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A symmetric frame PQR consists of two inclined members PQ and QR, connected at ‘Q’ with a rigid joint, and hinged at ‘P’ and ‘R’. The horizontal length PR is l. If a weight W is suspended at ‘Q’, the bending moment at ‘Q’ is
  • a)
    Wl/2
  • b)
    Wl/4
  • c)
    Wl/8
  • d)
    zero
Correct answer is option 'D'. Can you explain this answer?
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Their common endpoint Q. The frame is supported by a hinge at P and a roller at R. The lengths of the members PQ and QR are 4 ft and 6 ft, respectively.

To find the reactions at the hinge and roller, we can use the method of joints.

First, let's consider the forces acting at point Q. We have two unknown forces: the horizontal reaction at the hinge, H_P, and the vertical reaction at the roller, V_R. Since the frame is symmetric, we can assume that these forces are equal in magnitude but opposite in direction.

Applying the equilibrium equations in the x-direction:

ΣF_x = 0
H_P = 0

Therefore, the horizontal reaction at the hinge, H_P, is zero.

Now, let's consider the forces acting at point P. We have three unknown forces: the horizontal reaction at the roller, H_R, the vertical reaction at the hinge, V_P, and the force in member PQ, F_PQ. Again, due to symmetry, we can assume that the vertical reactions at P and Q are equal in magnitude but opposite in direction.

Applying the equilibrium equations in the x-direction:

ΣF_x = 0
H_R - F_PQ = 0
H_R = F_PQ

Applying the equilibrium equations in the y-direction:

ΣF_y = 0
V_P + V_Q - F_PQ = 0
V_P + (-V_R) - F_PQ = 0
V_P - V_R = F_PQ

Since we know that V_R is equal in magnitude but opposite in direction to V_P, we can rewrite the equation as:

V_P + (-V_P) = F_PQ
0 = F_PQ

Therefore, the force in member PQ, F_PQ, is zero.

Now, let's consider the forces acting at point R. We have two unknown forces: the vertical reaction at the roller, V_R, and the force in member QR, F_QR. Again, due to symmetry, we can assume that the vertical reactions at P and Q are equal in magnitude but opposite in direction.

Applying the equilibrium equations in the y-direction:

ΣF_y = 0
(-V_R) - F_QR = 0
V_R = -F_QR

Now, we can substitute the known values into the equation:

V_R = -F_QR
V_R = -(6 ft)
V_R = -6 ft

Therefore, the vertical reaction at the roller, V_R, is equal to -6 ft.

In summary, the reactions at the hinge and roller are as follows:

Hinge reaction at P, H_P = 0 ft
Vertical reaction at the hinge, V_P = -6 ft
Vertical reaction at the roller, V_R = -6 ft
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A symmetric frame PQR consists of two inclined members PQ and QR, connected at ‘Q’ with a rigid joint, and hinged at ‘P’ and ‘R’. The horizontal length PR is l. If a weight W is suspended at ‘Q’, the bending moment at ‘Q’ isa)Wl/2b)Wl/4c)Wl/8d)zeroCorrect answer is option 'D'. Can you explain this answer?
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