A erent calculations to ascertain the amounts of bending-mo-ments, the required moments of resistance and inertia, the amounts of reactions, vertical shearing on beam, d e fl e c-tions, etc., can be done graphically, as well as arithmetically. In cases of complic at cd loads, or where it is desired to economize by reducing size of flanges, the graphical method is to be preferred, but in cases of uniform loads, or where there are but one or two concentrated loads, the arithmetical method will probably save time. As a check, however, in important calculations, both methods might be used to advan tage.

Fig. 149.

If we have three concentrated loads w, w1, and wl1 on a beam A D (Fig. 149), as represented by the arrows, we can also represent the reactions p and q by arrows in opposite directions, and we know that the loads and reactions all counterbalance each other. The equilibrium of these forces will not be disturbed if we add at E a force = + y, providing that at F we add an equal force, in the same line, but in opposite direction or =

- y

We have now at E two forces, + y and p. If we draw at any scale a triangle a o x (or I) where a o parallel and =p, and where o x parallel and = + y, we get a force xa, which would just counterbalance them, or a x, which would be their resultant. That is, a force G E thrusting against E with an amount a x (or x1) and parallel a x would have the same effect on E as the two forces + y and p. Continuing xl till it intersects the vertical neutral axis through load w at G, we obtain the resultant x2 of the two forces acting at G, namely x1 and w (see triangle baxor II). Similarly we get resultant x2 at H, of load wlandx2, (see triangle cbxor III); also resultant xi at I of load w11 and x3 (see triangle dcx or IV); and finally resultant + y, at F of reaction of q and x4 (see triangle o d x or V). As this resultant is + y it must, of course, be resisted by a force - y that the whole may remain in equilibrium. By comparing the triangles I, II, III, IV and V, we see that they might all have been drawn in one figure (Fig. 150) for q + p = w11 + w1 + w, therefore: do+oa=dc+cb+ba,

Basis of Graphical Method.

 further both V and IV contain d x = x4 " " V " I " ox = y " " II " I " ax = x1 " " II " III " bx = x2 " " III " IV " cx = x3

We know further that the respective lines are parallel with each other.

 In Fig. 150 then, we have dc w11 cb = w1 ba = w ao = p and od = q

The distance xy of pole x from load line da being arbitrary, an:l the position of pole x the same. The figure E G H I F E (Fig 149) has many valuable qualities. If at any point K of beam we draw a vertical line KL M, then L M will represent (as compared with the other vertical lines) the proportionate amount of bending moment at A*. If we measure L M in parts of the length of A D and measure xy (the distance of pole, Fig. 150) in units of the load line da, then will the product of LM and xy represent the actual bending moment at K. That is, if we measure LM in inches and - (having laid out d c, cb, etc., in pounds) - measure x y in pounds, the bending moment at K will be = x y. L M (in pounds-inch.) Similarly at w the bending moment would be

 = xy. N G (in pounds-inch.) and at w1 it would be = xy. R H " " " and at w11 it would be = X V. SI " " "

measuring, in all cases, x y in pounds and N G, R H and Sim inches.

The area of EGHIFE, divided by the length of span in inches will give the average strain for the entire lengtn on extreme top or bottom fibres of beam, providing the beam is of uniform cross-section throughout. The area should be figured by measuring all horizontal dimensions in inches, and all vertical dimensions in parts of the longest vertical (R H in our case), this longest vertical being considered = (c/f) for top, or (t/f) for bottom fibres, or where these are practically equal = (k/f).

The greatest bending moment on the beam will occur at the point where the longest vertical can be drawn through the figure. From this figure can also be found the shearing strains and deflection of beam, as we shall see later.

If now instead of selecting arbitrarily the distance x y of the pole from load line d a (Fig. 150) we had made this distance equal the safe modulus of rupture of the material, or x y = (k/f) - measuring x y in pounds at same scale as the load line d a - it stands to reason that any vertical through the Figure E G HIF E (Fig. 140) measured in inches, will represent the required moment of resistance, for if LM. xy=m, we know from

Fig. 150.

Fig. 151.

Formula (18), that m = r. (k/f) and as we made x y=(k/f) we have, inserting values in above:

LM(k/f) = r(k/f)or

LM=r