Example T-8: Block Shear in Flat Plate

This example is similar to one done in class, January 23, 2020. Only the dimensions are different.

Compute the factored tension resistance, $T_r$, of the flat plate tension member shown in the figure. Assume CSA G40.21 350W steel and M20 bolts in punched holes. For this excercise, consider only block shear.

For all of the following failure modes, we will use the relationship from Clause 13.11 of S16-14:

$$ \begin{align} T_r &= \phi_u \left[ U_t A_n F_u + 0.6 A_{gv} \frac{(F_y + F_u)}{2} \right]\\ \phi_u &= 0.75\\ F_y &= 350\ \mathrm{MPa}\\ F_u &= 450\ \mathrm{MPa}\\ \frac{(F_y + F_u)}{2} &= \frac{(350 + 450)}{2}\ \mathrm{MPa} \\ &= 400\ \mathrm{MPa}\\ d &= (20 + 2 + 2)\ \mathrm{mm} = 24\ \mathrm{mm}\ \ \ \ \mathrm{\ \ \ (hole\ allowance)}\\ \end{align} $$

Failure Path 1-1

For failure path 1-1. above, we have:

$$ \begin{align} A_n &= (75 - 12 - 12)\ \mathrm{mm} \times 10\ \mathrm{mm} \\ &= 510\ \mathrm{mm}^2\\ A_{gv} &= (40 + 75)\ \mathrm{mm} \times 10\ \mathrm{mm} \times 2 \\ &= 2300\ \mathrm{mm}^2\\ U_t &= 1.0 \ \ \ \ \mathrm{(symmetric\ and\ concentric)}\\ T_r &= 0.75 \left[1.0 \times 510\ \mathrm{mm}^2 \times 0.45 \frac{\mathrm{kN}}{\mathrm{mm}^2} + 0.6\times 2300\ \mathrm{mm}^2\times 0.40 \frac{\mathrm{kN}}{\mathrm{mm}^2}\right]\\ T_r &= 586\ \mathrm{kN}\\ \end{align} $$

Failure Path 2-2

For failure path 2-2, above, we have:

$$ \begin{align} A_n &= (75 + 30 - 24 -12)\ \mathrm{mm} \times 10\ \mathrm{mm} \\ &= 690\ \mathrm{mm}^2\\ A_{gv} &= (40 + 75)\ \mathrm{mm} \times 10\ \mathrm{mm} \\ &= 1150\ \mathrm{mm}^2\\ U_t &= 0.8\\ T_r &= 0.75 \left[0.8 \times 690\ \mathrm{mm}^2 \times 0.45 \frac{\mathrm{kN}}{\mathrm{mm}^2} + 0.6\times 1150\ \mathrm{mm}^2\times 0.40 \frac{\mathrm{kN}}{\mathrm{mm}^2}\right]\\ T_r &= 393\ \mathrm{kN}\ \ \ \ \ \ \ \ \ \ \ \ \leftarrow \mathbf{governs}\\ \end{align} $$

Note: using a value of 0.8 for $U_t$ is a judgement call. The exact case isn't in Figure 2-24. Case 3 is similar, but the eccentricity of load here is less than it is for case 3, so we should use a value greater than 0.6. Our case is slightly more eccentric than case 8, so we should use a value slightly less than 0.9.

Failure Path 3-3

For failure path 3-3, above, we have:

$$ \begin{align} A_n &= 2 \times (30-12)\ \mathrm{mm} \times 10\ \mathrm{mm} \\ &= 360\ \mathrm{mm}^2\\ A_{gv} &= (40 + 75)\ \mathrm{mm} \times 10\ \mathrm{mm} \times 2 \\ &= 2300\ \mathrm{mm}^2\\ U_t &= 0.9 \mathrm{\ \ \ (case\ 8\ in\ Figure\ 2-24)}\\ T_r &= 0.75 \left[0.9 \times 360\ \mathrm{mm}^2 \times 0.45 \frac{\mathrm{kN}}{\mathrm{mm}^2} + 0.6\times 2300\ \mathrm{mm}^2\times 0.40 \frac{\mathrm{kN}}{\mathrm{mm}^2}\right]\\ T_r &= 523\ \mathrm{kN}\\ \end{align} $$

Failure Path 4-4 (tearout)

Failure path 4-4 above is called tearout. We have:

$$ \begin{align} A_n &= 0\\ A_{gv} &= (40 + 75)\ \mathrm{mm} \times 10\ \mathrm{mm} \times 4 \\ &= 4600\ \mathrm{mm}^2\\ U_t &= n.a.\\ T_r &= 0.75 \left[U_t \times 0\ \mathrm{mm}^2 \times 0.45 \frac{\mathrm{kN}}{\mathrm{mm}^2} + 0.6\times 4600\ \mathrm{mm}^2\times 0.40 \frac{\mathrm{kN}}{\mathrm{mm}^2}\right]\\ T_r &= 828\ \mathrm{kN}\\ \end{align} $$

Summary

Path 2-2 governs, and

$$ \begin{align} T_r &= 393\ \mathrm{kN}\ \ \ \ \ \ \ \ \ \ \ \ \ \leftarrow \mathbf{Ans.}\\ \end{align} $$