beam analysis by the direct stiffness method
Post on 09-Feb-2022
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ConsideraninclinedbeammemberwithamomentofinertiaI andmodulusofelasticityEsubjectedtoshearforceandbendingmomentatitsends.
x’axis(local1axisinSAP2000)i =initialendofelementj=terminalendelement
φ"
x’
y’ y
x
i
j
L
Δj
Δi
θi
θj
Vi
Vj Mj
EI
Mi
Note the sign convention
Beam Element Stiffness Matrix in Local Coordinates
Each Column of the Frame Element Matrix in Local Coordinates is derived from indeterminate fixed end moment solutions
φ
x
yy
x
i
j
L Δ
i =1
θi = 0
θj = 0
Δi = 0
Vi =12EIL3
M j =6EIL2
Mi =6EIL2
Vj = −12EIL3
set Δi = 1 and θi = θi = Δj = 0
First Column of the Frame ElementMatrix in Local Coordinates
ViMi
Vj
M j
!
"
##
$
##
%
&
##
'
##
=
12EIL3
6EIL2
−12EIL3
6EIL2
6EIL2
4EIL
−6EIL2
2EIL
−12EIL3
−6EIL2
12EIL3
−6EIL2
6EIL2
2EIL
−6EIL2
4EIL
)
*
++++++++++
,
-
.
.
.
.
.
.
.
.
.
.
Δi
θiΔ j
θ j
!
"
##
$
##
%
&
##
'
##
Frame Element Matrix in Local Coordinates
ConsiderthebeamcomprisedoftwoelementsNotethatsincethex’andxaxesarethesamethelocalandglobalDOFareinthesamedirectionsandso [k’] = [k] Eachbeamjointcanmoveintwo
directions2DegreesofFreedom(DOF)perjoint• Numbereachjoint;• NumbereachstructureDOF.
BeamStructureConnectivityTabletoAssemble
2.JointandDOFnumbering
1.Numbereachelement
Beam Structure System of Equations from Element Contributions
Labeltherowsandcolumnsofeach4x4elementstiffnessmatrixwithcorrespondingstructureDOFfromconnectivitytable;
Assemble6x6Structurestiffnessmatrixfromelementcontributions
Assembly of Beam Structure Stiffness Matrix from Element Contributions
DOF3,4and6arefreeDOF;
DOF1,2,and5arerestrained(support)DOF
AtrestrainedDOF(yellow), weknowthedisplacementsbuttheforces(supportreactions)areunknown
AtfreeDOF(grey),weknowtheforces(appliedjointloads)butthedisplacementsareunknown
FirstSolveFreeDOFEquationsettofindunknowndisplacementsK33 K34
K43
K63
K44
K64
K36
K46
K66
!
"
####
$
%
&&&&
Δ3θ4θ6
(
)*
+*
,
-*
.*=
−P00
(
)*
+*
,
-*
.*
solve free DOF equation set to find Δ3, θ4, and θ6
Structure System of Equations: Restrained and Free DOF
DOF3,4and6arefreeDOF;
DOF1,2,and5arerestrained(support)DOF
After displacements are found, multiply to find unknown forces (support reactions) V1
M2
V5
!
"#
$##
%
&#
'##
=
K13 K14 0K23 K24 0K53 K54 K56
(
)
****
+
,
----
Δ3θ4θ6
!
"#
$#
%
&#
'#
Δ3, θ4, and θ6 found from previous step
Find Support Reactions
Workinunitsofkipsandinches
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 !
y, y’
x, x’
!
12 k 3
4
2 2 !1 !3 !
!!
12 ft 6 ft
1 ! !
1
2 !!
6 !!
5
!!!
(144in) (72in)
Lab Problem
k[ ] = !k[ ] =
12EIL3
6EIL2
−12EIL3
6EIL2
6EIL2
4EIL
−6EIL2
2EIL
−12EIL3
−6EIL2
12EIL3
−6EIL2
6EIL2
2EIL
−6EIL2
4EIL
#
$
%%%%%%%%%%
&
'
((((((((((
Construct the 4x4 Element Stiffness Matrix for Each Element
Assemble the 6x6 Structure Stiffness Matrix from the 4x4 Element Stiffness Matrices and Connectivity Table
K33 K34
K43
K63
K44
K64
K36
K46
K66
!
"
####
$
%
&&&&
Δ3θ4θ6
(
)*
+*
,
-*
.*=
−P00
(
)*
+*
,
-*
.*
Multiplytoverifythatthesolutionis:
∆!= !−1.238155!!" !! = !0.0051590!!"# !! = !0.0232154!!"#
Solve the free DOF Equation Set to find the Unknown Displacements
–P = –12
1. Find Support Reactions (V1, M2, V5) from the joint displacement results and matrix multiplication;
2. Verify your results for V1, M2, and V5 satisfyequilibrium using statics.
V1M2
V5
!
"#
$##
%
&#
'##
=
K13 K14 0K23 K24 0K53 K54 K56
(
)
****
+
,
----
Δ3θ4θ6
!
"#
$#
%
&#
'#
∆!= !−1.238155!!" !! = !0.0051590!!"# !! = !0.0232154!!"#
Find the Support Reactions and Verify Equilibrium
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