Question

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension

Answer 1

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

               $=113.16$  GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

                    $\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

                       = 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

   = 0.016 mm