If the gap between c and the rigid wall at d is initially 0.15 mm, determine the support reactions at a and d when the force p = 200 kn is applied. the assembly is made of solid a-36 steel cylinders.

F(A)= 180kN; F(B)= 20.4 kN.

Explanation:

Note: check the free-body diagram in the attached file/picture. Also, the question's diagram is include with the written solution.

Equation of equilibrium can be represented by equation (1) below;

ΣFx = 0----------------------------------------------------------------------------------------(1).

Therefore; 200×10^3 -F(B)- F(A) = 0

Also, the compability equation can be represented by using superposition method in equation (2) below;

d = δ(P) - δF(B)------------------------------------------------------------------------------(2).

Therefore, we have that;

0.15 = (200 × 10^3 × 600)/{(π/4) × 0.05^2 × 200 × 10^9} - F(B) × 600/π/4×200×0.05^2×10^9 + F(B) × 600/(π/4× 0.0252× 200 × 10^9.

F(B) = 20365.1N

F(B)= 20.4kN.

Substituting the value of F(B) into equation (1). We then have that,

F(A)= 179 634.95 N = 180 kN

answer:

a

explanation: