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8/11/2019 Winkler Curved Beam Theory
http://slidepdf.com/reader/full/winkler-curved-beam-theory 1/21
Bending of Curved Beams – Strength of Materials Approach
N
M
V
r
θ
cross-section must besymmetric but does not have
to be rectangular
assume plane sections remain plane and just rotate about the
neutral axis, as for a straight beam, and that the only significant
stress is the hoop stress
θθ σ
θθ σ
8/11/2019 Winkler Curved Beam Theory
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N
M
centroidneutral axis
n R R
r
R = radius to centroid
Rn = radius to neutral axis
r = radius to general fiber in the beam
N, M = normal force and bending momentcomputed from centroid
θ ∆
A
B
A
B
P
P
n R r −
φ ∆
( )1
n n R r Rl
el r r
θθ
φ ω
θ
φ ω
θ
− ∆∆ ⎛ ⎞= = = −⎜ ⎟∆ ⎝ ⎠
∆=∆
Let ∆φ = rotation of the
cross-section
Reference: Advanced Mechanics of Materials : Boresi, Schmidt, and
Sidebottom
8/11/2019 Winkler Curved Beam Theory
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8/11/2019 Winkler Curved Beam Theory
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( )
( )
n m
n m
M E R RA A
N E R A A
ω
ω
= −
= −
from (1)
(1)
(2)
n
m
M E R
RA Aω =
−
from (2) ( )n m
m
m
N E R A E A
MA E A
RA A
ω ω
ω
= −
= −−
so solving for E ω ( )
m
m
MA N E RA A A
ω = −−
1
n
n
R
Ee E r
E R E
r
θθ θθ σ ω
ω ω
⎛ ⎞
= = −⎜ ⎟⎝ ⎠
= −
Recall, the stress is given by
so using expressions for ,n E R E ω ω
8/11/2019 Winkler Curved Beam Theory
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we obtain the hoop stress in the form
( )( )
m
m
M A rA N
Ar RA Aθθ σ
−= +−
axial
stress
bending
stress
nr R=
setting the total stress = 0 gives
0 N ≠
( )0
m m
AM r
A M N A RAθθ σ = =
+ −
0 N =setting the bending stress = 0 and gives n
m
A R
A=
which in general is not at the centroid
location of the
neutral axis
8/11/2019 Winkler Curved Beam Theory
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For composite areas
A1
A2
1 R2
R
i
m mi
i i
i
A A
A A
R A R
A
=
=
=
∑
∑∑∑
radii to centroids
areas
8/11/2019 Winkler Curved Beam Theory
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Example
P
P
100 mm30 mm
For a square 50x50 mm cross-section,
find the maximum tensile and
compressive stress if P = 9.5 kN and
plot the total stress across the cross-
section
8/11/2019 Winkler Curved Beam Theory
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P = 9500 N
M
N
155 mm
a
c
b
( )
2
lnm
A b c a
a c R
c A b
a
= −
+=
⎛ ⎞= ⎜ ⎟⎝ ⎠
a = 30 mm
b = 50 mm
c = 80 mm
so we have
( ) ( ) 250 50 2500
8050ln 49.04
30
80 30 552
m
A mm
A mm
R mm
= =
⎛ ⎞= =⎜ ⎟⎝ ⎠
+= =
250051
49.04n R mm= =
T C
8/11/2019 Winkler Curved Beam Theory
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max tensile stress is at r = 30 mm
( )
( )
( ) ( ) ( ) ( )( )( ) ( )( )155 9500 2500 30 49.049500
2500 2500 30 55 49.04 2500
106.2
m
m
M A rA N
A Ar RA A
MPa
θθ σ
−= +
−
−⎡ ⎤⎣ ⎦= +−⎡ ⎤⎣ ⎦
=
max compressive stress is at r = 80 mm
