ILD: shear at section of propped cantilever

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 (Muller-Breslau / equilibrium)

Beam AB: A fixed, B roller, $L = 10$ m, section 1-1 at 4 m from A. The beam is singly indeterminate; take $R_B$ as the redundant.

With support B removed, the cantilever deflection at B due to a unit load at $x$ from A is $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$, and due to a unit load at B it is $f_{BB} = \dfrac{L^3}{3EI}$. Compatibility (zero deflection at B) gives

$$R_B = \frac{\delta_{Bx}}{f_{BB}} = \frac{x^2(3L-x)}{2L^3}$$

Shear at the section (left part):

• load to the left of the section ($x < 4$): $V = R_A - 1 = -R_B$
• load to the right of the section ($x > 4$): $V = R_A = 1 - R_B$

So the ILD is the $R_B$ curve turned upside down ($-R_B$) to the left of the section and $1 - R_B$ to the right of it, with a unit jump at the section.

Sample: $x = 6$: $R_B = \dfrac{36(30-6)}{2000} = 0.432$, so $V = 1 - 0.432 = 0.568$. At the section: $R_B(4) = \dfrac{16\times26}{2000} = 0.208$, so $V = -0.208$ just left and $+0.792$ just right.

Ordinates

ILD shape

                   |****
                   |    ********
                   |            *******
                   |                   ******
*************------|-------------------------****
             ******|



^                                               ^

The maximum negative ordinate is $-0.208$ and the maximum positive ordinate is $+0.792$, both at the section (a jump of 1 unit). The ordinates are 0 at A and at B.