Combined bending of unsymmetrical beams

Internal forces and moments are defined positive as shown

y

z

x

My

Mz

Vy

Vz

Mx =T

Looking down the negative y -axis

O

dx

zV zz

dVV dxdx

+

yM yy

dMM dx

dx+

0

0

O

yy z y

M

dMM V dx M dx

dx

=

⎛ ⎞+ − + =⎜ ⎟

⎝ ⎠

∑

yz

dMV

dx=

+

Looking down the z -axis

O

dx

yV yy

dVV dx

dx+

zM zz

dMM dxdx

+

0

0

O

zz y z

M

dMM V dx M dxdx

=

⎛ ⎞− + + + =⎜ ⎟⎝ ⎠

∑

zy

dM Vdx

= −

+

Assume that plane sections remain plane but that is bending induced about both the y and z axes even if the moment might only be about one axis

xdw dvu z ydx dx

= − −

w is the displacement of the neutral axis in the z-directionv is the displacement of the neutral axis in the y-direction

Thus, the axial strain developed is2 2

2 2x

xxu d w d ve z yx dx dx

∂= = − −∂

curvatures of the neutral axis

If σxx is the only major stress 2 2

2 2xx xxd w d vEe E z ydx dx

σ⎛ ⎞

= = − −⎜ ⎟⎝ ⎠

Note that there are also strains

2 2

2 2yy zz xxd w d ve e e z ydx dx

ν ν⎛ ⎞

= = − = +⎜ ⎟⎝ ⎠

Relationship between the flexural stress and the internal moments

y

z

x

xxσy

z My

Mz

0xx

xx z

xx y

dA

y dA M

z dA M

σ

σ

σ

=

= −

=

∫∫∫

(1)

(2)

(3)

From (1) 2 2

2 2 0d w d vE zdA E ydAdx dx

− − =∫ ∫

0zdA ydA= =∫ ∫ i.e. neutral axis is at the centroidof the cross section

From (2) 2 22

2 2

2 2

2 2

z

yz zz z

d w d vE yzdA E y dA Mdx dxd w d vE I E I Mdx dx

+ =

+ =

∫ ∫

From (3)2 2

22 2

2 2

2 2

y

yy yz y

d w d vE z dA E yzdA Mdx dxd w d vE I E I Mdx dx

− − =

− − =

∫ ∫

2 2

2 2

2 2

2 2

z yz zz

y yy yz

d w d vM EI EIdx dx

d w d vM EI EIdx dx

= +

= − −

solving these two equations for the curvatures

( )( )( )( )

2

2 2

2

2 2

z yy y yz

yy zz yz

y zz z yz

yy zz yz

M I M Id vdx E I I I

M I M Id wdx E I I I

+=

−

− +=

−

and placing these curvature into the flexural stress relation gives

( ) ( )( )2

y zz z yz z yy y yzxx

yy zz yz

M I M I z M I M I y

I I Iσ

+ − +=

−

( ) ( )( )2

y zz z yz z yy y yzxx

yy zz yz

M I M I z M I M I y

I I Iσ

+ − +=

−

Note: If either the y or z axis is a plane of symmetry then Iyz =0 and we find

y zxx

yy zz

M z M yI I

σ = −

The neutral surface is located where σxx = 0

y

z

neutral surface

λ

tanz yλ=

where

( )( )/

tan/

z y yy yzz yy y yz

y zz z yz zz z y yz

M M I IM I M IM I M I I M M I

λ++

= =+ +

Although the bending moments may individually be functions of x, if the ratio Mz / My is independent of x (called single plane of loading), then the angle of the neutral axis will be a constant in the beam. Otherwise, the angle will vary with x.

