Consider a cylindrical specimen of a steel alloy (Figure below) 8.5 mm in diameter and 80 mm long that is pulled in tension. Determine its elongation when a load of 65,250 N is applied.
The Correct Answer and Explanation is:
To determine the elongation (ΔL) of the cylindrical steel specimen under a tensile load, we use the formula derived from Hooke’s Law for elastic deformation: ΔL=F⋅L0A⋅E\Delta L = \frac{F \cdot L_0}{A \cdot E}
Given:
- Initial length L0=80 mm=0.080 mL_0 = 80 \, \text{mm} = 0.080 \, \text{m}
- Diameter d=8.5 mm=0.0085 md = 8.5 \, \text{mm} = 0.0085 \, \text{m}
- Force (load) F=65,250 NF = 65,250 \, \text{N}
- Young’s modulus for steel E=200×109 Pa=200 GPaE = 200 \times 10^9 \, \text{Pa} = 200 \, \text{GPa}
Step 1: Calculate the cross-sectional area (A)
A=πd24=π(0.0085)24=3.1416×7.225×10−54=5.79×10−5 m2A = \frac{\pi d^2}{4} = \frac{\pi (0.0085)^2}{4} = \frac{3.1416 \times 7.225 \times 10^{-5}}{4} = 5.79 \times 10^{-5} \, \text{m}^2
Step 2: Plug into elongation formula
ΔL=F⋅L0A⋅E=65,250×0.0805.79×10−5×200×109=5,22011.58×106=0.0004507 m=0.4507 mm\Delta L = \frac{F \cdot L_0}{A \cdot E} = \frac{65,250 \times 0.080}{5.79 \times 10^{-5} \times 200 \times 10^9} = \frac{5,220}{11.58 \times 10^6} = 0.0004507 \, \text{m} = 0.4507 \, \text{mm}
✅ Correct Answer: ΔL=0.45 mm\Delta L = 0.45 \, \text{mm}
Explanation (300+ words):
To find the elongation of a steel specimen under tension, we rely on principles from materials science and mechanics of materials. When a material is loaded within its elastic limit (i.e., the load does not permanently deform it), its behavior can be described using Hooke’s Law, which relates stress and strain via the modulus of elasticity (Young’s modulus).
In this case, we are given the load (force), original length, and diameter of a cylindrical steel specimen. To calculate how much the material elongates under the given force, we first determine the cross-sectional area using the formula for the area of a circle. This is important because stress is defined as force per unit area, and it affects how much the material stretches.
Once we have the area, we plug all values into the elongation formula derived from Hooke’s Law: ΔL=F⋅L0A⋅E\Delta L = \frac{F \cdot L_0}{A \cdot E}
Here:
- FF is the applied force,
- L0L_0 is the original length,
- AA is the cross-sectional area,
- EE is the modulus of elasticity.
The modulus of elasticity EE for steel is very high (around 200 GPa), which reflects its stiffness—meaning it resists deformation under stress. Using all the correct units ensures the resulting elongation is in meters. We then convert meters to millimeters for clarity since the original length was given in millimeters.
The result, 0.45 mm, represents the amount the steel rod stretches under the load of 65,250 N, which is a typical tensile load for mechanical testing. This small deformation illustrates the strength and stiffness of steel under tension.