TABLE 104.-Of the Ratios of DEFLECTION in BEAMS VARIOUSLY

LOADED.

1.000

99

=

.00 .05 . 10 •15 • 20 •25 .30 .35 • 40 • 45

1.000
1:003
1:012
1:023
1:042
1.067
1.099
1.140
1.191
1.254

99

1.0000
.9950
.9801
.9555
•9216
.8789
.8281
• 7700
•7056
6360
•5625
• 4865
• 4096
.3335
•2601
•1914

1296
•0770
. 0361
.0095
0000
(2)

20 x 20 = 400
21 x 19 399
22 x 18 396
23 x 17 391
24 x 16 384
25 X 15 375
26 x 14 364
27 x 13 351
28 x 12

336
29 x 11

: 319 30 x 10 = 300 31 X 9 = 279 32 x 8 = 256 33 x 7 = 231 34 x 6 = 204 35 X 5 = 175 36 x 4 = 144 37 X 3 = 111

76 39 x 1 3 39

1.333

•50 •55 .60 • 65 • 70 • 75 .80 .85 .90 .95 1:00 (1)

1.0000

.9975 .9900 .9775 .9600 .9375 •9100 .8775 • 8400 • 7975 • 7500 .6975 .6400 •5775 •5100 • 4375 • 3600 • 2775 .1900 0975 0000 (5)

1.434 1.564 1.732 1.961 2.286 2.778 3.604 5.263 10.260 infinite

38 X

99

40 X

(3)

(4)

would be with that same weight in the centre: say we find by calculation or experiment it was 1] inch: then the half-length being 12 feet, our point is obviously 6 - 12 = 0·5, for which col. 2 gives 0.5625: hence the deflection at that point = -5625 x 1} = 0.7 inch.

(652.) “Safe Load.”—We have here taken the load as constant, at whatever point in the length it might be placed, but the safe load increases as it moves from the centre toward the end in inverse ratio of the product of the two parts into which the length is divided by the weight (420), for example with a beam 20 feet long, a weight in the centre divides it into two lengths, each 10 feet, and we have 10 x 10 = 100: now say

=

that the weight is 3 feet from one end, therefore 17 feet from the other: then, 3 x 17 = 51, or about half that with central load, showing that the equivalent load there is double the central load, the beam being equally strained in both cases, although the actual weights are in the ratio of 2 to 1, &c.

In Table 104 the whole length is divided into 40, and the half-length into 20 parts; col. 4 gives the ratio of the equivalent or safe load at each point, and col. 5 the ratio of the deflection at that point due to that load. Thus: the deflection with any central load being 1.0, at a point midway between the centre and the end, or .5 of the half-length, it would become • 5625 with that same weight at that point by col. 2: but the safe load would then become 1:0 x 400 = 300 = 1.333, as in col. 4, with which the deflection would be increased to .5625 x 1.333 = •75, as in col. 5. The curve C in Diagram, Fig. 213, has been obtained from col. 5, and this curve, it may be observed, is a parabola, but differs very slightly indeed from a simple spherical curve.

“ Curve of Flexure for Un-central Load.”—When the load is out of the centre, as in Fig. 175, the elastic curve may be found by the Rule:

(1 L) + 2 (653.) y = 38 x

x{L
(1 L) – z) x 2

Хz xx (1 L)

3
L– x
+

6

=

In which L = length between bearings: 8 = deflection in

8 centre which a given weight would produce if placed there : z = the distance of the same weight from the centre of the beam: x =

the distance from the weight towards the nearest support, of a point in the curve of flexure where the deflection is required: y = co-ordinate of the curve at that point.

(654.) The deflection produced by the weight at the point of application may be found by the Rule (650), &c. : for instance we have there calculated that with a certain beam 10 feet long, a weight which at the centre gave 8 3 inches, or foot, produced a deflection of 2·1168 inches at 2 feet from the centre when placed there. Now, if we make x = to the whole

* 2 x 2x3)

distance from the weight to the support, y would obviously be

y 2:1168, or equal to the deflection: then

5 + 2 15-2) x 2
y = 3 x į x Х
625

3
5 – 2) x 9

27

= .1764 foot,

2 or 2·1168 inches, as before. For the point B we have :

5+2 5 – 2) x 2 y = 3 x x Х

x 2 625

3

+

< 2 x 1)

{
6 – 2*1-}

) x

+

=:0448 foot,

2

or .5366 inch : hence the deflection at B = 2.1168 - .5366 = 1.5802 inch. Similarly for the point C we have:

5 + 2 15 – 2) x 2 y = 3 x x Х

x 2 x 2 625

3 2 8

= 1064 foot, 2

+(-3)* 4

}

=

or 1.2531 inch : hence the deflection at C = 2.1168 -1.2531 = .8637 inch, &c.

Thus the flexure at any number of points between the weight and 0 may be found, and by making z negative, the other part of the curve between the weight and n may be found also.

(655.) The curve of flexure has in all cases the shortest radius at the point where the weight is applied, showing that the strain is the greatest at that point, and that the beam will break there, but the deflection of the beam is not a maximum at that point except when the weight is at the centre. The deflection is a maximum between the weight and the centre of the beam, but much nearer the latter than the former: its position may be found by the Rule :

(1 L) X X 2

3 In which m = the distance of the weight from the point of

(656.) m = (} []+2) – (1 L) +

3)

{25 +

-
- {25+

maximum deflection and the rest as before: see Fig. 175. Thus in the case of the beam in (654) we obtain m = (5 + 2) 5 x 2 x 2 411

= 1.5 foot from the weight, therefore 3

3 2 1.5 = 0·5 foot, or 6 inches from the centre.

As the distance of the weight from the centre increases, so does m increase, and it becomes a maximum when the weight is at the support, when z = (4L), and it then becomes m = (5 + 5) 5 x 5 x 2 25

= 4.227 feet from the weight (and 3

3 the support), or 5 4.227 = 0.773 foot from the centre, which is equal to •773 - 5 = • 1546 of the half-length of the beam. From this it appears that wherever a weight is placed, the point of maximum deflection can never be more than 0.1546 of the half-length distant from the centre.

(657.) Comparing the curves A and B in the Diagram, Fig. 213, we observe this remarkable fact; that the deflection at any and every point of a beam with a central load is greater than would be produced by that same weight at any other point: for example, the curve B = the deflection at every point

: throughout the length by a given central weight, and A = the deflection at the same point by the self-same weight placed there. This of course is due to the fact that the beam is less strained by a weight out of the centre than by the same weight in the centre. When strained at every point to the same extent, or in proportion to the strength, the deflections are given by the line C.

(658.) “ Laws of Deflection.”—The Deflection of beams varies very much with the methods of fixing and loading (667): to simplify the matter we may take as a Standard case, that of a horizontal beam, supported at each end and loaded in the centre: other conditions may be considered afterwards. We then have the Rules

L' X W XC (659.)

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df x b