( )
( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
155 9500 2500 80 49.049500
2500 2500 80 55 49.04 2500
49.3
m
m
M A rA N
A Ar RA A
MPa
θθ σ −
= +
−−⎡ ⎤⎣ ⎦= +
−⎡ ⎤⎣ ⎦= −
8/11/2019 Winkler Curved Beam Theory
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>> r= linspace(30, 80, 100);
>> stress = 3.8 + 589*(2500 - 49.04.*r)./(197.2*r);
>> plot(r, stress)
30 35 40 45 50 55 60 65 70 75 80-60
-40
-20
0
20
40
60
80
100
120
radius, r
s t r e s s
8/11/2019 Winkler Curved Beam Theory
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Comparison with Airy Stress Function Results
Consider a rectangular beam under pure moment
a
b
r
MM
thickness = t
R
h
Note: ( )( )
( )
( )
/ 1/
2 / 1
2 / 1/
2 / 1
b a R h
b a
R hb a
R h
+=−
+=
−
y
8/11/2019 Winkler Curved Beam Theory
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2 2 2
2
4ln ln ln 1
M a b b b r a r b
Nta r a a a a b a a
θθ σ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
From Airy Stress Function
2 22 2
1 4 lnb b b
N a a a
⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎢ ⎥⎣ ⎦
Strength Approach
( )
( )
ln
/ 2
m
b A t
a
A t b a
R a b
⎛ ⎞= ⎜ ⎟⎝ ⎠
= −
= +
8/11/2019 Winkler Curved Beam Theory
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( )
( )( )
( ) ( )
( )
2
ln
ln
2
2 1 ln
1 2 1 1 ln
m
m
M A rA
Ar RA Ab
M b a r a
a b bt r b a b a
a
b r b M
a a a
r b b b bt a
a a a a a
θθ σ − −
=
−⎡ ⎤⎛ ⎞− − − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦=
+⎡ ⎤⎛ ⎞− − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦=⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − +
⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
If we had used the ordinary straight beam formula
instead
( )
( )
3
32
2
1
12
6 2 1
1
a b M r
My
I
t b a
r b
a a M
ta b
a
θθ σ
+⎡ ⎤−⎢ ⎥
⎣ ⎦= =
−⎡ ⎤⎛ ⎞− +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦=
⎛ ⎞−⎜ ⎟⎝ ⎠
minus on M since it is opposite to what
we had before
8/11/2019 Winkler Curved Beam Theory
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Comparison of the ratio of the max bending stresses
0.93310.99945.0
0.88810.99863.0
0.83130.99732.0
0.77370.99611.5
0.65450.99701.0
0.52621.01240.75
0.43901.04550.65
Strength/elasticity
Straight beam
formula My/I
Strength/elasticity
Curved beam
formula
R/h
R
h
y
8/11/2019 Winkler Curved Beam Theory
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Compare bending stress distributions at the smallest R/h value
1 2 3 4 5 6 7 8-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
r/a
n o r m
a l i z e d s t r e s s
solid curve – Airy stress function (elasticity)
dashed blue – Curved Strength formula
dashed red – Straight beam formula
R/h = 0.65 (b/a = 7.667)
8/11/2019 Winkler Curved Beam Theory
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% beam_compare.m
m=1;
Rhvals = [0.65 0.75 1.0 1.5 2.0 3.0 5.0]; % R/h ratios to consider
for Rh = Rhvalsba = (1+2*Rh)/(2*Rh -1); %corresponding b/a values
ra= linspace(1, ba, 100); % r/a values
