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 isGal = 3.8(103) ksi. 2. The elastic portion of the tension stress-straindiagram for an aluminum alloy is shown in thefigure. The specimen used for the test has a gaugelength of 2 in. and a diameter of 0.5 in. If the appliedload is 10 kip, determine the new diameter of thespecimen. The shear modulus isGal = 3.8(103) ksi.o (ksi)70e (in./in.)0.00614

Answered: Image /qna-images/answer/8ae802ec-7ccd-469d-a07f-b9d14f835cd8.jpg

2. 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. o (ksi) 70 e (in./in.) 0.00614

2. 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. o (ksi) 70 e (in./in.) 0.00614

Materials Science And Engineering Properties

1st Edition

ISBN:9781111988609

Author:Charles Gilmore

Publisher:Charles Gilmore

Chapter6: Introduction To Mechanical Properties

Section: Chapter Questions

Problem 6.2P

See similar textbooks

Related questions

Q: An aluminum alloy [E = 73 GPa; v = 0.33; a= 23.0 x 10-6/°C] plate is subjected to a tensile load P.…

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: Problem 11. [Concepts: Calculating stress from internal forces, nonlinear stress-strain response,…

A: a) Determine the yield strength assuming a 0.2% offset.b) If a bar of this material has a length of…

Q: Determine the following: a. Young's modulus within the linear portion b. Tangent modulus at a stress…

A: a).Young's Modulus : At the proportional limit, stress is directly proportional to strain stress α…

Q: Problem 4 The test data of the coupon tests of stainless steel alloys are given for three test…

A: Given, Stress strain data for three specimens To find : Elastic modulus, Yield stress and yield…

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

A: Given data Allowable tensile bendig stress = 18 MPa Allowable compressive bending stress = 9 MPa

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: Data taken from a stress-strain test for a brittle alloy are given in the table. The curve is linear…

A: , ksi, in/in0034.30.000843.50.001347.00.001749.00.002149.50.0023The rupture stress,

Q: Q.3 The following data were obtained from the tensile test of Aluminum alloy. The initial diameter…

A: Given data; Gauge length = 2 in Initial diameter = 0.505 in

Q: A steel alloy specimen having a rectangular cross section of dimensions 19.1 mm x 3.1 mm (0.7520 in.…

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

Q: An aluminum alloy rod has a circular cross section with a diameter of 7mm. The rod s subjected to a…

Q: 1. A cylindrical steel specimen, with an initial diameter of 7mm, has elongated by 40% at the point…

A: Given, Di = 7 mm EL = 40% P = 35 kN

Q: An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 580 mm.…

Q: Please answer d,e,f

Q: During a tension test, measurements for the applied load and the corresponding elongation are taken…

A: Shear stress is the force that tend to cause deformation of the material by sliding along the plane.…

Concept explainers

Material Properties

Construction in civil engineering demands high strength, durability, and longevity. These targets can be achieved by better design and analysis, along with the correct choice of materials. Different materials have different properties and hence, they respond and performs differently under the various condition of external loadings. It is therefore a critical task of an engineer to select the right material based on the property that the material is supposed to function in real-time. When speaking about the property, there are different categories of material properties like chemical property, physical property, mechanical property, electrical property, magnetic property, thermal property, etc. These material properties form the basis of material selection in various applications.

Question

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.

2. 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.
o (ksi)
70
e (in./in.)
0.00614

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

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

The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen with a circular cross section. The diameter was 3 8 inch and the gage length (The length over which the elongation is measured) was 2 inches. a. Use the data in Table 1.5.2 to produce a table of stress and strain values. b. Plot the stress-strain data and draw a best-fit curve. c. Compute the, modulus of elasticity from the initial slope of the curve. d. Estimate the yield stress.
Compare the engineering and true secant elastic moduli for the natural rubber in Example Problem 6.2 at an engineering strain of 6.0. Assume that the deformation is all elastic.
Urgent. a = 10 , b = 7 solve this question about those information.
a13 both
The specimen shown is made from a 25 mm diameter cylindrical steel rod with two 38 mm outer-diameter sleeves bonded to the rod as shown. Knowing that E = 200 GPa, determine (a) the load P so that the total deformation is 0.05 mm, (b) the corresponding deformation of the central portion BC. 38 mm diameter P' 25 mm diameter B 38 mm diameter C 50 mm 75 mm 50 mm
QL1
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)
I5 In a tensile test experiment of carbon steel. a standard specimen (D= 0.505 in. Gauge length=2.0 in & total length 8 in) had a 0.2% offset yield strength of 80.000 psi and engineering strain of 0.010 at the yield strength point Calculate: The load (force) at this point. The instantaneous area of the specimen at this point. c. The true stress at this point. d. The true strain at this point. e. The total length of the specimen. if the load is released at this point.
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= 6/8 v = -E(lateral)/(axial)
STRAIN
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.
ANSWER A PLS