FE1073 Lab Report C1

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NANYANG TECHNOLOGICAL UNIVERSITY

First Year Common Engineering Course

FE1072: Laboratory Experiment

Protective Engineering Laboratory (N1.1-B5-02)

FORMAL REPORT

Experiment C1: Equilibrium and Elasticity

Name: YEO SHI JING JACKIE

Matric No.: 083642A03

Group: AL08

Date: 20 FEB 2009



Page 2 of 24

Abstract

This experiment consisted of two parts. In the first experiment, equilibrium of concurrent

force systems was observed. This experiment used pulleys and hanging masses to setup 2

forces. Equilibrant force was determined from the setup. This force was used to compare with

the resultant force of the 2 forces. In theory, the 2 forces are equal in magnitude and opposite

in direction, so they cancel out each other. However, there may be possible sources of error

which result in equilibrant force do not exactly balance the resultant force. In the later part of

the report, we will discuss the possible sources of error in the measurements and construction,

and possible ways to improve it.

The second experiment was conducted to determine 3 results: strain in a truss member, elastic

modulus of material involved, and truss stiffness. During the experiment, strain increments

on one of the truss member while loading and unloading (increase P and decrease P) were

recorded. From the result, a graph of load P versus strain increment was plotted (See Page

24 of the report for the graph). It was found out that the load P was linearly proportional to

the strain increment . This showed that the truss member had undergone linear elastic

deformation, which meant the truss member obeyed Hooke’s law ( = ). The slope of the

graph was determined, and hence the elastic modulus and truss stiffness. The calculations

were shown on Page 12 of the report.



Page 3 of 24

Introduction

The purpose of the first experiment was to measure force vectors, force resultants, and

observe equilibrium of concurrent force systems.

Vector is defined both by its direction, the direction of arrow, and by its magnitude, which is

proportional to the length of arrow. An example is shown in Figure 1, a vector ⃗ makes an

angle with the horizontal axis (Direction), with a length λ (Magnitude).

Vector forces acting on the same point of the object are called concurrent forces. The

resultant forces can be determined by adding the vector forces together using the

parallelogram method (as shown in Figure 2).

Another vector shown in Figure 2 is the equilibrant of 1 and 2. is the force needed to

exactly offset the combined effect of 1and 2, which is . has the same magnitude as

, but is in the opposite direction.

Fig. 1 shows a vector ⃗ with length λ and

makes an angle with the horizontal axis.

Fig. 2 shows the resultant

vector force as the

result of 1 + 2. It also

shows, , the equilibrant

force which has the same

magnitude as but is inthe opposite direction.



Page 4 of 24

The purpose of the second experiment was to measure the deformation (strain) in a truss

member, determine the modulus of elasticity of the material involved and also determine the

stiffness of the truss model.

A truss (as shown in Figure 3(a)) is a structure composed of slender members joined together

at their end points. Each truss member acts as a two-force member. If the force tends to

elongate the member, it is a tensile force (T); whereas if it tends to shorten the member, it is a

compressive force (C).

In a statistically determinate truss, the forces (tension or compression) in all the members can

be calculated by considering equilibrium at joints.

The truss shown in Figure 3(a) has pin-joints at B, D and G. Deformations in DB are

neglected since the member DB is very rigid when compared with GB. There are two

unknowns acting on GB (Tensile force in GB and Compression force along DB) and

we can write two equilibrium equations to solve for and . However, we will be

interested only in . By equilibrium at joint B (Figure 3(b)), we obtain

=

sin =

ℎ/=

ℎ Equation (1)

ℎ

D

B

G

Fig. 3 (a) shows a

truss which

composed of

slender members

joined together at

their end points.

P

B

P

Fig. 3 (b) shows all

forces acting at joint B



Page 5 of 24

For truss member GB, we define normal strain (Consequences of the load) as deformation

per unit length,

=

where is the total deformation between the ends of the member (See Figure 5).

Note: (a) is the original length before loading.

(b) Strain is always dimensionless.

=

=

ℎ

Given a cross sectional area A of member GB, the stress (force per

unit area) in member GB will be

Note: (a) A is the original area before loading.

