UNIT III BENDING MOMENT AND SHEAR FORCE IN

UNIT –III BENDING 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

INTRODUCTION Ø Ø 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

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

� 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

Ø N. Ram Kumar, CUFE

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

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

Ø N. Ram Kumar, CUFE

SHEAR FORCE Ø Ø 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

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

BENDING MOMENT Ø 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

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

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

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

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

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

SFD AND BMD FOR THE STANDARD CASES Cantilever Subjected to (i) A Point or concentrated load at free end. (ii) Uniformly distributed load over entire span. (iii) Uniformly varying load over entire span. 2. Simply supported beam subjected to (i) A Point or concentrated load at free end. (ii) Uniformly distributed load over entire span. (iii) Uniformly varying load over entire span. (iv) An external moment. 3. Overhanging beam subjected to a Point or concentrated load at free end. 1. N. Ram Kumar, CUFE

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

EXAMPLES OF CANTILEVER BEAM PAMBAN BRIDGE Ø N. Ram Kumar, CUFE

CAR PARKING ROOF Ø N. Ram Kumar, CUFE

BUS STOP SHELTER Ø N. Ram Kumar, CUFE

TRAFFIC SIGNAL POST Ø N. Ram Kumar, CUFE

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

Crane Ø N. Ram Kumar, CUFE

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

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

SIMPLY SUPPORTED BEAM Ø Ø 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

EXAMPLE Ø Ø Ø 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

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

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

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

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

EXAMPLE N. Ram Kumar, CUFE

Common Relationships 0 Constant Linear Parabolic Cubic Load Shear Moment N. Ram Kumar, CUFE

Common Relationships Load 0 0 Constant Linear Parabolic M Shear Moment N. Ram Kumar, CUFE

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

12 k. N 8 k. N A C D B 1 m 3 m 8 7 V (k. N) 1 m 8 7 -15 -5 7 M (k. N-m) 2. 4 m -8 N. Ram Kumar, CUFE

Relationship between load, shear force and bending moment x x 1 dx w k. N/m L 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 x 1 -x 1 as shown. N. Ram Kumar, CUFE

x x 1 w k. N/m V+d. V M v M+d. M x dx O x 1 Fig. FBD of Differential element of the beam Taking moments about the point ‘O’ [Bottom-Right corner of the differential element ] - M + (M+d. M) – V. dx – w. dx/2 = 0 V. dx = d. M Neglecting the small quantity of higher order It is the relation between shear force and BM N. Ram Kumar, CUFE

x x 1 w k. N/m V+d. V M v M+d. M x dx O x 1 Fig. FBD of Differential element of the beam Considering the Equilibrium Equation ΣFy = 0 - V + (V+d. V) – w dx = 0 dv = w. dx It is the relation Between intensity of Load and shear force. N. Ram Kumar, CUFE

Variation of Shear force and bending moments for various standard loads are as shown in the following Table: Variation of Shear force and bending moments Type of load Between point Uniformly loads or for no distributed load varying load region SFD/BMD Shear Force Horizontal line Inclined line Two-degree curve Diagram (Parabola) Bending Moment Diagram Inclined line Two-degree curve (Parabola) N. Ram Kumar, CUFE Three-degree curve (Cubicparabola)

Example Problem 1 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. 10 N 5 N 8 N B A C 2 m D 2 m E 3 m N. Ram Kumar, CUFE 1 m

10 N 5 N 8 N B A C 2 m E D 2 m 3 m 1 m RA RB Solution: [Clockwise moment is Positive] 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 N. Ram Kumar, CUFE

Shear Force Calculation: 0 0 1 2 2 1 2 m 10 N 5 N 3 4 8 N 5 6 7 8 9 2 m 3 m RA = 9. 75 N 1 m RB=13. 25 N Shear Force at the section 1 -1 is denoted as V 1 -1 Shear Force at the section 2 -2 is denoted as V 2 -2 and so on. . . V 0 -0 = 0; V 1 -1 = + 9. 75 N V 6 -6 = - 5. 25 N V 2 -2 = + 9. 75 N V 7 -7 = 5. 25 – 8 = -13. 25 N V 3 -3 = + 9. 75 – 5 = 4. 75 N V 8 -8 = -13. 25 V 4 -4 = + 4. 75 N V 9 -9 = -13. 25 +13. 25 = 0 V 5 -5 = +4. 75 – 10 = - 5. 25 N (Check) N. Ram Kumar, CUFE

