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A gap of 0.02 in. exists between the composite rod and the wall when the temperature...
2- Knowing that a 0.5-mm gap exists between the rods shown below when the temperature is...
2- Knowing that a 0.5-mm gap exists between the rods shown below when the temperature is 20 °C, determine: (a) The temperature at which the normal stress in the stainless steel rod will be o = -150 MPa, (b) The corresponding exact length of the stainless steel rod 0.5 mm 300 mm- - 250 mm A В Aluminum Stainless steel A = 2000 mm2 E = 70 GPa 23 x 10-/C A = 800 mm2 E = 190 GPa 18...
.. At room temperature (20°C) a 0.5-mm gap exists between the of the rods shown. At...
.. At room temperature (20°C) a 0.5-mm gap exists between the of the rods shown. At a later time when the temperature has reached 140°C, determine (a) the normal stress in the aluminum rod, (b) the change in length of the aluminum rod. (Io 0.5 mm ー250 mm→ Aluminum Stainless steel A=2000min- E = 75 GPa α = 23 × 10-6PC A=soo E= 190 GPa α = 17.3 × 10-6/oC
3.- At the ambient temperature of 20 ° C, there is a 1.5 mm gap between...
3.- At the ambient temperature of 20 ° C, there is a 1.5 mm gap
between the ends of the rods shown below when the temperature
reaches 140 ° C. Find:
a) the normal stress on the aluminum rod,
b) changing length of aluminum rod.
Make the free body diagram and explain step by step
1.5 mm 325 mm 225 mm A B Aluminum A = 2000 mm E = 75 CPA a = 23 x 10-6/C Stainless steel Acero...
Chapter 2, Reserve Problem 036 At a temperature of 54°F, a 0.12 in. gap exists between...
Chapter 2, Reserve Problem 036 At a temperature of 54°F, a 0.12 in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy (a = 12.4 x 10-6/°F] and bar (2) is stainless steel [a = 8.3 x 10-6/F]. The supports at A and C are rigid. Determine the lowest temperature at which the two bars contact each other. Assume L1 = 45 in., L2 = 63 in. (1) Answer: Final temperature = the...
(3) Given yy, A-36 Steel Rod attached to the wall. E 29000 ksi, a 6.6 x...
(3) Given yy, A-36 Steel Rod attached to the wall. E 29000 ksi, a 6.6 x 10°poF Diameter 3 inches Rod Length 20 inches The gap around the beam is 0.004" (the gap above, below and to the right) These dimensions are at 55°F. Find: If the temperature is increased to 120 °F, what is the normal stress in the beam? Solution
The composite bar is firmly fixed at both ends. The bar is stress-free at 60°F Compute the stress in each material after the 50-kip force 42 in is applied and the temperature is increased to 120°...
The composite bar is firmly fixed at both ends. The bar is stress-free at 60°F Compute the stress in each material after the 50-kip force 42 in is applied and the temperature is increased to 120°F. Use α-6.5 x 10-6/°F for steel and α-1 2.8 x 10-6/°F for aluminum. At what temperature will the aluminum and steel have stresses of equal magnitude after the 50-kip force is applied? Problem 4(25 Points) The rigid bar ABCD is supported by a pin...
Problèm GIVEN b 2) Three rod elements shown above are stress free when they are assembled...
Problèm GIVEN b 2) Three rod elements shown above are stress free when they are assembled together and attached to the rigid walls A and D. Element (1) is steel with E1 30 x 103ksi, cross - sectional area Ai - 2.0 in, length L1-80 in and coefficient of thermal expansion α1=7x10-6/0F; the corresponding values for the aluminum element (2) are: E2 10 x 103ksi, A2-3.6 in, L,- 60 in, a2 13 x 10-6/oF; the corresponding values for the bronze...
(8%) Problem 5: A conducting rod spans a gap of length L = 0.015 m and...
(8%) Problem 5: A conducting rod spans a gap of length L = 0.015 m and acts as the fourth side of a rectangular conducting loop, as shown in the figure. A constant magnetic field with magnitude B= 0.15 T pointing into the paper is in the region. The rod is pulled to the right by an external force, and moves with constant speed v= 0.015 m/s. The resistance in the wire is R = 190 2. X X Х...
Need help on how to make an excel spreadsheet that will have drop down options that...
Need help on how to make an excel spreadsheet that will have drop
down options that reference to cells on another sheet if possible
or to the same sheet if not. The drop downs need to be for the
material type such as 6061-T6 Aluminum. When the material is
selected from the dropdown I need it to fill in values in the cells
for the Yield Strength and Mod of elast. at the same time based off
of the values...
