1

33. A beam 20 ft. span supports a load of 20 tons uniformly distributed. Find M. and M5.

and

Ans. Mm.W.l = × 20 × 20 = 50 ft.-tons;

M= = Mo

as may readily be seen from a bending moment diagram. 34. Find the equation to the bending moment in the previous example, and calculate Me from that equation. Also find the principal equation to the parabola which forms the bending moment diagram.

6

(c2x2) = (100-x); M = 32 ft.-tons.

Y = {X2.

35. In example 33, find M, directly, by taking a section at the point

= 6.

Beam divided at a number of points into equal intervals, and

loaded with equal weights at these points.-Fig. 28. Let W,

the total load, be distributed over the span l, in n parts each equal to w, and at equal intervals a apart; then

=

=

nw= W, (n+1)a 2cl, and P = nw. Take S, a point directly under one of the weights, and let BS = ra, then r will be a whole number; and if x be the distance of S from O the centre, then ra (c-x). On the portion of the beam to the left of S, there are in all forces, viz., P = nw acting upwards with a leverage about S of ra, and (r-1) forces each equal to w and acting downwards; the nearest to S has a leverage a, the next a leverage 2a, the next a leverage 3a, &c., and the last a leverage (r− 1)a; hence = :P.ra-w.a-w.2a-w.3a...-w.(r−1)a

which is the equation to the locus of T.

2c

2c

This equation gives the bending moment only at points where weights are, that is for values of x which are multiples of a, but not at intermediate points; it only differs from the equation we had for an uniform load, by the n+1

constant factor

n

The locus of T, the tops of the ordinates at the points where the weights are situated, is a

parabola with its axis vertical, and its apex above 0. Putting y instead of Mx for the ordinate to the curve at any point, we have

a maximum, and the height of the apex A in every case. On fig. 28, it will be seen that if n be odd, a weight comes exactly at the centre; so that yo, the ordinate of the parabola, is the maximum bending moment; hence the maximum bending moment

n+1

and the value of the constant is m = 108 n

When n is even, the maximum bending moment Mois less than Yo, and equals the ordinate of the parabola at the weight on either side of the centre; so that to find the value of M, it is only necessary to substitute for x half of an interval, that hence the maximum bending moment

C

is a, or

;

n+1

n+2

n+1'

and the value of the constant is m = }

The chords of the parabola give the bending moments at points intermediate between the weights, since the bending moment varies uniformly in these intervals.

COROLLARY. If the load be distributed in equal portions at equal intervals, the maximum bending moment exceeds that for the uniform distribution in the ratio

n+1 n+2

n

or

n+1'

according as n, the number of parts into which the load is divided, is odd or even.

Thus suppose M, is the maximum for uniform distribution; then if the load be concentrated at the centre, that is, if n = 1,

max. bending moment =

1+1

M2M,; see figs. 22 & 25;

if concentrated equally at two points dividing the three equal intervals, that is if n = 2,

span into

2+2
2+1

4

maximum bending moment = MU Mui
=

3

if concentrated equally at three points dividing the span into four equal intervals, that is if n = 3,

maximum bending moment =

4 3

The last two are each M, and are therefore equal to

each other; and for the same total load W placed on the span at equal intervals, the maximum bending moment M will be the same whether the load be divided into an even number of equal parts, or the next consecutive odd number of equal parts.

The Bending Moment Diagram is the polygon formed by the chords of the parabola.

COROLLARY. If n be great, the polygon nearly coincides n+1 with the parabola,

n

approaches unity, and the parabola

is nearly the same as that for the load uniformly distributed; that is, if the load be concentrated equally at a great number of points equally apart, as, for instance, when a girder supports, at equal intervals, the ends of cross girders which carry

equal loads, then the bending moments will be nearly the same as for the total load uniformly distributed.

Graphical Solution.-With a scale of feet for horizontals, lay off the span BC, fig. 28, and draw a vertical OA upwards through 0. Apply the parallel rollers to BC; place any parabolic segment cut on pear-tree against the rollers (see fig. 15) with its apex on the vertical through O; shift the rollers till the curved edge passes through B and C, which it will do simultaneously, and draw the dotted curve BAC. Draw up verticals to meet the parabola from the points at which the weights are, and draw the chords of the parabola. Construct a scale of ft.-lbs. for verticals such that W.l; where W = total load in lbs., l = span

OA =

18

n+1

n

in feet, and n the number of equal parts into which the load is divided.

Examples.

36. A beam 40 ft. span supports seven loads, each two tons, and placed symmetrically on the span at intervals of five feet. Calculate the maximum bending moment by substituting in the proper equation, and calculate at each load the height of the parabola which gives the bending moments.

Here W 14 tons, c = 20 ft., = 40 ft., and n = 7.
The maximum bending moment is

::

(c2 — x2) = 14 × ÷ (400-x2) = (400—x3).

at the weights, Mor-5=1(400-25)= 75 ft.-tons.