156 APPLIED MECHANICS 
a transverse strain. If a bar of length /, hreadth 0, and thickness ¢ be 
loaded, say in tension, in the direction of its length, the length will 
increase by an amount «, the breadth will decrease by an amount y, and 
the thickness will decrease by an amount z. The longitudinal strain 
is z/l, and the transverse strain is y/6 or 2/t. 
It is found that, within the elastic limit, the ratio of the transverse strain 
to the longitudinal strain is constant for any given material, and this con- 
stant ratio is generally called Povsson’s ratio. In this work Poisson’s ratio 
transverse strain 
longitudinal strain’ 
between 3 and 3, but values as low as 0°22 and as high as 0°45 are given 
by different authorities. For india-rubber, o is about $. 
_ 152. Relations between co, EB, C, and K.—Let a cube ABCD (Fig. 
220), whose edges are of length 7, be subjected to tensile stress of in- 
tensity f on the faces AD and BC, and all interfaces parallel to them. 
Also let the faces AB and DC, and all interfaces parallel to them, be 
subjected to a crushing stress of intensity 7. The cube will assume the 
rectangular shape A’B’C’D’. 
The tension will cause a strain in the direction EG amounting to 7/E, 
and the compression will increase this strain by the amount o//E. There- 
fore the total strain ;= ‘ (1+¢c). 
According to Art. 142, equal tensile and compressive stresses acting 
at right angles to one another, as in this case, are equivalent to shear 
stresses of the same intensity on planes inclined at 45° to the directions 
of the tensile and compressive stresses. If, therefore, a rectangular solid 
nll LL, 
is denoted by o; thus o — For metals o is generally 
7 
. 7 
4 
ae 
Hate = sa a SPR i: 
! 
To \E-$ 
mY 
ean} sbogee = 
Fig. 220. 
be formed whose front face is the square EFGH, having its angular points 
at the middle points of the sides of the square ABCD, the faces of this 
solid, which are perpendicular to EFGH, will be subjected to pure shear, 
and the intensity of the shear stress will be 7. If ¢ is the angle of dis- 
tortion, then ¢=//C. When the block EFGH is strained it assumes the 
form E’F’G’H’, and the angle EPG’ =5+ d. Let E’G’=/+2, and let 
EF be denoted by a, then 
; T 
tbe sin(5+4) cos p 
sin ie sin cos — cos 7 sin § 
