Imágenes de páginas
PDF
EPUB

((: L),X

[ocr errors]
[ocr errors]

nothing. In that case the elastic curve has its shortest radius at the centre, the curve becoming progressively flatter toward the ends, where it is a straight line.

When the strength at every point is proportional to the strain there, for example when the depth is uniform, and the breadth is reduced toward the ends in arithmetical ratio as in Fig. 116, the elastic curve is uniform in its radius from end-toend, that is to say, it is a simple spherical curve.

Curve with Central Load.—Let Fig. 174 be a parallel beam resting on two bearings and loaded in the centre: then we have the Rule :

3х8

x 2 (648.)

Х (1 L)

2 In which L= length of the beam between bearings.

8 deflection in centre by central strain.

distance from centre to a point whose deflection

is required.

y = co-ordinate of the curve at that point. Of course all the dimensions must be in the same terms, feet, inches, &c. (649.) Thus, with a beam 10 feet long, say d = 0.25 foot;

3 x .25 5 x 9 x = 3 feet: then by the Rule y=

Х 125

2 6 0.108 foot, or 1.296 inch: hence the deflection at that point is 3 - 1.296 = 1.704 inch. Calculating in this way we may obtain any number of points through which the entire curve of flexure may be drawn. Table 103 has been calculated by the rule, the half-length of the beam being divided into 10 parts, and the central deflection = 1.0, from which we may easily find the deflection at any point in a beam whose central deflection is known :-thus in our case, the point x being 3 = 5 = 0.6 of the half-length distant from the centre, we find the deflection at that point 0.568 x 3 = 1.704 inch as before. The curve B in Diagram, Fig. 213, has been obtained from Table 103.

Load out of Centre.”—Let Fig. 175 be a parallel beam with a load W out of the centre: knowing the deflection which any

[ocr errors]

=

[ocr errors]

TABLE 103.-Of the FORM of the CURVE of FLEXURE in a

PARALLEL BEAM DEFLECTED by a CENTRAL WEIGHT, Fig. 212.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

weight would produce at the centre, we may find the deflection at any other point by the same weight applied at that point, by the Rule: (650.)

Х (L)

[ocr errors]

x { 10 x 3) – 3")* =

3)*= 0.1764 foot, or 2-1168 inches.

=

54

In which 1 = the distance from the weight to the nearest bearing in the same terms as L; d = the deflection at the point of application of the weight W, and the rest as before. Thus taking the beam in (649) whose central deflection = 3 inches, and say we require the deflection at a point 3 feet from the end, therefore 2 feet from the centre :—then the Rule gives d = 0.25 Х

,

. Table 104 has been calculated by the Rule, taking the central deflection = 1.0, and dividing the half-length into 20 parts, col. 2 gives the deflection at each point in the length by that same constant weight.

(651.) For example, in our case, the point is 2:5 = 0.40 from the centre, hence the deflection there = .7056 x 3 = 2:1168 inches as before. The curve A in Diagram, Fig. 213, has been obtained from col. 2 of Table 104, and gives the deflection from centre to end for a rolling constant load. For example, say we have a beam 24 feet long, and we require the deflection at 6 feet from the centre by a certain weight applied at that point. We have first to find what the central deflection

[ocr errors]

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

LOADED.

[blocks in formation]

1.000

99

=

[ocr errors]

.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

[ocr errors]
[ocr errors]

=

[ocr errors]

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

+

[ocr errors]
[ocr errors]
[ocr errors]

< 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

[ocr errors]

+(-3)* 4

}

[ocr errors]

=

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)

« AnteriorContinuar »