Concept explainers
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 gage length of 2 in. and a diameter of 0,5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Calculate the shear modulus Gal for the aluminum.
Probs. R8-1/2
Find the shear modulus for an aluminum alloy
Answer to Problem 1RP
The shear modulus for an aluminum alloy is
Explanation of Solution
Given information:
Gage length is 2 in..
The diameter of the specimen is 0.5 in..
The axial load acts on the specimen is 9 kips..
The new diameter of the specimen is 0.49935 in.
Calculation:
Calculate the modulus of elasticity for aluminum
Here, the stress is
Refer the stress-strain diagram.
The value of stress is 70 ksi and the value of strain is
Substitute 70 ksi for
The expression to find the cross-sectional area of the specimen
Here, the diameter of the specimen is d.
Substitute 0.5 in. for d.
Find the value of stress when the specimen is loaded with a 9 kip load using the relation:
Here, the load is P.
Substitute 9 kip for P and
The expression to find the strain in the longitudinal or axial direction
Here, the Young’s modulus of the aluminum is E.
Substitute 45.84 ksi for
Find the strain in lateral direction
Here, the new diameter is
Substitute 0.49935 in. for
Find the Poisson’s ratio
Substitute
Calculate the modulus of rigidity for the specimen
Substitute 11,400.65 ksi for
Therefore, the shear modulus for an aluminum alloy is
Want to see more full solutions like this?
Chapter 8 Solutions
Statics and Mechanics of Materials (5th Edition)
- An element of material in plain strain is subjected to strains x = 0.0015, , y . = -0.0002, and xy = 0.0003. (a) Determine the strains for an element oriented at an angle = 20°. (b) Determine the principal strains of the element. Confirm the solution using Mohr’s circle for plane strain.arrow_forwardThe stress-strain diagram for a steel alloy having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support. o (ksi) 80 70 60 50 40 30 20 10 e (in./in.) 0 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0 .0.0005 0.0010.0015 0.002 0.0025 0.0030.0035arrow_forwardThe elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. If a load of P=60 kN is applied to the specimen, determine its new diameter and length. Take v = 0.35. o (MPa) 490 e (mm/mm) 0.007arrow_forward
- 8-21. The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. When the applied load on the specimen is 50 kN, the diameter is 12.67494 mm. Determine Poisson's ratio for the material. 8-22. The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. If a load of P = 60 kN is applied to the specimen, determine its new diameter and length. Take v = 0.35. σ (MPa) 490 0.007 Probs. 8-21/22 € (mm/mm)arrow_forward= 2-Consider a 45° off-axis tensile test coupon. Three strain gages attached as shown below are reading el 0.00647, 2= -0.00324, and 3 = 0.008095 at stress level of ox=100 MPa. Determine the off-axis modulus of elasticity Ex the off-axis major Poisson's ratio vxy and coefficient of mutual influence of the second kind nxy,x 2 √²-01-² y 45° €2 0₂arrow_forwardThe stress-strain diagram for an aluminum alloy specimen having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 60 ksi, determine the approximate amount of elastic recovery and the increase in the gage length after it is unloaded.arrow_forward
- The steel strip has a uniform thickness of 50 mm. Compute the elongation of the strip caused by the 500-kN axial force. The modulus of elasticity of steel is 200 GPa. 1000 mm- 1000 mm 500 kN 500 kN 50 mm -120 mm 50 mm Match each item to a choice: Choices: # 1.345 mm # 1.356 mm # 0.6725 mm #0.678 mm 1.251 mm # 0.6253 mmarrow_forward8-22. The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. If a load of P - 60 kN is applied to the specimen, determine its new diameter and length. Take v-0.35. a (MPa) 490 (mm/mm) 0.007arrow_forwardThe 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 gage length of 2 in. and a diameter of 0.5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Calculate the shear modulus Gal for the aluminum.arrow_forward
- A thin polymer plate PQR is deformed so that corner Q is displaced downward by a distance L = 0.50 mm to new position Q' as shown. Determine the shear strain at Q' associated with the two edges (PQ and QR). 120 mm 480 mm R 240 mm -891 urad -1160 μrad -811 urad -1248 urad -1436 prad Xarrow_forwardThe elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. If a load of P = 60 kN is applied to the specimen, determine its new diameter and length. Taken = 0.35.arrow_forwardThe stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 70 ksi, determine the approximate amount of elasticrecovery and the increase in the gage length after it is unloaded.arrow_forward
- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning