**Problem 5:** Starting with a 2-inch diameter rod of brass, we would like to process 0.2-inch diameter rods that possess minimum yield strength of 40 ksi and a minimum elongation to fracture of 40% (see the following figure). Design a process that achieves that.**Figure Explanation:**- The graph presents the relationship between tensile strength, yield strength, elongation percent, and the percent of cold work done on a brass alloy composed of 70% copper and 30% zinc. - **X-axis:** Represents the "Percent cold work" ranging from 0 to 60%.- **Left Y-axis:** Indicates "Tensile strength and yield strength" in ksi (kips per square inch).- **Right Y-axis:** Shows the "Elongation percent."- **Tensile strength curve:** This is the uppermost pink curve that increases steadily from approximately 58 ksi to 100 ksi as cold work increases from 0% to 60%.- **Yield strength curve:** The middle black curve starts at around 30 ksi at 0% cold work, increasing to approximately 90 ksi at 60% cold work.- **Elongation curve:** The bottom dark pink curve begins at about 70% elongation at 0% cold work, decreasing to close to 5% elongation as cold work approaches 60%.These curves illustrate how increasing the amount of cold work generally increases the tensile and yield strengths while decreasing the material's elongation percentage. The challenge is to design a manufacturing process for the brass rods to achieve a balance of 40 ksi yield strength and 40% elongation, based on these material properties and changes due to cold work.

Answered: Image /qna-images/answer/ebc4f139-f532-41ee-94be-328f612a821c.jpg

Problem 5. Starting with a 2-inch diameter rod of brass, we would like to process 0.2-inch diameter rods that possess minimum yield strength of 40 ksi and a minimum elongation to fracture of 40%, (see the following figure). Design a process that achieves that. 120 70% Cu-30% Zn 100 Tensile strength 80 60 60 Yield strength 40 40 20 Elongation 10 20 30 40 50 60 Tensile strength and yield strength (ksi) Elongation percent

Problem 5. Starting with a 2-inch diameter rod of brass, we would like to process 0.2-inch diameter rods that possess minimum yield strength of 40 ksi and a minimum elongation to fracture of 40%, (see the following figure). Design a process that achieves that. 120 70% Cu-30% Zn 100 Tensile strength 80 60 60 Yield strength 40 40 20 Elongation 10 20 30 40 50 60 Tensile strength and yield strength (ksi) Elongation percent

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: Some aircraft component is fabricated from an aluminium alloy that has a plane strain fracture…

A: Case 1) Stress=300MPa Internal crack length =4mm Case 2) Stress=260MPa Internal crack length =6mm

Q: In an engineering application, the material is a strip of iron with a fixed crystallographic…

A: Given: The yield strength of material for grain size of 30 microns, σ₁ = 110 MPa The yield strength…

Q: Using the equation Kic= of Vna , plot the fracture stress (MPa) for aluminium alloy 7075- T651…

Q: A steel plate, which has a fracture toughness in a plane strain state of 83 MPavm, is alternately…

A: Given Data: KIC=83 MPamσmax=750 MPaσmin=20 MPaf=1n=2.8ai=0.0015 mm To Find: Determine the value of…

Q: Iculate the activation energy (in J/mol) for creep for an alloy having the steady state creep havior…

Q: The compressive strength of alumina quartz (Al2O3) is 2.1 GPa. A quartz tube is being used for a…

Q: An Al2O3 specimen is being pulled in tension. The specimen contains flaws having a size of 100 µm.…

Q: 2- A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length…

A: Given: Dia-0.495 in. Gauge length = 2 in

Q: An aircraft component, made of 7075-T6 aluminum alloy, can be represented as a bar of diameter 20 mm…

Q: A tie rod made of quenched and tempered 4340 steel is used as a critical linkage in an industrial…

A: Given data as per question for the given 4340 steel Tensile stress =550 MpaCompressive stress…

Q: Example (6): A tensile specimen with a 12 mm initial djameter and 5 gage Jength reaches maximum load…

Q: A specimen of a steel alloy with a plane strain fracture toughness of 22.1 MPavm.The largest surface…

Q: Calculate the tensile modulus (G) of a 1-cm poly(methyl methacrylate) (CAS# 9011-14-7) rod with a…

Q: f 800 MPa, and a tensile this specimen experienc long? Why or why no

Q: Question-7. Steady-state creep data taken for an alloy at a stress level of 160 MPa are given below.…

A: Given, Data for stress level of 160 MPa The stress component , n =6.8 R = 8.3145 J/mol-K To find…

Q: A material has a yield strength of 1150Mpa and a fracture toughness of 47 Mpa m^1/2. If the applied…

A: Data Given : We need to find the internal crack that can lead to failure.

Q: Will fracture occur for a material exposed to a tensile stress of 900 MPa if the material has a…

Q: 4. The mechanical properties of a given titanium alloy (Ti-6Al-4V) are E = 114 GPa, σy = 880 MPa and…

Q: A mechanical component is fabricated from a metallic alloy. From the experiment, it is obtained…

A: Given , fracture stress of the alloy (σc)= 36ksi maximum internal crack length = 0.10 in plain…

Concept explainers

Thermal Properties of Materials

In mechanical engineering, the thermal property is represented as the physical quantity that is relevant to the application of heat. The thermal properties are evaluated by variations in any specific material due to low heat or high heat. The quantities that influence the thermal property of the materials are mass, density, temperature, and many others.

Question

100%

**Problem 5:** Starting with a 2-inch diameter rod of brass, we would like to process 0.2-inch diameter rods that possess minimum yield strength of 40 ksi and a minimum elongation to fracture of 40% (see the following figure). Design a process that achieves that.

