Question

In a T-section as shown figure below, the distance between elastic neutral axis and plastic neutral axis will be ________.

Answer 1

$\text{Let distance of elastic neutral axis is}{y}_c\text{from bottom}$

$\therefore y_c=\frac{(100\times 500\times \frac{500}{2})+(400\times 100\times 550)}{400\times 100\times +100\times 500}$

$=383.33mm$

$\text{Let}{y}_p\text{is distance of plastic neutral}\text{axis from bottom}$

$\therefore 100\times y_p=400\times 100+(500-y_p)\times 100$

$\Rightarrow 200y_p=400\times 100+500\times 100$

$y_p=450mm$

$\therefore |y_p-y_3|=450-383.33=66.67mm$