UNIT –III Bending Moment and Shear Force in Beams

1. UNIT –IIIBending Moment and Shear Force in Beams Subject : Mechanics of Materials N.RAM KUMAR M.E., ASSISTANT PROFESSOR DEPARTMENT OF MECHANICAL ENGINEERING CHRIST UNIVERSITY FACULTY OF ENGINEERING BANGALORE N.Ram Kumar, CUFE

2. INTRODUCTION • A beam is a structural member used for bearing loads. It is typically used for resisting vertical loads, shear forces and bending moments. • A beam is a structural member which is acted upon by the system of external loads at right angles to the axis. N.Ram Kumar, CUFE

3. TYPES OF BEAMS • Beams can be classified into many types based on three main criteria. They are as follows: Based on geometry: • Straight beam – Beam with straight profile • Curved beam – Beam with curved profile • Tapered beam – Beam with tapered cross section • Based on the shape of cross section: • I-beam – Beam with ‘I’ cross section • T-beam – Beam with ‘T’ cross section • C-beam – Beam with ‘C’ cross section • Based on equilibrium conditions: • Statically determinate beam – For a statically determinate beam, equilibrium conditions alone can be used to solve reactions. • Statically indeterminate beam – For a statically indeterminate beam, equilibrium conditions are not enough to solve reactions. Additional deflections are needed to solve reactions. N.Ram Kumar, CUFE

4. Based on the type of support: • Simply supported beam • Cantilever beam • Fixed beam • Overhanging beam • Continuous beam The last two beams are statically indeterminate beams. N.Ram Kumar, CUFE

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6. TYPES OF LOADS • Point Load or Concentrated Load. • Uniformly Distributed Loads. • Uniformly Varying Loads. • Externally Applied Moments. N.Ram Kumar, CUFE

7. TYPES OF SUPPORTS • Simple Support. • Roller Support. • Hinged Pin Support. • Fixed Support. N.Ram Kumar, CUFE

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9. SHEAR FORCE • Shear Force at a section in a beam is the force that is trying to shear off the section and is obtained as the algebraic sum of all forces including the reactions acting normal to the axis of the beam either to the left or right of the section. • The procedure to find shear force at a section is to imagine a cut in a beam at a section, considering either the left or the right portion and find algebraic sum of all forces normal to the axis. N.Ram Kumar, CUFE

10. F F + ve shear force - ve shear force F F SIGN CONVENTION FOR SHEAR FORCE N.Ram Kumar, CUFE

11. BENDING MOMENT • Bending moment at a section in a beam is the moment that is trying to bend it and is obtained as the algebraic sum of the moments about the section of all the forces(including the reaction acting on the beam either to the left or to the right of the section. N.Ram Kumar, CUFE

12. SIGN CONVENTION FOR BENDING MOMENTS The bending moment is considered as Sagging Bending Moment if it tends to bend the beam to a curvature having convexity at the bottom as shown in the Fig. given below. Sagging Bending Moment is considered as positive bending moment. Convexity Fig.Sagging bending moment(Positive bending moment) N.Ram Kumar, CUFE

13. SIGN CONVENTION FOR BENDING MOMENTS Similarly the bending moment is considered as hogging bending moment if it tends to bend the beam to a curvature having convexity at the top as shown in the Fig. given below. Hogging Bending Moment is considered as Negative Bending Moment. Convexity Fig. Hogging bending moment [Negative bending moment ] N.Ram Kumar, CUFE

14. Point of Contra flexure [Inflection point] It is the point on the bending moment diagram where bending moment changes the sign from positive to negative or vice versa. It is also called ‘Inflection point’. At the point of inflection point or contra flexure the bending moment is zero. N.Ram Kumar, CUFE

15. SHEAR FORCE DIAGRAM A diagram in which the ordinate represents shear force and the abscissa represents the position of the section is called shear force diagram. N.Ram Kumar, CUFE

16. BENDING MOMENT DIAGRAM Bending moment diagram may be defined as a diagram in which ordinate represents bending moment and abscissa represents the position of the section. N.Ram Kumar, CUFE

17. SFD AND BMD FOR THE STANDARD CASES • Cantilever Subjected to • A Point or concentrated load at free end. • Uniformly distributed load over entire span. • Uniformly varying load over entire span. 2. Simply supported beam subjected to • A Point or concentrated load at free end. • Uniformly distributed load over entire span. • Uniformly varying load over entire span. • An external moment. 3. Overhanging beam subjected to a Point or concentrated load at free end. N.Ram Kumar, CUFE

18. CANTILEVER BEAM • A cantilever is a beam whose one end is fixed and the other end is free. N.Ram Kumar, CUFE

19. EXAMPLES OF CANTILEVER BEAMPAMBAN BRIDGE N.Ram Kumar, CUFE

20. CAR PARKING ROOF N.Ram Kumar, CUFE

21. BUS STOP SHELTER N.Ram Kumar, CUFE

22. TRAFFIC SIGNAL POST N.Ram Kumar, CUFE

23. TRAFFIC SIGNAL POST The above traffic signal post is a best example of cantilever beam. Its a cantilever beam with three point loads. N.Ram Kumar, CUFE

24. Crane N.Ram Kumar, CUFE

25. CANTILEVER BEAM SUBJECTED TO POINT LOAD N.Ram Kumar, CUFE

26. CANTILEVER BEAM WITH UNIFORMLY DISTRIBUTED LOAD N.Ram Kumar, CUFE

27. SIMPLY SUPPORTED BEAM • A simply supported beam is one whose ends freely rest on walls or columns. • In all such cases, the reactions are always upwards. N.Ram Kumar, CUFE

