DEPARTMENT OF CIVILENGINEERING

SURVEYING
Surveying is the art of determining
the relative position of point on,
above or beneath the surface of the
earth by mean of direct or indirect
measurements of distance direction
and elevation. It also includes the art
of establishing point by
predetermined angular and linear
measurements
Principal of surveying
1. Working from Whole to Part
According to this principal the survey work must be carried out from
whole to part. This means that when an area is to be surveyed, first
of all, a system of control points is established such that it covers the
entire area with a higher degree of precision. After this, the minor
control points and details are further established with a lesser degree
of precision.
The main idea of this principle is to prevent the undue accumulation
of errors and thereby control and localize the minor errors.
If the survey is carried out from part to whole, the magnitudes of
errors accumulated would be very high.
Example: This principle is essential in International Boundary
Locating, Property Boundary Locating, etc.
Location of Point by Measurement From Two Points of Reference
On the basis of this principle, the relative position of the desired points to be
surveyed must be located by taking the measurement from at least two (preferably
three) points of reference, such that the position of the reference points has already
been fixed previously.
 Methods
There are 3 methods of making linear
measurements.
1- Direct Method.
2- Optical Method.
3- E.D.M Method.
Approximate Methods

Pacing.
Passometer.
Pedometer.
Odometer.
Speedometer.
Measuring Wheel.
 Speedometer pedometer

odometer

Measuring wheel passometer

Instruments used in Chain Surveying
1) Chains
2) Tapes
3) Arrows
4) Ranging Rods and Offset Rod
5) Pegs
6) Plumb- bob
 Types of Chains
Metric chain
Gunter’s chain or Surveyor’s chain
Engineer’s chain
Revenue chain
Steel band or Band chain
CHAINS
 Taking measurements on chain
Metric chains are made in lengths 20m and 30m.
Tallies are fixed at every five-meter length and brass
rings are provided at every meter length except
where tallies are attached.
 Testing and Adjustment of chain

If chain is found to be too If chain is found to be too

long short
1) Closing up the joints of the 1) Straightening the bent links.
rings if found to be opened 2) Opening the joints of the rings.
out. 3) Replacing one or more small
2) Reshaping the elongated circular rings by bigger ones.
rings. 4) Inserting new rings where
3) Replacing damaged rings. necessary.
4) Removing one or more small 5) Adjusting the links at the end.
rings.
5) Adjusting the links at the end.
 Types of Tapes
1) Cloth or linen tape.
2) Fibre tape.
3) Metallic tape.
4) Steel tape.
5) Invar tape.
Tapes
 Arrows
Arrows are made of tempered steel wire of diameter 4mm.
One end of the arrow is bent into a ring of diameter 50mm
and the other end is pointed.
Its overall length is 400mm.
An arrow is inserted into the ground after every chain
measured on the ground.
 Ranging Rods and Offset Rod
Ranging rods are 2 to 3 m in length.
Used for ranging some intermediate points on the
survey line.
Painted with alternate bands of black and white or
red and white colours.
With length of each equalising 20 cm.
Ranging
rods
 Pegs
Made of timber or steel.
Used to mark the position of stations.
Pegs are in length of 15 cm.
 Plumb-Bob
Used to transfer points on ground.
Used for fixing instrument exactly over the
stations.
 Principle of Chain Surveying

Triangulation is the principle.

In this area is divided into a number of
triangles with the suitable sides.
The plan of the area can be easily drawn.
As a triangle is the only simple plane
geometrical figure.
 Terms related with Chain
Surveying
1) Survey Stations:
(a) Main stations
(b) Subsidiary stations
(c) tie stations
2) Main survey lines
3) Base line
4) Check line
5) Tie-line
 Selection of Survey Stations
1) Survey lines should be minimum as far as possible and
should be taken on fairly level ground.
2) should be intervisible.
3) Should form well conditioned triangles.
4) Should be located that tie lines, check lines, baseline etc.
an be formed.
5) Should be selected within the boundary of the area to be
surveyed.
 Operations in Chain Surveying

Chaining
Ranging
Offsetting
 Chaining on Level Ground
It involves following operations
1) Fixing the stations.
2) Unfolding the chain.
3) Ranging.
4) Measuring the distance.
5) Folding the chain.
 Chaining on Sloping Ground
There are 2 methods:
1) Direct Method:
Also called as stepping. In this method, the
distance is measured in small horizontal
stretches. Say a1, a2 …an.
with suitable length of chain or tape.
finally the total horizontal distances are added to
get the required distances
 Chain surveying (“stepping”)

a
w
b
x c
y

z
 Indirect Method
It involves calculation from directly
measured lengths.
Method1:
Horizontal distance of the segment is calculated by
knowing sloping length of the segment and angle of
inclination of that with horizontal. The angle
The angle of the sloping surface with horizontal can be
known by instrument called Abney’s Level.
 Method2:
If the elevation difference between 2
terminals points and the sloping distance
between 2 terminal points is known the
horizontal distance D can be calculated as
D = (l²h²)
 Method3:
Also called as hypotenusal allowance method
Instead of putting the end arrow at actual end
of chain, it is put at some advanced distance
and that point is considered as the end of one
chain length
 Ranging
There are 2 methods:
1) Direct ranging
2) Indirect ranging or reciprocal ranging.
 Direct Ranging
When intermediate ranging rods are fixed on
a straight line by direct observation from end
stations, the process is known as direct
ranging.
Direct ranging also can be done with a line
ranger it consist of 2 right isosceles
triangular prisms. Placed one above other.
 Indirect or Reciprocal Ranging
When the end stations are not intervisible due
to there being high ground between them,
intermediate ranging rods are fixed on the
line in an indirect way.
The method is known as indirect ranging or
reciprocal ranging.
Indirect Ranging
 Offsetting
There are 2 types of offsets:
1) Perpendicular offsets:
the offsets which are taken perpendicular to the chain
are termed as above.
2) Oblique offsets:
Oblique distance is always greater than perpendicular
distance. All the offsets which are not taken at right
angle to chain line are known as above.
 Instruments for laying Offsets
Optical square.
Indian optical square.
Open cross staff.
Prism square.
 Errors in Chaining
There are 2 types
1) Compensating errors
2) Cumulating errors
Errors due to Incorrect Chain
If chain is too long Measured distance
will be less.
Correction:
positive

