6161103 7.2 shear and moment equations and diagrams

7.2 Shear and Moment Equations 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 end � A simply supported beam is pinned at one end

and roller supported at

the other

� A cantilevered beam is

fixed at one end and free

at the other


Equations and Diagrams� For actual design of a beam, apply

- Internal shear force V and the bending moment M analysis

- Theory of mechanics of materials- Theory of mechanics of materials

- Appropriate engineering code to determine beam’s required cross-sectional area

� Variations of V and M obtained by the method of sections

� Graphical variations of V and M are termed as shear diagram and bending moment diagram


Equations and Diagrams� Internal shear and bending moment

functions generally discontinuous, or their slopes will be discontinuous at points where a distributed load changes or where concentrated forces or couple moments where a distributed load changes or where concentrated forces or couple moments are applied

� Functions must be applied for each segment of the beam located between any two discontinuities of loadings

� Internal normal force will not be considered


Equations and Diagrams

� Load applied to a beam act perpendicular to the beam’s axis and hence produce only an internal shear force and bending momentforce and bending moment

� For design purpose, the beam’s resistance to shear, and particularly to bending, is more important than its ability to resist a normal force


Equations and DiagramsSign Convention� To define a positive and negative shear

force and bending moment acting on the beambeam

� Positive directions are denoted by an internal shear force that causes clockwise rotation of the member on which it acts and by an internal moment that causes compression or pushing on the upper part of the member



Sign Convention

� A positive moment would tend to bend the would tend to bend the member if it were elastic, concave upwards

� Loadings opposite to the above are considered negative



Procedure for AnalysisSupport Reactions

� Determine all the reactive forces and couple moments acting on the beam’couple moments acting on the beam’

� Resolve them into components acting perpendicular or parallel to the beam’s axis


Equations and DiagramsProcedure for Analysis

Shear and Moment Reactions

� Specify separate coordinates x having an origin at the beam’s left end and extending to regions at the beam’s left end and extending to regions of the beams between concentrated force and/or couple moments or where there is no continuity of distributed loadings

� Section the beam perpendicular to its axis at each distance x and draw the FBD of one of the segments



Procedure for Analysis

Shear and Moment Reactions

� V and M are shown acting in their positive sense

The shear V is obtained by summing the forces � The shear V is obtained by summing the forces perpendicular to the beam’s axis

� The moment M is obtained by summing moments about the sectioned end of the segment


Equations and DiagramsProcedure for Analysis

Shear and Moment Diagrams

� Plot the shear diagram (V versus x) and the moment diagram (M versus x)moment diagram (M versus x)

� If computed values of the functions describing V and M are positive, the values are plotted above the x axis, whereas negative values are plotted below the x axis

� Convenient to plot the shear and the bending moment diagrams below the FBD of the beam



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 a thrust bearing and the support at C is a

journal bearing.



Solution

Support Reactions

� FBD of the shaft� FBD of the shaft



Solution

mxkNMM

kNVFy

.5.2;0

5.2;0

==∑

==∑↑+

mxkNMM .5.2;0 ==∑



Solution

kNV

VkNkNFy

5.2

055.2;0

−=

=−−=∑↑+

mkNxM

xkNmxkNMM

kNV

.)5.210(

0)(5.2)2(5;0

5.2

−=

=−−+=∑

−=


Equations and DiagramsSolution

Shear diagram

� internal shear force is always positive within the shaft AB

Just to the right of B, the shear � Just to the right of B, the shear force changes sign and remains at constant value for segment BC

Moment diagram

� Starts at zero, increases linearly to B and therefore decreases to zero



� Graph of shear and moment diagrams is discontinuous at points of concentrated force points of concentrated force ie, A, B, C

� All loading discontinuous are mathematical, arising from the idealization of a concentrated force and couple moment



Example 7.8

Draw the shear and bending diagrams for

the beam.the beam.



Solution

Support Reactions

� FBD of the beam� FBD of the beam



� Distributed loading acting on this segment has an intensity of 2/3 x at its end and is replaced by a resultant its end and is replaced by a resultant force after the segment is isolated as a FBD

� For magnitude of the

resultant force,

½ (x)(2/3 x) = 1/3 x2



Solution

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

VxF 01

9;0 2 =−−=∑↑+

mkNx

xM

xx

xMM

kNx

V

VxFy

.9

9

0933

1;0

39

03

19;0

3

2

2

2

−=

=−

+=∑

−=

=−−=∑↑+



Solution

� For point of zero shear,

xV 0

39

3

=−=

� For maximum moment,

( ) ( )

mkN

mkNM

mx

.12.3

.9

20.520.59

20.53

3

max

=

−=

=

6161103 7.2 shear and moment equations and diagrams

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