Where λ is the charge density per unit length on the rod and εο is called the permittivity of fre...
where λ is the charge density per unit length on the rod and εο is called the permittivity of free space (it is a universal constant with the value 8.854 x 10-12 F/m (farads per metre)) The integral for the electric field can be evaluated exactly using a method called trigonometric substitution with the result AL We won't learn the method of trigonometric substitution in this course; however, you will approximate the value of the integral using methods we introduced...
2- Knowing that a 0.5-mm gap exists between the rods shown below when the temperature is 20 °C, determine: (a) The temperature at which the normal stress in the stainless steel rod will be o = -150 MPa, (b) The corresponding exact length of the stainless steel rod 0.5 mm 300 mm- - 250 mm A В Aluminum Stainless steel A = 2000 mm2 E = 70 GPa 23 x 10-/C A = 800 mm2 E = 190 GPa 18...
.. At room temperature (20°C) a 0.5-mm gap exists between the of the rods shown. At a later time when the temperature has reached 140°C, determine (a) the normal stress in the aluminum rod, (b) the change in length of the aluminum rod. (Io 0.5 mm ー250 mm→ Aluminum Stainless steel A=2000min- E = 75 GPa α = 23 × 10-6PC A=soo E= 190 GPa α = 17.3 × 10-6/oC
3.- At the ambient temperature of 20 ° C, there is a 1.5 mm gap
between the ends of the rods shown below when the temperature
reaches 140 ° C. Find:
a) the normal stress on the aluminum rod,
b) changing length of aluminum rod.
Make the free body diagram and explain step by step
1.5 mm 325 mm 225 mm A B Aluminum A = 2000 mm E = 75 CPA a = 23 x 10-6/C Stainless steel Acero...
Chapter 2, Reserve Problem 036 At a temperature of 54°F, a 0.12 in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy (a = 12.4 x 10-6/°F] and bar (2) is stainless steel [a = 8.3 x 10-6/F]. The supports at A and C are rigid. Determine the lowest temperature at which the two bars contact each other. Assume L1 = 45 in., L2 = 63 in. (1) Answer: Final temperature = the...
(3) Given yy, A-36 Steel Rod attached to the wall. E 29000 ksi, a 6.6 x 10°poF Diameter 3 inches Rod Length 20 inches The gap around the beam is 0.004" (the gap above, below and to the right) These dimensions are at 55°F. Find: If the temperature is increased to 120 °F, what is the normal stress in the beam? Solution
The composite bar is firmly fixed at both ends. The bar is stress-free at 60°F Compute the stress in each material after the 50-kip force 42 in is applied and the temperature is increased to 120°F. Use α-6.5 x 10-6/°F for steel and α-1 2.8 x 10-6/°F for aluminum. At what temperature will the aluminum and steel have stresses of equal magnitude after the 50-kip force is applied? Problem 4(25 Points) The rigid bar ABCD is supported by a pin...
Problèm GIVEN b 2) Three rod elements shown above are stress free when they are assembled together and attached to the rigid walls A and D. Element (1) is steel with E1 30 x 103ksi, cross - sectional area Ai - 2.0 in, length L1-80 in and coefficient of thermal expansion α1=7x10-6/0F; the corresponding values for the aluminum element (2) are: E2 10 x 103ksi, A2-3.6 in, L,- 60 in, a2 13 x 10-6/oF; the corresponding values for the bronze...
(8%) Problem 5: A conducting rod spans a gap of length L = 0.015 m and acts as the fourth side of a rectangular conducting loop, as shown in the figure. A constant magnetic field with magnitude B= 0.15 T pointing into the paper is in the region. The rod is pulled to the right by an external force, and moves with constant speed v= 0.015 m/s. The resistance in the wire is R = 190 2. X X Х...
Need help on how to make an excel spreadsheet that will have drop
down options that reference to cells on another sheet if possible
or to the same sheet if not. The drop downs need to be for the
material type such as 6061-T6 Aluminum. When the material is
selected from the dropdown I need it to fill in values in the cells
for the Yield Strength and Mod of elast. at the same time based off
of the values...
where λ is the charge density per unit length on the rod and εο is called the permittivity of free space (it is a universal constant with the value 8.854 x 10-12 F/m (farads per metre)) The integral for the electric field can be evaluated exactly using a method called trigonometric substitution with the result AL We won't learn the method of trigonometric substitution in this course; however, you will approximate the value of the integral using methods we introduced...
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