7.2 Shear and Moment Equations and Diagrams Beams – structural members designed to support...

7.2 Shear and Moment Equations

and Diagrams


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

and roller supported at the other

A cantilevered beam is fixed at one end and free at the other


and Diagrams


and Diagrams For actual design of a beam, apply- Internal shear force V and the bending

moment M analysis- 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


and Diagrams


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 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


and Diagrams


and DiagramsLoad applied to a beam act

perpendicular to the beam’s axis and hence produce only an internal shear force 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


and Diagrams


and DiagramsSign ConventionTo define a positive and negative shear

force and bending moment acting on the beam

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


and Diagrams


and DiagramsSign ConventionA positive moment

would tend to bend the member if it were elastic, concave upwards

Loadings opposite to the above are considered negative


and Diagrams


and DiagramsProcedure for AnalysisSupport ReactionsDetermine all the reactive forces and

couple moments acting on the beam’Resolve them into components

acting perpendicular or parallel to the beam’s axis


and Diagrams


and DiagramsProcedure for AnalysisShear and Moment Reactions Specify separate coordinates x having an

origin 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


and Diagrams


and DiagramsProcedure for AnalysisShear and Moment Reactions V and M are shown acting in their positive

sense 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


and Diagrams


and DiagramsProcedure for AnalysisShear and Moment Diagrams Plot the shear diagram (V versus x) and the

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


and Diagrams


and DiagramsExample 7.7Draw the shear and bending moments diagrams for the shaft. The support at A is

a thrust bearing and the support at C is a journal bearing.


and Diagrams


and DiagramsSolutionSupport ReactionsFBD of the shaft


and Diagrams


and DiagramsSolution

mxkNMM

kNVFy

.5.2;0

5.2;0


and Diagrams


and DiagramsSolution

mkNxM

xkNmxkNMM

kNV

VkNkNFy

.)5.210(

0)(5.2)2(5;0

5.2

055.2;0


and Diagrams


and DiagramsSolutionShear diagram internal shear force is always

positive within the shaft AB 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


and Diagrams


and DiagramsSolution Graph of shear and

moment diagrams is discontinuous at points of concentrated force ie, A, B, C

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


and Diagrams


and DiagramsExample 7.8Draw the shear and bending diagrams for the beam.


and Diagrams


and Diagrams

SolutionSupport ReactionsFBD of the beam

View Free Body Diagram


and Diagrams


and DiagramsSolution Distributed loading acting on this

segment has an intensity of 2/3 x at 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


and Diagrams


and DiagramsSolution Resultant force acts through the centroid

of the distributed loading area, 1/3 x from the right

mkNx

xM

xx

xMM

kNx

V

VxFy

.9

9

0933

1;0

39

03

19;0

3

2

2

2


and Diagrams


and DiagramsSolution For point of zero shear,

For maximum moment,

mkN

mkNM

mx

xV

.12.3

.9

20.520.59

20.5

03

9

3

max

3

7.2 Shear and Moment Equations and Diagrams Beams – structural members designed to support...

Documents

Transcript of 7.2 Shear and Moment Equations and Diagrams Beams – structural members designed to support...