Stress Correction Factor MCQ Quiz - Objective Question with Answer for Stress Correction Factor - Download Free PDF

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combine torsional, direct and curvature shear stress.

Torsional shear stress in the bar-

${\tau }_1=\frac{16M_t}{\pi d^3}$

${\mathrm{M}}_{\mathrm{t}}=\mathrm{T}\mathrm{o}\mathrm{r}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\Rightarrow \mathrm{P}\times \frac{\mathrm{D}}{2},\mathrm{D}=\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r},\mathrm{d}=\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{w}\mathrm{i}\mathrm{r}\mathrm{e}$

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar-

∴ τ = τ1 + τ2

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$
$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

When torsional shear stress, direct shear stress, and curvature shear stress is taken into consideration, then shear stress is given by-

$\tau =K\left(\frac{8PD}{\pi d^3}\right)\Rightarrow K\left(\frac{8PC}{\pi d^2}\right)K=Wah{l}^{\prime }sfactor$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combined direct shear and torsional shear stress in the coil wire cross-section.

Torsional shear stress in the bar:

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar:

${\tau }_2=\frac{P}{\frac{\pi }{4}{d}^2}=\frac{4P}{\pi d^2}$

Combining both equations

∴ τ = τ1 + τ2

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$

$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combine torsional, direct and curvature shear stress.

Torsional shear stress in the bar-

${\tau }_1=\frac{16M_t}{\pi d^3}$

${\mathrm{M}}_{\mathrm{t}}=\mathrm{T}\mathrm{o}\mathrm{r}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\Rightarrow \mathrm{P}\times \frac{\mathrm{D}}{2},\mathrm{D}=\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r},\mathrm{d}=\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{w}\mathrm{i}\mathrm{r}\mathrm{e}$

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar-

∴ τ = τ1 + τ2

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$
$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

When torsional shear stress, direct shear stress, and curvature shear stress is taken into consideration, then shear stress is given by-

$\tau =K\left(\frac{8PD}{\pi d^3}\right)\Rightarrow K\left(\frac{8PC}{\pi d^2}\right)K=Wah{l}^{\prime }sfactor$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$

Concept:

The torsional shear stress in the wire is given by, $\tau =\frac{8\mathrm{W}\mathrm{D}}{\pi {\mathrm{d}}^3}$

When the wire is bent in the form of coil, the length of the inside fiber is less than the length of the outside fiber. This results in the stress concentration at the inside fiber of the coil.

Therefore, the shear stress in fiber after stress concentration factor is $\tau =\frac{8\mathrm{W}\mathrm{D}}{\pi {\mathrm{d}}^3}\left(\frac{1}{2\mathrm{C}}+1\right)$

where $\left(\frac{1}{2\mathrm{C}}+1\right)$ is stress concentration factor (Kc).

Therefore, Kc =  $\left(\frac{1+2C}{2\mathrm{C}}\right)$ where C = $\frac{D}{d}$ (∵ D = diameter of coil, d = diameter of wire)

Fig. (a) Pure torsional stress

Fig. (b) Direct shear stress

Fig. (c) Combine torsional, direct and curvature shear stress 

Calculation:

Given:

D = 50 mm, d = 5 mm

$C=\frac{D}{d}=\frac{50}{5}=10$

Therefore, shear stress factor $K_c=\left(\frac{1+2C}{2\mathrm{C}}\right)=\frac{1+2\times 10}{20}=\frac{21}{20}=1.05$

Concept:

The spring index is defined as the ratio of the mean diameter of the coil to the diameter of the wire. Mathematically,

Spring index, $C=\frac{D}{d}$

Deflection of Helical Springs of Circular Wire:

$\delta =\frac{8W{D}^{3}n}{G{d}^4}=\frac{8W{C}^{3}n}{Gd}$

Stiffness of the spring or spring rate

$k=\frac{W}{\delta }=\frac{G{d}^4}{8W{D}^{3}n}$

$k=\frac{G{d}^4}{8D^{3}n}=\frac{G{d}^4}{8{\left(2R\right)}^{3}n}=\frac{G{d}^4}{64R^{3}n}$

Where D = Mean diameter of the spring coil, d = Diameter of the spring wire, n = Number of active coils, G = Modulus of rigidity for the spring material, W = Axial load on the spring

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combined direct shear and torsional shear stress in the coil wire cross-section.

