Problem 34.3 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 100 kN. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If the horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q.

The image is missing, so i have attached it.

A) P = 65.11 KN

B) Q = 30 KN

Explanation:

We are given;

The end reaction of the beam; F = 100KN

Coefficient of static friction between two steel surfaces;μ_ss = 0.3

Coefficient of static friction between steel and concrete;μ_sc = 0.6

So, F1 = μ_ss•F =0.3 x 100 = 30 KN

F2 = μ_ss•N_EF = 0.3N_EF

From the screen shot, we see that the angle is 12°

Sum of forces in the Y-direction gives;

F2•sin12 - N_EF•cos12 + 100 = 0

Rearranging gives;

N_EF•cos12 - F2•sin12 = 100

Let's put 0.3N_EF for F2 to give;

N_EF•cos12 - 0.3N_EF•sin12 = 100

Thus;

N_EF(0.9158) - 0.1247 = 100

N_EF(0.9781) = 100 + 0.1247

N_EF = 100.1247/0.9158

N_EF = 109.33 KN

Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN

Wedge will move if;

P = (F1 + F2cos12 + N_EFsin12)

Thus;

P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)

P ≥ 65.11 KN

B) For static equilibrium, Q = F1

Thus, Q = 30 KN

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