By calculating shear stress in leg AB, determine the shear centre location for the following section shown in Figure 14.35(a).

Question

Accepted Answer

Noting the symmetry of the section shown in Figure 14.35(a) and the geometry of the leg BC shown in Figure 14.35(b), we find

[latex] \bar{I}_{z z}=2\left[\left(\bar{I}_{ AB } ight)_{z z}+\left(\bar{I}_{ BD } ight)_{z z} ight] [/latex]

[latex] =2\left[(30 t)(40)^2+\frac{1}{12}\left\lgroup \frac{t}{0.8} ight group (40)^3+\left\lgroup \frac{t}{0.8} ight group (40)(20)^2 ight] mm ^4 [/latex]

or [latex] \bar{I}_{z z}=149333.33 t mm ^4 [/latex] (1)

In the next step, we calculate shear stress [latex] au_{x z} [/latex] in leg AB as

[latex] \left|\left( au_{x y} ight)_{ AB } ight|=\left|\frac{V_y Q_z}{\bar{I}_{z z}} ight| [/latex]

where [latex] Q_z [/latex] is the first moment about z-axis of the shaded area shown in Figure 14.36(a):

Thus, [latex] Q_z [/latex] is given by

[latex] Q_z=40 s t mm ^3 [/latex]

Therefore,

[latex] au_{x z}=\frac{40 V_y s t}{t \bar{I}_{z z}}=40 \frac{V_y}{\bar{I}_{z z}} s(\leftarrow) ; \quad 0 \leq s \leq 30 mm [/latex]

Now, force [latex] F_1 [/latex] in leg AB is

[latex] F_1=\int_0^{30}\left( au_{x z} ight) t d s(\leftarrow)=(40) \frac{V_y t}{\bar{I}_{z z}} \int_0^{30} s d s=20 \frac{V_y t}{\bar{I}_{z z}}(30)^2=18000 \frac{V_y t}{\bar{I}_{z z}} [/latex]

From Figure 14.36(b), applying Varignon’s theorem of moments, we obtain

[latex] V_y e=80 F_1=(18)(8)\left(10^4 ight) \frac{V_y t}{149333.33 t} [/latex]

or e = 9.64 mm

Question 14.7: By calculating shear stress in leg AB, determine the shear c...

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