2- What is the largest size (mm) internal through crack that a thick plate of aluminium alloy 7075-T651 can support at an applied stress of (a) three-quarters of the yield strength and (b) one-half of the yield strength? Assume Y = 1. for 7075-T651, KỊC = 24.2 MPa ym and oys = 495 MPa. %3D %3D

2- What is the largest size (mm) internal through crack that a thick plate of aluminium alloy 7075-T651 can support at an applied stress of (a) three-quarters of the yield strength and (b) one-half of the yield strength? Assume Y = 1. for 7075-T651, KỊC = 24.2 MPa ym and oys = 495 MPa. %3D %3D

Materials Science And Engineering Properties

1st Edition

ISBN:9781111988609

Author:Charles Gilmore

Publisher:Charles Gilmore

Chapter6: Introduction To Mechanical Properties

Section: Chapter Questions

Problem 6.4P

See similar textbooks

Related questions

Q: A single crystal of a metal is oriented for a tensile test such that its slip plane normal makes an…

A: Given tensile stress = 1.3 MPa We are asked to compute the critical resolved shear stress for the…

Q: A kinked edge crack has been detected in a large plate which is to be loaded by a remote tensile…

A: I have solved your problem

Q: The limit load of an elastic-plastic cracked three-point bend beam (Figure QA1) with S = 4W = 120…

A: Yield strength(fy)=300 Mpa Span length (S)=120 mm W=30 mm

Q: For a specimen of a steel alloy with a plane strain fracture toughness of 80 MPa√m, fracture results…

A: Given KIC=80MPamσ=510MPaa=6mm Consider the formula KIC=Yσπa80=Y×5103.14×6Y=0.036

Q: What are the consequences of strain hardening on ductility? Discuss the influence of…

A: Consequences of strain hardening on ductility: Strain hardening is the process of making the metal…

Q: A joint between two glass plates A and B is filled with a flexible epoxy that bonds securely to the…

Q: A steel plate is bent into a circular arc of radius 10 m. If the plate section be 120 mm wide and 20…

A: Given data: Radius, R = 10 m E=2×105 Nmm2 Area, A=120×10 mm2 y=10

Q: A single crystal of BCC iron is subjected to tensile stress of 100 MPa along the [001] direction.…

Q: In a laboratory test of a beam loaded by end couples, the fiber at layer AB as shown are found to…

A: Appreciate if there is detailed work with explanation and finds you helpful for ease of…

Q: Problem 4: A tensile test is carried out on a bar of a mild steel of diameter 2 cm. the bar yields…

A: Question is based on the tensile test which is performed on mild steel bar .Diameter of mild steel…

Q: When the maximum and minimum stresses of a fluctuating load are increased by 50%, the fatigue life…

A: Fatigue life is calculated using number of repeated cycles of loading to failure and applied stress…

Q: friction is zero. Calculate_ on crack earth pressure intensity at ba

Q: The support in Figure 14.21a is to be constructed from a 4340 steel plate tempered at 800◦F. The…

A: Given- σy=228KsiKIC=51kSIinW=4in

Q: 6) When iron is compressed hydrostatically (equal stresses in all directions) with 6 x 10? N/mm²,…

Q: Determine: a. The active force after the tensile crack occurs. (kN/m)

Q: A hot-rolled steel has a yield strength of Sy = Syc = 100 kpsi and a true strain at fracture of ɛf =…

A: Estimate the factor of safety for the following principal stresses

Q: As is shown in the Fig. 2, prove that there is no stress at all at the apex A (or a re-entrant,…

A: Resolving the forces N1 and N2 in the x and y direction - In x-direction - N1Cosα1andN2Cosα2 In…

Q: Compressive centric forces of 160 kN are applied at both ends of the assembly shown by means of…

Q: The ideal diameter of a steel tensile test specimen is O 1in O 2 in O 1/2 in • 1/4 in Question 2 How…

A: @1.The ideal diameter of a steel tensile test specimen is 1/2 in or 12.5mm Diameter doesn't effect…

Q: 4- Tensile strength of a material is obtained by dividing the maximum load during the test by the a)…

A: Given data in question Tensile strength To be discussed To get the tensile strength, load has to…

