EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 9780100793439
Author: KALPAKJIAN
Publisher: YUZU
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Textbook Question
Chapter 2, Problem 69QTP
The design specification for a metal requires a minimum hardness of 80 HRA. If a Rockwell test is performed and the depth of penetration is 60 μm, is the material acceptable?
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What are the different types of Indenters used in Rockwell Hardness Testing?
Determine the indentation diameter and the surface area of indentation of a copper material subjected to a Brinell hardness test with a test force of 9.4 kN using a hardened steel ball indentor of 10 mm. Take the Brinell Hardness Number as 955.
Solution:
i) Indentation Diameter (in mm) =
ii) Surface Area of Indentation (in mm2) =
Chapter 2 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 2 - Distinguish between engineering stress and true...Ch. 2 - In a stress-strain curve, what is the proportional...Ch. 2 - Describe the events that take place when a...Ch. 2 - What is ductility, and how is it measured?Ch. 2 - In the equation =Kn, which represents the true...Ch. 2 - What is strain-rate sensitivity, and how is it...Ch. 2 - What test can measure the properties of a material...Ch. 2 - What testing procedures can be used to measure the...Ch. 2 - Describe the differences between brittle and...Ch. 2 - What is hardness? Explain.
Ch. 2 - Describe the features of a Rockwell hardness test.Ch. 2 - What is a Leeb test? How is it different from a...Ch. 2 - Differentiate between stress relaxation and creep.Ch. 2 - Describe the difference between elastic and...Ch. 2 - Explain what uniform elongation means in tension...Ch. 2 - Describe the difference between deformation rate...Ch. 2 - Describe the difficulties involved in conducting a...Ch. 2 - What is Hookes law? Youngs modulus? Poissons...Ch. 2 - Describe the difference between transgranular and...Ch. 2 - What is the reason that yield strength is...Ch. 2 - Why does the fatigue strength of a specimen or...Ch. 2 - If striations are observed under microscopic...Ch. 2 - What is an Izod test? Why are Izod tests useful?Ch. 2 - Why does temperature increase during plastic...Ch. 2 - What is residual stress? How can residual stresses...Ch. 2 - On the same scale for stress, the tensile true...Ch. 2 - What are the similarities and differences between...Ch. 2 - Can a material have a negative Poissons ratio?...Ch. 2 - It has been stated that the higher the value of m,...Ch. 2 - Explain why materials with high m values, such as...Ch. 2 - With a simple sketch, explain whether it is...Ch. 2 - Explain why the difference between engineering...Ch. 2 - Consider an elastomer, such as a rubber band. This...Ch. 2 - If a material (such as aluminum) does not have an...Ch. 2 - What role, if any, does friction play in a...Ch. 2 - Which hardness tests and scales would you use for...Ch. 2 - Consider the circumstance where a Vickers hardness...Ch. 2 - Which of the two tests, tension or compression,...Ch. 2 - List and explain briefly the conditions that...Ch. 2 - List the factors that you would consider in...Ch. 2 - On the basis of Fig. 2.5, can you calculate the...Ch. 2 - If a metal tension-test specimen is rapidly pulled...Ch. 2 - Comment on your observations regarding the...Ch. 2 - Will the disk test be applicable to a ductile...Ch. 2 - What hardness test is suitable for determining the...Ch. 2 - Wire rope consists of many wires that bend and...Ch. 2 - A statistical sampling of Rockwell C hardness...Ch. 2 - In a Brinell hardness test, the resulting...Ch. 2 - Some coatings are extremely thinsome as thin as a...Ch. 2 - Select an appropriate hardness test for each of...Ch. 2 - A paper clip is made of wire 0.5 mm in diameter....Ch. 2 - A 250-mm-long strip of metal is stretched in two...Ch. 2 - Identify the two materials in Fig. 2.5 that have...Ch. 2 - Plot the ultimate strength vs. stiffness for the...Ch. 2 - If you remove the layer of material ad from the...Ch. 2 - Prove that the true strain at necking equals the...Ch. 2 - Percent elongation is always defined in terms of...Ch. 2 - You are given the K and n values of two different...Ch. 2 - A cable is made of two strands of different...Ch. 2 - On the basis of the information given in Fig. 2.5,...Ch. 2 - In a disk test performed on a specimen 1.00 in. in...Ch. 2 - A piece of steel has a hardness of 300 HB....Ch. 2 - A metal has the following properties: UTS = 70,000...Ch. 2 - Using only Fig. 2.5, calculate the maximum load in...Ch. 2 - Estimate the modulus of resilience for a highly...Ch. 2 - A metal has a strength coefficient K = 100,000 psi...Ch. 2 - Plot the true