(b) If force is in Newton (N) and the unit for length is in

millimetre (mm), the stress will be in / or /

or .

Equation (2)

Equation (3)

Fig. 5 shows the totaldeformation between

the ends of the

member GB.

Fig. 4 shows a small portion

of the member GB. It

indicates the variable and

that are needed to

calculate the stress .



Page 6 of 24

Most engineering structures are designed to function within the linear elastic range, i.e., the

stress is linearly proportional to the strain ,

=

This relation is known as Hooke’s law. Figure 6 shows a graph plotted against . The

coefficient E (Gradient of the graph) is called the modulus of elasticity (or Young’s

modulus). It is a measurement of stiffness of the material involved.

Substitute Equation (2) and Equation (4),

=

=

ℎ

Note: (a) The unit for E is / or .

(b) The strain equals to the measured strain increment ′ multiplied by a conversion factor

due to the instrumentation.

Fig. 6 shows graph plotted against

. The gradient of the graph is called

the modulus of elasticity.

Equation (4)

Equation (5)



Page 7 of 24

Another quantity which is of interest in this case is the deflection at joint B where force P is

applied. When the deformations in DB can be neglected, B will moved to B’ due to the

extension in the member GB and a rotation of GB about G (See Figure 7). Since the

deflections are small compared with the original length of the members, the circular arcs BB’

and LB’ due to rotation of DB and GL respectively, can be approximated to straight lines. It

follows that the right-angled triangle BLB’, shown in Figure 7, can represent the relationship

between and deflection of B. Therefore,

=

sin =

ℎ/ =

2

ℎ

Define a stiffness coefficient (truss stiffness), k, for the vertical deflection at B, as

=

=

ℎ

2

By plotting the relationship between P and the measured strain for a range of applied P

values, the Young’s modulus E and the stiffness coefficient k can be determined from the

experiment using Equations (5) and (7).

Equation (6)

Equation (7)

BD

G

L

B’

Fig. 7 shows deflection at joint B

where force P is applied.



Page 8 of 24

Equipments

Equipment was set up as in Experiment No. C1 laboratory manual. Refer to page 3 (For

experiment 1) and 6 (For experiment 2) of the laboratory manual.

Experimental Procedure

Experiment 1:

1) Use pulleys and hanging masses (1 = 55, 2 = 105) to setup the equipment as

shown in Figure 8, so that two known forces, 1 (=1) and 2 (=2), are pulling

the force ring.

2)

Use holding pin to prevent the ring from moving. The holding pin provides a force,

, that is exactly opposite to the resultant of 1 and 2.

1 2

1

2

1

2

Spring Balance

Pulley 1

Pulley 3

Pulley 2

Force Ring

HoldingPin

Figure 8 shows experiment 1 setup.

Experiment Board

Degree ScaleZero-degree line



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3) Adjust the spring balance to determine the magnitude of . As shown in Figure X,

keep the spring balance vertical and use Pulley 3 to direct the force from the spring in

the desired direction. Move the spring balance towards or away from the pulley to

vary the magnitude of the force. Adjust Pulley 3 and the spring balance so that the

holding pin is centred in the force ring.

Note: To minimize the effects of the friction in the pulleys, tap as needed on the

Experiment Board each time you re-position any component. This will help the force

ring come to its true equilibrium position.

4)

Record the values of the hanging masses (1 = 55, 2 = 105); the magnitude of

1, 2, (in Newton); the angles 1, 2and that each vector makes with respect

to the zero-degree line on the degree scale (See Figure 8). The results are shown on

page 21 of the report.

5) Change the hanging masses to 1 = 135, 2 = 205 and repeat step (1) to (4) one

more time.



Page 10 of 24

Experiment 2:

1)

Before the experiment is carried out, measurements of all necessary dimensions are

made. Use the calipers to measure the breadth and the thickness of the truss member

GB, and compute its cross sectional area. Repeat the procedure 2 more times along

the truss member GB, and find the average cross sectional area of GB.

2) Use a ruler to measure the value of (The length of truss member GB), and ℎ (The

vertical length from joint G to the member DB).

Note: The measurement, , should be measured from the joint point G to the other one, B,

whereas the measurement, ℎ, should be measured from the joint point G to the line that cuts

through both joint point D and B (See Figure 9).

