Shear Force in a Beam (Edited)1

BFC 2091 Structure Lab – Shear Force In A Beam

TITLE : SHEAR FORCE IN A BEAM________________________________________________________________________

1.0 OBJECTIVE

1.1 To examine how shear force varies with an increasing point load.

1.2 To examine how shear force varies at the cut position of the beam for

various loading conditions.

2.0 LEARNING OUTCOME

2.1 The application of engineering knowledge in practical application.

2.2 To enhance technical competency in structural engineering through

laboratory application.

2.3 To communicate effectively in group.

2.4 To identify problem, solving and finding out appropriate solution through

laboratory application.

3.0 INTRODUCTION AND THEORY

A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending The bending force induced into the material of the beam as a result of the external loads, own weight, span and external reactions to these loads is called a bending moment

. If the ends of a beam are restrained longitudinally by its support or if a

beam is a component of a continuous frame, axial force may also develop. If the

axial force is small, the typical situation for most beams can be neglected when

the member is designed. In the case, of reinforced concrete beams, small values

of axial compression actually produce a modest increase (on the order of 5 to 10

percent) in the flexural strength of the member.

To design a beam, the engineer must construct the shear and moment

curves to determine the location and magnitude of the maximum values of these

forces. Except for short, heavily loaded beams whose dimensions are controlled

Wong Siew Hung AF040176


by shear requirements, the proportion of the cross section are determined by the

magnitude of the maximum moment in the span. After a section is sized at the

point of maximum moment, the design is completed by verifying that the shear

stresses at the point of maximum shear usually adjacent to a support are equal to

or less than the allowable shear strength of the material. Finally, the deflection

produced by service loads must be checked to ensure that the member has

adequate stiffness. Limits on deflection are set by structural codes.

To provide this information graphically, we construct shear and moment

curves. These curves, which preferably should be drawn to scale, consist of values

of shear and moment plotted as ordinates against distance along the axis of the

beam. Although we can construct shear and moment curves by cutting free bodies

at intervals along the axis of a beam and writing equation of equilibrium to

establish the values of shear and moment at particular section, it is much simpler

to construct these curves from the basic relationships that exist between load,

shear and moment.

Bending moment at any section of a beam is defined to be the algebraic

sum of the moment at the sectioning developed by vertical components of

external forces applied on the beam by considering the left or the right of assumed

section, or unbalanced moment at the sectioning, to the left or the right of the

assumed section. Variation of bending moment along beam can be visualized by

Bending Moment Diagram (BMD), which is defined as a diagram that shows

variations of bending moment along the beam considered. The final step in the

design of a beam is to verify that it does not deflect excessively. Beams that are

excessively flexible undergo large deflections that can damage attached

nonstructural construction: plaster, ceiling, masonry walls, and rigid piping for

example may crack.

Since most beams are span short distances, say up to 30 or 40 ft, are

manufactured with a constant cross sections, to minimize cost, they have excess

flexural capacity at all sections except the one at which maximum moment



occurs. Beams are typically classified by the manner in which they are supported.

A beam supported by a pin at the one end and a roller at the other end is called a

simply supported beam. If the end of the simply supported beam extends over a

support, it is referred to as a beam with an overhang.

A cantilever beam is fixed at the one end against translation and rotation.

Beams are supported by several intermediate support are called continuous beam.

If both ends of a beam are fixed by the support, the beam is termed fixed ended.

Fixed ended beams are not commonly constructed in practice, but the values of

end moments in them produced by various types of load are used extensively as

the starting point in several methods of analysis for indeterminate structures.

Fig. 1 : Shear Force and Bending Moment



Fig. 2 : Change of Shape due to Shear Force

There are a number of assumptions that were made in order to develop the

Elastic Theory of Bending. These are:

3.1 The beam has a constant, prismatic cross-section and is constructed of a

flexible, homogenous material that has the same Modulus of Elasticity in

both tension and compression (shortens or elongates equally for same

stress).

3.2 The material is linearly elastic; the relationship between the stress and strain

are directly proportional.

3.3 The beam material is not stressed past its proportional limit.

3.4 A plane section within the beam before bending remains a plane after bending

(see AB & CD in the image below).

3.5 The neutral plane of a beam is a plane whose length is unchanged by the

beam's deformation. This plane passes through the centroid of the cross-

section.

