Diagonal shear test and flat-jack test calculations

(a) Diagonal compression (shear) test

For a square panel (ASTM E519), the load $P$ along the diagonal produces pure shear along the other diagonal. The shear stress is

$$f_v = \frac{0.707\,P}{A_n}, \qquad A_n = \frac{w + h}{2}\,t\,n$$

where $w$ and $h$ are the width and height of the panel, $t$ the thickness, and $n$ the fraction of solid area (1 for a solid wall).

$$\begin{aligned} A_n &= \frac{1200 + 1200}{2} \times 230 \times 1 = 276\,000\ \text{mm}^2 \\ f_v &= \frac{0.707 \times 160\,000}{276\,000} = 0.41\ \text{N/mm}^2 \end{aligned}$$

(b) Single flat-jack test

A slot is cut in a mortar bed joint, which relieves the vertical stress and reduces the distance between gauge points. A flat jack is inserted and pressurised until the original distance is restored. The in-situ stress is

$$\sigma_m = K_m K_a\, p$$

where $p$ is the jack pressure, $K_m$ the jack constant (stiffness of the jack) and $K_a$ the ratio of jack area to slot area.

$$\sigma_m = 0.74 \times 0.85 \times 0.62 = 0.39\ \text{N/mm}^2$$

(c) Push shear test

A brick is freed by removing the head joints, and a jack pushes it along the bed joints. The shear area is that of the two bed joints:

$$A = 2 \times 230 \times 110 = 50\,600\ \text{mm}^2, \qquad \tau = \frac{14\,000}{50\,600} = 0.277\ \text{N/mm}^2$$

If there is an overburden stress $\sigma_v$, the initial shear strength is corrected as $f_{v0} = \tau - \mu\,\sigma_v$ (with friction coefficient $\mu \approx 0.4$–$0.6$).

Answer: (a) 0.41 N/mm²; (b) 0.39 N/mm²; (c) 0.28 N/mm².