## 2 Web Bearing Capacity Design of Cold-Formed Austenitic Stainless-Steel Channels
### European Code  
In this code, a set of equations has been provided to determine the web-bearing capacity of unlipped channels under localised loading. Such equations are used to avoid crippling, crushing or buckling in a web subject to a support reaction or other local concentrated transverse force applied through the flanges.  
In the following, the design equations for single untiffened webs are given.  
The local transverse resistance of the web (\(R_{w,Rd}\)) for a cross-section with a single untstiffened web, is calculated from Equations (5)-(10) on a condition that the cross-section meets the criteria below:  
\[h_{w}/t\leq 200\]  
\[r/t\leq 6\]  
\[45^{\circ}\leq\varphi\leq 90^{\circ}\]  
where:  
\(h_{w}\) is the web height between the midlines of the flanges;  
\(r\) is the internal radius of the corners;  
\(\theta\) is the angle of the web relative to the flanges (degrees).  
From the EN 1993-1-3 standard, the local transverse resistance \(R_{w,Rd}\) of a single web under interior-one-flange (IOF) loading can be calculated from:  
For \(c>1.5h_{w}\) clear from a free end and for a cross-section with untstiffened flanges:  
If \(S_{S}/t\leq 60\):  
\[R_{w,Rd}=\frac{k_{3}k_{4}k_{5}\Big{[}14.7-\frac{h_{w}/t}{49.5}\Big{]}\Big{[}1 +0.007\frac{S_{S}}{t}\Big{]}t^{2}f_{yb}}{\gamma_{M1}} \tag{5}\]  
If \(S_{S}/t\geq 60\):  
\[R_{w,Rd}=\frac{k_{3}k_{4}k_{5}\Big{[}14.7-\frac{h_{w}/t}{49.5}\Big{]}\Big{[}0.75+0.011\frac{S_{t}}{t}\Big{]}t^{2}f_{yb}}{\gamma_{M1}} \tag{6}\]  
The local transverse resistance \(R_{w,Rd}\) of a single web under end-one-flange (EOF) loading can be calculated from:  
For \(c\leq 1.5h_{w}\) clear from a free end and for a cross-section with untstiffened flanges:If \(S_{S}/t\leq 60\):  
\[R_{w,Rd}=\frac{k_{1}k_{2}k_{3}\Big{[}5.92-\frac{h_{w}/t}{132}\Big{]}\Big{[}1+0.01 \frac{S_{S}}{t}\Big{]}t^{2}f_{yb}}{\gamma_{M1}} \tag{7}\]  
If \(S_{S}/t\geq 60\):  
\[R_{w,Rd}=\frac{k_{1}k_{2}k_{3}\Big{[}5.92-\frac{h_{w}/t}{132}\Big{]}\Big{[}0.71+ 0.015\frac{S_{S}}{t}\Big{]}t^{2}f_{yb}}{\gamma_{M1}} \tag{8}\]  
The local transverse resistance \(R_{w,Rd}\) of a single web under interior-two-flange (ITF) loading can be calculated from:  
For \(c>1.5h_{w}\) clear from a free end and for a cross-section with unstiffened flanges:  
\[R_{w,Rd}=\frac{k_{3}k_{4}k_{5}\Big{[}21.0-\frac{h_{w}/t}{16.3}\Big{]}\Big{[}1+ 0.0013\frac{S_{S}}{t}\Big{]}t^{2}f_{yb}}{\gamma_{M1}} \tag{9}\]  
The local transverse resistance \(R_{w,Rd}\) of a single web under end-two-flange (ETF) loading can be calculated from:  
For \(c\leq 1.5h_{w}\) clear from a free end and for a cross-section with unstiffened flanges:  
\[R_{w,Rd}=\frac{k_{1}k_{2}k_{3}\Big{[}6.66-\frac{h_{w}/t}{64}\Big{]}\Big{[}1+0. 01\frac{S_{S}}{t}\Big{]}t^{2}f_{yb}}{\gamma_{M1}} \tag{10}\]  
where:  
\(c\) is the distance from loading to the closest end of the beam;  
\(S_{s}\) is the nominal length of stiff bearing;  
\(f_{yb}\) is the yield strength of steel;  
\(\gamma_{M1}\) is the partial safety factor,  
\(\gamma_{M1}=1.00\).  
The values of the coefficients \(k_{1}\) to \(k_{5}\) should be determined as follows:  
\[k_{1}=(1.33-0.33k)\]  
\[k_{2}=(1.15-0.15\,r/t),\text{ but }k\geq 0.5\text{ and }k\leq 1.0\text{-change this to }K_{2}\]  
\[k_{3}=0.7+0.3\Big{(}\varphi/90)^{2}\]  
\[k_{4}=(1.22-0.22k)\]  
\[k_{5}=(1.06-0.06\,r/\,t)\text{ but }k_{5}\leq 1.0\]  
\[k=f_{yb}/228\ (f_{yb}\text{ in }\text{N}/\text{mm}^{2})\]