Theory

Part 1



W

a ‘cut’

Figure 1

Shear force at left of the section, Sc = W ( L-a ) …………..equation 1 L

Shear force at the right of the cut section, Sc = -Wa …………equation 2 L

Part 2

Use this statement :

“The shear force at the ‘cut’ is equal to the algebraic sum of the force

acting to the left or right of the cut”

4.0 APPARATUS


RA RBL


Figure 1 : Measuring Force Machine

Figure 2 : Load

Figure 3 : Data Analysis (Group members)

5.0 PROCEDURE

5.1 Part 1

5.1.1 Check the Digital Force Display meter reads zero with no load.



5.1.2 Place a hanger with a 100g mass to the left of the ‘cut’.

5.1.3 Record the Digital Force Display reading in Table 1. Repeat using

any masses between 200g and 500g. Convert the mass into a load

in Newton (multiply by 9.81).

Shear Force at the cut (N) = Displayed Force.

5.1.4 Calculate the theoretical Shear Force at the cut and complete the

Table 1.

5.2 Part 2

5.2.1 Check the Digital Force Display meter zero with no load.

5.2.2 Carefully load the beam with the hangers in any positions and

loads as example in Figure 2, Figure 3 and Figure 4 and complete

Table 2.

5.2.3 Record the Digital Force Display reading where :

Shear Force at the cut (N) = Displayed Force.

5.2.4 Calculate the support reaction (RA and RB) and calculated the

theoretical Shear Force at the cut.

140mm RA ‘cut’ RB



W1 = 200g (1.96N)

Figure 2

RA 220mm W1 W2 ‘cut’ RB

260mm

Where ;

W1 & W2 any load between 100g to 500g

Figure 3

RA 220mm W1 ‘cut’ RB

W2400mm

Where ;W1 & W2 any load between 100g to 500g

Figure 4

6.0 RESULT

Mass

*(g)

Load (N) Force

(N) Experimental Shear Force

(N)

Theoretical Shear Force

(N)



0 0 0 0 0

100 0.981 0.6 0.6 0.401

200 1.962 1.2 1.2 0.803

300 2.943 1.8 1.8 1.204

400 3.924 2.3 2.3 1.605

500 4.905 2.8 2.8 2.01

* Use any mass between 200g to 500g

Table 1

NoMass1

(g)

Mass2

(g)

W1

(N)

W2

(N)

Force

(N)

Experimental

Shear Force

(N)

RA (N) RB (N)

Theoretical

Shear Force

(Nm)

2 200 0 1.962 0 - 0.50 - 0.50 2.586 - 0.624 - 0.624

3 200 300 1.962 2.943 2.60 2.60 2.185 2.720 2.720

4 200 300 1.962 2.943 0.70 0.70 1.248 3.657 0.713

Table 2

7.0 DATA ANALYSIS

7.1 For Table 1 (Part 1)

From Figure 1;


W


a ‘cut’

For :Mass, g =100

Load, N =100 x 9.81 / 1000 = 0.981 NForce , N = 0.6 NExperimental shear force , N = displayed forced (shear force at a cut , N ) = 0.6

Theoretical shear force N, Sc = W (L-a) / L

= 0.981 x (0.44 – 0.26) / 0.44

= 0.401 N

For ; Mass, g = 200

Load, N = 200 x 9.81 / 1000 = 1.962 N

Force, N = 1.2 N

Experimental Shear Force, N = Displayed Force

(Shear Force at a cut, N) = 1.2 N

Theoretical Shear Force, N, Sc = W (L – a) / L

= 1.962 x (0.44 – 0.26) / 0.44

= 0.803 N

For ; Mass, g = 300

Load, N = 300 x 9.81 / 1000 = 2.943 N

Force, N = 1.8 N




RA RBL



= 2.943 x (0.44 – 0.26) / 0.44

= 1.204 N

For ; Mass, g = 400

Load, N = 400 x 9.81 / 1000 = 3.924 N

Force, N = 2.3 N




= 3.924 x (0.44 – 0.26) / 0.44

= 1.605 N

Fore ; Mass, g = 500

Load N = 500 x 9.81 / 1000 = 4.905 N Force , N = 2.8 N Experimental shear force , N = displayed force (shear force at a cut , N ) = 2.8 N Theoretical shear force , N, Sc = W (L-a) /L = 4.905x (0.44-0.26 )/0.44 = 2.01 N

