Shear Force and Bending Moment Diagrams SFD BMD

Shear Force and Bending Moment Diagrams [SFD & BMD] Mr. A. M. Kamble Visit for more Learning Resources

Shear Force and Bending Moments Consider a section x-x at a distance 6 m from left hand support A 6 m A 4 m RA = 8. 2 k. N 5 k. N x C x 5 m 10 k. N D 8 k. N B E 5 m 1 m RB=14. 8 k. N Imagine the beam is cut into two pieces at section x-x and is separated, as shown in figure

4 m A 8. 2 k. N 5 m 5 k. N 10 k. N 6 m 1 m 8 k. N B 9 m 14. 8 k. N To find the forces experienced by the section, consider any one portion of the beam. Taking left hand portion Transverse force experienced = 8. 2 – 5 = 3. 2 k. N (upward) Moment experienced = 8. 2 × 6 – 5 × 2 = 39. 2 k. N-m (clockwise) If we consider the right hand portion, we get Transverse force experienced = 14. 8 – 10 – 8 =-3. 2 k. N = 3. 2 k. N (downward) Moment experienced = - 14. 8 × 9 +8 × 8 + 10 × 3 = -39. 2 k. N-m = 39. 2 k. N-m (anticlockwise)

5 k. N A 8. 2 k. N 39. 2 k. N-m 10 k. N 39. 2 k. N-m 3. 2 k. N 8 k. N B 14. 8 k. N Thus the section x-x considered is subjected to forces 3. 2 k. N and moment 39. 2 k. N-m as shown in figure. The force is trying to shear off the section and hence is called shear force. The moment bends the section and hence, called bending moment.

Shear force at a section: The algebraic sum of the vertical forces acting on the beam either to the left or right of the section is known as the shear force at a section. Bending moment (BM) at section: The algebraic sum of the moments of all forces acting on the beam either to the left or right of the section is known as the bending moment at a section 3. 2 k. N 39. 2 k. N 3. 2 k. N F F Shear force at x-x M Bending moment at x-x

Moment and Bending moment Moment: It is the product of force and perpendicular distance between line of action of the force and the point about which moment is required to be calculated. Bending Moment (BM): The moment which causes the bending effect on the beam is called Bending Moment. It is generally denoted by ‘M’ or ‘BM’.

Sign Convention for shear force F F + ve shear force - ve shear force

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 ]

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 ]

Shear Force and Bending Moment Diagrams (SFD & BMD) Shear Force Diagram (SFD): The diagram which shows the variation of shear force along the length of the beam is called Shear Force Diagram (SFD). Bending Moment Diagram (BMD): The diagram which shows the variation of bending moment along the length of the beam is called Bending Moment Diagram (BMD).

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.

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.

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

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

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 Three-degree (Parabola) curve (Cubicparabola)

Sections for Shear Force and Bending Moment Calculations: Shear force and bending moments are to be calculated at various sections of the beam to draw shear force and bending moment diagrams. These sections are generally considered on the beam where the magnitude of shear force and bending moments are changing abruptly. Therefore these sections for the calculation of shear forces include sections on either side of point load, uniformly distributed load or uniformly varying load where the magnitude of shear force changes abruptly. The sections for the calculation of bending moment include position of point loads, either side of uniformly distributed load, uniformly varying load and couple Note: While calculating the shear force and bending moment, only the portion of the udl which is on the left hand side of the section should be converted into point load. But while calculating the reaction we convert entire udl to point load

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 1 m

10 N 5 N 8 N B A C 2 m E D 2 m RA Solution: 3 m 1 m RB [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

Shear Force Calculation: 0 0 1 2 2 1 2 m RA = 9. 75 N 10 N 5 N 3 4 2 m 8 N 5 6 7 8 9 3 m 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)

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

10 N 5 N 8 N B A C 2 m 9. 75 N 4. 75 N SFD E D 1 m 3 m 4. 75 N 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]

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

VM-34 A 10 N 5 N B C D 2 m 9. 75 N 8 N 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 13. 25 N

10 N 5 N 8 N B A C D 2 m 9. 75 N 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 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. 10 k. N 5 k. N C A 2 m 5 k. N 2 k. N/m B D 3 m 3 m 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

5 k. N 10 k. N 0 1 2 2 m 3 4 5 2 k. N/m 3 m 3 m RA=16 k. N Shear Force Calculation: V 0 -0 = 0 5 k. N 6 7 8 9 9 2 m RB = 16 k. N 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

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 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

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 10 k. Nm

10 k. N 5 k. N C 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 10 k. Nm 5 k. N SFD 14 k. Nm 5 k. N + 11 k. N BMD 10 k. Nm 5 k. N

10 k. N 5 k. N x C Ax 2 m x 10 k. Nm 5 k. N 2 k. N/m 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

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 5 k. N 2 k. N 10 k. N/m D B 1 m 2 m

10 k. N/m A 5 k. N 2 k. N B RA Solution : 4 m 1 m RB 2 m 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

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 5 -5 = - 19. 6 + 24. 6 = 5 k. N V 2 -2 = 22. 4 – 10 × 4 = -17. 6 k. N V 6 -6 = 5 k. N V 3 -3 = - 17. 6 – 2 = - 19. 6 k. N V 7 -7 = 5 – 5 = 0 (Check) V 4 -4 = - 19. 6 k. N

A 5 k. N 2 k. N 10 k. N/m 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 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

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 = MD = 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

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

A x = 2. 24 m RA=22. 4 k. N X 4 m 5 k. N 2 k. N 10 k. N/m X C D B 1 m 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. Nm BMD 19. 6 k. N Point of contra flexure 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

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

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

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 RAA = 46 k. N RA 46 k. N 3 m 2 m 60 k. N Parabola SFD 2 m 44 k. N 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 3 m 20 k. N 2 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

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 = MD = 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

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 Parabola 80 k. Nm

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

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 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

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 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 40 k. N 1 m E 2 m 30 k. N/m D C B 2 m 1 m 1 m 40 x 0. 5=20 k. Nm A D C B 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 ΣFy = 0 RA + 80 – 20 × 2 - 40 - ½ × 2 × 30 = 0 RA = 30 k. N

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

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 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 = ME = 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

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 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 20 k. Nm

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

20 k. N/m 14. 25 k. Nm HD 34. 64 k. N A 3 m B 1 m Calculation of Reactions (Here it is optional): D MD VD 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

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 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)

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

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)

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 BMD 150 k. Nm 164. 25 k. Nm 244. 25 k. Nm

W L/2 wk. N/m L W wk. N/m

VM-73 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 ]

VM-74 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 ]

Exercise Problems VM-75 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]

VM-76 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] B 2 m

VM-77 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 ]

VM-78 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. ]

VM-79 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]

VM-80 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. ]

VM-81 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

VM-82 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

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