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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 1 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 ๐๐ 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 ๐๐. This showed that the truss member had undergone linear elastic

deformation, which meant the truss member obeyed Hookeโs law (๐๐ = ๐ธ๐ธ๐ธ๐ธ). 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 2 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 ๐น๐นโ makes an angle ๐๐ with the horizontal axis (Direction), with a length ฮป (Magnitude).

Fig. 1 shows a vector ๐น๐นโ with length ฮป and makes an angle ๐๐ with the horizontal axis.

Vector forces acting on the same point of the object are called concurrent forces. The resultant forces ๐ญ๐ญ๐๐ can be determined by adding the vector forces together using the parallelogram method (as shown in Figure 2).

Fig. 2 shows the resultant vector force ๐ญ๐ญ๐๐ as the result of ๐ญ๐ญ1 + ๐ญ๐ญ2 . It also shows, ๐ญ๐ญ๐๐ , the equilibrant force which has the same magnitude as ๐ญ๐ญ๐๐ but is in the opposite direction.

Another vector ๐ญ๐ญ๐๐ shown in Figure 2 is the equilibrant of ๐ญ๐ญ1 and ๐ญ๐ญ2 . ๐ญ๐ญ๐๐ is the force needed to exactly offset the combined effect of ๐ญ๐ญ1 and ๐ญ๐ญ2 , which is ๐ญ๐ญ๐๐ . ๐ญ๐ญ๐๐ has the same magnitude as ๐ญ๐ญ๐๐ , but is in the opposite direction.

Page 3 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 ๐น๐น๐บ๐บ๐บ๐บ in GB and Compression force ๐น๐น๐ท๐ท๐ท๐ท along DB) and

we can write two equilibrium equations to solve for ๐น๐น๐บ๐บ๐บ๐บ and ๐น๐น๐ท๐ท๐ท๐ท . However, we will be interested only in ๐น๐น๐บ๐บ๐บ๐บ . By equilibrium at joint B (Figure 3(b)), we obtain

G

Fig. 3 (a) shows a truss which composed of slender members joined together at their end points.

D

โ

๐น๐น๐บ๐บ๐บ๐บ =

๐๐

sin ๐๐

๐๐

= โ /๐๐ =

๐๐๐๐

Equation (1)

โ

๐น๐น๐ท๐ท๐ท๐ท

๐๐

๐น๐น๐บ๐บ๐บ๐บ

๐๐

B

P ๐๐

B

Fig. 3 (b) shows all forces acting at joint B

P Page 4 of 24

Given a cross sectional area A of member GB, the stress (force per unit area) in member GB will be

๐๐ =

๐น๐น๐บ๐บ๐บ๐บ ๐๐๐๐ = ๐ด๐ด ๐ด๐ดโ

Equation (2)

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

Fig. 4 shows a small portion of the member GB. It indicates the variable ๐น๐น๐บ๐บ๐บ๐บ and ๐ด๐ด that are needed to calculate the stress ๐๐.

(b) If force ๐น๐น๐บ๐บ๐บ๐บ is in Newton (N) and the unit for length is in

millimetre (mm), the stress ๐๐ will be in ๐ต๐ต/๐๐๐๐๐๐ or ๐ด๐ด๐ด๐ด/๐๐๐๐

or ๐ด๐ด๐ด๐ด๐ด๐ด.

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

๐๐ =

๐ฟ๐ฟ ๐บ๐บ๐บ๐บ

Equation (3)

๐๐

where ๐ฟ๐ฟ๐บ๐บ๐บ๐บ is the total deformation between the ends of the member (See Figure 5). Note: (a) ๐๐ is the original length before loading. (b) Strain is always dimensionless.

๐๐

Fig. 5 shows the total deformation between the ends of the member GB.

Page 5 of 24

Most engineering structures are designed to function within the linear elastic range, i.e., the stress ๐๐ is linearly proportional to the strain ๐๐,

๐๐ = ๐ธ๐ธ๐ธ๐ธ

Equation (4)

This relation is known as Hookeโs law. Figure 6 shows a graph plotted ๐๐ against ๐๐. The coefficient E (Gradient of the graph) is called the modulus of elasticity (or Youngโs modulus). It is a measurement of stiffness of the material involved.

Fig. 6 shows graph plotted ๐๐ against ๐๐. The gradient of the graph is called the modulus of elasticity.

Substitute Equation (2) and Equation (4),

๐ธ๐ธ =

๐๐ ๐๐

Note: (a) The unit for E is ๐๐๐๐/๐๐๐๐๐๐ or ๐ฎ๐ฎ๐ฎ๐ฎ๐ฎ๐ฎ.

