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Castigliano: prop reaction of propped cantilever

Short answer

Treat the prop reaction R_B as redundant. Taking origin at B, moment M = R_B x – w x²/2, ∂M/∂R_B = x. Setting ∂U/∂R_B = 0 (zero deflection at B) gives ∫_0^L (R_B x² – w x³/2) dx = 0, yielding R_B = 3wL/8 upward.

Calculate the reaction at the prop of a propped cantilever with uniform distributed load throughout the span using Castigliano's theorem.

Take a propped cantilever of span LL, fixed at A and with a prop (roller) at B, carrying a UDL ww per unit length. Treat the prop reaction RBR_B as the redundant.

 w kN/m
 vvvvvvvvvvvvvvvvvvv
 ||==================o
 A (fixed)           B (prop, R_B)
 |<------- L ------->|

Take xx from B towards A.

Mx=RB x−wx22,∂Mx∂RB=xM_x = R_B\,x - \frac{w x^2}{2}, \qquad \frac{\partial M_x}{\partial R_B} = x

The prop is at the same level as the fixed end, so the deflection at B is zero:

∂U∂RB=∫0LMxEI∂Mx∂RB dx=0\frac{\partial U}{\partial R_B} = \int_0^L \frac{M_x}{EI}\frac{\partial M_x}{\partial R_B}\,dx = 0 ∫0L(RB x2−wx32)dx=0\int_0^L \left(R_B\,x^2 - \frac{w x^3}{2}\right)dx = 0 RBL33−wL48=0\frac{R_B L^3}{3} - \frac{w L^4}{8} = 0 RB=3wL8R_B = \frac{3wL}{8}

Other reactions

RA=wL−3wL8=5wL8R_A = wL - \frac{3wL}{8} = \frac{5wL}{8} MA=3wL8 L−wL22=−wL28M_A = \frac{3wL}{8}\,L - \frac{wL^2}{2} = -\frac{wL^2}{8}

The fixed-end moment is wL2/8wL^2/8 (hogging).

Answer: prop reaction RB=3wL8R_B = \dfrac{3wL}{8} upward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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