence, by definition

1 F K   66 33   2 F36 F63 F F  F F33   33 66 36  F66 

103 Similarly, F K  33 66 2 F33F66  F36 F K  63 63 2 F33F66  F36

104 HOW TO DERIVE K22 K22 INDETERMINATE 1

CONSIDER COMBINATION OF

Determine by K θ 1  virtual work method 33  A B Choose multipliers such that the net angle of

rotation on the left end of the beam is ZERO 105 OTHER ELEMENTS OF [K]L

K22

1

K62

Similarly derive K55, K35

106 HOW TO DERIVE K11 V=1 P Δ

External Virtual Work = External Virtual Work L 1.   FV  Rdx From actual 0 loads L  FR   FV  dx   AE  0 107 NON-PRISMATIC MEMEBERS How to obtain fixed ended forces?

Release the restraints and convert the structure into a determinate system Both rotations can θ be obtained using 1 θ2 the principle of virtual work

108 Superimpose the A, B, C:

θ1 θ 2 A

θ1 K63θ1 K33θ1 B

θ2 K66θ2 C

109 K36θ2 A+B+C

110 PRACTICE EXERCISE

2A A 2I I

Derive the stiffness matrix

111 INCLUSION OF SHEAR DEFORMATION EFFECT Necessary for deep sections L/D <= 6

Shear walls and lift cores 112 INCLUSION OF SHEAR DEFORMATION EFFECT

Treatment shall be restricted to prismatic sections only

In any deformed member, strain energy is given by

1 U  dV  2 V 113 WHY FORM FACTOR The shear stress varies across the height of the cross-section. A* y VQ b   N A Ib

Q(y)  y dA First moment of part of  cross-section above the A* section considered

114 To simplify the computation, shear stress is assumed to be uniform across the cross section, which is strictly not correct.

Form factor is introduced to apply correction for non-uniform shear stress, such that equivalent uniform stress gives same results as with actual non-uniform shear shear stress. V    S> 1  uniform  s   A   V  V  uniform       115  A/ s  Aeff FORM FACTOR (DEF.)

Form factor is defined as the ratio of the gross area of the section to the shear area of the section A s  S> 1 Aeff Alternately, the shear area of the member can defined as the area of the section which is effective in resisting shear deformation. 116 1 U  dV  2 V 1 M 2  1 F 2  U   dx   dx  2 EI   2 EA  L   L   1 V 2   s  dx L 2 GA  s = Form factor or shear correction factor or shear deformation coefficient 117 A  Q2  s   dA 2   2  I A  b 

A* y s = 1.2 b s = 10/9 N A s = 2

Practice: Find ‘s’ for a rectangular section bx2D

118 2 5 3 1 4 DEEP 6 1 F33

BMD m1(x) SFD 1/ L

F 66 1

m (x) BMD 2 SFD 1/ L 119 119 Applying the principle of Virtual Work, m (x)m (x) V (x)V (x) F  1 1 dx  s 1 1 dx 33   GA L EI L  x  1 m1(x)  1  V1(x)   L  L Similarly, m (x)m (x) V (x)V (x) F  1 2 dx  s 1 2 dx 63   L EI L GA

120 Similarly get F66 m (x)m (x) V (x)V (x) F  2 2 dx  s 2 2 dx 66   L EI L GA x 1 m (x)  V (x)  2 L 2 L and m (x)m (x) V (x)V (x) F  1 2 dx  s 1 2 dx 63   L EI L GA

121 m (x)m (x) V (x)V (x) F  2 2 dx  s 2 2 dx 66   L EI L GA

1 K33   F36 F63  F33    F66 

122 From slide18

In computer program, we need not store entire [T], we may simply store [R]

RT 0  K K  R 0 RT K R RT K R T A B  A B [k] = [T] [k] [T] =    T T  G L  T    0 R R K R R K R  0 R  K B KC     B C 

[KL]

123 124