17_defl_print.pdf

7/30/2019 17_defl_print.pdf

1/61

. . . . . .

Lecture : Bending (V) Deflection of Beams

Yubao Zhen

Nov ,

() Lecture : Bending (V) Deflection of Beams Nov , /


2/61

. . . . . .

Review: The Shear Formula

. fully stressed beams

. Transverse shear stress: =

VQ

Ib obtained via longitudinal shear stress location of neutral axis Calculation ofQ (using the outer part aboutNA) direction: in accord with V

. transverse shear stress varies along the height

. limitations on the application of shear formula

two assumptions: . V; . uniform over width

.

results on typical cross-sectionsrectangular: (max.

avg)

circular (solid): (max.

avg)

wide-flange: Q calculation



3/61

. . . . . .

Outline

. elastic/deflection curve (/)

characteristics, notations and conventions

. differential equations for elastic curve

bending-moment eqn., shear-force eqn. and load eqn.

. integration method for elastic curve ()



4/61

. Elastic/deflection curve

(/)

. . . . . .


5/61

. . . . . .

Shape of bent beams

Observations:

deformation of beams subjected to lateral loads

longitudinal axis becomes curved after deformation

continuous and differentiable (at least to the first order)

small slope assumption (angles between xand tangents to curve )

deflection (): vertical displacement of any point on the axis

static: structure design (deformation/stiffness control)

dynamic: response of dynamic loadings (vibrations, etc.)



6/61

. . . . . .

The elastic/deflection curve

On the deflected shape of the beamelastic curve ()

The deflection diagram of the longitudinal axis that passes through the centroid of each

cross-sectional area of the beam

deflection curve ()

the deformed shape of the originally straight longitudinal axis of a beamwhen it is

subjected to lateral forces

strategy for determination:

curve () curvature () bending moment ()



7/61

. . . . . .

Notation and sign convention for deflection

x

v

v(x)

(x)

v

v

(x

)

positive sensenegative sense

(x)

F

F

longitudinal axis: x, positive: to the right

deflection: v, positive: upward

deflection angle ()/angle of rotation: , positive: counter-clockwise

curvature and radius of curvature :

follow that of bending momentM



8/61

. . . . . .

Curvature and deflection angle

y

xA B

v dv

du

O

x dx

dsv

m1

m2

r

du

m1

m2

ds

u

u du

dxxx

v

v dv

d= ds, =

=

d

ds

d

dx(see below)

cos =dx

ds, dx= ds cos ds(

+

)

deflection angle: tan =dv

dx( ); curvature:

dv

dx



9/61

. . . . . .

Sign convention with xreversed

d

xx

ds

A

B

O

v v

v

v + dv

dx

dx

= dx d

ifxpoints to left, for a physical segment,

algebraically:

=

dv

dx=

dv

dx=

dv

dx=

=

d

dx

=

d

(

)dx=



10/61

. Differential equations for

deflection curve

. . . . . .


11/61

. . . . . .

Deflection angle and deflection

x dx

v

v + dv

A

B

ds

+ d

d

Relations between , deflection v, and

curvature

tan =dv

dx, = arctan

dv

dx, =

=

d

ds

ds =

+ dv

dxdx

For small deflection angles, ,

ds

dx =

=

d

ds=

d

dx

tan =dv

dx

d

dx=

dv

dx

=

=

dv

dx



12/61

. . . . . .

Deflection angle and deflection an alternative way

from calculus ()

=

dv

dx

+

(dv

dx

)

/

dv

dx, here

dv

dx

=dvdx

=

all the above relations arevalid for any material, provided that

consistent with the sign conventions



13/61

. . . . . .

Connecting deflection curve to applied moment

moment-curvature relation (review)

=

ds

ds

dsdx= ds = d

ds

=

( y

)d

=( y)ddd =

y

= y

=

E=

My

EI

=

=

M

EI

: radius of curvature

:curvature



14/61

. . . . . .

Differential equations of the deflection curve

combining =dv

dxand =

M

EI

dv

dx=

M

EI

(M(x): bending moment, EI: flexural rigidity/stiffness ())differential equations of the deflection curve (EI= const.)

()

EIdv

dx=M(x) bending-moment equation

dM

dx= V

(x

) EI

dv

dx= V

(x

)shear-force equation

dV

dx= q(x) EIdv

dx= q(x) load equation



15/61

. . . . . .

