This document provides guidance on calculating shear force and bending moment diagrams (SFD and BMD) for beams under different loading conditions. It begins by explaining the process for a sample problem, which involves a beam with uniform and point loads. The key steps are to determine support reactions, divide the beam into sections, then calculate the SFD and BMD for each section. Linear variation indicates a straight line SFD, while parabolic variation means a curved BMD. Interpretations are provided for different loading types and the shapes of the resulting diagrams. References for further reading are listed at the end.

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
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Important Note

3. 
Draw SFD and BMD of the beam shown below. Indicate the
numerical values at all important point
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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)
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Step 1
RA RB
8 4
2.5 2.52.5 1.25 1.255
5 4

5. 
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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
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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. 
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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. 
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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. 
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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. 
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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)
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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
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References

17. 17
Thank You
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