10 N 5 N 8 N B A C 2 m 9. 75 N E D 2 m 1 m 3 m 9. 75 N 4. 75 N SFD 5. 25 N N. Ram Kumar, CUFE 5. 25 N 13. 25 N

10 N 5 N 8 N B A C 2 m 9. 75 N E D 2 m 1 m 3 m 9. 75 N 4. 75 N SFD 5. 25 N N. Ram Kumar, CUFE 5. 25 N 13. 25 N

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

10 N 5 N 8 N A B C 2 m E D 2 m 3 m 1 m 29 Nm 19. 5 Nm 13. 25 Nm BMD N. Ram Kumar, CUFE

10 N 5 N VM-34 A B C 2 m 9. 75 N 8 N D 2 m SFD 1 m 3 m 9. 75 N 4. 75 N E 4. 75 N 5. 25 N Example Problem 1 5. 25 N 13. 25 N 29 Nm 19. 5 Nm 13. 25 Nm BMD N. Ram Kumar, CUFE 13. 25 N

10 N 5 N 8 N B A C 2 m 9. 75 N D 2 m E 1 m 3 m 9. 75 N 4. 75 N SFD 4. 75 N 5. 25 N 13. 25 N 29 Nm 19. 5 Nm 13. 25 Nm BMD N. Ram Kumar, CUFE 13. 25 N

Example Problem 2 2. Draw SFD and BMD for the double side overhanging beam subjected to loading as shown below. Locate points of contraflexure if any. C A 2 m 5 k. N 10 k. N 5 k. N 2 k. N/m B D 3 m 3 m N. Ram Kumar, CUFE E 2 m

10 k. N 5 k. N C A 2 m RA 5 k. N 2 k. N/m B D 3 m 3 m RB E 2 m Solution: Calculation of Reactions: Due to symmetry of the beam, loading and boundary conditions, reactions at both supports are equal. . `. RA = RB = ½(5+10+5+2 × 6) = 16 k. N N. Ram Kumar, CUFE

10 k. N 5 k. N 0 1 2 3 4 5 2 m 2 k. N/m 5 k. N 6 7 8 9 9 3 m 2 m 3 m RA=16 k. N RB = 16 k. N Shear Force Calculation: V 0 -0 = 0 V 1 -1 = - 5 k. N V 6 -6 = - 5 – 6 = - 11 k. N V 2 -2 = - 5 k. N V 7 -7 = - 11 + 16 = 5 k. N V 3 -3 = - 5 + 16 = 11 k. N V 8 -8 = 5 k. N V 4 -4 = 11 – 2 × 3 = +5 k. N V 9 -9 = 5 – 5 = 0 (Check) V 5 -5 = 5 – 10 = - 5 k. N N. Ram Kumar, CUFE

10 k. N 5 k. N C 5 k. N A 5 k. N 2 k. N/m B D 3 m 2 m 3 m E 2 m 11 k. N 5 k. N + 5 k. N SFD 11 k. N N. Ram Kumar, CUFE 5 k. N +

10 k. N 5 k. N C A 2 m 2 k. N/m B D 3 m 5 k. N 3 m RA=16 k. N E 2 m RB = 16 k. N Bending Moment Calculation: MC = ME = 0 [Because Bending moment at free end is zero] MA = MB = - 5 × 2 = - 10 k. Nm MD = - 5 × 5 + 16 × 3 – 2 × 3 × 1. 5 = +14 k. Nm N. Ram Kumar, CUFE

10 k. N 5 k. N C A 2 k. N/m B D 3 m 2 m 5 k. N 3 m E 2 m 14 k. Nm 10 k. Nm BMD N. Ram Kumar, CUFE 10 k. Nm

10 k. N 5 k. N C 2 m A 3 m 5 k. N 2 k. N/m B D 3 m 2 m E 11 k. N + 5 k. N 10 k. Nm SFD 5 k. N 14 k. Nm 5 k. N + 11 k. N BMD N. Ram Kumar, CUFE 10 k. Nm 5 k. N

10 k. N 5 k. N 2 k. N/m x C Ax 2 m x 10 k. Nm 5 k. N B D 3 m 3 m Points of contra flexure x E 2 m 10 k. Nm Let x be the distance of point of contra flexure from support A Taking moments at the section x-x (Considering left portion) x = 1 or 10. `. x = 1 m N. Ram Kumar, CUFE

