Shear and Bending Moment Diagram :The shear force diagram indicates the shear force withstood by the beam section along thelength of the beam.The bending moment diagram indicates the bending moment withstood by the beam sectionalong the length of the beam.It is normal practice to produce a free bodydiagram with the shear diagram and thebending moment diagram position below

For simply supported beams the reactions aregenerally simple forces. When the beam isbuilt-in the free body diagram will show therelevant support point as a reaction force and areaction moment....Sign ConventionThe sign convention used for shear forcediagrams and bending moments is onlyimportant in that it should be usedconsistently throughout a project. The sign convention used on this page .

Shearing Force :

The shearing force (SF) at any section of a beam represents the tendency for the portion of thebeam on one side of the section to slide or shearlaterally relative to the other portion.

The diagram shows a beam carrying loads . It issimply supported at two points where the reactionsare Assume that the beam is divided into two partsby a section XX The resultant of the loads andreaction acting on the left of AA is F verticallyupwards and since the whole beam is in equilibrium,the resultant force to the right of AA must be F downwards. F is called the Shearing Force at thesection AA. It may be defined as follows:-The shearing force at any section of a beam is the algebraic sum of the lateral components of theforces acting on either side of the section.Where forces are neither in the lateral or axial direction they must be resolved in the usual wayand only the lateral components use to calculate the shear force.

Bending Moment :In a similar manner it can that if the Bending moments (BM) of the forces to the left of AA areclockwise then the bending moment of the forces to the right of AA must be anticlockwise.Bending Moment at AA is defined as the algebraic sum of

A Concentrated LoadIs one which can be considered to act at a pointalthough of course in practice it must bedistributed over a small area like weight orreactions .

A Distributed Loadis one which is spread in some manner overthe length or a significant length of the beam.It is usually quoted at a weight per unit lengthof beam. It may either be uniform or vary frompoint to point.

the moments about the section of all forces acting oneither side of the sectionBending moments are considered positive when themoment on the left portion is clockwise and on the rightanticlockwise. This is referred to asa sagging bending moment as it tends to make the beamconcave upwards at AA. A negative bending moment istermed hogging.

Type of Loads :A beam is normally horizontal and the loads vertical. Other cases which occur are considered tobe exceptions.

Example Of Diagrams :A shear force diagram is simply constructed by moving a section along the beam from (say)theleft origin and summing the forces to the left of the section. The equilibrium condition statesthat the forces on either side of a section balance and therefore the resisting shear force of thesection is obtained by this simple operation

The bending moment diagram is obtained in the same way except that the moment is the sumof the product of each force and its distance(x) from the section. Distributed loads arecalculated buy summing the product of the total force (to the left of the section) and thedistance(x) of the centroid of the distributed load.

The sketches below show simply supported beams with on concentrated force.

Drawing of Shear Force and Bending MomentDiagrams:Consider a simple beam shown of length L that carriesa uniform load of w (N/m) throughout its length and isheld in equilibrium by reactions R1 and R2. Assumethat the beam is cut at point C a distance of x from theleft hand support and the portion of the beam to theright of C be removed. The portion removed must thenbe replaced by vertical shearing force V together with acouple M to hold the left portion of the bar inequilibrium under the action of R1 and wx.

Write shear and moment equations for the beams in the following problems. In each problem,let x be the distance measured from left end of the beam to the point under study. Also, drawshear and moment diagrams, specifying values at all change of loading positions and at pointsof zero shear. Neglect the mass of the beam in each problem.The vertical shear at C will beVC=R1wx (Linear variation)Where R1 = R2 = wL/2Vc=wL/2wxThe moment at C isMC= (wL/2)xwx x/2MC=wLx/2wx2/2 (Parabolic variation)If we differentiate M with respect to x:dM/dx=wL/2 wx=shear force at x

thus, dM/dx =VxThus, the rate of change of the bending moment with respect to x is equal to the shearing force RELATION BETWEEN BENDING MOMENT AND SHEAR FORCE: The slope of the bendingmoment diagram at the given point is the shear force at that point. dM/dx =Vx RELATION BETWEEN SHEAR FORCE and UDL: Differentiate V with respect to xgives dV/dx=0wthus, dV/dx =-UDL= -wdV/dx =-w Thus, the rate of change of the shearing force with respect to x is equal to the load (UDL)Properties of Shear Force and Bending Moment Diagrams .

The following are some important properties of shear and moment diagrams: 1. The area of the shear diagram to the left or to the right of the section is equal to the momentat that section.2. The slope of the moment diagram at a given point is the shear force at that point.3. The slope of the shear diagram at a given point equals the -UDL at that point.4. The maximum moment occurs at the point of zero shear. When the shear is zero, the slope ofmoment diagram is zero. Hence tangent drawn to the momentdiagram is horizontal.5. When the shear force is increasing, the moment diagram isconcave upward.

6. When the shear force is decreasing, the moment diagram isconcave downward.

Relations among Load, Shear, and Bending Moment:

When a beam carries more than two or three concentratedloads, or when it carries distributed loads, the method forplotting shear and bending moment can prove cumbersome.The construction of the shear diagram and, especially, of thebending-moment diagram will be greatly facilitated if certainrelations existing among load, shear, and bending momentare taken into consideration.

Relations between Load and Shear:

Fy =0=S-(S+dS)+F dx

0=S-S-dS+Fdx

F=dS/dx

S= F dx

MA=0=M-(M+dM)+(S+dS)-F dx (dx/2)

-dM+S dx + dS dx- (F dx dx )/2=0

dM/dx=S

M= S dx = F dx

S= F dx = F0 dx

=F0 x+ C1

S=F0 xM= S dx= F0x dx= F0 x2/2 +C2 M=(F0 x2)/2

Fy=F0LMz= (F0 L2)/2

S(L)=0 F0L+C1=0Then C1= - F0 LFy=F0 L

S+F0L=0 S=-F0 L

M= S dx = (F0 x-F0 L)dx=(F0 x2)/2 F0 L x + C2

At x=L M=0

0=(F0 L2)/2 F0 L2 + C2 C2=(F0 L2)/2