1 /29 M.Chrzanowski: Strength of Materials SM2-06: Bending deflections BENDING DEFLECTIONS

1/29M.Chrzanowski: Strength of Materials

SM2-06: Bending deflections

BENDING DEFLECTIONS



z

wz

x

1

c

0

limx1. Strain definition:

z 2. From picture:

zcx 5. Plane cross-section hypothesis

z

x 4. Linear strain

z

x0

lim

3. Substitution of 2 into 1

MyMy

Geometry of deformation



z

x 1. Normal strain:

zxx EE 2. From Hooke law:

zJ

M

y

yx 3. Normal stress:

1

EJ

M

y

y 4. From comparison of 2 and 3:

y

y

EJ

M

1

5. Beam curvature

Normal stresses and beam curvature



"wEJ

M

y

y

y

y

EJ

M1. Curavture-bending moment

relationship:

"//1

/ 222/32

22

wdxwddxdw

dxwd

2. Formula for curvature of a

curve:

3. Differential equation for beam

deflection w(x)

Differential equation for beam deflection



The signs of w(x), w’(x) i w’’(x) depend upon co-ordinate system:

w(x)

x

w(x)

x

w>0, w’<0, w’’>0

w(x)

x

w(x)x

Sign convention

w>0, w’>0, w’’>0

w>0, w’>0, w’’>0 w<0, w’<0, w’’<0



The sign of M(x) follows adopted convention (M is positive when „undersides” are under tension):

M>0

M>0

M<0

M<0

Sign convention



Combination of the previous two conventions will result in following form of the bending deflection equation: xw

EJ

xM

y

y "

PROVIDED that the positive direction of w axis will coincide with positive direction of bending moment axis pointing towards „undersides”:

The direction of x-axis only does not change the sign in the equation of deflection.However, one has to remember that in this case the first derivative of

deflection w’ will change the sign!

In the opposite case of the two directions discordance the minus sign in the bending equation has to be replaced by the positive sign

Sign convention

w(x)x

My

w(x)

xMy

or



To find beam deflection one has to integrate the deflection equation twice.

The first integration yields a tangent to the beam axis and, therefore, rotation of the beam cross-section

CdxEJ

Mw

y

y '

DCxdxdxEJ

Mw

y

y )(

The next integration results in finding beam deflection:

y

y

EJ

Mw "

w(x)

x

)'(warctg

w

To determine the values of integration constants C and D we need to formulate boundary conditions

Integration of the deflection equation



The boundary conditions have to represent supports of beam:

w=0w=0 w=0 w=0w=0 w=0

A

w=0 w’=0 w=0 w’=0w=0 w’=0

B

NOTICE: As we do not take into account normal forces all three cases shown in the row A or B are equivalent with respect to deflections calculation.




In a general case when moment bending equation cannot be given by in the analytical form for the whole beam it has to be formulated and integrated for all characteristic sections of a beam

As a consequence we need to find 2n constants of integration (where n denotes number of characteristic sections) and to write down 2n-2 (2 boundary conditions correspond to beam supports) additional adjustment conditions at the neighbouring characteristic sections

This procedure yields the set of 2n linear algebraic equations. The solution of this set can be cumbersome and it is advisable only if need an analytical form of a bending deflection for the whole beam.




Example of the boundary conditions adjustment

n 1 2 3 4 5 6

w1=w2

w’1=w’2

w2=w3

w’6=0

w4=w5 w5=w6 w6=0

w’2=w’3 w’5=w’6w’3=w’4

w3=w4

w1=0

Independent integration

Adjustment of first derivatives

Adjustment of the deflections



The Conjugate Beam Method(Mohr fictitious beam method)



STATICS DEFLECTIONS

xQdx

xdMz

xqdx

xdQz

This equation (7) is „integrated” on the basis of M(x) i Q(x) definitions when external loading q(x) and boundary conditions are known .

EJ

xM

dx

xwd

2

2

Why not to use the same method to determine the

deflections?

xqdx

xMd

2

2 )(

?

?

Mohr method



STATICS - Fictitious domain DEFLECTIONS - Real domain

FFFF DxCdxdxxqxM

dx xqdx

MdF

F 2

2

dx FFFF CdxxqQ

dx

dM

RR

RR DxCdxdx

EJ

xMxw

dx

RR

RR Cdx

EJ

xMw

dx

dw '

dx EJ

xM

dx

wd RR 2

2

EJ

xMxq R

F I F

T H E N

CF= CR , DF=DR

wR(x)MF(x) , w’R QF(x)



EJ

xMxq R

F

wR(x)MF(x) , w’R QF(x)

From the condition:

CF= CR , DF=DR

follows, that the only loading of the fictitious beams will be continuous loading of the dimension [Nm/(Nm-2m4)]=[ m-1] distributed exactly like bending moment distribution for the real beam. Therefore, the bending moment and shear force distributions in the fictitious beams cannot contain any discontinuities (no loading in form of concentrated moments or forces exists).

[1/m]

To satisfy the conditions:

the kinematical conditions have to bet set upon the fictitious beam in such a way that in characteristic points will be:

So, if for the real beam wR=0 in a given point, then for the fictitious beam has to be

MF=0 in this point. Similarly if w’R =0 then QF =0 etc.

Fundamental requirement to be satisfied is that fictitious and real beams have the same length (0 ≤ xR ≤ l, 0 ≤ xF ≤ l).



