Prof.Kodali Srinivas

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 and bending moment and we ca

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  = V (x) This relations implies that the first derivative of the bending moment with respect to the

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 is taken as the differe

  couple of Forces A couple is defined as two parallel forces with the same magnitude    but opposite in direction separated by a perpendicular distance d. The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals = F.d  The resultant of a couple at any point is always a pure constant moment  and equal to        M = F d Problem : 2.7 A 6 m-long simply supported beam carries a anti clockwise couple 400 N-m and a point load as shown in Fig. Calculate support reactions Solution: Let the vertical reaction at support A is Ra and at support B is Rb . Here the vertical forces are upwards reactive forces Ra ,Rb and downward point load of 1000 N. The horizontal pair of forces are formed as a couple and its resultant is a moment of 400N-m Apply the  equilibrium equation is: ∑ Fy = 0 R a  + R b  - 1000 = 0       R a  + R b    = 1000  -----(1) Taking moments of the forces about end A, ∑ M A  = 0 -  400  + 1000 x 4 - 6. R b   = 0 R b   = 3600/

If the beams are subjected to distributed loads like uniformly distributed load ( U.D.L. ) or distributed uniformly varying load ( U.V.L. ),in the analysis we can replace these loads by an equivalent load.  The magnitude of net equivalent point load is equal to total distributed load and acts at the centroid of the loading diagram. When solving for reactions, the following steps are recommended: Draw the beam free body diagram with support reactions and loads. Replace the uniform distributed load (if any) with the equivalent point load. If the loading (UDL) diagram is rectangle, the equivalent point load is W = wL, and its point of application is L/2 (mid point of the figure) If the loading (UVL) diagram is triangle, the equivalent point load is W = wL/2, and its point of application is L/3 as shown in the figure Solve ΣFx = 0 (sum of all horizontal forces) for finding the horizontal reaction (if any)  Solve ΣM A = 0 (sum of moments about support A) or  Solve ΣM B = 0 (sum of moments