4. A cast-iron built-in cantilever beam projects 8 feet from the wall. Its cross-section is represented in Fig. 40, and the moment of inertia with respect to the neutral axis is 50 inches4; the working strengths in tension and compression are 2,000 and 9,000 pounds per square inch respectively. Compute the safe uniform load which the beam can sustain, neglecting the weight of the beam.

The beam being convex up, the upper fibres are in tension and the lower in compression. The resisting moment (SI ÷ c), as determined by the compressive strength, is

Fig. 40.

9,000 X 50 /4.5 = inch-pounds; and the resisting moment, as determined by the tensile strength, is

2,000 X 50 / 2.5 = 40,000 inch-pounds.

Hence the safe resisting; moment is the lesser of these two, or 40,000 inch-pounds. The dangerous section is at the wall (see Table B, page 55), and the value of the maximum bending moment is ½ Wl, W denoting the load and I the length. If W is in pounds, then

M = ½ W X 8 foot-pounds = ½ W X 96 inch-pounds.

Equating bending and resisting moments, we have

½W X 96 = 40,000; or, W = 40,000 X 2 / 96 = 833 pounds.

## Examples For Practice

1. An 8 X 8-inch timber projects 8 feet from a wall. If its working strength is 1,000 pounds per square inch, how large an end load can it safely sustain?

Ans. 890 pounds.

2. A beam 12 feet long and 8 X 16 inches in cross-section, on end supports, sustains two loads P, each 3 feet from its ends respectively. The working strength being 1,000 pounds per square inch, compute P (see Table B, page 55).

Ans. 9,480 pounds.

3. An I-beam weighing 25 pounds per foot rests on end supports 20 feet apart. Its section modulus is 20.4 inches3, and its working strength 10,000 pounds per square inch. Compute the safe uniform load which it can sustain.

Ans. 10,880 pounds-66. Third Application. The loads, manner of support, and working strength of beam are given, and it is required to determine the size of cross-section necessary to sustain the load safely, that is, to "design the beam."

To solve this problem, we use the first beam formula (equation 6), written in this form,

I/C = M/S. (6'")

We first determine the maximum bending moment, and then substitute its value for M, and the working strength for S. Then we have the value of the section modulus (I ÷ c) of the required beam. Many cross-sections can be designed, all having a given section modulus. Which one is to be selected as most suitable will depend on the circumstances attending the use of the beam and on considerations of economy.

Examples. 1. A timber beam is to be used for sustaining a uniform load of 1,500 pounds, the distance between the supports being 20 feet. If the working strength of the timber is 1,000 pounds per square inch, what is the necessary size of cross-section ?

The dangerous section is at the middle of the beam; and the maximum bending moment (see Table B, page 55) is

1/8Wl = 1/8 x 1,500 X 20 = 3,750 foot-pounds, or 3,750 X 12 = 45,000 inch-pounds.

Hence I/C = 45,000/1,000 = 45 inches3.

Now the section modulus of a rectangle is 1/6 ba2 (see Table A, page 54, Part I); therefore, 1/6 ba2 = 45, or ba2 = 270.

Any wooden beam (safe strength 1,000 pounds per square inch) whose breadth times its depth square equals or exceeds 270, is strong enough to sustain the load specified, 1,500 pounds.

To determine a size, we may choose any value for b or a, and solve the last equation for the unknown dimension. It is best, however, to select a value of the breadth, as 1, 2, 3, or 4 inches, and solve for a. Thus, if we try b = 1 inch, we have a2 = 270, or a = 10.43 inches. This would mean a board 1 X 18 inches, which, if used, would have to be supported sidewise so as to prevent it from tipping or " buckling." Ordinarily, this would not be a good size.

Next try b = 2 inches: we have

2 X a2 = 270; or a = = 11.62 inches.

This would require a plank 2 X 12, a better proportion than the first. Trying b = 3 inches, we have

3 X a2 = 270; or a = = 9.40 inches.

This would require a plank 3 X 10 inches; and a choice between a 2 X 12 and a 3 X 10 plank would be governed by circumstances in the case of an actual construction.

It will be noticed that we have neglected the weight of the beam. Since the dimensions of wooden beams are not fractional, and we have to select a commercial size next larger than the one computed (12 inches instead of 11.62 inches, for example), the additional depth is usually sufficient to provide strength for the weight of the beam. If there is any doubt in the matter, we can settle it by computing the maximum bending moment including the weight of the beam, and then computing the greatest unit-fibre stress due to load and weight. If this is less than the safe strength, the section is large enough; if greater, the section is too small.

Thus, let us determine whether the 2 X 12-inch plank is strong enough to sustain the load and its own weight. The plank will weigh about 120 pounds, making a total load of

1,500 + 120 = 1,620 pounds.

Hence the maximum bending moment is

1/8 Wl = 1/8 1,620 X 20 X 12 = 48,000 inch-pounds.

Since I/C = 1/6 ba2 = 1/6 X 2 X 122 = 48, and S = M/I÷C'

S = 48,600 / 48 = 1,013 pounds per square inch.

Strictly, therefore, the 2 X 12-inch plank is not large enough; but as the greatest unit-stress in it would be only 13 pounds per square inch too large, its use would be permissible.

2. What size of steel I-beam is needed to sustain safely the loading of Fig. 9 if the safe strength of the steel is 16,000 pounds per square inch ?

The maximum bending moment due to the loads was found in example 1, Art. 43, to be 8,800 foot-pounds, or 8,800 X 12 = 105,600 inch-pounds.