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4.Bending Moment and ShearForce DiagramTheory at a Glance (for IES, GATE, PSU)4.1 Shear Force and Bending MomentAt first we try to understand what shear force is and what is bending moment?We will not introduce any other co-ordinate system.We use general co-ordinate axis as shown in thefigure. This system will be followed in shear force andbending moment diagram and in deflection of beam.Here downward direction will be negative i.e.negative Y-axis. Therefore downward deflection of theWe use above Co-ordinate systembeam will be treated as negative.Some books fix a co-ordinate axis as shown in thefollowing figure. Here downward direction will bepositive i.e. positive Y-axis. Therefore downwarddeflection of the beam will be treated as positive. Asbeam is generally deflected in downward flection is positive deflection.Consider a cantilever beam as shown subjected toexternal load ‘P’. If we imagine this beam to be cut bya section X-X, we see that the applied force tend todisplace the left-hand portion of the beam relative tothe right hand portion, which is fixed in the wall.This tendency is resisted by internal forces betweenthe two parts of the beam. At the cut section aresistance shear force (Vx) and a bending moment(Mx) is induced. This resistance shear force and thebending moment at the cut section is shown in theleft hand and right hand portion of the cut beam.Using the three equations of equilibrium Fx0 , FyWe find that Vx0 and M P and M xi0 P . xIn this chapter we want to show pictoriallyPage 125theof 429Some books use above co-ordinate system

Chapter-4Bending Moment and Shear Force DiagramS K Mondal’svariation of shear force and bending moment in abeam as a function of ‘x' measured from one end ofthe beam.Shear Force (V) ŋ equal in magnitude but opposite in directionto the algebraic sum (resultant) of the components in thedirection perpendicular to the axis of the beam of all externalloads and support reactions acting on either side of the sectionbeing considered.Bending Moment (M) equal in magnitude but opposite indirection to the algebraic sum of the moments about (thecentroid of the cross section of the beam) the section of allexternal loads and support reactions acting on either side ofthe section being considered.What are the benefits of drawing shear force and bending moment diagram?The benefits of drawing a variation of shear force and bending moment in a beam as a function of ‘x'measured from one end of the beam is that it becomes easier to determine the maximum absolutevalue of shear force and bending moment. The shear force and bending moment diagram gives aclear picture in our mind about the variation of SF and BM throughout the entire section of thebeam.Further, the determination of value of bending moment as a function of ‘x' becomes very importantso as to determine the value of deflection of beam subjected to a given loading where we will use theformula, EId 2ydx 2Mx .4.2 Notation and sign conventionx Shear force (V)Positive Shear ForceA shearing force having a downward direction to the right hand side of a section or upwardsto the left hand of the section will be taken as ‘positive’. It is the usual sign conventions to befollowed for the shear force. In some book followed totally opposite sign convention.Page 126 of 429

Chapter-4Bending Moment and Shear Force DiagramdownwardS K Mondal’sThe upward direction shearingThedirectionforce which is on the left handshearing force which is on theof the section XX is positiveright hand of the section XX isshear force.positive shear force.Negative Shear ForceA shearing force having an upward direction to the right hand side of a section or downwardsto the left hand of the section will be taken as ‘negative’.ThexdownwarddirectionThe upward direction shearingshearing force which is on theforce which is on the rightleft hand of the section XX ishand of the section XX isnegative shear force.negative shear force.Bending Moment (M)Positive Bending MomentA bending moment causing concavity upwards will be taken as ‘positive’ and called assagging bending moment.Page 127 of 429

Chapter-4Bending Moment and Shear Force DiagramS K Mondal’sSaggingIf the bending moment ofIf the bending moment ofA bending moment causingthe left hand of the sectionthe right hand of theconcavity upwards will beXX is clockwise then it is asectionpositive bending moment.clockwise then it is acalled as sagging bendingpositive bending ative Bending MomentHoggingIf the bending moment ofA bending moment causingthethe right hand of theconvexity upwards will beanti-section XX is clockwisetaken as ‘negative’ and calledIf the bending moment oftheleftsectionhandXXisofclockwise then it is athenpositive bending moment.bending moment.itisapositiveas hogging bending moment.Way to remember sign conventionxRemember in the Cantilever beam both Shear force and BM are negative (–ive).4.3 Relation between S.F (Vx), B.M. (Mx) & Load (w)xdVx -w (load) The value of the distributed load at any point in the beam isdxequal to the slope of the shear force curve. (Note that the sign of this rule may changedepending on the sign convention used for the external distributed load).xdM x Vx The value of the shear force at any point in the beam is equal to the slopedxof the bending moment curve.Page 128 of 429

