The aluminum shell is fully bonded to the brass core, and the assembly is unstressed at a temperature of 78°F. Thetemperature of the assembly reaches 190°F. Consider only axial deformations.1 in.-2.5 in.Brass coreE = 15 x 106 psia = 11.6 × 10-6°FAluminum shellE = 10.6 × 106 psia = 12.9 x 10-6°FDetermine stress in the aluminum shellThe stress in the aluminum shell is ____ksi.

Answered: Image /qna-images/answer/5790b0d1-66df-4b42-a4fa-3ce752d1fe2e.jpg

Answered: The aluminum shell is fully bonded to…

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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The copper pipe [G = 6,000 ksi] has an outside diameter of 3.375 in. and a wall thickness of 0.250 in. The pipe is subjected to a uniformly distributed torque of t = 85 lb-ft/ft along its entire length. Using a = 2.5 ft, b = 3.5 ft, and c =7.0 ft, determine (a) the maximum shear stress in the pipe. (b) the rotation angle pc at the free end of the shaft. b B C Answers: (a) Tmax psi i rad (b) ¤c = i
Tind the stresses in the steel and aluminum rods when the temperature rises to 60 C° and the distributed load is 10 kN/m. Given: Steel Rod E = 200 GPa L = 1.499 m 10KN/m esmott A = 80 mm? oc= 12 x 10-6 1/°c Gap 1 mm Aluminum Rod Steel Rod Aluminum Rod E = 70 GPa L = 1.5 m A = 200 mm? x= 22 x 10-6 1/°c Figure (2) !! 2 m 2 m
Determine the principal stresses at points A and B of the cylinder. Consider: L = 70 mm; d = 20 mm; F = 12 kN; P = 15 kN; T = 800 Nm.
The rectangular bar is connected to the support bracket with a 25-mm-diameter pin. The bar width is w = 65 mm and the bar thickness is 10 mm. Each side of the bracket has the same dimensions as the bar. The average shear stress in the pin cannot exceed 115 MPa, the bearing stress in the bar cannot exceed 120 MPa, and the bearing stress in the bracket cannot exceed 140 MPa. Determine the maximum value of Pmax that can be supported by the structure. Answer: Pmax i kN
* Your answer is incorrect. The solid circular rod has a cross-sectional area of 410 mm². It is subjected to a uniform axial distributed loading along its length of w= 9 kN/m. Two concentrated loads also act on the rod: P = 3 kN and Q = 5 kN. Determine the normal stress in the rod at x = 0.5 m. Assume a = 0.7 m and b = 1.3 m. W B >x 45012 MPa a Answer: 0 = b
As shown in the figure, when the temperature was 24 ° C, a 0.5 mm gap was left. Determine the approximate value (° C) of the required temperature so that the normal stress in the aluminum bar is 75 MPa.
The principal stresses shown below are calculated at two different cross sections of a brittle machine part. Determine if the cross section A or B is safer (Sy = 50 MPa, Sut = 100 MPa, and Suc = 200 MPa). 0,50 MPa + 0₂ = -20 MPa A 0₂=-20 MPa 0₁ = 50 MPa a, 20 MPa - ₂ =-120 MPa B 0-120 MPa a₁ = 20 MPa
Part 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.
The supporting wheel on a scaffold is held in place on the leg using a 4-mm-diameter pin as shown. (Figure 1) Figure 1 of 1 > Part A If the wheel is subjected to a normal force of P = 3.3 kN, determine the average shear stress developed in the pin. Neglect friction between the inner scaffold puller leg and the tube used on the wheel. Express your answer to three significant figures and include the appropriate units. Tavg= Submit Provide Feedback Value A Request Answer C Units ? Next >
For a steel rod with a circular cross section with a diameter of D = 20 mm, the following is required: 1. Draw a diagram of the longitudinal force 2. Draw a diagram of normal stresses 3. Determine the total elongation of the rod if Е = 2∙10^5 MPa When calculating, take: а = 2 m, b =1.2 m, F=10 кN - The point of application of force! The work must be done on one sheet of A4 paper, which must show: - Using the method of sections to determine the longitudinal forces in the rod. - Draw a diagram of longitudinal forces N - Determination of normal stresses based on the constructed plot of longitudinal forces. - Draw diagram a normal stresses - Determination of the full extension of the rod
-8 in.- -8 in.- (a) (b) Which of the following best explains why cylinder (b) has a higher longitudinal stress in its walls compared to cylinder (a)? Assume the gauge pressure of the fluid inside the cylinder is non-zero. O Cylinder (b) is supported at two locations Cylinder (a) is not free to deform in its longitudinal direction The vertical (longitudinal) pressure in cylinder (a) is transferred directly to the [bottom] support, bypassing the cylinder walls None of the above the longitudinal stress for both cylinders is the same

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