Experiment to Verify Shearing Force in Beams

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.


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.


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.


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


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


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


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


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


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



V =0RA + RB = 20NRA = 20 13.3 = 6.7N


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


V =0RA + RB =20NRA = 20 6.67 = 13.3N


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)




V =0RA + RB =20NRA = 20 3.33 = 16.67N


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


V =0RA + RB = 10NRA = 10 6.875 = 3.125N


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


V = 0RA + RB =10NRA = 10 3.75 = 6.257N


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)




V = 0RA + RB =10NRA = 10 1.88 = 8.12N


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


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


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


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


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]

Experiment to Verify Shearing Force in Beams

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