Select x distant x y= 1000 pounds from b a, (as 1000 = (k/f) for spruce, see Table IV). Draw xb, x h, xe, etc., and figure C D G. Draw xo parallel C G; it divides load line as follows: a o= 1580 pounds or reaction at A. ob = 1120 pounds or reaction at B. We find longest vertical through C D G, is at load w11, therefore greatest bending-moment on beam at w11; now D E scales 70 1/2", therefore Formula (93): mw11 = 70 1/2. 1000 = 70500 and Formula (92) r = 70, 5 From Table I, Section No. 2 r = b.d2/6 = 70, 5 and if b = 5, we have 5. d2 = 6. 70, 5 or d2 = 84,6 and or say 10" which is the nearest size larger than 9,2", and of course wooden beams are never ordered to fractions of inches.

Five Concenteated Loads 100259

Had we worked arithmetically we should have had practically the same results.

From Formulae (16) and (17) we should have had; reaction at A = 1580 pounds, reaction at B = 1120 pounds.

From rule for finding greatest bending-moment we should have located it at w11 and then had Formula (23) mw11 = 1580. 72 - 48.900 = 70560 and from Formula (18) r= 70560/1000 = 70,56

We now draw the shearing diagram O1 H I J K L M N P O and find as follows:

Cross-shearing

A

to

wl

=

HO

=

1580

pounds.

Cross-shearing

W1

to

w11

=

JK

=

680

pounds.

Cross-shearing

w11

to

w111

=

KL

=

40

pounds.

Cross-shearing

w111

to

WIv

=

MR

=

400

pounds.

Cross-shearing

WlV

to

wv

=

N S

=

580

pounds.

Cross-shearing

wv

to

B

=

P O

=

1120

pounds.

We need not bother with it, therefore. For deflection we now divide C G again into eight equal parts, (or l1, = 213/8 = 27") beginning with half parts at C and G. We now make lower load line g c = the sum of the eight verticals, putting the right vertical at the top from g down. We select pole z at a distance zj = 120" from g c and draw zg, zc, etc. We construct figure g11f1 c1 and draw zo parallel to c1g1 We now divide c1 g1 at f, so that g1f: fcl = co: og, carrying ff1, up to beam, we have the point

F, distant 102" from B, and 114" from A, which is the point of greatest deflection. We find that ff scales 102", remembering that e =850000 for spruce (Table IV), and that i=5.103/12 =417 (See

Table I, Section No. 2) we have, Formula (95).

δ = 102. 27.120. 1000 = 0,93" 850000.417

This would be too much for plastering, for if the girder supported plastering, the deflection should not exceed Formula (28) §=18. 0, 03 = 0,54"

We must therefore deepen the beam very materially. We use Formula (31), x = 1/b.d3 In our case it would be x= 1/5.103 = 1/5000 = 0,0002

Supposing we were to make the beam 4"xl2", then we should have x = 1/4.123 = 0, 000 144

The deflection of the latter, then, would be δ: 0,93 = 0,000144: 0, 0002 or

δ = 0,93.0.000144/0,0002 = 0,67" still too much deflection.

Were we to make the beam 3" x 14", we should have: x = 1/3.143 = 0,0001215

The corresponding deflection for this beam would be: δ: 0,93 = 0,0001215: 0,0002 or

δ = 0,93. 0,0001215 = 0,565" 0,0002 or just about what would be required in the way of stiffness.

Five Concenteated Loads 100260

Fig. 156.

Had we used Formula (95) we should have had, remembering that now i = 3.143/12 = 686

δ = 102. 27.120.1000

850000.686 = 0,568" showing that we have made no mistake in applying Formula (31).

If we have any doubts as to whether a 3" x 14" stick is as strong as a 5" x 10" we use Formula (30) and have for the former x - 3.142 = 588 while for the latter x = 5.102 = 500, so that the 3" x 14" stick is actually much stronger, as well as much stiffer than the 5" x 10". It is, however, a very thin beam, and would be apt to warp or twist, unless braced sideways about every five feet of its length.

To attempt to get the deflection of the girder arithmetically would be a very tedious operation. It could be done, however, by inserting in Formula (41) the different values for n and m, remembering every time to make n the distance from each weight to the nearer support to respective weight, and m the distance from same weight to the further support.

Example V.

Uniform Load. A. wrought-iron beam of 25-foot span (Figure 156) carries a uniform load of 800 pounds per running foot of beam, including weight of beam. The beam is thoroughly braced sideways. What beam should be used?

We draw A B = 300" at inch scale, and then divide our uniform load into a number of equal sections, say eight, each t1 = 300/8 = 37 1/2" long

The total load on beam is u = 25.800 = 20000 pounds.

Each section therefore carries: u/8 = 20000/8 = pounds.

We place our arrows w1, w11, etc., at the centre of each section, which will bring the end ones at l1/2 distant from each support, so that these same verticals will answer when obtaining deflection figure.

We now make b a = 20000 pounds at pound scale, and divide it into eight equal parts, each equal wl=w11 = w111 etc.,= 2500 pounds. We make xy = 12000 pounds, which is the (k/f) for wrought-iron, see

Table IV. We draw xb, xa, etc., and construct figure C E G, which will approach a parabola in outline. The more parts we take the nearer will it be to a parabola.

We draw x o parallel C G and find it bisects ba, or each reaction is one-half the load or= 10000 pounds. This we know is the case. The longest vertical will, of course, be at the centre D of C G, or greatest bending-moment will be at the centre, this we know is the case. D E scales (inch scale) 62 1/2" which will be the required r or moment of resistance (Formula 92). The bending-moment at the centre will be, Formula (93).

m= 62 1/2. 12000 = 750000 Had we used Formula (21) we should have had m =20000.300/8 = 750000 or same result, and from Formula (18) for r = 750000/12000 = 62 1/2 also the same as before. From

Table XX we find the nearest r to our required r (62,5) is 69,8 which calls for a 15" - 150 pounds beam; as the beam is braced sideways this will do, if sufficiently stiff.

In regard to shearing, we draw the figure O1 H IJ K L M N P R S O and find shearing on both sides of beam similar, increasing gradually from the centre to ends.1