A: 단면적	e: 중심축에서 가장 먼 거리	I: 단면 2차 모멘트
Z: 단면 계수 ($\displaystyle Z = \frac{I}{e}$)	k: 회전 반경 ($\displaystyle k = \sqrt{\frac{I}{A}}$)

$$ \begin{aligned} A &= a^2, \quad e = \frac{1}{2}a \\[8pt] I &= \frac{a^4}{12}, \quad Z = \frac{a^3}{6}, \quad k = \frac{a}{\sqrt{12}} = 0.289a \end{aligned} $$

$$ \begin{aligned} A &= a^2, \quad e = a \\[8pt] I &= \frac{a^4}{3}, \quad Z = \frac{a^3}{3}, \quad k = \frac{a}{\sqrt{3}} = 0.577a \end{aligned} $$

$$ \begin{aligned} A &= a^2, \quad e = \frac{a}{\sqrt{2}} = 0.707a \\[8pt] I &= \frac{a^4}{12}, \quad Z = \frac{a^3}{6\sqrt{2}} = 0.118a^3, \quad k = \frac{a}{\sqrt{12}} = 0.289a \end{aligned} $$

$$ \begin{aligned} A &= a^2 - b^2, \quad e = \frac{1}{2}a \\[8pt] I &= \frac{a^4 - b^4}{12}, \quad Z = \frac{a^4 - b^4}{6a}, \quad k = \sqrt{\frac{a^2 + b^2}{12}} \end{aligned} $$

$$ \begin{aligned} A &= a^2 - b^2, \quad e = \frac{a}{\sqrt{2}} = 0.707a \\[8pt] I &= \frac{a^4 - b^4}{12}, \quad Z = \frac{\sqrt{2}(a^4 - b^4)}{12a}, \quad k = \sqrt{\frac{a^2 + b^2}{12}} \end{aligned} $$

$$ \begin{aligned} A &= bd, \quad e = \frac{1}{2}d \\[8pt] I &= \frac{bd^3}{12}, \quad Z = \frac{bd^2}{6}, \quad k = \frac{d}{\sqrt{12}} = 0.289d \end{aligned} $$

$$ \begin{aligned} A &= bd, \quad e = d \\[8pt] I &= \frac{bd^3}{3}, \quad Z = \frac{bd^2}{3}, \quad k = \frac{d}{\sqrt{3}} = 0.577d \end{aligned} $$

$$ \begin{aligned} A &= bd-hk, \quad e = \frac{1}{2}d \\[8pt] I &= \frac{bd^3 - hk^3}{12}, \quad Z = \frac{bd^3 - hk^3}{6d}, \quad k = \sqrt{\frac{bd^3 - hk^3}{12(bd - hk)}} \end{aligned} $$

$$ \begin{aligned} A &= bd, \quad e = \frac{bd}{\sqrt{b^2 + d^2}} \\[8pt] I &= \frac{b^3d^3}{6(b^2+d^2)}, \quad Z = \frac{b^2d^2}{6\sqrt{b^2+d^2}}, \quad k = \frac{bd}{\sqrt{6(b^2+d^2)}} \end{aligned} $$

$$ \begin{aligned} A &= bd, \quad e = \frac{1}{2}(d\cos\alpha + b\sin\alpha) \\[8pt] I &= \frac{bd}{12}(d^2\cos^2\alpha + b^2\sin^2\alpha) \\[8pt] Z &= \frac{bd}{6}\left(\frac{d^2\cos^2\alpha + b^2\sin^2\alpha}{d\cos\alpha + b\sin\alpha}\right), \quad k = \sqrt{\frac{d^2\cos^2\alpha + b^2\sin^2\alpha}{12}} \end{aligned} $$

$$ \begin{aligned} A &= \frac{bd}{2}, \quad e = \frac{2}{3}d \\[8pt] I &= \frac{bd^3}{36}, \quad Z = \frac{bd^2}{24}, \quad k = \frac{d}{\sqrt{18}} = 0.236d \end{aligned} $$

$$ \begin{aligned} A &= \frac{bd}{2}, \quad e = d \\[8pt] I &= \frac{bd^3}{12}, \quad Z = \frac{bd^2}{12}, \quad k = \frac{d}{\sqrt{6}} = 0.408d \end{aligned} $$

$$ \begin{aligned} A &= \frac{d(a+b)}{2}, \quad e = \frac{d(a+2b)}{3(a+b)} \\[8pt] I &= \frac{d^3(a^2+4ab+b^2)}{36(a+b)}, \quad Z = \frac{d^2(a^2+4ab+b^2)}{12(a+2b)}\\[8pt] k &= \sqrt{\frac{d^2(a^2+4ab+b^2)}{18(a+b)^2}} \end{aligned} $$