If chain is too small Measured distance

will be more.
Correction:
negative
 Sources of Errors in Chaining
Instrumental Errors
Natural errors
Personal errors
EXAMPLE 2 :- FIND ALL INCLUDING ANGLES
Example 3 :- find all including angles and correct magnetic bearing oh the
line

LINE FB B.B
AB 71’05’ 250’20’

BC 110’20’ 292’35’

CD 161’35’ 341’45’

DE 220’50’ 40’05’

EA 300’50’ 121’10’
SOLUTION:-
1 st step draw dig.
2 nd step include column
3 rd step B.B-F.B
4 find included angle= BBBL-FBNL
5 Sum of angle = (2N-4)90’
6 Correction in angles
7 Corrected bearing
 B.B =F.B+-180 ANGLE =BBBL-FBNL
(+WHEN F. B IS LESS THAM 180/-WHEN F.B IS GREATRER THAN 180)

LINE FB LINE B.B DIFF(B.B-FB) ANGLE CORR.F.B CORRB.B

AB 71’05’ BA 250’20’ 180’45’=45’’ 49’40’ 72’10’ 252’10’

BC 110’20’ CB 292’35’ 180’30’=30’’ 139’45’ 112’25’ 292’25’

CD 161’35’ DC 341’45’ 180’10’=10’’ 130’45’ 161’40’ 341’40’

DE 220’50’ ED 40’05’ 180’45=45’’ 119’50’ 221’50’ 41’50’

EA 300’50’ AE 121’10’ 179’50’=-10’’ 100’ 301’50’ 121’50’

INCORRETED ANGLE CORRETED ANGLE
ANGLE A =(AE-AB) =121’10’-71’05’ =49’55’-0’15’’=49’40’
ANGLE B =(BA-BC) =250’20’-110’20’ =140’0’-0’15’’=139’45’
ANGLE C =(CB-CD) =292’35’-161’35’ =131’0’-0’15’’=130’45’
ANGLE D =(DC-DE) =341’45’-220’50’ =120’5’-0’15’’=119’50’
ANGLE E =(ED-EA) =40’5’-300’50’+360’ =100’ 15’-0’15’’=100’
Sum of all angles =49’55’+140’+131’0’+120’5’+100’15’=541’15’
Sum of angle = (2N-4)90’
N=5 THEN SUM OF ANGLES 540’
TOTAL EROR =541’15’-540=1’15’
THEN ERROR IN EACH NSIDE=1’15’’/5=0’15’’

CORRECTED BEARING
CORRECTION OF CD=180’-180’10’=0’10’/2=0’5’
CORRECTED F.B CD =161’35+0’5’=161’=161’40’
CORRETED B.B DC = 341’45’-0’5’=341’40’
INCUDING ANGLE=BBBL-FBNL
ANGLE D=DC-DE=119’50’=341’40’-DE
THEN DE=341’40’-119’50’=221’50’
BB OF ED=221’50’-180’=41’50’
ANGLE E=ED-EA=100’=41’50’-EA
THEN EA=41’50’-100’+360’’=301’50’
BB OF AE=301’50’-180’=121’50’
ANGLE A= AE-AB=49’40’=121’50’-AB
AB=121’50’-49’40’=72’10’
BA=72’10’+180’=252’10’
ANGLE OF B=BA-BC
139’45’=252’10’-BC
BC=252’10’-139’45’=112’25’
CB=112’25’+180’=292’25’
ANGLE OF C=CB-CD
130’45’=292’25’-CD
CD=292’25’-130’45’=161’40’
DC=161’40’+180’=341’40
Levelling & Contouring
Object of levelling
Theobjective of levelling is to
1) Find the elevation of given point with
respectto someassumedreferenceline called
datum.
2)Toestablish point at requiredelevation
with respectto datum.
Definitions used in levelling
Levelsurface:-It is the surface parallel to the
meanspheroidal surface of theearth
Levelline:- Line lying onlevelsurface.
Horizontal plane:- Horizontal plane through a
point is a plane tangential to levelsurface.
Horizontal line:- It is a straight line tangential
to level line.
Datum:-“Itis an arbitrary level surfacefromwhich
elevation of points maybereferred”.In India meansealevel
is consideredasdatum of zero elevation it is situated at
Karachi.
Mean sea level is the average height of sea for all stages of
tides it is derived by averaging the hourly tide height over a
periodof19 years.
Elevation or Reduced level:- It is height or depth of any
pointaboveorbelowanydatum.It isdenotedasR.L.
BenchMark(B.M.):-It is a fixed referencepoint of known
elevation with respectto datum.
Line of collimation:-It is a line joining the intersection of
crosshairs of diaphragmto the optical centre of object glass
and its continuation. It is also knownasline ofsight.
Height of instrument:-It is the elevation of line of
collimation with respectto datum
Backsight:- It is a staff reading taken at a known
elevation. It is the first staff reading taken after setup of
instrument.
Fore sight( F.S.):- It is the last staff reading taken denoting
the shifting of the instrument.
Intermediate sight.(I.S.):-It is staff reading taken ona
point whoseelevation is to bedetermined.All staff reading
betweenB.S. and F.S.areIntermediatesight.
ChangePoint:-It is a point onwhich both foreand back
sight aretaken.
Instruments for levelling
The following instruments areessentially
required
forlevelling
Level
Levellingstaff
Automatic level

It is also knownasself aligning level. It is a recentdevelopment.

