The Bending Moment at any point is the sum of the 8 6 W1 W2 802 10 16 ft.tons 4 tons W3 4 ft. M-100 B.M. DIAGRAM Verticals 100 Horizontals Fig.20. 100 ft,tons 10 ft. M=172 max. products got by multiplying each weight to the left of that point by its distance therefrom; the above is an arithmetical sum, since all the weights tend to bend the left portion downwards. Having found the bending moment at one point, we may derive the bending moment at another point nearer the fixed end and such that no weight intervenes, by adding the product of the sum of all the weights to the left into the distance between the two points, since no weights have to be considered, and all the leverages have increased by the distance between the two points. Hence the bending moments increase uniformly in each interval as you move towards the fixed end. new Bending Moment Diagram.-With a scale of feet for horizontals, lay off the length (fig. 20) and plot the positions of the loads; at each of these points, and at the fixed end, draw a vertical ordinate downwards, equal to the bending moment thereat, upon a suitable scale of ft.-tons for verticals; join the ends of these ordinates by straight lines. The ordinates are drawn downwards to signify that the moments on a cantilever are of a different sign, as compared to those on a beam. In either case the moments are all of one sign, which will always be reckoned as positive. Maximum Bending Moment.—It is evident that the maximum bending moment is at the fixed end, and is 3 M1 = Σ(Wx). 3 Graphical Solution.-Draw a vertical line through K, the fixed end, and draw the vertical lines W,, W, W, at the given horizontal distances apart upon a scale for dimensions (fig. 21). 21). Draw ab, bc, cd equal respectively to the forces W1, W, W, upon a scale for forces; that is, equal to 8, 6, and 4 tons in our example. Choose a pole O on the horizontal line passing through a, and such that the distance Oa is some convenient integral number upon the scale for dimensions; on the diagram Oa on the diagram Oa 10; and draw Ob, Oc, and Od. From A any point on W, draw AB parallel to Ob, and meeting W, at B; similarly, draw BC, CD, and AK parallel, respectively, to Oc, Od, and Oa. Then ABCDK is the bending moment diagram, its vertical ordinates give the bending moment at each point of the span upon a scale for verticals, obtained by subdividing 2 A B == 1 e K the divisions on the scale for forces by the number which Oa measures on the scale for dimensions; thus, on the figure this new scale is made by subdividing each division on the scale for forces into 10 equal parts, since Oa= 10 on the scale for dimensions. Proof-At any point e draw the ordinate eh, and produce those sides of the polygon that lie to the left till they meet it. On the figure, they meet it at e, f, g, h. As in the proof to fig. 19, the intercept on the vertical through e made by the pair of sides from any angle of the polygon, will, when measured upon the scale for verticals, give the moment about e of the force at that angle. Now the ordinate at e is the sum of the intercepts made by the pairs of sides from each angle to the left, and so will give on the vertical scale the sum of the moments of the forces to the left of e; that is, the bending moment at e. On fig. 21, ef the intercept by the pair of sides from A, gives the moment of W, about e; fg, the intercept by the pair of sides from B, gives the moment of W about e, and gh gives the moment of W, about e. Hence eh gives the sum of these three moments, that is the bending moment at e. = Beam loaded at the centre. Fig. 22. Let W be the load at the centre. By symmetry PQ W. For a section distant from the centre x towards the left, consider the forces on the left hand portion. The only force is P, and its leverage is (c-x); hence the equation to the bending moment. The value of Mx is zero at the end, that is where x = c; it increases uniformly as a decreases, that is, as you approach the centre, and it is greatest where x = 0, that is at the centre; by symmetry for the other half of the span, the value will decrease uniformly till it is again zero at the right hand end; hence the maximum bending moment In Rankine's "Applied Mechanics," the maximum bending moment in each case is given in the above form, viz., maximum bending moment = constant × total load × span. Mom. W.l. or Мо Throughout a great portion of that work, m stands for this constant, which he calls the numerical co-efficient of the maximum bending moment expressed in terms of the load and span. The value of m depends upon the manner of loading and of support. In the case we have just solved, we specify the mode of support by calling the piece a beam, and the manner of loading when we say that W is at the centre, and we find m = 1. W Bending Moment Diagram.-Upon a convenient scale of feet, lay off BC, fig. 22, equal to the span; construct a triangle with its apex above the centre O. Draw a scale of foot-lbs. to measure verticals upon, such that OA shall measure upon it one-fourth of the product of the load in lbs. into the span in ft. Graphical Solution. The simplest graphical solution is to draw the bending moment diagram as Q W PAW M≈ Fig 22. Draw ab vertical and equal above; one which is purely graphical may be made as in fig. 19, and it will not be necessary to use the first pole O', &c., as we know that P to the load W; from its middle point draw a horizontal line, choose O at a distance from ab equal to some convenient integral number on the scale for dimensions, and draw Oa and Ob. Then fig. 22 is constructed by drawing BA parallel to Oa, and AC parallel to Ob; and a scale for verticals is obtained by subdividing the scale for forces by the number chosen for the distance of O from ab. Cantilever loaded at the end.-Fig. 23. To find the bending moment at a section distant x from the fixed end K, consider the loads to the left of that section. The only force is W, and its leverage about the section at x is (c-x), and we have the equation to the bending moment. The value of Mx is zero when x equals c, that is at the free end; it uniformly increases as x decreases, and is a maximum when x = 0, that is at the fixed end; the maximum bending moment is Mo = Wc = W.l, and the value of the constant is m = 1. W Bending Moment Diagram.-Upon a convenient scale of feet, lay off KA, fig. 23, equal to the length; draw below KA the right angled triangle FAK, with the right angle at the fixed end, and construct a scale of ft.-lbs. for verticals, such that KF may measure upon it the product of the load in lbs. into the length in feet. Fig.23. Graphical Solution. The simplest graphical solution is to draw the bending moment diagram as above; one which is purely graphical may be made as on fig. 21. Make ab equal to W, and from a draw a horizontal line; choose a point ( such that Oa is a convenient integer on the scale for dimensions, and join Ob. Draw AF, fig. 23, parallel to Ob, and a scale for verticals is obtained by subdividing the scale for forces by the value of Oa. Beam uniformly loaded.-Fig. 24. Let w be the intensity of the uniform load in lbs. per foot of span. This is represented by a load area, consisting of a rectangle of height w feet standing on the span, and weighing one lb. per square foot. The total load W is the area of this rectangle, so that W2wc, W This weight may be considered to be concentrated at the centre of gravity of the area,-that is at its middle point; this gives a bending moment about the section equal to that for the actual distribution. We have two forces to the left |