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Castigliano: fixed-end moment, 15 kN/m, L = 5 m

Short answer

Using Castigliano's theorem with roller reaction as redundant, fixed-end moment M_A = 46.875 kNm (hogging). Roller reaction R_B = 28.125 kN, fixed-end reaction R_A = 46.875 kN.

Determine the moment at the fixed end of the propped cantilever beam shown in figure below using Castigliano's theorem. [Figure: beam, left end fixed, right end roller, span L = 5 m with UDL 15 kN/m.]

Propped cantilever: fixed at the left end A, roller at B, span L=5L = 5 m, UDL w=15w = 15 kN/m. Take the roller reaction RBR_B as the redundant and measure xx from B.

M=RBx−wx22=RBx−7.5x2,∂M∂RB=xM = R_Bx - \frac{wx^2}{2} = R_Bx - 7.5x^2,\qquad \frac{\partial M}{\partial R_B} = x

The deflection at B is zero:

∂U∂RB=1EI∫05(RBx2−7.5x3)dx=0\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^5\left(R_Bx^2 - 7.5x^3\right)dx = 0 RB533−7.5544=0 ⇒ 41.667 RB=1171.875 ⇒ RB=28.125 kN (=3wL8)R_B\frac{5^3}{3} - 7.5\frac{5^4}{4} = 0 \ \Rightarrow\ 41.667\,R_B = 1171.875 \ \Rightarrow\ R_B = 28.125\ \text{kN} \ \left(= \frac{3wL}{8}\right)

Moment at the fixed end

MA=RB(5)−7.5(5)2=140.625−187.5=−46.875 kNmM_A = R_B(5) - 7.5(5)^2 = 140.625 - 187.5 = -46.875\ \text{kNm}

The fixed-end moment is 46.875 kNm (hogging), equal to wL28=15(25)8\dfrac{wL^2}{8} = \dfrac{15(25)}{8}.

Check: RA=wL−RB=75−28.125=46.875R_A = wL - R_B = 75 - 28.125 = 46.875 kN.

Answer: MA=46.875M_A = 46.875 kNm (hogging); RB=28.125R_B = 28.125 kN and RA=46.875R_A = 46.875 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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