!Required informationThe 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 185°F. Consider only axial deformations.1 in.Brass coreE = 15 x 10° psia = 11.6 x 10-/°FAluminum shellE = 10.6 x 10° psia = 12.9 x 10-/°F- 2.5 in.-Determine the stress in the brass core.The stress in the brass core isksi.

Given data: Ti=78°FTf=185°FEBr=16×106 psiαBr=11.6×10-6/°FEAl=10.6×106 psiαAl=12.9×10-6/°F Need to…

Answered: Required information The aluminum shell…

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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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
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L b F What is the change in length (L) of the titanium bar (E = 16.5E6 psi) given: L = 36 in., b = 3 in., h = 0.25 in, and F = 3 kips Answer to above question is 8.72E-3 in, the shear modulus for the bar is 6.145E6 psi, poisons ratio is 0.343. My question is what is the strain (in macrostrain) in the bar in a direction perpendicular to the direction of the applied force? (Answer is 82.8, I need to understand how to reach this answer.)
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
PROBLEM NO. 4 In Fig. 4, determine the normal stress and change in length of the aluminum rod if the temperature was raised by 129 C. The 0.5-mm gap exists at 28 C (0.59 mm 399 mm A Aluminum A = 2888 mm² E = 75 GPa α = 23 x 10°/°C 258 mm B Stainless Steel A = 899 mm² E = 190 GPa α = 17.3 x 10/°C
The two angle bars are used to support a tensile load "T". Assume the tensile load to be equally resisted by each angle. 400 mm x 20 mm THK Gusset Plate The steel used is L 6 x 6 x 1. Properties: A = 7096.76 mm? y = 47.244 mm Fy = 248 MPa Fu = 400 MPa b = 152.4 mm b. t = 25.4 mm x = 47.244 mm Effective hole diameter = 25 mm y Use case 2 for shear lag factor. Use NSCP 2015 code. 1. Determine the maximum allowable load that can be carried by the two L6 x 6 x1 based on yielding of the gross section--- 2. Determine the maximum factored load that can be carried by two L6 x 6 x 1, based on the fracture of the effective net area.---- 3. Determine the maximum allowable tensile load that can be carried.------ 4. Determine the maximum factored tensile load that can be carried-- bo
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Consider the truss shown in (Figure 1). Each member has a cross-sectional area of 8.0(10³) mm² and is made of A- 36 steel. Take P = 150 kN. Figure P 4 m A -3 m B 1 of 1 Part A Determine the total strain energy stored in the truss. Express your answer to three significant figures and include the appropriate units. U₂ = Submit μA Value Provide Feedback Request Answer Units ?
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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