Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 6 (Q.No. 41)
41.
When a body is subjected to a direct tensile stress (σ) in one plane, then normal stress on an oblique section of the body inclined at an angle θ to the normal of the section is
Discussion:
5 comments Page 1 of 1.
Rutem Singh said:
2 years ago
P(X) = (Px+Py)/2+(Px-Py)/2×cos2θ+0 = (Px/2)+(Px/2)cos2θ = Pxcos^2θ.
Narendar said:
3 years ago
Pcosθ ÷ (A/cosθ) = P/A x cosθxcosθ = σCos^2θ.
Supriya said:
5 years ago
In case of obligue, normal stress=(p/a)Cos^2θ.
So, here we know that shear stress=(p/a). That means normal stress=(shear stress)(cos^2 θ).
So, here we know that shear stress=(p/a). That means normal stress=(shear stress)(cos^2 θ).
Shirin said:
5 years ago
When oblique section makes theta angle with the plane (lets say plane A) where σ acts. Then sin.
When oblique section makes theta angle with the normal of plane A then it will be cos.
When oblique section makes theta angle with the normal of plane A then it will be cos.
Rohit said:
5 years ago
Please, anyone, explain when it is sin & when it is cos?
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