7.3 relations between distributed load, shear and moment distributed load consider beam ad subjected...

7.3 Relations between Distributed Load, Shear and

Moment


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


Moment


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


Moment


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


Moment


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


Moment


MomentDistributed Load

Slope of the = Negative of shear diagram distributed load

intensity

Slope of = Shear moment diagram

Vdx

dM

xwdx

dV

)(


Moment


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


Moment


MomentDistributed Load

Change in = Area under shear shear diagram

Change in = Area undermoment shear diagram

VdxM

dxxwV

BC

BC )(


Moment


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


Moment


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


Moment


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


Moment


MomentExample 7.9Draw the shear and moment diagrams for

the beam.


Moment


MomentSolutionSupport ReactionsFBD of the beam


Moment


MomentSolutionShear DiagramV = +1000 at x = 0V = 0 at x = 2Since dV/dx = -w = -500, a straight negative

sloping line connects the end points


Moment


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


Moment



the cantilevered beam.


Moment




Moment


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


Moment


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


Moment


MomentSolutionSince the load between 1.2 ≤ x ≤ 2, w =

0, slope dV/dx = 0, at x = 2, V = +600

Shear Diagram


Moment


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


Moment


MomentSolutionMoment diagram is parabolic with a

linearly decreasing positive slope

Moment Diagram


Moment


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


Moment


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


Moment


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.


Moment


MomentSolutionSupport ReactionsFBD of the supports


Moment


MomentSolution At the ends of the beams,

when x = 0, V = +3.5when x = 8, V = -3.5

Shear Diagram


Moment


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


Moment


MomentSolutionBy Method of Sections, x = 2m and V

= 1.5kN


Moment


MomentSolutionAt the ends of the beams,

when x = 0, M = 0when x = 8, M = 0

Moment Diagram


Moment


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


Moment



the beam.


Moment



View Free Body Diagram


Moment


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


Moment


MomentSolutionShear Diagram

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


Moment


MomentSolutionMoment Diagram


Moment


MomentSolution Determine moment at C by Method of

Sections where MC = +1000kN or by computing area under the moment ∆MAC = (100kN)(10m) = 1000kN


Moment


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


Moment


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

7.3 relations between distributed load, shear and moment distributed load consider beam ad subjected...

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