The fundamental differencebetweenauto level and other levels is
that the levelling is not manually but it is levelled automatically.
It is achievedbyinclination compensatingdevice.
Levelling Staffs
Levelling staffs arescalesonwhich these distances
aremeasured.
Levelling staffs areof twotypes
Self reading staff
Targetstaff
 Bench Marks
Benchmarkis a point of knownelevation
Thereare4 kinds of benchmarks
GTS(Greattrigonometrically surveybenchmark)
Permanentbenchmark
Arbitrary benchmark
Temporarybenchmark
GTS Bench mark
They are the benchmarks established with veryhigh degreeof
precisionat regular intervals bythe surveyof India Department
all overthe country Their position and R.Ls valuesabovemean
seal level at Karachiaregiven in catalogueformedbythe
department.
Meansealevel
 Permanent Bench mark
Permanent bench marks are fixed in between GTS bench marks by
govt. agencies such as railways, PWD, etc. This bench marks are
written on permanent objects such as milestones, culverts, bridges
etctheirvalueareclearlywrittenandtheirpositionarerecorded for
futurereference.
Arbitrary benchmarks:-These arereferencepointswhose
R.L.s arearbitrarily assumed.They areusedin small works
suchbenchmarkmaybeassumedas100. or50m
Temporarybenchmarks:-They arethe referencepoints
established during the levelling operations whenthere is a
breakin work, orat the endof day’sworkthe value of
reducedlevels aremarkedonsomepermanent objectssuch
asstones, treesetc.
Temporary Adjustments of a level
These adjustments areperformed at everysetup of
instrument
Setting up oflevel
Levelling oftelescope
Focusingof the eyepeace
Focusingof objectglass
Setting up the level:-Thisincludes
A) Fixing the instrument ontripod
B)Levelling the instrument approximatelybyTripod
Levelling:- Levelling Levelling is done with the help of foot screws.
The purpose of levelling is to make vertical axis truly vertical. It is
donewiththehelpoffootscrews
A) Placethe telescopeparallel to a pair of foot screwthen hold the
foot screwsbetweenthumb and first finger and turn them either
inwardoroutwarduntil the longitudinal bubblecomesin the
centre.
B)Turnthe telescopethrough 900 sothat it lies parallel to third foot
screw,turn the screw until the bubble comes in thecentre.
Focusing the eye piece:- Tofocus the eye piece, hold a white
paper in front of object glass, and move the eye piece in or
outtill thecrosshairaredistinctly seen.
Focusingof object glass:-Direct the telescope to the
levelling staff and onlooking through the telescope, turn
the focusing screwtill the imageappearsclear andsharp.
Classification of levelling
Simple levelling
Differential leveling
Flylevelling
Check levelling
Profile levelling
Cross levelling
Reciprocal levelling
Precise levelling
Trignometric levelling
Barometric levelling
Hypersometric levelling
Simple levelling:- It is the simplest method used, whenit is
required to find the difference in elevation between2points.

Differential Levelling:-This method is usedto find the

difference in the elevation between points if they aretoo far
apart orthe difference in elevation between them is too much.
Flylevelling:-Flylevelling is just like differential levelling
carried out to checkthe accuracyof levelling work. In fly
levelling only B.S.and F.S.aretaken
Checklevelling:-This kind of levelling is carried out to
checkthe accuracyof work. It is doneat the end of the days
workin the form of fly levelling to connectthe finishing
point and starting point.
Profile levelling orL-Section:-This method is usedfor
taking levels along the centre line of any alignment like
road, railwaycanal etc.The object is to determine the
undulations of the ground surfacealong the alignment
Cross-sectioning:-This operation is carried out
perpendicular to alignment at an interval of 10, 20 ,30,
40 m.The ideais to makean estimate ofearthwork.
Preciselevelling:-It is usedfor establishing benchmarksfor
future public use. It is carried out with high degreeof
accuracyusing advancedinstruments
Trignometriclevelling:-In this method vertical distances
betweenpoints arecomputedbyobservinghorizontal
distances and vertical angle betweenpoints.
Barometriclevelling:-Inthis methodthealtitude difference is
determinedbymeansofabarometer.
Hyposometric levelling:- The working of Hyposometry for
determining the elevation depends upon the fact that the
temperatureat which waterboils varies with the atmospheric
pressure.Theboilingpointofwaterreducesat higheraltitude
thus knowing the boiling point of water, the atmospheric
pressure can be calculated and knowing the atmospheric
pressurealtitude orelevationcanbe determined
Reciprocal levelling
Reciprocal levelling:-This methodisadoptedtoaccurately determine
the difference of level between two points which are far apart. It is also
usedwhenit isnotpossibletosetuplevelin midwaybetweentwopoints
Let A and B be the two points on opposite banks of a river. It is
required to find out the level difference betweenA&B
Set up the level very near to Aand take the reading at Aand Blet the
reading be a1 and b1
Shift the level and set up very near to Band observe Aand Bto get
reading a2 and b2
Let d is the true difference of level between Aand B, and e= error
due to curvature, refrection and imperfectadjustment.
 Thus to eliminate the error take an average of the difference in
elevation taken from 2 points
i.e. from Athe true difference will be=(bb11-’e)-a1
From Bthe difference will be= b2-((aa2-’e)
Therefore dd=={{(bb11-’a1)+(b2-a‘2)}/2

Line of collimation
a1 b1’
e
level Line b1

A B

River
Thus to eliminate the error take an average of the difference in elevation
taken from 2 points
i.e. from Athe true difference will be=(b1’-e)-a1
From Bthe difference will be= b2-(a2’-e)
Therefore d={(b1-a1)+ (b2-a2)}/2

Line of collimation
a2’ b2
e
level Line
a2

A B

River
 Methods of Reducing levels
Height of Instrument Method:-This method consist of finding
H.I. for everysetup of instrument, and then obtaining the
R.L. of point of referencewith respectto H.I
Formula:-
New H.I= Old R.L +B.S
New R.L= OLD H.I – (IS OR F.S) =434.312-1.606=
CHECK ∑ B.S- ∑ F.S= LAST R.L- FIRST R.L= ∑ RISE-
∑FALL
 HI

B.S0.9m
B.M I.S F.S1.05m B.S1.45m F.S=1.55m
100
A100m B991..18mm
M. D 99.750m
C 99.85m C.P.

Station B.S I.S F.S H.I R.L Remark

A 0.9 100.9 100.00 B.M
B 1.1 99.800
C 1.450 1.05 101.3 99.850 C.P.
D 1.550 99.750
 Rise And Fall Method

B.S0.9m
B.M I.S 1.1m F.S1.05m B.S1.45m F.S=1.55m
100
A100m B99.8m
M.
C 99.85m C.P. D 99.750m

Station B.S I.S F.S Rise Fall R.L Remark

A 0.9 100.00 B.M
B 1.1 0.2 99.800
C 1.450 1.05 0.05 99.850 C.P.
D 1.550 0.1 99.750
Example 1
The following staff readings were observed successively with a level
the instrument is moved by third sixth and eighth readings.
2.228 :1.606 :0.988 :2.090 :2.864 :1.262 0.602:1.982:1.044
:2.684 m enter the reading in record book and calculate R.L. if the
first reading was taken at a B.M of 432.383m
 Hight of instrument methods
Station B.S I.S
Method
F.S HI RL REMAR
KS
1 2.228 434.612 432.384 M B.M.
2 1.606 433.006
3 2.090 0.988 435.714 433.624 3RD C.P.
4 2.864 432.850
5 0.602 1.262 435.054 434.452 6TH C.P
6 1.044 1.982 434.116 433.072 8TH C.P
7 2.684 431.432
5.964 6.916

CHECK ∑ B.S-∑ F.S= 5.964-6.916= -0.952 = LASTR.L- FIRST R.L= 431.432-432.384=-0.952

 Rise and fallmethod
Station B.S I.S F.S Rise Fall RL REMARKS

1 2.228 432.384 M B.M.