3) Before the weights are loaded, check the strain monitoring equipment is zeroed

(′ = 0). Then load a weight of 10N (which means P = 10N). Wait for the reading to

be stabilised. Record the value of ′ , read off from the strain monitoring machine.

4) Load another 10N weight (which means P = 20N) and record the value of ′ .

5) Repeat step (4) until P = 60N. The results are shown on page 23 of the report.

6) Now unload one 10N weight (which means P = 50N) and record the value of ′ .

Figure 9 shows how to measure

and ℎ for experiment 2 setup.

Line that cuts through both

joint point D and B



Page 11 of 24

7) Repeat step (6) until P = 0N. The results are shown on page 23 of the report.

8) Compute the values of strain for both loading and unloading.

Strain = strain increment ′ × conversion factor (CF)

Note: The conversion factor (CF) for this experiment = 0.5

Results

Experiment 1:

Appendix A1 shows two tables. Table 1 shows all the recorded values of 1, 2, , 1,

2and for the case where 1 = 55, 2 = 105. Table 2 shows all the recorded values

of 1, 2, , 1, 2and for the case where 1 = 135, 2 = 205. For each case, the

recorded values are used to construct 1, 2, with an appropriate scale (For 1st case:

10cm/Newton, For 2nd

case: 5cm/Newton). are also drawn on the same diagram for each

case using the parallelogram method. These diagrams are shown in Appendix A2.

From the diagrams,

Case 1: Length of = 6.9cm

∴ Magnitude of =6.9

10= 0.69

Case 2: Length of = 6.3cm

∴ Magnitude of =6.3

5= 1.26

From the above calculations, the equilibrant force vector, , does not exactly balance the

resultant vector, , for each case.



Page 12 of 24

Percentage error: −ℎ

ℎ × 100%

Case 1: Percentage error =0.70−0.69

0.69× 100% = 1.45%

Case 2: Percentage error =1.20−1.26

1.26× 100% = −4.76%

Experiment 2:

Table 3 in Appendix B1 shows the measurements of the breadth and the thickness along the

truss member GB, and the average cross sectional area is found to be 81.72. Also from

the experiment, h is determined to be 427 mm and l is determined to be 450 mm.

Another table, Table 4, shows all the recorded strain increments , while loading and

unloading. The table also includes computed strain , which can be found by

Strain = strain increment ′ × conversion factor (CF)

Note: The conversion factor (CF) for this experiment = 0.5

A graph of load P versus strain increment for the truss model is plotted and it is presented

in Appendix B2 of the report. From the graph, we can see that load P was linearly

proportional (Straight line graph) to the strain increment .

Slope of the straight line graph =

=

60

12×10−6= 5 × 1 06 N

From the slope of the graph, the elastic modulus E of the truss member GB, and the stiffness

k for the vertical deflection at B can be calculated.

Elastic modulus E =

ℎ= 5 × 1 06 ×

450

81.7×427= 64.5 /2

Truss stiffness k =

ℎ2 = 5 × 1 06 × 427

4502 = 10500 /



Page 13 of 24

Discussion

Experiment 1:

From the diagrams drawn in Appendix A2 and the calculations on Page 11 of the report, the

equilibrant force vector, , does not exactly balance the resultant vector, , for each case.

In next paragraph, we are going to discuss some possible sources of error in the

measurements and constructions that had caused the above result.

All experiments done by students, teachers, and even scientists are not perfect. There are

bound to have errors involved in the experiments. Some possible sources are:

1) There are friction between the contact surface of the string and pulleys. Although

tapping the experiment board minimizes the effect of friction, it does not remove the

effect of friction completely. Hence, the reading obtained from the spring balance is

not what we expected.

2) The portion of string that connects the spring balance is not parallel to the portion of

the string that hangs the 1 or 2 (See Figure 10). Hence, the spring balance may be

slanted at an angle, which results in an inaccurate reading of .

Fig. 10 shows the portion of string

that connects the spring balance

is not parallel to the portion of the

string that hangs the 1 or 2



Page 14 of 24

3) Even if it is parallel, there may be case where the spring balance is not upright (See

Figure 11), hence results in an inaccurate reading of .

