(c)A cylindrical metal specimen having an original diameter of 12.5 mm and gaugelength of 50.00 mm is pulled in tension until fracture occurs. The diameter at thepoint of fracture is 8.00 mm, and the fractured gauge length is 75.00 mm.Calculate the ductility in terms of percent reduction in area.(d) Compare the following attributes combination for two different metal alloys byusing a simple plot sketches on a single stress versus strain diagram.i.Metal A: High strength, high stiffness, no ductilityii.Metal B: High strength, moderate stiffness, high ductility QI (a)A tensile stress is to be applied along the axis of a cylindrical steel rod that has adiameter of 7.5 mm. Given the Poisson's ratio, v is 0.30 and the modulus ofelasticity, E of the steel is 207 GPa.Determine the magnitude of the load required to produce a 2.5 x10³ mm changein diameter if the deformation is entirely elastic.(b) Referring to the tensile test data tabulated in Table 1, answer the followingquestions:i.Select with justification the material that will experience the greatestpercent reduction in area.ii.Select with justification which material is the strongest.Table 1. Tensile stress-strain data for several hypothetical metalsMaterialYieldTensileStrain atFractureElasticStrengthStrengthFractureStrength Modulus(МРа)(MPa)(MPa)(GPa)A3103400.232652101001200.40105150C4155500.15500310D7008500.14720210EFracture before yielding650350

C) The property of the material to change its shape but without losing its properties such as…

(c) A cylindrical metal specimen having an original diameter of 12.5 mm and gauge length of 50.00 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.00 mm, and the fractured gauge length is 75.00 mm. Calculate the ductility in terms of percent reduction in area. (d) Compare the following attributes combination for two different metal alloys by using a simple plot sketches on a single stress versus strain diagram. i. Metal A: High strength, high stiffness, no ductility ii. Metal B: High strength, moderate stiffness, high ductility

(c) A cylindrical metal specimen having an original diameter of 12.5 mm and gauge length of 50.00 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.00 mm, and the fractured gauge length is 75.00 mm. Calculate the ductility in terms of percent reduction in area. (d) Compare the following attributes combination for two different metal alloys by using a simple plot sketches on a single stress versus strain diagram. i. Metal A: High strength, high stiffness, no ductility ii. Metal B: High strength, moderate stiffness, high ductility

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: Q/Figure i shows the stress and strain curve of a ... metal sample of standard length (250 mm) and…

A: Given: The stress-strain curve for the given sample is shown below:

Q: A three-point bending test was performed on an aluminum oxide specimen having a circular cross…

A: Given data:- Material aluminum oxide For circular cross-section Radius (r)=5.6 mm fracture load…

Q: A cold drawn 1144 Stressproof steel specimen with a diameter of 0.874 in. and a 1.0 in. gage length…

Q: A cylindrical specimen of steel having an original diameter of 12.8 mm (0.505 in.) is tensile tested…

A: Consider a specimen having initial cross-sectional and final cross-sectional area be A0 and Af…

Q: A cylindrical sample has a gauge length of 50.80 mm and original diameter of 12.8 mm. The sample is…

A: Given data as per question d0=12.8 mmdf=6.6 mmLO=50.8 mmLf=72.14 mm(1)calculating % change in…

Q: A UTS with a machine stiffness of 75 KN/mm is used to measure load/displacement points in the…

A: Given data:find:Engineering strain and true strain

Q: 3.33 The following data are obtained from a tensile test of a 12.7-mm-diametler aluminum specimen…

A: Given Specimen length = 50 mm Specimen diameter = 12.7 mm Specimen final diameter = 8.8 mm To find…

Q: 6. The following data were collected from a 12-mm diameter test specimen of magnesium (lo = 30.00…

Q: 3. A rod-shaped MgO specimen with round cross section is tested with the three-point bend method.…

Q: Problem: A tensile test specimen of stainless steel alloy having a with a diameter of 12.6 mm and a…

A: “Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for…

Q: Consider a cylindrical specimen of a steel alloy with 8.5 mm diameter and 80 mm long that is pulled…

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

(c)
A cylindrical metal specimen having an original diameter of 12.5 mm and gauge
length of 50.00 mm is pulled in tension until fracture occurs. The diameter at the
point of fracture is 8.00 mm, and the fractured gauge length is 75.00 mm.
Calculate the ductility in terms of percent reduction in area.
(d) Compare the following attributes combination for two different metal alloys by
using a simple plot sketches on a single stress versus strain diagram.
i.
Metal A: High strength, high stiffness, no ductility
ii.
Metal B: High strength, moderate stiffness, high ductility

