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1,009 1,009

16.7

498, 711, 925 lbs. Leakages

made good. 740, 796, 853, 896,782, 1,009 lbs. Flange bolts broke.

B

1:14

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20

5:51

142-213

:

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14.8

25.9

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9.77 R

0-142

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IV a

4.36

17.7

14.3

'R denotes that the washers are riveted to the front plate.

* The test pressures rose in steps of 71 and 142 lbs. Pressures at which the permanent set was very slight are printed in black figures, in all other cases the permanent set commenced between the two pressures given in this column.

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The maximum stress S exists on the concave side of the beam.

M
S=C.6.

bt2
In this formula M= bending moment and

m[m - log nat (1 + m) ]
(m + 2) log nat (1 +m) - 1

C=.

The following are a few numerical values :
10 5 1

0
C=d 2:889 2 103 1.287 1.154 1.033 1.000

m =

10

The practical conclusions to be drawn from these figures are that the bending stresses on the concave side of plates flanged to the usual radius, of, say, twice the thickness, are 15% greater than those in the adjoining flat plates; that when the inner radius is reduced to the thickness of the plate 29 % has to be added to the stress; while when there is no round on the inside of the flange the stresses are infinitely great. The end of every crack can also be looked upon as the inner surface of a curved beam, and the lightest external force would extend

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it, unless there are other stresses beyond the crack holding the plate together. When they are absent, or when tension stresses exist in the plate, there is nothing to stop the crack from extending with the usual explosive rapidity. Similarly, the concave sides of cold bend samples are very weak, on account of their small radii, and, as the residual stresses in the samples all tend to produce tensions at this point, it is not surprising that the majority of such test pieces crack as shown in fig. 155 some time after they have been put aside, and without any apparent cause.

Elastic Deformation of Curved Beams.—This subject is not of such importance as to warrant the development of the unavoidably long formulæ, but for those who wish to study the behaviour of flanged end plates or furnaces it is well to point out that there are two motions -a horizontal one a and a vertical one b in figs. 156, 157. In the one case it is assumed that the entire flange is tilted through an angle, a, and in the other a uniformly distributed bending moment is supposed to have reduced the radius r to ri. In the one case a = b = r. a, in the other a = r-ri, and b= (r-ry) (5 - 1).

Shearing Stresses in Beams.-No shearing stresses exist in a bent beam when it is exposed to only two external bending moments, m, m (fig. 158). In every other case the shearing force in any particular

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section of a beam is exactly equal to the otherwise unbalanced load at that point. Thus, in fig. 138, p. 174, the shearing force at the point x is Q-p (1-x).

If this force were evenly distributed over the sectional area of the beam, it would be easy to calculate it; but there are certain conditions which have to be fulfilled, and which lead to somewhat complicated results, particularly if the section is an irregular one. In rectangular beams, and therefore also in flat plates, the case is simple.

Let o represent the intensity of the shearing stress at a point situated at the height y above the neutral fibre op of a beam (fig. 159), while S is the longitudinal stress at tặat point; then it is easily proved that

do ds dy dx

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do

=C.y, dy

where C is a constant.

If, then, the total shearing force in the section at the distance x is we have

-4

2.7{

(7)"}

Fig. 161 represents the distribution of the longitudinal stresses, and fig. 160 that of the shearing stresses.

As a shearing stress o is a compound of a tension and compression stress S, acting at angles of 45° to the line of shear, as shown in fig. 125 (p. 166), it is easily proved that

o=+S.

2.

-P+S

+

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Y =P+S

✓ 4.

In order, then, to resolve the

various stresses to be met with in Fig. 160

Fig. 161 a beam into their respective right

angled resultants it is only necessary to apply the formulæ of p. 173; then

otototo
tg. 2x=
-P+S+o+o

S-P
4.09+(S—p)?,

4.0? +(S-p)2 2 2 2

2 Here P is the transverse pressure, but the signs of both S and p will have to be changed if they are compression stresses. The distribution of p is determined by o.

It is now possible to construct curves showing the direction and intensities of the resultant stresses at any point of a beam (see figs. 184, 185, 186, p. 216).

Plastic Beams.- When flanging or bending plates, the elastic limit of their material is always passed, and the resultant deformation is a permanent one. In the chapter on 'Strength of Materials' a formula has been explained which enables one to calculate the stresses in the outside fibres of narrow plastic beams, when their curvature is known. Roughly, S= ; so that S is only two-thirds of what it would be

t2 in an elastic beam, which means that a plastic beam is 50 % stronger than the elastic theory would make it.

Another very important difference between an elastic and a plastic beam is that the stresses of the former bear some relation to the deformation, while those of the latter are nearly, if not quite, independent of the same.

The work W, required to bend an elastic steel beam is proportional to the square of its final curvature, about

4.m

280. A. t3 W

foot tons.

| The axial shearing stresses on the threads of a screwed stay can be shown to be distributed in the same manner, and instead of making the thickness of the plate or nut half the diameter, it ought to be three-quarters.

Here A is the superficial area of the plate, measured in square feet, and p the radius of curvature, measured in feet. By the time that the elastic limit is reached this amounts to about

Ait
W

foot tons,
450

1

2

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and is independent of the radius. In other words, the maximum energy which can be stored in a flat spring is proportional to its volume.

The work W, required for bending a plastic beam or plate is proportional to the curvature

C.A. t
W

р The value of the constant C is about 6 for mild steel, cold.

5 for wrought iron, cold.
4
blue hot. 3!

blue hot.
0:3
red hot. 0.25

red hot. Thus to bend a cold steel shell plate whose dimensions are 14 feet x 7 feet x 1 inch, and whose radius is 6.7 feet, a power of 88 foot tons would be required. To do it at a red heat requires only 4? foot tons.

The above values have been estimated from the following experiments and other observations. Dr. J. Kollmann ( Ver. Gew.,' 1880, vol, lix. p. 107, &c.) states that he found the limits of elasticity (? plasticity) to be as follows:

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C. E. Stromeyer, ‘C. E.,' 1886, vol. Ixxxiv. p. 125, Nos. 4, 9, 10, 12, 18.

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From this it will be seen that at a blue heat and at a red heat iron, as well as steel, grows more pliable in the ratio respectively of 3 to 2 and 20 to 1. The power required to permanently bend an iron and a steel plate 12 inches wide and 1 inch thick is, therefore, as follows:

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