1.17 Figure P1.17 shows the stress-strain relations of metals A and B during tensiontests until fracture. Determine the following for the two metals (show all calcu-lations and units):a. Proportional limitb. Yield stress at an offset strain of 0.002 in./in.150 -- Metal A100• Metal B500.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14Strain, in./in.FIGURE P1.17c. Ultimate strengthd. Modulus of resiliencee. Toughnessf. Which metal is more ductile? Why?Stress, ksi

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

Answered: 1.17 Figure P1.17 shows the…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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At a temperature of 27.89 °C a 0.41-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 126.64°C, determine the magnitude of the normal stress (in MPa) in the steel rod if L = 320.36 mm and M = 231 mm. Round off the final answer to four decimal places. d ... M A B Aluminum Stainless steel A = 2000 mm² E = 75 GPa a = 23 × 10-6/C A = 800 mm² E = 190 GPa %3D a = 17.3 × 10-6/°C
An element in the plane stress is having the larger principal stress (tension) as 30 MPa and it is subjected to stresses, Ox = 50 MPa, 40 MPa then what will be and Txy the value of the stress oy? (in MPa) =
At a temperature of 28.42 °C a 0.4-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 137.38°C, determine the magnitude of the normal stress (in MPa) in the steel rod if L = 339.61 mm and M=240 mm. Round off the final answer to four decimal places. d ... В Aluminum Stainless steel A = 2000 mm² E = 75 GPa a = 23 × 10-6°C A = 800 mm² E = 190 GPa a = 17.3 × 10-6°C
1 It is the ratio of ultimate tensile stress to the permissible stress. Plastic Shear Plastic limit Proportional limit Factor of safety Poisson's ratio
o (MPa) 750 B -F 600 A parallel Ty = 450 300 150 D € (mm/mm) 0.01 | 0.02) 0.03 0.04 Ey = 0.006 - EOC- O7 0.023 The stress-strain diagram for an aluminum alloy that is used for making aircraft parts is shown in Fig. 3-19. If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, find the modulus of resilience both before and after the load application.
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.
An aluminium specimen with an initial gauge diameter d, = 10 mm and gauge length, 1, = 100 mm is %3D subjected to tension test. A tensile force P= 50 kN is applied at the ends of the specimen as shown, resulting in an elongation of 1 mm in gauge length. The Poisson's ratio (µ) of the specimen is Take shear modulus of material, G = 25 GPa. Consider engineering stress-strain conditions. P
At a point on the surface of a plane stressed region, measured strains reduced to the following: E=+900µ, ɛ,=-100µ, and yy=+400µ. The material on which strains were measured has the following properties: E=210 GPa and v=0.3. Determine the stresses, Ox, Oy, and tyy at the point. Note: u means micro radian or lx10-6 rad. Use biaxial stresses formulas and use txy=Gyxy:
Knowing that a 0.02in gap exists when the temperature is 75°F, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -11ksi, (b) the corresponding exact length (2 decimal places in 103 inches) of the aluminum bar. 0.02 in. 14 in. Bronze A = 2.4 in² E = 15 × 106 psi a = 12 x 10-6/°F 18 in. Aluminum A = 2.8 in² E = 10.6 × 106 psi a = 12.9 x 10-6/°F
Data taken from a stress-strain test for a brittle alloy are given in the table. The curve is linear between the origin and the first point. o (ksi) e (in./in.) 0 0 44.8 0.0008 54.0 0.0013 57.5 0.0017 59.5 0.0021 60.0 0.0023 Part A Determine approximately the modulus of toughness. The rupture stress is o, 30.0 ksi. Express your answer in inch-pounds per cubic inch as an integer. VAE Ivec U Submit Provide Feedback Request Answer in-lb/in³
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
2.27. The stress components at a given point in an engineering component are es- timated to be oxy=4.1 ksi, ơyy=0 ksi, ơz =0.9 ksi, ory =-3.1ksi, ơyz = 1.2 ksi, Ozx =0 ksi. Estimate the factor of safety against yielding using von Mises' theory if the uniaxial yield stress of the material is Sy =17.2 ksi.

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