=

๐๐ ๐๐

๐๐ ๐ด๐ดโ

Equation (5)

(b) The strain ๐๐ equals to the measured strain increment ๐๐ โฒ multiplied by a conversion factor due to the instrumentation.

Page 6 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 Bโ due to the extension ๐ฟ๐ฟ๐บ๐บ๐บ๐บ 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 BBโ and LBโ due to rotation of DB and GL respectively, can be approximated to straight lines. It follows that the right-angled triangle BLBโ, shown in Figure 7, can represent the relationship between ๐ฟ๐ฟ๐บ๐บ๐บ๐บ and deflection ๐ฟ๐ฟ of B. Therefore, ๐ฟ๐ฟ

G

๐๐๐๐

๐ฟ๐ฟ = sin๐บ๐บ๐บ๐บ๐๐ = โ /๐๐ =

๐๐ 2 ๐๐

Equation (6)

โ

Fig. 7 shows deflection at joint B where force P is applied.

๐๐

D

B ๐ฟ๐ฟ

Bโ

๐ฟ๐ฟ๐บ๐บ๐บ๐บ

L

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

๐๐ =

๐๐ ๐ฟ๐ฟ

=

๐๐ โ

๐๐ ๐๐ 2

Equation (7)

By plotting the relationship between P and the measured strain ๐๐ for a range of applied P

values, the Youngโs modulus E and the stiffness coefficient k can be determined from the experiment using Equations (5) and (7).

Page 7 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:

Spring Balance

Pulley 1

๐๐1

Force Ring

๐ญ๐ญ1

๐๐1

๐ญ๐ญ2

๐๐2

Degree Scale

Holding Pin

Pulley 2

๐ญ๐ญ๐๐

๐๐๐๐

Zero-degree line

๐๐2

Experiment Board

Pulley 3

Figure 8 shows experiment 1 setup.

1) Use pulleys and hanging masses (๐๐1 = 55๐๐, ๐๐2 = 105๐๐) to setup the equipment as

shown in Figure 8, so that two known forces, ๐ญ๐ญ1 (=๐๐1 ๐๐) and ๐ญ๐ญ2 (=๐๐2 ๐๐), are pulling the force ring.

2) Use holding pin to prevent the ring from moving. The holding pin provides a force, ๐ญ๐ญ๐๐ , that is exactly opposite to the resultant of ๐ญ๐ญ1 and ๐ญ๐ญ2 .

Page 8 of 24

3) Adjust the spring balance to determine the magnitude of ๐ญ๐ญ๐๐ . 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 (๐๐1 = 55๐๐, ๐๐2 = 105๐๐); the magnitude of ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ (in Newton); the angles ๐๐1 , ๐๐2 and ๐๐๐๐ 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 ๐๐1 = 135๐๐, ๐๐2 = 205๐๐ and repeat step (1) to (4) one more time.

Page 9 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 ๐๐ (The length of truss member GB), and โ (The vertical length from joint G to the member DB).

Figure 9 shows how to measure ๐๐ and โ for experiment 2 setup.

Line that cuts through both joint point D and B

Note: The measurement, ๐๐, should be measured from the joint point G to the other one, B,

whereas the measurement, โ, 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 (๐๐ โฒ = 0). Then load a weight of 10N (which means P = 10N). Wait for the reading to

be stabilised. Record the value of ๐๐ โฒ , read off from the strain monitoring machine.

4) Load another 10N weight (which means P = 20N) and record the value of ๐๐ โฒ .

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 ๐๐ โฒ .

Page 10 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 ๐๐ for both loading and unloading. Strain ๐๐ = strain increment ๐๐ โฒ ร 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 ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ , ๐๐1 ,

๐๐2 and ๐๐๐๐ for the case where ๐๐1 = 55๐๐, ๐๐2 = 105๐๐. Table 2 shows all the recorded values

of ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ , ๐๐1 , ๐๐2 and ๐๐๐๐ for the case where ๐๐1 = 135๐๐, ๐๐2 = 205๐๐. For each case, the

recorded values are used to construct ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ with an appropriate scale (For 1st case:

10cm/Newton, For 2nd case: 5cm/Newton). ๐ญ๐ญ๐๐ 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 ๐ญ๐ญ๐๐ = 6.9cm โด Magnitude of ๐ญ๐ญ๐๐ =

6.9 10

= 0.69๐๐

Case 2: Length of ๐ญ๐ญ๐๐ = 6.3cm โด Magnitude of ๐ญ๐ญ๐๐ =

6.3 5

= 1.26๐๐

From the above calculations, the equilibrant force vector, ๐ญ๐ญ๐๐ , does not exactly balance the

resultant vector, ๐ญ๐ญ๐๐ , for each case.

