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Chapter 1

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.

Screenshot 2022-10-07 213454

Step-by-Step

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