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ILD: bending moment at B, two-span beam

Short answer

Moment at B is hogging throughout. Using three-moment equation, ILD ordinates are negative, zero at A, B, C. Peak negative values: -0.667 in AB (at x=4 m), -1.067 in BC (at x=12 m from A).

Draw the ILD for bending moment at B. [Figure: two-span beam ABC, A hinged, B and C rollers, AB = 8 m, BC = 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

Two-span beam ABC (A hinged, B and C rollers), L1=8L_1 = 8 m, L2=10L_2 = 10 m, constant EIEI. Take MBM_B as the redundant; by Muller-Breslau the ILD for MBM_B is the deflected shape of the beam with a hinge inserted at B and a unit rotation (kink) applied there. Ordinates follow from the three-moment equation with zero moments at A and C.

For a unit load at xx from A in span AB:

MB=−x (L12−x2)2L1(L1+L2)M_B = -\frac{x\,(L_1^2 - x^2)}{2L_1(L_1+L_2)}

For a unit load at distance yy from C in span BC:

MB=−y (L22−y2)2L2(L1+L2)M_B = -\frac{y\,(L_2^2 - y^2)}{2L_2(L_1+L_2)}

Sample: x=4x = 4 m: MB=−4(64−16)2×8×18=−0.667M_B = -\dfrac{4(64-16)}{2\times8\times18} = -0.667. y=4y = 4 m (x = 14): MB=−4(100−16)2×10×18=−0.933M_B = -\dfrac{4(100-16)}{2\times10\times18} = -0.933.

Ordinates of MBM_B

x from A (m)024681012141618
MBM_B0-0.4167-0.6667-0.58330-0.8-1.0667-0.9333-0.53330

ILD shape





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

The ILD is negative (hogging) throughout and zero at A, B and C. The largest ordinate in AB is about −0.68-0.68 near x=4.6x = 4.6 m, and in BC about −1.07-1.07 near x=12.2x = 12.2 m (5.8 m from C); the tabulated value at 12 m is −1.067-1.067.

✓ Answer checked by Claude Opus 5.5 (high effort)

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