stresstrue strain curves for the...Ch. 2 - The design specification for a metal requires a...Ch. 2 - Calculate the major and minor pyramid angles for a...Ch. 2 - If a material has a target hardness of 300 HB,...Ch. 2 - A Rockwell A test was conducted on a material and...Ch. 2 - For a cold-drawn 0.5% carbon steel, will a...Ch. 2 - A material is tested in tension. Over a 1-in. gage...Ch. 2 - A horizontal rigid bar cc is subjecting specimen a...Ch. 2 - List and explain the desirable mechanical...Ch. 2 - When making a hamburger, you may have observed the...Ch. 2 - An inexpensive claylike material called Silly...Ch. 2 - In tension testing of specimens, mechanical and...Ch. 2 - Demonstrate the impact toughness of a piece of...Ch. 2 - Using a large rubber band and a set of weights,...Ch. 2 - Find or prepare some solid circular pieces of...Ch. 2 - Take several rubber bands and pull them at...Ch. 2 - Devise a simple fixture for conducting the bend...Ch. 2 - By pressing a small ball bearing against the top...Ch. 2 - Describe your observations regarding Fig. 2.14c.Ch. 2 - Embed a small steel ball in a soft block of...Ch. 2 - Devise a simple experiment, and perform tests on...Ch. 2 - Obtain some solid and some tubular metal pieces,...Ch. 2 - Explain how you would obtain an estimate of the...Ch. 2 - Without using the words stress or strain, define...Ch. 2 - We know that it is relatively easy to subject a...
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- Determine the surface area of indentation for a specimen that has Brinell hardness number as 851 and the applied force as 15 kN. The surface area of indentation (mm2) = %3Darrow_forwardWhich one is the correct answer please? Thank youarrow_forwardA tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 65 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 36.1 mm - Final Diameter after fracture = 4.25 mm & - Ultimate stress = 401 MPa SOLUTION: Initial Diameter (in mm) = Ultimate Load (in N) =arrow_forward
- Part A The stress-strain diagram for a steel alloy having an original diameter of 0.40 in. and a gage length of 9 in. is shown in the figure below. (Figure 1) Determine the modulus of elasticity for the material. Express your answer to three significant figures and include appropriate units. HA ? Eapprox = Value Units Submit Request Answer Figure < 1 of 1 Part B a (ksi) 80 Determine the load on the specimen that causes yielding. 70 Express your answer to three significant figures and include appropriate units. 60 50 40 HA 30 20 Py = Value Units 10 € (in./in.) 0.04 0,08 0.12 0,16 0,20 0,24 0,28 O 0.0005 0.001 0.0015 0.002 0.0025 0.0030.0035 Request Answer Submitarrow_forwardQuestion 6 What is the calculated fracture toughness, K, for the perspex specimen in MPa root m? Give your answer to 2 decimal places. P = 250 N, crack length = 5.5 mm W= 13.3 mm, B = 5.45 mm Note that the span length (s) is 38mm. Take f(a/W) as 0.78arrow_forwardA tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 66 % - Percentage of Reduction in Area = 38 % - Final length after fracture = 34.6 mm - Final Diameter after fracture = 4.43 mm & - Ultimate stress = 364 MPa Determine the Initial length, Initial diameter and Maximum load. ii) Final Area (in mm2) = iii) Initial Area (in mm2) =arrow_forward
- A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 65 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 36.1 mm - Final Diameter after fracture = 4.25 mm & - Ultimate stress = 401 MPa SOLUTION: Initial Length (in mm) = Final Area (in mm2) = Initial Area (in mm2) =arrow_forwardA mild steel tensile specimen of initial length 44 mm and initial diameter 6.4 mm is subjected to a tensile test and the following data are obtained. - Yield Strength as 88 MPa - Maximum Strength as 212 MPa - Fracture Strength as 152 MPa - Percentage of Elongation as 63 % - Percentage of Reduction in area as 39% Determine the Fracture load The Final length in mm = The Final area in mm2 =arrow_forwardDetermine the load on the specimen that causes yielding. Express your answer to three significant figures and include appropriate units. The stress-strain diagram for a steel alloy having an original diameter of 0.40 in. and a gage length of 9 in. is shown in the figure below. (Figure 1) ? Py = Value Units Submit Request Answer Part C Figure < 1 of 1 Determine the ultimate load the specimen will support. Express your answer to three significant figures and include appropriate units. o (ksi) 80 HA ? 70 60 Value Units 50 Pu = 40 30 Submit Request Answer 20 10 € (in./in.) 0 0.04 0.08 0.12 0,16 0,20 0,24 028 O 0.0005 0.001 0.0015 0.002 0.0025 0.0030.0035 Provide Feedbackarrow_forward
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