# Calculating Admissible Shear stress with a uniformly distributed load and a point load

I have the following equation for calculating the Admissible shear stress for a uniformly distributed load.

$$\tau_{adm} = \frac{3}{2}\cdot \frac{F\cdot L}{2b\cdot h}$$

where

• $$F$$ = kN/m
• $$L$$ = Effective length of beam
• $$b$$ = width of timber
• $$h$$ = depth of timber

I am struggling to find out how this formula was derived. And secondly how this formula would need to be amended to include a point load.

Shear stress in beam section coming from lateral force is not uniform across the section. Derivation comes from Zhuravskii. The local lateral force basically increases bending moment when you move a little bit along the beam axis. So you can investigate bending stresses in consecutive sections.

Not the trick is to cut the part between 2 sections horizontally, add the unknown shear stress $$\tau$$ and calculate it from the equilibrium.

Thea equilibrium would be: $$- \int\limits_{\gamma_\tau} \sigma_x(x, z)dydz + \int\limits_{\gamma_\tau} \sigma_x(x+dx, z)dydz - \tau\cdot b\cdot dx = 0$$

Stresses directly from moments $$M_o$$ cancel out, and what remains comes just from lateral force $$T$$:

$$\tau \cdot b\cdot dx = \frac{T\cdot dx}{I_y}\cdot \int\limits_{\gamma_\tau} z dydz$$

This can be simplified to: $$\tau = \frac{T\cdot U_\tau}{b\cdot I_y}$$

where:

• $$T$$ is the local lateral force
• $$U_\tau$$ is the first moment of the area above the horizontal section
• $$I_y$$ is the second moment of area for the whole section
• $$b$$ is local section width at the investigated coordinate $$z$$

which is basically the Zhuravskii shear stress formula.

In your case: $$I_y = \frac{b\cdot h^3}{12}$$ The highest shear stress will be at the section centroid, so you need static moment for half of the section: $$U_\tau = A_\tau\cdot z_\tau = \left(\frac{h}{2}\cdot b\right)\cdot \frac{h}{4} = \frac{b\cdot h^2}{8}$$ The width will be of course $$b$$, so: $$\tau = \frac{T\cdot U_\tau}{b\cdot I_y} = \frac{T\cdot \frac{b\cdot h^2}{8}}{b\cdot \frac{b\cdot h^3}{12}} = \frac{3}{2}\cdot \frac{T}{b\cdot h} = \frac{3}{2}\cdot \frac{T}{A}$$

$$T$$ is probably $$F\cdot \frac{L}{2}$$ in your formula, so the question is how will adding the point load change the critical lateral force.

Edit: Admissible stress $$\tau_{adm}$$ should be less than maximum $$\tau_{max}$$, so it is possible that in your case, 2 is safety factor. This would make sense, because in uniformly loaded cantilever beam, the maximum lateral force would be $$T_{max} = F\cdot L$$;

$$\tau_{adm} = \frac{\tau_{max}}{2} = \frac{\frac{3}{2}\cdot \frac{T_{max}}{A}}{2}$$