Example Problem 3 3. Draw SFD and BMD for the single side overhanging beam subjected to loading as shown below. Determine the absolute maximum bending moment and shear forces and mark them on SFD and BMD. Also locate points of contra flexure if any. A C 4 m D B 1 m N. Ram Kumar, CUFE 5 k. N 2 k. N 10 k. N/m 2 m

10 k. N/m A 5 k. N 2 k. N B RA 1 m RB 2 m 4 m Solution : Calculation of Reactions: ΣMA = 0 - RB × 5 + 10 × 4 × 2 + 2 × 4 + 5 × 7 = 0 RB = 24. 6 k. N ΣFy = 0 RA + 24. 6 – 10 x 4 – 2 + 5 = 0 RA = 22. 4 k. N N. Ram Kumar, CUFE

0 1 RA=22. 4 k. N 10 k. N/m 2 2 4 m 2 k. N 3 4 5 1 m 5 k. N 7 6 6 7 2 m RB=24. 6 k. N Shear Force Calculations: V 0 -0 =0; V 1 -1 = 22. 4 k. N V 2 -2 = 22. 4 – 10 × 4 = -17. 6 k. N V 3 -3 = - 17. 6 – 2 = - 19. 6 k. N V 4 -4 = - 19. 6 k. N V 5 -5 = - 19. 6 + 24. 6 = 5 k. N V 6 -6 = 5 k. N V 7 -7 = 5 – 5 = 0 (Check) N. Ram Kumar, CUFE

5 k. N 2 k. N 10 k. N/m A C RA=22. 4 k. N 4 m B 1 m D 2 m RB=24. 6 k. N 22. 4 k. N 5 k. N x = 2. 24 m SFD 17. 6 k. N 19. 6 k. N N. Ram Kumar, CUFE 19. 6 k. N 5 k. N

X A x RA=22. 4 k. N X 4 m 5 k. N 2 k. N 10 k. N/m C D B 1 m 2 m RB=24. 6 k. N Max. bending moment will occur at the section where the shear force is zero. The SFD shows that the section having zero shear force is available in the portion AC. Let that section be X-X, considered at a distance x from support A as shown above. The shear force at that section can be calculated as Vx-x = 22. 4 - 10. x = 0 x = 2. 24 m N. Ram Kumar, CUFE

A C RA=22. 4 k. N 4 m 5 k. N 2 k. N 10 k. N/m D B 1 m 2 m RB=24. 6 k. N Calculations of Bending Moments: MA = M D = 0 MC = 22. 4 × 4 – 10 × 4 × 2 = 9. 6 k. Nm MB = 22. 4 × 5 – 10 × 4 × 3 – 2 × 1 = - 10 k. Nm (Considering Left portion of the section) Alternatively MB = -5 × 2 = -10 k. Nm (Considering Right portion of the section) Absolute Maximum Bending Moment is at X- X , Mmax = 22. 4 × 2. 24 – 10 × (2. 24)2 / 2 = 25. 1 k. Nm N. Ram Kumar, CUFE

X A x = 2. 24 m RA=22. 4 k. N X C 4 m D B 1 m 2 m RB=24. 6 k. N Mmax = 25. 1 k. Nm 9. 6 k. Nm BMD 5 k. N 2 k. N 10 k. N/m Point of contra flexure 10 k. Nm N. Ram Kumar, CUFE

A x = 2. 24 m RA=22. 4 k. N 5 k. N 2 k. N 10 k. N/m X C X 1 m 4 m D B 2 m RB=24. 6 k. N 22. 4 k. N 5 k. N x = 2. 24 m SFD 17. 6 k. N 19. 6 k. N Point of contra flexure 9. 6 k. Nm BMD N. Ram Kumar, CUFE 10 k. Nm

X A x RA=22. 4 k. N X 4 m 5 k. N 2 k. N 10 k. N/m C D B 1 m 2 m RB=24. 6 k. N Calculations of Absolute Maximum Bending Moment: Max. bending moment will occur at the section where the shear force is zero. The SFD shows that the section having zero shear force is available in the portion AC. Let that section be X-X, considered at a distance x from support A as shown above. The shear force at that section can be calculated as Vx-x = 22. 4 - 10. x = 0 x = 2. 24 m Max. BM at X- X , Mmax = 22. 4 × 2. 24 – 10 × (2. 24)2 / 2 = 25. 1 k. Nm N. Ram Kumar, CUFE