2 1.606 0.622 433.006
3 2.090 0.988 0.618 433.624 3RD C.P.
4 2.864 0.774 432.850
5 0.602 1.262 1.602 434.452 6TH C.P
6 1.044 1.982 1.38 433.072 8TH C.P
7 2.684 1.64 431.432
5.964 6.916

CHECK ∑ B.S- ∑ F.S= 5.964-6.916= -0.952 = LASTR.L- FIRST R.L= 431.432-432.384=-0.952

∑RISE-∑ FALL=2.842-3.794=-0.952
in rise and fall =B.S-IS/IS-FS AND B.S-F.S NEVER F.S-B.S
NEW R.L=OLD R.L+-(RISE /FALL)
 Contour
Acontour is an imaginary line joining points of equal
elevation

120
100
105 110
 Contour Interval
The vertical distance betweenany two consecutive
contoursis knownascontourinterval
Contour interval
Uses of Contours
The nature of ground surfaceof a region canbeknown
Contour maphelps in locating propersite for bridges,
dams, reservoirsetc.
Capacity of a reservoircanbecalculated with the help of
contourmap
The quantity of cutting and filling canbedetermined
fromcontourmaps.
Routes for roads,railways,canals etc canbetraced.
Instrumental Error:
The Permanent adjustment of the instrument may
not be perfect. That is the line of collimation may
not be horizontal line
The internal arrangement of focusing tube may
not be correct
 The graduation of the staff may not be perfect
Curvature Correction
For long sights the curvature of earth can effect staff readings.
The line of sight is horizontal but the level line is curved and
parallel to the mean spheroidal surface of the earth. The
vertical distance between the line of sight and level line at
particular place is called the curvature correction
The effect of curvature is to cause the object sighted to appear
lower than they really are.
Curvature correction is alwaysSubtractive (-)
True staff reading= (Observed staff reading- 0.0785 D2 ) m
Where D= distance inKm.
Curvature Correction

(-)
Curvature
effect
 Refraction
The ray of light pass through layers of air of different densities
and refract or bent down. The effect of refraction is to make
the object appear higher then they really are. Refraction
varies considerably with climate conditions.
Howeverit is taken as
Cr= 0.0112 D2 m (+)
Refraction is always additive
True staff reading= Observed staff reading + Refraction
correction.
Refraction

(+)
Refraction
effect
 1

TOP = DISTANCE

BOTTOM = DISTANCE

CENTERLINE - ELEVATION
 Distance measurement

Distances measurement with the help of auto

leve
 1
The Auto Level:

The Stadia Rod

Or Surveyor’s Staff

The Instrument
AndTripod

= l (m) * 100(m)
 How to read the Stadia Rod:

11 meters on the dot

Each Black
Rectangle =
1 cm
10.8 meters on the dot
Same for
Eachwhite
one
So what’s
this height?

10.66 meters
 What you’ll see through the eyepiece:

Top Stadia = 10.91 m

StadiaLines Height = 10.85 m

Bottom Stadia = 10.79m

To Get Distance Away:

1) (Top - Bottom Stadia) = 0.12 m
2) Multiply difference by 100: Distance = 12 meters
 Engineering Mechanics
RADHESHYAM SAKRAVDIA
CIVIL DEPARTMENT
UJJAIN ENGINEERING COLLEGE UJJAIN
 Topic covered

 Introduction of Engineering Machines

 Composition and Resolution of Forces
 Equilibrium of Forces & Lamis thermos
 centre of gravity
 Moment of inertia
 Support Reactions
 Shear Force and Bending Moments
 Analysis of Perfect Frames (Analytical Method)
Centre of gravity formulas
MOMENT OF INERTIA OF A COMPOSITE SECTION
 Shear Force and Bending Moments
Consider a section x-x at a distance 6m from left hand support A

6m

5kN x 10kN 8kN

A B
C x D E
4m 5m 5m 1m

RA = 8.2 kN RB=14.8kN

Imagine the beam is cut into two pieces at section x-x and is separated, as
shown in figure
 4m 5kN 5m 1m
A
10kN 8kN B
6m
8.2 kN

9m 14.8 kN
To find the forces experienced by the section, consider any one portion of the
beam. Taking left hand portion
Transverse 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 get
Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward)
Moment experienced = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m
(anticlockwise)
 5kN 3.2 kN
A

8.2 kN 39.2 kN-m

10kN 8kN B

39.2 kN-m
3.2 kN 14.8 kN

Thus 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.
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
section

3.2 kN 39.2 kN

3.2 kN
F M

F
Shear force at x-x Bending moment at x-x
 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’.
Sign Convention for shear force

F
F

F
F

+ ve shear force - ve shear force

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.

