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# 6161103 7.3 relations between distributed load, shear and moment

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### 6161103 7.3 relations between distributed load, shear and moment

1. 1. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load Consider beam AD subjected to an arbitrary load w = w(x) and a series of concentrated forces and moments Distributed load assumed positive when loading acts downwards
2. 2. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load A FBD diagram for a small segment of the beam having a length ∆x is chosen at point x along the beam which is not subjected to a concentrated force or couple moment Any results obtained will not apply at points of concentrated loadings
3. 3. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load The internal shear force and bending moments shown on the FBD are assumed to act in the positive sense Both the shear force and moment acting on the right-hand face must be increased by a small, finite amount in order to keep the segment in equilibrium
4. 4. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load The distributed loading has been replaced by a resultant force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, where 0 < k <1+ ↑ ∑ Fy = 0;V − w( x)∆x − (V + ∆V ) = 0∆V = − w( x)∆x∑ M = 0;−V∆x − M + w( x)∆x[k (∆x )] + ( M + ∆M ) = 0∆M = V∆x − w( x)k (∆x) 2
5. 5. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load dV = −w(x) dx Slope of the = Negative of shear diagram distributed load intensity dM =V dx Slope of = Shear moment diagram
6. 6. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load At a specified point in a beam, the slope of the shear diagram is equal to the intensity of the distributed load Slope of the moment diagram = shear If the shear is equal to zero, dM/dx = 0, a point of zero shear corresponds to a point of maximum (or possibly minimum) moment w (x) dx and V dx represent differential area under the distributed loading and shear diagrams
7. 7. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load ∆VBC = − ∫ w( x)dx Change in = Area under shear shear diagram ∆M BC = ∫ Vdx Change in = Area under moment shear diagram
8. 8. 7.3 Relations between Distributed Load, Shear and MomentDistributed Load Change in shear between points B and C is equal to the negative of the area under the distributed-loading curve between these points Change in moment between B and C is equal to the area under the shear diagram within region BC The equations so not apply at points where concentrated force or couple moment acts
9. 9. 7.3 Relations between Distributed Load, Shear and MomentForce FBD of a small segment of the beam + ↑ ∑ Fy = 0; ∆V = − F Change in shear is negative thus the shear will jump downwards when F acts downwards on the beam
10. 10. 7.3 Relations between Distributed Load, Shear and MomentForce FBD of a small segment of the beam located at the couple moment ∑ M = 0; ∆M = M O Change in moment is positive or the moment diagram will jump upwards MO is clockwise
11. 11. 7.3 Relations between Distributed Load, Shear and MomentExample 7.9Draw the shear and moment diagrams for thebeam.
12. 12. 7.3 Relations between Distributed Load, Shear and MomentSolutionSupport Reactions FBD of the beam
13. 13. 7.3 Relations between Distributed Load, Shear and MomentSolutionShear DiagramV = +1000 at x = 0V = 0 at x = 2Since dV/dx = -w = -500, a straight negative slopingline connects the end points
14. 14. 7.3 Relations between Distributed Load, Shear and MomentSolutionMoment DiagramM = -1000 at x = 0M = 0 at x = 2dM/dx = V, positive yet linearly decreasing fromdM/dx = 1000 at x = 0 to dM/dx = 0 at x = 2
15. 15. 7.3 Relations between Distributed Load, Shear and MomentExample 7.10Draw the shear and moment diagrams for thecantilevered beam.