X A x = 2. 24 m RA=22. 4 k. N X C 4 m D B 1 m Mmax = 25. 1 k. Nm 9. 6 k. Nm BMD 5 k. N 2 k. N 10 k. N/m 2 m RB=24. 6 k. N Point of contra flexure 10 k. Nm N. Ram Kumar, CUFE

Let a be the distance of point of contra flexure from support B Taking moments at the section A-A (Considering left portion) A a = 0. 51 m Mmax = 25. 1 k. Nm 9. 6 k. Nm BMD N. Ram Kumar, CUFE A a Point of contra flexure 10 k. Nm

Example Problem 4 4. Draw SFD and BMD for the single side overhanging beam subjected to loading as shown below. Mark salient points on SFD and BMD. 60 k. N/m 20 k. N/m A B C 3 m 2 m N. Ram Kumar, CUFE 20 k. N D 2 m

60 k. N/m 20 k. N/m A B C RA 3 m 2 m RB 20 k. N D 2 m Solution: Calculation of reactions: ΣMA = 0 -RB × 5 + ½ × 3 × 60 × (2/3) × 3 +20 × 4 × 5 + 20 × 7 = 0 RB =144 k. N ΣFy = 0 RA + 144 – ½ × 3 × 60 – 20 × 4 -20 = 0 N. Ram Kumar, CUFE RA = 46 k. N.

60 k. N/m 20 k. N 0 1 2 3 4 5 6 0 1 2 3 4 5 RB = 144 k. N 6 RA = 46 k. N 3 m 2 m Shear Force Calculations: V 0 -0 =0 ; V 1 -1 = + 46 k. N V 2 -2 = +46 – ½ × 3 × 60 = - 44 k. N V 3 -3 = - 44 – 20 × 2 = - 84 k. N 2 m V 4 -4 = - 84 + 144 = + 60 k. N V 5 -5 = +60 – 20 × 2 = + 20 k. N V 6 -6= 20 – 20 = 0 (Check) N. Ram Kumar, CUFE

Example Problem 4 60 k. N/m 20 k. N 1 2 3 4 5 6 1 2 3 4 5 RB = 144 k. N 6 RA = 46 k. N 3 m 2 m 60 k. N Parabola 46 k. N SFD 2 m 44 k. N N. Ram Kumar, CUFE 20 k. N 84 k. N

60 k. N/m X 20 k. N/m A C x B RB=144 k. N X RA =46 k. N 20 k. N 2 m 3 m D 2 m Max. bending moment will occur at the section where the shear force is zero. The SFD shows that the section having zero shear force is available in the portion AC. Let that section be X-X, considered at a distance ‘x’ from support A as shown above. The shear force expression at that section should be equated to zero. i. e. , Vx-x = 46 – ½. x. (60/3)x = 0 x = 2. 145 m N. Ram Kumar, CUFE

60 k. N/m 20 k. N/m A C 20 k. N B RB=144 k. N RA =46 k. N 3 m 2 m D 2 m Calculation of bending moments: MA = M D = 0 MC = 46 × 3 – ½ × 3 × 60 × (1/3 × 3) = 48 k. Nm[Considering LHS of section] MB = -20 × 2 – 20 × 2 × 1 = - 80 k. Nm [Considering RHS of section] Absolute Maximum Bending Moment, Mmax = 46 × 2. 145 – ½ × 2. 145 ×(2. 145 × 60/3) × (1/3 × 2. 145) = 65. 74 k. Nm N. Ram Kumar, CUFE

60 k. N/m 20 k. N/m A C 20 k. N B RB=144 k. N RA =46 k. N 3 m 2 m D 2 m 48 k. Nm 65. 74 k. Nm Cubic parabola Parabola BMD Point of Contra flexure N. Ram Kumar, CUFE Parabola 80 k. Nm

60 k. N Parabola 46 k. N 44 k. N SFD 20 k. N 84 k. N 65. 74 k. Nm Cubic parabola Parabola BMD Point of Contra flexure N. Ram Kumar, CUFE Parabola 80 k. Nm