28. EXAMPLE • The above shaft could be a best example of a simply supported Beam. • In that shaft both ends are supported by bearings and two loads are given by the pulley which could be taken as point load acting downwards. N.Ram Kumar, CUFE

29. SIMPLY SUPPORTED BEAM WITH A POINT LOAD N.Ram Kumar, CUFE

30. SIMPLY SUPPORTED BEAM WITH UNIFORMLY DISTRIBUTED LOAD N.Ram Kumar, CUFE

31. SIMPLY SUPPORTED BEAM WITH UNIFORMLY VARYING LOAD N.Ram Kumar, CUFE

32. OVERHANGING BEAM • A overhanging beam is one which has the loads beyond the supports. • Eg: Car. N.Ram Kumar, CUFE

33. EXAMPLE N.Ram Kumar, CUFE

34. Common Relationships N.Ram Kumar, CUFE

35. Common Relationships M N.Ram Kumar, CUFE

36. Example: Draw Shear & Moment diagrams for the following beam 12 kN 8 kN A C D B 1 m 3 m 1 m RA = 7 kN  RC = 13 kN  N.Ram Kumar, CUFE

37. 12 kN 8 kN A C D B 1 m 3 m 1 m 8 7 8 7 V (kN) -15 -5 7 M (kN-m) 2.4 m -8 N.Ram Kumar, CUFE

38. wkN/m x x1 x x1 L dx Relationship between load, shear force and bending moment Fig. A simply supported beam subjected to general type loading The above Fig. shows a simply supported beam subjected to a general type of loading. Consider a differential element of length ‘dx’ between any two sections x-x and x1-x1 as shown. N.Ram Kumar, CUFE

39. w kN/m x x1 V+dV M+dM M v O x x1 dx Fig. FBD of Differential element of the beam • Taking moments about the point ‘O’ [Bottom-Right corner of the differential element ] • M + (M+dM) – V.dx – w.dx.dx/2 = 0 • V.dx = dM  Neglecting the small quantity of higher order It is the relation between shear force and BM N.Ram Kumar, CUFE

40. w kN/m x x1 V+dV M+dM M v O x x1 dx Fig. FBD of Differential element of the beam Considering the Equilibrium Equation ΣFy = 0 - V + (V+dV) – w dx = 0  dv = w.dx  It is the relation Between intensity of Load and shear force. N.Ram Kumar, CUFE

41. Variation of Shear force and bending moments Variation of Shear force and bending moments for various standard loads are as shown in the following Table Table: Variation of Shear force and bending moments N.Ram Kumar, CUFE

42. 10N 5N 8N B A C D 2m 2m 1m 3m E Example Problem 1 • Draw shear force and bending moment diagrams [SFD and BMD] for a simply supported beam subjected to three point loads as shown in the Fig. given below. N.Ram Kumar, CUFE

43. 10N 5N 8N B A C D 2m 2m 1m 3m E RA RB Solution: Using the condition: ΣMA = 0 - RB × 8 + 8 × 7 + 10 × 4 + 5 × 2 = 0  RB = 13.25 N Using the condition: ΣFy = 0 RA + 13.25 = 5 + 10 + 8  RA = 9.75 N [Clockwise moment is Positive] N.Ram Kumar, CUFE

44. 10N 5N 8N 2m 2m 1m 3m Shear Force Calculation: 0 1 9 8 3 2 4 6 7 5 9 8 0 2 3 1 1 4 5 7 6 RA = 9.75 N RB=13.25N Shear Force at the section 1-1 is denoted as V1-1 Shear Force at the section 2-2 is denoted as V2-2 and so on... V0-0 = 0; V1-1 = + 9.75 N V6-6 = - 5.25 N V2-2 = + 9.75 N V7-7 = 5.25 – 8 = -13.25 N V3-3 = + 9.75 – 5 = 4.75 N V8-8 = -13.25 V4-4 = + 4.75 N V9-9 = -13.25 +13.25 = 0 V5-5 = +4.75 – 10 = - 5.25 N (Check) N.Ram Kumar, CUFE

45. 10N 5N 8N B A C E D 2m 2m 1m 3m 9.75N 9.75N 4.75N 4.75N 5.25N SFD 5.25N N.Ram Kumar, CUFE 13.25N 13.25N

46. 10N 5N 8N B A C E D 2m 2m 1m 3m 9.75N 9.75N 4.75N 4.75N 5.25N SFD 5.25N N.Ram Kumar, CUFE 13.25N 13.25N

47. Bending Moment Calculation Bending moment at A is denoted as MA Bending moment at B is denoted as MB and so on…   MA = 0 [ since it is simply supported] MC = 9.75 × 2= 19.5 Nm MD = 9.75 × 4 – 5 × 2 = 29 Nm ME = 9.75 × 7 – 5 × 5 – 10 × 3 = 13.25 Nm MB = 9.75 × 8 – 5 × 6 – 10 × 4 – 8 × 1 = 0 or MB = 0 [ since it is simply supported] N.Ram Kumar, CUFE

48. 10N 5N 8N A B E C D 2m 2m 1m 3m 29Nm 19.5Nm 13.25Nm BMD N.Ram Kumar, CUFE

49. 10N 5N 8N B A C D 2m 2m 1m 9.75N 9.75N 3m 4.75N 4.75N 5.25N 5.25N 29Nm SFD E 19.5Nm 13.25Nm 13.25N 13.25N BMD VM-34 Example Problem 1 N.Ram Kumar, CUFE

50. 10N 5N 8N B A C D 2m 2m 1m 9.75N 9.75N 3m 4.75N 4.75N 5.25N 5.25N 29Nm SFD E 19.5Nm 13.25Nm 13.25N 13.25N BMD N.Ram Kumar, CUFE