Curved Beams A Theoretical and FEM Investigationimechanica.org/files/Final Presentation.pdfCurved...

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.