Convexity

Fig. Sagging bending moment [Positive bending moment

]
 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.
Convexity

Fig. Hogging bending moment [Negative bending moment ]

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).
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.
 Relationship between load, shear force and
bending moment
x x1 w kN/m

x x1
dx L

Fig. A simply supported beam subjected to general type loading

The above Fig. shows a simply supported beam subjected to a general

type of loading. Consider a differential element of length ‘dx’ between
any two sections x-x and x1-x1 as shown.
 w kN/m
x x1
V+dV

M M+dM
v
x dx O x1

Fig. FBD of Differential element of the beam

Taking moments about the point ‘O’ [Bottom-Right corner of the

differential element ]
- M + (M+dM) – V.dx – w.dx.dx/2 = 0
Neglecting the small quantity of higher order
dM
V.dx = dM  v
dx It is the relation between shear force and BM
 w kN/m
x x1
V+dV

M M+dM
v
x dx O x1

Fig. FBD of Differential element of the beam

Considering the Equilibrium Equation ΣFy = 0

- V + (V+dV) – w dx = 0  dv = w.dx 

dv
w
dx It is the relation Between intensity of Load and
shear force
 Variation of Shear force and bending moments

Variation of Shear force and bending moments for various standard

loads are as shown in the following Table
Table: Variation of Shear force and bending moments

Type of load Between point Uniformly Uniformly

loads OR for no distributed load varying load
SFD/BMD load region
Shear Force Horizontal line Inclined line Two-degree curve
Diagram (Parabola)
Bending Inclined line Two-degree curve Three-degree
Moment (Parabola) curve (Cubic-
Diagram parabola)
 Sections for Shear Force and Bending Moment Calculations:
Shear force and bending moments are to be calculated at various
sections of the beam to draw shear force and bending moment diagrams.
These sections are generally considered on the beam where the
magnitude of shear force and bending moments are changing abruptly.
Therefore these sections for the calculation of shear forces include
sections on either side of point load, uniformly distributed load or
uniformly varying load where the magnitude of shear force changes
abruptly.
The sections for the calculation of bending moment include position
of point loads, either side of uniformly distributed load, uniformly
varying load and couple
Note: While calculating the shear force and bending moment, only the
portion of the udl which is on the left hand side of the section should
be converted into point load. But while calculating the reaction we
convert entire udl to point load
 Example Problem 1

1. Draw 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.

5N 10N 8N
A B
C D E
2m 2m 3m 1m
 5N 10N 8N
A B
C D E
2m 2m 3m 1m

RA RB
Solution: [Clockwise moment is Positive]
Using the condition: ΣMA = 0
- RB × 8 + 8 × 7 + 10 × 4 + 5 × 2 = 0  RB = 13.25 N
Using the condition: ΣFy = 0
RA + 13.25 = 5 + 10 + 8  RA = 9.75 N
 Shear Force Calculation:

0 1 5N 10N 8N 9
2 3 4 5 6 7 8

0 1 2 3 4 5 7 8 9
6

2m 2m 3m 1m
RA = 9.75 N RB=13.25N

Shear Force at the section 1-1 is denoted as V1-1

Shear Force at the section 2-2 is denoted as V2-2 and so on...
V0-0 = 0; V1-1 = + 9.75 N V6-6 = - 5.25 N
V2-2 = + 9.75 N V7-7 = 5.25 – 8 = -13.25 N
V3-3 = + 9.75 – 5 = 4.75 N V8-8 = -13.25
V4-4 = + 4.75 N V9-9 = -13.25 +13.25 = 0
V5-5 = +4.75 – 10 = - 5.25 N (Check)
 5N 10N 8N
A B
C D E
2m 2m 3m 1m

9.75N 9.75N
4.75N 4.75N

SFD 5.25N 5.25N

13.25N 13.25N
 5N 10N 8N
A B
C D E
2m 2m 3m 1m

9.75N 9.75N
4.75N 4.75N

SFD 5.25N 5.25N

13.25N 13.25N
 Bending Moment Calculation

Bending moment at A is denoted as MA

Bending moment at B is denoted as MB
and so on…
MA = 0 [ since it is simply supported]
MC = 9.75 × 2= 19.5 Nm
MD = 9.75 × 4 – 5 × 2 = 29 Nm
ME = 9.75 × 7 – 5 × 5 – 10 × 3 = 13.25 Nm
MB = 9.75 × 8 – 5 × 6 – 10 × 4 – 8 × 1 = 0
or MB = 0 [ since it is simply supported]
 5N 10N 8N
A B
C D E
2m 2m 3m 1m
29Nm
19.5Nm
13.25Nm

BMD
VM-34 5N 10N 8N
A B
C D E
2m 2m 3m 1m
9.75N 9.75N
Example Problem 1
4.75N 4.75N

SFD 5.25N 5.25N

13.25N 13.25N
29Nm
19.5Nm 13.25Nm

BMD
 5N 10N 8N
A B
C D E
2m 2m 3m 1m
9.75N 9.75N
4.75N 4.75N

SFD 5.25N 5.25N

13.25N 13.25N
29Nm
19.5Nm 13.25Nm

BMD
 Example Problem 2
2. Draw SFD and BMD for the double side overhanging
beam subjected to loading as shown below. Locate points
of contraflexure if any.

5kN 10kN 5kN

2kN/m

C A D B E

2m 3m 3m 2m
5kN 10kN 5kN
2kN/m

C A D B E

2m RA 3m 3m RB 2m

Solution:
Calculation of Reactions:
Due to symmetry of the beam, loading and boundary
conditions, reactions at both supports are equal.
.`. RA = RB = ½(5+10+5+2 × 6) = 16 kN
 5kN 10kN 5kN
0 1 2 3 4 5 2kN/m 6 7 8 9

2 3 4 5 7 8 9
0 1 6
2m 3m 3m 2m
RA=16kN RB = 16kN
Shear Force Calculation: V0-0 = 0
V1-1 = - 5kN V6-6 = - 5 – 6 = - 11kN
V2-2 = - 5kN V7-7 = - 11 + 16 = 5kN
V3-3 = - 5 + 16 = 11 kN V8-8 = 5 kN
V4-4 = 11 – 2 × 3 = +5 kN V9-9 = 5 – 5 = 0 (Check)
V5-5 = 5 – 10 = - 5kN
 5kN 10kN 5kN
2kN/m

C A D B E

2m 3m 3m 2m

11kN
5kN 5kN 5kN
+
+

5kN 5kN 5kN

SFD 11kN
5kN 10kN 5kN
2kN/m

C A D B E

2m 3m 3m 2m
RA=16kN RB = 16kN

Bending Moment Calculation:

MC = ME = 0 [Because Bending moment at free end is zero]
MA = MB = - 5 × 2 = - 10 kNm
MD = - 5 × 5 + 16 × 3 – 2 × 3 × 1.5 = +14 kNm
5kN 10kN 5kN
2kN/m

C A D B E

2m 3m 3m 2m
14kNm

BMD
10kNm
10kNm
5kN 10kN 5kN
2kN/m

C A D B E
2m 3m 3m 2m
11kN
+ 5kN 5kN 5kN
+

5kN 5kN
SFD 14kNm 11kN

BMD
10kNm 10kNm
5kN 10kN 5kN
2kN/m
x

C Ax D B E
x x
2m 3m 3m 2m

10kNm 10kNm
Points of contra flexure

Let x be the distance of point of contra flexure from support A

Taking moments at the section x-x (Considering left portion)
2 x = 1 or 10
x
M x  x  5(2  x)  16 x  2 0
2 .`. x = 1 m
 Example Problem 3
3. Draw SFD and BMD for the single side overhanging beam
subjected to loading as shown below. Determine the
absolute maximum bending moment and shear forces and
mark them on SFD and BMD. Also locate points of contra
flexure if any.

