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

[math]A_1 = 200(40) = 8000 \, \text{mm}^2; \,\, y_1 = 20 \, \text{mm}[/math]
[math]A_2 = 20(100) = 2000 \, \text{mm}^2; \,\, y_2 = 90 \, \text{mm}[/math]

[math]A = A_1 + A_2 = 8000 + 2000[/math]
[math]A = 10 \, 000 \, \text{mm}^2[/math]

[math]A\bar{y} = \Sigma Ay[/math]
[math]10\,000\bar{y} = 8000(20) + 2000(90)[/math]
[math]\bar{y} = 34 \, \text{mm} \,\,[/math]            (ok!)

By transfer formula for moment of inertia
[math]I = \bar{I} + Ad^2[/math]
[math]I = \frac{1}{12}bh^3 + Ad^2[/math]

[math]I_1 = \frac{1}{12}(200)(40^3) + 8000(14^2) = 2\,634\,666.67 \, \text{mm}^4[/math]

[math]I_2 = \frac{1}{12}(20)(100^3) + 2000(56^2) = 7\,938\,666.67 \, \text{mm}^4[/math]

Thus,
[math]I_{NA} = I_1 + I_2[/math]
[math]I_{NA} = 2\,634\,666.67 + 7\,938\,666.67[/math]
[math]I_{NA} = 10\,573\,333.34 \, \text{mm}^4[/math]
[math]I_{NA} = 10.57 \times 10^6 \, \text{mm}^4 \,\,[/math]            (ok!)

(a) At the Neutral Axis
[math]Q_{NA} = 200(34)(17) = 115\,600 \, \text{mm}^3[/math]
[math]V = 60(1000) = 60\,000 \, \text{N}[/math]

[math](\,f_v\,)_{NA} = \dfrac{VQ}{Ib} = \dfrac{60\,000(115\,600)}{(10.57 \times 10^6)(200)}[/math]
[math](\,f_v\,)_{NA} = 3.28 \, \text{MPa} \,\, [/math]           answer

(b) At the junction between the two pieces of wood
[math]f_v = \dfrac{VQ}{Ib}[/math]

[math]Q = 100(20)(56) = 112\,000 \, \text{mm}^3[/math]
[math]V = 60(1000) = 60\,000 \, \text{N}[/math]

Flange:
[math]b = 200 \, \text{mm}[/math]

[math](\,f_v\,)_{flange} = \dfrac{60\,000(112\,000)}{(10.57 \times 10^6)(200)}[/math]

[math](\,f_v\,)_{flange} = 3.1788 \, \text{MPa} \,\,[/math]            answer

Web:
[math]b = 20 \, \text{mm}[/math]

[math](\,f_v\,)_{web} = \dfrac{60\,000(112\,000)}{(10.57 \times 10^6)(20)}[/math]

[math](\,f_v\,)_{web} = 31.7881 \, \text{MPa} \,\,[/math]            answer