Question 14.2: A cross-section of a slit rectangular tube of uniform thickn...

The above problem can be solved from the first principle as we did in our previous problem. Else, we can take the previous result and find the distance e by taking the length b_3 of Figure 14.21 as b_2 / 2 .
Putting b_3=b_2 / 2 in the previous example’s result, we get

e=\frac{3 b_1 b_2^2\left(b_1+b_2\right)-8 b_1\left(b_2 / 2\right)^3}{b_2^2\left(b_2+6 b_1+3 b_2\right)+4\left(b_2 / 2\right)^2\left(b_2-3 b_2\right)}

=\frac{3 b_1 b_2^2\left(b_1+b_2\right)-b_1 b_2^3}{b_2^2\left(6 b_1+4 b_2\right)+b_2^2\left(-2 b_2\right)}

=\frac{b_1 b_2^2\left(3 b_1+3 b_2-b_2\right)}{b_2^2\left(6 b_1+4 b_2-2 b_2\right)}

or              e=\frac{b_1\left(3 b_1+2 b_2\right)}{2\left(3 b_1+b_2\right)}