EXPERIMENT TO STUDY SHEARING FORCE IN BEAMS
UNIVERSITY OF MAURITIUSFACULTY OF ENGINEERINGCIVIL ENGINEERING
DEPARTMENTSTRUCTURES PRACTICALSSTRUCTURAL MECHANICS LABORATORYBEng
(HONS) CIVIL ENGINEERING P/T E411YEAR 1-SEMESTER 1MODULE:
INTRODUCTION TO STRUCTURES
SUBMITTED BY:GROUP 4LUCKEENARAIN Yogesh Kumar (Student
ID:1113061)MATTAPULLUT Mahendra Kumar (Student ID: 1118235)BHUNJUN
Vandana (Student ID:1115699)OLLITE Mohammad Jamiil Ally (Student
ID: 1100350)This experiment is designed to understand the action of
shear force in a beam.REPORT OF EXPERIMENT 1997/2: SHEARING FORCE
IN BEAMSDATE OF SUBMISSION: 18th NOVEMBER 2011
TABLE OF CONTENTS
INTRODUCTION1THEORY2-3EXPERIMENTATION3EXPERIMENTAL
PROCEDURES4TABULATION OF RESULTS5THEORETICAL RESULTSTABULATION OF
THEORETICAL RESULTSCOMPARISON OF THEORETICAL & EXPERIMENTAL
RESULTS6-8DISCUSSION9PRECAUTIONSLIMITATIONS AND
IMPROVEMENTSCONCLUSION10REFERENCES11
October 27, 2011EXPERIMENT TO STUDY SHEARING FORCE IN BEAMS
INTRODUCTION TO STRUCTURES | CIVE1105
INTRODUCTIONA forceis any influence that causes anobjectto
undergo a change in speed, a change in direction, or a change in
shape. Force can also be described by intuitive concepts such as a
push or pull that can cause an object withmassto change
itsvelocity(which includes to begin moving from astate of rest),
i.e., toaccelerate, or which can cause a flexible object todeform.
A force has bothmagnitudeand direction, making it
avectorquantity.Force is a quantity that is measured using the
standard metric unit known as the Newton.
There are different types of forces that act in different ways
on structures such as bridges, chairs, buildings, in fact any
structure. The main examples of forces are shown below. Static
load- the effect of gravity on an object or structure. Dynamic
load- the forces that move or change when acting on a structure.
Tension- the magnitude of the pulling force exerted by a string,
cable, chain, or similar object on another object. Compression- the
degree to which a substance has decreased in size (in volume,
length, or some other dimension) after being or while being subject
to stress that is when a force is applied to it. Shear Force- an
internal force in any material which is usually caused by any
external force acting perpendicular to the material, or a force
which has a component acting tangent to the material. Torsion- The
stress or deformation caused when one end of an object is twisted
in one direction and the other end is held motionless or twisted in
the opposite direction.
Shear force is an internal force in any material which is
usually caused by any force acting perpendicular to the material,
or a force which has a component acting tangent to the
material.
INTRODUCTION TO STRUCTURES | CIVE11051
A shear force diagram is simply constructed by moving a section
along the beam from (say) the left origin and summing the forces to
the left of the section. The equilibrium condition states that the
forces on either side of a section balance and therefore the
resisting shear force of the section is obtained by this simple
operation.
The bending moment diagram is obtained in the same way except
that the moment is the sum of the product of each force and its
distance(x) from the section.Distributed loads are calculated buy
summing the product of the total force (to the left of the section)
and the distance(x) of the centroid of the distributed load. The
sketches below show simply supported beams with on concentrated
force.
INTRODUCTION TO STRUCTURES | CIVE11052
THEORYIn this experiment, we will deal with the shearing forces
that exist in a split beam in two, part (A) and part (B), joined by
a ball bearing rolls at the normal section. Spring balances are
placed onto the system firstly to resist and avoid bending moments
and secondly to provide vertical shearing force.
The ends of the beam are supported by bearing stands, and to
introduce the forces shaped loads and hangers can be placed to the
beam at different distance from the section.For the first part of
the experiment, we would vary the applied loads only at the
different position on the beam keeping the distance between the
hangers fixed. And for the second part, the hangers and support
would be moved at different position on the beam.The springs used
to hold the split beam together must produce a system of forces
equivalent to those which would exist internally in the beam at
that section as if it was not split. Since the forces in part (A)
acting on part (B) is equal but opposite to those in (B) acting on
(A), the same values will be obtained by working either on the
right or left of the section plane.Given a horizontal beam with
vertical loading, the internal forces will be: For vertical
equilibrium, a shearing force in the section plane. For equilibrium
of moments, a moment of resistance due to compression in the top
half of beam section and tension in the bottom half.