4) A spring balance does not retain its accuracy permanently, for no matter how

carefully it is handled, the spring very gradually uncoils even though its limit of

elasticity has not been exceeded. Hence if the spring balance had been use dozens of

times for measurement, we may not obtain a very accurate reading of . Accuracy in

this experiment is important because the force to be measured, , is very small (<

2N).

Further Discussion:

For the given masses 1, 2 and the measured angles 1, 2, calculate the equilibrant

and its direction using the equilibrium conditions. Compare the calculated results with the

measured values.

Case 1:

1 : X-component = 0.54 cos168 = −0.5282

Y-component = 0.54 sin 168 = 0.1123

Fig. 11 shows spring balance is

not upright, hence results in an

inaccurate reading of .



Page 15 of 24

2 : X-component = 1.03 cos 26 = 0.9258

Y-component = 1.03 sin 26 = 0.4515

: X-component = 0.9258 – 0.5282 = 0.3976N

Y-component = 0.4515 + 0.1123 = 0.5638N

Under equilibrium conditions, | | = | | = 0.690

= 180° − = 180 − tan−1 0.5638

0.3976= 125°

Comparison:

() ()

Calculated 0.69 125

Measured 0.70 126

Case 2:

1 : X-component = 1.32 cos168 = −1.291

Y-component = 1.32 sin 168 = 0.2744

2 : X-component = 2.01 cos 26 = 1.807

Y-component = 2.01 sin 26 = 0.8811

: X-component = 1.807– 1.291 = 0.516N

Y-component = 0.2744 + 0.8811 = 1.156N

Under equilibrium conditions, | | = | | = 1.26

= 180° − = 180 − tan−1 1.156

0.516= 114°



Page 16 of 24

Comparison:

() ()

Calculated 1.26 114

Measured 1.20 117

Discuss the possible ways to improve the measurement accuracy.

1) Besides tapping the experiment board, we can also lubricate the contact surface of the

pulleys to minimize the effect of friction. This will further minimize error (1) stated in

Page 13 of the report.

2)

Before the experiment starts, place a meter ruler vertically upright in front of the

board (Shown in Figure 12). This is a check to make sure that the spring balance is

upright. Also when you conduct this check, make sure the board and the ruler are

placed on the same flat surface. This will eliminate error (2) & (3) stated in Page 13

&14 of the report.

3) Instead of using a spring balance, use a high tech digital force gauge to obtain a much

accurate reading of . This will minimize error (4) stated in Page 14 of the report.

Me

t

e

r

R

u

l

e

r

Fig. 12 shows a meter

ruler placed vertically

upright in front of the

board to eliminate error

(2) & (3).



Page 17 of 24

Experiment 2:

Discussion:

What is the significance of observed strains fitting into a straight line in the plot of P vs. ?

• Fitting all the points into a straight line signifies load P is linearly proportional to the

strain increment . This shows that the truss member had undergone linear elastic

deformation, which means the truss member obeys Hooke’s law ( = ).

If member GB is replaced by another member of the same length but different cross sectional

area, will the tensile force be different under the same load P (assuming small

deformation anyway)? What about the deflection at point B?

• From equation (1),

=

sin =

ℎ/=

ℎ

depends on 3 factors, P, l, and h. None of these factors are related to the breadth

and the thickness of member GB. Hence, tensile force will be the same under the

same load with different cross sectional area.

• From equation (6),

=

sin =

ℎ/=

2

ℎ

depends on 3 factors, , l, and h. l, and h are not related to the breadth and the thickness

of member GB. However, is dependant on cross sectional area. Based on equation (5),

=

=

ℎ



Page 18 of 24

Rearranging equation (5),

=

=

ℎ

Note that E is a constant, and P, l, and h are not related to the breadth and the thickness of

member GB. This shows that is dependent on cross sectional area and hence the deflection

at point B will be different under the same load but with different cross sectional area A of

member GB.

Further Discussion:

Discuss qualitatively how the rigidity of member DB affects the results of the experiment?

