**Problem 3 (25 points)**A steel disk with a 150 mm outer diameter is to be press fit to a hollow steel tube whose outer diameter is 90 mm and inner diameter is 45 mm. The maximum tangential stress in the disk is to be 180 MPa. What diametral interference is required between the shaft and the disk to cause this stress? The modulus of the steel is 207 GPa.

Answered: Image /qna-images/answer/715b5eaf-ff85-44e8-948e-9f1a37959ffc.jpg

Answered: A steel disk with a 150 mm outer…

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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A composite shaft consists of a brass sleeve securely bonded to half steel shaft, fixed atend A as shown in Figure 1. The brass sleeve and steel shaft have a modulus rigidity Gb= 40 GPa and Gs = x0 GPa. The length of the assembly is L = 400 mm. The outerdiameters of the shaft and sleeve are d1 = 70 mm and d2 = 90 mm, respectively. Giventhe shear stress of brass is τb = 70 MPa and shear stress of steel is τs = 110 MPa.[Note: x value - Use the last non-zero digit from your group number. If the group numberis zero, use the second last digit instead. For example, if your group number is 5, your xvalue is 5 and Gs = 50 GPa; if your group number is 10, your x value is 1 and Gs = 10GPa.] In this situation, Group number is 2. 1) Write the equations of equilibrium for AB and BC section of the shaft in terms of Twith the help of free body diagrams. 2) Determine the polar moment of inertia for brass sleeve and steel shaft.
Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown. The normal stress in aluminum rod (1) must be limited to 19 ksi, the normal stress in brass rod (2) must be limited to 22 ksi, and the normal stress in steel rod (3) must be limited to 14 ksi. Determine the minimum diameter required for each of the three rods. Assume P = 9 kips, Q = 5 kips, R = 20 kips and S = 30 kips. (1) VQ B R (2) C (3) D
69. A disc of 600 mm diameter and uniform thickness is rotating at 3000 rpm. Calculate the maximum stress developed in the disc. If a hole of 100 mm dianmeter is drilled at the centre of the disc, then determine maximum radial and hoop stresses developed. Take vs 0-28 and p = 7800 kg/m.
Please do it typed not handwriting. Make sure the answer is correct Don't copy from any other online answers please
Answer the ff correctly. The rectangular bar shown is connected to a support bracket with a 24 mm diameter pin. The width of the rectangular bar is 70 mm and its thickness is 15 mm. The bearing stress in the bar cannot exceed 120 MPa. Determine the maximum value of Pmax that can be supported based solely on consideration of the bearing stress in the bar. Note: This question is not asking about any stresses in the support bracket...just bearing stress in the rectangular bar.
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A cylinder of (320) mm inside diameter and (64) mm thickness is subjected to an internal pressure of 15 MPa. Find the variation of circumferential and radial stresses in the shell and plot them. What is the increase in the inside and outside diameters of the shell? Take E=200 GPa and v=0.3
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Coefficient of expansion αB= 165.10^-7 1/ 0C , solid copper bar diameter 20 mm , It is placed in a steel pipe with an outer diameter of 30 mm with an inner diameter of 2 mm larger than this bar and an expansion coefficient αÇ = 125.10^-7 1 / 0C, and is joined by a rigid plate from the top so that they move together. Find the stresses that will occur in steel and copper when the system is heated with a temperature difference of Δt = 80 0C. (EÇ = 200 GPa, EB = 100 GPa)
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The bronze bar 3.42 m long with a cross-sectional area of 627 mm2 is placed between two rigid walls. At a temperature of -14 °C, there is a gap Δ = 2.36 mm, as shown in the figure. Find the temperature at which the compressive stress in the bar will be 34.22 MPa. Use α = 18 × 10-6 /°C and E = 80.32 GPa. Round off your answer to two decimal places.

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