$A_s = 3 \times \frac{1}{4}\pi(32^2) = 768\pi ~ \text{mm}^2$

$nA_s = 9(768\pi) = 6912\pi ~ \text{mm}^2$

Assume NA is at the bottom of the flange

$Q_{\text{above NA}} = 600(80)(40) = 1\,920\,000 ~ \text{mm}^3$

$Q_{\text{below NA}} = nA_s(500 - 80) = 9\,201\,169 ~ \text{mm}^3$

Q_{above NA} < Q_{below NA}, therefore, NA is within the web.

$Q_{\text{above NA}} = Q_{\text{below NA}}$

$600x(\frac{1}{2}x) - 300(x - 80)[ \, \frac{1}{2}(x - 80) \, ] = nA_s(500 - x)$

$300x^2 - 150(x - 80)^2 = 6912\pi(500 - x)$

$x = 167 ~ \text{mm}$

$I_{NA} = \dfrac{600x^3}{3} - \dfrac{300(x - 80)^3}{3} + nA_s(500 - x)^2$

$I_{NA} = \dfrac{600(167^3)}{3} - \dfrac{300(167 - 80)^3}{3} + 6912\pi(500 - 167)^2$

$I_{NA} = 3\,273\,562\,384 ~ \text{mm}^4$

$\dfrac{f_s}{n} = \dfrac{M(d - x)}{I_{NA}}$

$\dfrac{f_s}{9} = \dfrac{100(500 - 167)(1000^2)}{3\,273\,562\,384}$

$f_s = 91.55 ~ \text{MPa}$

$T = f_s \, A_s = 91.55(768\pi)$

$T = 220.89 ~ \text{kN}$

$C = T$

$C = 220.89 ~ \text{kN}$ *answer*

**Another Solution for finding ***C*

$f_c = \dfrac{Mx}{I_{NA}} = \dfrac{100(167)(1000^2)}{3\,273\,562\,384}$

$f_c = 5.10 ~ \text{MPa}$

$f_{c1} = \dfrac{M(x - 80)}{I_{NA}} = \dfrac{100(167 - 80)(1000^2)}{3\,273\,562\,384}$

$f_{c1} = 2.66 ~ \text{MPa}$

$C = \frac{1}{2}(f_c + f_{c1})t_f \, b_f + \frac{1}{2}f_{c1}(x - t_f)b_w$

$C = \frac{1}{2}(5.10 + 2.66)(80)(600) + \frac{1}{2}(2.66)(167 - 80)(300)$

$C = 220.95 ~ \text{kN}$ (*okay*)