WIXLIB

Beam Analysis by the DirectStiffness MethodSteven VukazichSan Jose State University

Stiffness Based ApproachCan think of a structure acting as a multi-dimensional spring&Δ' ()Δ!*Δ*!)! &Δ!Δ, (! !-( !'!,! #

Beam Element Stiffness Matrix in Local CoordinatesConsider an inclined beam member with a moment of inertia I and modulus ofelasticity E subjected to shear force and bending moment at its ends.i initial end of elementj terminal end element-Note the sign convention*"!"x axis (local 1 axis in SAP 2000)! # #"Δ" ()',* We want to findΔ this 4x4 matrix%. 0 1

Each Column of the Frame Element Matrix in Local Coordinates isderived from Indeterminate Fixed End Moment Solutions4# %6# )% 6# )%2# %# %6# Δ)%12# Δ&%12# Δ&%Δ# %6# Δ)%

Find the First Column of the Frame ElementStiffness Matrix in Local Coordinatesset Δi 1and θi θi Δj 086,0* 3.Δ* 112,)* /.5* 06,.6,01 3.Δ1 07451 012,)1 /.

Frame Element Stiffness Matrix%&'&%('( 12-.6-. 12-. 6-. 002////26-.4-.6-. 2-. 22////12-. 6-. 12-. 6-. 0 2 20////6-.2-.6-. 4-. 22////! # &*& (*(

Structure Stiffness MatrixConsider a beam comprised of two elementsEach beam joint can move in two directions: 2Degrees of Freedom (DOF) per jointy31254EI1EI2L1L2! # 6xThe 6x6structurestiffness matrixcan be assembledfrom the elementstiffness matrices

Assemble Structure Stiffness Matrix1. Number eachelementy121n3 EI2 54 6EI11222nx312L12n - 12. Joint and DOFnumbering3EIi4L2Li3. Connectivity table to assemblestructure stiffness matrixElement DOF1234Associated global DOF for element 11234Associated global DOF for element 23456

Assembly of Structure Stiffness Matrixfrom Element ContributionsForm 4x4 element stiffness matrixfor element 1 from EI1 and L1Label the rows and columns of each 4x4element stiffness matrix withcorresponding structure DOF fromconnectivity tableAssemble 6x6 structure stiffness matrixForm 4x4 element stiffness matrixfor element 2 from EI2 and L2

Structure System of Equations: Free DOF12DOF 3 are 4 are free DOF;3 EI2 54 6EI1xDOF 1, 2, 5,and 6 are restrained (support) DOF("!""!"#!" !"%1!L"&!"'Δ")#!#"!##!# !#%!#&!#' #! "! #! ! %! &! 'Δ !%"!%#!% !%%!%&!%' %(&!&"!&#!& !&%!&&!&'Δ&)'!'"!'#!' !'%!'&!'' '( )% At free DOF (blue), we know the forces (applied jointloads) but the displacements are unknownAt restrained DOF, we know the displacements but the forces(support reactions) are unknown

Free DOF System of EquationsSuppose we have loads applied to joint 2V31DOF 3 are 4 are free DOF;M423465DOF 1, 2, 5,and 6 are restrained(support) DOFx("!""!"#!" !"%!"&!"'0)#!#"!##!# !#%!#&!#'0! "! #! ! %! &! 'Δ !%"!%#!% !%%!%&!%',%(&!&"!&#!& !&%!&&!&'0)'!'"!'#!' !'%!'&!''0( )% At free DOF (blue), we know the forces (applied jointloads) but the displacements are unknownAt restrained DOF, we know the displacements (all equal to zero)but the forces (support reactions) are unknown

Solve for Displacements at Free DOFV312M43465xFree DOF Equation Set!""!#"!"# "'" !## %#(#Solve for " and %#

Find Support ReactionsDOF 1, 2, 5 and 6 are restrained (support) DOF("!""!"#!" !"%!"&!"'0)#!#"!##!# !#%!#&!#'0! "! #! ! %! &! 'Δ !%"!%#!% !%%!%&!%',%(&!&"!&#!& !&%!&&!&'0)'!'"!'#!' !'%!'&!''0( )% After displacements arefound, multiply to findunknown forces (supportreactions)Δ3 and θ4, foundfrom previous step(")#(&)' !" !"%Δ !# !#%,%!& !&%!' !'%

Consider an inclined beam member with a moment of inertia Iand modulus of elasticity E subjected to shear force and bending moment at its ends. x axis (local 1 axis in SAP 2000) i initial end of element j terminal end element Note the sign convention Beam Element Stiffness Matrix in Local Coordinates!" #" # ! % Δ" Δ ' *" * ,-. 01 We .