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ILD: fixed-end moment, 12 m beam

Short answer

The ILD for M_A is entirely hogging. Ordinates every 3 m: 0, -1.97, -2.25, -1.41, 0 at 0,3,6,9,12 m. Maximum -2.31 occurs near 5.07 m.

Draw ILD for the support moment at A by computing the ordinates at 3 meter intervals. [Figure: propped cantilever AB, A fixed, B roller, span 12 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

Propped cantilever AB, L=12L = 12 m, A fixed, B roller. By the compatibility (or carry-over) argument, a unit load at xx from A gives the hogging fixed-end moment

MA=x(L−x)(2L−x)2L2M_A = \frac{x(L-x)(2L-x)}{2L^2}

(fixed-end moment ab2/L2ab^2/L^2 plus half of the far-end moment a2b/L2a^2b/L^2 carried over when the roller is released). The ILD is the rotated shape of the beam when A is changed to a hinge and given a unit rotation.

Sample: x=6x = 6: MA=6×6×182×144=2.25M_A = \dfrac{6\times6\times18}{2\times144} = 2.25.

Ordinates (3 m interval)

x from A (m)036912
MAM_A0-1.9687-2.25-1.40620

ILD shape





**--------------------------------------------***
  **                                     *****
    ****                            *****
        *****                *******
             ****************
^                                               ^

All ordinates are hogging; the maximum is about −2.31-2.31 at x=0.42L=5.07x = 0.42L = 5.07 m, and the value at mid-span is −2.25-2.25.

✓ Answer checked by Claude Opus 5.5 (high effort)

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