192 CHAPTER 6 Matrix Methods
SubstitutingEq.(v)intoEq.(iii),weobtain
{
v 2
θ 2 L}
=
L^3
9 EI
[
− 4 − 2
− 23
]{
−W
M/L
}
(vi)fromwhichtheunknowndisplacementsatnode2are
v 2 =−4
9
WL^3
EI
−
2
9
ML^2
EI
θ 2 =2
9
WL^2
EI
+
1
3
ML
EI
Inaddition,fromEq.(v)wefindthat
θ 1 =5
9
WL^2
EI
+
1
6
ML
EI
θ 3 =−4
9
WL^2
EI
−
1
3
ML
EI
Itshouldbenotedthatthesolutionhasbeenobtainedbyinvertingtwo2×2matricesratherthanthe
4 ×4matrixofEq.(ii).ThissimplificationhasbeenbroughtaboutbythefactthatM 1 =M 3 =0.
TheinternalshearforcesandbendingmomentscannowbefoundusingEq.(6.50).Forthebeam
element1–2,wehave
Sy,12=EI(
12
L^3
v 1 −6
L^2
θ 1 −12
L^3
v 2 −6
L^2
θ 2)
or
Sy,12=2
3
W−
1
3
M
L
and
M 12 =EI
[(
12
L^3
x−6
L^2
)
v 1 +(
−
6
L^2
x+4
L
)
θ 1+
(
−
12
L^3
x+6
L^2
)
v 2 +(
−
6
L^2
x+2
L
)
θ 2]
whichreducesto
M 12 =
(
2
3
W−
1
3
M
L
)
x