Muller-Breslau: ILD of shear force at B

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.

Reading of the figure

Beam: A fixed, B a roller at 5 m from A, and a free overhang BC of 1 m. The shear force just to the left of the support B (the section where the support reaction acts) is required. $EI$ is constant.

Muller-Breslau principle

Cut the beam just to the left of B and give the two faces a unit relative vertical displacement (the left face moves down by 1 relative to the right face, the faces stay parallel so the slope does not change). The resulting deflected shape is the ILD for the shear just left of B.

• Right of the cut: the support B holds the point at B, so the ordinate at B (right) is 0. The overhang BC continues with the slope of the left part: a straight line.
• Left of the cut: A is fixed, B moved down by 1: the shape is that of a propped-cantilever deflected curve, $y(x) = -\dfrac{x^2(3L-x)}{2L^3}$ with $L = 5$.

Ordinates

For a unit load at $x < 5$: $V_B^{left} = -R_B = -\dfrac{x^2(15-x)}{250}$. For a load on the overhang at $z = x - 5$ from B: AB acts as a propped cantilever with an applied moment $z$ at B, so $V_B^{left} = R_A = -\dfrac{3z}{2L} = -0.3z$.

Sample: $x = 3$: $-\dfrac{9\times12}{250} = -0.432$. Overhang end ($z = 1$): $-0.3$.

ILD shape

*************---------------------------|***-----
             **********                 |   *****
                       *******          |
                              *******   |
                                     ***|
^                                       ^

(The ordinate falls from 0 at A to -1 at B on the left, jumps up by 1 at B to 0, and then falls linearly to -0.3 at the free end C.)

The shear just to the right of B has the simple ILD $V_B^{right} = 0$ for a load on AB and $+1$ for a load on the overhang.