Bending of Curved Beams – Strength of Materials Approach

N

M

V

r

θ

cross-section must be symmetric 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

θθσ

θθσ

N

M

centroid neutral axis

nRR

r

R = radius to centroidRn = radius to neutral axisr = radius to general fiber in the beam

N, M = normal force and bending momentcomputed from centroid

θ∆

A

B

A

B

P

P

nR r−

φ∆

( ) 1n nR r Rlel r rθθ

φω

θφωθ

− ∆∆ ⎛ ⎞= = = −⎜ ⎟∆ ⎝ ⎠∆

=∆

Let ∆φ = rotation of thecross-section

Reference: Advanced Mechanics of Materials : Boresi, Schmidt, andSidebottom

From Hooke’s law

1nREe Erθθ θθσ ω ⎛ ⎞= = −⎜ ⎟

⎝ ⎠

Then the normal force is given by

( )

nA A A

n m

dAN dA E R dAr

E R A A

θθσ ω

ω

⎡ ⎤= = −⎢ ⎥

⎣ ⎦= −

∫ ∫ ∫

wherem

A

dAAr

= ∫ has the dimensions of a length

Similarly, for the moment ( )

( )

( )

1

A

n

A

n nA A A A

n m

M R r dA

RE R r dAr

dAE R R R dA R dA rdAr

E R RA A

θθσ

ω

ω

ω

= −

⎛ ⎞= − −⎜ ⎟⎝ ⎠

⎡ ⎤= − − +⎢ ⎥

⎣ ⎦= −

∫

∫

∫ ∫ ∫ ∫

( )( )

n m

n m

M E R RA A

N E R A A

ω

ω

= −

= −

from (1)

(1)

(2)

nm

ME RRA A

ω =−

from (2) ( )n m

m

m

N E R A E AMA E A

RA A

ω ω

ω

= −

= −−

so solving for Eω( )

m

m

MA NEA RA A A

ω = −−

1n

n

REe Er

E R Er

θθ θθσ ω

ω ω

⎛ ⎞= = −⎜ ⎟⎝ ⎠

= −

Recall, the stress is given by

so using expressions for ,nE R Eω ω

we obtain the hoop stress in the form

( )( )

m

m

M A rANA Ar RA Aθθσ

−= +

−

axialstress

bendingstress

nr R=

setting the total stress = 0 gives

0N ≠

( )0m m

AMrA M N A RAθθσ =

=+ −

0N =

setting the bending stress = 0 and gives nm

ARA

=

which in general is not at the centroid

location of theneutral axis

For composite areas

A1

A2

1R2R

i

m mi

i i

i

A A

A A

R AR

A

=

=

=

∑∑∑∑radii to centroids

areas

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

P = 9500 N

MN

155 mm

a

c

b

( )

2

lnm

A b c aa cR

cA ba

= −

+=

⎛ ⎞= ⎜ ⎟⎝ ⎠

a = 30 mmb = 50 mmc = 80 mm

so we have( )( ) 250 50 2500

8050ln 49.0430

80 30 552

m

A mm

A mm

R mm

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

+= =

2500 5149.04nR mm= =

T C

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 rANA Ar RA A

MPa

θθσ−

= +−

−⎡ ⎤⎣ ⎦= +−⎡ ⎤⎣ ⎦

=

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 rANA Ar RA A

MPa

θθσ−

= +−

−⎡ ⎤⎣ ⎦= +−⎡ ⎤⎣ ⎦

= −

>> 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

stre

ss

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 aR h

b a

R hb a

R h

+=

−

+=

−

y

2 2 2

2

4 ln ln ln 1M a b b b r a r bNta r a a a a b a aθθσ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

From Airy Stress Function

2 22 2

1 4 lnb b bNa a a

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎢ ⎥⎣ ⎦

Strength Approach

( )( )

ln

/ 2

mbA ta

A t b a

R a b

⎛ ⎞= ⎜ ⎟⎝ ⎠

= −

= +

( )( )

( )

( ) ( ) ( )

2

ln

ln2

2 1 ln

1 2 1 1 ln

m

m

M A rAAr RA A

bM b a ra

a b bt r b a b aa

b r bMa a a

r b b b bt aa a a a a

θθσ− −

=−

⎡ ⎤⎛ ⎞− − − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦=+⎡ ⎤⎛ ⎞− − −⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦=⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

If we had used the ordinary straight beam formulainstead

( )

