After a severe rainstorm, the ﬂat roof of the building shown in Fig. (a) is covered by 2.5 in. of rainwater. The speciﬁc weight of water is 62.4 lb/ft³, so water at a depth of 2.5in. causes a uniform pressure of 62.4(2.5/12) = 13 lb/ft². Determine the resultant force that the water exerts on

Question

After a severe rainstorm, the ﬂat roof of the building shown in Fig. (a) is covered by 2.5 in. of rainwater. The speciﬁc weight of water is 62.4 lb/ft³, so water at a depth of 2.5in. causes a uniform pressure of 62.4(2.5/12) = 13 lb/ft². Determine the resultant force that the water exerts on the roof.

Accepted Answer

The load diagram in Fig. (b) can be divided into three solid shapes: two right-triangular solids of volume [latex]V_{1}[/latex] and [latex]V_{3}[/latex] and a rectangular solid of volume [latex]V_{2}[/latex]. The resultant force corresponding to each of these shapes is equal to the volume of the shape. Letting [latex]P_{1}[/latex], [latex]P_{2}[/latex], and [latex]P_{3}[/latex] be the resultants, we have

[latex]P_{1}=V_{1}=13\left[\frac{1}{2}(12)(75) ight]=5850\:lb[/latex]

[latex]P_{2}=V_{2}=13[(40)(75)]=39000\:lb[/latex]

[latex]P_{3}=V_{3} =P_{1} =5850\:lb[/latex]

The lines of action of these forces pass through the centroids of the corresponding volumes. The points where these forces intersect the roof of the building are labeled [latex]C_{1}[/latex], [latex]C_{2}[/latex], and [latex]C_{3}[/latex] in Fig. (b).

The magnitude of the resultant force is given by

[latex]{ iny+}\big\downarrow\:\:\:\:R=P_{1}+P_{2}+P_{3}=5850+39000+5850=50700\:lb[/latex]

Because the load area (the roof of the building) is symmetrical about the [latex]y[/latex]-axis and the pressure is uniform, the resultant will lie along the [latex]y[/latex]-axis. Therefore, we need only calculate the distance [latex]\bar{y}[/latex] shown in Fig. (c).

Using Table 3.1, the coordinates of [latex]C_{1}[/latex], [latex]C_{2}[/latex], and [latex]C_{3}[/latex] in Fig. (b) are

A. Volumes

B. Volumes

Rectangular solid

[latex]\overline{x} = \frac{1}{2} b \qquad \overline{y} = \frac{1}{2} L \qquad \overline{z} = \frac{1}{2} h \qquad V= bLh[/latex]

Rectangle

[latex]\overline{x} = \frac{1}{2} b \qquad \overline{y} = \frac{1}{2} h \qquad A= bh[/latex]

Right-triangular solid

[latex]\overline{x} = \frac{1}{2} b \qquad \overline{y} = \frac{2}{3} L \qquad \overline{z} = \frac{1}{3} h \qquad V= \frac{1}{2} bhL[/latex]

Right triangle

[latex]\overline{x} = \frac{2}{3} b \qquad \overline{y} = \frac{1}{3} h \qquad A= \frac{1}{2} bh[/latex]

Table 3.1 Centroids of Some Common Geometric Shapes (Additional tables are found in Chapter 8.)

[latex]\bar{y}_{1}=\bar{y}_{3}=\frac{2}{3}(75)=50\:ft[/latex]

[latex]\bar{y}_{2}=\frac{1}{2}(75)=37.5\:ft[/latex]

We can now determine [latex]\bar{y}[/latex] using the moment equation

[latex]\sum{M_{x}}=−R\bar{y}[/latex] [latex]−5850(50)−39000(37.5)−5850(50)=−50700\:\bar{y}[/latex]

which yields

[latex]\bar{y}=40.4\:ft[/latex]

The resultant is shown in Fig. (c).

Question 3.12: After a severe rainstorm, the ﬂat roof of the building shown...

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