Mechanical Engineering - Strength of Materials - Discussion

@All.

That's correct.

This is a fundamental concept in solid mechanics often referred to as **complementary shear stress**.

Here's a breakdown of why this happens:

Equilibrium: For a body to be in equilibrium, all forces and moments acting on it must balance. This applies to internal stresses as well.
Shear Stress: Shear stress acts parallel to a surface. If you have shear stress acting on one plane, it creates an unbalanced moment (a twisting force).
Complementary Shear Stress: To counteract this moment and maintain equilibrium, equal and opposite shear stress must exist on a perpendicular plane.

For Eg;

Imagine holding a book between your hands and trying to slide the top cover to the right. To prevent the book from rotating, you instinctively apply an equal and opposite force with your other hand, sliding the bottom cover to the left. These opposing forces are analogous to complementary shear stresses.

Key Points:-

Magnitude: The complementary shear stresses are equal in magnitude.
Direction: They act in opposite directions.
Planes: They act on planes that are perpendicular to each other.

This principle of complementary shear stress is crucial in understanding how materials behave under load and is essential for analyzing and designing structures.

When a shear stess act on body, the body leads to stay stable. So he produces a complementary shear stress which is same value with previous shear stess. It is tangential to shear stress otherwise it can't counter the shear stress.

If Normal to it, then Total forces action on that body will be.

1) moment produced by couple of shear stress.
2) compressive or tensile stress action normal to the plane.

So balancing shear stress must be tangential to the plane.

I am not understanding. Please, anyone, give a perfect explanation for this.

Complimentary shear is acting on normal plane not across the same plane.

The force is tangential not normal to the plane.

I think true is the correct answer.

It's Complimentary shear stress.

Please explain the right answer.

Yes @Mopi, it should be true.