$A_1 = 200(40) = 8000 \, \text{mm}^2; \,\, y_1 = 20 \, \text{mm}$

$A_2 = 20(100) = 2000 \, \text{mm}^2; \,\, y_2 = 90 \, \text{mm}$

$A = A_1 + A_2 = 8000 + 2000$

$A = 10 \, 000 \, \text{mm}^2$

$A\bar{y} = \Sigma Ay$

$10\,000\bar{y} = 8000(20) + 2000(90)$

$\bar{y} = 34 \, \text{ mm}$ (*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*

Hello po mga ate's and kuya's, quick question lang po, if you where to get the shear stress at 50mm above the bottom of the beam, what value of b po ang gagamitin?

Ang gamitin mo na b sa pagkuha ng shear ay yung width kung saan tumama ang inyong plane, in your case, ang plane mo ay 50 mm above the bottom. Sa location na yan, ang width ng beam ay 20 mm, so gamitin mo ay b = 20 mm.