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 momentsforces 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 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 senseassumed 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� The distributed loading has been replaced by a resultant force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, fractional distance k (∆x) from the right end, where 0 < k <1

( )[ ]2)()(

0)()(;0

)(

0)()(;0

xkxwxVM

MMxkxxwMxVM

xxwV

VVxxwVFy

∆−∆=∆

=∆++∆∆+−∆−=∑

∆−=∆

=∆+−∆−=∑↑+



Distributed Load

Slope of the = Negative of

xwdx

dV−= )(

Slope of the = Negative of

shear diagram distributed load intensity

Slope of = Shear moment diagram

Vdx

dM=



Distributed Load

� At a specified point in a beam, the slope of the shear diagram is equal to the intensity of the distributed loaddistributed 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

Change in = Area under

∫−=∆ dxxwVBC )(


shear shear diagram


moment shear diagram

∫=∆ VdxM BC



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 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

FVF −=∆=∑↑+ ;0

� Change in shear is negative thus the shear will jump downwards when F acts downwards on the beam

FVFy −=∆=∑↑+ ;0



Force

� FBD of a small segment of the beam located at the couple momentmoment

� Change in moment is positive or the moment diagram will jump upwards MO is clockwise

OMMM =∆=∑ ;0



Example 7.9

Draw the shear and moment diagrams for the

beam.beam.



Solution

Support Reactions

� FBD of the beam� FBD of the beam



SolutionShear DiagramV = +1000 at x = 0V = 0 at x = 2Since dV/dx = -w = -500, a straight negative sloping 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 = 2M = 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.cantilevered beam.



Solution

Support Reactions




Solution

� At the ends of the beams,

when x = 0, V = +1080

when x = 2, V = +600when 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

6004801080)480(

480)2.1(400)(

02.1=−=−+=

−=−=−=∆

==

∫VV

dxxwV

xx

� Also, by Method of Sections, for equilibrium,

� Change in shear = area under the load diagram at x = 1.2, V = +600

600

6004801080)480(02.1

+=

=−=−+===

V

VVxx



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 = -100when 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� Moment Diagram



Solution

� Magnitude of moment at x = 1.2 = -580

� Trapezoidal area under the shear diagram = change in momentchange in moment

58010081588

1008

1008)2.1)(6001080(2

1)2.1(600

02.1

−=+−=

+=

+=−+=

=∆

==

∫

xxMM

VdxM



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.and the support at B is a journal bearing.



Solution

Support Reactions

� FBD of the supports� FBD of the supports



Solution


when x = 0, V = +3.5

when x = 8, V = -3.5when 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 forceeach 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 = 0when x = 8, M = 0

Moment Diagram



Solution

� 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

==

=+=+=

===∆

==

∫



Example 7.12


beam.beam.



Solution

Support Reactions




Solution

� At A, reaction is up,

vA = +100kN

� No load acts between A and C so shear remains � 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 � 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 slope dM/dx = -500 is maintained until the diagram closes to zero at B

6161103 7.3 relations between distributed load, shear and moment

Technology

internal shear force

arbitrary load w

length x

point x

beam ad subjected

aresultant force f

small segment

series of concentrated