To find the bending moment at the centre of the beam, sub stitute for x in equation (6), when we get 2 105. Diagram. When a = or a < the diagram of moments 2' a 4 2' may be constructed thus:-Draw the horizontal line, A, B1, (fig. 58a) to any scale, equal in length to the beam. From its centre, O, lay off OC=OD= ; the maximum bending moments will occur at C and D, and may be found by means of equation (8). Draw the verticals C C, D D1, equal to these moments on a scale of bending moments, and construct the parabolic curves A, C, E and B, D, F, intersecting at the point O,; then A, C, O, D1 B1 will represent the diagram of maximum bending moments for the beam during the passage of the load. The maximum moment at any point, such as G, is at once found by measuring the ordinate G G1 at this point. Example 4.-Two wheels of a loaded truck pass over a beam 30 feet span. If the load on each wheel be 5 tons, determine the maximum bending moment on the beam, and also the bending moment at its centre, the wheels being 10 feet apart points of the beam situated at a distance of 2.5 feet at each side of the centre. Their values are found from equation (8), viz. : The maximum moment at the centre of the beam is found from equation (9), viz. : Example 5.-In the last example, determine the greatest bending moment at a section 8 feet from the left abutment. The greatest bending moment at this section will occur when the left wheel rests upon it, and its value is found from the general equation— 106. Beam supported at Both Ends and Loaded with two Concentrated and Unequal Weights at a fixed interval apart, and moving over the Beam.-If A B represent the beam, W, and W, the two weights, and a = distance between them, then by adopting the same process of reasoning as that given in the last case, it may be shown that when both weights are on the beam the bending moment at any section of the beam is greatest when W1 is over it, provided that the section is situated between W1l A and F, where A F= W1+ W 1 1 1 2 When BF= section between F and B will occur when the weight W2 rests upon it. the maximum bending moment at any 2 1st. Suppose the section to lie between A and F; let x = its distance from A; then when W, rests upon it, let 21 = its 2nd. Suppose the section to lie between F and B; distance from A. The maximum bending moment will occur at this section when W, rests upon it; in which case 2 The loci of the bending moments, as represented by the two equations (10) and (11), are two parabolas, the axes of which are vertical and intersect the beam at two points, one at each side of the centre. At these points the moments are greatest for each half of the beam. The section where the greatest bending moment occurs in the left half of the beam is situated at a distance to the left of the W2 α centre = 2 (W1 + W1) The section in the right half, where the bending moment is a maximum, is at a distance from the centre = The distances x and x, of these points from A are— W1 a (12). (13). The values of these maximum bending moments may be found from equations (10) and (11) by substituting for x and x, the X1 values given in equations (12) and (13). An expression for the maximum bending at the centre of the beam may be found by putting x = 2 in equations (10) and (11). W2a (14), 1 2 (15). 4 2 W1a In the foregoing investigation the results arrived at are on the assumption that both weights rest on the span. If W2> W1, the maximum moment near the end, A, of the beam will occur when W, only is on the beam, the other weight resting on the abutment. Example 6.-A travelling load, concentrated on two wheels. 10 feet apart passes over a beam of 40 feet span. If 2 tons rest on the left wheel and 8 tons on the right, find the maximum bending moment at the centre of the beam, and also determine the section on the beam which has the greatest maximum moment, and find its amount. The bending moment at the centre will be a maximum when the weight of 8 tons rests on it. The maximum moment on the whole beam occurs at a section whose distance from the left abutment is found from equation (13), The section is, therefore, 21 feet from the left bearing, and the amount of the bending moment is found from equation (11) by putting x1 = 21 40 21. Thus The maximum bending moments at every 4 feet of the beam, reckoning from the left abutment, are 70, M16 = 84, M20 M, 0, M, 28.8, M. = 512, M12 70, M16 = = M21 88, M28 = 78, M2 = 60, M3 34, M40 = 0. = 90, The values, M, and Ms, are those produced when only one weight-viz., 8 tons-rests on the beam, the weight of 2 tons. resting on the left abutment. 107. Graphic Method of Finding the Position of the Rolling Load so as to produce the Maximum Bending Moment at any Section. A convenient method of finding by graphic construction what must be the position of the rolling load on the beam, in order that the bending moment at any section may be F Fig. 59. a maximum, is the following, which may be made to apply to any number of weights. over Let A B (fig. 59) represent a beam which passes a rolling load consisting, say, of four weights, W1,W,W, W4. We wish to determine the position of these loads on the beam so as to produce the maximum bending at any section, say at F. Draw a vertical line through A and on it set off Join x, B. Through F draw F F, parallel to x, B, meeting the vertical line through A at the point F1. The maximum moment at F will occur when the weight, represented by the division in which F is situated, rests on F. In the figure F1 lies in x1 x2, which represents the weight W2. The maximum bending at F will, therefore, occur when the weight W, rests upon it. 2 CHAPTER VIII. SHEARING FORCES ON BEAMS. FIXED LOADS. 108. Definition.-The shearing force at any transverse section of a beam is equal to the algebraic sum of all the external forces acting upon either segment of the beam into which the section divides it. We will illustrate this definition by a few simple examples. Suppose A B (fig. 60) to be a beam loaded with a single weight W placed at a distance a from A, then the proportion of W transmitted to the left abutment will be W -a will, therefore, represent the shearing force throughout. the segment A C. The shearing stress on the segment B C If W rest on the centre of the beam, the shearing stress will be constant throughout W and equal to 2 P Wa = W C ·1-a---> B, B2 Fig. 60. |