Mechanical Engineering - Strength of Materials - Discussion

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)].

Maximum normal stress : σ 1=1200 Mpa , σ2=600 Mpa, τ= 400 Mpa.

Maximum normal stress =(σ 1+σ2/2) +(1/2) √[(σ 1-σ2)^2 +4(T)^2].
=(1200+600/2) +(1/2) √ [(1200-600)^2 +4(400)^2],
=(1800/2)+(1000/2),
=1400 MPa.

Normal 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 ans 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)].

Assume stress in X direction(X) = 1200.
Stress in Y direction(Y) = 600.
Shear stress(S) = 400.

Max. Normal stress=(X+Y)/2+Sqrt(((X-Y)/2)square+S square).

=(1200+600)/2 + sqrt(300 square + 400 square).

=1400 mpa.

Principal stresses are the stresses in the planes, where shear stress is equal to zero! so the principal stresses are essentially maximum and minimum normal stresses, with an angle of 90 degrees between them!

Normal stress are the average stress in X and Y direction.

At particular plane that means angle of rotation the Normal stress will be maximum that plane/stress is called principle plane/stress.

Maximum normal stress : σ 1=1200 Mpa , σ2=600 Mpa, τ= 400 Mpa.

Maximum normal stress formula= (σ1 + σ2)/2 + √(σ1 + σ2)/2 + τ 1500 Mp.

Plane at which max normal stress & minimum normal stress present is called as principal plane. So, maximum principal stress is same as the maximum normal stress.

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.