Problem 572

The T section shown in Fig. P-572 is the cross-section of a beam formed by joining two rectangular pieces of wood together. The beam is subjected to a maximum shearing force of 60 kN. Show that the NA is 34 mm from the top and the I_NA = 10.57 × 10⁶ mm⁴. Using these values, determine the shearing stress (a) at the neutral axis and (b) at the junction between the two pieces of wood.

Solution 572

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} \,\,            (ok!)

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 \,\,            (ok!)

(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