Q: A structural component in the shape of a flat plate 20.8 mm thick is to be made from a metal with…

A: Given, Thickness of the flat plate = 20.8 mm Yield strength =σy=533 MPa Design stress=…

Q: A cylindrical specimen 7.5 mm in diameter of an S-590 alloy (Animated Figure) is to be exposed to a…

Q: A very large, steel plate of yield stress 200 MPa and Poisson's ratio 0.3 has a crack at the centre…

A: To determine the extent of plastic region at Crack tip : For plastic region we have,…

Q: 5- Tensile Failure Vertical creates a vertical fracture parallel with the OH 6- In Undrained…

A: In soil failure tests are done and observations are taken.

Q: 3. Calculate the cracking moment Mer, and the ultimate moment Mu = ÞMn for the following cross…

A: We have to determine the value of checking moment Mcr and the ultimate moment Mu =phi × Mn for the…

Q: A very large, steel plate of yield stress 200 MPa has a crack at the centre of length 18 mm,…

A: Solution:-The given values are as follows,Yield stress σys=200MPaγ=0.3Centre…

Q: ompare the effect of presence of a notch on ductile and brittle materials in terms of fracture…

A: The separation of atomic or molecule connections causes the structural fracture of a material. As…

Q: B. Determine the cracking moment of the section, in kN-m. 350 mm 390 mm 75 mm

Q: State the difference between cleavage and f

A: Solution: Difference between cleavage and fracture of minerals, using quartz and mica as examples.…

Q: (b) An aluminium plate (100 mm width x 300 mm height x 4 mm thick) with a centre-crack is subjected…

A: Given values: Aluminium plate size: 100 mm * 300 mm * 4 mm Service load = 40 KN Fracture toughness…

Q: A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450…

A: Concept: The internal crack occurs when the overheating is very fast because of surface unequal…

Q: :A member whose material properties remain unchanged under rotation of 90° about axes (x, y, z) is…

A: The expression of three principal strains is given as εxx=σxxC1-C2σyyC1-C2σzzC1 ...(1)…

Q: 4- Determine the critical crack length (mm) for a through crack in a thick 2042-T6 alloy plate that…

A: So using formulakIC=σπaYkIC=fracture toughnessMpamσ=stressmpaa=crack length=mY==Factor=1.2

Q: A steel specimen is tested in tension. The specimen is 25 mm wide by 12.5 mm thick in the test…

A: Area=25×12.5mm2=312.5mm2

Q: The R-curve for a certain material can be expressed as: K R =K + 3(a – a,)º E with R in MN/m, a and…

Q: u, When iron is compressed hydrostatically (equal stresses in all directions) with 6 x 102 N/mm',…

A: The given data is shown below: Hydrostatic compressive stress=6×102 N/mm2Change in…

Q: Which of the following is not a factor contributing to fatigue in a steel structure? oints) a)…

A: Given, Our aim is to select a option from given 4 option. The question is : Which of the…

Q: A ceramic part is used under a complete reverse cyclic stress with a stress amplitude (S) of 250…

A: Given data:

Q: A high-strength steel has a yield strength of 1380 MPa and fracture toughness of 91 MPAVM. A surface…

Q: A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa is exposed to a…

A: The formula to calculate the critical stress is given by, σc=KICγπa Here, KIC is the plane strain…

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

Q: SITUATION. Given a trapezoidal beam with width at the top of 300mm and width at the bottom of 500mm…

A: To Determine Cracking moment in kN-m

Q: . Calculate the strain at the centroid of the tension steel in single layer if the effective depth…

Q: Q.10 When a mild-steel specimen fails in a torsion test fracture looks like (a) (b) (c) (d)

A: Given, Mild steel specimen fails in torsion test fracture looks like

Q: Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs…

A: Given that: Crack length=0.31 mm Radius of curvature=0.003 mm Stress=1370 MPa It is asked to…

Q: An aluminum bar having a cross-sectional area of 160 square millimeters carries the axial loads at…

Q: A wing panel of a supersonic aircraft is made from a titanium alloy that has yield strength 1035 MPa…

Concept explainers

Crack Detection

Concrete materials are meant for supporting structures, building foundations and they act as several load-bearing structures. Over time, due to environmental conditions, and based on the constituent materials used, concrete materials are subjected to a great degree of distress. Corrosion of the reinforcement bars due to chemical influences, uneven base, earthquake, drying shrinkages, and alkali-aggregate reactions are some of the reasons for concrete distress.

