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1.18 A load P is applied to a stecl rod supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled. Knowing that the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can be applied to the rod. Problem 1.18 - 40 mm For the steel rod, A, = rd, t, = (T0.012 (0.010) 10 mm 8 mm 12 inm 2. = 376.99 x 10°“ m' P= 2A, %3D A, P. P, = (180x10ʻ)(376.99 10“) = 67.86 r10* N For the aluminum plate, A2= Td, t, = (TX0.040X0.00) = 1.00531x103 m² %3D P P = (70x10)(!.0053x10) 70. 372 * 10 N The limiting valve for the load P is the smallen of P, and Pa. P: 67.86x 10N P = 67.9 kN

1.18 A load P is applied to a stecl rod supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled. Knowing that the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can be applied to the rod. Problem 1.18 - 40 mm For the steel rod, A, = rd, t, = (T0.012 (0.010) 10 mm 8 mm 12 inm 2. = 376.99 x 10°“ m' P= 2A, %3D A, P. P, = (180x10ʻ)(376.99 10“) = 67.86 r10* N For the aluminum plate, A2= Td, t, = (TX0.040X0.00) = 1.00531x103 m² %3D P P = (70x10)(!.0053x10) 70. 372 * 10 N The limiting valve for the load P is the smallen of P, and Pa. P: 67.86x 10N P = 67.9 kN

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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can you draw the 2 Free body diagram where the shearing force is acting on the plate and the rod respectively. 

I do not understand why the shearing area is pi*t*d

Problem 1.18 1.18 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled, Knowing that the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can be applied to the rod. 40 mm For the steel rod, A, = rdt = (TYO. 012 (0.010) 10 mm 8 mm 12 inni = 376.99 x10 m P= ,A, P = (180 x10ʻ)(376.99 10“ ) = 67.86 10* N For the aluminum plate, T d, t, = (TX0.040X0.008) = Pa A Az= 1.00531x103 m² Pa = TAz Pz = (70x10)(1.0053× 10) : 70, 372 * 10 N The limiting va lve for the load P is the smallen of P, and Pa. P=67,86x10° N P = 67.9kN A
Transcribed Image Text:Problem 1.18 1.18 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled, Knowing that the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can be applied to the rod. 40 mm For the steel rod, A, = rdt = (TYO. 012 (0.010) 10 mm 8 mm 12 inni = 376.99 x10 m P= ,A, P = (180 x10ʻ)(376.99 10“ ) = 67.86 10* N For the aluminum plate, T d, t, = (TX0.040X0.008) = Pa A Az= 1.00531x103 m² Pa = TAz Pz = (70x10)(1.0053× 10) : 70, 372 * 10 N The limiting va lve for the load P is the smallen of P, and Pa. P=67,86x10° N P = 67.9kN A
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