Question

A rectangular block of aluminum30 mm×60 mm×90 mm is placed in apressurechamber and subjected to a pressure of 100 MPa. If the modulus of elasticity is 75GPa andPoisson's ratio is 0.35, what will be the decrease in the longest side of the block, assuming thatthe material remains within the linear elastic region? What will be the decrease in the volume oftheblock? g

Answer 1

Answer:

$\Delta V = - 216.415 mm^3$

Step-by-step explanation:

we know that change is length is calculated by following strain relation

$\Delta L = L \times \epsilon_x$

where strain is given as

$\epsilon_x = \frac{\sigma_x - \nu(\sigma_y + \sigma_z)}{E}$

$\epsilon_x = \frac{-10^8 - 0.35 ( -10^8 -10^8 (N/m^2))}{7.5 \times 10^10}$

$\epsilon_x = -4.453 \times 10^{-4}$

plugging strain value in change in length formula

$\Delta L = 90 \times -4.453 \times 10^{-4} = - 0.04008 mm$

calculate the length on the longer side

$L_{long} = L = \delta L$

              = 90 - 0.04008 = 89.95 mm

intial volume $= 90*60*30 = 1.62 \times 10^5 mm^3$

change in volume

$\Delta V =V ( \epsilon_x +\epsilon_y + \epsilon_z )$

$\Delta V = 1.62 \times 10^5 (-4.453 - 4.453 - 4.453) \times 10^{-4}$

$\Delta V = - 216.415 mm^3$