Concept explainers
Whether the elastic modulus of material decreases or increases with increase in temperature.
Answer to Problem 1CQ
The elastic modulus of material decreases with increase in temperature
Explanation of Solution
Elastic modulus can be defined as the ratio of applied stress to the obtained strain in the body up to proportional limit. Elastic modulus can also be defined as the slope of curve for applied stress versus strain in specimen up to the elastic limit of material.
Elastic modulus is a
When the temperature increases, the distance between the atoms in the material increases and they move farther from each other. Increasing temperature reduces the strength of material and its elastic modulus.
Conclusion:
Thus, the elastic modulus of material decreases with increase in temperature
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Chapter 9 Solutions
Materials Science And Engineering Properties
- 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)arrow_forwardIf it is constrained between two supports A and B and is stress-free at 20 ℃, what would be the stress in the two materials when it is heated to 70 ℃, For Steel: Es = 210 GPa, Coefficient of expansion = 12x10^6/℃. For Brass: Eb = 105 GPa, Coefficient of expansion = 19x10^-6/℃arrow_forwardThe stress in an elastic material is: A Inversely proportional to the force acting B Inversely proportional to the materials yield strength C Proportional to the displacement D Inversely proportional to the strainarrow_forward
- At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 12.5 x 10-6/°F] bar with a width of 3.0 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 9.6 x 10-6/°F] bar with a width of 2.0 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine (a) the lowest temperature at which the two bars contact each other. (b) the normal stress in the two bars at a temperature of 250°F. (c) the normal strain in the two bars at 250°F. (d) the change in width of the aluminum bar at a temperature of 250°F. (1) 3.0 in. 32 in. 2.0 in. B ↓ (2) 44 in. 0.04-in. gap Determine the lowest temperature, Tcontact, at which the two bars contact each other.arrow_forwardAt a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; α=α=12.5 x 10-6/°F] bar with a width of 2.5 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10-6/°F] bar with a width of 1.7 in. and a thickness of 0.75 in. The supports at A and C are rigid. Assume h1=2.5 in., h2=1.7 in., L1=31 in., L2=46 in., and Δ=Δ= 0.04 in. (A) Determine the lowest temperature, Tcontact, at which the two bars contact each other. (B) Find a geometry-of-deformation relationship for the case in which the gap is closed. Express this relationship by entering the sum δ1+δ2, where δ1 is the axial deflection of Bar (1), and δ2 is the axial deflection of Bar (2). δ1+δ2= _____in. (C) Find the force in the Bar (1), F1, and the force in Bar (2), F2, at a temperature of 225oF. By convention, a tension force is positive and a compression force is negative. IN KIPS (D) Find σ1 and σ2,…arrow_forwardAt a temperature of 60°F, a 0.02-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; α=α=12.5 x 10-6/°F] bar with a width of 2.8 in. and a thickness of 0.85 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10-6/°F] bar with a width of 1.6 in. and a thickness of 0.85 in. The supports at A and C are rigid. Assume h1=2.8 in., h2=1.6 in., L1=26 in., L2=40 in., and Δ=Δ= 0.02 in. Determine(a) the lowest temperature at which the two bars contact each other.(b) the normal stress in the two bars at a temperature of 225°F.(c) the normal strain in the two bars at 225°F.(d) the change in width of the aluminum bar at a temperature of 225°F.arrow_forward
- a=2 b=5 3xb=15x mm 12xb=60x mm 10xa kN = 20x kN ((3xb)-0,00735)) = (15x- 0,00735) mmarrow_forwardAn 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 mmarrow_forwardThe 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.arrow_forward
- 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 0.45marrow_forwardExplain in detail the effect of Temperature on the Elastic Modulus and Yield Strength ofEngineering materials.arrow_forwardAt a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 14.4 x 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 9.6 × 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and Care rigid. Determine the lowest temperature at which the two bars contact each other. (1) 3 in. 32 in. 105.3°F 75.3°F O 147.3°F 86.6°F 113.4°F B ↑ 2 in. ↓ (2) 44 in. 0.04-in. gaparrow_forward
- Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning