Mechanical Engineering - Strength of Materials - Discussion

Thank you all for the explanation.

The maximum and minimum normal stresses (σ1 and σ2) are known as the principal stresses.

Solution is very simple. Ans is 1400.

Explanation:

Normal stress =((x+y)÷2) + √ ((x - y)÷2) + (shear stress)^2 where x = 1200 & y= 600.

If shear stress is applied, maximum normal stress will always be greater than applied maximum stress(1200). So the answer is 1400. No calculation only observation.

Normal stress = (sum of stress/2) = 900 answer.

Thanks for all your explanation.

You can also use this formula,

Stress = [(x + y)/2] + 0.5 * (x - y)^2 + 4(shear stress)^2.

Simply use Mohr's circle to the equations.

Maximum. normal stress = T.S - S.S.
= T.S1 + T.S2 - S.S.
= 1200 + 600 - 400 .
= 1400 answer.

Principal stress are maximum and minimum normal stress which may be developed on a loaded bodyNormal Stress = (Shear Stress*sin2theta)+((Fx+Fy)/2)+(((Fx-Fy)/2)cos2theata).

In this Case theta = 90 because they are at right angle.
Other all values are given on calculation we get answer A)400.

Reference: Strength of materials by R.K.Rajput.
Chapter 2, Page NO.95 Eq.2.8..

Let, tensile stress on one plane(t.s1) = 1200mpa.

Tensile stress on another plane(t.s2) = 600mpa.

Shear stress (s.s) = 400mpa.

Now using major principal stress formula = [(t.s1+t.s2)/2]+sqrt[{(t.s1-t.s2)*(t.s1-t.s2)}/2 +(s.s*s.s)].