Analysis of Influence Lines for Beams With Overhangs

Influence lines can be used to obtain the static effects of moving loads on structures. The action effects that can be evaluated at any section of a structure using influence lines  are support reactions, internal stresses (bending, shear, axial), and deflections. As we shall see at the end of this article, if the influence line for a structure is adequately plotted with respect to a particular section, you can obtain the desired effect of the load on any part of the structure once the moving load is at that section. This post will be dedicated to simple and compound beams with overhangs.

Let us consider the beam loaded as shown below. It is desirous to obtain the influence line for the support reactions, and for the internal stresses with respect to section 1-1.

In all cases, we will be taking P as unity (i.e 1.0)

(1) Influence line for support reactions

Support A
Support reaction at point A (FA) = (L – x)/L
At x = -L1;
FA = (L + L1)/L

At x = 0;
FA = 1.0

At x = L + L2;
FA = (L – L – L2)/L = – L2/L

Support B
Support reaction at point B (FB) =  x/L
At x = -L1;
FB =  -L1/L

At x = 0;
FB = 0

At x = L;
FB = 1.0

At x = L + L2;
FB = (L + L2)/L

(2) Influence line for bending moment with respect to section 1-1

(0  ≤  x  ≤  a)
M1-1 = FA.a – P(a – x)
M1-1 = [P(L – x).a]/L – P(a – x)
But taking P = 1.0;
= [(L – x).a]/L – (a – x)

At x = -L1;
M1-1 =  [(L + L1).a]/L – (a + L1) = [L1(a – L)/L] =  -L1.b/L

At x = 0;
M1-1 =  [(L – 0).a]/L – (a – 0) = [L1(a – L)/L] =  0

At x = a;
M1-1 =  [(L – a).a]/L – (a – a) = [L1(a – L)/L] =  a.b/L

(a  ≤   ≤  L)
M1-1 =   [P(L – x)a]/L

At x = a;
M1-1 = [(L – a)a]/L   =  a.b/L

At x = L;
M1-1 = [(L – L)a]/L   =  0

At x = L + L2;
M1-1 = [(L – L – L2)a]/L   =  – L2a/L

(2) Influence line for shear with respect to section 1-1

(0  ≤  x  ≤  a)
Q1-1 = P(L – x)/L – P = – FB =  –x/L

At x = -L1;
Q1-1 =   L1/L

At x = 0;
Q1-1 =   0

At x = a;
Q1-1 = -a/L

(a  ≤  x  ≤  L)
Q1-1 = -(P.x)/L + P = (L – x)/L

At x = a;
Q1-1 = b/L

At x = L;
Q1-1 = 0

At x = L + L2;
Q1-1 = [(L – L – L2)]/L = -L2/L

In the influence line analysis of compound beams, it is always advantageous to decompose the beams into simple beams at the location of the internal hinges. A little consideration of the analysis below will show that it largely built off from the analysis of the simple versions as was done above. The simple logic is to pick the section under consideration and plot the influence line disregarding other adjacent beams. After drawing the diagram accurately, you can now use your ruler and draw your straight line towards the next support or internal hinge. All other values can be determined from similar triangles rule. Note that whenever your influence line gets a zero value at an internal hinge, the influence line terminates.

The figure below shows the influence line for internal stresses at section 1-1 when the load travels through the compound beam.

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  1. So nice site, so much to learn, or at least to find a information. So long I plan to do it something like this on my native anguage….. I like this site so much


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