Concept:

When a shaft is subjected to torsional or twisting moment T only, n shear stress induced in shaft is given by,

$\mathrm{Ï„}=\frac{16T}{\mathrm{Ï€}{d}^3}$

When shaft is subjected to bending moment only, n bending stress in shaft is given by,
${\mathrm{σ}}_b=\frac{32M}{\mathrm{π}{d}^3}$

When shaft is subjected to combined twisting moment and bending moment, n shaft is designed on basis of maximum shear stress ory.

According to maximum shear stress ory, maximum shear stress in shaft,

${\mathrm{τ}}_{max}=\frac{1}{2}\sqrt{{\mathrm{σ}}_b^2+4{\mathrm{τ}}^2}=\frac{16}{\mathrm{π}{d}^3}\sqrt{M^2+T^2}$

Concept:

Both are of equal area $⇒b^2=\frac{\mathrm{Ï€}{d}^1}{4}$

$⇒b=\frac{d}{2}\sqrt{\mathrm{π}}$

For circular section modulus,

$Z_c=\frac{\mathrm{Ï€}{d}^3}{32}$

For square section,

$\begin{array}{l}{Z}_{square}=\frac{b^3}{6}=\frac{\mathrm{π}\sqrt{\mathrm{π}}{d}^3}{48}\\ ∴Z_{square}=1.18Z_{circular}\end{array}$

So for same cross-section area, a square section is better than circular section in bending.

Note:-

For same cross-sectional area, order of sections in increasing bending strength

Explanation:

In general, beams have uniform cross-sections throughout ir length. When y are loaded, re is a variation in bending moment from section to section along length. The stress in extreme outer fiber (top and bottom) also varies from section to section along ir length.

The extreme fibers can be loaded to maximum capacity of permissible stress, but y are loaded to less capacity. Hence, in beams of uniform cross-section, re is a considerable waste of materials.

To avoid this waste of material, beam is suitably designed such that extreme fibers are loaded to maximum permissible stress by varying cross-section it will be known as a beam of uniform strength.

The beam is said to be in uniform strength if maximum bending stress is constant across varying section along its length.

The bending stress in extreme fiber can be calculated by using bending formula:

$\mathrm{σ}=\frac{M}{Z}$

For rectangular sections, Z = bd²/6

Based on above formula, to have beam of uniform strength, its cross-section varies in following way:

1. Changing depth (by keeping width constant) throughout length of beam
2. Changing width (by keeping depth constant) throughout length of beam.
3. By changing both width and depth suitably.

Concept:

Maximum Bending stress due to bending moment:

${σ}_b=\frac{32M}{\mathrm{π}{d}^3}$

Maximum Shear stress due to twisting moment/ torque:

$Ï„=\frac{16T}{\mathrm{Ï€}{d}^3}$

Calculation:

Given:

Maximum bending stress = Maximum shear stress

σ_b = Ï„  

$\frac{32M}{\mathrm{Ï€}{d}^3}$ = $\frac{16T}{\mathrm{Ï€}{d}^3}$

Hence, $M=\frac{T}{2}$

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