60 k. N/m X 20 k. N/m A C x=2. 145 m B RB=144 k. N X RA =46 k. N 3 m 20 k. N 2 m D 2 m Calculations of Absolute Maximum Bending Moment: Max. bending moment will occur at the section where the shear force is zero. The SFD shows that the section having zero shear force is available in the portion AC. Let that section be X-X, considered at a distance ‘x’ from support A as shown above. The shear force expression at that section should be equated to zero. i. e. , Vx-x = 46 – ½. x. (60/3)x = 0 x = 2. 145 m BM at X- X , Mmax = 46 × 2. 145 – ½ × 2. 145 ×(2. 145 × 60/3) × (1/3 × 2. 145)=65. 74 k. Nm. N. Ram Kumar, CUFE

60 k. N/m 20 k. N/m A C B RB=144 k. N RA =46 k. N 3 m 2 m D 2 m 48 k. Nm 65. 74 k. Nm Cubic parabola 48 k. Nm Parabola a BMD Point of Contra flexure N. Ram Kumar, CUFE Parabola 80 k. Nm

Point of contra flexure: BMD shows that point of contra flexure is existing in the portion CB. Let ‘a’ be the distance in the portion CB from the support B at which the bending moment is zero. And that ‘a’ can be calculated as given below. ΣMx-x = 0 a = 1. 095 m N. Ram Kumar, CUFE

Example Problem 5 5. Draw SFD and BMD for the single side overhanging beam subjected to loading as shown below. Mark salient points on SFD and BMD. 0. 5 m 40 k. N 30 k. N/m 20 k. N/m 0. 7 m A 2 m D C B 1 m N. Ram Kumar, CUFE 1 m E 2 m

0. 5 m 40 k. N 30 k. N/m 20 k. N/m 0. 7 m A 2 m 20 k. N/m B 2 m 1 m 1 m 40 x 0. 5=20 k. Nm A D C B 40 k. N 2 m 30 k. N/m D C 1 m N. Ram Kumar, CUFE E 1 m E 2 m

40 k. N 20 k. N/m A 20 k. Nm 2 m D C B RA 30 k. N/m 1 m 1 m RD E 2 m Solution: Calculation of reactions: ΣMA = 0 -RD × 4 + 20 × 2 × 1 + 40 × 3 + 20 + ½ × 2 × 30 × (4+2/3) = 0 RD =80 k. N ΣFy = 0 RA + 80 – 20 × 2 - 40 - ½ × 2 × 30 = 0 RA = 30 k. N N. Ram Kumar, CUFE

0 1 RA =30 k. N 20 k. N/m 2 m 20 k. Nm 40 k. N 2 3 1 m 4 30 k. N/m 5 6 7 4 1 m RD =80 k. N 2 m Calculation of Shear Forces: V 0 -0 = 0 V 1 -1 = 30 k. N V 5 -5 = - 50 k. N V 2 -2 = 30 – 20 × 2 = - 10 k. N V 6 -6 = - 50 + 80 = + 30 k. N V 3 -3 = - 10 k. N V 7 -7 = +30 – ½ × 2 × 30 = 0(check) V 4 -4 = -10 – 40 = - 50 k. N N. Ram Kumar, CUFE

20 k. N/m 1 20 k. Nm 40 k. N 1 RA =30 k. N 2 m 2 3 4 30 k. N/m 5 6 7 4 RD =80 k. N 1 m 1 m 30 k. N 2 m 30 k. N x = 1. 5 m 10 k. N SFD 50 k. N N. Ram Kumar, CUFE 50 k. N Parabola

40 k. N 20 k. N/m 20 k. Nm X A x = 1. 5 m RA X 2 m 30 k. N/m D C B 1 m 1 m RD E 2 m Calculation of bending moments: MA = M E = 0 MX = 30 × 1. 5 – 20 × 1. 5/2 = 22. 5 k. Nm MB= 30 × 2 – 20 × 2 × 1 = 20 k. Nm MC = 30 × 3 – 20 × 2 = 10 k. Nm (section before the couple) MC = 10 + 20 = 30 k. Nm (section after the couple) MD = - ½ × 30 × 2 × (1/3 × 2) = - 20 k. Nm( Considering RHS of the section) N. Ram Kumar, CUFE