7.2 For Table 2 (Part 2)

From Figure 2;

140mm RA ‘cut’ RB



W1 = 200g (1.962N)

Force, N = - 0.9N


(Shear Force at a cut, N) = - 0.9 N

∑M = 0, ∑Fx = 0, ∑Fy = 0

∑MB = 0 ; -2943 (0.58) + RA (0.44) = 0

RA = 3879.41N

∑Fx = 0, ∑Fy = 0 ; RB + 3879.41 –2943= 0

RB = -936.41 N

Theoretical Shear Force, N = - Wa/ L

= - (2943) x (0.14) / 0.44

= - 0.936 N

From Figure 3 ;

RA 220mm W1 W2 ‘cut’ RB

260mm



Force, N = 2.0 N



∑M = 0, ∑Fx = 0, ∑Fy = 0

∑MA = 0 ; RB (0.44) – 2.943(0.26) – 1.962(0.22) = 0

RB = 1.197 / 0.44

RB = 2.720 N

∑Fx = 0, ∑Fy = 0 ; RA – 1.962 – 2.943 + 2.720 = 0

RA = 1.962 – 2.943 – 2.720

RA = 2.185 N

Theoretical Shear Force,

= 0.981 + 1.739

= 2.720 N

From Figure 4 ;

RA 240mm W1 ‘cut’ W2 RB

400mm



Force, N = 0.70 N



∑M = 0, ∑Fx = 0, ∑Fy = 0

∑MB = 0 ; -1.962 (0.22) – 2.943(0.04) + RA (0.44) = 0

RA = 0.549 / 0.44

RA = 1.248 N

∑Fx = 0, ∑Fy = 0 ; RB + 1.248 – 1.962 – 2.943 = 0

RB = 1.962 + 2.943 – 1.248

RB = 3.657 N

Theoretical Shear Force,

= - 0.268 – (-0.981)

= 0.713 N

8.0 DISCUSSION

8.1 Part 1

8.11 Derive equation 1

From Figure 1;

W



a ‘cut’

Let ; ∑MB = 0

( RA x L ) – W ( L –a ) = 0

RA = W ( L –a ) L

Since the force at the cut is equal to the algebraic sum of the force acting

to the left or right of the cut;

Therefore,

SC = RA

Sc = W ( L –a ) L

Let ; ∑MA = 0

( -RB x L ) – ( W x a ) = 0

RB = ( - W x a ) L

Therefore ; SC = ( - W x a ) L

Where, W = Load

a = Cut section from RA

L = Length from RA to RB

This equation is used to determine the value of Shear Force by theory. W

is a load place upon the ‘cut’ section with the length of a. L is total length

from RA to RB.


RA RBL


8.12 Plot a graph, which compare your experimental result to those you

calculated using theory.

Please see graph 1, as attached.

8.13 Comment on the shape of the graph. What does it tell you about how

Shear Force varies due to an increased load?

From the Shear Force versus Load graph we plotted in this experiment, a

linear graph was obtained for both Experimental Shear Force and

Theoretical Shear Force values. Both graphs are linear and go through the

origin (0,0) which tell us that, Shear Force does not exist when no load

was applied on the beam. From the graph, we can notice that, when the

load applied on the beam was increase, the Shear Force will also increase.

This indicate that, Shear Force is linearly proportional (positive) to the

load apply on the beam :

Shear Force α Load

8.14 Does the equation you used accurately predict the behavior of the

beam?

Yes, the equation, Sc = W(L – a) / L that we used in this experiment for

Theoretical Shear Force calculation accurately predict the behavior of the

beam. This is because, from the Graph 1 plotted, we can notice that, when

the load we placed at the beam was increased, the value of Shear Force

also increased. This indicate that, Shear Force is linearly proportional

(positive) to the load apply on the beam.



Example ;

From the experiment, when a 2.453 N load was applied on the beam at the

‘cut’, the Experimental Shear Force obtained was 1.40 N. From the

calculation done for Theoretical Shear Force by using the Sc = W(L – a)/L

equation, the Shear Force we obtain was 1.45 N. This indicates that, this

equation can accurately predict the behaviors of the beam.

