Deflections - Lecture notes 2 PDF

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Chapter 9

Deflections of Beams

9.1 Introduction in this chapter, we describe methods for determining the equation of the deflection curve of beams and finding deflection and slope at specific points along the axis of the beam

9.2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection in the

y

v

is the displacement

direction

the angle of rotation



of the axis

(also called slope) is the angle between the x

axis and the tangent to the deflection

curve point

m1

is located at distance

x

point

m2

is located at distance

x + dx

slope at

m1

is



slope at

m2

is

 + d

denote 

O'

the center of curvature and

the radius of curvature, then  d

=

ds

and the curvature



is

1



1

=

=

C

d C



ds

the sign convention is pictured in figure slope of the deflection curve dv C

tan 

=



or

dv tan-1 C dx

=

dx for



small



=

ds j dx

1

d =

C

=

C

C

dx 2

d

d v

=

C



then

dv =

dx

1

=

tan  j ,



and

C

 

cos j 1

CC

dx2

dx

if the materials of the beam is linear elastic 1 

=

M =

C



[chapter 5]

C

EI

then the differential equation of the deflection curve is obtained d 

C

2

dv =

CC

dx

dx2

M =

C

EI

it can be integrated to find dM ∵

CC

then

CC

dx3

and v dV

=

V

dx d 3v



CC

=

-q

=

q -C EI

dx =

V

d 4v

C

CC

EI

dx4

2

sign conventions for

M,

V

and

q

are shown

the above equations can be written in a simple form EIv"

=

M

EIv"'

=

V

EIv""

=

-q

this equations are valid only when Hooke's law applies and when the slope and the deflection are very small for nonprismatic beam [I = I(x)], the equations are

CC d 2v

EIx

=

2

M

dx

C (EI CC)

dM CC

d 2v

d

x

dx

=

2

dx

d2

CC (EI CC) dx

2

dx

V

dx

d 2v

x

=

CC dV

=

2

=

-q

dx

the exact expression for curvature can be derived 

=

C1 

=

v" CCCCC

[1 + (v')2]3/2

9.3 Deflections by Integration of the Bending-Moment Equation substitute the expression of M(x)

into

the deflection equation then integrating to satisfy (i) boundary conditions (ii) continuity conditions (iii) symmetry conditions to obtain the slope



and the

3

deflection

v

of the beam

this method is called method of successive integration

Example 9-1 determine the deflection of beam supporting a uniform load of intensity max

also determine

A,

and

flexural rigidity of the beam is

AB q

B EI

bending moment in the beam is M

=

CC qLx

-

2

CC q x2 2

differential equation of the deflection curve

CC

EI v"

=

CC

q x2

qLx

-

2

2

Then qLx2

EI v'

=

CC

q x3

-

4

∵ the beam is symmetry,

CCCC

CC ∴

qL(L/2)2

0

=

4

+

C1

6

 = v' = 0

CCCC

at

q (L/2) 3

-

6

4

+

C1

x=L/2

then

= q L3 / 24

C1

the equation of slope is v'

q - CCC (L3 24 EI

=

6 L x2

-

+

4 x3)

integrating again, it is obtained v

q - CCC (L3 x 24 EI

=

boundary condition : v thus we have

2 L x3

-

= 0

C2

=

+

at

x4)

+

C2

x

=

0

0

then the equation of deflection is v

q - CCC (L3 x 24 EI

=

max

maximum deflection max

=

L

- v(C) 2

=

2 L x3

-

x4)

+

occurs at center 5 q L4

(x

=

L/2)

(↓)

CCC

384 EI

the maximum angle of rotation occurs at the supports of the beam q L3 A

=

v'(0)

=

- CCC 24 EI

( )

q L3 and

B

=

v'(L)

=

CCC

24 EI

5

( )

Example 9-2 determine the equation of deflection curve for a cantilever beam AB to a uniform load of intensity B

also determine

subjected

q B

and

at the free end

flexural rigidity of the beam is

EI

bending moment in the beam q L2 M

=

q x2

- CC 2

+

qLx

-

CC

2

q L2 EIv"

=

q x2

- CC 2

+

=

C1 v'

