15. The tube sheet shown in the sketch below is 36 inches in diameter and has thirty 3-inch holes drilled in it.
a. How many square inches of metal were removed?
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b. How many square inches of metal are left in the sheet?
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SOLID FIGURES
Plane surfaces have two dimensions: length and width. In solid figures, another dimension—thickness—occurs. Think of plane figures as two dimensional and solid objects as three dimensional. The amount of space a solid figure occupies is its volume. Volumes, which are obtained by dealing with all three dimensions, are expressed in cubic measures such as cubic inches (in.3), cubic feet (ft3), cubic yards (yd3), and cubic metres (m3). Just as in calculating dimensions of plane figures, all three dimensions in the calculation of volumes must be expressed in the same units. Feet cannot multiply inches nor can feet multiply yards or gal-lons multiply barrels.
Rectangular Solids
The rectangular solid shown in figure 6.29, called a cube when all sides are equal, illustrates cubic measure. Multiplying the three dimensions of the solid—length (l) times width (w) times height (h)—gives its volume (V). The formula is:
V = lwh.
In figure 6.29, each dimension of the cube is 3 feet. Thus:
V = 3 ft × 3 ft × 3 ft = 27 ft3.
Also notice in figure 6.29 that the cube is marked with lines 1 foot apart.
If you sawed the cube apart along these lines, the cube would be divided into twenty-seven smaller blocks, each 1 foot high, 1 foot wide, and 1 foot deep. In short, 27 cubic feet are in a cube that is 3 feet on each side.
A cube is a rectangular solid whose sides are all equal. When the sides are not equal (fig. 6.30), the solid is simply a rectangular solid. You can determine the volume with the same formula for finding the volume of a cube, which is
V = lwh.
w = 3'
l = 3'
h = 3' 1'
1'
1'
Figure 6.29 Cube
Figure 6.30 Rectangular solid 8' 2' 11/4'
160 PRACTICAL GEOMETRY
Example Problem: What is the volume of the rectangular solid shown in figure 6.30 where its length (l) is 8 feet, its width (w) is 2 inches, and its height (h) is 1¼ inches? Express the answer in cubic inches.
Solution: Using the formula V = lwh, plug in the values.
l = 8 ft = 96 in., w = 2, and h = 1º in.
V = 96 in. × 2 in. × 1.25 in. = 240 in.3
Sometimes, volumes are expressed in gallons or barrels, and to solve for them, you sometimes have to convert measurements. Table 6.3 lists several conversions. These conversions change cubic feet into cubic inches and barrels;
cubic metres into cubic feet, gallons, and barrels; cubic yards into cubic feet; and barrels into gallons and cubic feet.
Example Problem: How many barrels of water does it take to fill a rectangular tank that is 7 feet long, 4 feet wide, and 3½ feet high?
Solution:
V = lwh
V = 7 ft × 4 ft × 3½ ft = 98 ft.3
If 5.61 ft3 equals 1 barrel (bbl), then 98 ft3 equals 17.47 bbl because 98 ÷ 5.61 = 17.47 bbl.
To find the total surface area of all six sides of a rectangular solid, deter-mine the area of each side and add the areas together. The answer is in square measurements. For example, to find the total surface area of the rectangular solid in figure 6.30, which is 8 feet long, 2 inches wide, and 1.25 inches high, first find the area of the top and multiply it by 2 to find the area of the top and the bottom:
A = lw
A = 96 in. × 2 in. = 192 × 2 sides = 384 in.2 Then, find the area of the two sides:
A = 96 in. × 1.25 in. = 120 × 2 sides = 240 in.2 Next, find the area of the two ends:
A = 2 in. × 1.25 in. = 2.5 × 2 sides = 5 in.2
Finally, add the three dimensions to obtain the total surface area (T):
T = 384 + 240 + 5 = 629 in.2
Cylinders
In industry, it is necessary to find the volume of cylinders such as storage tanks, pipes, and pump cylinders. These volumes are usually expressed in gallons, barrels, or cubic inches and require the appropriate measurement conversions.
The volume of a cylinder (fig. 6.31) is equal to the product of the area of its base (B) times the height (h):
V = B × h.
Remembering that the area of a circle is equal to πr2, the area of the cylinder base is πr2, and the formula for finding the volume of a cylinder is
V = πr2h.
Solid Figures 161
Example Problem: Find the volume of the cylinder shown in figure 6.31. Its diameter is 10 feet, and its height is 8 feet.
Solution:
r = ½ of 10 ft = 5 ft; h = 8 ft.
V = πr2h
V = 3.1416 × 5 ft × 5 ft × 8 ft = 628.32 ft3.
Example Problem: How many barrels of oil can be stored in a tank that is 20 feet in diameter and 10 feet high?
