Nathan Thielen
Christian Wylonis
ES240
A Theoretical and FEM
Investigation
Curved Beams
• 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
• A beam is curved if the
line formed by the
centroids of all the cross
sections is not straight.
What are Curved Beams?
•
•
•
•
•
Constant Cross Section
Circular Arc Formed by the Centroid Curve
Isotropic
Homogeneous
Elastic
Assumptions
Force Balance:
σr
σθ
∂σ r (σ r − σ θ )
+
=0
∂r
r
σθ
dθ
θ
dr
dσ r
σr +
dr
dr
θ
r
Assume Shear Stresses are Zero (No Shear Stresses on the Surfaces)
Force Balance
)
(
)
2
 ∂2 1 ∂ 
∇ (σ r + σ θ ) =  2 +
σ r + σθ )= 0
(

r ∂r 
 ∂r
Converting to Polar Coordinates:
(
 ∂2
∂2 
2
+
σ
+
σ
=
∇
σx + σy = 0
y
 ∂x 2 ∂y 2  x
2-D Stress Compatibility (Hooke’s Law and Force Balance):
2
∂ 2ε x ∂ 2ε x ∂ γ xy
+ 2 =
2
∂y
∂y
∂x∂y
2-D Strain Compatibility (Plane Strain):
(Stress Boundary Conditions)
Stress Compatibility

 r   C3
σ θ (r ) = C1 + C2  1 + ln    − 2
 a r

(First Order Linear ODE)
∂σ r 1 

 r
+  2σ r − k1 ln   − k2  = 0
 a

∂r r 
 r  C3
σ r (r ) = C1 + C2 ln   + 2
 a r
Solve:
(Euler’s Differential Equation)
 r
 r
(σ r + σ θ ) = k1 ln  a  + k2 ⇒ −σ θ = σ r − k1 ln  a  − k2
Force Balance:
Solve:
Compatibility:
2
∂
σ r + σ θ ) 1 ∂ (σ r + σ θ )
(
2
∇ (σ r + σ θ ) =
+
=0
2
r
∂r
∂r
General Solution
a
M
a
h
b
M

 r   C3
σ θ (r ) = C1 + C2  1 + ln    − 2
 a r

σθ is not zero at r=a or r=b, therefore the distributed
normal stresses (σθ) cause the moment (M).
(iii) ∫ rtσ θ dr = M
b
a
(ii) ∫ tσ θ dr = 0
b
(i )σ r (r = a ) = σ r (r = b ) = 0
Boundary Conditions:
 r  C3
σ r (r ) = C1 + C2 ln   + 2
 a r
Boundary Conditions
C1 =
M
 b
a 2t ln  
 a
Solve:
b
C2 =
b
 b 
2 2 
a b t  ln   
  a 
2
(a + b )(a − b )M
C3 = −


