Question

A uniaxial steel bar is initially 200mm in length a square cross-section of 40mm X 40mm. The modulus of elasticity is 150Gpa and Poisson's ratio is 0.3, Determine the change in the dimensions of the bar when carrying a tensile load of 500KN

Answer 1

Answer of question #78565 -Physics- Mechanics - Relativity

A uniaxial steel bar is initially 200mm in length, a square cross-section of 40mm X 40mm. The modulus of elasticity is 150Gpa and Poisson's ratio is 0.3. Determine the change in the dimensions of the bar when carrying a tensile load of 500KN

Input Data:

Length: $L = 2m$ ;

Width: $= 0.04m$ ;

Modulus of elasticity (Young's): $E = 150 * 10^{9}Pa$ ;

Poisson's ratio: $\mu = 0.3$ ;

Tensile load $F = 500 * 10^{3}N$ ;

Solution:

According to Hooke's law, the mechanical stress is:

$$\sigma = \frac{F}{S}$$

where S is the cross-sectional area of a square the bar:

$$S = d^{2}$$

as is known, the increment of the length of the I bar $\Delta L$ is proportional to the mechanical stress $\sigma$

$$\Delta L = \frac{\sigma}{k}$$

the coefficient of proportionality $k$, in turn, is

$$k = \frac{E}{L}$$

we obtain the length increment $\Delta L$

$$\Delta L = \frac{FL}{Ed^{2}}$$ $$\Delta L = \frac{500 * 10^{3} * 2}{150 * 10^{9} * 16 * 10^{-4}} = 4.2 * 10^{-3}m$$

Poisson's ratio:

$$\mu = -\frac{\Delta d * L}{d * \Delta L}$$

hence we obtain a decrease in the thickness of the steel bar

$$\Delta d = -\frac{\mu * d * \Delta L}{L}$$ $$\Delta d = - \frac {0.3 * 0.04 * 0.0042}{2} = -2.52 * 10^{-5} m$$

**Answer:**

$$\Delta L = 4.2 mm$$ $$\Delta d = -0.0252 mm$$

Answer provided by https://www.AssignmentExpert.com