$$ \begin{aligned} A &= \frac{3d^2\tan30^\circ}{2} = 0.866d^2, \quad e = \frac{d}{2} \\[8pt] I &= \frac{A}{12}\left[\frac{d^2(1+2\cos^230^\circ)}{4\cos^230^\circ}\right] = 0.06d^4 \\[8pt] \end{aligned} $$ $$ Z = \frac{A}{6}\left[\frac{d(1+2\cos^230^\circ)}{4\cos^230^\circ}\right] = 0.12d^3 , \quad k = \sqrt{\frac{d^2(1+2\cos^230^\circ)}{48\cos^230^\circ}} = 0.264d $$

$$ \begin{aligned} A &= \frac{3d^2\tan30^\circ}{2} = 0.866d^2, \quad e = \frac{d}{2\cos30^\circ} = 0.577d \\[8pt] I &= \frac{A}{12}\left[\frac{d^2(1+2\cos^230^\circ)}{4\cos^230^\circ}\right] = 0.06d^4 \\[8pt] \end{aligned} $$ $$ Z = \frac{A}{6}\left[\frac{d(1+2\cos^230^\circ)}{4\cos^230^\circ}\right] = 0.104d^3, \quad k = \sqrt{\frac{d^2(1+2\cos^230^\circ)}{48\cos^230^\circ}} = 0.264d$$

$$ \begin{aligned} A &= 2d^2 \times \tan 22.5^\circ = 0.828d^2, \quad e = \frac{d}{2} \\[8pt] I &= \frac{A}{12}\left[\frac{d^2(1 + 2\cos^2 22.5^\circ)}{4\cos^2 22.5^\circ}\right] = 0.055d^4 \\[8pt] \end{aligned} $$ $$ Z = \frac{A}{6}\left[\frac{d(1 + 2\cos^2 22.5^\circ)}{4\cos^2 22.5^\circ}\right] = 0.109d^3, \quad k = \sqrt{\frac{d^2(1 + 2\cos^2 22.5^\circ)}{48\cos^2 22.5^\circ}} = 0.257d $$

$$ \begin{aligned} A &= \frac{\pi d^2}{4}, \quad e = \frac{d}{2} \\[8pt] I &= \frac{\pi d^4}{64}, \quad Z = \frac{\pi d^3}{32}, \quad k = \frac{d}{4} \end{aligned} $$

$$ \begin{aligned} A &= \frac{\pi(D^2 - d^2)}{4}, \quad e = \frac{D}{2} \\[8pt] I &= \frac{\pi(D^4 - d^4)}{64}, \quad Z = \frac{\pi(D^4 - d^4)}{32D}, \quad k = \frac{\sqrt{D^2 + d^2}}{4} \end{aligned} $$

$$ \begin{aligned} A &= \frac{\pi d^2}{8}, \quad e = \frac{(3\pi - 4)d}{6\pi} = 0.288d \\[8pt] I &= \frac{(9\pi^2 - 64)d^4}{1152\pi} = 0.007d^4, \quad Z = \frac{(9\pi^2 - 64)d^3}{192(3\pi - 4)} = 0.024d^3 \\[8pt] k &= \frac{\sqrt{(9\pi^2 - 64)d^2}}{12\pi} = 0.132d \end{aligned} $$

$$ \begin{aligned} A &= \frac{\pi(R^2 - r^2)}{2}, \quad e = \frac{4(R^3 - r^3)}{3\pi(R^2 - r^2)} \\[8pt] I &= 0.1098(R^4 - r^4) - \frac{0.283 R^2 r^2 (R - r)}{R + r}, \quad Z = \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= \pi a b, \quad e = a \\[8pt] I &= \frac{\pi a^3 b}{4}, \quad Z = \frac{\pi a^2 b}{4}, \quad k = \frac{a}{2} \end{aligned} $$

$$ \begin{aligned} A &= \pi (ab - cd), \quad e = a \\[8pt] I &= \frac{\pi}{4}(a^3 b - c^3 d), Z = \frac{\pi(a^3 b - c^3 d)}{4a}, \quad k = \frac{1}{2}\sqrt{\frac{a^3 b - c^3 d}{ab - cd}} \end{aligned} $$

