ILD: reactions, BM and SF beneath loads

Data: simply supported beam $0$–$1$, $L = 14$ m. $P_1 = 240$ kN at 3 m from support 0; $P_2 = 160$ kN at 4.5 m from support 1, i.e. 9.5 m from support 0.

Reaction at support 0

ILD for $R_0$: a straight line from 1 at support 0 to 0 at support 1; the ordinate at distance $s$ from 0 is $(L-s)/L$.

$$R_0 = 240\left(\frac{11}{14}\right) + 160\left(\frac{4.5}{14}\right) = 240.0\ \text{kN}$$

Reaction at support 1

Ordinate $s/L$:

$$R_1 = 240\left(\frac{3}{14}\right) + 160\left(\frac{9.5}{14}\right) = 160.0\ \text{kN}$$

(Check: $R_0 + R_1 = 400$ kN = total load.)

Bending moment beneath the loads

The ILD for $M$ at a section $x$ is a triangle with peak $x(L-x)/L$. For a load at $s$ the ordinate is $s(L-x)/L$ if $s\le x$, and $x(L-s)/L$ if $s\ge x$.

Under 240 kN ($x = 3$ m): peak ordinate $=3\times 11/14 = 2.357$; ordinate under the 160 kN load $=3\times 4.5/14 = 0.964$.

$$M_{3} = 240(2.357) + 160(0.964) = 720.0\ \text{kN m}$$

Under 160 kN ($x = 9.5$ m): peak ordinate $=9.5\times 4.5/14 = 3.054$; ordinate under 240 kN $=3\times 4.5/14 = 0.964$.

$$M_{9.5} = 240(0.964) + 160(3.054) = 720.0\ \text{kN m}$$

Shear force beneath the loads

The ILD for shear at $x$ has ordinate $-s/L$ for $s<x$ and $+(L-s)/L$ for $s>x$ (a jump of 1 at the section). A load standing exactly at the section counts as being on the right of the section for "just left" and on the left for "just right".

Answer: $R_0 = 240$ kN, $R_1 = 160$ kN; $M = 720$ kN m under the 240 kN load and $720$ kN m under the 160 kN load; the shear force is +240 kN from support 0 to the 240 kN load, 0 between the loads, and −160 kN from the 160 kN load to support 1.