The value of t will be one of the data, and d is then determined almost entirely by constructive reasons, into which we do not enter. Also in some cases the pitch thus found cannot be used if the joint is to be watertight, but this again is a purely practical consideration.

Efficiency of Joint. — By cutting a rivet hole through the plate, the strength of each strip is diminished from pt ×f to (p − d)t × ƒ or in the ratio p − d÷p. Hence the strength of the joint is (p–d)÷p times that of the solid plate, and this ratio is called the Efficiency of the Joint (see page 273). Evidently it is advantageous to have pas large as possible.

Normal and Tangential Stress.-Our next step would be to consider the change of shape produced by shearing, and its

connection with the stress. But it is convenient first to examine a little more closely into the nature of tangential or shearing stress.

Fig. 298 represents a bar of rectangular section, thickness t at right angles to the paper, subject to a load P. If we take a section AB transverse, then there is compressive strength of intensity P/(AB+) over that sec

E

A

B

Fig. 298.

section as

with AB.

But now take an inclined BC (Fig. 298), making an angle Then, considering the equilibrium of the piece EFBC (Fig. 299), the stress on the section BC must be parallel to P, and its total amount equal to P; also it will be uniformly distributed (page 264), hence

Fig. 299.

and its direction makes an angle with BC, so it is neither normal nor tangential.

We can, however, resolve this stress into two components, one tangential or shear along BC, and the other normal or direct compression on BC.

For this purpose resolve the total stress P along and perpendicular to BC. The first gives

Total tangential stress on BC= P sin 0.

Now P/AB.t is the intensity of the stress on the section AB, on which section it is purely normal; hence, denoting this by p, we have

Pn=p cos2 0, Pt=p . sin 0 . cos 0,

for the normal and tangential stresses on a section making an angle with the section AB on which the stress is normal.

The normal stresses are here compression; if P had been tensile the work would have been the same, but the signs of the normal stresses would have been altered, so they would be tensions; also the shearing stress would be in the opposite direction.

If we take a section at right angles to BC, the stresses 'n, l't on it will be obtained by writing π/2 – 0 for 0,

hence

P'n=p sin2 0, 't=p. cos 0. sin 0,

so that p't=pt, or the tangential stress on planes at right angles is equal, a fact of great importance in the theory of strength of materials, to which we shall again recur.

Pure Shear.-In Fig. 300 ABCD is a block of material of thickness t, and we apply to it a compressive load of intensity p on AD, and therefore on BC to balance, and a tensile load of equal intensity p on AB and CD. Then on all sections parallel to AD there is pure normal stress of intensity p, and on all sections parallel to AB pure normal stress of intensity -p, the negative sign denoting tension, as opposed to compression, which we will consider positive.

21 1 1 1 1 1 1 1°

Fig. 300.

Draw now on the side of the block a small square with sides inclined at 45° to the sides AB, BC, and consider what action takes place on this small square of material, supposing it to extend through the whole thickness t. Produce the sides as dotted.

Then, first, due to the compression p, there is, on each of the sections along the dotted lines, and thus on each side of the square,

The total effect of the tension and compression is now obtained by adding the separate results, whence we obtain the stress shown in Fig. 303, since the normal

1111

1111

Fig. 301.

Fig. 302.

Fig. 303.

stresses cancel, while the tangential p/2, p/2 being in each case (Figs. 301 and 302) in the same direction, produce a tangential stress of intensity p along each side of the

square.

Here then we have the square under pure shearing stress, and we see that this is of a dual nature, requiring for its production the existence of equal and opposite normal stresses in two directions at right angles, viz. directions parallel to AB and BC respectively; and hence in all cases of pure shear these two stresses p compressive and tensile necessarily accompany it.

[We have started with the + and stresses, and proved they produce pure shear, because this shows a practical way in which it may be produced; but we can if we please start by assuming pure shear along the edges of such a small square, and then taking sections of the square parallel to AB and BC, it will be found that the stresses on them are pure tension and pure compression respectively.]

Equality of Shearing on Planes at Right Angles. The preceding work proves the equality of shear on planes at right angles, which we have already noticed. For no other set of forces than those of Fig. 300 can produce pure tangential stress of intensity / on any side of the square, but this set produces equal tangential stress on all sides, which proves the result.

This being an important point we will consider it in another way.

Take ABCD, a small rectangle of material, thickness t, to which shear is applied along the edges AB and CD as shown (Fig. 304), the shear along

CD being necessary to balance that

tending to turn the piece clockwise. Now this cannot be balanced by the application of uniform normal stress to any of the faces, since the resultant stress on each face will pass through its centre. Hence to prevent the turning we must apply shear of intensity pt along the edges BC and AD. This will produce a turning moment

ptx (BCxt) x AB,

and since the two moments balance,

or

Fig. 304.

ptx (ABxt) × BC=p't × (BC xt) x AB,

Pt=P't

Distortion due to Shearing.-We can now see what change of shape or strain accompanies shearing, for, referring to Fig. 300, the block will, under those

A

B

Fig. 305.

D

stresses, elongate in the direction AD, and contract equally in the direction AB, hence taking the form of Fig. 305, in which the change of shape is shown in an exaggerated form.

Hence the square of Fig. 300 now distorts into the rhombus of Fig. 305, the lengths of its sides remaining unaltered.

[The student must remember that all these changes of shape are extremely small; otherwise the last statement would not