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Equation (12.47) can be solved for u ,
v , and
w
to obtain
N
EA +
I
A φ
u =
M y I yz
+ M z I zz
E I yy I zz
I yz I ω z
I zz I ω y
I yy I zz
v =
I yz φ
I yz
(12.48)
M y I yy
+ M z I yz
E I yy I zz
I yy I ω z
I yz I ω y
w =−
I yz φ
I yz +
I yy I zz
φ
However, as will be shown in the following section, the coefficients of
in the second
= A ω
and third equations of Eq. (12.48) are identically zero. This is the result of I
ydA
= A ω
ω
y
and I
zdA being equal to zero, which occurs if the shear center, which is defined
on the next page, and the origin of the coordinate system coincide. Thus, substitution of
Eq. (12.48) into Eq. (12.46) gives
ω
z
N
A
T
E 1
M y I yz + M z I zz
I yy I zz
M y I yy + M z I yz
I yy I zz
φ
σ x =
y
+
z
+
A I
ω ω
E
α
(12.49)
I yz
I yz
In Eq. (12.9), a bimoment is defined by M ω = A σ
ω
dA . With
σ
x of Eq. (12.49)
x
N
A
T
E 1
M y I yz + M z I zz
I yy I zz
M y I yy + M z I yz
I yy I zz
φ
M
ω =
y
+
z
+
A I
ω ω
E
α
ω
dA
I yz
I yz
A
E 1
N
A I
A I 2
φ
=
ω +
ω
I
E
α
M
(12.50)
ωω
ω
T
= A
is the warping constant. Note that I ω = A ω
where M ω T
T
ω
dA and I ωω =
dA
involves the integration of
along the coordinate s , which follows along the middle line
of the wall profile as shown in Fig. 12.11. The magnitude of this integral depends on the
location of the origin of s [Goodier, 1938]. There exist certain points on the contour of
the wall for which, when they are used for the origin of s, I ω vanishes. Let the origin of s be
chosen such that I
ω
0 (corresponding to Eq. 12.18b). Also, let M
ω =
ω =
M
ω +
E
α
M
T . Then
ω
Eq. (12.50) reduces to
M
ωω φ
ω =−
EI
(12.51)
Finally, an expression for the normal stress in a thin-walled beam is obtained by replacing
φ
in Eq. (12.49) by Eq. (12.51)
N
A
T
M y I yz + M z I zz
I yy I zz
M y I yy + M z I yz
I yy I zz
M
ω
σ x =
y
+
z
+
ωω ω
E
α
(12.52)
I yz
I yz
I
Note that the part of the normal stress due to the warping of the cross-section is
M
M
ω
ω
ω
σ ω =
ωω ω =
(12.53)
I
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