Question

For a beam of rectangular section of width b and depth d, the maximum shear stress in the cross section for a shear force v is

Answer 1

Answer:

3f/2bh

Explanation:

Shear stress for rectangle beam = 3/2 times of avg shear stress.

The avg shear stress = F/bh.

Shear stress =3/2,(F/bh

Answer 2

Answer:

The maximum shear stress in the cross-section for a shear force v is$\frac{\tau}{\sigma}=\frac{d}{4 a}\end{aligned}$

Explanation:

Concept:

Shear stress of beam for rectangular cross-section,

$\tau=\frac{3}{2} \times \frac{V_{\max }}{A}$

Where $V_{\max }=$Maximum shear force

Bending stress is given by,

$\sigma=\frac{M}{Z}$

Where, $M=$ maximum Bending moment, $Z$= Section modulus

Calculation:

From sheer force diagram

Maximum shear force, $V_{\max }=P$

So, Maximum shear stress, $\tau=\frac{3}{2} \times \frac{V_{\max }}{A}$

$\Rightarrow \tau=\frac{3}{2} \times \frac{P}{b d} \quad \cdots(\mathrm{i})$

From the bending moment diagram,

Maximum bending moment, $M=P a$

So, Bending stress, $\sigma=\frac{M}{Z}$

$\Rightarrow \sigma=\frac{P a}{\frac{b d^{2}}{6}}=\frac{6 P a}{b d^{2}} \quad \cdots(\text { ii) }$

The ratio of maximum shear stress to maximum bending stress

$\begin{aligned}&\frac{\tau}{\sigma}=\frac{\frac{3}{2} \times \frac{P}{b d}}{\frac{6 P a}{b d^{2}}} \\&\Rightarrow \frac{\tau}{\sigma}=\frac{d}{4 a}\end{aligned}$