198
6 Two-Dimensional Problems in Elasticity …
In straight beams, the rate of change of slope of the elastic curve is given by
d
2 y
dx 2 =
M
E I
. Whereas in initially curved beam, the rate of change of slope of the elastic
curve is
dθ
Rdθ
, which is the angle change per unit of arc length.
Now,
dθ
Rdθ
=
ω
R
=
M
E I
=
M mn
E I
for curved beams.
Or
ω =
R M mn
E I
Substituting the above in Eq. (b), we get
π
2
o
R.M mn
E I
dθ = 0
since R, E and I are constants,
∴
π
2
o
M mn dθ = 0
From Eq. (a), substituting the value of M mn , we obtain
−
π
2
0
M A dθ +
1
2
P R
π
2
0
dθ −
1
2
P R
π
2
0
cos θ dθ = 0
Integrating, we get
− M A [θ ]
π
2
0 +
1
2
P R[θ ]
π
2
0 −
1
2
P R[sin θ ]
π
2
0 = 0
− M A
π
2
+
1
2
P R
π
2
−
1
2
P R
sin
π
2
= 0
Thus M A =
P R
2
1 −
2
π
.
Therefore, knowing M A , the moment at any section such as MN can be computed
and then the normal stress can be calculated by curved beam formula at any desired
section.
6.15 Numerical Examples
Example 6.1 Given the following stress function.
6 Two-Dimensional Problems in Elasticity …
In straight beams, the rate of change of slope of the elastic curve is given by
d
2 y
dx 2 =
M
E I
. Whereas in initially curved beam, the rate of change of slope of the elastic
curve is
dθ
Rdθ
, which is the angle change per unit of arc length.
Now,
dθ
Rdθ
=
ω
R
=
M
E I
=
M mn
E I
for curved beams.
Or
ω =
R M mn
E I
Substituting the above in Eq. (b), we get
π
2
o
R.M mn
E I
dθ = 0
since R, E and I are constants,
∴
π
2
o
M mn dθ = 0
From Eq. (a), substituting the value of M mn , we obtain
−
π
2
0
M A dθ +
1
2
P R
π
2
0
dθ −
1
2
P R
π
2
0
cos θ dθ = 0
Integrating, we get
− M A [θ ]
π
2
0 +
1
2
P R[θ ]
π
2
0 −
1
2
P R[sin θ ]
π
2
0 = 0
− M A
π
2
+
1
2
P R
π
2
−
1
2
P R
sin
π
2
= 0
Thus M A =
P R
2
1 −
2
π
.
Therefore, knowing M A , the moment at any section such as MN can be computed
and then the normal stress can be calculated by curved beam formula at any desired
section.
6.15 Numerical Examples
Example 6.1 Given the following stress function.