Torsional shear stress in the bar:

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar:

${\tau }_2=\frac{P}{\frac{\pi }{4}{d}^2}=\frac{4P}{\pi d^2}$

Combining both equations

∴ τ = τ1 + τ2

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$

$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

Explanation:

Fig. (a) Pure torsional stress

Fig. (b) Direct shear stress

Fig. (c) Combine torsional, direct and curvature shear stress 

Torsional shear stress in the bar-

${\tau }_1=\frac{16M_t}{\pi d^3}$

M_t = Torque acting $\Rightarrow \mathrm{P}\times \frac{\mathrm{D}}{2}$, D = Mean coil diameter, d = Diameter of spring wire

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar –

${\tau }_2=\frac{P}{\frac{\pi }{4}{d}^2}=\frac{4P}{\pi d^2}$

∴ τ = τ₁ + τ₂

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

Important Points

When torsional shear stress, direct shear stress, and curvature shear stress is taken into consideration, then shear stress is given by 

$\tau =K\left(\frac{8PD}{\pi d^3}\right)\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{K}=\mathrm{W}\mathrm{a}\mathrm{h}\mathrm{l}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combine torsional, direct and curvature shear stress.

Torsional shear stress in the bar-

${\tau }_1=\frac{16M_t}{\pi d^3}$

${\mathrm{M}}_{\mathrm{t}}=\mathrm{T}\mathrm{o}\mathrm{r}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\Rightarrow \mathrm{P}\times \frac{\mathrm{D}}{2},\mathrm{D}=\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r},\mathrm{d}=\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{w}\mathrm{i}\mathrm{r}\mathrm{e}$

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar-

∴ τ = τ₁ + τ₂

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$
$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

When torsional shear stress, direct shear stress, and curvature shear stress is taken into consideration, then shear stress is given by-

$\tau =K\left(\frac{8PD}{\pi d^3}\right)\Rightarrow K\left(\frac{8PC}{\pi d^2}\right)K=Wah{l}^{\prime }sfactor$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$

Concept:

The spring index is defined as the ratio of the mean diameter of the coil to the diameter of the wire. Mathematically,

Spring index, $C=\frac{D}{d}$

Deflection of Helical Springs of Circular Wire:

$\delta =\frac{8W{D}^{3}n}{G{d}^4}=\frac{8W{C}^{3}n}{Gd}$

Stiffness of the spring or spring rate

$k=\frac{W}{\delta }=\frac{G{d}^4}{8W{D}^{3}n}$

$k=\frac{G{d}^4}{8D^{3}n}=\frac{G{d}^4}{8{\left(2R\right)}^{3}n}=\frac{G{d}^4}{64R^{3}n}$

Where D = Mean diameter of the spring coil, d = Diameter of the spring wire, n = Number of active coils, G = Modulus of rigidity for the spring material, W = Axial load on the spring

Concept:

The torsional shear stress in the wire is given by, $\tau =\frac{8\mathrm{W}\mathrm{D}}{\pi {\mathrm{d}}^3}$

When the wire is bent in the form of coil, the length of the inside fiber is less than the length of the outside fiber. This results in the stress concentration at the inside fiber of the coil.

Therefore, the shear stress in fiber after stress concentration factor is $\tau =\frac{8\mathrm{W}\mathrm{D}}{\pi {\mathrm{d}}^3}\left(\frac{1}{2\mathrm{C}}+1\right)$

where $\left(\frac{1}{2\mathrm{C}}+1\right)$ is stress concentration factor (K_c).

Therefore, K_c =  $\left(\frac{1+2C}{2\mathrm{C}}\right)$ where C = $\frac{D}{d}$ (∵ D = diameter of coil, d = diameter of wire)

Fig. (a) Pure torsional stress

Fig. (b) Direct shear stress

Fig. (c) Combine torsional, direct and curvature shear stress 

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combine torsional, direct and curvature shear stress.