10kN/m 2 kN 5kN

A
C B D

4m 1m 2m
 10kN/m 2 kN 5kN

A B
RA 4m 1m RB 2m

Solution : Calculation of Reactions:

ΣMA = 0
- RB × 5 + 10 × 4 × 2 + 2 × 4 + 5 × 7 = 0  RB = 24.6 kN
ΣFy = 0
RA + 24.6 – 10 x 4 – 2 + 5 = 0  RA = 22.4 kN
 10kN/m 2 kN 5kN
2 3 4 5 7
0 1 6

0 2 3 4 5 7
1 6

RA=22.4kN 4m 1m 2m
RB=24.6kN

Shear Force Calculations:

V0-0 =0; V1-1 = 22.4 kN V5-5 = - 19.6 + 24.6 = 5 kN
V2-2 = 22.4 – 10 × 4 = -17.6kN V6-6 = 5 kN
V3-3 = - 17.6 – 2 = - 19.6 kN V7-7 = 5 – 5 = 0 (Check)
V4-4 = - 19.6 kN
 10kN/m 2 kN 5kN

A
C B D
RA=22.4kN 4m 1m 2m
RB=24.6kN
22.4kN

5 kN 5 kN

x = 2.24m
17.6kN
19.6kN 19.6kN
SFD
 10kN/m 2 kN 5kN
X

A
x X C B D
RA=22.4kN
4m 1m 2m
RB=24.6kN

Max. bending moment will occur at the section where the shear force is
zero. The SFD shows that the section having zero shear force is available
in the portion AC. Let that section be X-X, considered at a distance x
from support A as shown above.
The shear force at that section can be calculated as
Vx-x = 22.4 - 10. x = 0  x = 2.24 m
 10kN/m 2 kN 5kN

A
C B D
RA=22.4kN
4m 1m 2m
RB=24.6kN
Calculations of Bending Moments:
MA = MD = 0
MC = 22.4 × 4 – 10 × 4 × 2 = 9.6 kNm
MB = 22.4 × 5 – 10 × 4 × 3 – 2 × 1 = - 10kNm (Considering Left portion
of the section)
Alternatively
MB = -5 × 2 = -10 kNm (Considering Right portion of the section)
Absolute Maximum Bending Moment is at X- X ,
Mmax = 22.4 × 2.24 – 10 × (2.24)2 / 2 = 25.1 kNm
 10kN/m 2 kN 5kN
X

A
x = 2.24m X C B D
RA=22.4kN
4m 1m 2m
RB=24.6kN
Mmax = 25.1 kNm

9.6kNm Point of
contra flexure

BMD 10kNm
 10kN/m 2 kN 5kN
X

A D
x = 2.24m X C B
RA=22.4kN
4m 1m 2m
RB=24.6kN
22.4kN
5 kN 5 kN

x = 2.24m
17.6kN
19.6kN 19.6kN
SFD
Point of
contra flexure
9.6kNm

BMD 10kNm
 10kN/m 2 kN 5kN
X

A
x X C B D
RA=22.4kN
4m 1m 2m
RB=24.6kN
Calculations of Absolute Maximum Bending Moment:
Max. bending moment will occur at the section where the shear force is
zero. The SFD shows that the section having zero shear force is available
in the portion AC. Let that section be X-X, considered at a distance x
from support A as shown above.
The shear force at that section can be calculated as
Vx-x = 22.4 - 10. x = 0  x = 2.24 m
Max. BM at X- X ,
Mmax = 22.4 × 2.24 – 10 × (2.24)2 / 2 = 25.1 kNm
 10kN/m 2 kN 5kN
X

A
x = 2.24m X C B D
RA=22.4kN
4m 1m 2m
RB=24.6kN
Mmax = 25.1 kNm

9.6kNm Point of
contra flexure

BMD 10kNm
Let a be the distance of point of contra flexure from support B
Taking moments at the section A-A (Considering left portion)

M A A  5( 2  a )  24.6a  0
A
a = 0.51 m
Mmax = 25.1 kNm

9.6kNm Point of
contra flexure

BMD 10kNm
a
A
 Example Problem 4
4. Draw SFD and BMD for the single side overhanging beam
subjected to loading as shown below. Mark salient points on
SFD and BMD.

60kN/m
20kN
20kN/m

A
C B D

3m 2m 2m
 60kN/m
20kN
20kN/m

A
C B D

RA 3m 2m RB 2m

Solution: Calculation of reactions:

ΣMA = 0
-RB × 5 + ½ × 3 × 60 × (2/3) × 3 +20 × 4 × 5 + 20 × 7 = 0  RB =144kN
ΣFy = 0
RA + 144 – ½ × 3 × 60 – 20 × 4 -20 = 0  RA = 46kN
 60kN/m
20kN
20kN/m
1 3 4 5
0 2 6

1 2 3 4 5 6
0
RB = 144kN
RA = 46kN 3m 2m 2m
RA

Shear Force Calculations:

V0-0 =0 ; V1-1 = + 46 kN V4-4 = - 84 + 144 = + 60kN
V2-2 = +46 – ½ × 3 × 60 = - 44 kN V5-5 = +60 – 20 × 2 = + 20 kN
V3-3 = - 44 – 20 × 2 = - 84 kN V6-6= 20 – 20 = 0 (Check)
 Example Problem 4
60kN/m
20kN
20kN/m
1 2 3 4 5 6

