- A cylindrical specimen of steel having a diameter of 15.2 mm (0.60 in.) and length of 250 mm (10.0 in.) is deformed elastically in tension with a force of 48,900 N (11,000 lbf). The modulus of elasticity and the Poisson’s ratio for this steel is 207 GPa (
psi) and 0.30. Determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Will the diameter increase or decrease?
The Correct Answer and Explanation is:
To solve this, we will break it into two parts: (a) calculating the elongation in the direction of applied stress, and (b) calculating the change in diameter using Poisson’s ratio.
Given:
- Diameter of the specimen, d=15.2 mmd = 15.2 \, \text{mm}d=15.2mm
- Length of the specimen, L=250 mmL = 250 \, \text{mm}L=250mm
- Force applied, F=48,900 NF = 48,900 \, \text{N}F=48,900N
- Modulus of elasticity, E=207 GPa=207×109 N/m2E = 207 \, \text{GPa} = 207 \times 10^9 \, \text{N/m}^2E=207GPa=207×109N/m2
- Poisson’s ratio, ν=0.30\nu = 0.30ν=0.30
(a) Elongation of the specimen
The elongation ΔL\Delta LΔL in the direction of the applied force is given by Hooke’s Law: ΔL=F⋅LA⋅E\Delta L = \frac{F \cdot L}{A \cdot E}ΔL=A⋅EF⋅L
Where:
- A=cross-sectional area=π(d2)2A = \text{cross-sectional area} = \pi \left( \frac{d}{2} \right)^2A=cross-sectional area=π(2d)2
First, calculate the cross-sectional area: A=π(15.2 mm2)2=π(7.6 mm)2=181.46 mm2A = \pi \left( \frac{15.2 \, \text{mm}}{2} \right)^2 = \pi \left( 7.6 \, \text{mm} \right)^2 = 181.46 \, \text{mm}^2A=π(215.2mm)2=π(7.6mm)2=181.46mm2
Now, plug in the values into the elongation formula: ΔL=48,900 N×250 mm181.46 mm2×207×109 N/m2\Delta L = \frac{48,900 \, \text{N} \times 250 \, \text{mm}}{181.46 \, \text{mm}^2 \times 207 \times 10^9 \, \text{N/m}^2}ΔL=181.46mm2×207×109N/m248,900N×250mm
Converting mm to meters: ΔL=48,900×0.25181.46×10−6×207×109\Delta L = \frac{48,900 \times 0.25}{181.46 \times 10^{-6} \times 207 \times 10^9}ΔL=181.46×10−6×207×10948,900×0.25 ΔL=0.000579 m=0.579 mm\Delta L = 0.000579 \, \text{m} = 0.579 \, \text{mm}ΔL=0.000579m=0.579mm
Thus, the elongation of the specimen is approximately 0.579 mm.
(b) Change in Diameter of the Specimen
Using Poisson’s ratio, we know that the lateral strain (change in diameter) is related to the longitudinal strain (elongation). The formula is: Δdd=−ν⋅ΔLL\frac{\Delta d}{d} = -\nu \cdot \frac{\Delta L}{L}dΔd=−ν⋅LΔL
Where:
- Δd\Delta dΔd is the change in diameter
- ddd is the initial diameter
- ν\nuν is Poisson’s ratio
- ΔL\Delta LΔL is the elongation
- LLL is the original length
Now, substitute the known values: Δdd=−0.30×0.579 mm250 mm=−0.30×0.002316=−0.000695\frac{\Delta d}{d} = -0.30 \times \frac{0.579 \, \text{mm}}{250 \, \text{mm}} = -0.30 \times 0.002316 = -0.000695dΔd=−0.30×250mm0.579mm=−0.30×0.002316=−0.000695
Thus, the change in diameter is: Δd=−0.000695×15.2 mm=−0.0105 mm\Delta d = -0.000695 \times 15.2 \, \text{mm} = -0.0105 \, \text{mm}Δd=−0.000695×15.2mm=−0.0105mm
So, the change in diameter is approximately -0.0105 mm, which means the diameter decreases.
Conclusion:
(a) The specimen will elongate by approximately 0.579 mm in the direction of the applied stress.
(b) The diameter of the specimen will decrease by approximately 0.0105 mm due to the lateral contraction resulting from the applied tension.
