Shear Force and Bending Moment

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SHEAR FORCE AND BENDING MOMENT

GEORGE KENJI PUTRA

0304559

Justin Moo Xian Yuen

Leong Yok Ben

Manish Kumar Sing Domun

Ng Yi Ming

School of Engineering

Taylors University

Malaysia

12 May 2014

1

TABLE OF CONTENTS

ABSTRACT 2

1.0 INTRODUCTION 2

2.0 EXPERIMENTAL DESIGN 2

2.1 Apparatus 3

2.2 Methods 3

2.3 Procedures 3

3.0 RESULTS AND CALCULATIONS 3

4.0 DISCUSSIONS 15

5.0 CONCLUSION 16

REFERENCES 16

2

ABSTRACT

The objective of this lab session is to acquire the experimental values of bending moment and

shear force that acted on simply supported beam that has some points to hang the load. After

conducting this experiment and analyzing the data, the results of experimental calculation will be

compared with theoretical calculation.

1.0 INTRODUCTION

Shear force is the internal resistance created in beam cross sections, in order to balance transverse

external load acting on beam, while bending moment is bending effect due to the forces that act

on the beam. There are several types of beams, such as: cantilever beam, simply supported beam,

overhanging beam, rigidly fixed beam and continuous beam. In this particular experiment, we were

dealing with simply supported beam.

Having better understanding about these two things are very important in engineering field, since

nowadays development of construction is growing rapidly.

In theory, when the loads apply to the beam, the beam hold those loads by giving off internal

stresses and strains inside its interior. The internal force that acts vertically to the longitudinal axis

is the shear force. The bending moment is the internal couple forces of the beam. The summation

of those internal forces need to be in equilibrium state in order to hold the external force that

applied to the beam.

2.0 EXPERIMENTAL DESIGN

Figure 1. Rough Design for Shear Force and Bending Moment Experiment

. . . . .

Aluminum

Profile Frame

Simply

Supported Beam

Console Box Loading Point

Holder

A B C

3

2.1 Apparatus

Aluminum profile frame

Simply supported beam (0.8 m length)

Loading Point Holder

Loads

Hangers

Console

2.2 Methods

Calibrate the console box and the beam before taking the reading. Put the loads to the hanger, and hang it

on the point holder. Wait until the values shown on console box are stable, then only take the reading. For

the next reading, make sure the console box is recalibrated back to zero and make sure the beam is flat by

using water level ruler.

2.3 Procedures

1. Level the beam by using water level ruler.

2. Set the console box to zero.

3. Add 2 N load to the hanger, and hang it at loading point holder A (0.095 m from left).

4. Write down the reading that shown on console box.

5. Repeat the step 14 by using load of 4 N and 6 N.

6. Repeat the step 1-5 by hanging the load at loading point holder B (0.245 m from left).

7. Hang the loads at both point A and B, with combination load of 2-2 N, 4-4 N, and 6-6

N, and dont forget to do step 1 and 2 before start the experiment.

8. Write down the reading separately according to which loading point holder the loads

were.

3.0 RESULTS AND CALCULATIONS

Table 1. Experimental Values for Loading Point Holder A (Distance: 0.095 m)

Load (N) Shear Force (N) Bending Moment (Nm)

2 0.202 0.100

4 0.405 0.200

6 0.603 0.160

Table 2. Experimental Values for Loading Point Holder B (Distance: 0.245 m)


2 0.505 0.090

4 1.013 0.280

6 1.534 0.730

4

Table 3. Experimental Values for Double Loading Point Holder (A and B)

Load 1 (N) Load 2 (N) Shear Force (N) Bending Moment (Nm)

2 2 0.708 0.360

4 4 1.422 0.700

6 6 2.135 0.800

Drawing the free body diagram will make our life easier to calculate the theoretical value of shear

force and bending moment. Look at next page for free body diagram and also calculation for the

theoretical values of shear force and bending moment. In this case, 4 N is used to be the example.

4 N Load Calculation Example at Point A

Figure 2. Free Body Diagram for 4 N Load at Point A (0.095 m)

Applying equilibrium conditions:

Fy = 0, taking up as positive

Ra + Rb 4 = 0 4 = Ra + Rb

MA = 0, taking clockwise as positive

4(0.095) Rb(L) = 0 0.38 0.8Rb = 0 Rb = 0.475 N

4 = Ra + 0.475

Ra = 3.525 N

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Section the beam right before the force (4 N).

Figure 3. Sectioned Beam for 0 x 0.095 m


Ra V = 0 V = 3.525 N

M = 0, taking clockwise as positive (at cut-off section)

Ra(x) - M = 0

M = 3.525(x)

At x = 0 m; M = 0

At x = 0.095 m; M = 0.335 Nm

Then make a section again, this time 0.095 x 0.8 m.

