61611037-2shearandmomentequationsanddiagrams-120122212823-phpapp02

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7.2 Shear and MomentEquations and Diagrams7.2 Shear and Moment

Equations and Diagrams Beams structural members designed to support

loadings perpendicular to their axes Beams straight long bars with constant cross-

sectional areas

A simply supported beam is pinned at one endand roller supported at

the other

A cantilevered beam isfixed at one end and free

at the other


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Equations and Diagrams For actual design of a beam, apply

- Internal shear force V and the bending momentM analysis

-

- Appropriate engineering code to determinebeams required cross-sectional area

Variations of V and M obtained by the method ofsections

Graphical variations of V and M are termed as

shear diagram and bending moment diagram


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Equations and Diagrams Internal shear and bending moment

functions generally discontinuous, or theirslopes will be discontinuous at pointswhere a distributed load changes or where

concentrated forces or couple momentsare applied

Functions must be applied for each

segment of the beam located between anytwo discontinuities of loadings

Internal normal force will not beconsidered


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Equations and Diagrams

Load applied to a beam actperpendicular to the beams axis andhence produce only an internal shear

For design purpose, the beamsresistance to shear, and particularly to

bending, is more important than itsability to resist a normal force


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Equations and DiagramsSign Convention To define a positive and negative shear

force and bending moment acting on the

Positive directions are denoted by aninternal shear force that causes clockwise

rotation of the member on which it actsand by an internal moment that causescompression or pushing on the upper part

of the member


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Sign ConventionA positive moment

would tend to bend the

member if it wereelastic, concave upwards

Loadings opposite to theabove are considerednegative


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Procedure for AnalysisSupport Reactions

Determine all the reactive forces and

couple moments acting on the beam Resolve them into components acting

perpendicular or parallel to the beams

axis


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Equations and DiagramsProcedure for Analysis

Shear and Moment Reactions

Specify separate coordinates x having an origin

of the beams between concentrated force and/orcouple moments or where there is no continuityof distributed loadings

Section the beam perpendicular to its axis ateach distance x and draw the FBD of one of thesegments


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Procedure for Analysis

Shear and Moment Reactions

V and M are shown acting in their positive sense

he shear V is obtained by summing the forcesperpendicular to the beams axis

The moment M is obtained by summing

moments about the sectioned end of thesegment


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Equations and DiagramsProcedure for Analysis

Shear and Moment Diagrams

Plot the shear diagram (V versus x) and the

If computed values of the functions describing Vand M are positive, the values are plotted abovethe x axis, whereas negative values are plottedbelow the x axis

Convenient to plot the shear and the bending

moment diagrams below the FBD of the beam


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Example 7.7

Draw the shear and bending moments

diagrams for the shaft. The support at A is a

thrust bearing and the support at C is ajournal bearing.


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Solution

Support Reactions

FBD of the shaft


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Solution

kNVFy 5.2;0 ==+

mx ..;==


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Solution

kNV

VkNkNFy

5.2

055.2;0

=

==+

mkNxM

xkNmxkNMM

.)5.210(

0)(5.2)2(5;0

=

=+=


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Equations and DiagramsSolution

Shear diagram internal shear force is always

positive within the shaft AB

Just to the right of B, the shearforce changes sign and remainsat constant value for segmentBC

Moment diagram

Starts at zero, increases linearlyto B and therefore decreases tozero


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Graph of shear and momentdiagrams is discontinuous at

ie, A, B, C

All loading discontinuous aremathematical, arising from

the idealization of aconcentrated force andcouple moment


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Example 7.8

Draw the shear and bending diagrams for

the beam.


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Solution

Support Reactions

FBD of the beam


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Distributed loading acting on thissegment has an intensity of 2/3 x at

force after the segment is isolated asa FBD

For magnitude of the

resultant force,

(x)(2/3 x) = 1/3 x2


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Solution

Resultant force acts through the centroid of thedistributed loading area, 1/3 x from the right

1

mkN

x

xM

xx

xMM

kNx

V

xy

.99

0933

1;0

39

3;

3

2

2

=

=

+=

=

==


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Solution

For point of zero shear,

xV 09

3

==

For maximum moment,

( ) ( )

mkN

mkNM

mx

.12.3

.9

20.520.59

20.5

3

max

=

=

=

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