Chapter 3-2[Compatibility Mode]

7/29/2019 Chapter 3-2[Compatibility Mode]

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DIFFERENTIAL EQUILIBRIUM

RELATIONSHIPS1. It is an alternative procedure for obtaining internal

2. Instead of cutting a beam in two and applying the

equilibrium conditions to one of the segments, very

small differential element of the beam will be considered

orces an momen s or e s en er mem ers

Mechanics of Solids 47

3. The conditions of equilibrium combined with a limitingconditions will lead us to differential equationsconnectingthe load, the shear force, and the bending moment


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4. Integration of these relationships for particular

cases furnishes us with an alternative method for

evaluating shear forces and bending moments.



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If the variation of q(x) is smooth and if x is very

small then R is very nearly given by qxand the

line of action of Rwill very nearly pass through the

midpoint o of the element.

The conditions of equilibrium applied to Fig. 14c

are then



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Integrating above equations with appropriateconditions will give values of shear forces andbending moments


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Example 3.5

In Fig.15aa beam carrying a uniformly distributed load

of intensity q= - wo is supported by a pinned joint at A

. -and bending-moment diagrams by integration of the

differential relationships (3.11) and (3.12).



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RA = RB = wo L / 2



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We have two boundar conditions available to find

External moments are absent at either end of the beam,hence

C1 and C2.

Mb= 0 at x= 0


b= =

Inserting these boundary conditions values of C1 and

C2 yield


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The shear force and bending moment are then


The bending moment is maximum when the shearforce is zero.


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Example 3.6

Consider the beam shown in Fig. 3.16a with simple

transverse supports at A and B and loaded with a

un orm y s r u e oa = - w0 over a por on o elength. It is desired to obtain the shear-force and

bending-moment diagrams.



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h r F r Bendin Moment

Part AC

1 0O

dVw

dx =

1 1OV w x C =

1 110 0

b bO

dM dM V w x C

dx dx+ = + + =

2

1 1 2

1

2b Ow x C x C + + =

Part CB


2 0dV

dx=

2 3V C=

2 230 0

b bdM dM V Cdx dx

+ = + =

2 3 4b C x C+ =


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For 4 Constants i.e. C1, C2, C3 and C4 we need

to have 4 boundary conditions

At x = 0; MA = 0 At x = L; MB = 0

At x = a; V1 = V2 = VC and

Mb1 = Mb2 = MC

By applying these BC we get values of C1, C2,C and C as follows


2 0C =

1

1( 2)

2O

aC w a

L=

2

3

1

2

Ow aCL

=

2

4

1

2OC w a=


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Shear Force

1

1( 2)

2O O

aV w x w a

L= +

Bending Moment2

11 1 ( 2) 02 2

b O OaM w x w a x

L+ + =

Part AC

1 1 ( )( 2)A O O

a L bV w a w a

+= = 0AM =

Part CBShear Force

2

Bending Moment2 21 1

2

2

OC

w aV

L=

21( )

2C O

bM w a

L=


2 2

O

V L= 2

2 2b O O

L

=

Shear force will beconstant in betn C t o B

C BV V= =

21 ( )2

C O

bM w a

L=

0B

M =


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1

1( 2) 0

2O O

aV w x w a

L= + =

( )

2

a L bx

L

+=


2 2

max 2

( )

8

Ob

w a L bM

L

+=


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SINGULARITY FUNCTION METHOD

n

n axxf >=


01

1

+

>


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is called unit concentrated moment2


1axfunction >


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2 2

2

2A A O

L bV R w

L

aV w

= =

=

2

2B O

aV w

L=


2

0

2

0

A

C O

B

a bM w

L

M

=

=

=


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1 ( 2) 02O OV w x w a L= + =

( )2

a L bxL+=


2 2

max 2

( )

8

Ob

w a L bM

L

+=

12kN


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20kN

20kN

12kN

RA RB

A BC D

Distance betn 20kNforces is 0.6m

kNRR BA 12=+

06.0204126 =BR

=


A

kNRB 10=


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4321 )()()()()( xqxqxqxqxq +++=

11 0)( >


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[-V = 002 >< x 1212 >< x kN212 ==

V for CD


[-VCD = ]020

>< x kN212 ==

V for DB


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V for DB

[-VDB =0

02 >< x ]412 0>< x 1212 >


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=M 002[ >< x 0212 >< x 0212 >


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M for DB

=DBM 102 >< x 0212 >


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20kNRA RB

A BC D


Exercise Problems


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Exercise Problems

Find the values of shear force and bending

.force and bending moment diagrams. Usethe general method for analysis.


E 1


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Ex. 1

Ex. 2


Ex 3


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Ex. 3

Ex. 4


Ex 5


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Ex. 5

Ex. 6


Ex 7


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Ex. 7


Solve the Exercise problems 1 7 using


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Solve the Exercise problems 1- 7 using

singularity function method.Solve the Exercise problems 2, 3, 6, and

relationships. Ignore the values of pointloads and concentrated moments given in

those problems.Note: The values of SF & BM for the problem solved by General


.

SF and BM of the problem solved by differential equilibrium methodwill not be same as we are considering only uniform distributed loadsand neglecting the point loads and concentrated moments.

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