Generalization of the diff. eqn. of deflection curve

For non-uniform EI(e.g. non-prismatic bars)

EI

(x

)dv

dx=M

(x

)bending-moment equation

dMdx

= V(x) ddxEI(x)dvdx = V(x) shear-force equationdV

dx= q

(x

)

d

dx

EI

(x

)dv

dx

= q

(x

)load equation



16/61

. . . . . .

Remarks on the diff. eqns.

EI

dv

dx,dv

dx,dv

dx=

(M

(x

),V

(x

), q

(x

))all q(x),V(x),M(x) may containjumps, hence only dv

dxand vare

continuous.

For piece-wise q(x),V(x),M(x), diff. eqn. has a piece-wise () formfor given load, any form of the three is applicableEqn. solving requires integration to get vand

dv

dxon the integration constants

.

source boundary conditions () and continuity conditions ().

. number of integration constants () M, ; V, ; q,



17/61

. Sketching deflection curves

. . . . . .


18/61

. . . . . .

Characteristic values on the deflection curve

maximum deflection ()

conditions: inner points: dv/dx= , or at outer boundary pointsinflection point ()

where curvature changes its sign

max. deflection angle ()



19/61

. . . . . .

Sketching () the deflection curve rules

Sketching the shape before actual solving:

. local curvature followsM application ofmoment diagram

sign of internal moment convex/concave shape

. respect the constraints.

support that resists a force displacement constraint support that resists a moment rotation constraint combination of the above


Th l l d fl i


20/61

. . . . . .

The local curvature on a deflection curve

with =M

EI

local curvature follows strictly the sense of localM

xwhereM(x) = inflection point ( = )


Sk hi h d fl i ill i


21/61

. . . . . .

Sketching the deflection curve an illustration


F th ill t ti


22/61

. . . . . .

Further illustration



23/61

. Integration method

for deflection curve

()

. . . . . .

Solving the diff eqn of deflection curve


24/61

. . . . . .

Solving the diff. eqn. of deflection curve

basic idea and procedure

choose one type of eqn.integrate consecutively () up to v

apply any of the known conditions on shear, moment, slope or

displacement to solve integration constants

substitute back

conditions for integration constants determination

. boundary conditions (usually at the supports)

. continuity conditions (usually at places of jumps)

matching continuous quantities at connections of ranges (usually vanddv

dx)

. symmetry conditions (subset in the above two types)


Typical constraints in terms of v M and V


25/61

. . . . . .

Typical constraints in terms ofv, , Mand V

correspondence:

v, dv

dx, M

dv

dx, V

dv

dx


The continuity conditions


26/61

. . . . . .

The continuity conditions

A BC

C

A B

(v)AC=(v)CB

(v)AC=(v)CB

At point C:

FIG. 9-7 Continuity conditions at

point C

ensure continuity of

. deflection v

. deflection angledv

dx


Application of the continuity conditions


27/61

. . . . . .

Application of the continuity conditions

x1

x2

v

v

A

B

dv

dx1

dx2

opposite xcoordinates

conditions matching in uniform/opposite xsystems:

right-going x: v,dv

dx,dv

dx ,dv

dxpiece-wise function in x(common origin, same direction)

left-going x: v, dv

dx,dv

dx,

dv

dxpiece-wise function in x(different origins, opposite directions)


Example


28/61

. . . . . .

Example

Determine elastic curve for the cantilever beam (EI= const.)

P

A B

xL

vA A

. set up coordinate system

. FBDmomentM= Px

. Apply bending-moment equation

EIdv

dx= Px

EIdv

dx=

Px

+ C

EIv= Px

+ Cx+ C


Example (cont )


29/61

. . . . . .

Example (cont.)

P

A B

xL

vA A

. apply B.C.s at B (x= L)

v x=L = vx=L = =

PL

+ C, =

PL

+ CL + C

C =PL

, C =

PL

. substituting constants into results

=P

EI(L x),

v=P

EI(x + Lx L

) check the max/min valuesmax = A =

PL

EI, vmax = vA =

PL

EI


Example : example - page


30/61

. . . . . .

Example : example , page

Determine the equation of the deflection curve for a simple beamAB supporting

a uniform load of intensity q acting throughout the span of the beam. Also,determine the maximum deflection max at the midpoint of the beam and the

angles of rotation A and B at the supports. (Note: The beam has lengthL and

constant flexural rigidity EI.)

B

q

L


Example : example -, page (cont.)


31/61

. . . . . .

Example : example , page (cont.)

A

x

M

V

q

qL2

(b)

dmax

y

xA B

uA uB

L2

L2

trick: determine the integration

constants ASAP

solution:

. bending moment through balance eqn.

