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

SolutioN
We seek the horizontal deflection of the truss at joint D.
Because no external load acts horizontally at joint D,
a dummy load will be applied in the horizontal direction
at that joint. A free-body diagram of the truss
with load P applied at joint D is shown.
We perform a truss analysis, using the loadings
shown in the free-body diagram, to find the axial force in
each truss member. With these loads, the member forces
F will each be expressed as a unique function of P.
In the table that follows, expressions for each
member’s internal force in terms of the variable P are
listed in column (2). The partial derivative ∂F/ ∂P
is
listed in column (3). The actual force F in each member
is calculated by substituting the value P = 0 kips into each force expression listed in
column (2). The member forces are listed in column (4). Finally, the length of each member
is shown in column (5).
Castigliano’s second theorem applied to trusses is expressed by Equation (17.39).
For this particular truss, each member has a cross-sectional area A = 3.7 in. 2 and an elastic
modulus E = 29,000 ksi; therefore, the calculation process can be simplified when both A
and E are moved outside of the summation operation:
F
∆=
1 ∑
⎛ ∂ ⎞
⎜ ⎟
AE ⎝∂P ⎠
FL
For each truss member, the terms in columns (3), (4), and (5) are multiplied together
and recorded in column (6). Here is the table with all results shown:
(1) (2) (3) (4) (5) (6)
Member
F
(kips)
∂F
∂P
F
(for P = 0 kips)
(kips)
L
(ft)
⎛∂F
⎞
⎜
⎝ ∂P ⎠
⎟ FL
(kip ⋅ ft)
AB 0.447P + 48.3 0.447 48.3 17.9 386.46
AC 0.8P - 21.6 0.8 -21.6 20 -345.60
BC -0.778P - 84.0 -0.778 -84.0 20 1,307.04
BD 0.703P + 75.9 0.703 75.9 25.3 1,349.95
CD 0.401P - 86.5 0.401 -86.5 14.4 -499.49
⎛ F
∑ ∂ ⎞
⎜ ⎟
⎝∂P ⎠
FL = 2,198.36
Now apply Equation (17.39) to compute the horizontal deflection of joint D from the
tabulated results:
(2,198.36 kip⋅ft)(12in./ft)
D D =
2
(3.7 in. )(29,000 ksi)
= 0.246 in. → Ans.
Since the dummy load was applied rightward at D, the positive value of the result
confirms that joint D is in fact displaced to the right.
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