Fic

titi

ou

s b

eam

Deflection wR

Rotation w’R

Rea

l b

eam

Bending moment MF

Shear force QF

Supports (KBC)F

ww0

0 0

0 0

0

0 0

0

Example #1 Example #2

Supports (KBC)R

0

0

0

0

0

0

0

0



Fic

titi

ou

s b

eam

Deflection wR

Rotation w’R

Rea

l b

eam

Bending moment MF

Shear force QF

Supports (KBC)F

ww0

Example #3 Example #4

Supports (KBC)R

0

0

0

0 0

0

0

0

0

0

0

0

0

0

0

PL ww ''

PL QQ

0

0PL ww

0PL ww

PL MM

PL QQ 0

0

0

0

0

0



0 PL ww

0 PL QQ

0FM0FQ

0FM

0 PL MM

0 PL QQ

0'' PL ww

0FM0FQ

Built-in support Free endPin-pointed supportHingeBELKA FIKCYJNA

0'w 0 PL ww0w 0w 0w

0'w

Rea

l b

eam

Fic

titi

ou

s b

eam



Ben

ding

mom

ents

usi

ng t

he s

uppe

rpos

ition

prin

cipl

e



Bending moment diagram for fictitious beam

Real beam

Bending moment for the fictitious beam

Fictitious beam loading

Elementary but troublesome !

Fictitious beam

w

≡ deflections of the real beam



Mohr method applied to the evaluation of deflections and rotations in chosen points

l

aw

xEJ

BA

Pa Bending moments for the fictitious beam

MBF=wB=

2a/3a/3

(Pa/EJ)(a/2) l-a/3

wA=

w’B=

w’A=

wB=

MAF=wA=

QAF=QB

F=w’A=w’B=

Shear forces for the fictitious beam:

(Pa/EJ)(a/2)(2a/3)= Pa3/3EJ

(Pa/EJ)(a/2)[l-a/3]= [Pa2/2EJ][l-a/3]

(Pa/EJ)(1/2)a=Pa2/2EJ

Pa3/3EJ Pa2/2EJ

[Pa2/2EJ][l-a/3] Pa2/2EJ

Straight line

3rd order parabola

P

Pa/EJ

w



B

w

xEJ

A

P

a

Pl

Bending moment for the fictitious beam

MBF=wB=

l/3 2l/3

Pl/EJ)(1/2)l

w’B= Pl2/2EJ

QAF=QB

F=w’A=w’B=

Shear forces for the fictitious beam

Pl/EJ)(1/2)l[2l/3]= Pl3/3EJ

Pl/EJ)(1/2)l= Pl2/2EJ

a = l

P

wB= Pl3/3EJ = wmax

Pl/EJ

Special case: a=l



[Pa2/2EJ][l-a/3]

x

l

a

w

EJ

BA

P

wwB=

2

22

3

3

31

2 l

ll

l

a

EJ

PawB

2

23

2

3

31

3 l

a

l

a

EJ

PlwB

23

2

3

31

3

EJ

PlwB

0,50,25

0,75

1,0

f

0,25

0,75

0,5

1,0

l

a

0,01

0,31

0,63

0,61

Special example: end beam deflection as the function of force position

wmax

fwB wmax



Area and centre of gravity for parabolic shapes

Parabola of nth degree

Horizontal tangenta

b

c

A= ab/(n+1)

Area:

Position of the centre of gravity:

c= b/(n+2)

An c

0

1

2

3

…

ab

ab/2

ab/3

ab/4

…

b/2

b/3

b/4

b/5

…



y

y

EJ

M

dx

wd

2

2

Let us differentiate twice the equation:

zy Q

dx

dM q

dx

dQz making use of M-Q-q relationships

y

y

EJ

M

dx

dw

dx

d"

y

y

EJdx

dM

dx

wd 13

3

y

z EJQ

dx

wd 13

3

First differentiation yields:

yz EJ

Qdx

d

dx

wd

dx

d 13

3

yEJ

qdx

wd 14

4

Second differentiation:

CdxEJ

Mw

y

y ' DCxdxdxEJ

Mw

y

y )(

Double integration of the equation w”= - M/EJ allows for finding rotations and deflections:

General bending equation



yz EJ

Qdx

wd 13

3

yEJq

dx

wd 14

4

CdxEJ

Mw

y

y ' DCxdxdxEJ

Mw

y

y )(y

y

EJ

M

dx

wd

2

2

DIF

FE

RE

NT

IAT

IONI N T E G R A T I O N

Loading

RotationBending moment; Curvature

Shear force

Deflection

The overall picture of bending problem

General bending equation



xEJ

xM

dx

xwd

2

2Therefore the points in which bending stiffness changes are also a characteristic point.

If EJ changes along beam axis EJ(x), then differential equation for displacement becomes:



Deflection of a beam of variable stifness

EJ2

Pa2/EJ2

Pa2/EJ1

wK= Pl3/3EJ

x

l

{Pl/EJ2} {1-J2/J1}

wK= ?

Pl

Pl/EJ2

xEJ

l

P

K

w

Pl

Pl/EJ1

Pl/EJ2

OR…

EJ2

{Pa2/EJ2} {1-J2/J1}

{Pl/EJ2} {1-J2/J1}

EJ1

K

w

P

a1 a2

> EJ2

a1

{P(l-a2)/a1} {1-J2/J1}

{Pa2} {1-J2/J1}

Pl/EJ



EJ2

Pa2/EJ2

Pa2/EJ1

wK= Pl3/3EJ

x

l

{Pl/EJ2} {1-J2/J1}

Pl

Pl/EJ2

xEJ

l

P

K

w

Pl

Pl/EJ1

Pl/EJ2

OR…

EJ2

{Pa2/EJ2} {1-J2/J1}

{Pl/EJ2} {1-J2/J1}

EJ1

K

w

P

a1 a2

> EJ2

a1

{P(l-a2)/a1} {1-J2/J1}

{Pa2} {1-J2/J1}

Pl/EJ



stop

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1 /29 M.Chrzanowski: Strength of Materials SM2-06: Bending deflections BENDING DEFLECTIONS