θ

θ

θ

single plane of loading

general loading

/ 2π

P

P

q

q

Flexure Stress Expressions

( ) ( )( )2

y zz z yz z yy y yzxx

yy zz yz

M I M I z M I M I y

I I Iσ

+ − +=

−(1)

(2) can use ( )

( )/

tan/

z y yy yz

zz z y yz

M M I I

I M M Iλ

+=

+to eliminate explicit dependency

on Mz in equation (1) and obtain

( )tantan

yxx

yy yz

M z yI I

λσ

λ−

=− Note: Mz is still in here

(3) can use principal axes of the cross section

y

z

YZ

pθ

sin coscos sin

Y z yZ z y

θ θθ θ

= += −

Y Zxx

YY ZZ

M Z M YI I

σ = −

yθ

z

y’z’

( ) ( )

( ) ( )

( ) ( )

cos 2 sin 22 2

cos 2 sin 22 2

sin 2 cos 22

yy zz yy zzy y yz

yy zz yy zzz z yz

yy zzy z yz

I I I II I

I I I II I

I II I

θ θ

θ θ

θ θ

′ ′

′ ′

′ ′

+ −= + −

+ −= − +

−= +

transformation relations for area moments

parallel axis theorem

y

z

G

yG

zG

y’

z’

( ) ( )22 2zz G z z GG

I y dA y y dA I Ay′ ′′= = + = +∫ ∫similarly

( )( )

2yy y y GG

yz y z G GG

I I Az

I I Ay z

′ ′

′ ′

= +

= +

G is the centroid of A

principal area moments and principal directions

( )2

21, 2

2tan 2

2 2

yzp

yy zz

yy zz yy zzp p yz

II I

I I I II I

θ−

=−

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

(4) can use neutral surface coordinates

y

z

λ

η

ζ

(eta)

(zeta)

can write ( )2 tany zz z yzxx

yy zz yz

M I M Iz y

I I Iσ λ

+= −

−

but cos sintan

cos cosz yz y λ λ ζλ

λ λ−

− = =

so

2 cosy zz z yz

xxyy zz yz

M I M II I I

ζσλ

+=

−

displacements of the neutral axis

( )( )( )( )

2

2 2

2

2 2

z yy y yz

yy zz yz

y zz z yz

yy zz yz

M I M Id vdx E I I I

M I M Id wdx E I I I

+=

−

− +=

−

we must integrate

twice to obtain the displacements, using boundary conditions on both v and w

( )( )

2

2

2

2

2 2

2 2

tan

tan

z yy y yz

y zz z yz

d vM I M Idx

d w M I M Idxd v d wdx dx

λ

λ

− += = −

+

= −

If λ is a constant and we let ( )2

2

d w F xdx

= − we find, on integration

( )

( )

1 2

3 4tan

w F x dxdx C x C

v F x dxdx C x Cλ

= − + +

−= − + +

∫∫

∫∫

If the boundary conditions on v, w are the same so that

1 3 2 4,C C C C= = then

1tan

wv λ

−= or tan v

wλ −=

y

z

λw

-v

λ

δ = total displacement

Thus, in this case the total displacement

2 2w uδ = + is perpendicular to the

neutral surface.

Example 1

3 m

60o P = 12 kN (through the shear center)

140 mm

200 mm

20 mm all around

Determine:

1. The maximum tensile and compressive stresses and their location

2. The total deflection at the load P

E = 72 GPa

140 mm

200 mm

20 mm all around

1

2

3 y

z

zc

( )( ) ( ) ( )

1 1 2 2 3 3

1 2 3

2 100 4000 10 200010000

82

cz A z A z Az

A A A

mm

+ +=

+ +

× +=

=

( )( ) ( )( )( ) ( )( )

( )( ) ( )( )( )

( )( ) ( )( )( )