Solution:
r = ½ of 20 ft = 10 ft; h = 10 ft.
V = πr2h
V = 3.1416 × 10 ft × 10 ft × 10 ft = 3,141.6 ft3.
Because 1 cubic foot equals 7.48 gallons, then 3,141.6 cubic feet equals 23,499 gallons. And because 42 gallons equals1 barrel, then
23,499 ÷ 42 = 559.5 bbl.
Elliptical Solids
An elliptical solid is similar to a cylinder except that its base is an ellipse instead of a circle. The equation for finding the area of an ellipse (A = πab) is used to determine the volume of an elliptical solid. Referring to figure 6.32, the formula for finding the volume of an elliptical solid can be written as
V = πabh where
a = one-half the major axis b = one-half the minor axis h = height.
Example Problem: Find the volume of an elliptical tank whose major axis is 8 feet, whose minor axis is 6 feet, and whose height is 10 feet.
Solution:
a = 4 ft; b = 3 ft; h = 10 ft.
V = πabh
V = 3.1416 × 4 × 3 × 10 = 376.992, or 377 ft3.
Example Problem: A tank truck has an elliptical tank that is 14 feet long. The major and minor axes of the elliptical ends of the tank are 8 feet and 6 feet. How many gallons of gasoline does the tank hold?
Solution:
a = 4 ft; b = 3 ft; h = 14 ft.
V = πabh
V = 3.1416 × 4 × 3 × 14 = 527.789 ft3. If 1 cubic foot equals 7.48 gallons, then
527.79 × 7.48 = 3,947.86 gal.
MAJOR AXIS
ELLIPSE
MINOR AXIS b
a
h Figure 6.32 Elliptical solid
162 PRACTICAL GEOMETRY
Cones
The most common type of cone—the right circular cone—is a solid generated by rotating a right triangle about one of its legs (fig. 6.33). The circle at the bot-tom of the cone is its base, and its altitude, or height, is the distance between the center of the base and the apex. It also has a slant height, which is measured from the apex to the edge of the cone’s circular base.
The volume of a cone is equal to one-third of the area of the base times the altitude:
V = ⅓(Bh) or
V = ⅓(πr2h).
Example Problem: Find the volume of a cone whose base has a 6-inch diameter and whose altitude is 7 in.
Solution:
V = ⅓(πr2h)
= ⅓(3.1416 × 32 × 7)
V = 197.9208 ÷ 3 = 65.9736 in.3
Another important cone measurement is its lateral area, or conical surface (fig. 6.34). Unfolding a cone and placing it flat on a surface shows its lateral area.
The lateral area of a right circular cone equals half the product of the slant height and the circumference of the base:
L = ½sC
Example Problem: Find the lateral area of a cone whose slant height is 5 mil-limetres and whose base radius is 2 milmil-limetres.
Solution:
L = πrs
L = 3 .1416 × 2 × 5 = 31.416 mm2.
If you know the altitude and radius of the cone’s base, you can calculate the slant height by using the formula for finding the hypotenuse of a triangle:
a2 + b2 = c2.
Example Problem: What is the lateral area of the cone depicted in figure 6.35, where the height (h) is 12 inches and the radius (r) of the base is 6 inches?
Solution: First, find the slant height (s) of the cone, substituting h for a and r for b in the equation above:
Figure 6.35 Dimensioned cone r
6"
12" h
Solid Figures 163
Then solve for the lateral area:
L = πrs
L = 3.1416 × 6 × 13.4164 = 252.894 in.2
Pyramids
A pyramid is a solid figure that has a polygon for its base and triangles with a common vertex for its faces. Figure 6.36 shows three pyramids: one has a square as its base, another a pentagon, and the other a hexagon. These pyramids are regular pyramids; indeed, any pyramid that has a regular polygon as its base is a regular pyramid. Remember: a regular polygon has equal sides and equal angles.
The volume of a pyramid equals one-third the product of its base area (B) and its altitude (h):
V = ⅓(Bh).
Example Problem: Find the volume of the pyramid shown in figure 6.37, which has a height (h) of 3 feet and whose base (b) is a square each side of which is 2 feet.
Solution:
B = 2 ft × 2 ft = 4 ft2; h = 3 ft.
V = ⅓(Bh).
V = (4 × 3) ÷ 3 = 4 ft3.
To find the lateral area of a pyramid, find the areas of the triangles that make up the lateral surface and add them together. Or, you can use a formula that involves the pyramid’s slant height and perimeter. In a regular pyramid (one whose base is a regular polygon), the altitudes of these triangles from their common vertex to their bases are the same as the pyramid’s slant height (fig.
