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ESE Mechanical 2015 Paper 2: Official Paper

Option 3 : 50√7 MPa and 100 MPa (compressive)

CT 1: Ratio and Proportion

3536

10 Questions
16 Marks
30 Mins

__Concept:__

Centre \(H = \frac{{{\sigma _1} + {\sigma _2}}}{2}\)

Radius = \(\frac{{{\sigma _1}\; - \;{\sigma _2}}}{2}\)

__Calculation:__

Given, σ_{1} = + 250 MPa, σ_{2} = - 100 MPa

\(H = \frac{{250\; + \;\left( { - 150} \right)}}{2} = 50\) MPa

Radius = \(\frac{{{\sigma _1}\; - \;{\sigma _2}}}{2}\)

\(R = \frac{{250 - \left( { - 150} \right)}}{2} = 200\;MPa\)

The plane at right angle to the plane HC will be shown by 2 × 90° = 180 it is shown by the plane HC.

From the similiar triangle HCF and HEG

\(\frac{{HF}}{{HC}} = \frac{{HG}}{{HE}}\)

As HC = HE = R

So, HF = HG

150 = HO + OG

150 = 50 + OG

OG = 100 MPa (Compressive)

Shear stress = \(FC = \sqrt {H{C^2} - H{F^2}}\)

⇒ Shear stress = \(\sqrt {{{200}^2} - {{150}^2}}\) = \(50\sqrt 7\) MPa