Generally we may say, within the elastic limit, there is little difference between the laws of tension and compression. The values of E and the proof stress are about the same (see chap. xxi.), and the calculations of stress and work done will be the same, if the bar be such as to be in simple compression. The strength of columns of ordinary dimensions, however, can only be properly determined by a formula derived from special experiments, and which is given in chap. xxi.

EXAMPLES.

1. Find the stress produced in a pump rod 4 ins. diameter, lifting a bucket 28 ins. diameter; the pressure on top of the bucket being 6 lbs. per square inch in addition to the atmosphere, and a vacuum below of 26 ins. by gauge; taking each inch aslb. Ans. 925 lbs. per square inch.

2. The rod in the preceding is 5 ft. long, find its extension. Material of rod brass, for which E-9,000,000.

Ans. .0061 ins.

3. Assuming the strength of a chain to be double that of the bar from which the links are made, find the proper size of chain for a 20-ton crane, using three sheaved blocks; allowing ƒ=6000 lbs. Ans. I inch.

4. A steel piston rod is 8 ins. diameter; the diameter of cylinder being 88 ins., and effective pressure 40 lbs. per square inch. Find the stress produced, and the total alteration of length during a revolution. Length of rod 9 ft. E=29,000,000.

Ans. 4840 lbs. per square inch, .036 ins.

5. A cylindrical boiler, 12 ft. diameter, is constructed of " steel plate. The test pressure applied is 245 lbs. per square inch. Find the stress produced in the solid metal, and hence deduce the stress in the metal of the joints, the sectional area being there reduced to .77 of the solid. Find also the increase of diameter under the test, neglecting the joints.

Ans. 19,500, 25,300 lbs. per square inch, .097 ins.

6. A cylindrical vessel 6 ft. diameter, with hemispherical ends, is exposed to internal pressure of 200 lbs. per square inch above the atmosphere. It is constructed of solid steel rings riveted together. Find the necessary thickness of metal, taking f=7 tons per square inch. Also find the longitudinal stress in

the metal of the ring joints, the sectional area being reduced to of the solid plate. Ans.in., 10 tons.

7. In question 4 find the work done in extending or compressing the rod; and also find the resilience of the rod. f= 12 tons. Ans. 2, 60 inch-tons.

8. In question 5 the flat ends of the boiler are stayed by steel bar stays, pitched 16 inches apart, both vertically and horizontally. Find the necessary diameter of stay that the stress per square inch at the test pressure may not exceed 18,000 lbs. per square inch. Ans. 2 ins.

9. Find the area of the base of a stone column carrying a load of 5 tons, allowing a crushing stress of 150 lbs. per square inch. Ans. 75 square ins.

10. An iron rod is suspended by one end. Draw a curve showing the stress at any section, and find the length of a rod which can just carry its own weight, allowing f=9000 lbs. per square inch. Ans. Curve is a straight line; 2700 ft.

II. An iron bar 18 feet long, 1 in. diameter, is heated to 400° F.; nuts on its ends are then screwed up so as to bear against the walls of a house which have fallen away from the perpendicular. Find the pull on the walls when the bar has cooled to 300° F. Coefficient of expansion of iron .0000068 per degree Fahrenheit. Ans. 35,000 lbs.

12. A bar of iron is placed between two bars of copper of the same section and length, and the ends are rigidly connected together when at a temperature of 60° F. Find the stresses in the bars when the temperature is raised to 200° F. Coefficient of expansion of copper .0000095, E=17,000,000.

Ans. Iron, 5920; copper, 2960 lbs. per square inch.

13. A bar of 2-inch round iron, 3 ft. long, is turned down to I inch diameter over the centre foot length. Compare its resilience-Ist, with that of the original bar; 2d, with that of a uniform bar of the same weight. Ans. 1:8, 1:6.

CHAPTER XIV

TRANSVERSE LOADS-BENDING AND SHEARING

WHEN the forces acting on a bar of material act, not along the axis, but transversely to it, the bar is called a Beam. In most cases the forces all act in or parallel to one plane passing through the axis of the bar, and we shall confine our work to this case.

Our problem then is-To investigate the straining actions in a beam, subjected to forces all in one plane.

The term straining action here used means any action tending to strain-i.e. alter the shape of the piece of material. Thus tension and compression already considered are two cases, and the simplest, of straining actions.

K

B

AB is a beam, which for definiteness we will suppose to be fixed at A to a wall, and loaded with a weight W at B. Any other case might be taken, as what we are going to say will apply generally. Take now a transverse section of the beam at any point as KK. Then the body BK is in equilibrium under the action of W and of the forces which the end KK of AK exerts on the end KK of BK. the stresses on the section KK; and we see that the action of the load W is to produce stresses which can just balance the effect of W, and so keep the piece BK in equilibrium.

Fig. 184.

These forces are

K

K

(a)

K

B

Now

We first then inquire, What is the effect of W? For this purpose imagine the beam actually cut through at KK; then BK will fall to the position shown in (a). there is here a double effect, viz. falling, keeping parallel to itself, and turning round K; each of (b) these is shown separately, the first in (c) and the second in (b). The combination of the effects (b) and (c) would evidently produce (a).

K

K

K

K

K

Fig. 185.

B(c)

For reasons which will appear plainly as our work progresses in the present and succeeding chapters, it is convenient to take these effects separately; and we will commence then with effect (b).

Bending Moment.—The effect (b) may be expressed as a turning or bending round K as a joint; and it would occur even if the two parts AK, BK were not separated but united by a pin joint. In this case effect (c) could not take place. Now the magnitude of the actions or stresses on the end KK of BK must be such as to enable them to prevent this turning; and hence we must have

Moment of stresses resisting bending=moment of W about K (page 64),

W being the effort and the stresses the resistances.

We are not now going to inquire into the magnitude of these stresses, but since they are due to bending, and their magnitude depends on the moment of the load W, we call this moment the Bending Moment at KK.

We have taken for simplicity one load W; but all we have said will apply equally well, no matter how many forces we apply to BK; some may act upwards, and others downwards. In all cases we have BK in equilibrium, under the actions of all the loads acting on it, and of the stresses on the end KK, and hence

=

Moment of internal Sum of the moments of all the external stresses forces acting on BK about K.

Sum here means of course algebraic sum, allowing for sign, e.g. in Fig. 186 the sum is, counting clockwise turning plus,

W.BK-P.CK+Q.DK-R.EK,

and this is the bending moment at K.

A

K

B

E

с

K

K'

D

Fig. 186.

Since now KK is any section, we can define, for any section, the term bending moment as the moment about the section of all the forces on the right of the section. For example, for the section through K' (Fig. 186), the moments about K' of W, P, and Q only would be taken, since R is not a force affecting the equilibrium of K'B, which would now be the piece considered.

[There may appear to be an ambiguity in writing of the moment of a force about KK, since strictly we can only have moments about a point; but all points in KK are equidistant from W, KK and W being parallel, so the moment about KK is really about any point in KK.]

Shearing Force.-Let us now return to the effect (c), Fig. 185. This is a bodily vertical movement, and to prevent it the end KK of AK must exert on KK of BK forces or stresses sufficient to keep BK in equilibrium vertically. The total amount of this stress must therefore be, in Fig. 185, a vertical force W, in Fig. 186, a vertical force W-P+Q-R; then the forces balance vertically, and BK is in vertical equilibrium. The magnitude then, generally, of this action is the algebraic sum of the forces acting on BK, or to the right of the section; and, since the effect produced is that which would be caused by the jaws of a shearing machine, this sum is called the Shearing Force at the section.

Both with bending moment and shearing force we have spoken of the forces on the right of the section