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Influence Lines
Influence Lines for a Simple Beam by Developing the Equations


problem statement

Draw the influence lines for the reactions YA, YC, and the shear and bending moment at point B, of the simply supported beam shown by developing the equations for the respective influence lines.

Beam structure to analyze
Figure 1 - Beam structure to analyze


The influence line for a reaction at a support is found by independently applying a unit load at several points on the structure and determining, through statics, what the resulting reaction at the support will be for each case. In this example, one such equation for the influence line of YA can be found by summing moments around Support C.

Unit Load Applied
Figure 2 - Application of unit load

MC = 0  (Assume counter-clockwise positive moment)
-YA(25)+1(25-x) = 0
YA = (25-x)/25 = 1 - (x/25)

The graph of this equation is the influence line for YA (See Figure 3). This figure illustrates that if the unit load is applied at A, the reaction at A will be equal to unity. Similarly, if the unit load is applied at B, the reaction at A will be equal to 1-(15/25)=0.4, and if the unit load is applied at C, the reaction at A will be equal to 0.

Influence line for the reaction at support A
Figure 3 - Influence line for YA, the support reaction at A

The fact that YA=1 when the unit load is applied at A and zero when the unit load is applied at C can be used to quickly generate the influence line diagram. Plotting these two values at A and C, respectively, and connecting them with a straight line will yield the the influence line for YA. The structure is statically determinate, therefore, the resulting function is a straight line.


The equation for the influence line of the support reaction at C is found by developing an equation that relates the reaction to the position of a downward acting unit load applied at all locations on the structure. This equation is found by summing the moments around support A.

Unit Load Applied
Figure 4 - Application of unit load

MA = 0  (Assume counter-clockwise positive moment)
YC(25)-1(x) = 0
YC = x/25

The graph of this equation is the influence line for YC. This shows that if the unit load is applied at C, the reaction at C will be equal to unity. Similarly, if the unit load is applied at B, the reaction at C will be equal to 15/25=0.6. And, if the unit load is applied at A, the reaction at C will equal to 0.

Influence line for the reaction at support C
Figure 5 - Influence line for the reaction at support C

The fact that YC=1 when the unit load is applied at C and zero when the unit load is applied at A can be used to quickly generate the influence line diagram. Plotting these two values at A and C, respectively, and connecting them with a straight line will yield the the influence line for YC. Notice, since the structure is statically determinate, the resulting function is a straight line.


The influence line for the shear at point B can be found by developing equations for the shear at the section using statics. This can be accomplished as follows:

a) if the load moves from B to C, the shear diagram will be as shown in Fig. 6 below, this demonstrates that the shear at B will equal YA as long as the load is located to the right of B, i.e., VB = YA. One can also calculate the shear at B from the Free Body Diagram (FBD) shown in Fig. 7.

Shear diagram for load located between B and C
Figure 6 - Shear diagram for load located between B and C

Free body diagram for section at B with a load located between B and C
Figure 7 - Free body diagram for section at B with a load located between B and C

b) if the load moves from A to B, the shear diagram will be as shown in Fig. 8, below, this demonstrates that the shear at B will equal -YC as long as the load is located to the left of B, i.e., VB = - YC. One can also calculate the shear at B from the FBD shown in Fig. 9.

Shear diagram for load located between A and B
Figure 8 - Shear diagram for load located between A and B

Free body diagram for section at B with a load located between A and B
Figure 9 - Free body diagram for section at B with a load located between A and B

The influence line for the Shear at point B is then constructed by drawing the influence line for YA and negative YC. Then highlight the portion that represents the sides over which the load was moving. In this case, highlight the the part from B to C on YA and from A to B on -YC. Notice that at point B, the summation of the absolute values of the positive and negative shear is equal to 1.

Influence line for shear at point B
Figure 10 - Influence line for shear at point B


The influence line for the moment at point B can be found by using statics to develop equations for the moment at the point of interest, due to a unit load acting at any location on the structure. This can be accomplished as follows.

a) if the load is at a location between B and C, the moment at B can be calculated by using the FBD shown in Fig. 7 above, e.g., at B, MB = 15 YA - notice that this relation is valid if and only if the load is moving from B to C.

b) if the load is at a location between A and B, the moment at B can be calculated by using the FBD shown in Fig. 9 above, e.g., at B, MB = 10 YC - notice that this relation is valid if and only if the load is moving from A to B.

The influence line for the Moment at point B is then constructed by magnifying the influence lines for YA and YC by 15 and 10, respectively, as shown below. Having plotted the functions, 15 YA and 10 YC, highlight the portion from B to C of the function 15 YA and from A to B on the function 10 YC. These are the two portions what correspond to the correct moment relations as explained above. The two functions must intersect above point B. The value of the function at B then equals (1 x 10 x 15)/25 = 6. This represents the moment at B if the load was positioned at B.

Influence line for moment at point B
Figure 11 - Influence line for moment at point B


Contact Dr. Fouad Fanous for more information.