Timber beam: value of W for stress limit

Given: 160 mm × 300 mm timber beam, $L = 5$ m, UDL $w = 1$ kN/m, three equal point loads $W$ at 1.25 m, 2.5 m and 3.75 m (quarter points and mid-span). Permissible $\sigma = 10\ \text{N/mm}^2$. Self-weight is ignored.

         W     W     W
 1 kN/m  v     v     v
 vvvvvvvvvvvvvvvvvvvvvvvv
 A ^-----|-----|-----|-----^ B
     1.25  1.25  1.25  1.25

Moment of resistance

$$Z = \frac{bd^2}{6} = \frac{160\times300^2}{6} = 2.4\times10^6\ \text{mm}^3$$ $$M_r = \sigma Z = 10\times2.4\times10^6 = 24\times10^6\ \text{Nmm} = 24\ \text{kNm}$$

Maximum bending moment (at mid-span, by symmetry)

Reaction:

$$R_A = R_B = \frac{1\times5 + 3W}{2} = 2.5 + 1.5W$$ $$M_{max} = R_A\times2.5 - W\times1.25 - \frac{1\times2.5^2}{2} = 6.25 + 3.75W - 1.25W - 3.125$$ $$M_{max} = 3.125 + 2.5W$$

Value of W

$$3.125 + 2.5W = 24 \Rightarrow W = \frac{24 - 3.125}{2.5} = 8.35\ \text{kN}$$

Answer: $W = 8.35$ kN.