Figure 4. Sectioned Beam for 0.095 x 0.800 m


Ra 4 V = 0 V = -0.475 N

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Ra(x) 4(x 0.095) M = 0

At x = 0.095 m; 3.525(0.095) 4(0.095-0.095) = M M = 0.335 Nm

At x = 0.800 m; 3.525(0.800) 4(0.800-0.095) = M M = 0 Nm

The shear force and bending moment diagram for calculation above would be like figures below:

Figure 5. Shear Force Diagram for 2 N Load at Point A (0.095 m)

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Figure 6. Bending Moment Diagram for 2 N Load at Point A (0.095 m)

4 N Load Calculation Example at Point B

Figure 7. Free Body Diagram for 4 N Load at Point B (0.245 m)



Ra + Rb 4 = 0

8

4 = Ra + Rb


4(0.245) Rb(L) = 0 0.98 0.8Rb = 0 Rb = 1.225 N

4 = Ra + 1.225

Ra = 2.775 N




Ra V = 0 V = 2.775 N


Ra(x) - M = 0

M = 2.775(x)

At x = 0 m; M = 0

At x = 0.245 m; M = 0.680 Nm

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Ra 4 V = 0 V = -1.225 N


Ra(x) 4(x 0.245) M = 0

At x = 0.245 m; 2.775(0.245) 4(0.245-0.245) = M M = 0.680 Nm

At x = 0.800 m; 2.775(0.800) 4(0.800-0.245) = M M = 0 Nm

The shear force and bending moment diagram for calculation above would be like figures on the

next page:

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Figure 10. Shear Force Diagram for 4 N Load at Point B (0.245 m)

Figure 11. Bending Moment Diagram for 4 N Load at Point B (0.245 m)

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4 N Load Calculation Example at Point A and B (Double Point)

Figure 12. Free Body Diagram for 4-4 N Load at Point A and B (Double Point)



Ra + Rb 4 4 = 0 8 = Ra + Rb


4(0.095) + 4(0.245) Rb(L) = 0 0.38 + 0.98 0.8Rb = 0 Rb = 1.700 N

8 = Ra + 1.700

Ra = 6.300 N



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Ra V = 0 V = 6.300 N


Ra(x) - M = 0

M = 6.300(x)

At x = 0 m; M = 0

At x = 0.095 m; M = 0.599Nm




Ra 4 V = 0 V = 2.300 N


Ra(x) 4(x 0.095) M = 0

At x = 0.095 m; 6.300(0.095) 4(0.095-0.095) = M M = 0.599 Nm

At x = 0.245 m; 6.300(0.245) 4(0.245-0.095) = M M = 0.944 Nm

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Once again make a section, this time 0.245 x 0.800 m.

Figure 15. Sectioned Beam for 0.245 x 0.800m


Ra 4 4 V = 0 6.300 8 = V V = -1.700 N

M = 0, taking clockwise as positive

Ra .x - 4(x 0.095) 4(x 0.245) M = 0 6.300(x) 4(x 0.095) 4(x 0.245) = M

At x = 0.245 m; 6.300(0.245) 4(0.245 0.095) 4(0.245 0.245) M = 0.944 Nm

At x = 0.800 m; 6.300(0.800) 4(0.800 0.095) 4(0.800 0.245) = M M = 0 Nm

The shear force and bending moment diagram for calculation above would be like figures on the

next page:

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Figure 15. Shear Force Diagram for 4-4 N Load at Point A and B (Double Point)

Figure 16. Bending Moment Diagram for 4-4 N Load at Point A and B (Double Point)

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Table 4. Theoretical Values for Loading Point Holder A (Distance: 0.095 m)


2 -0.238 0.167

4 -0.475 0.335

6 -0.712 0.502

Table 5. Theoretical Values for Loading Point Holder B (Distance: 0.245 m)


2 1.388 0.340

4 2.775 0.680

6 4.162 1.020

Table 6. Theoretical Values for Double Loading Point Holder (A and B) with 2-2 N Load

Distance (m) Shear Force (N) Bending Moment (Nm)

0 3.150 0

0.095 1.150 0.299

0.245 -0.850 0.472

0.800 -0.850 0



0 6.300 0

0.095 2.300 0.599

0.245 -1.700 0.944

0.800 -1.700 0



0 9.450 0

0.095 3.450 0.898

0.245 -2.550 1.415

0.800 -2.550 0

4.0 DISCUSSIONS

The tables and the graphs above show that the values of the load is proportional to the values of

shear force as well as bending moment. Which means, the bigger the load applied, the bigger shear

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force and bending moment will produced. This statement is valid to both conditions, one point

loaded and also double point loaded.

Nonetheless, the experimental results are not really accurate since the console box somehow were

only showing one direction (in this case is positive). As we all can see on theoretical values, there

are some negative forces which means the force acting downwards. If only the console box worked

how it supposed to be, the experimental values would not have big different compared to the

theoretical ones.

Other than that, the shocks from another table that we sharing with affected the number that coming

out of the console box.

5.0 CONCLUSION

In conclusion, this experiment show us the values of theoretical calculation are larger than the

actual values that measured by a console box. Both single loaded and double loaded experienced

the same condition. In outline, from the experiment, when the load is added, the number of shear

force and bending moment will get higher too.

REFERENCES

1. Analysis of Beams | Shear Force & Bending Moment Diagram ~ Learn Engineering.

[ONLINE] Available at: http://www.learnengineering.org/2013/08/shear-force-bending-

moment-diagram.html. [Accessed 11 May 2014].

2. Shear Force and bending diagrams. [ONLINE] Available at:

http://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html. [Accessed 11

May 2014].

3. Civil Engineering: TYPES OF BEAMS & TYPES OF LOADINGS. [ONLINE] Available

at: http://civilengineerworks.blogspot.com/2011/12/types-of-beams-types-of-

loadings.html. [Accessed 11 May 2014].

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Shear Force and Bending Moment