curved beams a theoretical and fem investigation presentation.pdffem theoretical stress (pa) 0 500 0...
TRANSCRIPT
![Page 1: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/1.jpg)
Curved Beams
A Theoretical and FEM
Investigation
Nathan Thielen
Christian W
ylonis
ES240
![Page 2: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/2.jpg)
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
![Page 3: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/3.jpg)
Assumptions
•Constant Cross Section
•Circular Arc Form
ed by the CentroidCurve
•Isotropic
•Isotropic
•Homogeneous
•Elastic
![Page 4: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/4.jpg)
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
![Page 5: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/5.jpg)
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:
![Page 6: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/6.jpg)
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)
![Page 7: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/7.jpg)
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
![Page 8: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/8.jpg)
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
![Page 9: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/9.jpg)
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:
![Page 10: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/10.jpg)
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
![Page 11: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/11.jpg)
Curved Beams FEM
Models compared to theoretical
results
![Page 12: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/12.jpg)
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
![Page 13: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/13.jpg)
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
)
![Page 14: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/14.jpg)
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
![Page 15: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/15.jpg)
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
)
![Page 16: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/16.jpg)
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
)
![Page 17: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/17.jpg)
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
)
![Page 18: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/18.jpg)
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
![Page 19: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/19.jpg)
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
![Page 20: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/20.jpg)
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
![Page 21: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/21.jpg)
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
![Page 22: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/22.jpg)
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
![Page 23: Curved Beams A Theoretical and FEM Investigation Presentation.pdfFEM Theoretical Stress (Pa) 0 500 0 0.2 0.4 0.6 0.8 1 1.2 Curvature (a/b) •Ascurvature increases along the top increasesand](https://reader034.vdocuments.site/reader034/viewer/2022051607/6033ea33ca64066c8b17f746/html5/thumbnails/23.jpg)
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.