What is the magnitude of axial load that cause 2 planks of 12 mm thick and 225 mm wide to fail if joined by dry mortise and used shears along its grain with average shearing stress is 8 MPa?

Answer 1

To determine the axial load that would cause failure, we can use the formula for shear stress, which is given by ( \tau = \frac{F}{A} ), where ( \tau ) is the shear stress, ( F ) is the force (axial load), and ( A ) is the cross-sectional area. The area ( A ) for two planks joined by dry mortise is calculated as ( A = 2 \times \text{width} \times \text{thickness} = 2 \times 225 , \text{mm} \times 12 , \text{mm} = 5400 , \text{mm}^2 ) or ( 5.4 \times 10^{-3} , \text{m}^2 ). Given the average shearing stress is 8 MPa, we can rearrange the formula to find ( F = \tau \times A = 8 , \text{MPa} \times 5.4 \times 10^{-3} , \text{m}^2 = 43.2 , \text{kN} ). Thus, the axial load that would cause failure is approximately 43.2 kN.