Explain briefly how did we get s=sP-sFD
Chapter2: Loads On Structures
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Problem 1P
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Explain briefly how did we get s=sP-sFD

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4-46. If the gap between C and the rigid wall at D is
initi ly 0.15 mm, determine the support reactions at A and
D when the force P = 200 kN is applied. The assembly
-600 mm
600 mm
-0.15 mm
P.
is made of A36 steel.
25 mm
50 mm
Equation of Equilibrium: Referring to the free-body diagram of the assembly
shown in Fig.a,
EF, = 0;
200(10°) – FD - FA = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
(+)
8 = Sp – dF,
FD (600)
200(10°)(600)
(0.052)(200)(10°)
FD (600)
0.15 =
(0.05)(200)(10°) "{(0.025°)(200)(10°)
Ans.
Fp = 20 365.05 N = 20.4 kN
Substituting this result into Eq. (1),
100CO OF N
1001-AT
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