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ILD: reaction at B, propped cantilever 10 m

Short answer

Use Muller-Breslau: RB = x^2(3L-x)/(2L^3) for a propped cantilever of span 10 m. Ordinates at 2 m intervals are 0, 0.056, 0.208, 0.432, 0.704, 1, positive throughout and zero at A.

Draw influence line diagram for the reaction at B of the propped cantilever beam shown in figure below. Find the ordinates at 2 m interval. [Figure: beam AB, A fixed, B roller, span 10 m.]

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method (Muller-Breslau)

Propped cantilever AB of span L=10L = 10 m: A fixed, B roller, constant EIEI. Remove the support at B and give B a unit upward displacement. The ILD for RBR_B is the deflected shape of the cantilever AB (fixed at A) scaled so that the ordinate at B equals 1.

By Maxwell's reciprocal theorem, the deflection at B due to a unit load at xx from A equals the deflection at xx due to a unit load at B. So

RB(x)=δBxfBB=x2(3L−x)/6EIL3/3EI=x2(3L−x)2L3R_B(x) = \frac{\delta_{Bx}}{f_{BB}} = \frac{x^2(3L-x)/6EI}{L^3/3EI} = \frac{x^2(3L-x)}{2L^3}

Sample: x=8x = 8 m: RB=82(3×10−8)2×103=0.704R_B = \dfrac{8^2(3\times10 - 8)}{2\times10^3} = 0.704.

Ordinates (2 m interval)

x from A (m)0246810
RBR_B00.0560.2080.4320.7041

ILD shape

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The ordinate is 0 at A (a load at the fixed support is carried by A), increases with zero slope at A and reaches 1 at B. It is positive throughout, so a downward load always increases RBR_B. A UDL over the whole span gives RB=w×R_B = w\times (area of the ILD) =3wL8= \dfrac{3wL}{8}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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