Unsymm Bending

8/12/2019 Unsymm Bending

1/29

Combined bending of unsymmetrical beams

Internal forces and moments are defined positive as shown

y

z

x

My

Mz

Vy

Vz

Mx =T


2/29

Looking down the negative y -axis

O

dx

zV z

z

dVV dx

dx+

y

y

y

dM

M dxdx+

0

0

O

y

y z y

M

dMM V dx M dx

dx

=

+ + =

y

z

dMV

dx=

+


3/29

Looking down the z -axis

O

dx

yV y

y

dVV dx

dx+

z

z

z

dM

M dxdx+

0

0

O

zz y z

M

dMM V dx M dx

dx

=

+ + + =

zy

dMV

dx=

+


4/29

Assume that plane sections remain plane but that is bending

induced about both the y and z axes even if the moment might onlybe about one axis

x

dw dvu z y

dx dx

=

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

v is the displacement of the neutral axis in the y-direction

Thus, the axial strain developed is

2 2

2 2

x

xx

u d w d ve z y

x dx dx

= =

curvatures of the neutral axis

If xx is the only major stress 2 2

2 2xx xx

d w d v

Ee E z ydx dx

= =


5/29

Note that there are also strains

2 2

2 2yy zz xx

d w d ve e e z y

dx dx = = = +

Relationship between the flexural stress and the internal moments

y

z

x

xxy

zMy

Mz

0xx

xx z

xx y

dA

y dA M

z dA M

=

=

=

(1)

(2)

(3)


6/29

From (1) 2 2

2 2

0d w d v

E 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 M

dx dx

d 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 dx

d w d vE I E I M

dx dx

=

=


7/29


8/29

( ) ( )

( )2y zz z yz z yy y yz

xx

yy zz yz

I M I z M I M I y

I I I

+ +

=

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 y

I I =

The neutral surface is located where xx = 0

y

z

neutral surface

tanz y=

where

( )( )/

tan/

z y yy yzz yy y yz

y zz z yz zz z y yz

M M I IM I M I

M I M I I M M I

++= =+ +


9/29


10/29

Flexure Stress Expressions

( ) ( )( )2

y zz z yz z yy y yz

xx

yy zz yz

I M I z M I M I y

I I I

+ +=

(1)

(2) can use( )

( )

/tan

/

z y yy yz

zz z y yz

M M I I

I M M I

+=

+to eliminate explicit dependency

on Mz

in equation (1) and obtain

( )tan

tan

y

xx

yy yz

z y

I I

=

Note: Mz is still in here


11/29

(3) can use principal axes of the cross section

y

z

Y

Z

p

sin cos

cos sin

Y z y

Z z y

= +

=

Y Zxx

YY ZZ

Z M Y

I I =


12/29

y

z

yz

( ) ( )

( ) ( )

( ) ( )

cos 2 sin 2

2 2

cos 2 sin 22 2

sin 2 cos 22

yy zz yy zz

y y yz

yy zz yy zz

z z yz

yy zz

y z yz

I I I II I

I I I II I

I II I

+ = +

+ = +

= +

transformation relations for area moments


13/29

parallel axis theorem

y

z

G

yG

zG

y

z

( ) ( )22 2

zz G z z GGI y dA y y dA I Ay = = + = +

similarly

( )

( )

2

yy y y GG

yz y z G GG

I I Az

I I Ay z

= +

= +

G is the centroid of A


14/29

principal area moments and principal directions

( )

2

2

1, 2

2tan 2

2 2

yz

p

yy zz

yy zz yy zz

p p yz

I

I I

I I I II I

=

+ = +


15/29

(4) can use neutral surface coordinates

y

z

(eta)

(zeta)

can write ( )2 tany zz z yz

xx

yy zz yz

M I M I z yI I I

+=

butcos sin

tan

cos cos

z yz y

= =

so

2 cos

y zz z yz

xx

yy zz yz

M I M I

I I I

+=


16/29

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 v

dx E I I I

M I M Id w

dx 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 v M I M Idx

d w M I M I

dx

d v d wdx dx

+= =

+

=

If is a constant and we let ( )2

2

d wF x

dx= we find, on integration


17/29

( )

( )

1 2

3 4tan

w F x dxdx C x C

vF x dxdx C x C

= + +

= + +

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

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

1

tan

w

v

= or tan v

w

=

y

z

w

-v

= total displacement

Thus, in this case the total displacement

2 2w u= + is perpendicular to the

neutral surface.


18/29

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 andcompressive stresses and their location

2. The total deflection at the load P

E = 72 GPa


19/29

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 2000

1000082

c

z A z A z Az

A A A

mm

+ +=

+ +

+=

=

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

( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

3 2 3

6 4

3 2

3 2

6 4

01 1

2 200 20 200 20 70 10 20 10012 12

30.73 10

12 20 200 200 20 100 8212

1100 20 100 20 82 10

12

39.69 10

yz

zz

yy

I

I

mm

I

mm

=

= + +

=

= +

+ +

=


20/29

Maximum bending moments are at the wall

y

z

sin 60P o

cos60P o

6

6

300 sin 60 31.17 10

3000 cos 60 18 10

y

z

P N mm

M P N mm

= =

= =

o

o

cot 60 0.577z

y

M

M= = o

( )( )

( )

/tan/

39.690.577 0.746

30.73

z y yy yz

zz z y yz

M M I II M M I

+=+

= =

0.640

36.7

rad=

= o36.7o


21/29

( )

( )

tan

tan

31.170.746

39.69

0.785 0.585

y

xx

yy yz

M z y

I I

z y

z y

=

= +

=

N-mm mm

mm4

N/mm2 = MPa

36.7o

A

B

At point A: 200 82 118

70

z mm

y mm

= =

=

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

133.6

xx A

MPa

=

=

At point B: 82

70

z mm

y mm

=

=

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

105.3

xx B

MPa

=

=


22/29

alternately, with principal values

31.17 18

39.69 30.73

0.785 0.585

y zxx

yy zz

M z M y

I I

z y

z y

=

=

= which is same as before


23/29

Now, compute deflection at the load

yz

sin 60P o

cos60P o

2

2

y

yy

Md w

dx EI

=

P

3

3

PLw

EI=

L

( )

( )( )( )( )

3

3

3 6

sin 60

3

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 mm= = =

2 2

40.8w v mm= + =


24/29

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


25/29

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


26/29

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 0500 1500 1125

zo

z

z

M

M x xM x x N m

=

+ == +

0

550 1500 01500 500

yo

y

y

M

M xM x N m

=

+ == +


27/29

For 1.5 < x < 3

x

0.75 m

500 N 1500 N1125 N-m

1500 n-m

1500 N

My

Mz

o

0

1500 1500( 0.75) 1125 0

0

zo

z

z

M

M x x

M

=+ =

=

01500 500 0

500 1500

yo

y

y

M

M x

M x N m

=+ =

=


28/29

39.69tan 1.29

30.73

yy z z z

zz y y y

I M M M

I M M M= = =

0 < x < 1.5

2500 1500 1125tan 1.29

500 1500

x x

x +=

1.5 < x < 3

0tan 1.29 0

500 1500

0

x

= =

=


29/29

% 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 degrees

x_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

angle,

degrees

Unsymm Bending

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