=

=

qx - CC 6EI

CC

qLx2

- CC 2

boundary condition

-

2

qL2x EIy'

qLx

+

q x3 -

CC

v'

= 

+

CC

2

C1

6 =

0

at

x =

0

0 (3 L2

-

3Lx

+

x2)

integrating again to obtain the deflection curve qx2 v

=

- CC 24EI

boundary condition C2

=

(6 L2

v

-

4Lx

= 0

+

at

0

6

x2)

x

+

= 0

C2

then qx2 v

=

(6 L2

- CC 24EI

-

4Lx

x2)

+

q L3 max

=

B

=

v'(L)

=

- CC ( ) 6 EI q L4

max

=

- B

=

- v(L)

=

CC

(↓)

8 EI

Example 9-4 determine the equation of deflection curve,

A,

B,

max

and

flexural rigidity of the beam is

C EI

bending moments of the beam M

Pbx

=

CC

(0 ≦ x ≦ a)

L Pbx M = CC - P (x - a) (a ≦ x ≦ L) L differential equations of the deflection curve EIv"

=

Pbx CC

(0 ≦ x ≦ a)

L Pbx EIv" = CC - P (x - a) (a ≦ x ≦ L) L integrating to obtain

7

CC

Pbx2

EIv'

=

+

C1 (0 ≦ x ≦ a)

-

CCCC

+

C1 x

C3

2L

CC

Pbx2

EIv'

=

P(x - a)2

2L

C2

(a ≦ x ≦ L)

2

2nd integration to obtain EIv

CC

Pbx3

=

+

+

(0 ≦ x ≦ a)

6L

EIv

CC - CCCC + C x

=

Pbx3

P(x - a)3

6L

6

2

(a ≦ x ≦ L)

+ C4

boundary conditions (i)

v(0)

=

0

(ii)

y(L)

=

0

continuity conditions (iii)

(i) (ii)

v'(a-)

v(0)

=

=

v(L)

v'(a+)

(iv)

v(a-)

=>

C3

= 0

=>

CC - CC + C L + C

0

=

0

PbL3

Pb3

2

6

(iii)

-

v'(a )

=

+

v'(a )

= v(a+)

=>

4

(iv)

-

v(a )

= =

CC Pba2

+

C1

=

=

CC

Pba2

+

C2

2L

C2 +

v(a )

=>

CC + C a + C = CC + C a + Pba3

Pba3

1

6L

C3

0

6

2L

C1

=

C4 8

3

2

6L

C4

then we have Pb (L2 - b2) C1

=

C2

=

- CCCCC 6L

C3

=

C4

=

0

thus the equations of slope and deflection are v'

v'

v

v

Pb - CC (L2 - b2 - 3x2) 6LEI

=

Pb

=

=

=

2

2

(0 ≦ x ≦ a)

2

- CC (L - b - 3x ) 6LEI

P(x - a )2

-

2EI

Pbx - CC (L2 - b2 - x2) 6LEI Pbx

2

2

2

- CC (L - b - x ) 6LEI

(a ≦ x ≦ L)

CCCC

(0 ≦ x ≦ a)

-

P(x - a )3

(a ≦ x ≦ L)

CCCC

6EI

angles of rotation at supports A

=

v'(0)

=

Pab(L + b) - CCCCC 6LEI

B

=

v'(L)

=

CCCCC

( )

Pab(L + a) ( )

6LEI ∵

A

is function of

a (or b), to find

(A)max,

set

dA / db = 0

Pb(L2 - b2) A

=

dA / db

- CCCCC 6LEI =

0

=> L2 - 3b2

9

=

0

=> b

=

L/ 3

PL2 3 (A)max

=

for maximum 

- CCCC 27 EI occurs at

x1,

if

a

>

b,

x1

(a

≧

L2 - b2

dv =

C

0

=>

x1

=

CCC

dx

<

a

b)

3

max

=

- v(x1)

Pb(L2 - b2)3/2

=

(↓)

CCCCC

9 3 LEI at

x

=

C

L/2

=

- v(L/2)

Pb(3L2 - 4b2)

=

CCCCCC

(↓)

48 EI ∵ the maximum deflection always occurs near the midpoint, ∴ C gives a good approximation of the max in most case, the error is less than an important special case is P v'

