Prof.Kodali Srinivas

1. The Shear Force at a section may be defined as "the algebraic sum of the total vertical forces on either side of the cross section".  The unit of shear force is :  N , or  kN   2.  The Bending moment at a section may be defined as "the algebraic sum of the moments about that section of all external forces acting to one side of that section". The unit of bending moment is :  N-mm ,or  kN-m 3. A Shear Force diagram ( SFD ) is one which shows the variation of the shear force along the length of the beam.  A bending moment diagram ( BMD ) is one which shows the variation of the bending moment along the length of the beam. 4.Draw the shearing force, and bending moment diagram for the structure, noting the sign conventions. Sign convention for Shear and Bending moments To draw  SFD  or  BMD  the negative values are plotted to scale below a horizontal reference line and positive values are plotted above the line. 5.Key points in Sh...

A concentrated moment has no “up and down” force, it does not cause any change in the magnitude of the shear diagram at its point of application. That does not mean that they do not influence the shear diagram, because they do. They influence of concentrated moment is in the support reactions, which in turn influences the shear diagram. Thus you will see no change in the shear diagram at the point of application of a concentrated moment.  Concentrated moments changes the magnitude of the moment diagram to a sudden “jump” at their points of application and equal to the magnitude of moments.   Problem 1.  Draw the shear force and bending moment diagrams in a simply supported beam carrying a concentrated moment M  as shown in figure. Solution :  Step.1 Find the unknown support reactions Ax,   Ay  and  B y at supports as shown in the figure. Along direction x, there is no imposed loads applied to the structure. Thus the first equilibrium equation...

A simple beam extending beyond its support on one end. is called as overhanging beam. The overhanging may be one side or both sides of the supports. Q1.  Draw the shear force and bending moment diagrams in a simply supported overhanging beam carrying loads as shown in figure. locate point of contraflexure  if any. Solution :  Step.1 Find the unknown support reactions at support A (Ax, Ay) and support C (Cy) at supports as shown in the figure. Along direction x, there is no imposed force applied to the structure. Thus the first equilibrium equation is: ∑ Fx = 0 Ax = 0 Along direction y, the imposed loads as well as support reactions. Thus the second equilibrium equation  ∑ Fy = 0 Ay + Cy - 10x4 - 20 = 0 Ay + Cy = 60 --- (a) Taking moments about A and apply the third equilibrium equation ∑ Mxy = 0 10x4x2 + 20x10 - Cy. 8 = 0 --- (b) Cy = 280/8 = 35 kN Substitute the Cy value in equation (a) Ay = 60 - Cy = 60 - 35 = 25 kN Step.2 : Segment AB Take a section x-x in between...

Q.1. Draw the shear force and bending moment diagrams in a simply supported beam carrying a point load W  as shown in figure. Solution :  Step.1 Find the unknown support reactions  H A ,  R A and R B at supports as shown in the figure . H A = 0 ;  R A = Wb/L R B = Wa/L Step.2  Take a section x-x in between A and C points at a distance x form support A Using the following sign convention for shear force and bending moments and write shear force (Fx) and bending moment (Mx) equations. +ve Sign Convention for Shear Force and Bending Moment  Consider left side of the section, the limits of  x is valid in between A and just left side of C only.  ie.  0 <  x < a considering sign convention (Left side), the S.F and B.M. equations are Vx = + R A = + Wb/L                  ---  (1) Mx = + ( R A ).x  =  + (Wb/L). x     ---  (2) substituting  ...

Q.1 Draw the shear force and bending moment diagrams of a cantilever beam carrying a point load W at its free end as shown in figure.  Solution: For cantilevered beams we need not find support reactions, if we considering the free-end of the beam as the initial starting point of the analysis. Let us consider one section XX at a distance x from free end i.e. from end B.  Now we will have two portions of the beam AB i.e. left portion and right portion.  Let us deal with the right portion and draw the free body diagram. (you can also go with left portion of the beam, you required to find the support reactions )  The section is valid in between B and A, becouse of there is no load in between the points. Let us assume that Vx is shear force and Mx is the bending moment at section xx.  Shear force at section xx will be equal to the resultant force acting to the right portion of the section.  Vx = + W ---(1) We will recall here the sign conventions for shear force...

There is a relation among intensity of load 'w' , Shear Force 'F' and Bending Moment 'M'.   For the derivation of the relations among these three we consider a simply supported beam subjected to a uniformly distributed load w(x) throughout its length (L), as shown in Figure. consider a small element of length dx at a distance of x form left support. Let the shear force and bending moment at a section located at a distance of x from the left support be V and M , respectively, and at a section x + dx be V + dV and M + dM , respectively.  The total load acting in the element is wdx, acts at the center of the element length dx. Considering the element is in equilibrium ,take moments on right side of  element. ∑ M x+dx  =  ∑ M R =  0                    = -(M+ dM) + M + Vdx +  wdx.dx/2  = 0  (Neglecting the small term wdx.dx/2 ) - dM + Vdx =0 dM = V dx   or   dM/dx...

Beam  Normally beams stand for horizontal structural members which transmit loads horizontally along their length to the supports. where the loads are generally transformed into vertical forces. The objective of beams is to withstand vertical loads, shear forces and bending moments, couples that lie in a plane containing the longitudinal section of the bar.  According to determent, a beam may be determinate or indeterminate. Statically Determinate Beams  Statically determinate beams are those beams in which the reactions of the supports may be determined by the use of the three static equilibrium equations .  Statically Indeterminate Beams  If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate. In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam. The degree of indeterminacy...