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

elastic modulus E = 200 GPa; therefore, the calculation process can be simplified by moving
both A and E outside of the summation operation:
1 ⎛∂F
⎞
∆= ∑⎜
⎟
AE ⎝∂P
⎠
FL
For each truss member, multiply the terms in column (3) by those in columns (4)
and (5), and record the result in column (6). Then, sum these values for all of the members.
Here is the table with all results shown:
(1) (2) (3) (4) (5) (6)
Member
F
(kN)
∂F
∂P
F
(for P = 75 kN)
(kN)
L
(m)
⎛∂F
⎞
⎜
⎝ ∂P ⎠
⎟ FL
(kN ⋅ m)
AB -1.25P -1.25 -93.75 7.5 878.91
AC 0.75P 0.75 56.25 4.5 189.84
BC 1.00P 1.00 75.00 6.0 450.00
BD -0.75P -0.75 -56.25 7.5 316.41
CD -1.60P - 200.10 -1.60 -320.10 9.605 4,919.30
CE 2.00P + 156.25 2.00 306.25 7.5 4,593.75
DE 1.00P + 125.00 1.00 200.00 6.0 1,200.00
⎛ F
∑ ∂ ⎞
⎜ ⎟
⎝∂P ⎠
FL = 12,548.20
Now apply Equation (17.39) to compute the deflection of joint A from the tabulated
results:
(12,548.20 kN⋅m)(1,000 N/kN)(1, 000 mm/m)
D A =
2 2
(1,100 mm )(200,000 N/mm )
= 57.0 mm ↓ Ans.
Since the load P was applied in a downward direction at A, the positive value of the
result confirms that joint A is in fact displaced downward.
ExAmpLE 17.17
Calculate the horizontal deflection at joint D for the truss shown.
For all members, the cross-sectional area is A = 3.7 in. 2 and the
elastic modulus E = 29,000 ksi.
Plan the Solution
The horizontal deflection is to be determined at joint D. Since there
is no external load in the horizontal direction at D, a dummy load P
will be required. Apply such a load and include it in the truss analysis.
Follow the procedure outlined in Example 17.16. However,
when calculating the actual member force, substitute the value
P = 0 kips in the member-force expressions. Finally, apply Equation
(17.39) to determine the horizontal deflection D of joint D.
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