sm7 172

________________________________________________________________________
PROBLEM (*7.172) The trapezoidal cross section of a curved beam is shown in Fig. P7.172. Derive
the expression for the radius R along the neutral axis. Compare the result with that given for Fig. D in
table 7.1.
*SOLUTION
12
1()
2
Abb=+
h
cr=+
w
dr
b2
O
b
1
r
ο
r
hri
The section width w varies
linearly with r as
wc (1)
01
We have
(2)
1
2
()
()
i
o
wb atrr
wb atrr
==
==
Introduce Eq.(1) into Eq.(2), then solve for c and
0 1
c
12 12
01
oi
rb rb bb
cc
hh
−
−
==− (3)
Then, we write
01 01
ln ( )
oo
ii
rr ooi
rr i
ccr r
dA wdr dr c c r r
rr r r
+
== =+−
∫∫ ∫
This gives, substituting Eqs.(3):
12 12
ln ( )
oi o
i
rb rb r
dA bb
rhr
−
=−
−
∫
Hence
212
12 12
1()
2
()ln(
o
oi i
hb b
A
RdA r
rb rb b b
rr
+
==
−−−
∫)
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