Using the method of virtual work, determine the reaction at E

Using the method of virtual work, determine the reaction at E.

Firstly to solve, let us assume the support at E is released.

From similar triangles,

\( \frac{\delta Y_D}{2.1} = \frac{\delta Y_E}{1.2} \)

Or, \( \delta Y_D = 1.75 \delta Y_E \)

\( \frac{\delta Y_C}{2.7} = \frac{\delta Y_E}{1.2} \)

Or, \( \delta Y_C = 2.25 \delta Y_E \)

\( \frac{\delta Y_B}{0.5} = \frac{\delta Y_C}{0.9} \)

Or, \( \delta Y_B = 0.55 \delta Y_C = 1.25 \delta Y_E \)

From the principle of virtual work,

\( \delta U = 0 \)

Or, \( 2 \delta Y_B + 3 \delta Y_D - E \delta Y_E = 0 \)

Or, \( 2 \times 1.25 \delta Y_E + 3 \times 1.75 \delta Y_E - E \delta Y_E = 0 \)

Or, \( 7.75 \delta Y_E - E \delta Y_E = 0 \)

∴ \( E = 7.75 \text{ KN} \uparrow \)