N=(ba^2-1)^2 -4*ba^2*(log(ba))^2;
% Airy function flexure stress expression
pa =4*(-(ba./ra).^2.*log(ba)+(ba)^2.*log(ra./ba) - log(ra) +(ba)^2 -1)./N;
% Curved beam strength expression for flexure stress
ps = 2*((ba-1)-ra.*log(ba))./(ra.*(ba-1).*(2.*(ba-1) -(ba+1).*log(ba)));% Straight beam flexure formula My/I
pb = 6*(2*ra-(ba+1))./((ba-1)^3);
%obtain ratio of max stresses Curved beam Strength formula/Airy
ratio1(m) = max(abs(ps))/max(abs(pa));
%obtain ratio of max stresses: Straight beam strength formula/Airy
ratio2(m) = max(abs(pb))/max(abs(pa));
m=m+1;
end% Now plot stress distributions for smallest R/h value
Rh=0.65
ba = (1+2*Rh)/(2*Rh -1); %corresponding b/a values
ra= linspace(1, ba, 100);
N=(ba^2-1)^2 -4*ba^2*(log(ba))^2;
pa =4*(-(ba./ra).^2.*log(ba)+(ba)^2.*log(ra./ba) - log(ra) +(ba)^2 -1)./N;
ps = 2*((ba-1)-ra.*log(ba))./(ra.*(ba-1).*(2.*(ba-1) -(ba+1).*log(ba)));pb = 6*(2*ra-(ba+1))./((ba-1)^3);
plot(ra, pa)
hold on
plot(ra, ps, '--b')
plot(ra, pb, '--r')
xlabel('r/a')
ylabel( 'normalized stress')
hold off
8/11/2019 Winkler Curved Beam Theory
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Comparison with Bickford’s expression (pure bending)
( ) 21
1/
kM My
A ky I
k R
θθ σ = − −+
=
R
r
centroid
y
y−
First note that
0
A
ydA =∫
Here, y is distance from the centroid
so ( ) 2
2
1
1 / 10
1 / 1 / 1 / A A A
y y R I ydA y dAdA I
y R y R R y R R
+= + = + =
+ + +∫ ∫ ∫ 1
2
2
1 /
1 /
A
A
ydA I y R
y dA I
y R
=+
=+
∫
∫2
1
I I
R= −
R r y− = −
or r y R= +
8/11/2019 Winkler Curved Beam Theory
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m A A
dA dA RA R R
r y R= =
+∫ ∫( )
( )
1
A A A A
m
y R y dA A dA dA dA R
y R y R y R
I RA R
+= = = +
+ + +
= +
∫ ∫ ∫ ∫
1 / I R
Thus,( ) 1 2
/m R RA A I I R− = − =
8/11/2019 Winkler Curved Beam Theory
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( ) 21
1/
kM My
A ky I
k R
θθ σ = − −+
=
Now, start with Bickford’s expression
( )
( )
( )
( ) ( )
( ) ( )
( )
( )
( ) ( )
( )
2
2
2
2
2 32
2 2
2 2
2
2 2
222
2 2
2 3
2
2
2
1
/
/
yR M
AR y R I
y R I yAR M
y R ARI
y R I y R AR AR
M y R ARI y R ARI
I AR R M
ARI y R I
I AR R M
y R I ARI
A y R I AR R M
y R AI R
θθ σ ⎡ ⎤
= − +⎢ ⎥+⎣ ⎦
⎡ ⎤+ += − ⎢ ⎥
+⎣ ⎦
⎡ ⎤+ + +
= − −⎢ ⎥+ +⎣ ⎦
⎡ ⎤+= − −⎢ ⎥
+⎣ ⎦
⎡ ⎤+= −⎢ ⎥+⎣ ⎦
⎡ ⎤− + +⎢ ⎥=
+⎢ ⎥⎣ ⎦
same terms added in and subtracted out
8/11/2019 Winkler Curved Beam Theory
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( ) ( )( )
2 3
2
2
2
/
/
A y R I AR R
M y R AI Rθθ σ
⎡ ⎤− + +
⎢ ⎥= +⎢ ⎥⎣ ⎦
but 2
2 /m RA A I R− = so ( )2 3
2 /m A I AR R= +
and we find
y R r + =
( )m
m
A rA
r RA A Aθθ σ
⎡ ⎤−= ⎢ ⎥
−⎣ ⎦
which agrees with our previous expression
8/11/2019 Winkler Curved Beam Theory
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from Bickford’s expression
( ) 21
1/
kM My
A ky I
k R
θθ σ = − −+
=
it is easy to see, as , R → ∞ 0k →
and 2
2
2
1 / A A
y dA I y dA I
y R
= → =
+∫ ∫
and we recover the straight beam flexure expression
My
I θθ σ = −