len A material has the stress-strain behavior shown in the Figure below. Calculate the toughness (T) of this material. given that S1=195 Mpa; S2=243.75 Mpa; El=0.19; E2=0.38 S2- S1 E1 Strain E2 Stress- Strain diagram of a metal alloy Stress (Mpa)

len A material has the stress-strain behavior shown in the Figure below. Calculate the toughness (T) of this material. given that S1=195 Mpa; S2=243.75 Mpa; El=0.19; E2=0.38 S2- S1 E1 Strain E2 Stress- Strain diagram of a metal alloy Stress (Mpa)

Materials Science And Engineering Properties

1st Edition

ISBN:9781111988609

Author:Charles Gilmore

Publisher:Charles Gilmore

Chapter6: Introduction To Mechanical Properties

Section: Chapter Questions

Problem 15ETSQ

See similar textbooks

Related questions

Q: An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 860 mm.…

Q: An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 780 mm.…

A: Allowable tension bending stress = 16 MPa Allowable compression bending stress = 10 MPa

Q: 400 200 Strain 0.05 0.10 0.15 0.20 From your graph, plotting the yield stress against 1/vd, you…

Q: Determine the tensile yield strength (0.2% offset) and the maximum strength of a metal alloy having…

A: Tensile yield strength :-yield strength of the material is the stress at a point at which material…

Q: An aluminum alloy cylinder with a diameter of 76 mm and a height of 150 mm. is subjected to a…

Q: he stress-strain relation shown in figure below was obtained during the tensile test of an aluminum…

Q: B. C E Strain FIGURE Q1 (a) Stress -

Q: Question 3 o(MPa) The elastic portion of the stress-strain diagram for a steel alloy is shown in the…

A: GIVEN DATA:- ORIGINAL DIAMETER=15MM ORIGINAL LENGTH=60MM INITIAL APPLIED LOAD=20KN FINAL DIAMETER…

Q: The bending stresses on an element are: om 25,00 psi or= 21,667 psi. There is a stress raiser with…

A: Given:σm = 2500 psiσr = 21667 psiSy = 80000psiSu = 90000 psiSe = 40000 psiKf = 1.4

Q: An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 840 mm.…

Q: A circular bar of magnesium alloy is 750mm long. The stress-strain diagram for the material is shown…

A: Given: The length of the magnesium alloy is 750 mm. The bar is loaded to an elongation of 6.0 mm.

Q: Given that a steel cyInder of dimenslons 12.0 mm in dlameter and 70 mm bng is subjected to tensicn.…

A: a) According to Hook's law within the elastic limit, stress is directly proportional to strain.…

Q: The poisson ratio of an elastic material is 0.3, then the ratio of its young's modulus and bulk…

Q: 60,000 40,000 20,000 0.002 0.004 0.006 0.008 Strain, in./in. Stress, psi

A: The given stress-strain graph of aluminum alloy is-

Q: An elastoplastic material with strain hardening has the stress–strain relationship shown in Figure .…

A: a) In order to find strain. first, calculate E: Substitute the value in the formula:

Q: 1.17 Figure P1.17 shows the stress-strain relations of metals A and B during tension tests until…

A: Any pressure of a solid object creates a stretched internal stress, similar to the spring-turning…

Q: A material obeys hook's law up to a. Elastic limit O b. Plastic limit O c. Limit of proportionality…

A: Following answer is in step 2

Q: An aluminum alloy rod has a circular cross section with a diameter of 8 mm. This rod is subjected to…

A: It is required to calculate the stress in the circular rod. Here, A0 is the original area of the…

Q: In the figure shown below, determine: 1) The final temperature if the normal stress at aluminium is…

Q: An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 680 mm.…

A: FIND NA :y=36×95(952)-22×22(95-7-222)36×95-22×22y=22.637mmFIND MOI…

Q: A) A specimen of alloy having the following engineering stresses produces the corresponding plastic…

A: Given :- To find :- Engineering stress corresponding to engineering strain of 0.4

Q: STRENGTH OF MATERIALS. Please show the complete solutions. Rate will be given. Answer in 3 decimal…

Q: An aluminum alloy bar with a radius of 7 mm was subjected to tension until fracture and produced…

A: Since we answer Three subpart at a time so we will answer the first three. Please resubmit another…

Q: The deformation per unit of the original length of a material is called as a. Stress O b. Strain O…

Q: The failure stress of a homogeneous steel bar having a diameter of 10 mm is 250 MPa. Using a factor…

Q: For a material with poisons ratio 0.25, the ratio of modulus of rigidity to modulus of elasticity…

A: Given, Poisson's ratio, v = 0.25 Find, The ratio of modulus of rigidity to the modulus of elasticity…

Q: Question No.2 Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion…

A: It is that point on stress-strain curve, till which stress is linearly varying with strain. As per…

Q: The stresses shown in the figure act at a point on the free surface of a machine component. If o= 45…

Q: The strength-to-weight ratio of a structuralmaterial is defined as its load-carrying capacitydivided…

A: Given:- RS/W =σ/γ To find:- strength to weight ratio

Q: Stress-Strain Behavior 6.3 A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm…

A: Given: Tension force, F= 44,500 N Cross-section=15.2 mm×19.1 mm

Q: 2. The elastic portion of the tension stress-strain diagram for an aluminum alloy is shown in the…

Q: Q-5 Consider an S-590 alloy component as shown in following figure that is subjected to a stress of…

A: Solution: Given data: stress=200 Mpa

Q: An elastoplastic material with strain hardening has the stress-strain relation shown in the figure.…

A: Permanent strain is calculated as: εpermanent=εtotal-εelastic--------------------------------(1)…

Q: An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 500 mm.…

A: Determine the largest moment M that can be applied as shown to the beam.

