(Table IV; e= 800000

δ = 44.24180.750/800000.32 = 0,535" Had we calculated the deflection arithmetically from Formula (40) we should have had:

δ = 1/48. 1000.1923/ 800000.333 = 0,548" or practically the same result.

If the beam supported plastered work the deflection should not exceed, Formula (28)

δ = 16.0,03 = 0,48" Still, unless we were very particular, the beam could be passed as practically stiff enough.

Example III.

A white pine beam A B Fig. 154, of 12-foot span carries two loads, one w1 = 800 pounds, four feet from left support, the other w11 = 1200 pounds, two feet from right support. What size should the beam be?

Make A B at inch scale = 144 inches, locate w1 so that A w1,= 48", and w11 so that B w11 = 24". At any (pounds scale) make b c= 1200 pounds and ca=800 pounds. Select pole x distant from ba; a;xy= 900 pounds, the safe modulus of rupture per square inch of white pine; draw xb, xc and xa. Construct CBE G parallel to these lines. Draw C G, and parallel to same xo, then will a o = 733 pounds be reaction at A, and ob = 1267 pounds be reaction at B. We scale vertical DN at wl = 39" and TE at w11=35", therefore greatest bending-moment is at w1 and Formula (93) mWl = 39. 900 = 35100

Further, the required moment of resistance at w1 Formula (92) will be: r = DN=39.

Now from Table I, Section No. 2, r = b.d2/6, or b.d2 = 6.39 = 234

Now if b = 3" we should have d2 = 234/3 = 78 and or the beam would need to be 3" x 9".

Two Loads 100256

Two concentrated loads,

We should have obtained practically the same results arithmetically, for: Formulae (16) and (17):

Reaction A = 800.96/144 + 1200.24/144 = 733

Reaction at B = 800.48/144 + 1200.120/144 = 1267 check: A + B = w1 + w11 = 800 + 1200 = 2000 pounds and 733 + 1267 = 2000 pounds.

Beginning at B we have to pass over load w11 (1200 pounds) and on to w1, before passing amount of reaction B (1267 pounds) therefore greatest bending-moment at w1,. We know from Formula (24) it would be: mWl = 1267.96 - 72.1200= 35232 and check from Formula (23) mWl = 733.48 - 0.800 = 35184 being near enough for practical purposes. From Formula (18) we should have had: r = 35184/900 = 39,09

We now draw the shearing diagram O1 H IJ K L MO, as shown in Figure 154, and find the amount of shearing

from

A

to

w1

=

01H

=

733

pounds,

from

w1

to

W11

=

J S

=

67

pounds,

from

w11

to

B

=

MO

=

1267

pounds.

We can overlook it, tor even at the weakest point of beam for resisting cross-shearing we have half the area, or 3.9/2 = = 13 1/2 square inches.

White pine will safely resist 250 pounds per square inch in cross-shearing (Table IV) or the beam would resist.

13 1/2. 250 = 3375 pounds at its weakest point for cross-shearing, (viz.: at to,) and twice as much at the reactions.

To find the deflection we divide G C into eight equal parts, beginning with half parts (or l1 = 144/8 = 18") and draw the verticals through C D E G. We now make the lower load line g c equal the sum of these verticals, beginning at the top with the right vertical.

Select z distant from g c (the load line) zj= 108". Draw zg,zc, etc., and construct g1 c1f1 as before.

We draw z o parallel r, g1. Now g o measures 116 inches and o c

108 inches, therefore divide c1 g1 at f so that: c1f::fg1 = 116: 108 Carrying the vertical ff up to point F of beam, we find the point of greatest deflection F, where

B F= 69 1/2" and A F= 74 1/2" We find ff1 scales 42", remembering that (Table I, Section No. 2) i = 3.93/12 182, and that for white pine Table IV e - 850000 pounds we have Formula (95):

δ = 42. 18. 108. 900= 0,475"

850000 .182

Had we attempted to get this result arithmetically by inserting the values in Formula (41) (and remembering that n is always the nearer support, or in our case respectively 48" and 24", while m respectively 96" and 120") we should realize the advantage of the graphical method, for:

Two Loads 100257

If we figure out the above tedious formula we should have δ = 0,422" or practically the same result as we obtained graphically.

The safe deflection, were the beam to carry plastering, should not exceed Formula (28)

δ= 12.0,03 = 0,36" Our beam is therefore not nearly stiff enough, and we must make it thicker; or else if we wish to save material, we will make it thinner, but deeper; and then brace it sideways, see Formula (31).

Example IV.

A spruce girder A B of 18-foot span carries five loads, as shown in Figure 155. What size should the girder be?

Five Concentrated Loads.

We draw A B

=

216" (inch scale); further

ba

=

2700 pounds (sum of loads at pounds scale); make

bh

=

wv

=

540

pounds,

he

=

w1v

=

180

pounds,

e d

=

w111

=

360

pounds,

dc

=

w11

=

720

pounds, and

c a

=

w1

=

900

pounds.

Two Loads 100258

Fig. 155.