The slope-deflection equation at the end B of member BC for the frame shown in the given figure will be

M_{B C} = \frac{4 E I}{8} (2 θ_{C} - θ_{B})

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M_{B C} = \frac{4 E I}{8} (2 θ_{C} + θ_{B})

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M_{B C} = \frac{4 E I}{8} (2 θ_{B} + θ_{C})

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M_{B C} = \frac{4 E I}{8} (2 θ_{B} - θ_{C})

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Solution

The correct option is C $M_{B C} = \frac{4 E I}{8} (2 θ_{B} + θ_{C})$
$M_{B C} = \frac{2 (2 E I)}{8} (2 θ_{B} + θ_{C}) = \frac{4 E I}{8} (2 θ_{B} + θ_{C})$

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