1 2 3 4 5 6
RB = 144kN
RA = 46kN 3m 2m 2m
RA

46kN Parabola 60kN

20kN

44kN
SFD 84kN
 X 60kN/m
20kN
20kN/m

A x C B D
X RB=144kN
RA =46kN 3m 2m 2m

Max. bending moment will occur at the section where the shear force is
zero. The SFD shows that the section having zero shear force is available
in the portion AC. Let that section be X-X, considered at a distance ‘x’
from support A as shown above. The shear force expression at that section
should be equated to zero. i.e.,
Vx-x = 46 – ½ .x. (60/3)x = 0  x = 2.145 m
 60kN/m
20kN
20kN/m

A
C B D
RB=144kN
RA =46kN 3m 2m 2m

Calculation of bending moments:

MA = M D = 0
MC = 46 × 3 – ½ × 3 × 60 × (1/3 × 3) = 48 kNm[Considering LHS of
section]
MB = -20 × 2 – 20 × 2 × 1 = - 80 kNm [Considering RHS of section]
Absolute Maximum Bending Moment, Mmax = 46 × 2.145 – ½ × 2.145
×(2.145 × 60/3) × (1/3 × 2.145) = 65.74 kNm
 60kN/m
20kN
20kN/m

A
C B D
RB=144kN
RA =46kN 3m 2m 2m
48kNm

65.74kNm

Cubic
parabola Parabola

Point of
BMD Contra flexure Parabola
80kNm
46kN Parabola 60kN
20kN

44kN
SFD 84kN

65.74kNm

Cubic
parabola Parabola

Point of
BMD Contra flexure Parabola
80kNm
 X 60kN/m
20kN
20kN/m

A x=2.145m C B D
X RB=144kN
RA =46kN 3m 2m 2m

Calculations of Absolute Maximum Bending Moment:

Max. bending moment will occur at the section where the shear force is
zero. The SFD shows that the section having zero shear force is available
in the portion AC. Let that section be X-X, considered at a distance ‘x’
from support A as shown above. The shear force expression at that section
should be equated to zero. i.e.,
Vx-x = 46 – ½ .x. (60/3)x = 0  x = 2.145 m
BM at X- X , Mmax = 46 × 2.145 – ½ × 2.145 ×(2.145 × 60/3) × (1/3 × 2.145)=65.74
kNm
 60kN/m
20kN
20kN/m

A
C B D
RB=144kN
RA =46kN 3m 2m 2m
48kNm
65.74kNm 48kNm
Cubic
parabola Parabola

Point of
BMD Contra flexure Parabola
80kNm
Point of contra flexure:
BMD shows that point of contra flexure is existing in the
portion CB. Let ‘a’ be the distance in the portion CB from the
support B at which the bending moment is zero. And that ‘a’
can be calculated as given below.
ΣMx-x = 0

(2  a) 2
144a  20(a  2)  20 0
2

a = 1.095 m
 Example Problem 5
5. Draw SFD and BMD for the single side overhanging beam
subjected to loading as shown below. Mark salient points on
SFD and BMD.
40kN
0.5m
30kN/m
20kN/m
0.7m
A
B C D E

2m 1m 1m 2m
 40kN
0.5m
30kN/m
20kN/m
0.7m
A
B C D E

2m 1m 1m 2m

40x0.5=20kNm
40kN 30kN/m
20kN/m

A
B C D E

2m 1m 1m 2m
 40kN 30kN/m
20kN/m
20kNm
A
B C D E
RA
2m 1m 1m RD 2m

Solution: Calculation of reactions:

ΣMA = 0
-RD × 4 + 20 × 2 × 1 + 40 × 3 + 20 + ½ × 2 × 30 × (4+2/3) = 0  RD =80k
ΣFy = 0
RA + 80 – 20 × 2 - 40 - ½ × 2 × 30 = 0  RA = 30 kN
 20kNm 40kN
30kN/m
0 1 20kN/m 2 4 6 7
3 5

0 1 2 4 5 6 7
3
RD =80kN
RA =30kN
2m 1m 1m 2m

Calculation of Shear Forces: V0-0 = 0

V1-1 = 30 kN V5-5 = - 50 kN
V2-2 = 30 – 20 × 2 = - 10kN V6-6 = - 50 + 80 = + 30kN
V3-3 = - 10kN V7-7 = +30 – ½ × 2 × 30 = 0(check)
V4-4 = -10 – 40 = - 50 kN
 20kNm 40kN
30kN/m
1 20kN/m 2 4 6 7
3 5

1 2 4 5 6 7
3
RD =80kN
RA =30kN
2m 1m 1m 2m

30kN Parabola
30kN

x = 1.5 m
10kN 10kN

SFD
50kN 50kN
 40kN 30kN/m
20kN/m
X 20kNm
A
B C D E
x = 1.5 m X
RA
2m 1m 1m RD 2m

Calculation of bending moments:

MA = M E = 0
MX = 30 × 1.5 – 20 × 1.5 × 1.5/2 = 22.5 kNm
MB= 30 × 2 – 20 × 2 × 1 = 20 kNm
MC = 30 × 3 – 20 × 2 × 2 = 10 kNm (section before the couple)
MC = 10 + 20 = 30 kNm (section after the couple)
MD = - ½ × 30 × 2 × (1/3 × 2) = - 20 kNm( Considering RHS of the section
 40kN 30kN/m
20kN/m
X 20kNm
A
B C D E
x = 1.5 m X
RA
2m 1m 1m RD 2m
22.5kNm 30kNm
Parabola 20kNm
10kNm Point of contra flexure

BMD Cubic parabola

20kNm
 30kN Parabola
30kN

x = 1.5 m
10kN 10kN

SFD
50kN 50kN

Parabola 20kNm
10kNm Point of contra flexure

BMD Cubic parabola

20kNm
6. Draw SFD and BMD for the cantilever beam subjected
to loading as shown below.