Chapter-4Bending Moment and Shear Force DiagramS K Mondal’s4.4 Procedure for drawing shear force and bending moment diagramConstruction of shear force diagramxFrom the loading diagram of the beam constructed shear force diagram.xFirst determine the reactions.xThen the vertical components of forces and reactions are successively summed from the leftend of the beam to preserve the mathematical sign conventions adopted. The shear at asection is simply equal to the sum of all the vertical forces to the left of the section.xThe shear force curve is continuous unless there is a point force on the beam. The curve then“jumps” by the magnitude of the point force ( for upward force).xWhen the successive summation process is used, the shear force diagram should end up withthe previously calculated shear (reaction at right end of the beam). No shear force actsthrough the beam just beyond the last vertical force or reaction. If the shear force diagramcloses in this fashion, then it gives an important check on mathematical calculations. i.e. Theshear force will be zero at each end of the beam unless a point force is applied at the end.Construction of bending moment diagramxThe bending moment diagram is obtained by proceeding continuously along the length ofbeam from the left hand end and summing up the areas of shear force diagrams using propersign convention.xThe process of obtaining the moment diagram from the shear force diagram by summation isexactly the same as that for drawing shear force diagram from load diagram.xThe bending moment curve is continuous unless there is a point moment on the beam. Thecurve then “jumps” by the magnitude of the point moment ( for CW moment).xWe know that a constant shear force produces a uniform change in the bending moment,resulting in straight line in the moment diagram. If no shear force exists along a certainportion of a beam, then it indicates that there is no change in moment takes place. We alsoknow that dM/dx Vx therefore, from the fundamental theorem of calculus the maximum orminimum moment occurs where the shear is zero.xThe bending moment will be zero at each free or pinned end of the beam. If the end is builtin, the moment computed by the summation must be equal to the one calculated initially forthe reaction.4.5 Different types of Loading and their S.F & B.M Diagram(i) A Cantilever beam with a concentrated load ‘P’ at its free end.Page 129 of 429

Chapter-4Bending Moment and Shear Force DiagramS K Mondal’sShear force:At a section a distance x from free end consider the forces tothe left, then (Vx) - P (for all values of x) negative in signi.e. the shear force to the left of the x-section are in downwarddirection and therefore negative.Bending Moment:Taking moments about the section gives (obviously to the leftof the section)Mx -P.x (negative sign means that theS.F and B.M diagrammoment on the left hand side of the portion is in theanticlockwise direction and is therefore taken as negativeaccording to the sign convention) so that the maximumbending moment occurs at the fixed end i.e. Mmax - PL(at x L)(ii) A Cantilever beam with uniformly distributed load over the whole lengthWhen a cantilever beam is subjected to a uniformlydistributed load whose intensity is given w /unit length.Shear force:Consider any cross-section XX which is at a distance of x fromthe free end. If we just take the resultant of all the forces onthe left of the X-section, thenVx -w.xfor all values of ‘x'.At x 0, Vx 0At x L, Vx -wL (i.e. Maximum at fixed end)Plotting the equation Vx -w.x, we get a straight linebecause it is a equation of a straight line y (Vx) m(- w) .xBending Moment:Bending Moment at XX is obtained by treating the load to theleft of XX as a concentrated load of the same value (w.x)S.F and B.M diagramacting through the centre of gravity at x/2.Therefore, the bending moment at any cross-section XX isMx w .x .x2 w .x 22Therefore the variation of bending moment is according to parabolic law.The extreme values of B.M would beat x 0,Mx 0and x L, Mx wL22Page 130 of 429

Chapter-4Bending Moment and Shear Force DiagramMaximum bending moment,MmaxwL22S K Mondal’sat fixed endAnother way to describe a cantilever beam with uniformly distributed load (UDL) over it’s wholelength.(iii) A Cantilever beam loaded as shown below draw its S.F and B.M diagramIn the region 0 x aFollowing the same rule as followed previously, we getVx - P; and Mx - P.xIn the region a x LVx - P P 0; and Mx - P.x P x aP.aS.F and B.M diagram(iv) Let us take an example: Consider a cantilever bean of 5 m length. It carries a uniformlydistributed load 3 KN/m and a concentrated load of 7 kN at the free end and 10 kN at 3 meters fromthe fixed end.Draw SF and BM diagram.Page 131 of 429

