In tension the load is applied directly to the material, and it is, therefore, evident that no matter of what shape the material may be, the strain will always be the same. This strain, of course, will be just equal to the load, and we have, therefore: s=w. Where s = the amount of strain. Where w = the amount of load.

The weakest point of the piece under tension will, of course, be where it has the smallest area of cross-section; and the stress at such point will be equal to the area of cross-section, multiplied by the amount of resistance its fibres are capable of.1

The amount of resistance to tension the fibres of a material are capable of is found by experiments and tests, and is given for each material per square inch of cross-section. A table of constants for the ultimate and safe resistances to tension of different materials will be given later; in all the formulae these constants are represented by the letter t.

We have, then, for the stress: - v = a. t

Where v = the amount of ultimate stress.

Where a = the area of cross-section.

Where t = the ultimate resistance to tension, per square inch of the material.

Therefore, the fundamental formula (1), viz.: v=s.f, becomes for pieces under tension: a. t = w. f or: w = a.(t/f) (6)

Where w = the safe load or amount of tension the piece will stand. Where a = the area of cross-section at the weakest point (in square inches).

Where (t/f)= the safe resistance to tension per square inch of the material.

Example.

A weight is hung at the lower end of a vertical wrought-iron rod, which is firmly secured at the other end. The rod is 3" at one end and tapers to 2" at the other end. How much weight will the. rod safely carry?

The smallest cross-section of the rod, where it would be likely to break, would be somewhere very close to the 2" end, or, say, 2" in diameter. Its area of cross-section at this point will be: a = 22/7. 22/4 = 3 1/7 square inches.

The ultimate resistance to tension of wrought-iron per square inch is, from forty-eight thousand pounds to sixty thousand pounds. We do not know the exact quality, and, therefore, take the lower figure; lThis, again, is not theoretically correct, as a piece under tension is apt to stretch and so reduce, the area of its cross-section; but the above is sufficiently correct for all practical purposes.

Using a factor-of-safety of four, we have for the safe resistance to tension per square inch: (t/f)=48000/4 =12000 pounds.

Therefore, the safe load will be.: w = 3 1/7. 12 000 =37 714 pounds. Shearing.

In compression the fibres are shortened by squeezing; in tension they are elongated by pulling. In shearing, however, the fibres are not disturbed in their individualities, but slide past each other.

When this sliding takes place across the grain of the fibres, the action of shearing is more like cutting across. When this sliding takes place along the grain, the action of shearing is more like splitting. Thus, if a very deep, but thin, beam is of short span and heavily loaded, it might not break transversely, nor deflect excessively, but shear off at the supports, as shown in Figure 6, the action of the loads and supports being like a large cutting-machine, the weights cutting off the central part of beam and forcing it downwards past the support. This would be shearing across the grain.

If the foot of a main rafter is toed-in to the end of a tie-beam, and the foot forces its way outwardly, pushing away the block or part of tie-beam resisting it (splitting it out as it were), this would be shear-ing along; the grain.

In most cases (except in transverse strains) the load is directly applied to the point being sheared off; the strain will, therefore, just equal the load, and we have: - s=w. Where s = the amount of the shearing strain.

The stress will be equal to the area of cross-section (affected by the shearing strain) multiplied by the amount of resistance to separation from each other that its fibres are capable of.

This amount of resistance is found by tests and experiments, and is given for each material per square inch of cross-section. A table of constants for resistance to shearing of different materials will be given later; in the formulae these constants are represented by the letter g for shearing across the grain, and gl for shearing along the grain.

Fig. 6.

We have, then, for the stress: v = a. g. Where u = the amount of ultimate stress. Where a = the area of cross-section in square inches. Where g = the ultimate resistance to shearing across the grain per square inch.

Therefore, the fundamental formula (1) v=s.f, becomes for pieces under shearing strains across the grain: - a. g = w. f, or: w=a.(g/f) (7)

And similarly, of course, we shall find: w=a.(g/f) (8)

Where a = the area of cross-section in square inches, at the point where there is danger of shearing.

Where (g/f) = the safe-resistance to shearing across the fibres per square inch.

Where (g1/f) = the safe-resistance to shearing along the fibres per square inch.

Example.

At the lower end of a vertical wrought-iron flat bar is suspended a load of eight thousand pounds. The bar is in two lengths, riveted together with one rivet. What diameter should the rivet be?

The strain on the rivet will, of course, be a shearing strain across the grain, and will be equal to the amount of tension on the bar, which we know is equal to the load. We use Formula (7), and have: - w = 8000 pounds.

The safe shearing for wrought-iron is about ten thousand pounds per square inch; inserting this in formula, we have: 8000 = a. 10000, or a. = 8000/10000 =4/5.

The area of rivet must, therefore, be four-fifths of a square inch.

To obtain diameter, we know that: -

This is, practically, equal to one; therefore, the diameter of rivet should be 1".