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

Christ University Faculty of Engineering 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 N.Ram Kumar, Assistant Professor, Department of Mechanical Engineering

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

N.Ram Kumar, CUFE

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

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

N.Ram Kumar, CUFE

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

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

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

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 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 An external moment. 3. Overhanging beam subjected to a Point or concentrated load at free end. N.Ram Kumar, CUFE

CANTILEVER BEAM 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 N.Ram Kumar, CUFE

CAR PARKING ROOF N.Ram Kumar, CUFE

BUS STOP SHELTER N.Ram Kumar, CUFE

TRAFFIC SIGNAL POST 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 N.Ram Kumar, CUFE

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

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

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

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

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

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

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

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

EXAMPLE N.Ram Kumar, CUFE

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

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

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

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

Relationship between load, shear force and bending moment L w kN/m x x1 dx 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

w kN/m M+dM M v x x1 dx Fig. FBD of Differential element of the beam V+dV M M+dM Fig. FBD of Differential element of the beam x x1 w kN/m O 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

w kN/m M+dM M v x x1 dx Fig. FBD of Differential element of the beam V+dV M M+dM Fig. FBD of Differential element of the beam x x1 w kN/m O 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

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 Type of load SFD/BMD Between point loads or for no load region Uniformly distributed load Uniformly varying load Shear Force Diagram Horizontal line Inclined line Two-degree curve (Parabola) Bending Moment Diagram Three-degree curve (Cubic-parabola) N.Ram Kumar, CUFE

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

Using the condition: ΣMA = 0 B 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

5N 10N 8N 2m 3m 1m Shear Force Calculation: 1 8 9 2 3 4 5 6 7 8 9 1 1 2 3 4 5 6 7 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

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

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

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

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

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

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

2. Draw SFD and BMD for the double side overhanging 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. 2m 3m 5kN 10kN 2kN/m A B C D E 5kN N.Ram Kumar, CUFE

Calculation of Reactions: 2m 3m 5kN 10kN 2kN/m A B C D E RA RB 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 kN N.Ram Kumar, CUFE

Shear Force Calculation: V0-0 = 0 V1-1 = - 5kN V6-6 = - 5 – 6 = - 11kN 2kN/m 4 5 2 3 6 7 1 8 9 8 9 2 3 4 5 7 1 6 2m 3m 3m 2m RA=16kN RB = 16kN Shear Force Calculation: V0-0 = 0 V1-1 = - 5kN V6-6 = - 5 – 6 = - 11kN V2-2 = - 5kN V7-7 = - 11 + 16 = 5kN V3-3 = - 5 + 16 = 11 kN V8-8 = 5 kN V4-4 = 11 – 2 × 3 = +5 kN V9-9 = 5 – 5 = 0 (Check) V5-5 = 5 – 10 = - 5kN N.Ram Kumar, CUFE

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

Bending Moment Calculation: 5kN 10kN 2kN/m A B C D E RA=16kN RB = 16kN Bending Moment Calculation: MC = ME = 0 [Because Bending moment at free end is zero] MA = MB = - 5 × 2 = - 10 kNm MD = - 5 × 5 + 16 × 3 – 2 × 3 × 1.5 = +14 kNm N.Ram Kumar, CUFE

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

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

2m 3m 5kN 10kN 2kN/m A B C D E 10kNm 10kNm x = 1 or 10 Points of contra flexure 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. 5kN 10kN/m 2 kN A D C B 4m 1m 2m N.Ram Kumar, CUFE

Solution : Calculation of Reactions: ΣMA = 0 5kN 10kN/m 2 kN A B RA 4m RB 1m 2m Solution : Calculation of Reactions: ΣMA = 0 - RB × 5 + 10 × 4 × 2 + 2 × 4 + 5 × 7 = 0  RB = 24.6 kN ΣFy = 0 RA + 24.6 – 10 x 4 – 2 + 5 = 0  RA = 22.4 kN N.Ram Kumar, CUFE

