A cylindrical bar of metal having a diameter of 20.8 mm and a length of 194 mm is deformed elastically in tension with a force of 53,100 N. Given that the elastic modulus and Poisson’s ratio of the metal are 61.2 GPa and 0.34, respectively, 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: Indicate an increase in diameter with a positive number and a decrease with a negative number.
The Correct Answer and Explanation is:
To solve this problem, we will need to use the formulas related to elasticity, specifically for tension and the relationship between stress, strain, and the material properties.
Given:
- Diameter of the cylindrical bar d=20.8 mmd = 20.8 \, \text{mm}d=20.8mm
- Length of the cylindrical bar L=194 mmL = 194 \, \text{mm}L=194mm
- Force applied F=53,100 NF = 53,100 \, \text{N}F=53,100N
- Elastic modulus (Young’s modulus) E=61.2 GPa=61.2×109 PaE = 61.2 \, \text{GPa} = 61.2 \times 10^9 \, \text{Pa}E=61.2GPa=61.2×109Pa
- Poisson’s ratio ν=0.34\nu = 0.34ν=0.34
Step 1: Calculate the cross-sectional area
The cross-sectional area of the cylindrical bar can be calculated using the formula for the area of a circle: A=π(d2)2=π(20.8 mm2)2=π×(10.4 mm)2A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{20.8 \, \text{mm}}{2} \right)^2 = \pi \times (10.4 \, \text{mm})^2A=π(2d)2=π(220.8mm)2=π×(10.4mm)2 A=π×(108.16 mm2)=339.45 mm2=339.45×10−6 m2A = \pi \times (108.16 \, \text{mm}^2) = 339.45 \, \text{mm}^2 = 339.45 \times 10^{-6} \, \text{m}^2A=π×(108.16mm2)=339.45mm2=339.45×10−6m2
Step 2: Calculate the tensile stress
Tensile stress σ\sigmaσ is defined as the force per unit area: σ=FA=53,100 N339.45×10−6 m2=156,000,000 Pa=156 MPa\sigma = \frac{F}{A} = \frac{53,100 \, \text{N}}{339.45 \times 10^{-6} \, \text{m}^2} = 156,000,000 \, \text{Pa} = 156 \, \text{MPa}σ=AF=339.45×10−6m253,100N=156,000,000Pa=156MPa
Step 3: Calculate the elongation (change in length)
The elongation ΔL\Delta LΔL of the specimen can be calculated using Hooke’s Law for elastic deformation in tension: ΔL=FLAE\Delta L = \frac{F L}{A E}ΔL=AEFL
Substitute the known values: ΔL=53,100 N×0.194 m339.45×10−6 m2×61.2×109 Pa=0.000264 m=0.264 mm\Delta L = \frac{53,100 \, \text{N} \times 0.194 \, \text{m}}{339.45 \times 10^{-6} \, \text{m}^2 \times 61.2 \times 10^9 \, \text{Pa}} = 0.000264 \, \text{m} = 0.264 \, \text{mm}ΔL=339.45×10−6m2×61.2×109Pa53,100N×0.194m=0.000264m=0.264mm
So, the specimen elongates by 0.264 mm in the direction of the applied stress.
Step 4: Calculate the change in diameter
The lateral strain (change in diameter) is related to the longitudinal strain by Poisson’s ratio. The longitudinal strain ϵL\epsilon_LϵL is given by: ϵL=ΔLL=0.264 mm194 mm=0.00136\epsilon_L = \frac{\Delta L}{L} = \frac{0.264 \, \text{mm}}{194 \, \text{mm}} = 0.00136ϵL=LΔL=194mm0.264mm=0.00136
The lateral strain ϵD\epsilon_DϵD in the direction perpendicular to the applied stress (change in diameter) is: ϵD=−ν×ϵL=−0.34×0.00136=−0.0004624\epsilon_D = -\nu \times \epsilon_L = -0.34 \times 0.00136 = -0.0004624ϵD=−ν×ϵL=−0.34×0.00136=−0.0004624
Now, the change in diameter Δd\Delta dΔd is: Δd=ϵD×d=−0.0004624×20.8 mm=−0.0096 mm\Delta d = \epsilon_D \times d = -0.0004624 \times 20.8 \, \text{mm} = -0.0096 \, \text{mm}Δd=ϵD×d=−0.0004624×20.8mm=−0.0096mm
Thus, the diameter decreases by 0.0096 mm.
Final Answer:
(a) The elongation in the direction of the applied stress is 0.264 mm.
(b) The change in diameter is a decrease of 0.0096 mm.
