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- 1. 7.2 Shear and Moment Equations and DiagramsBeams – structural members designed to supportloadings perpendicular to their axesBeams – straight long bars with constant cross-sectional areasA simply supported beam is pinned at one endand roller supported atthe otherA cantilevered beam isfixed at one end and freeat the other
- 2. 7.2 Shear and Moment 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- 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
- 3. 7.2 Shear and Moment Equations and DiagramsInternal shear and bending momentfunctions generally discontinuous, or theirslopes will be discontinuous at pointswhere a distributed load changes or whereconcentrated forces or couple momentsare appliedFunctions must be applied for eachsegment of the beam located between anytwo discontinuities of loadingsInternal normal force will not beconsidered
- 4. 7.2 Shear and Moment Equations and DiagramsLoad applied to a beam actperpendicular to the beam’s axis andhence produce only an internal shearforce and bending momentFor design purpose, the beam’sresistance to shear, and particularly tobending, is more important than itsability to resist a normal force
- 5. 7.2 Shear and Moment Equations and DiagramsSign Convention To 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
- 6. 7.2 Shear and Moment Equations and DiagramsSign Convention A positive moment would tend to bend the member if it were elastic, concave upwards Loadings opposite to the above are considered negative
- 7. 7.2 Shear and Moment Equations and DiagramsProcedure 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 beam’s axis
- 8. 7.2 Shear and Moment Equations 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
- 9. 7.2 Shear and Moment Equations 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
- 10. 7.2 Shear and Moment Equations 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
- 11. 7.2 Shear and Moment Equations and DiagramsExample 7.7Draw the shear and bending momentsdiagrams for the shaft. The support at A is athrust bearing and the support at C is ajournal bearing.
- 12. 7.2 Shear and Moment Equations and DiagramsSolutionSupport Reactions FBD of the shaft
- 13. 7.2 Shear and Moment Equations and DiagramsSolution+ ↑ ∑ Fy = 0;V = 2.5kN∑ M = 0; M = 2.5 xkN .m
- 14. 7.2 Shear and Moment Equations and DiagramsSolution+ ↑ ∑ Fy = 0;2.5kN − 5kN − V = 0V = −2.5kN∑ M = 0; M + 5kN ( x − 2m) − 2.5kN ( x) = 0M = (10 − 2.5 x)kN .m
- 15. 7.2 Shear and Moment Equations 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 BCMoment diagram Starts at zero, increases linearly to B and therefore decreases to zero
- 16. 7.2 Shear and Moment Equations 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
- 17. 7.2 Shear and Moment Equations and DiagramsExample 7.8Draw the shear and bending diagrams forthe beam.
- 18. 7.2 Shear and Moment Equations and DiagramsSolutionSupport Reactions FBD of the beam
- 19. 7.2 Shear and Moment Equations 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
- 20. 7.2 Shear and Moment Equations and DiagramsSolution Resultant force acts through the centroid of the distributed loading area, 1/3 x from the right 1 2 + ↑ ∑ Fy = 0;9 − x − V = 0 3 x2 V = 9 − kN 3 1 2 x ∑ M = 0; M + x − 9 x = 0 3 3 x3 M = 9 x − kN .m 9
- 21. 7.2 Shear and Moment Equations and DiagramsSolution For point of zero shear, 3 x V = 9− = 0 3 x = 5.20m For maximum moment, M max = 9(5.20 ) − (5.20) kN .m 3 9 = 3.12kN .m

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