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Curved Beams
A Theoretical and FEM
Investigation
Nathan Thielen
Christian W
ylonis
ES240
What are Curved Beams?
•A beam is curved if the
line form
ed by the
centroids of all the cross
sections is not straight.
•Curved beams loaded in
bending appear as
arches, chains, hooks, an
d many other connectors.
http://www.timyoung.net/contrast/images/chain02.jpg
http://www.photogen.com/free-photos/data/media/10/179_7908.jpg
Assumptions
•Constant Cross Section
•Circular Arc Form
ed by the CentroidCurve
•Isotropic
•Isotropic
•Homogeneous
•Elastic
Force Balance
Assume Shear Stresses are Zero (No Shear Stresses on the Surfaces)
r
θ θθθ
σr+dσr
drdr
∂σr
∂r+σr−σθ
()
r=0
Force Balance:
θ θθθ
dθ θθθdr
σ σσσr
σ σσσθ θθθ
σ σσσθ θθθ
Stress Compatibility
2-D Strain Compatibility (Plane Strain):
∂2ε x
∂y2+∂2ε x
∂y2=∂2γxy
∂x∂y
2-D Stress Compatibility (Hooke’s Law and Force Balance):
(Stress Boundary Conditions)
2-D Stress Compatibility (Hooke’s Law and Force Balance):
∂2 ∂x2+∂2 ∂y2
σx+σy
()=
∇2σx+σy
()=0
∇2σr+σθ
()=
∂2 ∂r2+1 r
∂ ∂r
σr+σθ
()=0
Converting to Polar Coordinates:
General Solution
Solve:
∇2σr+σθ
()=
∂2σr+σθ
()
∂r2
+1 r
∂σr+σθ
()
∂r=0
Compatibility:
(Euler’s Differential Equation)
σr+σθ
()=k 1lnr a
+k 2⇒
−σθ=σr−k 1lnr a
−k 2
σrr ()=C1+C2lnr a
+C3
r2
σθr ()=C1+C21+lnr a
−C3
r2
Solve:
Force Balance:
a
a
∂σr
∂r+1 r2σr−k 1lnr a
−k 2
=0
(First Order Linear ODE)
Boundary Conditions
σrr ()=C1+C2lnr a
+C3
r2
σθr ()=C1+C21+lnr a
−C3
r2
i ()σrr=a
()=
σrr=b
()=0
b
Boundary Conditions:
i ()σrr=a
()=
σrr=b
()=0
(ii)tσ
θdr=0
ab ∫
(iii)rtσθdr=M
ab ∫
σ θis not zero at r=a or r=b, therefore the distributed
norm
al stresses (σ θ) cause the m
oment (M
).
h
a
MM
Applying Boundary Conditions
i ()σrr=a
()=C1+C3
a2=0
(ii)tσ
θdr=tC1+C21+lnr a
−C3
r2
dr=
ab ∫0
ab ∫
(iii)rtσdr=rtC+C
1+lnr
−C3
dr=
b ∫M
b ∫(iii)rtσθdr=rtC1+C21+lnr a
−C3
r2
dr=
a∫M
a∫
Solve:
C1=
M
a2tlnb a
C2=a+b
()a−b
()M
a2b2tlnb a
2
C3=−
M
tlnb a
Curved Beam Stress Results
hb
a
MM
σrr ()=
4M
tb2N
1−a2
b2
lnr a
−1−a2
r2
lnb a
N=1−a2
b2
2
−4a2
b2
lnb a
=const.
