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**Consider a section x-x at a distance 6m from left hand support A**

Shear Force and Bending Moments Consider a section x-x at a distance 6m from left hand support A x 6 m 10kN 5kN 8kN B A C D E 4m 5m 5m 1m RA = 8.2 kN RB=14.8kN Imagine the beam is cut into two pieces at section x-x and is separated, as shown in figure

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5kN A 8.2 kN 10kN 8kN B 14.8 kN 4 m 6 m 9 m 1 m 5 m 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 kN (upward) Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise) If we consider the right hand portion, we get Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward) Moment experienced = × 9 +8 × × 3 = kN-m = 39.2 kN-m (anticlockwise)

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5kN A 8.2 kN 10kN 8kN B 14.8 kN 3.2 kN 39.2 kN-m Thus the section x-x considered is subjected to forces 3.2 kN and moment 39.2 kN-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.

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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 39.2 kN 3.2 kN M F Shear force at x-x Bending moment at x-x

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**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’.

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**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 ]

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**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 ]

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**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).

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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.

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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

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5N 10N 8N 2m 3m 1m A C D B E RA RB Solution: Using the condition: ΣMA = 0 - RB × × × × 2 = RB = N Using the condition: ΣFy = 0 RA = RA = 9.75 N [Clockwise moment is Positive]

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**Shear Force Calculation:**

2m 3m 1m 1 9 2 3 8 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 = N V6-6 = N V2-2 = N V7-7 = 5.25 – 8 = N V3-3 = – 5 = 4.75 N V8-8 = V4-4 = N V9-9 = = 0 V5-5 = – 10 = N (Check)

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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 13.25N 13.25N

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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 13.25N 13.25N

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**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 = Nm MB = 9.75 × 8 – 5 × 6 – 10 × 4 – 8 × 1 = 0 or MB = 0 [ since it is simply supported]

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10N 5N 8N A B C D E 2m 2m 1m 3m 29Nm 19.5Nm 13.25Nm BMD

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**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

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**E 5N 10N 8N 2m 3m 1m A C D B 9.75N 4.75N 5.25N SFD 13.25N 29Nm BMD**

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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. 10kN 16kN 4kN/m 600 B A 1m 2m 1m 1m 1m [Ans: Absolute maximum bending moment = kNm Its position is 3.15m from Left hand support ]

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CHAPTER OBJECTIVES Use various methods to determine the deflection and slope at specific pts on beams and shafts: Integration method Discontinuity functions.

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