40kN
0.5m
300
20kN/m
0.7m

3m 1m 1m
 40kN
0.5m
300
20kN/m
0.7m

A
3m 1m 1m

40Sin30 = 20kN

0.5m

20kN/m 40Cos30 =34.64kN

0.7m

3m 1m 1m
 40Sin30 = 20kN

0.5m

20kN/m 40Cos30 =34.64kN

0.7m

3m 1m 1m

20x0.5 – 34.64x0.7=-14.25kNm
20kN
20kN/m
34.64kN

3m 1m 1m
 20kN
20kN/m 14.25kNm

34.64kN HD
A 3m B 1m C 1m D
MD
VD
Calculation of Reactions (Here it is optional):
ΣFx = 0  HD = 34.64 kN
ΣFy = 0  VD = 20 × 3 + 20 = 80 kN
ΣMD = 0  MD - 20 × 3 × 3.5 – 20 × 1 – 14.25 = 244.25kNm
 20kN
1 20kN/m 14.25kNm 6
2 3 4 5
34.64kN HD
2 3 4
1 3m 1m 1m 5 6 MD
VD=80kN

Shear Force Calculation:

V1-1 =0
V2-2 = -20 × 3 = - 60kN
V3-3 = - 60 kN
V4-4 = - 60 – 20 = - 80 kN
V5-5 = - 80 kN
V6-6 = - 80 + 80 = 0 (Check)
 20kN
1 20kN/m 14.25kNm 6
2 3 4 5
34.64kN
HD
2 3 4
1 3m 1m 1m 5 6 MD
VD=80kN

60kN 60kN
SFD
80kN 80kN
 20kN
20kN/m 14.25kNm

34.64kN
A B
3m 1m C 1m D
MD

Bending Moment Calculations:

MA = 0
MB = - 20 × 3 × 1.5 = - 90 kNm
MC = - 20 × 3 × 2.5 = - 150 kNm (section before the couple)
MC = - 20 × 3 × 2.5 – 14.25 = -164.25 kNm (section after the couple)
MD = - 20 × 3 × 3.5 -14.25 – 20 × 1 = -244.25 kNm (section before MD)
moment)
MD = -244.25 +244.25 = 0 (section after MD)
 20kN
20kN/m 14.25kNm

34.64kN
A B
3m 1m C 1m D

90kNm
150kNm
BMD
164.25kNm

244.25kNm
 W

L/2 L/2

wkN/m

wkN/m
 Exercise Problems VM-73
1. Draw SFD and BMD for a single side overhanging beam
subjected to loading as shown below. Mark absolute
maximum bending moment on bending moment diagram and
locate point of contra flexure.
10kN 15kN/m
20kN/m
5kNm

1m 1m 3m 1m 1m 2m

[Ans: Absolute maximum BM = 60.625 kNm ]

 Exercise Problems VM-74

2. 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.
10kN 16kN
4kN/m
600 B
A

1m 1m 2m 1m 1m

[Ans: Absolute maximum bending moment = 22.034kNm

Its position is 3.15m from Left hand support ]
 Exercise Problems VM-75

3. Draw shear force and bending moment diagrams [SFD

and BMD] for a single side overhanging beam subjected
to loading as shown in the Fig. given below. Locate
points of contra flexure if any.
50kN 25kN/m
10kN/m
10kNm
A
B
1m 1m 3m 2m

[Ans : Position of point of contra flexure from RHS = 0.375m]

 Exercise Problems VM-76

4. Draw SFD and BMD for a double side overhanging beam

subjected to loading as shown in the Fig. given below.
Locate the point in the AB portion where the bending
moment is zero.
8kN 16kN
4kN/m 8kN

A B
2m 2m 2m 2m

[Ans : Bending moment is zero at mid span]

 Exercise Problems VM-77
5. A single side overhanging beam is subjected to uniformly distributed
load of 4 kN/m over AB portion of the beam in addition to its self
weight 2 kN/m acting as shown in the Fig. given below. Draw SFD
and BMD for the beam. Locate the inflection points if any. Also locate
and determine maximum negative and positive bending moments.
4kN/m
2kN/m

A
B
6m 2m

[Ans :Max. positive bending moment is located at 2.89 m from LHS.

and whose value is 37.57 kNm ]
Example A beam AB 8.5 m long is hinged at A and supported on rollers over
a smooth surface inclined at 30° to the horizontal at B. The beam is loaded as
shown in Fig. 12.21. Fig. Determine graphically, or otherwise, the reactions at
A and B
 Unit II Shear Force and Bending Moment
Diagrams
CO2. DRAW Shear force and bending moment diagram for various types of
transverse loading and support.
SFD & BMD: Introduction to SFD, BMD with application, SFD & BMD for
statically determinate beam due to concentrated load, uniformly distributed
load, uniformly varying load, couple and combined loading, Relationship
between rate of loading, shear force and bending moment, Concept of zero
shear force, Maximum bending moment, point of contra-flexure

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 Types of Beams

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 Types of Loads

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 Shear Force Diagram (SFD)
• A shear force (SF) is defined as the A shear force which tends to rotate
algebraic sum of all the vertical forces, the beam in clockwise direction is
either to the left or to the right hand positive and vice versa
side of the section
• Shear Force Diagram: is graph
connecting Shear Forces at various
locations on the beam.

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 Bending Moment Diagram (BMD)
• A bending moment (BM) is defined as the algebraic
sum of the moments of all the forces either to the
left or to the right of a section.
• BMD: Diagram is graph connecting bending
moments at various locations

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 General Guidelines on Construction of
SFD and BMD
• The load, shear and bending moment diagrams should be constructed one below
the other, in that order, all with the same horizontal scale.
• The dimension on the beam need not be scaled but should be relative and
proportionate
• Ordinates (i.e., BM and SF values) need not be plotted to scale but should be
relative. Curvature may need to be exaggerated for clarity.
• Principal ordinates (BM and SF values at salient points) should be labeled on both
SFD and BMD.
• A clear distinction must be made on all straight lines as to whether the line is
horizontal or has a positive or negative slope.
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 Sign Conventions for the Bending Moment:
a) Positive Bending Moment (Sagging)

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 b) Negative Bending Moment(Hogging)

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 Point of Contraflexure:
• It corresponds to a point where the bending moment changes the sign,
hence in order to find the point of contraflexures obviously the B.M
would change its sign when it cuts the X-axis therefore to get the points
of contraflexure equate the bending moment equation equal to zero. The
fibre stress is zero at such sections

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 Interpretations
Types of Loading Shape of Shear Force Diagram Shape of Bending Moment
Diagram

Point Load Sudden Jump at load and constant at Linearly Varying Graph (Straight
other locations Line)

Uniformly Distributed Load (UDL) Linearly Varying Graph (Straight Parabolic Graph (Smooth Curve)
Line)

Uniformly Varying Load (UDL) Parabolic Graph (Smooth Curve) Cubically varying Graph (Curve)

Moment Couple No effect Sudden Jump at Moment and

constant at other locations
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 Relationship between shear force, bending
moment

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