Consider the circular section in Table 3–4 with r_{c} = 3 in and R = 1 in. Determine e by using the formula from the table and approximately by using Eq. (3–66). Compare the results of the two solutions.
e \dot {=} \frac {I}{r_{c} A } (3.66)
Table 3–4 Formulas for Sections of Curved Beams
r_{c} = r_{i} +\frac{h}{2}
r_{n} =\frac{h}{ln (r_{o}/r_{ i} )} |
r_{c} = r_{i} +\frac {h}{3} \frac {b_{i} + 2b_{o}}{b_{i} + b_{o}} |
r_{c} = r_{i} +\frac{b_{i}c^{2}_{1}+ 2b_{o}c_{1}c_{2} + b_{o}c^{2}_{2}}{ 2(b_{o}c_{2} +b_{i}c_{1})} |
r_{c} = r_{i} + R |
r_{c} = r _{i} +\frac {\frac {1}{2} h^{2}t + \frac {1}{2} t^{2}_{i} (b_{i} − t ) + t_{o}(b_{o} − t )(h − t_{o}/2)}{t_{i} (b_{i} − t ) + t_{o}(b_{o} − t ) + ht}
r_{n} =\frac {t_{i} (b_{i} − t ) + t_{o}(b_{o} − t ) +ht_{o}}{b_{i} ln \frac {r_{i} + t}{r_{i}} ln \frac {r_{o}− t_{o}}{r_{i} + t_{i}} + b_{o} ln \frac {r_{o}}{r_{o} − t_{o}}} |
r_{c} = r _{i} +\frac {\frac {1}{2} h^{2}t + \frac {1}{2} t^{2}_{i} (b − t ) + t_{o}(b − t )(h − t_{o}/2)}{ht + (b − t )(t_{i} + t_{o})}
r_{n} =\frac {(b − t )(t_{i} + t_{o}) + ht}{b (ln \frac {r_{i} + t_{i}}{r_{i}} + ln \frac {r_{o}}{r_{o} − t_{o}}) + t ln \frac {r_{o} − t_{o}}{r_{i} + t_{i}}} |