Beam Analysis by the Direct Stiffness Method Steven Vukazich San Jose State University Beam Element Stiffness Matrix in Local Coordinates
Consider an inclined beam member with a moment of inertia I and modulus of elasticity E subjected to shear force and bending moment at its ends.
i = initial end of element θ j = terminal end element j x’ axis (local 1 axis in SAP 2000)
Vj x’ Mj y Δj y’ θi j Note the sign convention Vi EI
Mi Δi L i
φ" x Each Column of the Frame Element Matrix in Local Coordinates is derived from indeterminate fixed end moment solutions First Column of the Frame Element Matrix in Local Coordinates set Δi = 1 6EI x M j = 2 and θ = θ = Δ = 0 = 0 L
i i j θj = 0 Δi y y 12EI j V = − j L3 θ = 0 12EI i
V = i L3 =1 Δi 6EI L M = i L2 i
φ
x
Frame Element Matrix in Local Coordinates
) 12EI 6EI 12EI 6EI , + 3 2 − 3 2 . + L L L L . ! V % ! Δ % # i # + 6EI 4EI 6EI 2EI .# i # + 2 − 2 . # Mi # L L L L # θi # " & = + ." & Vj + 12EI 6EI 12EI 6EI . Δ j # # − − − # # + 3 2 3 2 . # M # L L L L # θ # $ j ' + .$ j ' + 6EI 2EI 6EI 4EI . 2 − 2 *+ L L L L -. Beam Structure System of Equations from Element Contributions Consider the beam comprised of two elements Note that since the x’ and x axes are the same the local and global DOF are in the same directions and so [k’] = [k] Each beam joint can move in two directions 1. Number each 2 Degrees of Freedom (DOF) per joint element • Number each joint; • Number each structure DOF.
2. Joint and DOF numbering Beam Structure Connectivity Table to Assemble Assembly of Beam Structure Stiffness Matrix from Element Contributions
Label the rows and columns of each 4x4 element stiffness matrix with corresponding structure DOF from connectivity table;
Assemble 6x6 Structure stiffness matrix from element contributions Structure System of Equations: Restrained and Free DOF DOF 3, 4 and 6 are free DOF; At free DOF (grey), DOF 1, 2, and 5 are restrained (support) DOF we know the forces (applied joint loads) but the displacements are unknown At restrained DOF (yellow), we know the displacements but the forces (support reactions) are unknown First Solve Free DOF Equation set to find unknown displacements solve free DOF equation ! $( , ( , K33 K34 K36 Δ3 −P # &* * * * set to find Δ3, θ4, and θ6 # K43 K44 K46 &) θ4 - = ) 0 - # & K K K * * * 0 * "# 63 64 66 %&+ θ6 . + . Find Support Reactions
DOF 3, 4 and 6 are free DOF; DOF 1, 2, and 5 are restrained (support) DOF
Δ3, θ4, and θ6 found from previous step After displacements are found, multiply to find unknown forces ! V % ( K K 0 +! Δ % (support reactions) # 1 # * 13 14 -# 3 # " M 2 & = * K23 K24 0 -" θ4 & * - # V # K K K # # $# 5 '# )* 53 54 56 ,-$ θ6 ' Lab Problem
y, y ’ Work in units of kips and inches
12 k 1 3 5
! 2 4 6 ! ! ! ! ! ! ! ! ! x, x’ 1 ! ! 1! 2! 2 3! 12 ft 6 ft
(144 in) (72 in) For the beam shown the properties of the elements are:
Member Section I E 1 W8x10 30.8 in4 29000 ksi 2 W8x10 30.8 in4 29000 ksi ! Construct the 4x4 Element Stiffness Matrix for Each Element
# 12EI 6EI 12EI 6EI & % 3 2 − 3 2 ( % L L L L ( % 6EI 4EI 6EI 2EI ( % 2 − 2 ( L L L L [k] = [k!] = % ( % 12EI 6EI 12EI 6EI ( − − − % L3 L2 L3 L2 ( % ( % 6EI 2EI 6EI 4EI ( 2 − 2 $% L L L L '( Assemble the 6x6 Structure Stiffness Matrix from the 4x4 Element Stiffness Matrices and Connectivity Table Solve the free DOF Equation Set to find the Unknown Displacements
Multiply to verify that the –P = –12 solution is: ∆ = ! −1.238155!!" ! $( , ( , ! K K K36 Δ P # 33 34 &* 3 * − * * !! = !0.0051590!!"# # K43 K44 K46 &) θ4 - = ) 0 - # &* * * * K K K θ 0 !! = !0.0232154!!"# "# 63 64 66 %&+ 6 . + . Find the Support Reactions and Verify Equilibrium
1. Find Support Reactions
(V1, M2, V5) from the joint displacement results and matrix multiplication;
2. Verify your results for V1, M2, and V5 satisfy equilibrium using statics. ! V % ( K K 0 +! Δ % # 1 # * 13 14 -# 3 # ∆ = ! −1.238155!!" ! " M 2 & = * K23 K24 0 -" θ4 & * - # # # # !! = !0.0051590!!"# V5 K53 K54 K56 θ6 $# '# )* ,-$ '
!! = !0.0232154!!"#