Page 11 of 24

Percentage error:

๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ โ๐๐โ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ ๐๐โ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ

Case 1: Percentage error =

0.70โ0.69

Case 2: Percentage error =

1.20โ1.26

0.69 1.26

ร 100%

ร 100% = 1.45% ร 100% = โ4.76%

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 81.7๐๐๐๐2 . 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 ๐๐ , while loading and unloading. The table also includes computed strain ๐๐, which can be found by Strain ๐๐ = strain increment ๐๐ โฒ ร conversion factor (CF)

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

A graph of load P versus strain increment ๐๐ 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 ๐๐. Slope of the straight line graph =

๐๐ ๐๐

60

= 12ร10 โ6 = 5 ร 106 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 =

Truss stiffness k =

๐๐ ๐๐

๐๐ ๐ด๐ดโ

๐๐ โ ๐๐

๐๐ 2

450

= 5 ร 106 ร 81.7ร427 = 64.5 ๐๐๐๐/๐๐๐๐2 427

= 5 ร 106 ร 450 2 = 10500 ๐๐/๐๐๐๐

Page 12 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, ๐ญ๐ญ๐๐ , does not exactly balance the resultant vector, ๐ญ๐ญ๐๐ , 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 ๐๐1 or ๐๐2 (See Figure 10). Hence, the spring balance may be

slanted at an angle, which results in an inaccurate reading of ๐ญ๐ญ๐๐ .

Fig. 10 shows the portion of string that connects the spring balance is not parallel to the portion of the string that hangs the ๐๐1 or ๐๐2

Page 13 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 ๐ญ๐ญ๐๐ .

Fig. 11 shows spring balance is not upright, hence results in an inaccurate reading of ๐ญ๐ญ๐๐ .

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 ๐ญ๐ญ๐๐ . Accuracy in

this experiment is important because the force to be measured, ๐ญ๐ญ๐๐ , is very small (< 2N).

Further Discussion: For the given masses ๐๐1 , ๐๐2 and the measured angles ๐๐1 , ๐๐2 , calculate the equilibrant ๐ญ๐ญ๐๐

and its direction ๐๐๐๐ using the equilibrium conditions. Compare the calculated results with the

measured values. Case 1:

๐ญ๐ญ1 : X-component = 0.54 cos 168 = โ0.5282๐๐ Y-component = 0.54 sin 168 = 0.1123๐๐

Page 14 of 24

๐ญ๐ญ2 : X-component = 1.03 cos 26 = 0.9258๐๐

Y-component = 1.03 sin 26 = 0.4515๐๐

๐ญ๐ญ๐๐ : X-component = 0.9258 โ 0.5282 = 0.3976N Y-component = 0.4515 + 0.1123 = 0.5638N Under equilibrium conditions, |๐ญ๐ญ๐๐ | = |๐ญ๐ญ๐๐ | = 0.690๐๐

๐๐๐๐ = 180ยฐ โ ๐๐๐๐ = 180 โ tanโ1

0.5638

0.3976

= 125ยฐ

Comparison:

Calculated Measured

๐ญ๐ญ๐๐ (๐๐) 0.69 0.70

๐๐๐๐ (๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท) 125 126

Case 2: ๐ญ๐ญ1 : X-component = 1.32 cos 168 = โ1.291๐๐ Y-component = 1.32 sin 168 = 0.2744๐๐

๐ญ๐ญ2 : X-component = 2.01 cos 26 = 1.807๐๐

Y-component = 2.01 sin 26 = 0.8811๐๐

๐ญ๐ญ๐๐ : X-component = 1.807โ 1.291 = 0.516N

Y-component = 0.2744 + 0.8811 = 1.156N

Under equilibrium conditions, |๐ญ๐ญ๐๐ | = |๐ญ๐ญ๐๐ | = 1.26๐๐

๐๐๐๐ = 180ยฐ โ ๐๐๐๐ = 180 โ tanโ1

1.156

0.516

= 114ยฐ

Page 15 of 24

Comparison: ๐ญ๐ญ๐๐ (๐๐) 1.26 1.20

Calculated Measured

๐๐๐๐ (๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท) 114 117

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.

M e 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).

