Question

a hollow cylinder 2m long has an outside diameter of 50mm and internal diameter of 20mm. if the cylinder is carrying a load of 25kN, find the stress in the cylinder also find the deformation of the cylinder, if the volume of modulus of elasticity for the cylinder material is 100giga Pascal​

Answer 1

The large modulus of elasticity and the magnitude of the thickness of the

cylinder allows it to support the heavy load with minimal deformation.

• The stress on the cylinder is approximately 15.158 MPa

• The deformation of the cylinder is approximately 0.303 mm

Reasons:

Length of the cylinder, L = 2 m

Outside diameter of the cylinder, $D_o$ = 50 mm

Internal diameter of the cylinder, $D_i$ = 20 mm

Load applied to the cylinder, P = 25 kN = 25,000 N

Modulus of elasticity of the material, E = 100 GPa

First part; To find the stress in in the cylinder, σ

Solution:

$\displaystyle Tensile \ stress, \ \sigma = \mathbf{\frac{Load \ applied}{Cross\ sectional \ area}} = \frac{P}{A}$

The cross sectional area of the cylinder, A = $\displaystyle \mathbf{\frac{\pi}{4} \cdot \left(D_o^2 - D_i^2 \right)}$

Therefore;

$A = \displaystyle \frac{\pi}{4} \cdot \left(\left(50 \, mm \right)^2 - \left(20 \, mm\right)^2 \right) \approx \mathbf{ 1,649.336 \, mm^2}$

$\displaystyle Tensile \ stress, \ \sigma = \frac{25,000 \, N}{1,649.336 \, mm^2} \approx 15.158 \, N/mm^2 = \mathbf{ 15.158 \, MPa}$

• The tensile stress in the cylinder, σ ≈ 15.158 MPa

Second part;

The deformation of the cylinder is found as follows;

$\displaystyle Modulus \ of \ Elasticity, \, E = \mathbf{\frac{Tensile \ stress}{Tensile \ strain}} = \frac{\sigma}{\epsilon}$

$\displaystyle \epsilon = \mathbf{\frac{\delta L}{L}}$

Therefore;

$\displaystyle E = \frac{\sigma}{\frac{\delta L}{L} }$

$\displaystyle \frac{\delta L}{L} = \frac{\sigma}{E}$

Which gives;

$\displaystyle \delta L = \mathbf{\frac{\sigma}{E} \times L}$

Where;

δL = The deformation of the cylinder

$\displaystyle \delta L = \frac{15.158 \, N/mm^2}{100 \, GPa} \times 2,000 \, mm = \frac{15.158 \, MPa}{100 \times 10^3 \, MPa} \times 2,000 \, mm \approx \mathbf{0.303 \, mm}$

• The deformation of the cylinder, δL ≈ 0.303 mm

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