Shear force and bending moment Solved Numerical
- 1. Shear Force and Bending Moment (Solved Numerical) Kirtan Adhikari Assistant Lecturer College of Science and Technology Royal University of Bhutan adhikari.cst@rub.edu.bt COLLEGE OF SCIENCE AND TECHNOLOGY ROYAL UNIVERSITY OF BHUTAN RINCHENDING BHUTAN 18/3/2015
- 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 8/3/2015 2 Important Note
- 3. Draw SFD and BMD of the beam shown below. Indicate the numerical values at all important point 8/3/2015 3 Question 1
- 4. a) For any type of questions start by computing support reactions. b) Draw Free Body Diagram of the beam c) Convert UDL, UVL to point load to compute support reaction. (only to compute support reaction) 8/3/2015 4 Step 1 RA RB 8 4 2.5 2.52.5 1.25 1.255 5 4
- 5. 8/3/2015 5 Step 2: Compute Support Reactions ΣMA = 0 5 * 2.5 + 8 * 5 + 4 * 7.5 - Rb * 12.5 + 4 * 13.75 = 0 Rb = 11 kN Therefore, Ra = (5 + 8 + 4 + 4 -11) Ra = 10 kN RA RB 8 45 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 8/3/2015 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 = Ra – 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 * 𝑥 2 Where (0 ≤ x ≥ 5) BM = f(x2)…………….This function Produces a Parabolic curve x
- 8. 8/3/2015 8 SFD & BMD for AC Shear Force = Ra – 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 – 𝑥2 2 @ x = 0 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. 8/3/2015 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. 8/3/2015 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) * (x−12.5) 2 BM = Ra*x + Rb * (x – 12.5) – 17x + 82.5 - 0.8 * (x – 12.5)2 Where (12.5 ≤ x ≥ 1.5) BM = f(x2)……Parabolic variation x(X- 5) m(X- 7. 5) m(X- 12. 5) m
- 14. 8/3/2015 14 SFD & BMD for BE Shear Force = 24 - 1.6x @ x = 12.5 (B) 4 kN @ x = 14 1.6 kN @ x = 15 (E) 0 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) 0 kNm
- 15. Types of Loading Shape of Shear Force Diagram Shape of Bending Moment Diagram Point Load Linearly Varying Graph (Straight Line) Linearly Varying Graph (Straight Line) Uniformly Distributed Load (UDL) Linearly Varying Graph (Straight Line) Parabolic Graph (Smooth Curve) Uniformly Varying Load (UDL) Parabolic Graph (Smooth Curve) Cubically varying Graph (Curve) 8/3/2015 15 Interpretations
- 16. 1. R.S.Khurmi. Strength of Materials. New Delhi: S.Chand & Company Ltd. 2. Timoshenko, S.P., and D.H. Young (1993). Elements of Strength of Materials.(5th Ed.).East West Press. 3. Bhavikari, S. S., 2008. Strength of Materials. 3rd ed. Delhi: Vikas Publishing House Pvt Ltd. 4. Ramamrutham, S. & Narayan, R., 2009. Setrength of Materials. Noida: Dhanpat Rai Publishing Company (P) Ltd 5. A.R.Jain and B.K.Jain (1987). Theory and Analysis of Structures, Vol. Roorkee: Nemchand and Bros. 6. B.C.Punmia (1994). Strength of Materials and Theory of Structures, Vol. 1. New Delhi: Laxmi publications. 7. M. M. Ratwani & V.N.Vazirani (2008). Analysis of Structure, Vol.1. New Delhi: Khanna Publishers. 8. R.K. Bansal (1994). A Text Book on Strength of Materials. New Delhi: Laxmi Publications. 9. R.K. Rajput.(2007). Strength of Materials. New Delhi: S.Chand & Company Ltd 8/3/2015 16 References