M(x) = qLx

qx

. bending-moment equation

EIv

=

qLx

qx

. integrate once angle of rotation

EIv

=

qLx

qx

+ C

symmetry condition: v

(x=

L

)=

C = qL

EIv

=

qLx

qx

qL




32/61

. . . . . .

Example : example , page (cont.)

A B

q

L

(b)

dmax

y

xA B

uA uB

L

2

L

2

. integrate again deflection

EIv=qLx

qx

qL

x+ C

B.C.: v(x= ) = C = . Hence

v= qx

EI

(L Lx + x

) the maxdue to symmetry: max = v(L

) = qL

EI(downward)

. angles of rotation at the supports

A = v() = qLEI (clockwise)

B = v

(L

)=

qL

EI(counter-clockwise)


Example : example -, page


33/61

. . . . . .

p p , p g

Determine the equation of the deflection curve for a cantilever beamAB subjected

to a uniform load of intensity q. Also, determine the angle of rotation B and thedeflection B at the free end. (Note: The beam has length L and constant flexural

rigidity EI.)

AB

q

L




34/61

. . . . . .

p p , p g ( )

A

x

M

V

q

qL

qL2

2

y

x

(b)

A B

dB

uB

trick: determine the integration

constants ASAP

solution:

. bending moment through balance eqn.

M(x) = qL

+ qLxqx

. bending-moment equation

EIv

=

qL

+

qLx

qx

. integrate once angle of rotation

EIv

=

qLx

+

qLx

qx

+ C

Boundary condition: v

(x=

)=

C =

EIv

=

qLx

+

qLx

qx




35/61

. . . . . .

p p p g

AB

q

L

y

x

(b)

A B

dB

uB

. integrate twice deflection

EIv= qLx

+

qLx

qx

+ C

Boundary condition: v

(x=

)=

C =

Hence: v= qx

EI(L Lx+ x) at the free end (x= L)

B = v

(x= L

)=

qL

EI(clockwise)

B = v(x= L) = qLEI (downward)


Example : Hibbelers book, example -, p.


36/61

. . . . . .

p p pProblem: For the given load, determine the vmax, EI=const.


. symmetric, work with half. w = wL

x.

M= M+wx

Lx wL

x=

internal moment: M= wx

L

+

wL

x

. apply bending-moment equation

EIdv

dx=

wx

L+

wL

x

integrate once:

EIdvdx

= wx

L+ wL

x + C

apply the symmetry condition:dv

dxx=L/

=

C = wL


Example (cont.)


37/61

. . . . . .

.

integrate again:

EIv= wx

L+

wL

x

wL

x+ C apply

B.C. v

x= = C =

. substituting constants into results (x [,L

])EIv = wL

x +wL

x

wL

EIv= w

Lx +

wL

x

wL

x

. vmax = vx=L/ = wLEI() Lecture : Bending (V) Deflection of Beams Nov , /

Example : simple beam with a concentrated load


38/61

. . . . . .

A concentrated loadPis applied on a simply supported beam.

Solve:

. elastic curve. A, B. vmax. vC (midpoint)

a b

L

P

A BC

RB =Pa

LRA =

Pb

L

x . set up coordinate system

. bending-moment in segments

M=

PbLx, x [, a]

Pa(L x)L

, x [a, L]() Lecture : Bending (V) Deflection of Beams Nov , /

Example (cont.)


39/61

. . . . . .

a b

L

P

A BC

RB =Pa

LRA =

Pb

L

x

. apply the bending-moment equation to segments

the left part: x [, a]EI

dv

dx =Pb

L x

EIdv

dx=

Pb

Lx + C

EIv=Pb

L

x + Cx+ C

the right part: x [a, L]EI

dv

dx =Pa

(L x

)LEIdv

dx=

Pa

L(Lx x

) + C

EIv=Pa

L (Lx

x

)+ Cx+ C


Example (cont.)


40/61

. . . . . .

. Continuity and Boundary conditions

continuity ofdv

dxx=a

Pba

L

+ C =Pba

(L a

)L+ C

continuity of vx=a

Pba

L+ Ca + C =

Pa(L a)L

+ Ca + C

vx= = C =

v

x=L

= PaL

+ CL + C =

C =Pab(a L)

L, C = ,

C = Pa(L + a)

L, C =

Pa

. substituting into the results

v=

Pbx

LEI(L b x), x [, a]

Pa

LEI

(L x

)[(x a

)

bx

], x

[a, L

]() Lecture : Bending (V) Deflection of Beams Nov , / Example an alternative way

P (L )


41/61

. . . . . .

