Take Poisson’s ratio as 0.3 and E as 210 GPa. AB = 500mm, BC = 200mm and AE = 400mm. -

Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 10M

Year: June 2014

$L(AB) = 500mm \ \ \ L(BC) = 200mm$

$L(AE) = 400mm \ \ \ E = 210GPa$

Poisson’s Ratio = $\frac{1}{m} = 0.3$

Stress in the x – direction

$σ_x = \frac{Force}{C/SArea} = \frac{1000 × 10^3}{200 × 400} = 12.5N/mm^2 (Tensile)$

Stress in the y – direction

$σ_y = \frac{Force}{C/SArea} = \frac{800 × 10^3}{500 × 400} = 4N/mm^2 (Tensile)$

Stress in the z – direction

$σ_z = \frac{Force}{C/SArea} = \frac{1500 × 10^3}{500 × 200} = 15N/mm^2 (Compressive)$

Strain in the x – direction

$e_x = \frac{σ_x}{E} - \frac{1}{m}\frac{σ_y}{E} - \frac{1}{m}\frac{σ_z}{E}$

$e_x = \frac{1}{210 × 10^3}[12.5 - 0.3(4) - 0.3(-15)]$

$e_x = 7.52 × 10^-5$

Strain in the y – direction

$e_y = \frac{σ_y}{E} - \frac{1}{m}\frac{σ_x}{E} - \frac{1}{m}\frac{σ_z}{E}$

$e_y = \frac{1}{210 × 10^3}[4 - 0.3(12.5) - 0.3(-15)]$

$e_y = 2.26 × 10^-5$

Strain in the z – direction

$e_z = \frac{σ_z}{E} - \frac{1}{m}\frac{σ_x}{E} - \frac{1}{m}\frac{σ_y}{E}$

$e_z = \frac{1}{210 × 10^3}[-15 - 0.3(12.5) - 0.3(4)]$

$e_z = -9.5 × 10^-5$

Change in the dimension AB = $L(AB) ×e_x$

$= 500 × 7.52 × 10^-5$

$3.76 × 10^-2 mm (Increase)$

Change in the dimension BC = $L(BC) ×e_y$

$= 200 × 2.26 × 10^-5$

$0.452 × 10^-2 mm (Increase)$

Change in the dimension AE = $L(AE) × e_z$

$= 400 × -9.5 × 10^-5$

$3.8 × 10^-2 mm (Decrease)$

Old Volume of the Block = AB × BC × AE

= 500 × 200 × 400

= $40000000mm^3$

New Volume of the Block = (500.0376) × (200.00452) × (399.962)

= $40000111.696mm^3$

Change in the Volume = New Volume – Old Volume

= $111.696mm^3$