For a fixed beam loaded as shown below, if the support, B rotates + θB radian anticlockwise, the fixed end moment at ‘B’ is

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Option 2 :
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Explanation:

For the given fixed beams:

Fixed end moments can be thought of as the superposition of moments induced due to uniformly distributed load and due to rotational slip of end B calculated separately.

Thus net fixed end moment at B –

MBA = MBA1 – MBA2 {Taking clockwise as positive}

Calculation for moment uniformly distributed load:-

By using standard result,

Calculation for moment due to rotational slip:-

Compatibility Equation:

θB = -θ0

⇒ On conjugate Beam - S.F at B = - θ0  

δB = 0

⇒ On conjugate Beam - B.M at B = 0

From BMD

         ---(i)

&

⇒ M1 = 2M2

From (1)

Thus MBA2 = M1 as explained above

Hence

∵ Our assumed direction is not similar to those given in the option hence the adopted sign convention must be opposite to that given. Adopting proper sign convention.

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