Question

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 18 kips. 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

Answer 1

Answer:

a) P ≥ 22.164 Kips

b) Q = 5.4 Kips

Explanation:

GIven

W = 18 Kips

μ₁ = 0.30

μ₂ = 0.60

a) P = ?

We get F₁  and F₂ as follows:

F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips

F₂ = μ₂*Nef = 0.6*Nef

Then, we apply

∑Fy = 0   (+↑)

Nef*Cos 12º -  F₂*Sin 12º = W

⇒   Nef*Cos 12º -  (0.6*Nef)*Sin 12º = 18

⇒   Nef = 21.09 Kips

Wedge moves if

P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º

⇒  P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º

⇒  P ≥ 22.164 Kips

b) For the static equilibrium of base plate

Q = F₁ = 5.4 Kips

We can see the pic shown in order to understand the question.

Answer 2

Answer:

a; P = 15.25 kips

b) Q = 5.4 kips

Explanation:

To understand scenerio, see attached image also.

Consider the forces acting on the system, F1 is the force acting against the direction of P, F2 acting in the along direction of P, N is at angle 12 degrees

Calculating F1, we have;

F1 = µ x 18 = 0.30 x 18 = 5.4 kips

F2 = = µ x N = 0.3N

To be in static condition, sum of forces would be equal to 0

18 = N cos 12 – F2 sin 12

Putting value of F2 and solving above equation for N;

N x 0.915 = 18

N = 19.65 kips

Now the wedge will move if, P > ( F1  + F2 cos 12 + N sin 12)

Putting the values;

P >= 15.25 kips

The corresponding force Q will be equal to F1 = 5.4 kips