3) Instead of using a spring balance, use a high tech digital force gauge to obtain a much accurate reading of ๐ญ๐ญ๐๐ . This will minimize error (4) stated in Page 14 of the report. Page 16 of 24

Experiment 2: Discussion: What is the significance of observed strains fitting into a straight line in the plot of P vs.๐๐? โข

Fitting all the points into a straight line signifies load P is linearly proportional to the strain increment ๐๐. This shows that the truss member had undergone linear elastic deformation, which means the truss member obeys Hookeโs law (๐๐ = ๐ธ๐ธ๐ธ๐ธ).

If member GB is replaced by another member of the same length but different cross sectional area, will the tensile force ๐น๐น๐บ๐บ๐บ๐บ be different under the same load P (assuming small

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

From equation (1), ๐น๐น๐บ๐บ๐บ๐บ =

๐๐ ๐๐ ๐๐๐๐ = = sin ๐๐ โ/๐๐ โ

๐น๐น๐บ๐บ๐บ๐บ 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 ๐น๐น๐บ๐บ๐บ๐บ will be the same under the

same load with different cross sectional area. โข

From equation (6), ๐ฟ๐ฟ =

๐ฟ๐ฟ๐บ๐บ๐บ๐บ ๐๐๐๐ ๐๐ 2 ๐๐ = = sin ๐๐ โ/๐๐ โ

๐ฟ๐ฟ depends on 3 factors, ๐๐, l, and h. l, and h are not related to the breadth and the thickness

of member GB. However, ๐๐ is dependant on cross sectional area. Based on equation (5),

๐ธ๐ธ =

๐๐ ๐๐

=

๐๐ ๐๐

๐๐ ๐ด๐ดโ

Page 17 of 24

Rearranging equation (5), ๐๐ =

๐๐ ๐๐ ๐๐ = ๐ธ๐ธ ๐ธ๐ธ ๐ด๐ดโ

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 ๐๐ 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? โข

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 โ 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 ๐น๐น๐ท๐ท๐ท๐ท .

Hence, Equation (6) (on Page 7 of the report) is not valid because circular arcs BBโ and LBโ due to rotation of DB and GL respectively, cannot be approximated to straight lines. Therefore, making our calculation much more difficult.

Page 18 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? โข

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.

โข

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.

โข

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 19 of 24

Conclusion Experiment 1: In theory, the equilibrant vector force ๐ญ๐ญ๐๐ exactly offsets the combined effect of ๐ญ๐ญ1 and ๐ญ๐ญ2 , which is the resultant force ๐ญ๐ญ๐๐ . However, in this experiment we observed that ๐ญ๐ญ๐๐ does not

exactly balance out ๐ญ๐ญ๐๐ . 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 ๐๐. This showed that the truss member had undergone linear elastic deformation,

which meant the truss member obeyed Hookeโs law (๐๐ = ๐ธ๐ธ๐ธ๐ธ). The slope of the graph was determined,

๐๐

Slope of the straight line graph = = And hence, the elastic modulus Elastic modulus E = and truss stiffness. Truss stiffness k =

๐๐ ๐๐

๐๐ ๐ด๐ดโ

๐๐ โ

๐๐ ๐๐ 2

๐๐

60

12ร10โ6

= 5 ร 106 ร

= 5 ร 106 ร

450

= 5 ร 106 N

81.7ร427

427

450 2

= 64.5 ๐๐๐๐/๐๐๐๐2

= 10500 ๐๐/๐๐๐๐

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, โMaterials Science and Engineering, An Introductionโ, 7th Edition, John Wiley & Sons, Inc. 4) A/P Tan Ming Jen, 2009, FE1005 Materials Science Lecture Notes. Page 20 of 24

Appendix A1

Appendices

Table 1: (Case 1) ๐๐1 = 55๐๐, ๐๐2 = 105๐๐ ๐๐1 (g) 55

๐๐2 (g)

105

๐๐1 (degree) 168

๐๐2 (degree) 26

Table 2: (Case 2) ๐๐1 = 135๐๐, ๐๐2 = 205๐๐ ๐๐1 (g)

135

๐๐2 (g)

205

๐๐1 (degree) 168

๐๐2 (degree) 26

๐ญ๐ญ1 = ๐๐1 ๐๐ (N)

๐ญ๐ญ2 = ๐๐2 ๐๐ (N)

๐ญ๐ญ1 = ๐๐1 ๐๐ (N)

0.54

1.32

1.03

๐ญ๐ญ๐๐ (N)

0.70

๐๐๐๐ (degree)

๐ญ๐ญ2 = ๐๐2 ๐๐ (N)

๐ญ๐ญ๐๐ (N)