Question

2- What is the largest size (mm) internal through crack that a thick plate of aluminium alloy
7075-T651 can support at an applied stress of (a) three-quarters of the yield strength and (b)
one-half of the yield strength? Assume Y = 1.
for 7075-T651, KỊC = 24.2 MPa ym and oYS = 495 MPa.

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

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

A component made of Aluminium 6061-T651 has an edge crack with length equal to 15 mm, as shown The yield stress of the material is oy = 280 MPa and its fracture toughness (Kic) ranges from 40 MN/m3/2 to 60 MN/m³/2 Consider now the same edge crack in a semi-infinite plate. Determine the critical stress to avoid fracture. P I I I I ← 15 mm P↓ 100 mm 30 mm
An aluminum alloy [E = 70 GPa; v = 0.33; a = 23.0×10-6/°C] bar is subjected to a tensile load P. The bar has a depth of d = 260 mm, a cross-sectional area of A = 14720 mm2, and a length of L = 5.5 m. The initial longitudinal normal strain in the bar is zero. After load P is applied and the temperature of the bar has been increased by AT = 46°C, the longitudinal normal strain is found to be 1680 µɛ. % D Calculate the change in bar depth d after the load P has been applied and the temperature has been increased. L P Answer: Ad = i mm
A structural component in the shape of a flat plate 24.3 mm thick is to be fabricated from a metal alloy for which the yield strength and plane strain fracture toughness values are 533 MPa and 22.0 MPa-m1/2, respectively. For this particular geometry, the value of Y is 1.3. Assuming a design stress of 0.4 times the yield strength, calculate the critical length of a surface flaw. What formulas do i use ? And how do i use them?
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.
The following information about the o-e curve is given for a steel alloy. E = 0.001527 at o = 300 MPa and ɛ = 0.003054 for o = 600 MPa. (a) Draw the stress-strain diagram and calculate the E for this alloy.
Narrow bars of aluminum are bonded to the two sides of a thick steel plate as shown. Initially, at T₁ = 70°F, all stresses are zero. Knowing that the temperature will be slowly raised to T₂ and then reduced to T₁, determine (a) the highest temperature T₂ that does not result in residual stresses, (b) the temperature T₂ that will result in a residual stress in the aluminum equal to 58 ksi. Assume aa = 12.8 x 10-6/°F for the aluminum and a = 6.5 × 10-6/°F for the steel. Further assume that the aluminum is elastoplastic with E = 10.9 × 106 psi and ay = 58 ksi. (Hint: Neglect the small stresses in the plate.) Fig. P2.121
For a specimen of a steel alloy with a plane strain fracture toughness of 80 MPa√m, fracture results at a stress of 510 MPa when the maximum (or critical) internal crack length is 6 mm. For the same alloy, will fracture occur at a stress level of 380 MPa when the maximum internal crack is 9.0 mm? Why or why not? Select the most appropriate answer based on your calculation. Select one: a. It will not fracture b. Not enough information c. It will fracture
Problem 4: A tensile test is carried out on a bar of a mild steel of diameter 2 cm. the bar yields under a load of 150 kN and breaks finally at a load of 70 kN. Estimate; 1-the tensile stress at the yield point 2-the ultimate tensile stress 3-the average stress at the breaking point, if the diameter of the fractured neck is 1 сm.
A cylindrical specimen of cold-worked steel has a Brinell hardness of 240. If the specimen remained cylindrical during deformation and its original radius was 11.8 mm, determine its radius after deformation. For steel, the dependence of tensile strength on percent cold work is shown in Animated Figure 7.19b. i mm
An aluminum alloy [E = 67 GPa; ν = 0.33; α = 23.0 × 10–6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 225 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 4.1 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by ΔT = 63°C, the longitudinal normal strain in the plate is found to be 2900 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Δd.