Question

1. Determine the rotation at point 1 (i.e., Qi) due to support movement for the following rigid frame. 2 3 14 3ml PI To.06m 2m 2m * 0.03m

Answer 1

Equilibrium of Rigid body -

Total work done on the entire rigid body is zero, Therefore Virtual work done on each particle which is in equilibrium is zero.

Principle of virtual work -

Total work done on any ideal body which is in equilibrium by an active external force is zero.

2 4 3m 3 P 10.06m 2m 2m -Rx + 0.03m Try

Taking Moment about fixed point 1 -

Rx x 0.06 = Ry x 4.03

Rx = 67.2 Ry

Apply the principle of virtual work

Work done by 2 - ∂U2 = M ∂φ1

Work done by 5 in x direction (Fixed point) - ∂U5x = - Rx x 0.06 x ∂X

Work done by 5 in y direction (Fixed point) - ∂U5y = Ry x 4.03 x ∂Y

Total work-done = M ∂φ1 - Rx x 0.06 x ∂X + Ry x 4.03 x ∂Y= 0

M ∂φ1 - Rx x 0.06x ∂X + Ry x 4.03 x ∂Y= 0

M ∂φ1 = Rx x 0.06 x ∂X - Ry x 4.03 x ∂Y

∂φ1 = (Rx x 0.06 x ∂X - Ry x 4.03 x ∂Y)/ M

φ1 = (Rx x 0.06 x 4.03 - Ry x 4.03 x 0.06)/ M

From above Rx = 67.2 Ry

φ1 = (67.2 Ry x 0.06 x 4.03 - Ry x 4.03 x 0.06)/ M

φ1 = 16 Ry / M

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