Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- 4 in. P 3 in. В 1200 N 3. A bar, show to the right, has a rectangular cross section of 3 in. by 4 in. and is fixed to the floor at D, and is loaded with three forces: PA = 120 kips, PB = 360 kips, and Pc = 540 kips. Determine: (a) the reaction at the floor, RD; specify the direction ( ↑ or I ), and (b) the normal stresses in each of the three pieces, AB, BC, and CD; specify tension (T) or compression (C) for each. Ans: Rp = 300 kips T, oaB = 10 ksi (C), obC = 20 ksi (T), OCD = 25 ksi (C) В D 7 in. 6 in. 8 in.arrow_forwardThe rigid beam BC is supported by rods (1) and (2). The cross-sectional area of rod (1) is 9 mm². The cross-sectional area of rod (2) is 18 mm². For a uniformly distributed load of w = 2.3 kN/m, determine the length a so that the normal stress is the same in each rod. Assume L = 3.00 m. (1) B Answer: a = i L W a (2) marrow_forwardTwo solid cylindrical rods (1) and (2) are joined together at flange B and loaded as shown. If F₁ = 13 kips, F₂ = 34 kips, and the normal stress in each rod must be limited to 22 ksi, determine the minimum diameter d₁ required for rod (1). A (2) d₁ F₂ B C O 1.087 in. O 0.823 in. O 0.541 in. O 0.867 in. O 1.003 in. d₂arrow_forward
- 5arrow_forward(8) The homogeneous bar is supported by a smooth pin at C a cable that runs from A to B around the smooth peg at D as shown. Find the stress in the cable if its diameter is 15mm and bar weighs 6,000 kg. 5m B 5m Aarrow_forwardPart 1. A composite assembly of length, L-375 mm, consisting of a steel (G-77 GPa) core (2) connected by rigid plates at the ends of an aluminum [G-35 GPa) tube (1) is shown. The aluminum tube has outside diameter D₁ - 50 mm and inside diameter d₁-30 mm. = 100 MPa, and the The solid steel core has diameter d₂-25 mm. The allowable shear stress of aluminum tube (1) is (rallow) allowable shear stress of steel core (2) is (Tallow)2= 135 MPa. Determine (a) the allowable torque T that can be applied to the composite shaft. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced by the allowable torque T. T D₁ L Calculate the polar moment of inertia, J₁, of the aluminum tube. Answer: J₁ (106) mm².arrow_forward
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