7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentDistributed LoadConsider 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
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
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MomentDistributed Load The distributed loading has been replaced
by a resultant force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, where 0 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
2)()(
0)()(;0
)(
0)()(;0
xkxwxVM
MMxkxxwMxVM
xxwV
VVxxwVFy
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentDistributed Load
Slope of the = Negative of shear diagram distributed load
intensity
Slope of = Shear moment diagram
Vdx
dM
xwdx
dV
)(
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentDistributed Load
Change in = Area under shear shear diagram
Change in = Area undermoment shear diagram
VdxM
dxxwV
BC
BC )(
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentForce 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
FVFy ;0
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentForce 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
OMMM ;0
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentExample 7.9Draw the shear and moment diagrams for
the beam.
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionSupport ReactionsFBD of the beam
7.3 Relations between Distributed Load, Shear and
Moment
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
sloping line connects the end points
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionMoment DiagramM = -1000 at x = 0M = 0 at x = 2dM/dx = V, positive yet linearly decreasing from
dM/dx = 1000 at x = 0 to dM/dx = 0 at x = 2
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentExample 7.10Draw the shear and moment diagrams for
the cantilevered beam.
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionSupport ReactionsFBD of the beam
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionAt the ends of the beams,
when x = 0, V = +1080when x = 2, V = +600
Uniform load is downwards and slope of the shear diagram is constantdV/dx = -w = - 400 for 0 ≤ x ≤ 1.2
The above represents a change in shear
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolution
Also, by Method of Sections, for equilibrium,
Change in shear = area under the load diagram at x = 1.2, V = +600
600
6004801080)480(
480)2.1(400)(
02.1
V
VV
dxxwV
xx
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionSince the load between 1.2 ≤ x ≤ 2, w =
0, slope dV/dx = 0, at x = 2, V = +600
Shear Diagram
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolution At the ends of the beams,
when x = 0, M = -1588when x = 2, M = -100
Each value of shear gives the slope of the moment diagram since dM/dx = Vat x = 0, dM/dx = +1080at x = 1.2, dM/dx = +600
For 0 ≤ x ≤ 1.2, values of the shear diagram are positive but linearly increasing
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionMoment diagram is parabolic with a
linearly decreasing positive slope
Moment Diagram
7.3 Relations between Distributed Load, Shear and
Moment
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
58010081588
1008
1008)2.1)(6001080(2
1)2.1(600
02.1
xxMM
VdxM
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentExample 7.11Draw 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.
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionSupport ReactionsFBD of the supports
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolution At the ends of the beams,
when x = 0, V = +3.5when x = 8, V = -3.5
Shear Diagram
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionBy Method of Sections, x = 2m and V
= 1.5kN
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionAt the ends of the beams,
when x = 0, M = 0when x = 8, M = 0
Moment Diagram
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolution Area under the shear diagram =
change in moment
Also, by Method of Sections,
mkNMmx
MM
VdxM
xx
.7,2
7707
7)2(5.3
02
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentExample 7.12Draw the shear and moment diagrams for
the beam.
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionSupport ReactionsFBD of the beam
View Free Body Diagram
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionShear Diagram
Slope of moment from A to C is constant since dM/dx = V = +100
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionMoment Diagram
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
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
7.3 Relations between Distributed Load, Shear and
Moment
7.3 Relations between Distributed Load, Shear and
MomentSolutionAt x = 15m, MD = - 1500 + 4000 =
2500kN.mAlso, by Method of Sections, from point D,
slope dM/dx = -500 is maintained until the diagram closes to zero at B