Problem 3.10. Shear correction factor

The shear correction factor is, n =

Step 1. Calculate section properties, A, I, y_c.

Check section propertiesHide section properties

Step 2. Determine shear stress distribution the height of the cross section.

Hint: For simplicity we treat separately the web and the flange using different coordinate systems, the origin of which are located at the top or at the bottom of the cross section, respectively.

Hide shear stress funtcion

Shear stress can be determined by the Zhuravskii formula.

Eq.(3-34)

The function is different in the flange and web of the T section. For simplicity we use different coordinate systems.

For the flange:

$$\begin{gathered} S_y=b_{\mathrm{f}}y(y_{\mathrm{c}}–\frac{y}{2}) \\ {\tau }_{xy}=\frac{V{S}_y}{b_{\mathrm{f}}I}=V\frac{y(y_{\mathrm{c}}–{\displaystyle \frac{y}{2}})}{I} \end{gathered}$$

For the web:

$$\begin{gathered} S_{\eta }=b_{\mathrm{w}}\eta (h–y_{\mathrm{c}}–\frac{\eta }{2}) \\ {\tau }_{\xi \eta }=\frac{V{S}_y}{b_{\mathrm{f}}I}=V\frac{\eta (h–y_{\mathrm{c}}–\frac{\eta }{2})}{I} \end{gathered}$$

Step 3.  Sketch shear stress distribution.

The shear stress is uniform along the width, the distribution given above is shown in the Figure.

Check shear stress diagramHide shear shear stress diagram

Step 4. Integrate the shear stress function and the square of the function over the area of the cross section. Determine the shear correction factor. 

Check shear correction factorHide shear correction factor

Shear stiffness is determined using the shear correction factor. The shear correction factor is calculated as:

See Eqs.(3-59) and (3-60)

$$\begin{gathered} S=\frac{GA}{n} \\ n=\frac{A{\int }_A{\tau }_{xy}^{2}dA}{{\left({\int }_A{\tau }_{xy}^{}dA\right)}^2}=A\frac{b_{\mathrm{f}}{\int }_0^{h_{\mathrm{f}}}{\tau }_{xy}^{2}dy+b_{\mathrm{w}}{\int }_0^{h–h_{\mathrm{f}}}{\tau }_{\xi \eta }^{2}d\eta }{{\left(b_{\mathrm{f}}{\int }_0^{h_{\mathrm{f}}}{\tau }_{xy}^{}dy+b_{\mathrm{w}}{\int }_0^{h–h_{\mathrm{f}}}{\tau }_{\xi \eta }^{}d\eta \right)}^2}=A\frac{I_1+I_2}{{\left(I_3+I_4\right)}^2} \end{gathered}$$

Where we perform the integrations separately over the flange and the web:$$\begin{gathered} I_1=b_{\mathrm{f}}{\int }_0^{h_{\mathrm{f}}}{\tau }_{xy}^{2}dy=b_{\mathrm{f}}\left(y_{\mathrm{c}}^2\frac{h_{\mathrm{f}}^3}{3}+\frac{h_{\mathrm{f}}^5}{20}–y_{\mathrm{c}}\frac{h_{\mathrm{f}}^4}{4}\right)=2.15\times {10}^{11} {\mathrm{mm}}^6 \\ {I}_2=b_{\mathrm{w}}{\int }_0^{h–h_{\mathrm{f}}}{\tau }_{\xi \eta }^{2}d\eta =b_{\mathrm{w}}\left({\left(h–y_{\mathrm{c}}^{}\right)}^2\frac{{\left(h–h_{\mathrm{f}}^{}\right)}^3}{3}+\frac{{\left(h–h_{\mathrm{f}}^{}\right)}^5}{20}–\left(h–y_{\mathrm{c}}\right)\frac{{\left(h–h_{\mathrm{f}}^{}\right)}^4}{4}\right)=6.33\times {10}^{12} {\mathrm{mm}}^6 \\ {I}_3=b_{\mathrm{f}}{\int }_0^{h_{\mathrm{f}}}{\tau }_{xy}^{}dy=b_{\mathrm{f}}\left(y_{\mathrm{c}}^{}\frac{h_{\mathrm{f}}^2}{2}–\frac{h_{\mathrm{f}}^3}{6}\right)=5.77\times {10}^7 {\mathrm{mm}}^4 \\ {I}_4=b_{\mathrm{w}}{\int }_0^{h–h_{\mathrm{f}}}{\tau }_{\xi \eta }^{}d\eta =b_{\mathrm{w}}\left(\left(h–y_{\mathrm{c}}^{}\right)\frac{{\left(h–h_{\mathrm{f}}^{}\right)}^2}{2}–\frac{{\left(h–h_{\mathrm{f}}^{}\right)}^3}{6}\right)=5.28\times {10}^{8 }{\mathrm{mm}}^4 \end{gathered}$$