In step : bending moment for the right part,Pa(L x)

L, x [a, L]

M=Pb

L

x, x

[, a

]For left part: x [, a]EIdv

dx=

Pb

Lx

EIdv

dx=

Pb

Lx + C

EIv=Pb

Lx + Cx+ C

M=Pbx

L

P

(x a

), x

[a, L

]For

right part: x [a, L]EIdv

dx=

Pbx

L P

(x a

)EIdvdx=Pbx

L P(x a)

+ C

EIv=Pbx

L P

(x a)

+ Cx+ C

continuity ofdv

dxx=a Pba

L+ C

=

Pba

L+ C

continuity of vx=a

Pba

L+ Ca + C =

Pba

L+ Ca + C

C = C and C = C


Example an alternative way (cont.)


42/61

. . . . . .

vx=

= C = C =

v

x=L

= PbL

Pb

+ CL =

C = C = Pb(L b)

L. const. solving done, substituting back

v=

Pbx

LEI(L b x), x [, a]

Pa

LEI(L x

)[(x a

) bx

], x

[a, L

]dvdx

=

Pb

LEI(L b x), x [, a]

Pa

LEI((x a)(a + L x) + b(x L)), x [a, L]

A =dv

dxx==

Pab

(L + b

)LEI, B =

dv

dxx=L=Pab

(L + a

)LEIAt vmax, dv/dx= x =L b

, (a b)vmax =

Pb(L b)/LEI

, vC = Pb(L b)

EI, (a b)


Example : Special case


43/61

. . . . . .

load at the midpoint: a = b = L

/

symmetry only consider x [, L/]dv

dx=

P

EI(L x)

v= Px

EI(L x

)A =B =

PL

EI

vmax =vC =PL

EI

check with Appendix G, page


Example : Hibbelers book, example -, p.


44/61

. . . . . .

Problem: a overhanging beam subjected toP.Determine vC.

a2a

P

A B C

x vc

RA =P

2

RB =3

2P

. setting up the coordinate system

. bending moment in segments:

M=

P

x, x [, a]

P

(x a

)x

[a, a

]


Example (cont.)


45/61

. . . . . .

a2a

P

A B C

x vc

RA =P

2

RB =3

2P

. the bending-moment equation

the left part: x [, a]EI

dv

dx=

P

x

EIdv

dx=

P

x + C

EIv= P

x + Cx+ C

the right part: x [a, a]EI

dv

dx=

P(x a)EIdv

dx=P(x

ax) + C

EIv=P

(x

a

x

)+ Cx+ C


Examples : (cont.)


46/61

. . . . . .

. boundary and continuity conditions:

B.C. vx= = C = B.C. v

x=a =

Pa + aC =

aP+ Ca + C =

continuity ofdv

dxx=a Pa + C = Pa

+ C

C =

Pa, C = , C =

Pa, C = Pa

. substituting into results for the right part

v=P

EI(x

a

x

+

a

x a), x [a, a]

at C, x= a vC = Pa

EI


Example (cont.) using multiple xcoordinates


47/61

. . . . . .

a2a

P

A B C

x1vc

x2

RA =P

2

RB =3

2

P

the left part: x [, a]EIdv

dx=

P

x

EIdv

dx=

Px + C

EIv = P

x + Cx + C


.

bending moment:

M=

P

x, x

[, a

]Px x [

, a

]the right part: x [, a]EIdv

dx= Px

EIdv

dx=

P

x + C

EIv = P

x + Cx + C


Examples (cont.) using multiple xcoordinates (cont.)


48/61

. . . . . .

. boundary and continuity conditions:

B.C. v

x== C =

B.C. vx=a = vx=a Pa + aC = Pa + Ca + C = continuity of

dv

dx

x=a

=

dv

dx

x=a

Pa + C =

(

P

a + C

) C =

Pa, C

= , C

=

Pa , C

= Pa

. substituting into results for the right part

v = P

EI

(x

a

x + a

), x

[, a

](compare with the result using a single xcoordinate:

v= PEI(x a x + ax a), x [a, a])

. at C, x = vC = Pa

EI


Examples (cont.) double check


49/61

. . . . . .

with Matlab, note x = a x

>> syms x x2 a P %()

>> v2=-(x2^3/6-7*x2*a^2/6+a^3);

>> v=x^3/6-3*a/2*x^2+10/3*a^2*x-2*a^3;

>> v2new=subs(v2,x2, 3*a-x)

>> simple(v2new-v)

0



50/61

. Integration of the

shear-force and load equations

. . . . . .