6.38). The sum of the sides of the polygon that form the pyramid base equals its perimeter. Thus, using the slant height and the perimeter, the lateral area of a pyramid can be found using the formula
L = ½(sP) where
L = lateral area s = slant height
P = perimeter of the base.
Example Problem: Find the lateral area of a pyramid whose slant height is 10 inches and whose base is a regular hexagon with 6-inch sides.
Solution:
L = ½(sP)
L = 10 in. × 6 in. × 6 sides √ 2 = 180 in.2
The total area of a pyramid, like the total area of a cone, is found by adding the lateral area and the base area. For regular pyramids and right circular cones, the total area can be calculated with the formula
T = πr2 + πrs
Figure 6.37 Dimensioned pyramid
Figure 6.38 Lateral area of
164 PRACTICAL GEOMETRY
where
T = total area
r = radius of the base s = slant height.
This formula combines the formulas for calculating the lateral area and the base area.
Example Problem: Find the total area of a cone with a base radius of 1.5 feet and a slant height of 30 inches.
Solution: First, convert the measurements to inches or feet. Using feet as the measurement,
When a pyramid or a cone is cut off at any point below the apex by a plane that is parallel to the base, the portion below the cutting plane is a frustum of a cone (fig. 6.39) or a frustum of a pyramid (fig. 6.40). (Frustum comes from the Latin word for piece or bit.) Frustums are commonly seen in concrete bases or in the welded portion of a pipe swedge.
The base of a pyramid or cone (B) and the section of the intersecting plane (B') are the bases of the frustum; the rest of the surface is the lateral surface (L).
The altitude (h) of a frustum is the distance between the centers of the two bases.
This altitude line is perpendicular to both bases.
The formula for finding the lateral area of a frustum of a cone is half the slant height (s) times the sum of the circumferences (C and C') of the bases:
L = ½s(C + C') where
L = lateral area
s = slant height of the frustum C = circumference of the cone base C' = circumference of the section base.
Example Problem: Find the lateral area of the frustum shown in figure 6.41 where the slant height (s) is 52 millimetres, the radius of the larger base (C) is 21 mil-limetres, and the radius of the smaller base (C') is 12 millimetres.
Solution: Using the radii given, find circumferences C and C':
C = π2r
C = 3.1416 × 2 × 21 mm = 131.95 mm.
C' = 3.1416 × 2 × 12 mm = 75.40 mm.
Then solve for lateral area (L):
L = ½s(C + C')
Figure 6.40 Frustrum of a pyramid
Solid Figures 165
The formula for finding the lateral area of the frustum of a pyramid is the same as that for a cone, except perimeters are used instead of circumferences:
L = ½s(P + P')
The volume of a frustum of a cone or a pyramid is obtained by the formula V = ½h(B + B' + √ BB')
This formula can be used whether the frustum is regular or not—that is, the bases do not have to be perfect circles or even-sided polygons.
Example Problem: Find the volume of the frustum shown in figure 6.42, where the base (B) is a square that is 15 inches on each side, the section (B') is a square that is 6 inches on each side, and the height (h) is 24 inches. Give the answer in cubic feet.
Solution: First, solve for base areas:
B = 15 in. × 15 in. = 225 in.2 B' = 6 in. × 6 in. = 36 in.2 Then use the formula to find volume:
V = ½h(B + B' + √ BB')
V = 24 ÷ 3 × (225 + 36 + √ 225 × 36) V = 8 × (261 + 90) = 8 × 351 = 2,808 in.3
Since 1 ft3 equals 1,728 in.3, then 2,808 ÷ 1,728 = 1.625 ft3.
Spheres
A sphere is a closed solid whose points are all equally distant from a point within called its center (fig. 6.43). A sphere is different from a circle in that a sphere exists in space while a circle is flat. It is helpful to think of a sphere as the shape of a basketball or a globe. Both the sphere and the circle have centers, diameters, and radii.
The volume of a sphere is equal to four-thirds π times the radius cubed:
V = 4⁄₃πr3.
Example Problem: Find the volume of a sphere that has an 8-inch diameter.
Solution:
V = 4⁄₃πr3, or 4 × π × r3 ÷ 3
V = 4 × 3.1416 × 4 × 4 × 4 ÷ 3 = 268.0832 in.3 Figure 6.43 Sphere
Figure 6.42 Frustrum pyramid calculations
166
A sphere is not a ruled surface. Thus, it cannot be rolled out into a flat surface without distorting its shape and area. Nevertheless, the area of a sphere can be determined by using the formula
S = πd2 where
S = area of the sphere d = diameter of the sphere.
Example Problem: Find the area of a sphere that has a 10-inch diameter.
Solution:
S = πd2
S = 3.1416 × 10 × 10 = 314.16 in.2