6161103 7.3 relations between distributed load, shear and moment

7.3 Relations between Distributed
Load, Shear and Moment
Distributed 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

Distributed 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

Distributed 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

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The distributed loading has been replaced by a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

Distributed Load
dV
= −w(x)
dx
Slope of the = Negative of
shear diagram distributed load intensity
dM
=V
dx
Slope of = Shear moment diagram

Distributed 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

Distributed Load
∆VBC = − ∫ w( x)dx
Change in = Area under
shear shear diagram
∆M BC = ∫ Vdx

Change in = Area under
moment shear diagram

Distributed 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

Force
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

Force
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

Example 7.9
Draw the shear and moment diagrams for the
beam.

Solution
Support Reactions
FBD of the beam

Solution
Shear Diagram
V = +1000 at x = 0
V = 0 at x = 2
Since dV/dx = -w = -500, a straight negative sloping
line connects the end points

Solution
Moment Diagram
M = -1000 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

Example 7.10
cantilevered beam.

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 = - 400 for 0 ≤ x ≤ 1.2
The above represents a change in shear

Solution
∆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

Solution
Since the load between 1.2 ≤ x ≤ 2, w =
0, slope dV/dx = 0, at x = 2, V = +600
Shear Diagram

Solution
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

Solution
Moment diagram is parabolic with a linearly
decreasing positive slope
Moment Diagram

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

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

Example 7.11
shaft. The support at A is a thrust bearing
and the support at B is a journal bearing.

Solution
Support Reactions
FBD of the supports

Solution
when x = 0, V = +3.5
when x = 8, V = -3.5
Shear Diagram

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

Solution
By Method of Sections, x = 2m and V =
1.5kN

Solution
when x = 0, M = 0
when x = 8, M = 0
Moment Diagram

Solution
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

Example 7.12
beam.

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

Solution
Shear Diagram

Slope of moment from A to C is constant
since dM/dx = V = +100

Solution
Moment Diagram

Solution
Determine moment at C by Method of Sections
where MC = +1000kN or by computing area
under the moment
∆MAC = (100kN)(10m) = 1000kN


Solution
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

Solution
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

6161103 7.3 relations between distributed load, shear and moment

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6161103 7.3 relations between distributed load, shear and moment