The shear correction factor results in:

$\mathit{n}=A\frac{I_1+I_2}{{\left(I_3+I_4\right)}^2}=\mathbf{1}\mathbf{.}\mathbf{334}$

The shear stiffness of the cross section is:

$S=\frac{GA}{n}=\frac{GA}{1.334}$

First the properties, A, I, y_{c are calculated.}

$$\begin{gathered} A=b_{\mathrm{w}}h+\left(b_{\mathrm{f}}–b_{\mathrm{w}}\right)h_{\mathrm{f}}=200\times 300+200\times 50=70000{\mathrm{mm}}^2 \\ {y}_{\mathrm{c}}=\frac{b_{\mathrm{w}}h\times h/2+\left(b_{\mathrm{f}}–b_{\mathrm{w}}\right)h_{\mathrm{f}}\times h_{\mathrm{f}}/2}{A}=\frac{200\times 300\times 150+200\times 50\times 25}{70000}=132.14 \mathrm{mm} \\ I=\frac{b_{\mathrm{w}}{h}^3}{12}+b_{\mathrm{w}}h{\left(\frac{h}{2}–y_{\mathrm{c}}\right)}^2+\frac{\left(b_f\mathit{–}{b}_w\right)h_f^{\mathit{3}}}{\mathit{12}}+\left(b_{\mathrm{f}}\mathit{–}{b}_{\mathrm{w}}\right)h_f^{}{\left(y_{\mathrm{c}}–\frac{h_{\mathrm{f}}}{2}\right)}^2= \\ =\frac{200\times {300}^3}{12}+200\times 300\times 17.{86}^2+\frac{200\times {50}^3}{12}+200\times 50\times 107.{14}^2=5.86\times {10}^8{\mathrm{mm}}^4 \end{gathered}$$

Shear stress can be determined by the Zhuravskii formula.

Eq.(3-34)

The moment of area, S_y and the shear stress, τ are expressed parametrically along the height of the cross section. The function is different in the flange and web of the T section. 

Hint: For simplicity we treat separately the web and the flange using different coordinate systems, the origin of which are located at the top or at the bottom of the cross section, respectively.

For the flange:

$$\begin{gathered} S_y=b_{\mathrm{f}}y(y_{\mathrm{c}}–\frac{y}{2}) \\ \tau =\frac{V{S}_y}{b_{\mathrm{f}}I}=V\frac{y(y_{\mathrm{c}}–{\displaystyle \frac{y}{2}})}{I} \end{gathered}$$

For the web:

$$\begin{gathered} S_{\eta }=b_{\mathrm{w}}\eta (h–y_{\mathrm{c}}–\frac{\eta }{2}) \\ \tau =\frac{V{S}_y}{b_{\mathrm{f}}I}=V\frac{\eta (h–y_{\mathrm{c}}–\frac{\eta }{2})}{I} \end{gathered}$$

The shear stress is uniform along the width, the distribution given above is shown in the Figure.

Shear stiffness is determined using the shear correction factor. To determine the shear correction factor the shear stress function and the square of the function must be integrated over the area of the cross section.  

See Eqs.(3-59) and (3-60)

We perform the integrations separately over the flange and the web:undefined

The shear correction factor results in:

$\mathit{n}=A\frac{I_1+I_2}{{\left(I_3+I_4\right)}^2}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{334}$

The shear stiffness of the cross section is:

$S=\frac{GA}{n}=\frac{GA}{1.334}$