$A_1 = 200(40) = 8000 \, \text{mm}^2; \,\, y_1 = 20 \, \text{mm}$
$A_2 = 20(100) = 2000 \, \text{mm}^2; \,\, y_2 = 90 \, \text{mm}$
$A = A_1 + A_2 = 8000 + 2000$
$A = 10 \, 000 \, \text{mm}^2$
$A\bar{y} = \Sigma Ay$
$10\,000\bar{y} = 8000(20) + 2000(90)$
$\bar{y} = 34 \, \text{ mm}$ (okay!)
By transfer formula for moment of inertia
$I = \bar{I} + Ad^2$
$I = \frac{1}{12}bh^3 + Ad^2$
$I_1 = \frac{1}{12}(200)(40^3) + 8000(14^2) = 2\,634\,666.67 \, \text{mm}^4$
$I_2 = \frac{1}{12}(20)(100^3) + 2000(56^2) = 7\,938\,666.67 \, \text{mm}^4$
Thus,
$I_{NA} = I_1 + I_2$
$I_{NA} = 2\,634\,666.67 + 7\,938\,666.67$
$I_{NA} = 10\,573\,333.34 \, \text{mm}^4$
$I_{NA} = 10.57 \times 10^6 \, \text{ mm}^4$ (okay!)
(a) At the Neutral Axis
$Q_{NA} = 200(34)(17) = 115\,600 \, \text{mm}^3$
$V = 60(1000) = 60\,000 \, \text{N}$
$(\,f_v\,)_{NA} = \dfrac{VQ}{Ib} = \dfrac{60\,000(115\,600)}{(10.57 \times 10^6)(200)}$
$(\,f_v\,)_{NA} = 3.28 \, \text{ MPa}$ answer
(b) At the junction between the two pieces of wood
$f_v = \dfrac{VQ}{Ib}$
$Q = 100(20)(56) = 112\,000 \, \text{mm}^3$
$V = 60(1000) = 60\,000 \, \text{N}$
Flange:
$b = 200 \, \text{mm}$
$(\,f_v\,)_{flange} = \dfrac{60\,000(112\,000)}{(10.57 \times 10^6)(200)}$
$(\,f_v\,)_{flange} = 3.1788 \, \text{ MPa}$ answer
Web:
$b = 20 \, \text{mm}$
$(\,f_v\,)_{web} = \dfrac{60\,000(112\,000)}{(10.57 \times 10^6)(20)}$
$(\,f_v\,)_{web} = 31.7881 \, \text{MPa}$ answer
