Influence Line for Simple-Beam Reaction

IL_{R_{A}} (x) = \frac{L - x}{L} IL_{M_{C}} (x) = {b x / L a (L - x) / L x < a x > a

An influence line gives the value of a reaction, shear, or moment as a unit load moves. For RA on a simple span, the IL is 1 at A and 0 at B (straight). Place distributed load where the IL is positive to maximize that effect.

$x$: Position of the unit load from A
$L$: Span

Step-by-step solved example

L=10 m. Unit load at 4 m from A. RA from the IL?

Influence line for RA is 1 at A and 0 at B (straight).

1. IL
RA=(L−x)/L=6/10=0.60 (force per unit load).

Answer: 0.60 (same units as the load)

10 practice questions

0/10 correct

1.IL for RA at A is

2.IL for RA at B is

3.To maximize RA, place live load where IL is

4.A concentrated load P at x contributes

5.Uniform w on a segment contributes

6.IL for midspan moment of a simple beam is a triangle peaking at

7.Müller-Breslau: the IL is the deflected shape from

8.IL for shear just right of A is

9.Dead load uses the IL

10.Units of IL for a reaction are