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Chapter 4: Problem 59

Represent the surface stresses on a Mohr circle of an internally pressurized section of round steel tubing that is subjected to tangential and axial stresses at the surface of 400 and \(250 \mathrm{MPa}\), respectively. Superimposed on this is a torsional stress of \(200 \mathrm{MPa}\).

Short Answer

Expert verified
The center of the Mohr's Circle is at \(325 \mathrm{MPa}\) (average of axial and tangential stresses) and the radius of the Mohr's Circle is \(225 \mathrm{MPa}\) (square root of sum of squares of half the difference of axial and tangential stresses and the torsional stress). Hence, the points representing the given stresses on the Mohr's circle are (550, 200) and (100, -200).

Step by step solution

01

Determine the Center and Radius of Mohr's Circle

We determine the center (C) and the radius (R) of the Mohr's circle as follows:Center, C = \(\frac{(\sigma_{axial} + \sigma_{tangential})}{2} = \frac{(400 + 250)}{2} \mathrm{MPa} = 325 \mathrm{MPa}\)Radius, R = \(\sqrt{(\frac{\sigma_{axial} - \sigma_{tangential}}{2})^2 + \tau_{torsional}^2} = \sqrt{(\frac{400 - 250}{2})^2 + (200)^2} \mathrm{MPa} = 225 \mathrm{MPa}\)
02

Draw Mohr's Circle

With the center 'C' at \(325 \mathrm{MPa}\) along the horizontal axis (representing normal stress) and the radius 'R' equal to \(225 \mathrm{MPa}\), we can draw Mohr's Circle on a graph. The circle will intersect the horizontal axis at points (C - R) and (C + R) which are \(100 \mathrm{MPa}\) and \(550 \mathrm{MPa}\), respectively.
03

Represent Stresses on Mohr's Circle

The given stresses can be represented on Mohr's circle with the help of coordinates:Axial Stress: \(325 + 225 = 550 \mathrm{MPa}\) Tangential Stress: \(325 - 225 = 100 \mathrm{MPa}\) Torsional Stress: \(200 \mathrm{MPa}\)So, the points are represented as (550, 200) and (100, -200) on the Mohr's Circle.

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