3 2 3

6 4

3 2

3 2

6 4

0

1 12 200 20 200 20 70 10 20 10012 12

30.73 1012 20 200 200 20 100 82

121 100 20 100 20 82 10

1239.69 10

yz

zz

yy

I

I

mm

I

mm

=

⎡ ⎤ ⎡ ⎤= × + − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= ×

⎡ ⎤= × + −⎢ ⎥⎣ ⎦⎡ ⎤+ + −⎢ ⎥⎣ ⎦

= ×

Maximum bending moments are at the wall

yz

sin 60P o

cos60P o

6

6

300 sin 60 31.17 10

3000 cos60 18 10y

z

M P N mm

M P N mm

= − = − × −

= = × −

o

o

cot 60 0.577z

y

MM

= − = −o

( )( )

( )

/tan

/

39.690.577 0.74630.73

z y yy yz

zz z y yz

M M I I

I M M Iλ

+=

+

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

0.64036.7

radλ = −

= − o

36.7o

( )

( )

tantan

31.17 0.74639.690.785 0.585

yxx

yy yz

M z yI I

z y

z y

λσ

λ−

=−

−= +

= − −

N-mm mm

mm4

N/mm2 = MPa

36.7o

A

B

At point A: 200 82 11870

z mmy mm= − ==

( ) ( )( ) ( )( )0.785 118 0.585 70

133.6xx A

MPa

σ = − −

= −

At point B: 8270

z mmy mm= −= −

( ) ( )( ) ( )( )0.785 82 0.585 70

105.3xx B

MPa

σ = − − − −

=

alternately, with principal values

31.17 1839.69 30.730.785 0.585

y zxx

yy zz

M z M yI I

z y

z y

σ = −

= − −

= − − which is same as before

Now, compute deflection at the load

yz

sin 60P o

cos60P o

2

2y

yy

Md wdx EI

−=

P

3

3PLwEI

=

L( )

( )( )( )( )

3

3

3 6

sin 603

12,000sin 60 3000

3 72 10 39.69 10

32.72

yy

P Lw

EI

mm

=

=× ×

=

o

o

tan 0.745 24.38v w w mmλ= − = =

2 2 40.8w v mmδ = + =

Example 2

1.5 m

P = 500 N1.5 m

q = 1000 N/m

Determine the angle that the neutral surface makes with respect to the y-axis as a function of x. Use the cross section properties of example 1.

y

z

First, we need the reactions at the wall

3.0 m

P = 500 N

0.75 m

500 N 1500 N1125 N-m

1500 N-m

1500 N

500 N1500 N

1125 N-m 1500 n-m

My

Mz

x

1000 N/m

For 0 < x < 1.5

o

2

2

0

1500 1000 / 2 1125 0

500 1500 1125

zo

z

z

M

M x x

M x x N m

=

+ − − =

= − + −

∑ 0

550 1500 0

1500 500

yo

y

y

M

M x

M x N m

=

− + =

= − + −

∑

For 1.5 < x < 3

x

0.75 m

500 N 1500 N1125 N-m

1500 n-m

1500 N

My

Mz

o

01500 1500( 0.75) 1125 00

zo

z

z

MM x xM

=

+ − − − ==

∑ 0

1500 500 0

500 1500

yo

y

y

M

M x

M x N m

=

+ − =

= − −

∑

39.69tan 1.2930.73

yy z z z

zz y y y

I M M MI M M M

λ = = =

0 < x < 1.5

2500 1500 1125tan 1.29500 1500

x xx

λ⎛ ⎞− +

= ⎜ ⎟−⎝ ⎠

1.5 < x < 3

0tan 1.29 0500 1500

0x

λ

λ

⎛ ⎞= =⎜ ⎟−⎝ ⎠=

% script neutral_axis

x=linspace(0,1.5,200);num=1.29.*(500*x.^2-1500.*x+1125);denom = 500.*x-1500;ang=atan(num./denom);ang = ang*180/pi; % change angle to degreesx_plot=linspace(0,3,400);ang_plot=[ang zeros(size(x))];plot(x_plot, ang_plot)xlabel('x-distance')ylabel('angle, degrees')

0 0.5 1 1.5 2 2.5 3-45

-40

-35

-30

-25

-20

-15

-10

-5

0

x-distance

angl

e, d

egre

es