40 k. N 20 k. N/m 20 k. Nm X A x = 1. 5 m RA Parabola X 2 m 22. 5 k. Nm 30 k. N/m D C B 1 m 20 k. Nm 1 m RD E 2 m 30 k. Nm Point of contra flexure Cubic parabola BMD N. Ram Kumar, CUFE 20 k. Nm

30 k. N Parabola 30 k. N x = 1. 5 m 10 k. N SFD 50 k. N Parabola 20 k. Nm 10 k. Nm 50 k. N Point of contra flexure Cubic parabola BMD N. Ram Kumar, CUFE 20 k. Nm

6. Draw SFD and BMD for the cantilever beam subjected to loading as shown below. 0. 5 m 40 k. N 300 20 k. N/m 0. 7 m A 3 m 1 m N. Ram Kumar, CUFE 1 m

40 k. N 0. 5 m 300 20 k. N/m 0. 7 m A 3 m 1 m 1 m 40 Sin 30 = 20 k. N 0. 5 m 20 k. N/m 0. 7 m 40 Cos 30 =34. 64 k. N A 3 m N. Ram Kumar, CUFE 1 m 1 m

40 Sin 30 = 20 k. N 0. 5 m 20 k. N/m 0. 7 m 3 m 1 m 40 Cos 30 =34. 64 k. N 1 m 20 x 0. 5 – 34. 64 x 0. 7=-14. 25 k. Nm 20 k. N/m 34. 64 k. N 3 m 1 m N. Ram Kumar, CUFE 1 m

20 k. N/m 14. 25 k. Nm HD 34. 64 k. N A 3 m B 1 m C 1 m D VD MD Calculation of Reactions (Here it is optional): ΣFx = 0 HD = 34. 64 k. N ΣFy = 0 VD = 20 × 3 + 20 = 80 k. N ΣMD = 0 MD - 20 × 3. 5 – 20 × 1 – 14. 25 = 244. 25 k. Nm N. Ram Kumar, CUFE

1 20 k. N/m 2 3 14. 25 k. Nm 5 4 6 HD 34. 64 k. N 1 2 3 m 1 m 3 4 1 m 5 6 MD VD=80 k. N Shear Force Calculation: V 1 -1 =0 V 2 -2 = -20 × 3 = - 60 k. N V 3 -3 = - 60 k. N V 4 -4 = - 60 – 20 = - 80 k. N V 5 -5 = - 80 k. N V 6 -6 = - 80 + 80 = 0 (Check) N. Ram Kumar, CUFE

1 20 k. N/m 20 k. N 2 3 14. 25 k. Nm 5 4 6 HD 34. 64 k. N 1 3 m 2 1 m 3 4 1 m 5 6 MD VD=80 k. N SFD 60 k. N 80 k. N N. Ram Kumar, CUFE 80 k. N

20 k. N 14. 25 k. Nm 20 k. N/m A 34. 64 k. N 3 m B 1 m C 1 m D MD Bending Moment Calculations: MA = 0 MB = - 20 × 3 × 1. 5 = - 90 k. Nm MC = - 20 × 3 × 2. 5 = - 150 k. Nm (section before the couple) MC = - 20 × 3 × 2. 5 – 14. 25 = -164. 25 k. Nm (section after the couple) MD = - 20 × 3. 5 -14. 25 – 20 × 1 = -244. 25 k. Nm (section before MD) moment) MD = -244. 25 +244. 25 = 0 (section after MD) N. Ram Kumar, CUFE

20 k. N 14. 25 k. Nm 20 k. N/m A 34. 64 k. N 3 m B 1 m C 1 m D 90 k. Nm 150 k. Nm BMD 164. 25 k. Nm N. Ram Kumar, CUFE 244. 25 k. Nm

W L/2 wk. N/m L W wk. N/m N. Ram Kumar, CUFE

Exercise Problems 1. Draw SFD and BMD for a single side overhanging beam subjected to loading as shown below. Mark absolute maximum bending moment on bending moment diagram and locate point of contra flexure. 15 k. N/m 10 k. N 20 k. N/m 5 k. Nm 1 m 1 m 3 m 1 m 1 m 2 m [Ans: Absolute maximum BM = 60. 625 k. Nm ] N. Ram Kumar, CUFE

