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ILD: fixed-end reaction, propped cantilever 5 m

Short answer

For a 5 m propped cantilever, RB = x^2(3L-x)/(2L^3) and RA = 1 - RB. Ordinates for RA at 1 m intervals are 1, 0.944, 0.792, 0.568, 0.296, 0.

Draw ILD at 1 m interval for the support reaction at the fixed end of a propped cantilever beam of span 5 m. Take EI is constant.

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=5L = 5 m, A fixed, B roller, constant EIEI. The structure is indeterminate to the first degree; take RBR_B as the redundant. For a unit load at xx from A, the deflection at B of the released cantilever is δBx=x2(3L−x)6EI\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}, and fBB=L33EIf_{BB} = \dfrac{L^3}{3EI}, so

RB=x2(3L−x)2L3,RA=1−RB,MA=x−RBL (hogging)=x(L−x)(2L−x)2L2R_B = \frac{x^2(3L-x)}{2L^3},\qquad R_A = 1 - R_B,\qquad M_A = x - R_BL\ \text{(hogging)} = \frac{x(L-x)(2L-x)}{2L^2}

Sample: x=2x = 2: RB=4×13250=0.208R_B = \dfrac{4\times13}{250} = 0.208, so RA=0.792R_A = 0.792; MA=2−0.208×5=0.96M_A = 2 - 0.208\times5 = 0.96 (hogging).

ILD for the vertical reaction RAR_A at the fixed end (1 m interval)

x from A (m)012345
RAR_A10.9440.7920.5680.2960
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Muller-Breslau: release the vertical restraint at A (keeping the fixed rotation) and displace A upward by 1; the shape equals 1−RB(x)1 - R_B(x).

ILD for the reaction moment MAM_A at the fixed end (1 m interval; hogging shown negative)

x from A (m)012345
MAM_A0-0.72-0.96-0.84-0.480

The moment ILD is negative (hogging) throughout, with a maximum of about −0.96-0.96 at x≈0.42L=2.1x \approx 0.42L = 2.1 m, and it is zero at A and B.

✓ Answer checked by Claude Opus 5.5 (high effort)

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