**Figure Explanation:**

- The graph presents the relationship between tensile strength, yield strength, elongation percent, and the percent of cold work done on a brass alloy composed of 70% copper and 30% zinc.

- **X-axis:** Represents the "Percent cold work" ranging from 0 to 60%.

- **Left Y-axis:** Indicates "Tensile strength and yield strength" in ksi (kips per square inch).

- **Right Y-axis:** Shows the "Elongation percent."

- **Tensile strength curve:** This is the uppermost pink curve that increases steadily from approximately 58 ksi to 100 ksi as cold work increases from 0% to 60%.

- **Yield strength curve:** The middle black curve starts at around 30 ksi at 0% cold work, increasing to approximately 90 ksi at 60% cold work.

- **Elongation curve:** The bottom dark pink curve begins at about 70% elongation at 0% cold work, decreasing to close to 5% elongation as cold work approaches 60%.

These curves illustrate how increasing the amount of cold work generally increases the tensile and yield strengths while decreasing the material's elongation percentage. The challenge is to design a manufacturing process for the brass rods to achieve a balance of 40 ksi yield strength and 40% elongation, based on these material properties and changes due to cold work.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Types of Properties of Engineering Materials

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

(2) Consider a homogeneous structure made out of metal subjected to homogeneous boundary conditions t = σn. For o corresponding to uniaxial, biaxial and shear stress state, calculate the stress at which the material yields.
In designing a ceramic part for a jet engine, to ensure that the part does not fail, we plan to ensure that the maximum applied stress is only one-third of the yield strength. The yield strength of this CMC part for a jet engine is 520 MPa, and its plane strain fracture toughness is 5.5 MPa√?. We utilize a nondestructive test to find any internal faults that are longer than 1.27 mm. Does our non-destructive test have the requisite sensitivity if Y = 1.4? Explain.
What would be the compressive stress in MPa developed in the silica glass furnace tube constrained in the way illustrated in the figure if it were heated to 1000C?
1. For the stress-strain curve shown below, please estimate the properties indicated. (a) Fracture Strain Please do your work on a separate sheet of paper, and put your answers in the boxes on the right. Be sure to include the proper symbol and units. Stress Strain 70 60 50 Stress (ksi) 240 30 20 10 70 0 0.000 60 50 Stress (ksi) 40 20 10 KULL 0 0.000 0.010 0.050 0.100 Strain (in/in) Stress Strain 0.020 0.030 Strain (in/in) 0.040 0.150 0.050 (b) Ultimate Tensile Stress (c) Fracture Stress (d) Proportional Limit (e) Elastic Modulus (1) Yield Stress (g) Tensile Toughness (Modulus of Toughness) (h) Modulus of Resilience
(a) An AISI steel plate has width W = 30 cm and a central crack with size of 3 mm. The plate is under a uniform stress. Find the maximum value of the stress is KIc = 50 MPa m1/2. (b) If the part has to operate at a stress of 1,500 MPa, compute the maximum crack size that the plate can have.
A tie rod made of quenched and tempered 4340 steel is used as a critical linkage in an industrial apparatus. The rod is subjected to an alternating cyclic tensile/compressive stress from 550 MPa to -550 MPa. An inspection of the rod revealed a 3.5 mm deep edge crack on the surface of the rod. Please answer the following question for the rod given the properties in the table and curve below and fracture toughness of 55 MPa√m. (NOTE: Y=1.12 for edge crack and use “a” not “2a” for the length of the crack) 1. What is the minimum crack size that would cause instantaneous fast fracture at the current maximum axial tensile loading, assuming plane strain conditions apply? If the resolution limit of the flaw detection apparatus were 1.5 mm, would this flaw be subject to detection? Show calculations to back up your conclusions.
A tie rod made of quenched and tempered 4340 steel is used as a critical linkage in an industrial apparatus. The rod is subjected to an alternating cyclic tensile/compressive stress from 550 MPa to -550 MPa. An inspection of the rod revealed a 3.5 mm deep edge crack on the surface of the rod. Please answer the following question for the rod given the properties in the table and curve below and fracture toughness of 55 MPa√m. (NOTE: Y=1.12 for edge crack and use “a” not “2a” for the length of the crack) 1. Given the S-N curve for 4340 steel, is there an endurance limit? If so, what is the stress value for the endurance limit? 2. If there were no cracks and the stress on the rod still alternated from 550 MPa to -550 MPa with mean stress of zero, what would be the cycles to failure?
A high-strength steel has yield strength of 1165 MPa and fracture toughness of 95 MPavm. Calculate the size of an internal crack that will lead to catastrophic failure at an applied stress of 80% yield strength. (Y = 1.00 for internal crack, and Y=1.12 for surface crack)
5. How much more or less likely is it for a brittle material with volume V to fail compared to the same material with volume Vo/2 at the same applied stress? Why is this the case? Talk specifically in the context of flaw landscape.
Draw two schematic graphs using pencil showing a typical stress-strain curve for aluminum. The first graph should show engineering stress vs engineering strain, and the second graph should show true stress vs true strain. Label the showing: (i) elastic modulus (ii) proportional limit (iii) yield stress (iv)yield strain (v) fracture stress (vi) fracture strain on each graph. You may showboth graphs on one plot. Explain the difference between engineering stress and true stress.
A 3105 aluminum plate is reduced from 4.38 cm to 2.88 cm. Determine the final properties of the plate. See attached figure
(a) An AISI 4340 steel plate has a width W of 30 cm and has a central crack 2a of 3 mm. The plate is under a uniform stress σ. This steel has a KIc value of 50 MPa m1/2. Find the maximum stress for this crack length. (b) If the operating stress is 1,500 MPa, compute the maximum crack size that the steel may have without failure