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Chapter 6: Problem 4

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa \(\left(30 \times 10^{6}\right.\) psi) and an original diameter of \(10.2 \mathrm{~mm}(0.40 \mathrm{in}\).) experiences only elastic deformation when a tensile load of \(8900 \mathrm{~N}\left(2000 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{~mm}\) \((0.010 \mathrm{in} .)\)

Short Answer

Expert verified
Answer: To find the maximum length of the specimen before deformation, follow these steps: Step 1: Calculate the original cross-sectional area of the specimen A = π(5.1)^2 ≈ 81.712 mm² Step 2: Calculate the engineering stress σ = 8900 N / 81.712 mm² ≈ 108.94 MPa Step 3: Apply Hooke's Law ε = 108.94 MPa / 207 GPa ≈ 5.26 x 10^-4 Step 4: Calculate the original length L₀ = 0.25 mm / 5.26 x 10^-4 ≈ 475.29 mm The maximum length of the specimen before deformation is approximately 475.29 mm.

Step by step solution

01

Calculate the original cross-sectional area of the specimen

We are given the original diameter (\(D\)) of the specimen, which is 10.2 mm. The original cross-sectional area can be found using the formula for the area of a circle. $$A = \pi \left(\frac{D}{2}\right)^{2}$$ Substitute the given diameter value and calculate the original cross-sectional area (A).
02

Calculate the engineering stress

The engineering stress (\(\sigma\)) can be calculated using the formula: $$\sigma = \frac{F}{A}$$ where \(F\) is the tensile load applied to the specimen and \(A\) is the original cross-sectional area calculated in Step 1. Substitute the given tensile load (8900 N) and the calculated cross-sectional area to find the engineering stress.
03

Apply Hooke's Law

Hooke's law describes the relationship between stress (\(\sigma\)) and strain (\(\epsilon\)) in an elastic material: $$\sigma = E \cdot \epsilon$$ where \(E\) is the elastic modulus of the material (207 GPa in this case). We can rearrange this formula to find the strain: $$\epsilon = \frac{\sigma}{E}$$ Substitute the values of engineering stress and elastic modulus to find the strain.
04

Calculate the original length

The relationship between strain (\(\epsilon\)), change in length (\(\Delta L\)), and original length (\(L_0\)) is: $$\epsilon = \frac{\Delta L}{L_0}$$ We are given the maximum allowable elongation which is equal to the change in length (\(\Delta L = 0.25\) mm). We can rearrange the formula to find the original length: $$L_0 = \frac{\Delta L}{\epsilon}$$ Substitute the values of change in length and strain to find the maximum length of the specimen before deformation.

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Most popular questions from this chapter

Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 15 and \(12 \mathrm{~mm}\), respectively. The second specimen, with an initial radius of \(11 \mathrm{~mm}\), must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation

Briefly explain why HCP metals are typically more brittle than \(\mathrm{FCC}\) and \(\mathrm{BCC}\) metals.

Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along, a [112] direction. If slip occurs on a (111) plane and in a [011] direction, and the crystal yields at a stress of \(5.12 \mathrm{MPa}\), compute the critical resolved shear stress.

(a) Equations \(6.4 a\) and \(6.4 \mathrm{~b}\) are expressions for normal \(\left(\sigma^{\prime}\right)\) and shear \(\left(\tau^{\prime}\right)\) stresses, respectively, as a function of the applied tensile stress \((\sigma)\) and the inclination angle of the plane on which these stresses are taken ( \(\theta\) of Figure 6.4). Make a plot showing the orientation parameters of these expressions (i.e., \(\cos ^{2} \theta\) and \(\sin \theta \cos \theta\) ) versus \(\theta\). (b) From this plot, at what angle of inclination is the normal stress a maximum? (c) At what inclination angle is the shear stress a maximum?

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{6}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 lb_{f } ). If the length of the rod is \(500 \mathrm{~mm}\) (20.0 in.), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015 \mathrm{in} .)\) ?
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