Shear Force Calculations: 5kN 10kN/m 2 kN 5 2 4 1 3 6 7 4 5 1 2 3 6 7 RA=22.4kN 4m 1m 2m RB=24.6kN Shear Force Calculations: V0-0 =0; V1-1 = 22.4 kN V5-5 = - 19.6 + 24.6 = 5 kN V2-2 = 22.4 – 10 × 4 = -17.6kN V6-6 = 5 kN V3-3 = - 17.6 – 2 = - 19.6 kN V7-7 = 5 – 5 = 0 (Check) V4-4 = - 19.6 kN N.Ram Kumar, CUFE

5kN 10kN/m 2 kN 4m 1m 2m SFD A C B D RA=22.4kN RB=24.6kN 22.4kN 5 kN x = 2.24m 17.6kN 19.6kN 19.6kN SFD N.Ram Kumar, CUFE

4m 1m 2m 2 kN 5kN 10kN/m A B C D RA=22.4kN RB=24.6kN X x 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

MB = -5 × 2 = -10 kNm (Considering Right portion of the section) A B C D RA=22.4kN RB=24.6kN Calculations of Bending Moments: MA = MD = 0 MC = 22.4 × 4 – 10 × 4 × 2 = 9.6 kNm MB = 22.4 × 5 – 10 × 4 × 3 – 2 × 1 = - 10kNm (Considering Left portion of the section) Alternatively MB = -5 × 2 = -10 kNm (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 kNm N.Ram Kumar, CUFE

5kN 10kN/m 2 kN A D C B 4m 1m 2m BMD X x = 2.24m X RA=22.4kN RB=24.6kN Mmax = 25.1 kNm 9.6kNm Point of contra flexure BMD 10kNm N.Ram Kumar, CUFE

4m 1m 2m 2 kN 5kN 10kN/m A B C D SFD BMD X x = 2.24m RA=22.4kN RB=24.6kN X x = 2.24m 22.4kN 19.6kN 17.6kN 5 kN SFD x = 2.24m 9.6kNm 10kNm BMD Point of contra flexure N.Ram Kumar, CUFE

4m 1m 2m 2 kN 5kN 10kN/m A B C D X x RA=22.4kN RB=24.6kN 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 kNm N.Ram Kumar, CUFE

5kN 10kN/m 2 kN A D C B 4m 1m 2m BMD X x = 2.24m X RA=22.4kN RB=24.6kN Mmax = 25.1 kNm 9.6kNm Point of contra flexure BMD 10kNm N.Ram Kumar, CUFE

BMD 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 kNm 9.6kNm Point of contra flexure BMD 10kNm a N.Ram Kumar, CUFE A

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

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

60kN/m 20kN 20kN/m 3m 2m 2m Shear Force Calculations: 1 4 2 3 5 6 3 4 1 2 5 6 RB = 144kN RA RA = 46kN 3m 2m 2m Shear Force Calculations: V0-0 =0 ; V1-1 = + 46 kN V4-4 = - 84 + 144 = + 60kN V2-2 = +46 – ½ × 3 × 60 = - 44 kN V5-5 = +60 – 20 × 2 = + 20 kN V3-3 = - 44 – 20 × 2 = - 84 kN V6-6= 20 – 20 = 0 (Check) N.Ram Kumar, CUFE

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

60kN/m 3m 20kN/m 20kN 2m A B RA =46kN C D RB=144kN X x 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

60kN/m 20kN 20kN/m A C B D 3m 2m Calculation of bending moments: RA =46kN C D RB=144kN Calculation of bending moments: MA = MD = 0 MC = 46 × 3 – ½ × 3 × 60 × (1/3 × 3) = 48 kNm[Considering LHS of section] MB = -20 × 2 – 20 × 2 × 1 = - 80 kNm [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 kNm N.Ram Kumar, CUFE

