Question

7.28 Show that for a quasi-isotropic laminate the fol- lowing relation holds among axial modulus, shear modulus and Poisson's ratio: 2(1 Vy)

Answer 1

Consider a cube of a material of side ‘a' subjected to the action of the shear and complementary shear stresses as shown in the figure and producing the strained shape as shown in the figure below.

Assuming that the strains are small and the angle A C B may be taken as 45⁰.

Therefore strain on the diagonal OA = Change in length / original length

Since angle between OA and OB is very small hence OA @ OB, therefore, BC, is the change in the length of the diagonal OA

So,

strain in diagonal OA =$= \frac{BC}{OA}=\frac{ACcos45}{OA}$

Now, $OA=\frac{a}{sin45}=\frac{a}{\frac{1}{\sqrt{2}}}=a\sqrt{2}$

So,

$Strain=\frac{AC}{a\sqrt{2}\sqrt{2}}=\frac{AC}{2a}$

But,

$AC = a*\gamma$

where, $\gamma$ =shear strain

Thus, strain on diagonal = $\frac{a\gamma}{2a}=\frac{\gamma}{2}$

Now,

$G=\frac{\tau}{\gamma}$

$\therefore \gamma=\frac{\tau}{G}$

Therefore,

Strain on diagonal = $\frac{\gamma}{2}=\frac{\tau}{2G}$

Now this shear stress system is equivalent or can be replaced by a system of direct stresses at 45⁰as shown below. One set will be compressive, the other tensile, and both will be equal in value to the applied shear strain.

Thus, for the direct state of stress system which applies along the diagonals:

strain on diagonal =$\frac{\sigma_1}{E} - \frac{\mu \sigma_2}{E} = \frac{\tau}{E} - \frac{\mu (-\tau)}{E} =\frac{\tau}{E}(1+\mu)$

equating the two strains,

$\frac{\tau}{2G} =\frac{\tau}{E}(1+\mu)$

$\therefore G= \frac{E}{(1+\mu)}$