Resources for Structural Engineers and Engineering Students

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How can we deal with these types of uniform or other distributed loads when performing equilibrium calculations? The way to do this is to consider a equivalent total load or effective force caused by the distributed load, which acts at the centroid of the distribution. The location of this centroid is different depending on the type of load distribution as shown on the right side of Figure 4.1. For a uniform load, the effective force is equal to the total load given by the load per unit length multiplied by the total length (or $wL$). This is also equal to the area under the distributed load diagram, in this case a rectangle. For the uniform load, the centroid is at the center of the distribution ($L/2$). This is the location where you would put the effective force in order to use it in equilibrium calculations. For the triangular load, the effective load is again the total load which is equal to the area under the distribution, in this case $wL/2$ ($\frac{1}{2}bh$) and this acts at the centroid of the triangle, which is located at one-third of the length from the high side. For the trapezoidally distributed load, the case is slightly more complex as shown in Figure 4.1.

The point load and point moment do not have any equivalent total load, since they already act at a single point.

Effective forces are only used for calculating the effects of distributed loads with equilibrium calculations. Do not replace distributed loadings with effective forces, the rest of the analysis will be incorrect when determining internal shear and moment diagrams.

Interactive Quiz