16. 16. 7.3 Relations between Distributed Load, Shear and MomentSolutionSupport Reactions FBD of the beam
17. 17. 7.3 Relations between Distributed Load, Shear and MomentSolution At the ends of the beams, when x = 0, V = +1080 when x = 2, V = +600 Uniform load is downwards and slope of the shear diagram is constant dV/dx = -w = - 400 for 0 ≤ x ≤ 1.2 The above represents a change in shear
18. 18. 7.3 Relations between Distributed Load, Shear and MomentSolution ∆V = − ∫ w( x)dx = −400(1.2) = −480 V x =1.2 =V x =0 + (−480) = 1080 − 480 = 600 Also, by Method of Sections, for equilibrium, V = +600 Change in shear = area under the load diagram at x = 1.2, V = +600
19. 19. 7.3 Relations between Distributed Load, Shear and MomentSolution Since the load between 1.2 ≤ x ≤ 2, w = 0, slope dV/dx = 0, at x = 2, V = +600Shear Diagram
20. 20. 7.3 Relations between Distributed Load, Shear and MomentSolution At the ends of the beams, when x = 0, M = -1588 when x = 2, M = -100 Each value of shear gives the slope of the moment diagram since dM/dx = V at x = 0, dM/dx = +1080 at x = 1.2, dM/dx = +600 For 0 ≤ x ≤ 1.2, values of the shear diagram are positive but linearly increasing
21. 21. 7.3 Relations between Distributed Load, Shear and MomentSolution Moment diagram is parabolic with a linearly decreasing positive slope Moment Diagram
22. 22. 7.3 Relations between Distributed Load, Shear and MomentSolution Magnitude of moment at x = 1.2 = -580 Trapezoidal area under the shear diagram = change in moment ∆M = ∫ Vdx 1 = 600(1.2) + (1080 − 600)(1.2) = +1008 2 M x =1.2 = M x =0 + 1008 = −1588 + 1008 = −580
23. 23. 7.3 Relations between Distributed Load, Shear and MomentSolution By Method of Sections, at x = 1.2, M = -580 Moment diagram has a constant slope for 1.2 ≤ x ≤ 2 since dM/dx = V = +600 Hence, at x = 2, M = -100
24. 24. 7.3 Relations between Distributed Load, Shear and MomentExample 7.11Draw the shear and moment diagrams for theshaft. The support at A is a thrust bearingand the support at B is a journal bearing.
25. 25. 7.3 Relations between Distributed Load, Shear and MomentSolutionSupport Reactions FBD of the supports
26. 26. 7.3 Relations between Distributed Load, Shear and MomentSolution At the ends of the beams, when x = 0, V = +3.5 when x = 8, V = -3.5Shear Diagram
27. 27. 7.3 Relations between Distributed Load, Shear and MomentSolution No distributed load on the shaft, slope dV/dx = - w=0 Discontinuity or “jump” of the shear diagram at each concentrated force Change in shear negative when the force acts downwards and positive when the force acts upwards 2 kN force at x = 2m changes the shear from 3.5kN to 1.5kN 3 kN force at x = 4m changes the shear from 1.5kN to -1.5kN
28. 28. 7.3 Relations between Distributed Load, Shear and MomentSolution By Method of Sections, x = 2m and V = 1.5kN
29. 29. 7.3 Relations between Distributed Load, Shear and MomentSolution At the ends of the beams, when x = 0, M = 0 when x = 8, M = 0Moment Diagram
30. 30. 7.3 Relations between Distributed Load, Shear and MomentSolution Area under the shear diagram = change in moment ∆M = ∫ Vdx = 3.5(2) = 7 M x=2 =M x =0 +7 = 0+7 = 7 Also, by Method of Sections, x = 2m, M = 7 kN .m
31. 31. 7.3 Relations between Distributed Load, Shear and MomentExample 7.12Draw the shear and moment diagrams for thebeam.
32. 32. 7.3 Relations between Distributed Load, Shear and MomentSolutionSupport Reactions FBD of the beam
33. 33. 7.3 Relations between Distributed Load, Shear and MomentSolution At A, reaction is up, vA = +100kN No load acts between A and C so shear remains constant, dV/dx = -w(x) = 0 600kN force acts downwards, so the shear jumps down 600kN from 100kN to -500kN at point B No jump occur at point D where the 4000kN.m coupe moment is applied since ∆V = 0
34. 34. 7.3 Relations between Distributed Load, Shear and MomentSolutionShear Diagram Slope of moment from A to C is constant since dM/dx = V = +100
35. 35. 7.3 Relations between Distributed Load, Shear and MomentSolutionMoment Diagram
36. 36. 7.3 Relations between Distributed Load, Shear and MomentSolution Determine moment at C by Method of Sections where MC = +1000kN or by computing area under the moment ∆MAC = (100kN)(10m) = 1000kN
37. 37. 7.3 Relations between Distributed Load, Shear and MomentSolution Since MA = 0, MC = 0 + 1000kN.m = 1000kN.m From C to D, slope, dM/dx = V = -500 For area under the shear diagram between C and D, ∆MCD = (-500kN)(5m) = -2500kN, so that MD = MC + ∆MCD = 1000 – 2500 = -1500kN.m Jump at point D caused by concentrated couple moment of 4000kN.m Positive jump for clockwise couple moment
38. 38. 7.3 Relations between Distributed Load, Shear and MomentSolution At x = 15m, MD = - 1500 + 4000 = 2500kN.m Also, by Method of Sections, from point D, slope dM/dx = -500 is maintained until the diagram closes to zero at B