60kN/m 20kN 20kN/m A C B D 3m 2m BMD RB=144kN RA =46kN 48kNm 65.74kNm Parabola Cubic parabola Point of Contra flexure BMD Parabola N.Ram Kumar, CUFE 80kNm

60kN 46kN 20kN 44kN SFD 84kN BMD Parabola 65.74kNm Parabola Cubic Point of Contra flexure BMD Parabola N.Ram Kumar, CUFE 80kNm

60kN/m 3m 20kN/m 20kN 2m A B RA =46kN C D RB=144kN X x=2.145m 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 kNm. N.Ram Kumar, CUFE

60kN/m 20kN 20kN/m A C B D 3m 2m BMD RB=144kN RA =46kN 48kNm 65.74kNm Parabola Cubic parabola a Point of Contra flexure BMD Parabola N.Ram Kumar, CUFE 80kNm

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

5. Draw SFD and BMD for the single side overhanging beam 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. 40kN 0.5m 30kN/m 20kN/m 0.7m A B C D E 2m 1m 1m 2m N.Ram Kumar, CUFE

20kN/m 30kN/m 40kN 2m A D 1m B C E 20kN/m 30kN/m 40kN 2m A D 1m B C E 40x0.5=20kNm 20kN/m 30kN/m 40kN 2m A D 1m B C E N.Ram Kumar, CUFE

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

20kNm 40kN 30kN/m 20kN/m 1 2 3 4 6 5 7 7 1 2 5 3 4 6 RD =80kN RA =30kN 2m 1m 1m 2m Calculation of Shear Forces: V0-0 = 0 V1-1 = 30 kN V5-5 = - 50 kN V2-2 = 30 – 20 × 2 = - 10kN V6-6 = - 50 + 80 = + 30kN V3-3 = - 10kN V7-7 = +30 – ½ × 2 × 30 = 0(check) V4-4 = -10 – 40 = - 50 kN N.Ram Kumar, CUFE

20kNm 40kN 30kN/m 20kN/m 2m 1m SFD 1 2 7 5 4 6 3 RD =80kN RA =30kN Parabola SFD x = 1.5 m N.Ram Kumar, CUFE

40kN 30kN/m 20kN/m 20kNm A B C D E 2m 1m 1m 2m X x = 1.5 m X RA RD Calculation of bending moments: MA = ME = 0 MX = 30 × 1.5 – 20 × 1.5 × 1.5/2 = 22.5 kNm MB= 30 × 2 – 20 × 2 × 1 = 20 kNm MC = 30 × 3 – 20 × 2 × 2 = 10 kNm (section before the couple) MC = 10 + 20 = 30 kNm (section after the couple) MD = - ½ × 30 × 2 × (1/3 × 2) = - 20 kNm( Considering RHS of the section) N.Ram Kumar, CUFE

40kN 30kN/m 20kN/m 20kNm A B C D E 2m 1m 1m 2m BMD X x = 1.5 m X RA RD Parabola 20kNm 10kNm Point of contra flexure BMD Cubic parabola 20kNm N.Ram Kumar, CUFE

SFD BMD 30kN 10kN 50kN Parabola x = 1.5 m Parabola 20kNm 10kNm Point of contra flexure BMD Cubic parabola 20kNm N.Ram Kumar, CUFE

6. Draw SFD and BMD for the cantilever beam subjected to loading as shown below. 40kN 0.5m 300 20kN/m 0.7m A 3m 1m 1m N.Ram Kumar, CUFE

40kN 20kN/m A 3m 1m 1m 20kN/m A 3m 1m 1m 0.5m 300 0.7m 40Sin30 = 20kN 40Cos30 =34.64kN 0.7m A 3m 1m 1m N.Ram Kumar, CUFE

20kN/m 3m 1m 20kN/m 3m 1m 40Sin30 = 20kN 0.5m 40Cos30 =34.64kN 0.7m 20x0.5 – 34.64x0.7=-14.25kNm 20kN/m 3m 1m 20kN 34.64kN N.Ram Kumar, CUFE

