ILD: bending moment at fixed support, 10 m beam

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

Propped cantilever AB, span $L = 10$ m, A fixed, B roller, constant $EI$. The bending moment at the fixed support A is the reaction moment (hogging). Muller-Breslau: replace the fixed support by a hinge and apply a unit rotation at A; the deflected shape of the beam is the ILD of $M_A$.

Derivation of the ordinates

Take the roller reaction $R_B$ as the redundant. For a unit load at $x$ from A:

1. Deflection at B of the released cantilever (fixed at A) under the unit load: $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$.
2. Deflection at B under a unit upward force at B: $f_{BB} = \dfrac{L^3}{3EI}$.
3. Compatibility $\delta_{Bx} = R_Bf_{BB}$ gives $R_B = \dfrac{x^2(3L-x)}{2L^3}$.
4. Moment at A (hogging): $M_A = x\cdot 1 - R_BL = x - \dfrac{x^2(3L-x)}{2L^2} = \dfrac{x(L-x)(2L-x)}{2L^2}$.

Sample: $x = 5$: $R_B = \dfrac{25\times25}{2000} = 0.3125$, $M_A = 5 - 0.3125\times10 = 1.875$ (hogging). $x = 3.75$: $R_B = \dfrac{14.0625\times26.25}{2000} = 0.1846$, $M_A = 3.75 - 1.846 = 1.904$.

Ordinates (1.25 m interval, hogging = negative)

ILD shape

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The maximum ordinate is about $-1.925$ at $x = 4.22$ m ($0.42L$). A concentrated load $P$ at $x = 5$ m gives $M_A = 1.875P$ hogging; a UDL $w$ over the full span gives $M_A = w\times$ area $= wL^2/8 = 12.5w$.