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
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
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