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Examples on rectangular and square beams.

1. A rectangular beam of elm 10 inches deep and 8 inches broad, rests on two supports 10 feet asunder at its extremities, which are in the same horizontal line; find what weight suspended from its centre would produce fracture, the coefficient of fracture being 7850.

The formula of solution is (Arts. 38, 40),

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7 = 120 inches, b = 8, d = 10, and S = 7850.

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2. If in the last example the beam were fixed in two walls 10 feet asunder, what weight suspended 3 feet from one extremity would produce fracture?

If the beam merely rested on two supports in the same horizontal plane, the formula of solution (Arts. 38, 40) would be,

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7850, b = 8, d

=

10, 7 =

120, as before; and

of which S =
p = 3 × 12 = 36 inches, q = 7 x 12 = 84 inches.

Wherefore in this case

W = 41534.4 lbs.

But by Art. 33, the resistance of a beam when fixed in a wall is greater than when it is merely supported, in the ratio of 3 : 2. Increasing therefore W in this ratio, we get for the weight which produces fracture

W' = 62301 lbs.

3. In example (1) find the breaking weight when the weight of the uniform beam is taken into account, its weight per linear foot being taken at 18 lbs.

It is shown in Art. 35, that the moment of the strain on any point of a beam by a weight uniformly distributed along the beam, is the same as that of half the weight suspended from that point. Wherefore if W be the breaking weight, W' the weight of the beam, then by Ex. 1,

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But W' = 10 × 18 = 185 lbs., and therefore

W = 34888

185
2

7

= 34796, the weight of fracture.

18

Obs. In this example the point of suspension of the load, and the centre of gravity of the beam are coincident, and therefore

the plane of fracture is at the middle of the beam. The following example is one of a different kind.

4. A wrought-iron axle 12 feet in length, and of square section, is loaded at the distances of 2, 3, 6, and 8 feet from one end, with weights of 1200, 3000, 1800 and 4,000 lbs. The axle being placed on two supports in the same horizontal line, its coefficient of safety taken to be 7,000 lbs,, and the weight of a cubic foot of it 480lbs., it is required to find the length of a side (a) of the section, in order that the weights may be securely sustained in the two following cases:—

(1.) When the weight of the axle is neglected; (2.) When the weight of the axle is considered.

Let us first find the resultant R and point of application of the weights, as in Art. 36. Denote the distance of R from one end by h. Then by Statics,

R = 1200 + 3000 + 1800 + 4000 = 10000,

h =

1200 × 2 + 3000 × 3 + 1800 × 6 + 4000 × 8

10000

= 5.42 feet.

Hence, when the weight of the axle is neglected, the plane of fracture is 5.42 feet or 65.04 inches from one end, and 12 - 5.42 = 6.58 feet or 78.96 inches from the other. Where

fore (Art. 33),

=

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l

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10000, h=65·04, 7-h=78.96, l = 144, S 7000,

=

Now R

1

1

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a*, 12

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Substituting these values in (1) we readily get

a3 = 305-688, or a = 6-7 inches.

This is the side of the square section when the weight of the axle is neglected.

Let us now find the same when the weight of the axle is taken into account.

The plane of fracture evidently corresponds (Art. 36) to the point of application of the resultant R, and the weight W of the axle. The solution will hence be effected by equating the sum of the moments of the straining forces R and W, taken with respect to this plane, with the moment of resistance of the axle, in reference to the same plane. But this method is long and tedious, as the resulting equation is a biquadratic. The following approximative method, which is sufficiently accurate for all practical purposes, is to be preferred in all similar cases.

Neglecting the weight of the axle, we have seen that the side. of the square section is

α = 6.7 inches.

By means of this value of a, we can find an approximate value for the weight of the axle, viz.

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Let d be the distance of the plane of fracture from the end of the axle; then

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Substituting in this the values of 1, d, R, W, and h, we get

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a3 = 329 · 488, or a = 6 9 inches.

If, again, we find from this an approximate value for the weight of the axle, and proceed as before, a still more accurate result will be obtained.

5. From a three-sided prism a beam is to be cut, whose section shall be a rectangle, and whose moment of resistance shall be a maximum.

Let b denote the base AB (Fig. 19) of the triangular section of the prism, h its height, the breadth, and y the height of the rectangular section of the beam to be determined. Then the expression

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C

Fig. 19.

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is constant, viz. 2h; and therefore their product will be a maximum when they are equal to one another. This gives

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6. A beam AB, 6 feet in length, is supported at its extremities on two props in the same horizontal line, and is loaded by a weight of 500 lbs. at a distance of 2 feet from the end A; find the moment of the strain on the beam, neglecting its weight. Ans. 666.

7. Find the moment of the strain in the last example when the extremities of the beam are fixed in walls. Ans. 4444.

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