A solid cube of side 1 m is kept at a room temperature of 32°C. The coefficient of linear Q.4 thermal expansion of the cube material is 1 × 10^–5/°C and the bulk modulus is 200 GPa. If the cube is constrained all around and heated uniformly to 42°C, then the magnitude of volumetric (mean) stress induced due to heating is __________MPa.

The correct answer is 60.

1. Identify the given parameters:
- Coefficient of linear thermal expansion of the cube, $\alpha =1\times {10}^{-5}/°C$
- Bulk modulus of the material, $K=200\text{GPa}=200\times {10}^3\text{MPa}$
- Initial temperature, $T_1=32°C$
- Final temperature, $T_2=42°C$

2. Calculate the change in temperature:
$\mathrm{\Delta T}=T_2-T_1=42-32=10°C$

3. Understanding the constraints:
Since the cube is completely constrained in all directions and heated uniformly, it cannot expand in any direction. This creates a state of hydrostatic (or volumetric) stress. The total volumetric strain (${\epsilon }_v$) must be zero because the constraints prevent any physical change in volume.

The total volumetric strain is the sum of the volumetric strain due to thermal expansion and the volumetric strain due to the induced compressive stress:
${\epsilon }_v={\epsilon }_{v,\text{thermal}}+{\epsilon }_{v,\text{stress}}=0$

4. Calculate the thermal volumetric strain:
The free volumetric strain due to temperature change is given by:
${\epsilon }_{v,\text{thermal}}=3\alpha \mathrm{\Delta T}$

5. Relate stress to strain using Bulk Modulus:
The bulk modulus ($K$) is defined as the ratio of volumetric stress (${\sigma }_v$) to volumetric strain (${\epsilon }_{v,\text{stress}}$):
$K=\frac{{\sigma }_v}{{\epsilon }_{v,\text{stress}}}$
Rearranging for volumetric strain due to stress gives:
${\epsilon }_{v,\text{stress}}=\frac{{\sigma }_v}{K}$

6. Solve for volumetric stress (${\sigma }_v$):
Substituting the strains into the compatibility equation:
$3\alpha \mathrm{\Delta T}-\frac{{\sigma }_v}{K}=0$
Therefore, the magnitude of the volumetric stress induced is:
${\sigma }_v=3K\alpha \mathrm{\Delta T}$

Substitute the given values into the formula:
${\sigma }_v=3\times (200\times {10}^3\text{MPa})\times (1\times {10}^{-5}/°C)\times 10°C$
${\sigma }_v=3\times 200\times {10}^3\times {10}^{-5}\times 10$
${\sigma }_v=3\times 200\times {10}^{-1}$
${\sigma }_v=60\text{MPa}$