sm7 171

________________________________________________________________________
PROBLEM (*7.171) The circular cross section of a curved beam is shown in Fig. P7.171. Derive
the expression for the radius R along the neutral axis and compare the result with that given for Fig. B in
Table 7.1.
*SOLUTION
2
Ac
π
=
Through use of the polar coordinates we write:
2sin coswc rrc
α
α ==−
sindr c d
α
α =
22
2sindA wdr c d
α
α ==
22
0
2sin
cos
dA c d
rrc
π
α
α
α
=−
∫∫
22 2222
00
(1 cos ) cos ( )
22
cos cos
crc
rc rc
ππ
αα
αβ
−−−
==
−−
∫∫ 2
rc−
22
00
2( cos) 2( ) cos
d
rc d r c rc
ππ
α
αα
α
=+ −− −
∫∫
22
22 1
00 22
0
tan
22
22sin2() tan
rc
rc rc rc
rc
π
ππ
α
αα
−−
=+ −− +
−
w
O
dr
α
r
c
ccos
α
r
C
This gives
22
2( )
dA rrc
r
π =−−
∫
Hence, it can be shown that
22
1()
2
A
R
rrc
dA
r
==++
∫
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