δ = v1.l1,.zj.xy/e.i. (97)

Amount of Deflection, Definite Pole Distance.

Deflection varying Cross-section.

Deflection Pole Distance arbitrary.

1 This would be the greatest possible deflection. If the beam were not so proportioned, but of uniform cross-section throughout, the deflection would be less.

## Deflection

Where δ, v1, l1 e, z j, and i same value as in Formula (95).

Where x y = the length of pole distance from load line in upper strain diagram, measured in pounds.

The same formulas and methods could be applied to cantilevers, but for these the arithmetical calculations are so very simple that it would be taking unnecessary trouble.

A few practical examples will make all of the foregoing more clear.

Example I.

A Georgia pine girder A B of 20-foot span carries a load wl of 2000 pounds 5' 0" from right reaction B. What size should the girder be?

We draw (Figure 152) A .5 = 240" at inch-scale, and locate w1 at 60" to the left of B. Now draw a vertical line b a = 2000 pounds at pounds-scale. Select point x anywhere, but distant x y = 1200 pounds. (1200 pounds being = (k/f) or the safe modulus of rupture, per square-inch, of Georgia pine). Draw x b and x a. Draw verticals through A, wl and B. On vertical A begin at any point C, draw C E parallel x a, till it intersects verticals w1 at E; then draw E G till it intersects vertical B at G. Draw G C and o x parallel to G C. We scale o b, it scales 1500 pounds, so this, is the reaction at B. We scale a o, it scales 500 pounds and this is the reaction at A. The longest vertical through C E G is vertical w1, therefore greatest bending-moment is at w1 which we know is the case. We scale E D at inch-scale, it scales 75 inches, therefore the (greatest) required moment of resistance will be at w1 and will be Formula (92). r = 75. From Table I, section No. 2, we know for rectangular beams, r =b.d2/6, therefore: b.d2/6 = 75, or b. d2 = 450. We will suppose the girder is not braced sideways, and needs to be pretty broad; let us try b = 5", we have then: 5. d2 = 450 or d2=450/5 = 90 and d = 9, 5" or the girder would have to be 5" x 9 1/2" or say 5"x 10". The bending-moment at to, is, of course, Formula (93) = E D. x y = 75.1200 = 90000 (pounds-inch).

Had we calculated arithmetically, we should have had, Formulae (14) and (15):

 reaction A = 60/240.2000 = 500 pounds. " B = 180/240.2000 = 1500 pounds.

Beading moment at wl would be (right side) Formulas (23) and (24). mW1 = 1500.60 - 0.2000 = 90000 (pounds-inch) or check (left) side mW1 = 500.180 - 0.2000 = 90000 (pounds-inch.) Therefore required moment of resistance, Formula (18)

R=90000/1200=75 or same result as graphically.

By drawing the horizontals from b between verticals B and wl; from a between verticals A and wt; and from o between verticals A and B we get the etched figure for measuring vertical shearing strains. We see at a glance that the shearing to the right of load is equal to the right reaction, and is constant at all points of right side of beam; while on the left side of load it is equal to the left reaction, and is constant at all points of the left side of beam. And this we know is the case. We need not bother with shearing, however, for we can readily see there is no danger. For even immediately to the right of the load, the weakest point in our case, we know that one-half of the fibres of cross-section are not strained at all, or we should

5 10 have one-half of area or5.10/2= 25 square-inches to resist 1500 pounds of shearing, or1500/25 = 60 pounds per square-inch, while the safe resistance, per square-inch, of Georgia pine to shearing across the grain is (Table IV) (g/f) = 570 pounds.

