1. 7.3 Relations between Distributed Load, Shear and Moment
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. 7.3 Relations between Distributed Load, Shear and Moment
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. 7.3 Relations between Distributed Load, Shear and Moment
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. 7.3 Relations between Distributed Load, Shear and Moment
The distributed loading has been replaced by a resultant force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, where 0 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

5. 7.3 Relations between Distributed Load, Shear and Moment
Slope of the = Negative of
shear diagram distributed load intensity
Slope of = Shear moment diagram

6. 7.3 Relations between Distributed Load, Shear and Moment
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. 7.3 Relations between Distributed Load, Shear and Moment
Change in = Area under
shear shear diagram
moment shear diagram

8. 7.3 Relations between Distributed Load, Shear and Moment
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. 7.3 Relations between Distributed Load, Shear and Moment
Force
FBD of a small segment of the beam
Change in shear is negative thus the shear will jump downwards when F acts downwards on the beam

10. 7.3 Relations between Distributed Load, Shear and Moment
Force
FBD of a small segment of the beam located at the couple moment
Change in moment is positive or the moment diagram will jump upwards MO is clockwise

11. 7.3 Relations between Distributed Load, Shear and Moment
Example 7.9
Draw the shear and moment diagrams for the
beam.

12. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Support Reactions
FBD of the beam

13. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Shear Diagram
V = at x = 0
V = 0 at x = 2
Since dV/dx = -w = -500, a straight negative sloping
line connects the end points

14. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Moment Diagram
M = at x = 0
M = 0 at x = 2
dM/dx = V, positive yet linearly decreasing from
dM/dx = 1000 at x = 0 to dM/dx = 0 at x = 2

15. 7.3 Relations between Distributed Load, Shear and Moment
Example 7.10
Draw the shear and moment diagrams for the
cantilevered beam.

16. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Support Reactions
FBD of the beam

17. 7.3 Relations between Distributed Load, Shear and Moment
Solution
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 = for 0 ≤ x ≤ 1.2
The above represents a change in shear

18. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Also, by Method of Sections, for equilibrium,
Change in shear = area under the load diagram at x = 1.2, V = +600

19. 7.3 Relations between Distributed 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 between Distributed 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 between Distributed Load, Shear and Moment
Solution
Moment diagram is parabolic with a linearly decreasing positive slope
Moment Diagram

22. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Magnitude of moment at x = 1.2 = -580
Trapezoidal area under the shear diagram = change in moment

23. 7.3 Relations between Distributed Load, Shear and Moment
Solution
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. 7.3 Relations between Distributed Load, Shear and Moment
Example 7.11
Draw the shear and moment diagrams for the
shaft. The support at A is a thrust bearing
and the support at B is a journal bearing.

25. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Support Reactions
FBD of the supports

26. 7.3 Relations between Distributed Load, Shear and Moment
Solution
At the ends of the beams,
when x = 0, V = +3.5
when x = 8, V = -3.5
Shear Diagram

27. 7.3 Relations between Distributed Load, Shear and Moment
Solution
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. 7.3 Relations between Distributed Load, Shear and Moment
Solution
By Method of Sections, x = 2m and V = 1.5kN

29. 7.3 Relations between Distributed Load, Shear and Moment
Solution
At the ends of the beams,
when x = 0, M = 0
when x = 8, M = 0
Moment Diagram

30. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Area under the shear diagram = change in moment
Also, by Method of Sections,

31. 7.3 Relations between Distributed Load, Shear and Moment
Example 7.12
Draw the shear and moment diagrams for the
beam.

32. 7.3 Relations between Distributed Load, Shear and Moment
View Free Body Diagram
Solution
Support Reactions
FBD of the beam

33. 7.3 Relations between Distributed Load, Shear and Moment
Solution
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. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Shear Diagram
Slope of moment from A to C is constant since dM/dx = V = +100

35. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Moment Diagram

36. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Determine moment at C by Method of Sections where MC = +1000kN or by computing area under the moment
∆MAC = (100kN)(10m) = 1000kN

37. 7.3 Relations between Distributed Load, Shear and Moment
Solution
Since MA = 0, MC = kN.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. 7.3 Relations between Distributed Load, Shear and Moment
Solution
At x = 15m, MD = = 2500kN.m
Also, by Method of Sections, from point D, slope dM/dx = -500 is maintained until the diagram closes to zero at B