Question 3.11: A square foundation mat supports the four columns shown. Det......
A square foundation mat supports the four columns shown. Determine the magnitude and point of application of the resultant of the four loads.
We first reduce the given system of forces to a force-couple system at the origin O of the coordinate system. This force-couple system consists of a force R and a couple vector {M}_{0}^{R} defined as follows:
{ R}\,=\,\Sigma{ F}\qquad{ M}_{ O}^{R}\,=\,{\Sigma}({ r}\,\times\,{ F})The position vectors of the points of application of the various forces are
determined, and the computations are arranged in tabular form.
Since the force R and the couple vector {{{ M}_{O}^{R}}} are mutually perpendicular, the force-couple system obtained can be reduced further to a single force R . The new point of application of R will be selected in the plane of the mat and in such a way that the moment of R about O will be equal to {{{ M}_{O}^{R}}}. Denoting by r the position vector of the desired point of application, and by x and z its coordinates, we write
{r}\times{R}={{{ M}_{O}^{R}}} \\ \\ (x\mathrm{i}+\mathrm{z}\mathrm{k})\times(-80\mathrm{j})\,=\,240\mathrm{i}\,-\,280\mathrm{k} \\ \\ -80x{\mathrm k}\,+\,80z{\mathrm i}\,=\,240{\mathrm i}\,-\,280{\mathrm k}from which it follows that
-80x = -280 80z = 240
x = 3.50 ft z = 3.00 ft
We conclude that the resultant of the given system of forces is
R = 80 kips w at x = 3.50 ft, z = 3.00 ft
| r, ft | F, kips | r × F, kip · ft |
| 0 | -40 j | 0 |
| 10 i | -12 j | -120 k |
| 10 i + 5 k | -8 j | 40 i – 80 k |
| 4 i + 10 k | -20 j | 200 i – 80 k |
| R = -80 j | {{{ M}_{O}^{R}}}=240\mathrm{i}\ -\ 280\mathrm{k} |