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## Q. 1.23

Figure 1.33 shows metal cubes A, B and C of side 100 mm in direct contact resting on rigid base and confined in x direction between two rigid-end plates. Young’s modulus of elasticity and Poisson’s ratio for cubes A and C are 150 GPa and 0.25 respectively and for cube B are 200 GPa and 0.3 respectively.

If the upper face of the centre cube B is subjected to uniform compressive stress of 0.5 kN/mm², compute for cube B, the following:

(i) the direct stress $\sigma_x$ in x-direction,

(ii) the direct strain in x, y and z direction,

(iii) the volumetric strain.

## Verified Solution

Using suffix 1 for blocks A and C and suffix 2 for block B,

Equilibrium condition: $\sigma _{x2}= \sigma _{x1}= \sigma _x$  (comp.)

Compatibility condition:

Displacement of block B = Total displacement of blocks A and C

or, $\Delta L _{x2} = 2 \Delta L _{x1}$

or, $\varepsilon _{x2} = 2\varepsilon _{x1}$ (i)

Assuming that there is no friction at any interface, Hooke’s law gives extension for block B,

$\varepsilon_{x 2}=\frac{1}{E_2}\left(-\sigma_x+\mu \times \sigma_z\right)$

Contraction for blocks A and C,

$\varepsilon_{x 1}=\frac{\sigma_x}{E_1}$
Substituting values of $\varepsilon_{x 1}$  and $\varepsilon_{x 2}$ in Eq. (i),

$\frac{1}{E_2}\left(-\sigma_x+\mu \times \sigma_z\right)=2 \frac{\sigma_x}{E_1}$

or,  $\frac{1}{200}\left(-\sigma_x+0.3 \times 0.5\right)=\frac{2 \sigma_x}{150}$
or, $\sigma_x=0.09 kN / mm ^2 \$ (comp.)

For block B,

$\varepsilon_x =\frac{1}{E}\left(-\sigma_x+\mu \times \sigma_z\right)$ $=\frac{1}{200}(-0.09+0.3 \times 0.5)=0.0003$

$\varepsilon_y =\frac{-\mu}{E}\left(\sigma_x+\sigma_z\right)=-\frac{0.3}{200}(-0.09-0.5) =0.000885$

$\varepsilon_z =\frac{1}{E}\left(-\sigma_z+\mu \times \sigma_x\right)$

$=\frac{1}{200}(-0.5+0.3 \times 0.09)=-0.002365$

and volumetric strain,

$\varepsilon_v= \varepsilon_x+ \varepsilon_y+ \varepsilon_z$ = 0.0003+0.000885-0.002365=-0.00118