A force of 500 N is applied to one end of a cylindrical steel rod of diameter 50cm, the tensile stress is;
Correct answer: C. 2.5 x 10^3Nm- 2
- A. 1.5 x 10^5Nm- 2
- B. 2.5 x 10^5Nm- 2
- C. 2.5 x 10^3Nm- 2
- D. 1 x 10^5Nm- 2
Explanation
We use the formula: σ= F/A Given:Force (F) = 500 NDiameter of the rod = 50 cm = 0.5 mThe cross-sectional area of a cylinder is calculated using the formula for the area of a circle ( A = πr2 ), where ( r ) is the radius of the circle.The radius is half of the diameter, so:r =0.5m/2 =0.25mNow, we can calculate the area:A = π(0.25)2 = π(0.0625 = 0.19635m² (rounded to 5 decimal places)Finally, we can calculate the tensile stress: σ = 500/0.19635 = 2546.48 Nm-2So, the tensile stress is approximately 2546.48 or 2.5×103Nm- 2.
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About Young's Modulus and Stress-Strain
Stress and strain describe how a solid deforms under an applied force, while Young's modulus is the ratio of tensile stress to tensile strain within the elastic limit. The topic includes elastic and plastic behaviour, Hooke's law, the stress strain graph and the difference between stiffness, strength and breaking stress.
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