=

CC

(L2 - 4x2)

a

3% =

b

=

L/2

(0 ≦ x ≦ L/2)

16EI P v

=

CC

(3L2 - 4x2) (0 ≦ x ≦ L/2)

48EI v'

and A

v

are symmetric with respect to B

=

PL2

=

CC

16EI max

=

C

PL3 =

CC

48EI

10

x

=

L/2

9.4 Deflections by Integration of Shear-Force and Load Equations the procedure is similar to that for the bending moment equation except that more integrations are required if we begin from the load equation, which is of fourth order, four integrations are needed

Example 9-4 determine the equation of deflection curve for the cantilever beam

AB

supporting a

triangularly distributed load of maximum intensity

q0 B

also determine

B

and

flexural rigidity of the beam is

EI

CCCC q0 (L - x)

q

=

L

CCCC q0 (L - x)

EIv""

=

-q

=

-

L

the first integration gives EIv"'

=

-

CCCC

+

V

=> C1

q0 (L - x)

2

C1

2L

∵

thus

v"'(L)

EIv"'

=

=

=

-

0

CCCC

q0 (L - x)2 2L

11

=

0

2nd integration EIv" ∵

q0 (L - x)3

=

- CCCC 6L

v"(L)

thus

=

EIv"

M

=

+

C2

0

=>

C2

=

0

q0 (L - x)3

=

- CCCC 6L

3rd and 4th integrations to obtain the slope and deflection EIv'

=

q0 (L - x)

4

+

C3

- CCCC 120L

+

C3 x

+

C4

boundary conditions : v'(0)

=

v(0)

=

0

EIv

q0 (L - x)5

=

the constants C3

=

- CCCC 24L

C3

and

q0L3

- CC 24

C4

can be obtained

C4

q0L4

=

CC

120

then the slope and deflection of the beam are v'

=

q0x - CCC (4L3 24LEI

6L2x

-

4Lx2

+

x3)

-

2

v

=

q0x

- CCC (10L3 120LEI

10L2x

-

+ 5Lx2

3

B

q0L =

v'(L)

=

- CCC 24 EI

( )

12

-

x3)

B

=

- v(L)

q0L4

=

(↓ )

CCC

30 EI

Example 9-5 an overhanging beam concentrated load

P

ABC

with a

applied at the end

determine the equation of deflection C

curve and the deflection

at the end

flexural rigidity of the beam is

EI

the shear forces in parts AB and BC are V

=

P -C 2

(0 < x < L)

V

=

P

3L (L < x < C) 2

the third order differential equations are EIv'"

=

P - C (0 < x < L) 2

EIv'"

=

P

3L (L < x < C) 2

bending moment in the beam can be obtained by integration M

=

EIv"

=

Px -C + 2

M

=

EIv"

=

Px

+

C1 C2

13

(0 ≦ x ≦ L) (L ≦ x ≦

3L ) 2

C

boundary conditions : v"(0) we get

C1

=

0

=

v"(3L/2)

C2

=

0

3PL - CC 2

=

therefore the bending moment equations are M

=

EIv"

=

Px -C 2

M

=

EIv"

=

P(3L - 2x) - CCCCC 2

(0 ≦ x ≦ L)

(L ≦ x ≦

3L ) 2

C

2nd integration to obtain the slope of the beam EIv'

=

EIv'

=

Px2

- CC 4

+

Px(3L - x) - CCCCC 2

continuity condition : PL2 - CC 4 then

C4

(0 ≦ x ≦ L)

C3

+

=

C3 C3

=

+

C4 (L ≦ x ≦

+

v'(L-)

=

- PL2

+ C4

3L ) 2

C

v'(L+)

3PL2 CC

4 the 3rd integration gives EIv

EIv

=

=

Px3

- CC 12

+

C3 x

Px2(9L - 2x)

- CCCCC 12

+

+

C4 x

14

(0 ≦ x ≦ L)

C5

+

C6

(L ≦ x ≦

3L C) 2

boundary conditions :

v(0)

v(L-)

=

=

0

we obtain C5

=

0

C3

PL2

=

CC

12 and then

C4

=

5PL2 CC

6 the last boundary condition : v(L+) then
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