A uniform rectangular bar of area of cross-section A is fixed at one end and on other end forces F is applied as shown in figure. Find the shear stress at a plane through the bar making an angle θ with the vertical as shown in figure.

Question

A uniform rectangular bar of area of cross-section A is fixed at one end and on other end forces F is applied as shown in figure. Find the shear stress at a plane through the bar making an angle θ with the vertical as shown in figure. (A) (F/2 A)( cos 2 θ) (B) (F/2 A) (C) (F/2 A)( sin 2 θ) (D) (F/2 A) cos θ

Tardigrade Admin · Accepted Answer

Consider A prime is the area of cross-section of plane shown in figure <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/biology/bda54dc985479b2962a922a42cbac4d6-.png" /> ∴ A'=(A/ cos θ) Now the tangential force on plane is F sin θ Shear stress =(F sin θ/(A) cos θ) =(F/A) sin θ cos θ=(F/2 A) 2 sin θ cos θ =(F/2 A)( sin 2 θ)

Q. A uniform rectangular bar of area of cross-section $A$ is fixed at one end and on other end forces $F$ is applied as shown in figure. Find the shear stress at a plane through the bar making an angle $θ$ with the vertical as shown in figure.

$\frac{F}{2 A} (cos 2 θ)$

$\frac{F}{2 A}$

$\frac{F}{2 A} (sin 2 θ)$

$\frac{F}{2 A} cos θ$

Consider $A^{'}$ is the area of cross-section of plane shown in figure

$∴ A^{'} = \frac{A}{c o s θ}$
Now the tangential force on plane is $F sin θ$
Shear stress $= \frac{F s i n θ}{\frac{A}{c o s θ}}$
$= \frac{F}{A} sin θ cos θ = \frac{F}{2 A} 2 sin θ cos θ$
$= \frac{F}{2 A} (sin 2 θ)$

Q. A uniform rectangular bar of area of cross-section A is fixed at one end and on other end forces F is applied as shown in figure. Find the shear stress at a plane through the bar making an angle θ with the vertical as shown in figure.

2AF​(cos2θ)

2AF​

2AF​(sin2θ)

2AF​cosθ