# Example 01: Maximum bending stress, shear stress, and deflection

**Problem**

A timber beam 4 m long is simply supported at both ends. It carries a uniform load of 10 kN/m including its own weight. The wooden section has a width of 200 mm and a depth of 260 mm and is made up of 80% grade Apitong. Use dressed dimension by reducing its dimensions by 10 mm.

Bending and tension parallel to grain = 16.5 MPa

Shear parallel to grain = 1.73 MPa

Modulus of elasticity in bending = 7.31 GPa

- What is the maximum flexural stress of the beam?
- What is the maximum shearing stress of the beam?
- What is the maximum deflection of the beam?

**Solution**

$F_v = 1.73 ~ \text{MPa}$

$E = 7.31 ~ \text{GPa}$

**Part 1: Maximum flexural stress**

$M = \dfrac{10(4^2)}{8}$

$M = 20 ~ \text{kN}\cdot\text{m}$

$f_b = \dfrac{6M}{bd^2}$

$f_b = \dfrac{6(20)(1000^2)}{190(250^2)}$

$f_b = 10.105 ~ \text{MPa}$ *answer*

$f_b \lt F_b$ (okay)

**Part 2: Maximum flexural stress**

$f_v = \dfrac{3V}{2bd}$

$f_v = \dfrac{3(20)(1000)}{2(190)(250)}$

$f_v = 0.6316 ~ \text{MPa}$ *answer*

$f_v \lt F_v$ (okay)

**Part 3: Maximum deflection**

$I = \dfrac{190(250^3)}{12}$

$I = 247\,395\,833 ~ \text{mm}^4$

$\delta = \dfrac{5w_o L^4}{384EI}$

$\delta = \dfrac{5(10)(4)(1000^4)}{384(7\,310)(247\,395\,833)}$

$\delta = 18.43 ~ \text{mm}$ *answer*