$$ \begin{aligned} A &= dt + 2a(s+n), \quad e = \frac{d}{2} \\[8pt] k &= \sqrt{\frac{\frac{1}{12}\left[bd^3 - \frac{1}{4g}(h^4 - \ell^4)\right]}{dt + 2a(s+n)}} \end{aligned} $$ $$ I = \frac{1}{12}\left[bd^3 - \frac{1}{4g}(h^4 - \ell^4)\right], \quad Z = \frac{1}{6d}\left[bd^3 - \frac{1}{4g}(h^4 - \ell^4)\right], \quad g = \frac{h-\ell}{b-t} $$

$$ \begin{aligned} A &= dt + 2a(s+n), \quad e = \frac{b}{2} \\[8pt] I &= \frac{1}{12}\left[b^3(d - h) + \ell t^3 + \frac{g}{4}(b^4 - t^4)\right], \quad g = \frac{h - \ell}{b - t} \\[8pt] Z &= \frac{1}{6b}\left[b^3(d - h) + \ell t^3 + \frac{g}{4}(b^4 - t^4)\right], \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= dt + a(s+n), \quad e = \frac{d}{2} \\[8pt] I &= \frac{1}{12}\left[bd^3 - \frac{1}{8g}(h^4 - \ell^4)\right], \quad Z = \frac{1}{6d}\left[bd^3 - \frac{1}{8g}(h^4 - \ell^4)\right] \\[8pt] k &= \sqrt{\frac{\frac{1}{12}\left[bd^3 - \frac{1}{8g}(h^4 - \ell^4)\right]}{dt + a(s+n)}}, \quad g = \frac{h - \ell}{2(b - t)} \end{aligned} $$

$$ \begin{aligned} A &= dt + a(s+n) \\[8pt] e &= b - \left[b^2s + \frac{ht^2}{2} + \frac{g}{3}(b-t)^2 \cdot (b+2t)\right] \div A, \quad g = \frac{h-\ell}{2(b-t)} \\[8pt] I &= \frac{1}{3}\left[2sb^3 + \ell t^3 + \frac{g}{2}(b^4 - t^4)\right] - A(b-e)^2, \quad Z = \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= \frac{\ell(T-t)}{2} + Tn + a(s+n) \\[8pt] \end{aligned} $$ $$ \begin{aligned} e &= d - \left[3s^2(b-t) + 2am(m+3s) + 3Td^2 - \ell(T-t)(3d-\ell)\right] \div 6A \\[8pt] I &= \frac{1}{12}\left[\ell^3(T+3t) + 4bn^3 - 2am^3\right] - A(d - e - n)^2, \quad Z = \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= \frac{\ell(T-t)}{2} + Tn + a(s+n), \quad e = \frac{b}{2} \\[8pt] Z &= \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$ $$ I = \frac{sb^3 + mT^3 + \ell t^3}{12} + \frac{am\left[2a^2 + (2a+3T)^2\right]}{36} + \frac{\ell(T-t)}{144}\left[(T-t)^2 + 2(T-2t)^2\right] $$

$$ \begin{aligned} A &= t(2a - t) \\[8pt] e &= a - \frac{a^2 + at - t^2}{2(2a - t)}, I = \frac{1}{3}\left[te^3 + a(a - e)^3 - (a - t)(a - e - t)^3\right] \\[8pt] Z &= \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= t(2a - t) \\[8pt] e &= \frac{a^2 + at - t^2}{2(2a - t)\cos 45^\circ}, \quad I = \frac{1}{3}\left[2x^4 + 2(x - t)^4 + t(a - 2x + \frac{1}{2}t)^3\right] \\[8pt] x &= \frac{a^2 + at - t^2}{2(2a - t)}, \quad Z = \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= bd - h(b - t), \quad e = \frac{d}{2} \\[8pt] I &= \frac{bd^3 - h^3(b - t)}{12}, \quad Z = \frac{bd^3 - h^3(b - t)}{6d} \\[8pt] k &= \sqrt{\frac{bd^3 - h^3(b - t)}{12[bd - h(b - t)]}} \end{aligned} $$

$$ \begin{aligned} A &= bd - h(b - t), \quad e = \frac{b}{2} \\[8pt] I &= \frac{2sb^3 + ht^3}{12}, \quad Z = \frac{2sb^3 + ht^3}{6b} \\[8pt] k &= \sqrt{\frac{2sb^3 + ht^3}{12(bd - h(b - t))}} \end{aligned} $$