σθr ()=
4M
tb2N
1−a2
b2
1+lnr a
−1+a2
r2
lnb a
Where:
Simple Example
h=0.1mm
M=10N⋅mm
a=1mm
b=3mm
hb
a
MM
The neutral axis is NOT the centroid axis (i.e. the center plane is not a neutral
axis). This is the PRIM
ARY difference between a curved beam and a straight
beam.σrr=2mm
()=
−36MPa
σθr=2mm
()=23MPa
Curved Beams FEM
Models compared to theoretical
results
The model
Constants:
Case 1:
h= 0.4 m
M = 40 Nm
t = 1 m
Straight beam
a = ∞
a
b
hM
Case 2:
Case 3:
Case 4:
a = ∞
b = ∞
a = 0.6 m
b = 1.0 m
a = 0.4 m
b = 0.8 m
a = 0.2 m
b = 0.6 m
Straight Beam
-2000
-1500
-1000
-5000
500
1000
1500
2000
00.1
0.2
0.3
0.4
•Neutral axis = 0.2 m
(0.2 m
)
•along the bottom = -1368 Pa (-1500 Pa)
•along the top = 1382 Pa (1500 Pa)
ϑσ
ϑσ
3
100
10.4
12
x
My
y
Iϑ
σσ
−×
−×
==
=×
ϑσ
(Pa)
r (m
)
Things to keep in mind
•Point loads not as good as distributed
loads and could is introducing some
error
100 N
100 N
•Theory does not take into account
bending due to the m
oment
Material: Steel
Density = 7800 kg/m^3
Modulus = 2 x 10^11 Pa
Case 1 -1500
-1000
-5000
500
1000
1500
2000
00.1
0.2
0.3
0.4
0.5
Neutral axis = 0.175 m
(0.165 m
)
along the bottom = -1200 Pa (-1287Pa)
along the top = 1540 Pa (1804 Pa)
ϑσ
ϑσ
ϑσ
(Pa)
r (m
)
Case 2
Neutral axis = 0.163 m
(0.180 m
)
along the bottom = -1141 Pa (-1229Pa)
along the top = 1600 Pa (1939 Pa)
ϑσ
ϑσ
-1500
-1000
-5000
500
1000
1500
2000
2500
00.1
0.2
0.3
0.4
0.5
ϑσ
(Pa)
r (m
)
Case 3
Neutral axis = 0.144 m
(0.165 m
)
along the bottom = -1063 Pa (-1130Pa)
along the top = 1875 Pa (2292 Pa)
ϑσ
ϑσ
-1500
-1000
-5000
500
1000
1500
2000
2500
00.1
0.2
0.3
0.4
0.5
ϑσ
(Pa)
r (m
)
Placement of the neutral axis
0.1
0.15
0.2
0.25
FEM
Theoretical
r (m)
0
0.05
00.2
0.4
0.6
0.8
11.2
Curvature (a/b)
•As curvature increases the neutral axis shifts farther away from
the center axis
•The results converge to the straight beam case
Magnitude of along bottom
ϑσ
-1000
-800
-600
-400
-2000
00.2
0.4
0.6
0.8
11.2
FEM
Theoretical
Stress (Pa)
-1600
-1400
-1200
Curvature (a/b)
•As curvature increases along the bottom decreases
•Stress highest for the straights beam
ϑσ
Magnitude of along top
ϑσ
1000
1500
2000
2500
FEM
Theoretical
Stress (Pa)
0
500
00.2
0.4
0.6
0.8
11.2
Curvature (a/b)
•As curvature increases along the top increases and causes a
stress concentration along this edge
•As the beam gets flatter it approaches the stress in a straight
beam
ϑσ
Influence on .
rσ
-300
-250
-200
-150
-100
-500
50
00.1
0.2
0.3
0.4
0.5
Stress (Pa)
•From top to bottom
•Low curvature, medium curvature, high curvature
•is zero for a straight beam
•Order of magnitude lower than
rσ
-500
-450
-400
-350
-300
r (m
)
ϑσ
Conclusion
•Beams with low curvature are closely
approximated by straight beam theory
•High curvature beams see large stresses
in the theta direction on the inside edge
in the theta direction on the inside edge
•The neutral axis shifts farther away from
the center as curvature increases
•When designing hooks, chains, and
arches this should be kept in mind
References
Haslach, H. Arm
strong, R. ”Deform
able Bodies and Their
Material Behavior”. (2004). John W
iley & Sons: USA. pp.
125-127, 137-138, 183-187.
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