### Influence Lines for Beams

Constructing influence lines for beam is very important since the beam is the main load-carrying elements of a floor system or abridges deck. Constructing the influence lines will enable determining the reaction, shear, or a moment that can be generated by a moving load at a specific point. Once the influence lines have been constructed as a function of reaction, shear, or moment. Then it is possible to place the moving loads to produce the maximum influence over the beam. Influence lines will be used by designers to determine the worst cases of loading and design the beam accordingly. Beams will be subjected to two types of loading:

Example: Determine the maximum positive shear that can be developed at point C in the beam shown in Figure no:3 due to a concentrated moving load of 4000 lb and a uniform moving load of 2000 lb-ft.

Concentrated load: from the influence line shown in figure no:4. The maximum positive shear is located at x=2.5, the ordinate of the influence lines=0.75.

V@c=0.75*4000=3000lb.

Uniform load: the maximum positive shear will be generated by a uniform load if the uniform load placed between x=2.5 to x=10.

Vc=0.5*(10-2.5)*0.75*2000=5,625lb

Total maximum shear at C

Vmax=3000+5625=8625lb

- Concentrated loading: we used a dimensionless unit of the force to construct the influence lines, therefore, if it needs to know the influence of a force F at a specific point on the beam. Assuming the point is located at x from the specific point, by the using of influence lines diagram, we can find the effects by multiplying the magnitude of force F by the corresponding ordinate at the distance X. for example, let consider the influence line shown in figure no:1. Calculate the reaction at A if we place a force equal F at X=L/2. The reaction at Ay=(1/2)*F, if we assume X=L, then Ay=(1)*F=F.

Figure 1

- Uniform load: if a portion of the beam is subjected to a uniform load W0 , as shown in figure no:2, each segment of dx creates a concentrated load of dF=dx* W0 on the beam. If dF is located at x from a specific point over the beam, assuming that the influence line is constructed for this particular point. The influence of dF at a distance x is dF*y= dx* W0 *y (the ordinate of the influence lines at x). this the influence by a segment of the uniform load. If we want to calculate the influence of the remaining segments of the uniform load, we need to sum it in other words to integrate it. In general, the value of a function caused by a uniform distributed load is simply the area under the influence line for the function multiplied by the intensity of the uniform load. The below example will illustrate everything

Figure 2

Example: Determine the maximum positive shear that can be developed at point C in the beam shown in Figure no:3 due to a concentrated moving load of 4000 lb and a uniform moving load of 2000 lb-ft.

Figure 3

Concentrated load: from the influence line shown in figure no:4. The maximum positive shear is located at x=2.5, the ordinate of the influence lines=0.75.

Figure 4

V@c=0.75*4000=3000lb.

Uniform load: the maximum positive shear will be generated by a uniform load if the uniform load placed between x=2.5 to x=10.

Vc=0.5*(10-2.5)*0.75*2000=5,625lb

Total maximum shear at C

Vmax=3000+5625=8625lb

## Comments

## Post a Comment