Question

The shape factor of the T section shown in figure below is

• A. 1.8

Answer 1

The correct option is A 1.8
Depth of N.A. from top
$\overline{y}=\frac{A_1{Y}_1+A_2{Y}_2}{A_1+A_2}$
$\overline{y}=\frac{10\times 1\times 5+(10-1)\times 1\times 9.5}{10\times 1+(10-1)\times 1}$
$\overline{y}=\frac{50+85.5}{10+9}$
$=\frac{135.5}{19}=7.13cm$
$I=\frac{1\times {10}^3}{12}+1\times 10\times (7.13-5{)}^2+\frac{9\times 1^3}{12}+9\times 1\times {2.37}^2$
$I=\frac{1\times {10}^3}{12}+10\times {2.13}^2+0.75+9\times {2.37}^2$
$I=\frac{1\times {10}^3}{12}+1\times 10\times (7.13-5{)}^2\times 9\times \frac{13}{12}+79\times 1\times {2.37}^2$
$I=\frac{250}{3}+10\times {2.13}^2+0.75+9\times {2.37}^2$
$I=83.33+45.37+0.75+50.55$
$I=180.00c{m}^4$
$My=fy.\frac{I}{{\overline{y}}_{max}}=fy\times \frac{180}{7.13}$
Half area will be $9.5c{m}^2$ the N.A. will pass through the horizontal at $0.95cm$ from bottom.
$My=9\times f_y\times (4.5+0.05+\frac{0.95}{2})+10\times 0.05$
$(\frac{0.05}{2}+\frac{0.95}{2})fy$
$My=\frac{9\times 10.05}{2}\times fy+\frac{0.5\times 1}{2}fy$
$My=45.475fy$
$=\frac{45.475fy}{\frac{180}{7.3}fy}$
$S.F.=1.84$