Shear force and bending moment Solved Numerical

Step wise explanation to draw SFD and BMD

1. Shear Force and Bending Moment (Solved Numerical) Kirtan Adhikari Assistant Lecturer College of Science and Technology Royal University of Bhutan [email protected] COLLEGE OF SCIENCE AND TECHNOLOGY ROYAL UNIVERSITY OF BHUTAN RINCHENDING BHUTAN 1 02-Aug-15

2. To understand the fundamental concept on Bending Moment and Shear Force please read through text books (References). This ppt contains step wise procedure to draw SFD and BMD 02-Aug-15 2 Important Note

3. Draw SFD and BMD of the beam shown below. Indicate the numerical values at all important point 02-Aug-15 3 Question 1

4. For any type of questions start by computing support reactions. Draw Free Body Diagram of the beam Convert UDL, UVL to point load to compute support reaction. (only to compute support reaction) 02-Aug-15 4 Step 1 R A R B 8 4 2.5 2.5 2.5 1.25 1.25 5 5 4

5. 02-Aug-15 5 Step 2: Compute Support Reactions Σ M A = 0 5 * 2.5 + 8 * 5 + 4 * 7.5 - R b * 12.5 + 4 * 13.75 = R b = 11 kN Therefore, Ra = (5 + 8 + 4 + 4 -11) R a = 10 kN R A R B 8 4 5 4 Sign Convention: Clockwise Moment = Positive 2.5 5 7 .5 12.5 13.75

6. Mark the point where Point load/support reaction acts or points can be marked at the starting and ending of UDL/UVL In this case the beam is divided into 4 portions AC, CD, DB and BE. Each section has to be considered independently when calculating SF and BM 02-Aug-15 6 Step 3 Divide The Beam into portions

7. 7 Step 5 Draw a sectional line anywhere in portion AC x Shear force at a sectional point (at a distance x from A) F = R a – 1*x Where (0 ≤ x ≥ 5) F = f(x)……….. This function produces straight line graph (Linear Variation) Bending Moment at a sectional point (at a distance x from A ) BM = Ra*x – 1*x * Where (0 ≤ x ≥ 5 ) BM = f(x 2 )…………….This function Produces a Parabolic curve   x

8. 02-Aug-15 8 SFD & BMD for AC Shear Force = R a – x @ x = 0 10 kN @ x = 0.25 9.75 kN @ x = 0.5 9.5 kN @ x = 2.5 7.5 kN @ x = 5 5 kN Bending M = Ra*x – @ x = 0 kNm @ x = 0.25 2.47 kNm @ x = 0.5 4.88 kNm @ x = 2.5 21.88 kNm @ x = 5 37.5 kNm @ x = 0 kNm @ x = 0.25 2.47 kNm @ x = 0.5 4.88 kNm @ x = 2.5 21.88 kNm @ x = 5 37.5 kNm

9. 9 For Portion CD x x SF = Ra – 1*5 – 8 BM = Ra*x – 5*( x-2.5) – 8*(x-5) Where (5 ≤ x ≥ 7.5) BM = f(x)……Linear variation

10. 02-Aug-15 10 SFD & BMD for CD Shear Force = Ra – 5 – 8 @ x = 5 (C) -3 kN @ x = 7.5 (D) -3 kN Bending M = Ra*x – 5*( x-2.5) – 8*(x-5) @ x = 5 (C) 37.5 kNm @ x = 5.25 36.75 kNm @ x = 5.75 35.25 kNm @ x = 6 34.5 kNm @ x = 7.5 (D) 30 kNm

11. 11 For Portion DB x SF = Ra – 1*5 – 8 - 4 BM = Ra*x – 5*( x-2.5) – 8*(x-5) – 4(x-7.5) Where (7.5 ≤ x ≥ 12.5) BM = f(x)……Linear variation x

12. 02-Aug-15 12 SFD & BMD for DB Shear Force = Ra – 5 – 8 - 4 @ x = 7.5 (D) -7 kN @ x = 12.5 (B) -7 kN Bending M = Ra*x – 17x + 82.5 @ x = 7.5 (D) 30 kNm @ x = 8.5 23 kNm @ x = 9.5 16 kNm @ x = 11.5 2 kNm @ x = 12.5 (B) -5 kNm

13. 13 For Portion BE SF = Ra – 5 – 8 – 4 + Rb – 1.6*(x-12.5) BM = Ra*x – 5*( x-2.5) – 8*(x-5) – 4(x-7.5) + Rb * (x – 12.5) – 1.6 * (x-12.5) * BM = Ra*x + Rb * (x – 12.5) – 17x + 82.5 - 0.8 * (x – 12.5) 2 Where (12.5 ≤ x ≥ 1.5) BM = f(x 2 )……Parabolic variation   x (X- 5) m (X- 7. 5) m (X- 12. 5) m

14. 02-Aug-15 14 SFD & BMD for BE Shear Force = 24 - 1.6x @ x = 12.5 (B) 4 kN @ x = 14 1.6 kN @ x = 15 (E) kN BM = 4x – 55 - 0.8 * (x – 12.5) 2 @ x = 12.5 (B) -5 kNm @ x = 13 -3.2 kNm @ x = 13.5 -1.8 kNm @ x = 14 -0.8 kNm @ x = 15 (E) kNm

15. Types of Loading Shape of Shear Force Diagram Shape of Bending Moment Diagram Point Load Linear ly Varying Graph (Straight Line) Linear ly Varying Graph (Straight Line) Uniformly Distributed Load (UDL) Linear ly Varying Graph (Straight Line) Parabolic Graph (Smooth Curve) Uniformly Varying Load (UDL) Parabolic Graph (Smooth Curve) Cubically varying Graph (Curve) 02-Aug-15 15 Interpretations

16. R.S.Khurmi . Strength of Materials. New Delhi: S.Chand & Company Ltd. Timoshenko, S.P., and D.H. Young (1993). Elements of Strength of Materials.( 5 th Ed.).East West Press . Bhavikari , S. S., 2008. Strength of Materials. 3rd ed. Delhi: Vikas Publishing House Pvt Ltd. Ramamrutham , S. & Narayan, R., 2009. Setrength of Materials. Noida: Dhanpat Rai Publishing Company (P) Ltd A.R.Jain and B.K.Jain (1987). Theory and Analysis of Structures, Vol. Roorkee : Nemchand and Bros. B.C.Punmia (1994). Strength of Materials and Theory of Structures, Vol. 1. New Delhi: Laxmi publications. M. M. Ratwani & V.N.Vazirani (2008). Analysis of Structure, Vol.1. New Delhi: Khanna Publishers. R.K. Bansal (1994). A Text Book on Strength of Materials. New Delhi: Laxmi Publications. R.K. Rajput.(2007). Strength of Materials. New Delhi: S.Chand & Company Ltd 02-Aug-15 16 References

17. 17 Thank You 02-Aug-15