QI (a)
A tensile stress is to be applied along the axis of a cylindrical steel rod that has a
diameter of 7.5 mm. Given the Poisson's ratio, v is 0.30 and the modulus of
elasticity, E of the steel is 207 GPa.
Determine the magnitude of the load required to produce a 2.5 x10³ mm change
in diameter if the deformation is entirely elastic.
(b) Referring to the tensile test data tabulated in Table 1, answer the following
questions:
i.
Select with justification the material that will experience the greatest
percent reduction in area.
ii.
Select with justification which material is the strongest.
Table 1. Tensile stress-strain data for several hypothetical metals
Material
Yield
Tensile
Strain at
Fracture
Elastic
Strength
Strength
Fracture
Strength Modulus
(МРа)
(MPa)
(MPa)
(GPa)
A
310
340
0.23
265
210
100
120
0.40
105
150
C
415
550
0.15
500
310
D
700
850
0.14
720
210
E
Fracture before yielding
650
350

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

Step 2

Step 3

Step 4

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 1 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

A 12mm thick steel tire has a width of 110mm and has an internal diameter of 800mm. The tire is heated and shrunk to a steel wheel 800.5 mm in diameter. The modulus of elasticity E = 200 MPa. a. Determine the tensile stress in the tire. b. Determine the compressive pressure between the tire and the wheel. c. Determine the thickness of the tire to resist a pressure of 1.5 MPa if it has an allowable stress of 124 MPa. I need answer ASAP. Thank you!
(d) Define Modulus of Resilience 2. Draw the stress-strain diagram for a typical ductile material and locate the salient points onit 3. A bar of 25 mm diameter is tested in tension. It is observed that when a load of 60 kN is applied, the measured over a gauge length of 200 mm is 0.12 mm and contraction in diameter is 0.0045 mm. Find Poissa 4. A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile force 40 kN. The modulus of e steel may be taken as 200 kN/mm?. Find the elongation and strainin the bar due to applied load. 5. Define (a) Resilience (b) Proof resilience PART -B structions: Part B consists of 3 Units. Answer any one full question from each unit. Each questioncarries 8 marks and may have su 1. A steel circular bar has three segments as shown in the fig. Determine the total elongation ofthe bar. Take E = 210 GPa. 300 kNI 330 kN 30 mm 50KN 15 mm 20 mm 80 kN A + 150 mm 200 mm 250 mm B. D (OR)
The principal plane stresses and associated strains in a 35 ksi, 02 = 15 ksi, plane at a point are 01 1 €1 = 1.02(10-3), 2 = 0.180(10-³). ▼ Determine the modulus of elasticity. Express your answer using three significant figures and include the appropriate units. E= Submit Part B V= μA Value Request Answer Submit Determine the Poisson's ratio. Express your answer using three significant figures. ΠΑΠΙ ΑΣΦ | Η VE Units Request Answer ? vec POSSIA space ?
Please fast
The data shown in the accompanying table are From a tensile test of high-strength steel. The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.5-3). At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (the slope of the initial part of thestress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area. TENSILE-TEST DATA FOR PROB. L.5-7 Laid (lb) Elongation (in) 1000 0.0002 2000 0.0006 6000 0.0019 10,000 0.0033 12,000 0.0039 12,900 0.0041 13,400 0.0047 13,600 0.0054 13,800 0.0063 14,000 0.0090 14,400 0.0102 15,200 0.0130 16,800 0.0230 18,400 0.0336 20,000 0.0507 22,400 0.1108 22,600 Fracture
A stress-strain diagram for a tension test of an alloy steel sample is shown in the diagram below. The initial sample diameter is 0.502”, the diameter of the fractured sample is 0.412”, the original gage length is 2.00” and the final length after fracture is 2.78”. Determine/ calculate the following: (a) The stress at the proportional limit. (b) The modulus of elasticity (E). (c) The yield stress using the 0.2% offset method. (d) Theultimatetensilestrength. (e) The rupture stress. (f) The percent reduction in area. (g) The percent elongation. (h) Is the material considered ductile or brittle?
a)Draw the stress-strain curve of a ductile material such as that of a ductile steel and discuss the various regions of the curve. Explain the difference between a ductile and brittle material and the concept of ductility and its measurement. b)A machine part 10 mm thick, having the dimensions shown in the figure, is to be subjected to cyclic loading. If the maximum stress is limited to 60 MPa, determine the allowable force P. Approximate the stress concentration factors form the graph given. Where might a potential fracture occur?
A tensile stress is applied along the long axis of a cylindrical brass rod thathas a diameter of 10 mm. Given the Poisson’s ratio for brass is 0.35,determine the magnitude of the load required to produce a 2.5 x 10-3 mmchange in diameter if the deformation is entirely elastic. Take the Young’sModulus (E) for brass as 10.1x104 MPa.
Please can you help me with the attached question?
3. A mass of 102kg falls 12mm directly onto the end of a bar 25 mm in diameter and 1.5m long. If E for the material is 207 GPa, find the instantaneous stress produced.
please includde scetch
I need the answer quickly