Exercise Problems 2. Draw shear force and bending moment diagrams [SFD and BMD] for a simply supported beam subjected to loading as shown in the Fig. given below. Also locate and determine absolute maximum bending moment. 10 k. N 16 k. N 4 k. N/m 600 B A 1 m 1 m 2 m 1 m 1 m [Ans: Absolute maximum bending moment = 22. 034 k. Nm Its position is 3. 15 m from Left hand support ] N. Ram Kumar, CUFE

Exercise Problems 3. Draw shear force and bending moment diagrams [SFD and BMD] for a single side overhanging beam subjected to loading as shown in the Fig. given below. Locate points of contra flexure if any. 25 k. N/m 50 k. N 10 k. N/m 10 k. Nm A B 3 m 1 m 1 m 2 m [Ans : Position of point of contra flexure from RHS = 0. 375 m] N. Ram Kumar, CUFE

Exercise Problems 4. Draw SFD and BMD for a double side overhanging beam subjected to loading as shown in the Fig. given below. Locate the point in the AB portion where the bending moment is zero. 16 k. N 8 k. N 4 k. N/m A 2 m 2 m 2 m [Ans : Bending moment is zero at mid span] N. Ram Kumar, CUFE B 2 m

Exercise Problems 5. A single side overhanging beam is subjected to uniformly distributed load of 4 k. N/m over AB portion of the beam in addition to its self weight 2 k. N/m acting as shown in the Fig. given below. Draw SFD and BMD for the beam. Locate the inflection points if any. Also locate and determine maximum negative and positive bending moments. 4 k. N/m A 2 k. N/m B 6 m 2 m [Ans : Max. positive bending moment is located at 2. 89 m from LHS. and whose value is 37. 57 k. Nm ] N. Ram Kumar, CUFE

Exercise Problems 6. Three point loads and one uniformly distributed load are acting on a cantilever beam as shown in the Fig. given below. Draw SFD and BMD for the beam. Locate and determine maximum shear force and bending moments. 10 k. N 5 k. N 2 k. N/m 20 k. N A 1 m 1 m 1 m B [Ans : Both Shear force and Bending moments are maximum at supports. ] N. Ram Kumar, CUFE

Exercise Problems 7. One side overhanging beam is subjected loading as shown below. Draw shear force and bending moment diagrams [SFD and BMD] for beam. Also determine maximum hogging bending moment. 200 N 100 N 30 N/m A 3 m B 4 m 4 m [Ans: Max. Hogging bending moment = 735 k. Nm] N. Ram Kumar, CUFE

Exercise Problems 8. A cantilever beam of span 6 m is subjected to three point loads at 1/3 rd points as shown in the Fig. given below. Draw SFD and BMD for the beam. Locate and determine maximum shear force and hogging bending moment. 10 k. N 5 k. N 0. 5 m 8 k. N 5 k. N 300 A 2 m 2 m 2 m B [Ans : Max. Shear force = 20. 5 k. N, Max BM= 71 k. Nm Both max. shear force and bending moments will occur at supports. ] N. Ram Kumar, CUFE

Exercise Problems 9. A trapezoidal load is acting in the middle portion AB of the double side overhanging beam as shown in the Fig. given below. A couple of magnitude 10 k. Nm and a concentrated load of 14 k. N acting on the tips of overhanging sides of the beam as shown. Draw SFD and BMD. Mark salient features like maximum positive, negative bending moments and shear forces, inflection points if any. 14 k. N 40 k. N/m 20 k. N/m 10 k. Nm 600 A 1 m B 4 m 2 m [Ans : Maximum positive bending moment = 49. 06 k. Nm N. Ram Kumar, CUFE

Exercise Problems 10. Draw SFD and BMD for the single side overhanging beam subjected loading as shown below. . Mark salient features like maximum positive, negative bending moments and shear forces, inflection points if any. 24 k. N 0. 5 m 1 m 1 m 4 k. N/m 6 k. N/m 3 m 2 m 3 m Ans: Maximum positive bending moment = 41. 0 k. Nm N. Ram Kumar, CUFE

ACKNOWLEDGEMENT My Acknowledgement to innumerable websites in the internet, and to all those who have uploaded their knowledge, imaginations, ideas, graphic skills etc. , on these websites. Also, to all those(including my parents, teachers, friends, and relatives), from prehistoric days to to-day, who have registered their knowledge, imaginations, thoughts etc. , through different means and mediums. N. Ram Kumar, CUFE

THANK YOU N. Ram Kumar, CUFE