Example T-7: Capacity of Double Channel Tension Member

Compute the factored tension resistance, $T_r$, of the double channel tension member shown in the figure.
Assume CSA G40.21 300W steel and M20 bolts in punched holes.

Calculate the capacity of one channel. We have the following data:

$$ \begin{align} F_y &= 300~\mathrm{MPa}\\ F_u &= 450~\mathrm{MPa}\\ A &= 2840~\mathrm{mm}^2\\ w &= 7.2~\mathrm{mm}\\ \end{align} $$

I) Gross Section Yielding:

$$ \begin{align} T_r &= \phi A_g F_y \\ &= 0.9 \times 2840~\mathrm{mm}^2 \times 0.300~\mathrm{kN/mm}^2 \\ &= 767~\mathrm{kN}\\ \end{align} $$

II) Net Section Fracture:

$$ \begin{align} A_n &= 2840~\mathrm{mm}^2 - 2\times 24~\mathrm{mm}\times 7.2~\mathrm{mm} \\ &= 2494~\mathrm{mm}^2\\ A_{ne} &= 0.75 A_n ~~~~~~~~~~~~~~~~~~~~~~~\S \text{12.3.3.2 c) ii)}\\ &= 0.75 \times 2494~\mathrm{mm}^2 \\ &= 1871~\mathrm{mm}^2\\ T_r &= \phi_u A_{ne} F_u \\ &= 0.75 \times 1871~\mathrm{mm}^2 \times 0.450~\mathrm{kN/mm}^2 \\ &= 631~\mathrm{kN}\\ \end{align} $$

III) Block Shear:

$$ \begin{align} T_r &= \phi_u \left[ U_t A_n F_u + 0.6 A_{gv} {(F_y+F_u)\over 2} \right]\\ {(F_y+F_u)\over 2} &= {(300~\mathrm{MPa} + 450~\mathrm{MPa})\over 2} \\ &= 375~\mathrm{MPa}\\ \end{align} $$

III-1) Path 1-1:

$$ \begin{align} A_n &= (100~\mathrm{mm} - 24~\mathrm{mm})\times7.2~\mathrm{mm} \\ &= 547.2~\mathrm{mm}^2\\ A_{gv} &= 2\times(60~\mathrm{mm}+90~\mathrm{mm})\times 7.2~\mathrm{mm} \\ &= 2160~\mathrm{mm}^2\\ U_t &= 1.0\\ T_r &= 0.75\left[1.0\times547.2~\mathrm{mm}^2\times0.450~\mathrm{KN/mm}^2 \right.\\ &\qquad\left. {} + 0.6\times2160~\mathrm{mm}^2\times0.375~\mathrm{KN/mm}^2\right] \\ &= 549~\mathrm{kN}\\ \end{align} $$

III-2) Path 2-2:

$A_n$ is calculated by subtracting the area between the holes, including the width of the holes, from the tabulated gross cross-sectional area of the channel. That way we don't have to worry about the area of fillets, tapered flanges, etc.

$$ \begin{align} A_n &= 2840~\mathrm{mm}^2 - (100~\mathrm{mm} + 24~\mathrm{mm})\times7.2~\mathrm{mm} \\ &= 1947~\mathrm{mm}^2\\ A_{gv} &= 2\times(60~\mathrm{mm}+90~\mathrm{mm})\times 7.2~\mathrm{mm} \\ &= 2160~\mathrm{mm}^2\\ U_t &= 0.9\\ T_r &= 0.75 \left[0.9\times1947~\mathrm{mm}^2\times0.450~\mathrm{KN/mm}^2 \right.\\ &\qquad\left. {} + 0.6\times2160~\mathrm{mm}^2\times0.375~\mathrm{KN/mm}^2\right] \\ &= 1147~\mathrm{kN}\\ \end{align} $$

We use $U_t=0.9$ above as there is no comparable in Figure 2-26 in the handbook. However the stress distribution on the tension face will be similar to that of case 6a in that table.

III-3) Path 3-3:

$A_n$ is calculated by subtracting 1/2 of the net area calculated above from the tabulated gross cross-sectional area of the channel. That gives the area remaining, which is then adjusted for the hole widths.

$$ \begin{align} A_n &= 2840~\mathrm{mm}^2 - 1947~\mathrm{mm}^2/2 - 2\times24~\mathrm{mm}\times7.2~\mathrm{mm} \\ &= 1521~\mathrm{mm}^2\\ A_{gv} &= 1\times(60~\mathrm{mm}+90~\mathrm{mm})\times 7.2~\mathrm{mm} \\ &= 1080~\mathrm{mm}^2\\ U_t &= 0.6\\ T_r &= 0.75 \left[0.6\times1521~\mathrm{mm}^2\times0.450~\mathrm{KN/mm}^2 \right.\\ &\qquad\left. {} + 0.6\times1080~\mathrm{mm}^2\times0.375~\mathrm{KN/mm}^2\right] \\ &= 490~\mathrm{kN}~~~~~~~~~~~~~~~~~~~~~~\leftarrow\mathbf{governs}\\ \end{align} $$

I think $U_t = 0.6$ is too conservative. It is not reasonable that the strength of this failure path is so much less than net section fracture (case II above). Compared to that, this case loses $1947/2 = 974mm^2$ in tension area but gains $1080mm^2$ in shear area.

III-4) Path 4-4:

$$ \begin{align} A_n &= 0\\ A_{gv} &= 4\times(60~\mathrm{mm}+90~\mathrm{mm})\times 7.2~\mathrm{mm} \\ &= 4320~\mathrm{mm}^2\\ U_t &= n.a.\\ T_r &= 0.75 \left[ 0.6\times4320~\mathrm{mm}^2\times0.375~\mathrm{KN/mm}^2\right] \\ &= 729~\mathrm{kN}\\ \end{align} $$

Summary

For 2 channels:

$$ \begin{align} T_r &= 2 \times 490~\mathrm{kN} = 980~\mathrm{kN}~~~~~~~~~~\leftarrow\mathbf{Ans.}\\ \end{align} $$