Shear

The loads and reactions, besides causing bending or flexure, create shearing stresses in the beam by their opposing tendency; that is, as the reactions act upwards and the loads downwards, the effect is to shear the fibers of the beam vertically. At any section of a beam, the shear is equal to either reaction minus the sum of the loads between that reaction and the section considered. The maximum shear is always equal to the greatest reaction. For a simple beam with a uniformly distributed load, the maximum shear is at the supports, and is equal to one-half the load, or to the reaction; the shear changes at every point of the loaded length, the minimum shear being zero at the center of the span. The maximum shear in a simple beam having a single load concentrated at the center is equal to one-half the load, and is uniform throughout the beam. Where a beam supports several concentrated loads, changes in the amount of shear occur only at the points where the loads are applied.

For example, in the beam loaded as shown in Fig. 16, the shear on the line a b is equal to the reaction R1 of 40 lb. minus the load n of 10 lb., or 30 lb. The shear between o and p, on the line cd, is equal to the reaction R1 of 40 lb. minus the sum of the loads n, m, and o, or zero. Working from the same reaction R1 the shear on the beam between p and q is equal to R1 - (n+m + o + p), or 401b. - 55 lb. = -15 lb). Thus, all the beam to the left of o is in what may be called positive shear, while to the right of p the shear is in the opposite direction, and may be called negative shear. It is evident that then a section in the beam - for instance, cd - where the shear changes from positive (+) to negative (-); that is, it passes through zero, or changes sign.

Example

(a) What is the maximum shear on the beam shown in Fig. 15? (b) What is the shear 11 ft. from the right support? (c) Where does the shear change sign?

Solution

(a) Taking the center of moments at R2, the reaction at R1 is as follows:

Theory Of Beams 243

Fig.16.

9,000 lb. X 10 ft. =

90,000 ft.-lb.

3,000 lb. X 10 ft. X 17 ft. =

510,000 ft.-lb.

4,000 lb. X 26 ft. =

104,000 ft.-lb.

Total =

704,000 ft.-lb.

Then, 704,000 / 30 = 23,466 2/3 lb., the reaction R1. The total load being 43,000 lb., the reaction R2 is 43,000 - 23,466 2/3 = 19,533 1/3 lb.; hence the maximum shear is R1. (6) The reac-tion R2, being 19533 1/3 lb. and then being but a load of 9,000 lb, between it and a section 11 ft. distant, the shear at the latter point is 19,533 1/3 - 9,000 = 10,533 1/3 lb. (c) Working from R1 the first load is 4,000 lb., and the shear at this- point if 23.466 2/3 -

4,000 = 19,466 2/3 lb. Then enough of the uniform load must be taken to equal this amount. It is clear that the shear becomes negative somewhere in the uniform load, since the latter is 30,000 lb., or more than 19,466 2/3 lb. Dividing 19,466 2/3 by 3,000, the load per foot, the result is 6.48 ft., the distance from left end of uniform load to point where the shear changes sign; hence the distance from R1 is 4 + 4 + 6.48 = 14.48 ft.

Bending Moment - The algebraic sum of the moments of the external forces about any point in a beam is the bending moment at that point; that is, the bending moment at any point is the moment about that point of either reaction minus the sum of the moments of the intermediate loads about the same point. For example, the bending moments at several points on the beam shown in Fig. 17 are as follows: At W1=R1a; at W2=R1(a+b)-W1b; at W3 = R1(a + b + c) - [W2c+ W1(b+ c)], or R2 d, etc.

The bending moment varies, depending on the shear, and attains a maximum value at the point where the shear changes sign. If the loads are concentrated at several points, the maximum bending moment will be under the load at which the sum of all the loads between one support up to and including the load in question first becomes equal to, or greater than, the reaction at the support. Hence, to find the maximum bending moment in any simple beam:

Rule

Compute the reactions and determine the point where the shear changes sign. Calculate the moment about this point of either reaction, and of each load between the reaction and the point, and subtract the sum of the latter moments from the former. Example. - What is the maximum bending moment in inch-pounds of the beam loaded as shown in Fig. 18?

Solution

Taking moments about R1 and remembering that a uniform load has the same moment as an equal load •concentrated at the center of gravity of the uniform load, the total moment is found to be 358,250 ft.-lb. The span being •25 ft., the reaction R2 is 358,250 / 25=14,330 lb. As the sum of the loads is 32,500 lb., the reaction R1 is 32,500 - 14330

Theory Of Beams 244Theory Of Beams 245

= 18,1701b. This is the greatest reaction and the greatest shear.

Beginning at R1 and adding the loads in succession, it is found that the load of 10,000 lb. plus the uniform load between the reaction and the load d is 10,000 + (500 X 13) = 16,500 lb., which is less than the left reaction, or R1; when, however, the load d is added, the sum of the loads is greater than the reaction; hence, the shear changes sign under the load <I, and the greatest bending moment is also at that point. Taking moments about the point under the load d, the moment of R1 is 18,170 X13 = 236,210 ft.-lb. The moments of the loads between d and R1 are:

6,500 lb. X 6 1/2 ft. = 42,250 ft.-lb. 10,0001b. X 7 ft.= 70,000 ft.-lb.

Total = 112,250 ft.-lb.