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COMPOUND STRAINS AND STRESSES 139 
of a normal stress on CD, and the force Q is the resultant of a tan- 
ntial or shear stress on CD. By the triangle of forces N = P cos 6, and 
=P sin 0. 
_If x is the intensity of the normal stress, and q the intensity of the 
tangential stress on CD, then N =—,=P cos = pa cos 0, therefore 
n=p cos? 0. When @=0, n is a maximum, and is then equal to p. 
Also Q--55 =P sin 0=pa sin 0, therefore g=p sin @ cos 0, but 
sin 0 cos 0=}sin 20, therefore g=}p sin 20. When sin 20=1, 
i.e. when 0 = 45°, ¢ is a maximum and is then equal to }p. 
If a section be taken perpendicular to CD its inclination to the cross 
section will be 90°-— 9, and the shear stress on this section will be 
kp sin 2 (90 - 0) = hp sin (180 — 20) = $p sin 20, which is the same as the 
shear stress on CD. It will be shown in the next Article that in all cases 
where there is a shear stress on one section there is always an equal shear 
stress on a section perpendicular to it. 
The fact that there is a shear stress g on the section CD having a 
maximum value equal to 4p when @ is 45°, suggests that if the resistance 
_ of a material to rupture by shearing be less than half its resistance to 
rupture by direct tension or compression, it will give way by shear- 
ing when subjected to tension or compression. This is what really 
erm with several materials, and examples will be found in Art. 166, 
p. 175 
; 141. Equality of Shear Stresses on Planes at Right Angles.— 
- Consider an indefinitely small rectangular portion ABCD (Fig. 183) of a 
strained body, and let h be the height, b the breadth, 
and ¢ the thickness of this portion. Assume that 
there is no stress on the face ABCD or on any inter- 
face parallel to it. The portion of material ABCD 
being at rest, the stresses on the faces which are 
perpendicular to the face ABCD must balance 6ne 
another. The stresses on the faces AD and BC 
may be resolved into normal stresses p, and shear 
stresses g. In Fig. 183 the arrows representing the 
normal stresses are omitted. The normal stresses on 
AD and BC must evidently balance one another. 
The resultant of the shear stress on AD equals Fig. 183. 
qht, and this will also be the magnitude of the result- 
ant shear stress on BC. These two resultants will form a couple whose 
moment is gitb, Now no system of forces but a couple will balance a 
couple, therefore the stresses on AB and CD must have components which 
are shear stresses s on these faces. The normal components of the 
stresses on AB and CD must balance one another. The resultants 
of the shear stresses on AB and CD will form a couple whose moment 
is sbth, and if this couple is to balance the other couple, then sbth = ghtb, 
therefore s=g. Hence if at any point of a strained body there is a 
shear stress in one plane there must be a shear stress of equal in- 
tensity in another plane at right angles to the first, but these two planes 
must be perpendicular to a plane which is parallel to the directions of the 
stresses. 
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