INTRODUCTION TO STRUCTURES | CIVE11053
OBJECTIVES To understand the action of the shear in the beam and
to measure the shearing force of a normal section of a loaded beam
and to compare with the theory.Equipments Shear force apparatus
with supports, 3 hangers, weights (10.0N 20.0N) and a spirit
level.Set UpThe apparatus is set up shown as below
INTRODUCTION TO STRUCTURES | CIVE11054
EXPERIMENTAL PROCEDURESPART 1 The beam is set up so that the
face of the normal section is 300mm, labeled as (A) from the left
hand support and 600mm from the right hand support, labeled as (B).
One load hanger is positioned on the middle of the smaller part (A)
of the beam, one in the middle of the part (B) Place a third hanger
on the groove just to the right of the normal section, which is on
the start of section (B). Measure the distance between the hangers
and the supports and label them x and y for the smaller section and
longer section, respectively. Make the necessary adjustment to the
under hang and spring balance so that the two parts of the system
are aligned and level, and the reading on the spring balance is
noted.(Note: A spirit level is used for the leveling process) A
load of 10N is hanged on the part (B), the beam is re-aligned and
the new reading indicated on the spring balance is noted. The
difference between the two readings is the effect of applying the
1N weight on the beam. Record the distance from this weight to the
right hand support. The procedure is repeated using the hanger just
to the right of the normal section. The hanger is then moved just
to the left of the normal section and both the spring balance and
distance from left hand support were noted. The 10N load is
transferred to the hanger at the middle of part (A) and the reading
is again noted. Finally the procedures above are repeated now using
the load 20N and the differences in the spring balance readings is
again calculated.
PART 2 Remove all loads and realign and level the beam again.
Return all the hangers to the initial position. For the second part
of this experiment, all the 3 hangers must be loaded, and the beam
system must be realigned and leveled. The spring balance reading,
the applied loads, and the length between the hanger and the
supports are noted for the new configuration of the loads. The
procedure is repeated with different distance between supports with
the load hangers in different position and with different
spans.
TABULATION OF RESULTSPART 1Note: Middle hanger on the right
position of the split Breakage/Split
A B C X Y Exp. No.Distance between spansLoad applied on
hangerSpring BalancereadingShearing Force(N)
X(mm)Y(mm)A(N)B(N)C(N)Initial(N)(Without
Load)Final(N)(WithLoad)
1300600102.53.00.5
2300600102.56.03.5
3300600102.5-2.5-1.5
43006001010102.5129.5
5300600202.56.03.5
6300600202.5-0.9-3.4
7300600202.5No readingNo reading
Note: Shear Force = Measured value with load Initial Measured
value without load
PART 2Note: Middle hanger on the left position of the split
Breakage/Split
A B C X Y Exp. No.Distance between spansLoad applied on
hangerSpring Balance readingShearing Force (N)
X(mm)Y(mm)A(N)B(N)C(N)Initial(N)(Without Load)Final(N)(With
Load)
1300500101.02.01.0
2300500101.02.31.3
3300500101.0 No readingNo reading
43005001010101.06.05.0
53005002020201.012.011.0
Note: Shear Force = Measured value with load Initial Measured
value without load
THEORETICAL RESULTSTo allow determination of all of the external
loads a free-body diagram is construction with all of the loads and
supports replaced by their equivalent forces.A typical free-body
diagram is shown below.
The unknown forces (generally the support reactions) are then
determined using the equations for plane static equilibrium.