8.2 Part 2

8.21 Comment on how the results of the experiments compare with those

calculated using the theory?

From the experiments done by our group, we found that, there is only a

small difference between the values of Experimental Shear Force and the

Theoretical Shear Force. For figure 2 and figure 3, the value of the

Experimental Shear Force is almost the same compare to the Theoretical

Shear Force. While for the figure 4, the value of the Theoretical Shear

Force is higher than the value of the Experimental Bending Moment.

Referring to this results, we conclude that the differences between the

value of the experiment and theory was probably cause by the mistake

done by our group member when taking the value for the force when it

was hang on the beam.

8.22 Does the experiment proof that the shear force at the ‘cut’ is equal to

the algebraic sum of the forces acting to the left or right of the cut. If

not, why?

Yes, the experiment proof that the shear force at the ‘cut’ is equal to the

algebraic sum of the forces acting to the left or right of the cut. This is

because, from the value of W1, W2, RA and RB , we can conclude that,

W1 + W2 = RA + RB

For the example, from data in the table 2,



Figure 2

W1 + W2 = RA + RB

1.962 N + 0 = 2.586 N + (-0.624 N)

= 1.962 N

Figure 3

W1 + W2 = RA + RB

1.962 N + 2.943 = 2.185 N + 2.720 N

= 4.905 N

Figure 4

W1 + W2 = RA + RB

1.962 N + 2.943 = 1.248 N + 3.657 N

= 4.905 N

8.23 Plot the shear force diagram for load cases in Figure 2,3 and 4.

Please see graph 2 and 3 as attached.

8.24 Comment on the shape of the graph. What does it tell you about how

Shear Force varies due to various loading condition?

From GDR Graph for Figure 2 we obtained in Graph 2, we can noticed

that when a loading, -1.962 N is put at the end of the beam (left side of

RA), the value of the shear force cause by this load is negative. Reaction

Force at A is equal to 2.586 N and therefore the total Shear Force at this

point is + 0.624 N. Negative force of -0.624 N at B balances the Shear

Force at A and thus, total Shear Force at B is zero.



From GDR Graph for Figure 3 we obtained in Graph 2, when a loading, -

1.962 N and -2.943 N are both place at the length of 220 mm and 260 mm

from the right side of RA, calculation reveal that reaction force at A is +

2.185 N and reaction force at B is + 2.720 N. The graph also indicates that

Shear Force on the negative part is equivalent to the positive part, that is

equal to zero.

From GDR Graph for Figure 4 we obtained in Graph 3, we can conclude

that, when a loading of 1.962 N and 2.943 N are both place 240 mm and

400 mm from the right side of RA, calculation reveal that reaction force at

A is + 1.248 N and reaction force at B is + 3.657 N. The graph also tells us

that Shear Force on the negative part is equilibrium to the positive part,

that is zero.

From both GDR graph obtained from the Graph 2 and Graph 3, the shape

of the graph is close at the both end of the origin. This indicate that Shear

Force will change according to the load apply to the beam. This happens

to ensure that Shear Force at left side is equal to the Shear Force at the

right side to create equilibrium.

9.0 CONCLUSION

From this experiment, our group managed to examine how shear force

varies with an increasing point load. We also managed to examine how shear

force varies at the cut position of the beam for various loading conditions.

For part one experiment, we conclude that, when the load we place at

beam is increase, the Shear Force will also increase. Thus, we conclude that,

Shear Force is linearly proportional (positive) to the load apply on the beam.



While for the part two experiment, we conclude that, from the GDR graph

draw by our group in this experiment, we noticed that, Shear Force normally will

happen at any point on the beam when a load is apply at the ‘cut’. The result from

the experiment also indicate that Shear Force at the ‘cut’ section is equal to the

forces acting at both right and left side of the ‘cut’ section on the beam.

10.0 REFERENCES

Yusof Ahamad (2001). “Mekanik Bahan Dan Struktur.” Malaysia: Universiti

Teknologi Malaysia Skudai Johor Darul Ta’zim.

R. C. Hibbeler (2000). “Mechanic Of Materials.” 4th. ed. England: Prentice Hall

International, Inc.


Shear Force in a Beam (Edited)1

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