6161103 7.3 relations between distributed load, shear and moment

1. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load 7
Consider beam AD subjected to an arbitrary
load w = w(x) and a series of concentrated
forces and moments forces and moments
7
Distributed load assumed positive when loading
acts downwards

2. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load 7
A FBD diagram for a small
segment of the beam having a
length
∆x is chosen at point x
length
∆x is chosen at point x
along the beam which is not
subjected to a concentrated force
or couple moment
7
Any results obtained will not apply
at points of concentrated loadings

3. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Distributed Load 7
The internal shear force and bending
moments shown on the FBD are
assumed to act in the positive sense assumed to act in the positive sense
7
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
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The distributed loading has been replaced by a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, fractional distance k (∆x) from the right end, where 0 < k <1
( )
[ ]
2
)()(
0) ( )( ;0
)(
0) ( )( ;0
x kxwxV M
M M xkxxw MxV M
xxw V
V V xxw V F
y
∆ −∆= ∆
= ∆+ + ∆ ∆ + −∆− = ∑
∆ −= ∆
= ∆+ −∆ − = ∑↑+

5. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load
Slope of the
=
Negative of
xw
dx
dV
−=)(
Slope of the
=
Negative of
shear diagram distributed load intensity
Slope of = Shear moment diagram
V
dx
dM
=

6. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Distributed Load 7
At a specified point in a beam, the slope of the
shear diagram is equal to the intensity of the
distributed load distributed load
7
Slope of the moment diagram = shear
7
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
7
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
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load
Change in = Area under
∫
−= ∆dxxw V
BC
)(
Change in = Area under
shear shear diagram
Change in = Area under
moment shear diagram
∫
= ∆Vdx M
BC

8. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Distributed Load 7
Change in shear between points B and C is
equal to the negative of the area under the
distributed
-
loading curve between these
distributed
-
loading curve between these
points
7
Change in moment between B and C is equal
to the area under the shear diagram within
region BC
7
The equations so not apply at points where
concentrated force or couple moment acts

9. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Force (
FBD of a small segment of
the beam
F
V
F
−
=
∆
=
∑
↑
+
;
0
(
Change in shear is negative
thus the shear will jump
downwards when F acts
downwards on the beam
F
V
F
y
−
=
∆
=
∑
↑
+
;
0

10. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Force (
FBD of a small segment of the
beam located at the couple
moment moment
(
Change in moment is positive
or the moment diagram will
jump upwards M
O
is clockwise
O
M M M
=
∆
=
∑
;0

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

12. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution Support Reactions 7
FBD of the beam
7
FBD of the beam

13. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Solution Shear Diagram
V = +1000 at x = 0
V = 0 at x = 2
Since dV/dx =
-
w =
-
500, a straight negative sloping
Since dV/dx =
-
w =
-
500, a straight negative sloping
line connects the end points

14. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Solution Moment Diagram
M = -1000 at x = 0
M =
0
at x =
2
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
7.3 Relations between Distributed
Load, Shear and Moment
Example 7.10 Draw the shear and moment diagrams for the cantilevered beam. cantilevered beam.

16. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution Support Reactions 7
FBD of the beam
7
FBD of the beam

17. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
At the ends of the beams,
when x = 0, V = +1080
when x = 2, V = +600 when x = 2, V = +600
7
Uniform load is downwards and slope of
the shear diagram is constant
dV/dx = -w = -400for 0 ≤ x≤1.2
7
The above represents a change in shear

18. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Solution
600
480
1080
)
480
(
480 )2.1( 400 )(
0
2.
1
=
−
=
−
+
=
−= −= −= ∆
=
=
∫ V
V
dxxw V
x
x
(
Also, by Method of Sections, for equilibrium,
(
Change in shear = area under the load
diagram at x = 1.2, V = +600
600
600
480
1080
)
480
(
0
2.
1 +=
=
−
=
−
+
=
=
=
V
V
V
x
x

19. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
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
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
At the ends of the beams,
when x = 0, M = -1588
when x = 2, M =
-
100
when x = 2, M =
-
100
7
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
7
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
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
Moment diagram is parabolic with a linearly
decreasing positive slope
Moment Diagram
7
Moment Diagram

22. 7.3 Relations between Distributed
Load, Shear and Moment
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 change in moment
580 1008 1588
1008
1008 )2.1)( 600 1080 (
2
1
)2.1(600
0 2.1
−= + −=
+ =
+= − + =
= ∆
= =
∫
x x
M M
Vdx M

23. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
By Method of Sections,
at x = 1.2, M = -580
7
Moment diagram has a constant slope for 1.2 ≤
x ≤ 2 since dM/dx = V = +600 7
Hence, at x = 2, M = -100

24. 7.3 Relations between Distributed
Load, Shear and Moment
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. and the support at B is a journal bearing.

25. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution Support Reactions 7
FBD of the supports
7
FBD of the supports

26. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
At the ends of the beams,
when x = 0, V = +3.5
when x = 8, V =
-
3.5
when x = 8, V =
-
3.5
Shear Diagram

27. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
No distributed load on the shaft, slope dV/dx = -
w = 0 7
Discontinuity or “jump” of the shear diagram at each concentrated force each concentrated force
7
Change in shear negative when the force acts
downwards and positive when the force acts
upwards
7
2 kN force at x = 2m changes the shear from
3.5kN to 1.5kN 7
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
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
By Method of Sections, x = 2m and V =
1.5kN

29. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Solution 7
At the ends of the beams,
when x = 0, M = 0
when x = 8, M = 0 when x = 8, M = 0
Moment Diagram

30. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Solution (
Area under the shear diagram = change in
moment
=
=
=
∆
∫
(
Also, by Method of Sections,
mkN Mm x
M M
Vdx M
x x
. 7 ,2
7707
7)2(5.3
0 2
= =
=+=+ =
=
=
=
∆
= =
∫

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

32. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution Support Reactions 7
FBD of the beam
7
FBD of the beam

33. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
At A, reaction is up,
v
A
= +100kN
7
No load acts between A and C so shear remains
7
No load acts between A and C so shear remains constant, dV/dx = -w(x) = 0
7
600kN force acts downwards, so the shear jumps
down 600kN from 100kN to -500kN at point B
7
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
7.3 Relations between Distributed
Load, Shear and Moment Solution Shear Diagram 7
Slope of moment from A to C is constant
since dM/dx = V = +100

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

36. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
Determine moment at C by Method of Sections
where M
C
= +1000kN or by computing area
under the moment
∆M
AC
= (100kN)(10m) = 1000kN

37. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment Solution 7
Since M
A
= 0, M
C
= 0 + 1000kN.m = 1000kN.m
7
From C to D, slope, dM/dx = V = -500 For area under the shear diagram between C and
7
For area under the shear diagram between C and D, ∆M
CD
= (-500kN)(5m) = -2500kN, so that M
D
= M
C
+ ∆M
CD
= 1000 –2500 = -1500kN.m
7
Jump at point D caused by concentrated couple
moment of 4000kN.m
7
Positive jump for clockwise couple moment

38. 7.3 Relations between Distributed
Load, Shear and Moment
7.3 Relations between Distributed
Load, Shear and Moment
Solution 7
At x = 15m, M
D= -1500 + 4000 = 2500kN.m
7
Also, by Method of Sections, from point D, slope dM/dx =
-
500
is maintained until the
slope dM/dx =
-
500
is maintained until the
diagram closes to zero at B