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 21 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 31 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 0F1 f4 cos sin 0F4 f2 sin cos 0 F2 f sin cos 0 F 5 5 f3 0 0 1F3 f6 0 0 1F6 16 f1 cos sin 0F1 F1 f2 sin cos 0 F2 F 1 R 2 f3 0 0 1F3 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 0d 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 0F1 f4 cos sin 0F4 f2 sin cos 0 F2 f sin cos 0 F 5 5 f3 0 0 1F3 f6 0 0 1F6 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).