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An aluminum alloy rod 1.5 m long required a force of 60,000 N to reach its...
1. An aluminum alloy rod 1.5 m long required a force of 60,000 N to reach...
1. An aluminum alloy rod 1.5 m long required a force of 60,000 N to reach its yield strength of 280 MPa. E = 7.0 x 104 MPa. Ts = 490 MPa. a) What is its diameter (in cm)? b) What was the total strain? c) What was its final length after the load was released? d) What is the maximum elastic strain this rod must withstand?
HotLW Problem: Giving the engineering stress - engineering strain carve for an aluminum alloy, Calalate: 400FTTTTTTTTTT...
HotLW Problem: Giving the engineering stress - engineering strain carve for an aluminum alloy, Calalate: 400FTTTTTTTTTT 10 psi 300 MPa 50 $ Stress (MPa) Stress (10 psi) 8 100- 10/ bd = 110m 1oof lo=50 mm 0 0 0.005 0.010 o 0.05 0. 100 . 150 .20 6 percent elongation at fracture (% elongation = ?) 2 yield strength (Tyield = ?) 3 Tensile strength (TS = ?) ♡ True stress at an engineering strain of 0.04. (I=?) 5 True...
Consider a dilute Cu alloy, with a Young's modulus of 100 GPa, a strain hardening exponent...
Consider a dilute Cu alloy, with a Young's modulus of 100 GPa, a strain hardening exponent of n=0.5, strength coefficient of 33 MPa. If the initial length of a rod of Cu is 1 m, and the Cu is pulled in a tensile test to failure, what is the length of the rod at its maximum load? Group of answer choices 1.35 m 1.01 m 2.00 m 1.5 m 1.65 m
A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must...
A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected to a 3.0-kN axial load. Knowing that E= 70 GPa, determine the required diameter of the rod. The required diameter of the rod is mm.
A rod is to be used in an aerospace application. The defense contractor has mandated a...
A rod is to be used in an aerospace application. The defense
contractor has mandated a factor of safety of 2.6. In its
application, the maximum tensile axial load on the rod ranges
from = 2.1 kN to 2.8 kN. The
metal alloy that the rod is made of has a tensile yield strength of
480 MPa and an ultimate strength of 560 MPa. The specification for
the radius of the rod is given on the drawing.
If the rod...
Problem #1 A cylindrical rod of copper (E 110 GPa) having a yield strength of 240...
Problem #1 A cylindrical rod of copper (E 110 GPa) having a yield strength of 240 MPa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must the diameter be to allow an elongation of 0.50 mm? A cylindrical specimen of some alloy 8 mm in diameter is stressed elastically in tension. A force of 15,700 N produces a reduction in the specimen diameter of 5 10-3 mm. Compute...
question 6 of 25 mm is pulled in tension with a 44,500 N force producing only...
question 6
of 25 mm is pulled in tension with a 44,500 N force producing only elastic detormation . What is oreonding strain and the length between the gage marks when subjected to this force? Given: elastic modulus of copper is 105 GPa. 5. A 10-mm diameter bar of 1040 carbon steel is subjected to a tensile load of 50,000 N, taking it beyond its yield point. Calculate the elastic recov removal of the tensile load. Given: elastic modulus of...
Strength of materials please fast Problem 3.28 Part A The elastic portion of the stress-strain diagram...
Strength of materials please fast
Problem 3.28 Part A The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure below Figure 1). The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.3 mm. Take -0.35 If a load of P 60 kN is applied to the specimen, determine its new diameter Express your answer to four significant figures and include appropriate units |d- 1...
Problem #1 (35 pts.) A solid 5 cm by 5 cm square aluminum rod, with a...
Problem #1 (35 pts.) A solid 5 cm by 5 cm square aluminum rod, with a 70 GPa modulus of elasticity, is stretched by an unknown axial load. The initial rod length is 1 m, and its final length (after applying the axial load) is 1.0005 m. Assume that a strain gauge (with a gauge factor of 2.2) is firmly affixed to the rod, its active axis aligned with the length of the rod. You may assume the strain gauge...
