Question

What is the distance between the kerns for the section given below?

• A. 100 units
• B. > 100 units, < 150 units
• C. < 100 units
• D. > 150 units

Answer 1

The correct option is C < 100 units


$\begin{array}{cc}{k}_b& =\text{bottom kern distance}\\ k_t& =\text{top kern distance}\\ k_t& =\frac{I}{A{y}_b}and{k}_b=\frac{I}{A{y}_t}\\ & \text{But as the}I\text{beam is symmetrical hence}\\ y_t& =y_b=y\\ \therefore k_t+k_b& =\frac{I}{Ay}+\frac{I}{Ay}=\frac{2I}{Ay}\\ & =2\times \frac{[\frac{1}{12}\times 100\times (300{)}^3-\frac{1}{12}\times (100-50)\times \left(200{)}^3\right]}{\left[\right(2\times 50\times 100)+(200\times 50\left)\right]\times 150}\\ & =127.77units\end{array}$