For the Beam shown below,( Assume there is no sink of support C)

For the Beam shown below,( Assume there is no sink of support C) Which of the following is/are true?

A. Moment at support A is 135 kN-m

B. Moment at support C is 180 kN-m

C. Ratio of moment at A to the moment at C is 0.75

D. Ratio of moment at A to the moment at C is 1.33

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Calculation

Given,

Support A is fixed θ_A = 0°

CD is free at end D ∴ BM at C = 60 × 3 = 180 kN-m (Clockwise)

Fixed end moments

\({F_{AB}} = - \frac{{PL}}{8} = - \frac{{120 \times 6}}{8} = - 90\;kN - m\)

\({F_{BA}} = \frac{{PL}}{8} = \frac{{120 \times 6}}{8} = 90\;kN - m\)

From slope deflection equation

As there is no sink of member deflection Δ = 0

\({M_{AC}} = {F_{AC}} + \frac{{2EI}}{L} \times \left( {2{\theta _A} + {\theta _C}} \right)\)

\({M_{AC}} = - 90 + \frac{{2EI}}{L} \times \left( {{\theta _C}} \right)\) –-(I)

\({M_{CA}} = {F_{CA}} + \frac{{2EI}}{L} \times \left( {2{\theta _C} + {\theta _A}} \right) + 180\)

\( - 180 = 90 + \frac{{2EI}}{L} \times \left( {2{\theta _C}} \right)\)

\( - 90 = \frac{{2EI}}{L} \times \left( {2{\theta _C}} \right)\)

\(- \frac{{90}}{2} = \frac{{2EI}}{L} \times \left( {{\theta _C}} \right)\) ---(II)

(II) in (I)

\({M_{AC}} = - 90 - \frac{{90}}{2} = - 135kN - m\)

∴ The moment at support A is 135 kN-m

(Or)

Using Moment Distribution Method