Integrating the shear-force and load equations


51/61

. . . . . .

different forms of differential equation: order of derivative ofv

deflection (), slope (), moment (), shear force (), load q ()less effort for the R.H.S. lower order derivative

same procedure, but with more integration constants

conditions available:. load-equation th order eqn.

. shear-force, . moment, . slope and . deflection. shear-force equation rd order eqn.

. moment, . slope and . deflection.

bending moment equation nd order eqn.. slope and . deflection


Example application of load equation


52/61

. . . . . .

Problem: cantilevered beam, distributed load

Solve: () equation for deflection curve; () vB; () B

A B

q0

x

(a)

y

L

y

xA B

uB

dB


. loading intensityw =q

(L x

)L

. apply the load equation

EIdv

dx= w =

q(L x)L

EI

dv

dx =qL(L x) + C


Example , load equation (cont.)

( d) EIdv q (L ) C


53/61

. . . . . .

(repeated) EIdx

=

qL

(L x) + C. shear force B.C.: V

(x= L

)=

so EId

vdx (x= L) = C =

the shear force V= EIdv

dx=

qL

(L x). integrate again

EId

vdx

= qL(L x) + C

now, the bending moment B.C.: M(x= L) = so EI

dv

dx

(x= L

)= C =

the bending moment: M= EId

vdx

= qL(L x)

. integrate again twice

EIdv

dx=

qL

(L x

)

+ C, EIv= q

L

(L x

)

+ Cx+ C


Example application of load equation (cont.)

q L q L


54/61

. . . . . .

. B.C.s at x= : v

= v= C = qL

, C =

qL

. substitute into the results

dv

dx=

qx

LEI(L Lx+ Lx x)

v= qx

LEI

(L Lx+ Lx x

) deflection angle and deflection at the free endB =

dv

dxx=L

=

qL

EI, vB = v(L) = qL

EI

y

xA B

uB

dB


Example (same as example ), with shear-force equation


55/61

. . . . . .

Problem: overhanging beam, concentrated load

Determine: () equation for deflection curve; () vC

A C

P

3P

B

2

P2

L L

2

A

y

B C

xdC

. RA = P

, RB =

P

V=

P

, x (, L)

P, x (L, L)


Example (cont.)


56/61

. . . . . .

A C

P

3P

B

2

P2

L L

2

A

y

B C

xdC

. apply the shear-force equation

EIdv

dx=

P, x (, L)

P, x (L, L)

M= EId

vdx

=

Px

+

C,

x (, L)Px+ C, x (L, L

)

BCs: MA =MC = dv

dx () = ,dv

dx (L

) = C = , C =

PL

we may verifyMat this point


Example (cont.)

Bending moment (final results)


57/61

. . . . . .

Bending moment (final results)

M=

EI

dv

dx=

Px

, x

(, L

)P(L x) , x (L, L )Slopes:

EIdv

dx=

Px

+ C, x

(, L

)Px(L x) + C, x (L, L )Continuity of slope at B

PL

+ C = PL

+ C C = C +

PL

Deflections:

EIv=

Px

+ Cx+ C, x (, L)

Px(L x)

+ Cx+ C, x (L, L)


Example (cont.)BCs: v() = v(L) = applied to AB


58/61

. . . . . .

C = , C =PL

; C =

PL

v(L) = applied to BC C = PLsubstitute into the results

v=

Px

EI

(L x

), x

(, L

) PEI(L Lx+ Lx x), x (L, L )deflection at the free end vC = v(L

) = PL

EI

A

y

B Cx

dC


Remarks


59/61

. . . . . .

. which equation to use:

a matter of conveniencedepends on the quantities most easily available: V,Mor qand balance of numbers of the integration constants

personal preference

the bending-moment equation is a good starting point.

. when to use the conditions:

way : separately as early as possible to simplify expressions way : postponed to final stage, solved together in a linear system

. for the ease of use (to determine constants), form of expression is important

while integrating. Best practicing resource Appendix G, p.


Summary


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deflection/elastic curve, sketching

sign conventions for v, , , ,Mdifferential equations of the deflection curve (EI=const.)

bending-moment equation: EIdv

dx=M

(x

)shear-force equation: EIdv

dx=

V(x)load equation: EI

dv

dx= q(x)

Integration method for deflection curve

type of eqn., number of required integration constants application ofboundary/continuity conditions coordinate system selection, form of expression


Homework


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Chap. ,,, (all with integration method)

Due: .. (Fri.)


17_defl_print.pdf

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