Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown. A concentrated load P is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of d1 = 0.40 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of d2 = 0.70in. Aluminum rod (3) has a diameter of d3 = 1.00in. The yield strength of the bronze is 48 ksi and the yield strength of the aluminum is 40 ksi. Assume a = 2.5 ft, b = 1.5 ft, L1 = 6 ft, L2 = 8 ft, and L3 = 3 ft. L₁ABronze(1)BAluminum(3)DAluminum(2)PbL3CL₂ Part 4Based on the allowable stress for the bronze, calculate the maximum allowable force thatcan be supported by rod (1), and the corresponding maximum value for the applied load P.Answers:Fmax,1 =Pmax, 1Part 5=Fmax, 2 =kipsBased on the allowable stress for the aluminum, calculate the maximum allowable forcethat can be supported by rod (2), and the corresponding maximum value for the appliedload P.Answers:Pmax, 2 =kipskipskips

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Answered: Based on the allowable stress for the…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown. A concentrated load P is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of d1 = 0.40 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of d2 = 0.70in. Aluminum rod (3) has a diameter of d3 = 1.00in. The yield strength of the bronze is 48 ksi and the yield strength of the aluminum is 40 ksi. Assume a = 2.5 ft, b = 1.5 ft, L1 = 6 ft, L2 = 8 ft, and L3 = 3 ft.

L₁
A
Bronze
(1)
B
Aluminum
(3)
D
Aluminum
(2)
P
b
L3
C
L₂

Part 4
Based on the allowable stress for the bronze, calculate the maximum allowable force that
can be supported by rod (1), and the corresponding maximum value for the applied load P.
Answers:
Fmax,1 =
Pmax, 1
Part 5
=
Fmax, 2 =
kips
Based on the allowable stress for the aluminum, calculate the maximum allowable force
that can be supported by rod (2), and the corresponding maximum value for the applied
load P.
Answers:
Pmax, 2 =
kips
kips
kips

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