Page 41 - Aircraft Stuctures for Engineering Student

26 Basic elasticity

For the case of plane stress they simplify to

(1.47)

It may be seen from the third of Eqs (1.47) that the conditions of plane stress and
plane strain do not necessarily describe identical situations.
Changes in the linear dimensions of a strained body may lead to a change in
volume. Suppose that a small element of a body has dimensions Sx, Sy and Sz.
When subjected to a three-dimensional stress system the element will sustain a
volumetric strain e (change in volume/unit volume) equal to

(1 + &,)ax( 1 + &,,)by( 1 + EZ)SZ - SxSySz
e=
SxSySz
Neglecting products of small quantities in the expansion of the right-hand side of the
above equation yields

e = E, + E,, + E, (1.48)
Substituting for E,, E~ and E~ from Eqs (1.42) we find, for a linearly elastic, isotropic
body
1
e = - [a, + a,, + az - 2v(ax + ay + 41
E
or

In the case of a uniform hydrostatic pressure, a, = a,, = az = -p and
3(1 - 2~)
e=- P (1.49)
E
The constant E/3( 1 - 2v) is known as the bulk modulus or modulus of volume
expansion and is often given the symbol K.
An examination of Eq. (1.49) shows that v < 0.5 since a body cannot increase in
volume under pressure. Also the lateral dimensions of a body subjected to uniaxial
tension cannot increase so that v > 0. Therefore, for an isotropic material
0 < v < 0.5 and for most isotropic materials v is in the range 0.25 to 0.33 below
the elastic limit. Above the limit of proportionality v increases and approaches 0.5.

Example 1.2
A rectangular element in a linearly elastic isotropic material is subjected to tensile
stresses of 83 N/mm2 and 65 N/mm2 on mutually perpendicular planes. Determine
the strain in the direction of each stress and in the direction perpendicular to both