• The rigidity of member DB indicates whether there is deformation in DB. In this

experiment, DB is rigid with respect to GB. DB is non-deformable – that is, for ideal

rigid body, the relative locations of all particles of which the object is composed

remain constant. Hence, there will not be any deflection shown (in DB with respect to

GB) at joint B which will make our calculation much simple. However, if the DB

used in the experiment is not rigid, then the deformations in DB cannot be neglected

and hence deflections will be significant and we will have another unknown .

Hence, Equation (6) (on Page 7 of the report) is not valid because circular arcs BB’

and LB’ due to rotation of DB and GL respectively, cannot be approximated to

straight lines. Therefore, making our calculation much more difficult.



Page 19 of 24

Why is it preferable to measure the strain along GB rather than the deflection with a dial

gauge mounted on the truss at B?

•

A dial gauge is able to detect smallest dimensional variations (Up to millimetres).

However, the dial gauge is not preferable. Due to the moving parts (Levers), friction

is generated within the gauge, hence reducing the accuracy of the reading obtained.

The accuracy of the reading is important because the deflection of B is very small in

this experiment.

• It can also generate error due to parallax with the dial gauge as pointer moves over a

fixed scale. However we do not have this problem if we use strain monitoring

equipment (digital) to measure strain along GB.

• The dial gauge is also sensitive to small vibration because the mechanisms in these

gauges have more inertia. It takes a longer time to get an accurate reading because the

weights tend to sway a little in the air while we were loading and unloading (increase

P and decrease P respectively) during the experiment, and hence the reading fluctuates

before it is stabilized.



Page 20 of 24

Conclusion

Experiment 1:

In theory, the equilibrant vector force exactly offsets the combined effect of 1and 2,

which is the resultant force . However, in this experiment we observed that does not

exactly balance out . This may due to possible sources of error in our measurements and

constructions (This had been discussed in Page 13 &14 of the report).

Experiment 2:

From the graph, it was found out that the load P was linearly proportional to the strain

increment . This showed that the truss member had undergone linear elastic deformation,

which meant the truss member obeyed Hooke’s law ( = ). The slope of the graph was

determined,

Slope of the straight line graph = = 60

12×10−6 = 5 × 1 06 N

And hence, the elastic modulus

Elastic modulus E =

ℎ= 5 × 1 06 ×

450

81.7×427= 64.5 /2

and truss stiffness.

Truss stiffness k =

ℎ

2= 5 × 106 ×

427

4502= 10500 /

References

1) Laboratory Manual Experiment No. C1.

2) R.A. Serway & J.W. Jewett, 2008, "Physics for Scientists and Engineers with Modern

Physics", 7th Edition, Thomson Brooks/Cole Publishing.

3) William D. Callister, Jr., 2007, “Materials Science and Engineering, An

Introduction”, 7th Edition, John Wiley & Sons, Inc.

4) A/P Tan Ming Jen, 2009, FE1005 Materials Science Lecture Notes.



Page 21 of 24

Appendices

Table 1: (Case 1) 1 = 55, 2 = 105

1 (g)

2 (g)

1 (degree)

2 (degree)

1 = 1 (N)

2 = 2 (N)

(N)

(degree)

55 105 168 26 0.54 1.03 0.70 126

Table 2: (Case 2) 1 = 135, 2 = 205

1 (g)

2 (g)

1 (degree)

2 (degree)

1 = 1

(N)

2 = 2

(N) (N)

(degree)

135 205 168 26 1.32 2.01 1.20 117

Appendix A1



Page 22 of 24

Appendices

Appendix A2



Page 23 of 24

Appendices

Table 3: Cross sectional area of truss member GB

1 2 3

Breadth (mm) 12.75 12.75 12.70

Thickness (mm) 6.40 6.45 6.40

Area (2) 81.6 82.2 81.3

Average Area, A =81.6+82.2+81.3

3= 81.72

Other measurements: h = 427mm l =

P

(N)

450mm

Table 4:

Strain Increment

Loading Unloading

, (× 10−6) (× 10−6) , (× 10−6) (× 10−6)

0 0 0 0 0

10 4 2 4 2

20 8 4 8 4

30 12 6 12 6

40 16 8 16 8

50 20 10 20 10

60 24 12 24 12

Appendix B1



Appendices

Appendix B2

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