( )3

32

21

12

6 2 1

1

a bM r

MyI t b a

r ba aM

ta ba

θθσ

+⎡ ⎤−⎢ ⎥

⎣ ⎦= =−

⎡ ⎤⎛ ⎞− +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦=⎛ ⎞−⎜ ⎟⎝ ⎠

minus on M since it is opposite to what we had before

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/elasticityStraight beam formula My/I

Strength/elasticityCurved beamformula

R/h

R

h

y

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

norm

aliz

ed s

tress

solid curve – Airy stress function (elasticity)dashed blue – Curved Strength formuladashed red – Straight beam formula

R/h = 0.65 (b/a = 7.667)

% beam_compare.mm=1;Rhvals = [0.65 0.75 1.0 1.5 2.0 3.0 5.0]; % R/h ratios to considerfor Rh = Rhvalsba = (1+2*Rh)/(2*Rh -1); %corresponding b/a valuesra= linspace(1, ba, 100); % r/a valuesN=(ba^2-1)^2 -4*ba^2*(log(ba))^2;% Airy function flexure stress expressionpa =4*(-(ba./ra).^2.*log(ba)+(ba)^2.*log(ra./ba) - log(ra) +(ba)^2 -1)./N; % Curved beam strength expression for flexure stressps = 2*((ba-1)-ra.*log(ba))./(ra.*(ba-1).*(2.*(ba-1) -(ba+1).*log(ba)));% Straight beam flexure formula My/Ipb = 6*(2*ra-(ba+1))./((ba-1)^3);%obtain ratio of max stresses Curved beam Strength formula/Airyratio1(m) = max(abs(ps))/max(abs(pa));%obtain ratio of max stresses: Straight beam strength formula/Airyratio2(m) = max(abs(pb))/max(abs(pa));m=m+1;end% Now plot stress distributions for smallest R/h valueRh=0.65ba = (1+2*Rh)/(2*Rh -1); %corresponding b/a valuesra= 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 onplot(ra, ps, '--b')plot(ra, pb, '--r')xlabel('r/a')ylabel( 'normalized stress')hold off

Comparison with Bickford’s expression (pure bending)

( ) 211/

kM MyA ky I

k R

θθσ = − −+

=

R

r

centroid

y

y−

First note that

0A

ydA =∫

Here, y is distance from the centroid

so ( ) 22

1

1 / 1 01 / 1 / 1 /A A A

y y R IydA y dAdA Iy R y R R y R R+

= + = + =+ + +∫ ∫ ∫

1

2

2

1 /

1 /

A

A

ydAIy R

y dAIy R

=+

=+

∫

∫21

IIR

= −

R r y− = −

or r y R= +

mA A

dA dARA R Rr y R

= =+∫ ∫

( )( )

1

A A A A

m

y R y dAA dA dA dA Ry R y R y R

I RAR

+= = = +

+ + +

= +

∫ ∫ ∫ ∫

1 /I R

Thus, ( ) 1 2 /mR RA A I I R− = − =

( ) 211/

kM MyA ky I

k R

θθσ = − −+

=

Now, start with Bickford’s expression

( )( )( )

( ) ( )( ) ( )

( )

( )( )( )( )

2

22

2

2 32

2 2

2 22

2 2

222

2 2

2 32

22

1

//

yRMAR y R I

y R I yARM

y R ARI

y R I y R AR ARMy R ARI y R ARI

I AR RMARI y R I

I ARRMy R I ARI

A y R I AR RM

y R AI R

θθσ⎡ ⎤

= − +⎢ ⎥+⎣ ⎦⎡ ⎤+ +

= − ⎢ ⎥+⎣ ⎦⎡ ⎤+ + +

= − −⎢ ⎥+ +⎣ ⎦⎡ ⎤+

= − −⎢ ⎥+⎣ ⎦⎡ ⎤+

= −⎢ ⎥+⎣ ⎦⎡ ⎤− + +⎢ ⎥=

+⎢ ⎥⎣ ⎦

same terms added in and subtracted out

( ) ( )( )

2 32

22

//

A y R I AR RM

y R AI Rθθσ⎡ ⎤− + +⎢ ⎥=

+⎢ ⎥⎣ ⎦

but 22 /mRA A I R− = so ( )2 3

2 /mA I AR R= +

and we find

y R r+ =

( )m

m

A rAr RA A Aθθσ⎡ ⎤−

= ⎢ ⎥−⎣ ⎦

which agrees with our previous expression

from Bickford’s expression

( ) 211/

kM MyA ky I

k R

θθσ = − −+

=

it is easy to see, as , R →∞ 0k →

and 22

2 1 /A A

y dAI y dA Iy R

= → =+∫ ∫

and we recover the straight beam flexure expression

MyIθθσ = −