Calculation of Reactions (Here it is optional): 20kN 20kN/m 14.25kNm HD 34.64kN A B C 3m 1m D 1m MD VD Calculation of Reactions (Here it is optional): ΣFx = 0  HD = 34.64 kN ΣFy = 0  VD = 20 × 3 + 20 = 80 kN ΣMD = 0  MD - 20 × 3 × 3.5 – 20 × 1 – 14.25 = 244.25kNm N.Ram Kumar, CUFE

20kN/m HD 3m 1m 1m MD Shear Force Calculation: V1-1 =0 6 2 3 4 5 HD 34.64kN 3m 2 1m 3 4 1m 1 5 MD 6 VD=80kN Shear Force Calculation: V1-1 =0 V2-2 = -20 × 3 = - 60kN V3-3 = - 60 kN V4-4 = - 60 – 20 = - 80 kN V5-5 = - 80 kN V6-6 = - 80 + 80 = 0 (Check) N.Ram Kumar, CUFE

20kN/m HD 3m 1m 1m SFD 20kN 1 14.25kNm 6 2 3 4 5 34.64kN 2 3 4 1 5 MD VD=80kN 60kN 60kN SFD 80kN 80kN N.Ram Kumar, CUFE

20kN/m 3m 1m A B C D Bending Moment Calculations: MA = 0 MD Bending Moment Calculations: MA = 0 MB = - 20 × 3 × 1.5 = - 90 kNm MC = - 20 × 3 × 2.5 = - 150 kNm (section before the couple) MC = - 20 × 3 × 2.5 – 14.25 = -164.25 kNm (section after the couple) MD = - 20 × 3 × 3.5 -14.25 – 20 × 1 = -244.25 kNm (section before MD) moment) MD = -244.25 +244.25 = 0 (section after MD) N.Ram Kumar, CUFE

20kN/m A B C D 3m 1m 1m BMD 20kN 14.25kNm 34.64kN 90kNm 150kNm N.Ram Kumar, CUFE

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

Exercise Problems 20kN/m 5kNm 15kN/m 10kN 3m 1m 2m 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. 20kN/m 5kNm 15kN/m 10kN 3m 1m 2m [Ans: Absolute maximum BM = 60.625 kNm ] 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. 10kN 16kN 4kN/m 600 B A 1m 2m 1m 1m 1m [Ans: Absolute maximum bending moment = 22.034kNm Its position is 3.15m 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. 25kN/m 50kN 10kN/m 10kNm A B 1m 3m 1m 2m [Ans : Position of point of contra flexure from RHS = 0.375m] 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. 16kN 8kN 8kN 4kN/m A B 2m 2m 2m 2m [Ans : Bending moment is zero at mid span] N.Ram Kumar, CUFE

Exercise Problems 5. A single side overhanging beam is subjected to uniformly distributed load of 4 kN/m over AB portion of the beam in addition to its self weight 2 kN/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. 4kN/m 2kN/m A B 6m 2m [Ans :Max. positive bending moment is located at 2.89 m from LHS. and whose value is 37.57 kNm ] 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. 10kN 5kN 20kN 2kN/m A 1m 1m 1m 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. 200N 100N 30N/m A B 4m 4m 3m [Ans: Max. Hogging bending moment = 735 kNm] N.Ram Kumar, CUFE

Exercise Problems 8. A cantilever beam of span 6m is subjected to three point loads at 1/3rd 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. 10kN 5kN 8kN 5kN 0.5m 300 A 2m 2m 2m B [Ans : Max. Shear force = 20.5kN, Max BM= 71kNm 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 kNm and a concentrated load of 14 kN 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. 14kN 40kN/m B 2m 10kNm 1m A 4m 20kN/m 600 [Ans : Maximum positive bending moment = 49.06 kNm 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. 24kN 4kN/m 0.5m 6kN/m 1m 1m 3m 2m 3m Ans: Maximum positive bending moment = 41.0 kNm 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 pre-historic 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