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A wrought-iron bar 24 ft. long, 1} in. in diameter, would elongate, under a tensile stress of 15 tons,
(15 X 2,000) X (24 X 12) e =
.196 in. T (14)X 25,000,000 To find the breaking strength of a beam, use the formula
M= SR. (3) Obtain M and R from the two following tables, according to the kind of beam and nature of cross-section. A simple beam is one merely supported at its ends. In the expression for R, d is always understood to be the vertical side or depth; hence, that beam is the stronger which always has its greatest depth or longest side vertical. The moment of inertia I is taken about an axis perpendicular to d, and lying in the same plane.
Thus, the breaking strength of a cast-iron simple beam uniformly loaded and 20 ft. long between the supports, having a hollow rectangular cross-section 8 in. by 6 in. outside and 6 in. by 4 in. inside, is given by the formula M SR, or swi
bd3 – 61 dis; 36,000 x
6 d 36,000 X 8 X (6 X 83 — 4 X 63) whence, W
55,200. (20 X 12) X (6 X 8) Using a factor of safety of 6, the beam should support
= 9,200 lb.
6 with perfect safety. The value of S for beams should be taken from the flexure column of Table II.
To find the amount of deflection in a beam due to a load, substitute the values of W, 1, E, and I in the different expressions for the deflection 8 in Table IV.
The value of I is to be taken from Table V.
EXAMPLE.—What is the deflection of a wrought-iron beam fixed at both ends, 7 ft. long between the supports, having a solid rectangular cross-section 6 in. wide and 21 in. deep, carrying a load of 21,000 lb. in the middle ? SOLUTION.–From the table, W 13 W 13 21,000 X (7 X 12)5 X 12
.249'. 192 EI
bd3 192 X 25,000,000 X 6X (21) 192 EX
12 EXAMPLE.-It is desired to calculate the depth (d) of a cast-iron cantilever 36 in. in length (= 1) that will sustain at its end a weight of 4,000 lb. (= W), the lever to be of rectangular section and 2 in. in width.
SOLUTION.—The ultimate stress per square inch for cast iron in flexure is given in Table II as 36,000 lb. (= S). The weight will be a steady load, and therefore, according to Table I, a factor of safety of 6 should be used. By formula (3), M = SR. For a cantilever beam carrying a load at the end, M WI (Table IV); and for a rectangular sec
bd2 tion, R
6 Then, as W 4,000, 1 36, 6 = 2, f 6, we have
The value of d is found by substituting in this equation the known values of S, b, W, 1, and f, as follows: 36,000 X 2 x d2
4,000 X 36; whence, d 8.49 in. 6 X 6 At the point where the beam is supported, the required depth is found to be 8.49, or, practically, 8. in. At a point 6 in. from the support, the depth may again be calculated by substituting in the equation the value of l (the overhanging length beyond this point); 1 = 30, and the equation becomes 36,000 X 2 x d2
4,000 X 30. 6 X 6
d 7.75 in. At a point 12 in. from the support, 1 =
24, and 36,000 X 2 x d2
4,000 X 24; whence, d 6.93 in. 6 X 6 At a point 18 in. from the support, 1 18; and from the equation, a 6 in.; at 24 in. from the support, 1 12 and d 4.9 in.; at 30 in. from the support, 1 6 and d - 3.46 in.; at 36 in. from the support, or at the end of the beam, l = 0 and d = 0.
The depths required to be given to the lever or beam at the point of support and at intervals of 6 inches along its
length, are found to be 8.49, 7.75, 6.93, 6, 4.90, and 3.46 inches, respectively.
The lever is shown in the figure; theoretically, it would taper to nothing at the end, as indicated by dotted lines, but practically sufficient metal must be added at that point to provide means of attaching the weight.
NOTE.-In the preceding examples the weight of the beam has been neglected. If, however, this weight is large in comparison with the weight or weights carried by the beam, it should be taken into account, considering it (when the crosssection of the beam is the same throughout) as a load uniformly distributed over the whole length of the beam.
COLUMNS. To find the breaking strength of a column, use the following formula:
G2 S is taken from Table II, in the column for compression, G2 from Table V, and q from the following table, according to the character of the ends.
The breaking load of an elliptical wooden column 18 ft. long, having rounded ends, the diameters of the cross-section being 12 in. and 8 in., is
SA 8,000 X (IT X 12 X 8)
16 Using a factor of safety of 8, the column should support 36,442
4,565 ? b with perfect safety. 8