For example considering the simple beam above the reaction R2 is
determined by Summing the moments about R1 to zero R2. L - W.a = 0
Therefore R2 = W.a / LR1 is determined by summing the vertical
forces to 0 W - R1 - R2 = 0 Therefore R1 = W - R2
Part 1 (i) 10N A B RA 300 mm 600 mm RB Let RA be the vertical
reaction at A and let RB be the vertical reaction at BAnd let X/mm
be the distance along the span.Using Equation of staticsFor
equilibrium of the beam at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0(RA X 0)+ (10 X 600) =
(RB X 900)RB = 6.67N
V =0RA + RB = 10NRA = 10 6.67 = 3.33N
S.FShear Force Diagram
3.33
X/mm +ve0 300 600 900 -ve
-6.67Therefore, from the shear force diagram, Shear force at
split (300 mm from point A) = 3.33 KN(ii) 10N A B RA 300 mm 600 mm
RB Let RA be the vertical reaction at A and let RB be the vertical
reaction at B
Using Equation of staticsFor equilibrium of the beam at any
point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0(RA X 0)+ (10 X 300) =
(RB X 900)RB = 3.33NV =0RA + RB =10NRA = 10 3.33 = 6.67N
S.FShear Force Diagram
6.67
X/mm +ve 0 300 -ve 900 3.33
Therefore, from the shear force diagram, Shear force at split
(300 mm from point A) = 0.0 KN
(iii)
10N A B RA 300 mm 600 mm RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (10 X 150) =
(RB X 900)RB = 1.67N
V =0RA + RB =10NRA = 10 1.67 = 8.33N
S.FShear Force Diagram
8.33
X/mm +ve
-1.67 0 150 300 -ve 900
Therefore, from the shear force diagram, Shear force at split
(300 mm from point A) = -1.67 KN
(iv) 10N 10N 10N A B RA 300 mm 600 mm RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0(RA X 0)+ (10 X 150) +
(10 X 300) + (10 X 600) = (RB X 900)RB = 11.67N
V =0RA + RB =30NRA = 30 11.67 = 18.33N
S.FShear Force Diagram
18.33
8.33
X/mm +ve
-1.67 0 150 300 -ve 900
-11.67 Therefore, from the shear force diagram, Shear force at
split (300 mm from point A) = 8.33 KN (v) 20N A B RA 300 mm 600 mm
RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (20 X 600) =
(RB X 900)RB = 13.3N
V =0RA + RB = 20NRA = 20 13.3 = 6.7N
S.FShear Force Diagram
6.7
X/mm +ve0 300 600 900 -ve
-13.3
Therefore, from the shear force diagram, Shear force at split
(300 mm from point A) = 6.7 KN (vi) 20N A B RA 300 mm 600 mm RB Let
RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (20 X 300) =
(RB X 900)RB = 6.67N
V =0RA + RB =20NRA = 20 6.67 = 13.3N
S.FShear Force Diagram
13.3
X/mm +ve 0 300 -ve 900
-6.67
Therefore, from the shear force diagram, Shear force at split
(300 mm from point A) = 0 KN (vii)
20N A B RA 300 mm 600 mm RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (20 X 150) =
(RB X 900)RB = 3.33N
V =0RA + RB =20NRA = 20 3.33 = 16.67N
S.FShear Force Diagram
16.67
X/mm +ve
-3.33 0 150 300 -ve 900 Therefore, from the shear force diagram,
Shear force at split (300 mm from point A) = -3.33 KN Part 2(i) 10N
A B RA 300 mm 500 mm RB Let RA be the vertical reaction at A and
let RB be the vertical reaction at BUsing Equation of staticsFor
equilibrium of the beam at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (10 X 550) =
(RB X 800)RB = 6.875N
V =0RA + RB = 10NRA = 10 6.875 = 3.125N
S.FShear Force Diagram
3.13
X/mm +ve0 300 550 800 -ve
-6.88
Therefore, from the shear force diagram, Shear force at split
(300 mm from point A) = 3.13 KN.(ii) 10N A B RA 300 mm 500 mm RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (10 X 300) =
(RB X 800)RB = 3.75N
V = 0RA + RB =10NRA = 10 3.75 = 6.257N
S.FShear Force Diagram
6.25
X/mm +ve
-3.75 0 300 -ve 800
Therefore, from the shear force diagram, Shear force at split
(300 mm from point A) =0.0 KN (iii)
10N A B RA 300 mm 500 mm RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (10 X 150) =
(RB X 800)RB = 1.88N
V = 0RA + RB =10NRA = 10 1.88 = 8.12N
S.FShear Force Diagram
8.12
X/mm +ve
-1.88 0 150 300 -ve 800 Therefore, from the shear force diagram,
Shear force at split (300 mm from point A) = -1.88 KN (iv) 10N 10N
10N A B RA 300 mm 500 mm RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (10 X 150) +
(10 X 300) + (10 X 550) = (RB X 800)RB = 12.5NV = 0RA + RB =30NRA =
30 12.5 = 17.5N
S.FShear Force Diagram
17.5
7.5
X/mm +ve
-2.5 0 150 300 -ve 800
-12.5 Therefore, from the shear force diagram, Shear force at
split (300 mm from point A) = 7.5 KN (v) 20N 20N 20N A B RA 300 mm
500 mm RB
Let RA be the vertical reaction at A and let RB be the vertical
reaction at BUsing Equation of staticsFor equilibrium of the beam
at any point, M =0, V = 0 & H=0
Taking point A as moment centreM at A =0 (RA X 0)+ (20 X 150) +
(20 X 300) + (20 X 550) = (RB X 800)RB = 25N
V = 0RA + RB =60NRA = 60 25 = 35N
S.FShear Force Diagram
35
15
X/mm -5 150 300 550 800 -ve
-25 Therefore, from the shear force diagram, Shear force at
split (300 mm from point A) = 15.0 KN TABULATION OF THEORETICAL
RESULTSPart 1Exp. No.Distance between spansLoad applied on
hangerVertical ReactionShearing Force (At split) (N)
X(mm)Y(mm) A(N) B(N) C(N)RARB
1300600103.336.673.33
2300600106.673.330.00
3300600108.331.67-1.67
430060010101018.311.78.33
5300600206.713.36.70
63006002013.36.700.00
73006002016.73.33-3.33
Part 2Exp. No.Distance between spansLoad applied on
hangerVertical ReactionShearing Force(At Split) (N)
X(mm)Y(mm)A(N)B(N)C(N)RARB
1300500103.136.883.13
2300500106.253.750.00
3300500108.121.88-1.88
430050010101017.512.57.5
5300500202020352515.0
COMPARISON OF THEORETICAL & EXPERIMENTAL RESULTSPART 1 Exp.