Chapter-4Bending Moment and Shear Force DiagramAnswer: In the region 0 x 2 mConsider any cross section XX at a distance x from free end.Shear force (Vx) -7- 3xSo, the variation of shear force is linear.at x 0,Vx -7 kNat x 2 m , Vx -7 - 3 u 2 -13 kNat point ZVx -7 -3 u 2-10 -23 kNBending moment (Mx) -7x - (3x).x2 3x 2 7x2So, the variation of bending force is parabolic.at x 0,Mx 0at x 2 m,Mx -7 u 2 – (3 u 2) u2 - 20 kNm2In the region 2 m x 5 mConsider any cross section YY at a distance x from freeendShear force (Vx) -7 - 3x – 10 -17- 3xSo, the variation of shear force is linear.at x 2 m, Vx - 23 kNat x 5 m, Vx - 32 kN§x· - 10 (x - 2) 2¹Bending moment (Mx) - 7x – (3x) u 3 x 2 17 x 202So, the variation of bending force is parabolic.at x 2 m, Mx3 u 22 17 u 2 20 - 20 kNm2at x 5 m, Mx - 102.5 kNmPage 132 of 429S K Mondal’s

Chapter-4(v)Bending Moment and Shear Force DiagramS K Mondal’sA Cantilever beam carrying uniformly varying load from zero at free end and w/unitlength at the fixed endConsider any cross-section XX which is at a distance of x from the free end.At this point load (wx) Therefore total load (W)w.xLLL³ w x dx³ L .xdx 0Shear force Vxw0wL2area of ABC (load triangle)1 §w ·wx 2 . x .x 2 L ¹2L? The shear force variation is parabolic.at x 0, Vx 0at x L, Vx WLi.e. Maximum Shear force (Vmax )2Page 133 of 429 WLat fixed end2

Chapter-4Bending Moment and Shear Force DiagramBending moment Mxload u distance from centroid of triangle ABCS K Mondal’swx 2 § 2x ·wx 3. 2L 3 ¹6L? The bending moment variation is cubic.at x 0, Mx 0 at x L,Mx wL2i.e. Maximum Bending moment (Mmax )6Alternative way :We know thatIntegration methodd Vx loaddxor d(Vx ) w.xLw.x .dxLIntegrating both sideVxx³ d Vx ³ w x2.L 20or Vx0w. x .dxLAgain we know thatd MxVxdxord Mx--wx 22Lwx 2dx2LPage 134 of 429wL2at fixed end.6

Chapter-4Bending Moment and Shear Force DiagramIntegrating both side we get at x 0,Mx 0Mx³xd(Mx )0or Mx ³0-S K Mondal’swx 2.dx2Lw x3 2L 3-wx 36L(vi) A Cantilever beam carrying gradually varying load from zero at fixed end andw/unit length at the free endConsidering equilibrium we get, MAwL2and Reaction R A3wL2Considering any cross-section XX which is at a distance of x from the fixed end.At this point load (Wx )Shear force VxW.xLRA area of triangle ANMwL 1 § w ·wL wx 2- . .x .x 2 2 L ¹22L? The shear force variation is parabolic.wLat x 0, Vx i.e. Maximum shear force, Vmax2at x L, Vx 0Bending moment Mx R A .x - wL2wx 2 2x.- MA2L 3wLwx 3wL2.x 26L3? The bending moment variation is cubic at x 0, Mx at x0L, MxwL2i.e.Maximum B.M. Mmax3Page 135 of 429 wL2.3

Chapter-4Bending Moment and Shear Force DiagramS K Mondal’s(vii) A Cantilever beam carrying a moment M at free endConsider any cross-section XX which is at a distance of x from the free end.Shear force: Vx 0 at any point.Bending moment (Mx) -M at any point, i.e. Bending moment is constant throughout thelength.(viii) A Simply supported beam with a concentrated load ‘P’ at its mid span.Considering equilibrium we get, R A RB P2Now consider any cross-section XX which is at a distance of x from left end A and section YY ata distance from left end A, as shown in figure below.Page 136 of 429Shear force: In the region 0 x L/2

Chapter-4Bending Moment and Shear Force DiagramS K Mondal’sVx RA P/2 (it is constant)In the region L/2 x LVx RA – P P- P - P/22(it is constant)Bending moment: In the region 0 x L/2Mx P.x2(its variation is linear)at x 0, Mx 0and at x L/2 Mx Maximum bending moment,PLi.e. maximum4PL4Mmaxat x L/2 (at mid-point)In the region L/2 x LMx PL PP .x (its variation is linear).x – P(x - L/2) 222at x L/2