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![7.3 Relations between Distributed
Load, Shear and Moment
Distributed 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The distributed loading has been replaced by a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](https://image.slidesharecdn.com/61611037-3relationsbetweendistributedloadshearandmoment-120122212826-phpapp01/75/6161103-7-3-relations-between-distributed-load-shear-and-moment-4-2048.jpg)
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6161103 7.3 relations between distributed load, shear and moment
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- 5. 7.3 Relations betweenDistributed Load, Shear and Moment Distributed Load dV = −w(x) dx Slope of the = Negative of shear diagram distributed load intensity dM =V dx Slope of = Shear moment diagram
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- 7. 7.3 Relations betweenDistributed Load, Shear and Moment Distributed Load ∆VBC = − ∫ w( x)dx Change in = Area under shear shear diagram ∆M BC = ∫ Vdx Change in = Area under moment shear diagram
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- 12. 7.3 Relations betweenDistributed Load, Shear and Moment Solution Support Reactions FBD of the beam
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- 15. 7.3 Relations betweenDistributed Load, Shear and Moment Example 7.10 Draw the shear and moment diagrams for the cantilevered beam.
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- 19. 7.3 Relations betweenDistributed Load, Shear and Moment Solution Since the load between 1.2 ≤ x ≤ 2, w = 0, slope dV/dx = 0, at x = 2, V = +600 Shear Diagram
- 20. 7.3 Relations betweenDistributed Load, Shear and Moment Solution 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. 7.3 Relations betweenDistributed Load, Shear and Moment Solution Moment diagram is parabolic with a linearly decreasing positive slope Moment Diagram
- 22. 7.3 Relations betweenDistributed Load, Shear and Moment Solution 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
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