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Chapter 6: Problem 8

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 \(0.38 \mathrm{~mm}(0.015 \mathrm{in} .)\) ?

Short Answer

Expert verified
Answer: Approximately 13.6 mm.

Step by step solution

01

Convert given values to consistent units

We will use SI units in this problem. The given values are: Load (P) = 11,100 N Elastic Modulus (E) = 207 GPa = 207 × 10^9 Pa Yield Strength = 310 MPa = 310 × 10^6 Pa Length (L) = 500 mm = 0.5 m Elongation (ΔL) = 0.38 mm = 0.00038 m Now that all values have been converted to consistent units, we can proceed to the next step.
02

Calculate the cross-sectional area (A) using elongation formula

Using the given elongation formula: ΔL = (P * L) / (A * E) Rearrackan elongation form for the area (A): A = (P * L) / (ΔL * E) Now, substitute the given values: A = (11,100 * 0.5) / (0.00038 * 207 × 10^9) A ≈ 1.457 × 10^-4 m²
03

Find the diameter (d) using the cross-sectional area formula

The cross-sectional area of a cylindrical rod is given by: A = (πd²) / 4 Rearrange this formula for the diameter (d): d = 2 * sqrt(A / π) Substitute the cross-sectional area (A) we found in the previous step: d = 2 * sqrt(1.457 × 10^-4 / π) d ≈ 0.0136 m Now, convert the diameter back to millimeters: d = 0.0136 * 1000 d ≈ 13.6 mm So, the required diameter of the cylindrical rod of steel to allow an elongation of 0.38 mm when subjected to a load of 11,100 N is approximately 13.6 mm.

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Most popular questions from this chapter

Explain the differences in grain structure for a metal that has been cold worked and one that has been cold worked and then recrystallized.

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

(a) What is the approximate ductility (\%EL) of a brass that has a yield strength of \(345 \mathrm{MPa}\). \((50,000 \mathrm{psi}) ?\) (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 620 MPa \((90,000\) psi)?

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 \(0.25 \mathrm{~mm}\) \((0.010 \mathrm{in} .)\)

Consider a hypothetical material that has a grain diameter of \(2.1 \times 10^{-2} \mathrm{~mm}\). After a heat treatment at \(600^{\circ} \mathrm{C}\) for \(3 \mathrm{~h}\), the grain diameter has increased to \(7.2 \times 10^{-2} \mathrm{~mm}\). Compute the grain diameter when a specimen of this same original material (i.e., \(d_{0}=2.1 \times 10^{-2} \mathrm{~mm}\) ) is heated for \(1.7 \mathrm{~h}\) at \(600^{\circ} \mathrm{C}\). Assume the \(n\) grain diameter exponent has a value of 2
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