Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Following is the table produced by the preceding instructions for the truss considered
here:
Member
L
(mm)
A
(mm 2 )
F
(kN)
f
(kN)
⎛ FL ⎞
f ⎜ ⎟
⎝ A ⎠
(kN 2 /mm)
AB 3,000 1,050 -9.375 0.500 -13.393
AF 5,000 1,050 15.625 0.833 61.979
BC 3,000 1,050 -9.375 0.500 -13.393
BF 4,000 750 75.000 0 0
CD 3,000 1,050 -18.750 0 0
CF 5,000 750 -109.375 -0.833 607.396
CG 5,000 750 -93.750 0 0
DE 3,000 1,050 -18.750 0 0
DG 4,000 750 50.000 0 0
EG 5,000 1,050 31.250 0 0
FG 6,000 1,050 75.000 1.000 428.571
FL
∑ f
⎛ ⎝ ⎜ ⎞
⎟ = 1,071.161
A ⎠
Equation (17.30) can now be applied:
1⋅D =
∑
j
⎛ FL ⎞
j j
f j ⎜
⎝ E
⎟ =
j ⎠
1
E
∑
j
⎛ FL ⎞
j j
f j ⎜
⎝ A
⎟
j ⎠
From the tabulated results,
2
(1,071.161 kN /mm)(1, 000 N/kN)
(1 kN) ⋅D G =
2
(70,000 N/mm )
D = 15.30 mm → Ans.
G
Since the virtual load was applied horizontally to the right at G, the positive value of
the result confirms that joint G does become displaced to the right.
(b) Vertical Deflection of Joint G: Again, remove all loads from the truss. The vertical
deflection of the truss at joint G is to be determined next; therefore, apply a vertical
virtual load of 1 kN at joint G. For this analysis, the virtual load will be directed
downward. Perform a third analysis of the truss in which the vertical virtual load is
the only load acting on the truss. This analysis produces a different set of virtual internal
forces f. Replace the previous values of f with these results, and recalculate the
final column of the table:
760