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

F 1
(1)
B
SolutioN
The internal axial forces in members (1) and (2) can be calculated from equilibrium equations
based on a free-body diagram of joint B. The sum of forces in the horizontal (x)
direction can be written as
42.61°
Σ Fx = −F1 − F2cos 42.61°=
0
(2)
and the sum of forces in the vertical (y) direction can be expressed as
F 2
50 kN
Σ F = − F sin 42.61° − 50 kN = 0
∴ F
y 2
2
= −73.85 kN
Substituting this result back into the preceding equation gives
F
1 =
54.36 kN
The strain energy in tie rod (1) is
U
1
F1 2 L
2 2
1 (54.36 kN) (1.25 m)(1,000 N/kN)
= = = 14.2068 Nm ⋅
2 2
2A E 2(650 mm )(200,000 N/mm )
1
The length of inclined pipe strut (2) is
Thus, its strain energy is
2 2
L = (1.25 m) + (1.15 m) = 1.70 m
2
U
2
F2 2 L
2 2
2 ( 73.85 kN) (1.70 m)(1,000 N/kN)
= = - = 25.0581 Nm ⋅
2 2
2A E 2(925 mm )(200,000 N/mm )
2
The total strain energy of the two-bar assembly is, therefore,
U = U1 + U2
= 14.2068 Nm ⋅ + 25.0581 Nm ⋅ = 39.2649 Nm ⋅
The work of the 50 kN load can be expressed in terms of the downward deflection D
of joint B as
1
W = 2
(50 kN)(1,000 N/kN) D = (25,000 N) D
From the conservation-of-energy principle, W = U; thus,
(25,000 N) D= 39.2649 Nm ⋅
-3
∴D= 1.571 × 10 m = 1.571 mm
Ans.
Compare this calculation method with the method demonstrated in Example 5.4, in
which the same two-member assembly was considered. By the work–energy method, the
downward deflection at B can be determined in a much simpler manner. However, the
work–energy method cannot be used to determine the horizontal deflection of B.
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