sm12 74

________________________________________________________________________
PROBLEM (12.74) Applying the unit-load method, determine the horizontal displacement of point
C of the truss shown in Fig. P12.74. Each member has the same axial rigidity AE.
Given: P = 40 kips E = 30 x 106 psi A = 1.2 in.2 .
12.806 10 16
AC CB AB
LftLftL== ft=
SOLUTION
543
4
P
A
2
P
B
5
41
C
P
2
P
8 ft
16 ft
P
FAB
5
4
P
/2
F
AC
A
Joint A
41
0: 8
yAC
FF P==
∑
3
0: 8
xAB
FF==P
∑
5
8
BC
FP=−
Joint B
For a horizontal unit load applied at C:
35
88
AB BC AC
ff f==−=
41
8
Therefore,
1
Cjjj
f
FL
AE
δ =∑
3
6
12(40 10 ) 41 41
[( )( )(12.806)
1.2(30 10 ) 8 8
×
=×
55 33
( )( )(10) ( )( )(16)]
88 88
+− − +
=→ 0.191 .in
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