Q: A copper rod is deformed using a uniaxial tensile force of 16000 N. Deformation continues until…

A: P=16000N Diameter (d) =0.01m n=0.54

Q: 2. Shown below are pieces of unsaid material. They are combined and locked in each other and a force…

Q: Piano strings are made of a fibre which is pulled taut. When the string is tapped in the centre by a…

A: Given, e = 0.2% σ = 100 MPa

Q: Ox =10 MPa (compressive) Figure below shows the state of stress at a critical point on a component…

A: Stress is the internal resistance force that develops in the body due to the applied force that…

Q: The stress-strain diagram for a metal alloy having an original diameter of 16 mm and a gauge length…

A: In this question , We calculate approximately Elastic modulus , yielding load and ultimate load…

Q: Your answer is incorrect. An extruded polymer beam is subjected to a bending moment M. The length of…

Q: An aluminum rod of length 100 mm and diameter 15 mm is subjected to a normal load of 10 kN. The new…

A: Introduction- Shear stress occurs when a force operates parallel to an object's surface. Consider a…

Q: A sample of tin alloy has the following stress-strain curve 400 200 Strain 0.05 0.10 0.15 0.20 From…

Q: A tension test was conducted for a material called MatX and the following stress-strain curve was…

A: Given Data: Stress strain curve Load = 8 KN Diameter = 20mm length = 200mm

Q: Below is the stress-strain curves of four different materials (A, B, C, and D). Make a brief…

A: On the bases of curves following things can b infer Meterial W has highest yield stress hence is the…

Q: Fatigue strength for non- ferrous materials in defined at stress cycles 10^3 10^5 10^7 10^9

Q: The following information about the o-e curve is given for a steel alloy. E = 0.001527 at o = 300…

Q: For a material, stress at proportionality limit is 200 MPa and modulus of elasticity is 200 GPa. The…

A: Stress at proportionality limit = 200 MPa Modulus of elasticity = 200GPa It has been asked in the…

Q: 1.17 Figure P1.17 shows the stress-strain relations of metals A and B during tension tests until…

A: a) Proportional limit : Stress is proportional to strain in this limit. The gradient will be equal…

Question

Time len
A material has the stress-strain behavior shown in the Figure below. Calculate the toughness (T) of
this material. given that S1=195 Mpa; S2=243.75 Mpa; El=0.19; E2=0.38
E1
Strain
E2
Stress - Strain diagram of a metal alloy
OA.T= 18.52 Mpa
O B.T= 92.62 Mpa
O C.T= 37.05 Mpa
O D.T= 60.21 Mpa
Stress (Mpa) 3

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

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

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

Similar questions

1.17 Figure P1.17 shows the stress-strain relations of metals A and B during tension tests until fracture. Determine the following for the two metals (show all calcu- lations and units): a. Proportional limit b. Yield stress at an offset strain of 0.002 in./in. 150 - - Metal A 100 • Metal B 50 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 Strain, in./in. FIGURE P1.17 c. Ultimate strength d. Modulus of resilience e. Toughness f. Which metal is more ductile? Why? Stress, ksi
S Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion tests until fracture. Determine the following for the two metals (show all calculations and units): a. Proportional limit b. Yield stress at an offset strain of 0.002 m/m. c. Ultimate strength d. Modulus of resilience e. Toughness f. Which metal is more ductile? Why? 900 Metal A 600 Metal B 300 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 Strain, m/m FIGURE P1.16 Stress, MPa
Strain = 600 Stress = Strain = 500 Stress = 400 500 300 400 300 200 200 100 100 0.000 0.002 0.004 0.006 Strain 0.00 0.04 0.08 0.12 0.16 0.20 Strain Stress (MPa)
Fatigue strength for non- ferrous materials in defined at stress cycles 10^3 10^5 10^7 10^9
Problem 1 An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown. Axial loads are applied at the positions indicated. Aluminum Steel A=800mm2 A=1000mm2 Bronze A=700mm? 4P 2P 500mm 600mm 700mm 1. What is the maximum value of P that will not exceed the axial stress bronze of 105 MPa? 2. What is the maximum value of P that will not exceed the axial stress in aluminum of 90 MPa? 3. If P=10KN, what is the axial force to be carried by the aluminum in KN? 4. If P is 5KN, what is the axial stress of steel?
In the figure shown below, determine: 1) The final temperature if the normal stress at aluminium is Og = -90 MPa and the initial temperature %3D 20°C. 2) The final length of the aluminium member. Aluminum A=1800mm? Bronze A=1500mm? E=73GPA E=105GPA a=23.2x10-6/°C a=21.6x10-6/°C Gap=0.5mm 0.35m 0.45m
A steel 0.6 inch×1.2 inch steel 90 m long is subjected to a 45 KN tensile load along its lenght.If poison's ratio is 0.3 Find: A. The deformation along its lenght. B. The deformation along its thickness. C. The defirmation along uts width. D. The lateral strain.
30. An aluminum alloy rod has a circular cross section with a diameter of 8mm. The rod is subjected to a tensile load of 5kN. Assume that the material is in the elastic region and E=69 GPa If Poisson's Ratio is 0.33, what will be the lateral strain? E=68 v=-e(lateral) e(axial)
The elastic portion of the tension stress-strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is G al =3.811032 ksi
The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is Gal = 3.8(103) ksi.