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Chapter 7: Problem 20

A brass alloy is known to have a yield strength of \(240 \mathrm{MPa}(35,000\) psi), a tensile strength of \(310 \mathrm{MPa}(45,000 \mathrm{psi})\), and an elastic modulus of 110 GPa \(\left(16.0 \times 10^{6}\right.\) psi). A cylindrical specimen of this alloy \(15.2 \mathrm{~mm}(0.60 \mathrm{in}\).) in diameter and \(380 \mathrm{~mm}(15.0\) in.) long is stressed in tension and found to elongate \(1.9 \mathrm{~mm}(0.075\) in.). On the basis of the information given, is it possible to compute the magnitude of the load necessary to produce this change in length? If so, calculate the load; if not, explain why.

Short Answer

Expert verified
If so, determine the magnitude. Answer: No, it is not possible to compute the magnitude of the load necessary to produce the elongation of 1.9 mm in the given brass alloy cylindrical specimen due to the calculated stress (550 MPa) being outside of the acceptable range between the yield strength (240 MPa) and tensile strength (310 MPa).

Step by step solution

01

Calculate the cross-sectional area

The cross-sectional area of the cylindrical specimen (A) can be calculated using the formula for the area of a circle: \(A = \pi r^{2}\) where r is the radius of the cylinder. In this case, the diameter of the cylinder is given as 15.2 mm, so the radius (r) is: \(r = \frac{15.2}{2} = 7.6 \; \text{mm}\) Now we can calculate the cross-sectional area (A): \(A = \pi(7.6^2) = 181.6 \; \text{mm}^2\)
02

Calculate the elongation ratio

To find the elongation ratio (ER), we need to divide the elongation of the cylinder (1.9 mm) by its original length (380 mm): \(\text{ER} = \frac{1.9}{380} = 0.005\)
03

Calculate the stress

Now we will use the relationship between stress, elongation ratio, and elastic modulus: \(\text{Stress (}\sigma\text{)} = \text{Elongation Ratio (ER)} \cdot \text{Elastic Modulus (E)}\) Plugging in the given values, we get: \(\sigma = 0.005 \cdot 110 \cdot 10^{3} = 550\; \text{MPa}\)
04

Check if the stress is within the acceptable range

We need to check if the calculated stress (550 MPa) falls within the acceptable range, which is between the yield strength (240 MPa) and the tensile strength (310 MPa) of the brass alloy. In this case, the calculated stress (550 MPa) is not within the acceptable range, so it is not possible to compute the magnitude of the load necessary to produce this change in length. If the stress had been within the acceptable range, we would proceed to step 5 to calculate the load, but in this case, the problem can't be solved as the calculated stress is higher than the tensile strength of the alloy.

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Most popular questions from this chapter

For each of edge, screw, and mixed dislocations, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.

(a) What is the driving force for recrystallization? (b) What is the driving force for grain growth?

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(10.0\) \(\mathrm{mm}(0.39 \mathrm{in} .) .\) A tensile force of \(1500 \mathrm{~N}\left(340 \mathrm{lb}_{\mathrm{f}}\right)\) produces an elastic reduction in diameter of \(6.7 \times 10^{-4} \mathrm{~mm}\left(2.64 \times 10^{-5}\right.\) in.). Compute the elastic modulus of this alloy, given that Poisson's ratio is \(0.35\).

For a brass alloy, the stress at which plastic deformation begins is \(345 \mathrm{MPa}(50,000 \mathrm{psi})\), and the modulus of elasticity is \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\) (a) What is the maximum load that can be applied to a specimen with a cross- sectional area of \(130 \mathrm{~mm}^{2}\left(0.2 \mathrm{in} .{ }^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(76 \mathrm{~mm}(3.0\) in.), what is the maximum length to which it can] be stretched without causing plastic deformation?

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