2. A compound thin cylinder has a common diameter of 100 mm, and the inner cylinder has a thickness of 2,5 mm. The radial pressure between the two cylinders is 200 kPa and the difference between the two common diameters before shrinkage was 4,305x 10-3 mm. E = 200 GPa. Determine a. Thickness of the outer cylinder b. Resultant hoop stresses in both cylinders if the compound cylinder is subjected to an internal pressure of 180 kPa.

2. A compound thin cylinder has a common diameter of 100 mm, and the inner cylinder has a thickness of 2,5 mm. The radial pressure between the two cylinders is 200 kPa and the difference between the two common diameters before shrinkage was 4,305x 10-3 mm. E = 200 GPa. Determine a. Thickness of the outer cylinder b. Resultant hoop stresses in both cylinders if the compound cylinder is subjected to an internal pressure of 180 kPa.

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

See similar textbooks

Similar questions

An aluminum [E-9600 ksi] control rod with a circular cross section must not stretch more than 0.34 in. when the tension in the rod is 2070 lb. If the maximum allowable normal stress in the rod is 12.9 ksi, determine: (a) the smallest diameter d that can be used for the rod. (b) the corresponding maximum length L of the rod. Answers: (a) d- i in. (b) L- in.
3) At room temperature (20 °C) a 0.8 mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 120 °C, determine (a) the normal stress in the aluminum rod, (b) the change in length of the aluminum rod Bronze: Aluminum: Steel: Area: 1000 mm2 Area: 2000 mm2 Area: 1500 mm2 a = 18.4 x 10-6 E = 103 GPa a = 23.6x 10-6 E = 75 GPa a = 17.3 x 10-6 E = 190 GPa %3D %3D 0.8 mm 100 mm 150 mm 300 mm Bronze Al Steel
1. The frame shown is used to support a window-type air conditioning unit. It weighs 100 lbs and its equivalent weight is applied as shown. Determine the stress in the brace, the deformation of the brace when the load is applied, and the shear stress in the ¼ inch diameter bolts. 14 in 12 in 12 in Brace 1% x % inch steel angles ¼ inch bolt in single shear
PROBLEM 4. A piece of rubber, which was originally a square shape, has been deformed into a shape shown with the dashed lines in the figure. After the deformation, the CD line remained parallel to the direction before the deformation. (a) Calculate the average normal strains along AB and diagonal AD. (b) Calculate the average shear strain yxy at point A. 6 mm 84 mm 4 mm 80 mm 2 mm 80 mm mm
An aluminium processing hot extrusion steel cylinder has an internal diameter of 0.18 m and an outer diameter of 0.72 m. Determine the stresses across the cylinder for the following conditions. (a) During cold extrusion of aluminium at an internal pressure of 600 MPa. 1.
5. Problem: From the figure as shown, determine the stress in psi at the top surface of the second cylinder. Assume the specific weight of the material is 460 lbf/ft³. A. B. C. D. 4225 psi 4620 psi 4840 psi 4510 psi F=3000 lbf h=1.5ft| d=1ft.
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.
Question 4 A steel rod is subjected to a gradually applied load (F) which gave a rise to a maximum stress of 200 MPa. The rod is 250 mm long and one part of its length is square and the remainder is cireular with a diameter of 25 mm. If the total strain energy in the rod and modulus elasticity of the material is 1.3 J and 200 GPa, determine the following: 4.1 The applied load F 4.2 The total extension of the bar 4.3 The length of the square portion of the bar 4.4 The suddenly applied load that will induce the same amount of energy 4.5 The load that falls from a height of 8 mm induces 1,3 J in the bar.
The rigid beam BC is supported by rods (1) and (2). The cross-sectional area of rod (1) is 11 mm². The cross-sectional area of rod (2) is 14 mm2. For a uniformly distributed load of w = 4.1 kN/m, determine the length a so that the normal stress is the same in each rod. Assume L = 5.35 m. A (1) (2) a L. Answer: a = i m
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
[4] In thin walled sphere, the wall is subjected to ..... O A. Hoop stress only B. Longitudinal stress only. C. Both A & B O D. Radial stress only
3. A 1.40 m long steel circular bar is acted upon by forces as shown in fig below. determine the total elongation of the bar. Take E steel=200Gpa and E brass = 85GN/m² 4 cm 3 cm 3 cm 40 kN 60 kN Steel 100 kN 80 kN Steel Brass 40 cm 60 cm 40 cm