Recitation 5 Understanding Shear Force and Bending Moment Diagrams Shear force and bending moment are examples of interanl forces that are induced in a structure when loads are applied to that structure Loading tends to cause failure in two main ways a by shearing the beam across its cross section b by bending the beam to an excessive amount Shear force may be defined as the algebraic sum of the loads to the left or right of a point such that the addition of this force restores vertical equilbrium The accepted sign convention is ve shear sum of forces to the left is upwards ve sum of forces to the right is downwards ve ve shear sum of forces to the left is downwards ve sum of forces to the right is upwards ve A shear force diagram is one which shows variation in shear force along the length of the beam Bending moment may be defined as the sum of moments about that section of all external forces acting to one side of that section Moments at any point are calculated by multiplying the magnitude of the external forces loads or reactions by the distance between the point at which moment is being determined and the point at which the external forces loads or reactions are being applied The accepted sign convention is ve bending moment sum of moments to the left is clockwise sum of moments to the right is anticlockwise ve bending moment sum of moments to the left is anti clockwise sum of moments to the right is clockwise Note For beams spanning between two simple pin jointed supports i e no cantilevers moment will always be positive and although the beam sags moment is drawn above the axis The maximum bending moment occurs at the point of zero shear force Before shear force and bending moments can be calculated the reactions at the supports must be determined This is usually done by first taking moment s about one of the supports sum of moments about support 0 to determine the reaction at the other support and secondly by resolving vertically sum of vertical reactions 0 and determining the reaction at the other support A bending moment diagram is one which shows variation in bending moment along the length of the beam Example 1 Draw the shear force and bending moment diagrams for the beam shown below a determine the reactions at the supports Taking moments about A clockwise moments anti clockwise moments 10 x2 5RC 5RC 20 RC 20 5 4kN Resolving vertically RA RC 10kN Substituting in RC 4kN RA 4 10 RA 10 4 6kN b draw the shear force diagram working from left to right the reaction at A 6kN the shear force remains constant between A and B i e 6kN and so the shear force diagram is horizontal between these points at point B where the 10kN point load is applied the shear force changes from 6kN to 610 4kN the shear force remains constant between B and C i e 4kN and so the shear force diagram is horizontal between these points at support C the reaction of 4kN brings the value of shear force back to zero and the diagram is complete c draw the bending moment diagram The moments at the supports A and C are zero The maximum moment occurs at B point of zero shear force Taking moments at B to the left MB 6 x 2 12kNm Bending moment varies uniformly between A and B and between B and C i e the bending moment diagram is a straight line For example the bending moment midway between A and B 6 x1 6kNm which is half the value of the bending moment at B Therefore the bending moment diagram is Example 2 Draw the shear force and bending moment diagrams for the beam show below a determine the reactions at the supports Taking moments about A clockwise moments anti clockwise moments 10 x 6 x 3 6RC where 10 x 6 60kN total load and 3m distance from A to where the load is acting 6RC 180 RC 180 6 30kN Resolving vertically RA RC 10 x 6 60kN Substituting in RC 30kN RA 30 60 RA 60 30 30kN Note In this example because the loading is applied symmetrically the reactions at each support must be equal and are therefore half the load It is acceptable to assume this and not necessary to calculate the reactions b draw the shear force diagram Working from left to right the reaction at A 30kN Moving from left to right the shear force reduces by 10kN per metre and so changes from 30kN at A to 30 6x10 30kN at C The shear force diagram is sloping between A and C and the point aof zero shear force occurs at B the centre point At support C the reaction of 30kN brings the value of shear force back to zero and the diagram is complete c draw the bending moment diagram The moments at the supports A and C are zero The maximum moment occurs at B point of zero shear force Taking moments about B to the left MB 30 x3 10x3x 3 2 90 45 45kNm where 10 x3 30kN is the value of uniform load contributing to the moment at B and 3 2 is the distance from B to the point where this load is assumed to act Bending moment does not vary uniformly between A and B and between B and C but the bending moment diagram is parabolic curved This can be demonstrated by calculating the bending moment at 1m intervals measured from A BM1 30 x1 10x1 x 1 2 25kNm BM2 30x2 10x2 x 2 2 40kNm BM3 30x3 10x3 x 3 2 45kNm BM4 30x4 10x4 x 4 2 40kNm BM5 30x5 10x5 x 5 2 25kNm Note It is usually sufficient to calculate the maximum moment only Note The Bending Moment is Parabolic CURVED when you are applying a uniform load q to the beam SO Now you can combine these two types of loads uniform q and your P loads don t forget the loads at the pins need to be taken into account SO SUMMARY Now you know that for your simple beam system Use the Free Body Diagram of section at x you define your system SHEAR DIAGRAMS Use Fy LoadA LoadB V 0 Concentrated Load P flat line Uniform Load q some slope on your line either decreasing or increasing if you have a flat uniform q like the example above we haven t discussed the shear diagram if your q increases by some slope over your beam BENDING MOMENT DIAGRAMS M LoadA distance LoadB distance M 0 Solve for M Concentrated Load P inc or decrease linearly to the applied P Uniform Load q Parabolic