There is, however, some danger of excessive deflection; we draw, therefore, the figure c1f1 g1 by dividing the beam into ten equal parts, beginning and ending with half parts at the reaction, (each whole part being 24" long, or l1 = 240/10=24")

We draw the verticals through these parts and get their lengths through figure C E G. These lengths we carry down in their proper succession on the load line gj c of the lower strain diagram, beginning at the top with the right vertical 1, putting immediately under this the length of second vertical 2, then 3 and so on till gc = sum of lengths of all ten verticals through C E G. We now select z at random (in our case 120 inches from load line or zj = 120"). We now draw lines from z to g I, II, III, etc., to c. Construct figure g1f c1 by beginning at g1 drawing line parallel to zg until it intersects prolongation of first vertical 1; then line parallel to zI till it intersects prolongation of second vertical 2, etc. We now draw z o parallel c1 g1,. We scale g o and find it scales 222", also c o which scales 1G2"; we divide c1 g1 a f, so that c1f;g1 = 222: 162.

Carrying vertical ff through figure we find it scales (v1) = 117" continuing ff1, up to beam it gives us point Fas the point of greatest deflection, we find A F scales 138". Had we used Formula (43) we should have located F at a distance from A or A F=

= 134, 17". So that we have a sufficiently accurate result.

For the amount of deflection at F we use Formula (95); we know that (Table I, Section No. 2) i = b.d3/12 = 5.103/12= 417, further for Georgia pine (k/f) = 1200 pounds.

e = 1200000 (inch-pounds.) l1 = 24" v1=ff1, = 117" zj=120", therefore: δ = 117. 24.120. 1200 1200000. 417 = 0.808"

Had we calculated the deflection by Formula (41) we should have had: remembering that m = 180" and n = 60" and l+ n = 240+60 = 300" ______

Which proves the accuracy of the graphical method.

For a beam of 20 feet span the deflection not to crack plastering should not exceed, Formula (28). δ = 20.0,03 = 0,6"

Therefore, if our beam supports a plastered ceiling, it must be redesigned to be stiffer. Either made deeper, in which case it can be thinner, if braced sideways, or it can be thickened sufficiently to reduce the deflection, see Formula (31).

Example II.

A hemlock girder A B (Fig. 153) of 16-foot span, arries,a centre load w of 1000 pounds. What size should the girder be?

We make A 5 = 192" at inch scale; locate w at its centre F; make b a at any scale - (pounds-scale) - equal 1000 pounds. Select pole x distant, xy=750 pounds, from load line b a, (as 750 pounds = (k/f) the safe modulus of rupture per square inch of hemlock). Draw xb and x a. Begin at G, draw G E parallel b x to vertical through load, and then draw E C parallel ax. Draw CG and then x o parallel C G, we find that o bisects b a or a o=0o b =500 pounds. Each reaction is therefore one-half of the load; this we know is the case. Greatest line through C G E we find is at D E, so that greatest bending-moment is at load; this we know is the case. DE scales 64" at inch-scale, therefore the required moment of resistance for the beam is, Formula (92): r = 64. and the greatest bending-moment at load, Formula (93): mw = 64. xy= 64.750 = 48000 Had we calculated arithmetically we should have obtained the same results, for Formula (22) mw = 1000.192/4= 48000 and Formula (18): r = 48000/750= 64

Now from Table I, Section No. 2, we know that for rectangular sections: r = b.d2/6or b.d2= 64.6 = 384. If we assume the beam as 4" thick, we have then: 4.d2 = 384.

d2 = 384/4 = 96 or or we will make the beam 4" x 10".

We draw the figure 0, H J K N O for shearing and find it is constant throughout the whole length of beam and equal to length O1 H or N O measured at pounds scale, or 500 pounds. This is so small we need not bother with it.

To obtain the deflection diagram we divide G C into eight equal parts, each part l1 = 192/8 = 24" and begin at each end with half parts, drawing the eight verticals through C E G.

We lay off their exact lengths in proper succession on the lower load line g c, beginning at the top with the right vertical. Select pole z at random, in this case distant from load line zj= 180". We now draw the figure c1 g1 fl and find greatest deflection is at its centre fft; for z o parallel c, g1 bisects gc. We scale ff1 at inch scale

Fig. 154.

= 44". therefore greatest deflection of beam at centre, Formula (95), remembering that (Table I, Section No. 2) 1= 4.103/12 = 33 and