6. The data in the table below was obtained from room temperature tensile tests of two Mg alloy s...
6. The data in the table below was obtained from room temperature tensile tests of two Mg alloy samples (A and B): Determine: elastic modulus, yield strength, UTS, and total elongation of these samples. Calculate the strain hardening exponent (n). Sa 35 120 158 333 203.333 228 333 245 0.33 0.63 0.83 0.28 0.4 0.61 0.85 1.75 3.75 5.16 7.64 0.78 11.25 11.9 1225 12.25 12.25 1225 123 203.333 235 260 83.333 295 305 311.66 18.333 321.667 325 1.9 263.333...
1. An aluminum alloy rod 1.5 m long required a force of 60,000 N to reach its yield strength of 280 MPa. E = 7.0 x 104 MPa. Ts = 490 MPa. a) What is its diameter (in cm)? b) What was the total strain? c) What was its final length after the load was released? d) What is the maximum elastic strain this rod must withstand?
HotLW Problem: Giving the engineering stress - engineering strain carve for an aluminum alloy, Calalate: 400FTTTTTTTTTT 10 psi 300 MPa 50 $ Stress (MPa) Stress (10 psi) 8 100- 10/ bd = 110m 1oof lo=50 mm 0 0 0.005 0.010 o 0.05 0. 100 . 150 .20 6 percent elongation at fracture (% elongation = ?) 2 yield strength (Tyield = ?) 3 Tensile strength (TS = ?) ♡ True stress at an engineering strain of 0.04. (I=?) 5 True...
A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected to a 3.0-kN axial load. Knowing that E= 70 GPa, determine the required diameter of the rod. The required diameter of the rod is mm.
A rod is to be used in an aerospace application. The defense
contractor has mandated a factor of safety of 2.6. In its
application, the maximum tensile axial load on the rod ranges
from = 2.1 kN to 2.8 kN. The
metal alloy that the rod is made of has a tensile yield strength of
480 MPa and an ultimate strength of 560 MPa. The specification for
the radius of the rod is given on the drawing.
If the rod...
Problem #1 A cylindrical rod of copper (E 110 GPa) having a yield strength of 240 MPa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must the diameter be to allow an elongation of 0.50 mm? A cylindrical specimen of some alloy 8 mm in diameter is stressed elastically in tension. A force of 15,700 N produces a reduction in the specimen diameter of 5 10-3 mm. Compute...
question 6
of 25 mm is pulled in tension with a 44,500 N force producing only elastic detormation . What is oreonding strain and the length between the gage marks when subjected to this force? Given: elastic modulus of copper is 105 GPa. 5. A 10-mm diameter bar of 1040 carbon steel is subjected to a tensile load of 50,000 N, taking it beyond its yield point. Calculate the elastic recov removal of the tensile load. Given: elastic modulus of...
Strength of materials please fast
Problem 3.28 Part A The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure below Figure 1). The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.3 mm. Take -0.35 If a load of P 60 kN is applied to the specimen, determine its new diameter Express your answer to four significant figures and include appropriate units |d- 1...
Problem #1 (35 pts.) A solid 5 cm by 5 cm square aluminum rod, with a 70 GPa modulus of elasticity, is stretched by an unknown axial load. The initial rod length is 1 m, and its final length (after applying the axial load) is 1.0005 m. Assume that a strain gauge (with a gauge factor of 2.2) is firmly affixed to the rod, its active axis aligned with the length of the rod. You may assume the strain gauge...
6. The data in the table below was obtained from room temperature tensile tests of two Mg alloy samples (A and B): Determine: elastic modulus, yield strength, UTS, and total elongation of these samples. Calculate the strain hardening exponent (n). Sa 35 120 158 333 203.333 228 333 245 0.33 0.63 0.83 0.28 0.4 0.61 0.85 1.75 3.75 5.16 7.64 0.78 11.25 11.9 1225 12.25 12.25 1225 123 203.333 235 260 83.333 295 305 311.66 18.333 321.667 325 1.9 263.333...
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