No.Distancebetween spansLoad applied on hangerShearing Force(N)
X(mm)Y(mm) A(N) B(N) C(N)ExperimentalTheoretical
1300600100.53.33
2300600103.50.00
330060010-1.5-1.67
43006001010109.58.33
5300600203.56.70
630060020-3.40.00
730060020No reading-3.33
PART 2Exp. No.Distance between spansLoad applied
onhangerShearing Force(N)
X(mm)Y(mm)A(N)B(N)C(N)ExperimentalTheoretical
1300500101.03.13
2300500101.30.00
330050010No reading-1.88
43005001010105.07.50
530050020202011.015.0
DISCUSSIONWhen comparing the theoretical values of shear force
with experimental values from the tables above, the ratio of
experimental shear force to theoretical shear force is more or less
equal to 1. This indicates the experimental results are compatible
with the theoretical/calculated results.
PRECAUTIONS1. For each condition of loading, we should ensure
that the beam is horizontal by using the spirit level.2. Excessive
loads need to be avoided.3. The foot of the tripods is held with
two boxes so as to avoid the beam from sliding when adding and
adjusting the load.4. The screw above the spring balance is
adjusted carefully to ensure the beam is horizontal.5. The screw in
the under slung spring is adjusted to remove any bending effect
because if there is a bending effect, the reaction obtained on the
spring balance will be different.6. Care should be taken all
throughout the experiment as it deals with sufficiently large loads
which are able to cause injuries (for example, if it falls on the
foot).7. We should avoid parallax error when reading the spring
balance.
LIMITATIONS AND IMPROVEMENTS
The spring balance was not so accurate; instead a Digital Force
Display meter could has been used instead to obtain more precise
values. At the normal section where there is a pair of ball bearing
rollers pinned in B, running on a flat vertical track fixed in A,
friction could has existed and hence affecting the values noted.
The ball bearing should be lubricated continuously during the
experiment.
CONCLUSION
There is a slight difference between the experimental results
and theoretical values, which were perhaps due to the laboratory
equipments, especially the spring balance, and due to friction
which we neglected while doing this experiment. The experiment was
a time consuming one and requiring maximum attention.
This experiment helped us to understand the action of shear
force in a beam. When the load was doubled in part 1 of the
experiment, the shear force was also doubled as shown in both
experimental and theoretical values. This is because shear force
does not vary with the distance from the point of application of
the force but it depends on the magnitude of the force on the
application. The shear force is always maximum at supports and it
creates a shearing effect there.
REFERENCESTextbook R.S.KURMI, S.CHANDStrength of materials
(Mechanics of solids)(Pages 355-357)
Er.R.K.Rajput 5th Edition 2010 Reprint 2011(Pages 206-214)
Websites Douglas R. Rammer - United States Department of
Agriculture - Research Paper FPL-RP527
http://www.fpl.fs.fed.us/documnts/fplrp/fplrp527.pdf[Accessed on:
14/11/11]
Forces - V. Ryan 2002 2010
http://www.technologystudent.com/forcmom/force1.htm[Accessed on:
14/11/11]
The physics classroom-Newtons law-Lesson
2http://www.physicsclassroom.com/class/newtlaws/u2l2b.cfm[Accessed
on: 14/11/11]
Wikipedia, the free encyclopediaShear force diagram and
definition http://en.wikipedia.org/wiki/Force[Accessed on:
14/11/11]Other websites http://wiki.answers.com
http://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html[Accessed
on: 14/11/11]