 r   C3 
(iii) ∫ rtσ θ dr = ∫ rt C1 + C2  1 + ln    − 2  dr =M
 a r 


a
a
C3
(i )σ r (r = a ) = C1 + 2 = 0
a
b
b


 r   C3 
(ii) ∫ tσ θ dr = ∫ t C1 + C2  1 + ln    − 2  dr = 0
 a r 

a
a 
M
 b
t ln  
 a
Applying Boundary Conditions
h
M
Where:
2

 a2   b 
a2 
N =  1 − 2  − 4  2  ln   = const.
b 

 b   a
a2  
a2   b  
 4 M  
 r  
σ θ (r ) =  2   1 − 2   1 + ln    −  1 + 2  ln   
 tb N  
 a 
b 
r   a 
a2   r  
a2   b  
 4 M  
σ r (r ) =  2   1 − 2  ln   −  1 − 2  ln   
 tb N  
b   a 
r   a 
M
a
b
Curved Beam Stress Results
M
a
h
b
M
The neutral axis is NOT the centroid axis (i.e. the center plane is not a neutral
axis). This is the PRIMARY difference between a curved beam and a straight
beam.
σ θ (r = 2mm ) = 23MPa
σ r (r = 2mm ) = −36MPa
b = 3mm
M = 10N ⋅ mm
a = 1mm
h = 0.1mm
Simple Example
Models compared to theoretical
results
Curved Beams FEM
h
b
a
M
Straight beam
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
Case 1:
Case 2:
Case 3:
Case 4:
Constants: h = 0.4 m
M = 40 Nm
t=1m
The model
• σ ϑ along the top = 1382 Pa (1500 Pa)
• σ ϑ along the bottom = -1368 Pa (-1500 Pa)
• Neutral axis = 0.2 m (0.2 m)
− M × y −100 × y
σϑ = σ x =
=
1
I
× 0.43
12
-2000
-1500
-1000
-500
0
500
1000
1500
2000
0
σ ϑ (Pa)
0.1
Straight Beam
r (m)
0.2
0.3
0.4
Material: Steel
Density = 7800 kg/m^3
Modulus = 2 x 10^11 Pa
100 N 100 N
• Theory does not take into account
bending due to the moment
• Point loads not as good as distributed
loads and could is introducing some
error
Things to keep in mind
σ ϑ along the top = 1540 Pa (1804 Pa)
σ ϑ along the bottom = -1200 Pa (-1287Pa)
Neutral axis = 0.175 m (0.165 m)
-1500
-1000
-500
0
500
1000
1500
2000
0
σ ϑ (Pa)
Case 1
0.1
0.2
r (m)
0.3
0.4
0.5
σ ϑ along the top = 1600 Pa (1939 Pa)
σ ϑ along the bottom = -1141 Pa (-1229Pa)
Neutral axis = 0.163 m (0.180 m)
-1500
-1000
-500
0
500
1000
1500
2000
2500
0
σ ϑ (Pa)
Case 2
0.1
0.2
r (m)
0.3
0.4
0.5
σ ϑ along the top = 1875 Pa (2292 Pa)
σ ϑ along the bottom = -1063 Pa (-1130Pa)
Neutral axis = 0.144 m (0.165 m)
-1500
-1000
-500
0
500
1000
1500
2000
2500
0
σ ϑ (Pa)
Case 3
0.1
0.2
r (m)
0.3
0.4
0.5
r (m)
0
0.2
0.4
Curvature (a/b)
0.6
0.8
1
1.2
Theoretical
FEM
• As curvature increases the neutral axis shifts farther away from
the center axis
• The results converge to the straight beam case
0
0.05
0.1
0.15
0.2
0.25
Placement of the neutral axis
Stress (Pa)
0
0.2
0.4
Curvature (a/b)
0.6
0.8
1
• As curvature increases σ ϑ along the bottom decreases
• Stress highest for the straights beam
-1600
-1400
-1200
-1000
-800
-600
-400
-200
0
1.2
Theoretical
FEM
Magnitude of σ ϑ along bottom
Stress (Pa)
0
0.2
0.4
Curvature (a/b)
0.6
0.8
1
1.2
Theoretical
FEM
• As curvature increases σ ϑ 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
0
500
1000
1500
2000
2500
Magnitude of σ ϑ along top
Stress (Pa)
0
0.1
r (m)
0.2
0.3
0.4
0.5
• Order of magnitude lower than σ ϑ
•σ r is zero for a straight beam
• From top to bottom
• Low curvature, medium curvature, high curvature
-500
-450
-400
-350
-300
-250
-200
-150
-100
-50
0
50
Influence on σ r .
• 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
• The neutral axis shifts farther away from
the center as curvature increases
• When designing hooks, chains, and
arches this should be kept in mind
Conclusion
Haslach, H. Armstrong, R. ”Deformable Bodies and Their
Material Behavior”. (2004). John Wiley & Sons: USA. pp.
125-127, 137-138, 183-187.
References