Page:Encyclopædia Britannica, Ninth Edition, v. 4.djvu/336

292 &quot;We must, therefore, on a beam of this kind ascertain whether -j- or is the greater, and replace - iu equations J c Jt 2 3 and 4 by the value of y in the larger of the two expres sions. Having selected y and / in this manner, we have—

6. y

{{ti|1em|§15. Modulus of {{nowrap|Rupture.—}}If the above hypothesis of a constant modulus of elasticity in a given material under both descriptions of stress, and under stresses of all magnitudes, were accurate, we should require no fresh experiments to determine the values of / and /, ; these would be already known from direct experiments on tension or compression. For both wrought and cast iron beams of the maximum strength, such as will be described hereafter, experiment gives results closely in accordance with strengths calculated by equation 6 ; and for wrought iror. beams the hypothesis appears to be approximately true when tested by experiment with any form of beam. But it is usual, in calculating the strengths of beams or bars of simple rectangular or circular cross section, to assume that y is equal to ^d and to employ equation 5 instead of equation 6. The imperfection of the theory is then to some extent corrected by determining/ from direct experiments on solid rectangular bars. The value of M I} I, and d being known, / is calculated from equation 5, and the number thus determined has received the name of modulus of rupture.}} VIII. Modulus of Rupture ? f (Rankine). Name of Material. Lbs. per sq. inch. Wrought Iron Plate Beams... .... ....... 42,000 Cast-iron Solid Bars ......................... 33,000 to 43,000 Fagersta Steel (Kirkaldy) .................. 110,000 ro 191,000 Red Pine ....................................... 7100 to 9540 Larch ........................................... 5000 to 10,000 Oak (British and Russian) .................. 10,000 to 13,600 Indian Teak .................................... 14,770 Sandstone .......... . ........................... ?100 to 2360 Experiments on the modulus of rupture have generally been made by hanging weights at the centre of a rect angular bar, supported at both ends, and increasing the weights till the bar breaks. Then let b and d and I be the breadth, depth, and length of the bar in inches, and W the breaking weight. M is a maximum at the centre, and, neglecting the weight of the bar, is equal to %Wl; substitut ing the value of I for a rectangle in equation 3, we get

and equating M and p. we have—

or, if the span is measured in feet and called L x, while I and d are measured in inches, we have—

2. /-18 W 2 Hence the modulus of rupture is stated by Rankine to be &quot; eighteen times the load required to break a bar of 1 inch square, supported at two points 1 foot apart, and loaded in the middle between the points of support.&quot; The use of a modulus of rupture determined by experi ment on a special form of beam is not based on any satis factory principle. The employment of this modulus is an imperfect means of correcting a defective theory. A dif ferent value is found for / when bars of different sizes or cross sections are tested. Even the same bar broken in different ways will give a sensibly different value for /. The use of this modulus is, however, convenient when great accuracy is not required.

§16. Expressions for the Bending Moment caused by Special Distributions of Load.—From equation 1, 14, it is easy to deduce the following values for the bending moments in beams subject to various simple distributions of load:— Case 1. Let a single load W be placed o.t the centre of a beam ; let M x be the moment at any section taken at the distance x from the nearest pier ; let M c be the moment at the centre of the span, we have—

1 ...... M = M c is the maximum moment at any section of the beam. Case 2. Let a single load W be placed at the distance x l from the nearest pier ; let M_,. be the moment at the sec tion where the load is applied—

M^-W 1 1 M^ is the greatest moment which the load at that place produces at any part of the beam, and increases as the load rolls from the end to the centre in proportion to the product (i-Xj) x : of the parts into which the load divides the beam. Case 3. Let the load be uniformly distributed along the beam so that each unit of length bears an equal load W ^v? ) when the unit of length adopted is the foot, the quantity iv is called the load per foot run. Let M c and M,( be as before the moments at the centre of span, and at any distance x from the nearest pier, AYC have—

4 ...... M e = wP,

When a uniform passing load, such as is approximately represented by a train of locomotives, of length at least equal to the span of the bridge, comes on to a bridge at one end and passes over to the other, gradually covering the whole span, the bending moment reaches a maximum for all sections when the bridge is wholly covered. The bending moment at any section for a combination of loads is the sum of the moments at that section due to each load taken separately. When many different distributions of load have to be provided for engineers are in the habit of representing the bending moment for each load by a line, the ordinates of which are the bending moments at the sections, and the abscissce the distances of the several sec tions from one point of support. The lines having been drawn for the several loads, it is easy by superposition to find the bending moment due to the combination, and thus to pick out for each section of the girder the maximum bending moment which any combination gives. Figs. 15, 15a, 156, and 15c show diagrams giving the curve of bending moments for some simple distributions. Fig. 15. Fig. 15 shows the line of bending moments when a single load of 10 tons hangs at the centre of a span of 50 feet. The vertical ordinates measured on the vertical scale give the bending moments at each section. Fig. 15 shows the curve of bending moments for a uniformly distributed load 