Torsional shear stress in the bar-

${\tau }_1=\frac{16M_t}{\pi d^3}$

${\mathrm{M}}_{\mathrm{t}}=\mathrm{T}\mathrm{o}\mathrm{r}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\Rightarrow \mathrm{P}\times \frac{\mathrm{D}}{2},\mathrm{D}=\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r},\mathrm{d}=\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{w}\mathrm{i}\mathrm{r}\mathrm{e}$

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar-

∴ τ = τ1 + τ2

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$
$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

When torsional shear stress, direct shear stress, and curvature shear stress is taken into consideration, then shear stress is given by-

$\tau =K\left(\frac{8PD}{\pi d^3}\right)\Rightarrow K\left(\frac{8PC}{\pi d^2}\right)K=Wah{l}^{\prime }sfactor$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combined direct shear and torsional shear stress in the coil wire cross-section.

Torsional shear stress in the bar:

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar:

${\tau }_2=\frac{P}{\frac{\pi }{4}{d}^2}=\frac{4P}{\pi d^2}$

Combining both equations

∴ τ = τ1 + τ2

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$

$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

Explanation:

Fig. (a) Pure torsional stress

Fig. (b) Direct shear stress

Fig. (c) Combine torsional, direct and curvature shear stress 

Torsional shear stress in the bar-

${\tau }_1=\frac{16M_t}{\pi d^3}$

M_t = Torque acting $\Rightarrow \mathrm{P}\times \frac{\mathrm{D}}{2}$, D = Mean coil diameter, d = Diameter of spring wire

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar –

${\tau }_2=\frac{P}{\frac{\pi }{4}{d}^2}=\frac{4P}{\pi d^2}$

∴ τ = τ₁ + τ₂

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$

$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

Important Points

When torsional shear stress, direct shear stress, and curvature shear stress is taken into consideration, then shear stress is given by 

$\tau =K\left(\frac{8PD}{\pi d^3}\right)\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{K}=\mathrm{W}\mathrm{a}\mathrm{h}\mathrm{l}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$

Explanation:

Since the spring rod is bent into a curve, the shear stress on the inner surface is more than the shear stress developed on the outer surface. To take into account the curvature effect. Wahl’s correction factor may be applied and maximum shear stress can be calculated as follows.

${\tau }_{max}=K_w\left(\frac{16WR}{\pi d^3}\right)$

Where Kw is Wahl’s correction factor and is given by

$K_w=\left(\frac{4c-1}{4c-4}+\frac{0.615}{c}\right)$

Where ‘c’ is the spring index, D = Mean coil diameter, d = Diameter of spring wire

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combine torsional, direct and curvature shear stress.

Torsional shear stress in the bar-

${\tau }_1=\frac{16M_t}{\pi d^3}$

${\mathrm{M}}_{\mathrm{t}}=\mathrm{T}\mathrm{o}\mathrm{r}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\Rightarrow \mathrm{P}\times \frac{\mathrm{D}}{2},\mathrm{D}=\mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r},\mathrm{d}=\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{w}\mathrm{i}\mathrm{r}\mathrm{e}$

$\Rightarrow {\tau }_1=\frac{8PD}{\pi d^3}$

Direct shear stress in the bar-

∴ τ = τ₁ + τ₂

$\Rightarrow \frac{8PD}{\pi d^3}+\frac{4P}{\pi d^2}$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{d}{2D}\right)$
$\Rightarrow \frac{8PD}{\pi d^3}\left(1+\frac{1}{2C}\right),\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{C}=\frac{\mathrm{D}}{\mathrm{d}}=\mathrm{S}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{x}$

$\Rightarrow \frac{8PD}{\pi d^3}\left(\frac{2C+1}{2C}\right)$
$\Rightarrow K_s\left(\frac{8PD}{\pi d^3}\right)$

${\mathrm{K}}_{\mathrm{s}}=\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\Rightarrow \frac{2C+1}{2C}$

When torsional shear stress, direct shear stress, and curvature shear stress is taken into consideration, then shear stress is given by-

$\tau =K\left(\frac{8PD}{\pi d^3}\right)\Rightarrow K\left(\frac{8PC}{\pi d^2}\right)K=Wah{l}^{\prime }sfactor$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$