$$ \begin{aligned} A &= bd - h(b - t), \quad e = \frac{d}{2} \\[8pt] I &= \frac{bd^3 - h^3(b - t)}{12}, \quad Z = \frac{bd^3 - h^3(b - t)}{6d} \\[8pt] k &= \sqrt{\frac{bd^3 - h^3(b - t)}{12[bd - h(b - t)]}} \end{aligned} $$

$$ \begin{aligned} A &= bd - h(b - t) \\[8pt] e &= b - \frac{2b^2s + ht^2}{2bd - 2h(b - t)}, \quad I = \frac{2sb^3 + ht^3}{3} - A(b - e)^2 \\[8pt] Z &= \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= dt + s(b - t), \quad e = \frac{d}{2} \\[8pt] I &= \frac{td^3 + s^3(b - t)}{12}, \quad Z = \frac{td^3 + s^3(b - t)}{6d} \\[8pt] k &= \sqrt{\frac{td^3 - s^3(b - t)}{12[td + s(b - t)]}} \end{aligned} $$

$$ \begin{aligned} A &= bs + ht + as \\[8pt] e &= d - [d^2t + s^2(b - t) + s(a - t)(2d - s)] \div 2A \\[8pt] I &= \frac{1}{3}[te^3 + b(d - e)^3 - (b - t)(d - e - s)^3], \quad Z = \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= bs + ht, \quad e = d - \frac{d^2t + s^2(b - t)}{2(bs + ht)} \\[8pt] I &= \frac{1}{3}[b(d - e)^3 + ae^3 - (b - t)(d - e - s)^3 - (a - t)(e - s)^3] \\[8pt] Z &= \frac{I}{e} \\[8pt] k &= \sqrt{\frac{1}{3(bs + ht)}[te^3 + b(d - e)^3 - (b - t)(d - e - s)^3]} \end{aligned} $$

$$ \begin{aligned} A &= bs + \frac{h(T + t)}{2} \\[8pt] e &= d - [3bs^2 + 3ht(d + s) + h(T - t)(h + 3s)] \div 6A \\[8pt] I &= \frac{1}{12}[4bs^3 + h^3(3t + T)] - A(d - e - s)^3 \\[8pt] Z &= \frac{I}{e}, \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$

$$ \begin{aligned} A &= (a + b - t), \quad e = b - \frac{t(2d + a) + d^2}{2(d + a)} \\[8pt] I &= \frac{1}{3}[te^3 + a(b - e)^3 - (a - t)(b - e - t)^3], \quad Z = \frac{I}{e} \\[8pt] k &= \sqrt{\frac{1}{3t(a + b - t)}[te^3 + a(b - e)^3 - (a - t)(b - e - t)^3]} \end{aligned} $$

$$ \begin{aligned} A &= (a + b - t), \quad e = b - \frac{t(2c + b) + c^2}{2(c + b)} \\[8pt] I &= \frac{1}{3}[te^3 + b(a - e)^3 - (b - t)(a - e - t)^3], \quad Z = \frac{I}{e} \\[8pt] k &= \sqrt{\frac{1}{3t(a + b - t)}[te^3 + b(a - e)^3 - (b - t)(a - e - t)^3]} \end{aligned} $$

$$ \begin{aligned} A &= t[b + 2(a - t)], \quad e = \frac{b}{2} \\[8pt] I &= \frac{ab^3 - c(b - 2t)^3}{12}, \quad Z = \frac{ab^3 - c(b - 2t)^3}{6b} \\[8pt] k &= \sqrt{\frac{ab^3 - c(b - 2t)^3}{12t[b + 2(a - t)]}} \end{aligned} $$

$$ \begin{aligned} A &= t[b + 2(a - t)], \quad e = \frac{2a - t}{2} \\[8pt] I &= \frac{b(a + c)^3 - 2c^3d - 6a^2cd}{12} \\[8pt] Z &= \frac{b(a + c)^3 - 2c^3d - 6a^2cd}{6(2a - t)}, \quad k = \sqrt{\frac{b(a + c)^3 - 2c^3d - 6a^2cd}{12t[b + 2(a - t)]}} \end{aligned} $$

$$ \begin{aligned} A &= \frac{\pi}{2}d^2 + 2b(h - d), \quad e = \frac{h}{2} \\[8pt] I &= \frac{1}{12}\left[\frac{3\pi}{16}d^4 + b(h^3 - d^3) + b^3(h - d)\right] \\[8pt] Z &= \frac{1}{6h}\left[\frac{3\pi}{16}d^4 + b(h^3 - d^3) + b^3(h - d)\right], \quad k = \sqrt{\frac{I}{A}} \end{aligned} $$