Presentation on theme: "Consider a section x-x at a distance 6m from left hand support A"— Presentation transcript:
1 Consider a section x-x at a distance 6m from left hand support A Shear Force and Bending MomentsConsider a section x-x at a distance 6m from left hand support Ax6 m10kN5kN8kNBACDE4m5m5m1mRA = 8.2 kNRB=14.8kNImagine the beam is cut into two pieces at section x-x and is separated, as shown in figure
2 5kNA8.2 kN10kN8kNB14.8 kN4 m6 m9 m1 m5 mTo find the forces experienced by the section, consider any one portion of the beam. Taking left hand portionTransverse 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 getTransverse 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)
3 5kNA8.2 kN10kN8kNB14.8 kN3.2 kN39.2 kN-mThus 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.
4 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 section39.2 kN3.2 kNMFShear force at x-xBending moment at x-x
5 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’.
6 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.ConvexityFig. Sagging bending moment [Positive bending moment ]
7 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.ConvexityFig. Hogging bending moment [Negative bending moment ]
8 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).
9 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.
10 Example Problem 1Draw 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.E5N10N8N2m3m1mACDB
11 5N10N8N2m3m1mACDBERARBSolution:Using the condition: ΣMA = 0- RB × × × × 2 = RB = NUsing the condition: ΣFy = 0RA = RA = 9.75 N[Clockwise moment is Positive]
12 Shear Force Calculation: 2m3m1m1923845678911234567RA = 9.75 NRB=13.25NShear Force at the section 1-1 is denoted as V1-1Shear Force at the section 2-2 is denoted as V2-2 and so on...V0-0 = 0; V1-1 = N V6-6 = NV2-2 = N V7-7 = 5.25 – 8 = NV3-3 = – 5 = 4.75 N V8-8 =V4-4 = N V9-9 = = 0V5-5 = – 10 = N (Check)
15 Bending Moment Calculation Bending moment at A is denoted as MABending moment at B is denoted as MBand so on… MA = 0 [ since it is simply supported]MC = 9.75 × 2= 19.5 NmMD = 9.75 × 4 – 5 × 2 = 29 NmME = 9.75 × 7 – 5 × 5 – 10 × 3 = NmMB = 9.75 × 8 – 5 × 6 – 10 × 4 – 8 × 1 = 0or MB = 0 [ since it is simply supported]
17 E 5N 10N 8N 2m 3m 1m A C D B VM-34 9.75N Example Problem 1 4.75N 5.25N BMD19.5Nm29Nm13.25Nm9.75N4.75N5.25N13.25NSFDExample Problem 1
18 E 5N 10N 8N 2m 3m 1m A C D B 9.75N 4.75N 5.25N SFD 13.25N 29Nm BMD
19 VM-74Exercise Problems2. 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.10kN16kN4kN/m600BA1m2m1m1m1m[Ans: Absolute maximum bending moment = kNmIts position is 3.15m from Left hand support ]