๐๐๐๐ (degree)

2.01

1.20

126

117

Page 21 of 24

Appendix A2

Appendices

Page 22 of 24

Appendix B1

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 (๐๐๐๐2 )

81.6

82.2

81.3

Average Area, A =

Other measurements: h = 427mm

81.6+82.2+81.3 3

= 81.7๐๐๐๐2

l = 450mm

Table 4:

Strain Increment P Loading (N) 0

๐๐ , (ร 10โ6 ) 0

Unloading

๐๐ (ร 10โ6 ) 0

๐๐ , (ร 10โ6 )

๐๐ (ร 10โ6 )

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

Page 23 of 24

Appendix B2

Appendices

Page 24 of 24

View more...
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 1 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 ๐๐ 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 ๐๐. This showed that the truss member had undergone linear elastic

deformation, which meant the truss member obeyed Hookeโs law (๐๐ = ๐ธ๐ธ๐ธ๐ธ). 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 2 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 ๐น๐นโ makes an angle ๐๐ with the horizontal axis (Direction), with a length ฮป (Magnitude).

Fig. 1 shows a vector ๐น๐นโ with length ฮป and makes an angle ๐๐ with the horizontal axis.

Vector forces acting on the same point of the object are called concurrent forces. The resultant forces ๐ญ๐ญ๐๐ can be determined by adding the vector forces together using the parallelogram method (as shown in Figure 2).

Fig. 2 shows the resultant vector force ๐ญ๐ญ๐๐ as the result of ๐ญ๐ญ1 + ๐ญ๐ญ2 . It also shows, ๐ญ๐ญ๐๐ , the equilibrant force which has the same magnitude as ๐ญ๐ญ๐๐ but is in the opposite direction.

Another vector ๐ญ๐ญ๐๐ shown in Figure 2 is the equilibrant of ๐ญ๐ญ1 and ๐ญ๐ญ2 . ๐ญ๐ญ๐๐ is the force needed to exactly offset the combined effect of ๐ญ๐ญ1 and ๐ญ๐ญ2 , which is ๐ญ๐ญ๐๐ . ๐ญ๐ญ๐๐ has the same magnitude as ๐ญ๐ญ๐๐ , but is in the opposite direction.

Page 3 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 ๐น๐น๐บ๐บ๐บ๐บ in GB and Compression force ๐น๐น๐ท๐ท๐ท๐ท along DB) and

we can write two equilibrium equations to solve for ๐น๐น๐บ๐บ๐บ๐บ and ๐น๐น๐ท๐ท๐ท๐ท . However, we will be interested only in ๐น๐น๐บ๐บ๐บ๐บ . By equilibrium at joint B (Figure 3(b)), we obtain

G

Fig. 3 (a) shows a truss which composed of slender members joined together at their end points.

D

โ

๐น๐น๐บ๐บ๐บ๐บ =

๐๐

sin ๐๐

๐๐

= โ /๐๐ =

๐๐๐๐

Equation (1)

โ

๐น๐น๐ท๐ท๐ท๐ท

๐๐

๐น๐น๐บ๐บ๐บ๐บ

๐๐

B

P ๐๐

B

Fig. 3 (b) shows all forces acting at joint B

P Page 4 of 24

Given a cross sectional area A of member GB, the stress (force per unit area) in member GB will be

๐๐ =

๐น๐น๐บ๐บ๐บ๐บ ๐๐๐๐ = ๐ด๐ด ๐ด๐ดโ

Equation (2)

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

Fig. 4 shows a small portion of the member GB. It indicates the variable ๐น๐น๐บ๐บ๐บ๐บ and ๐ด๐ด that are needed to calculate the stress ๐๐.

(b) If force ๐น๐น๐บ๐บ๐บ๐บ is in Newton (N) and the unit for length is in

millimetre (mm), the stress ๐๐ will be in ๐ต๐ต/๐๐๐๐๐๐ or ๐ด๐ด๐ด๐ด/๐๐๐๐

or ๐ด๐ด๐ด๐ด๐ด๐ด.

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

๐๐ =

๐ฟ๐ฟ ๐บ๐บ๐บ๐บ

Equation (3)

๐๐

where ๐ฟ๐ฟ๐บ๐บ๐บ๐บ is the total deformation between the ends of the member (See Figure 5). Note: (a) ๐๐ is the original length before loading. (b) Strain is always dimensionless.

๐๐

Fig. 5 shows the total deformation between the ends of the member GB.

Page 5 of 24

Most engineering structures are designed to function within the linear elastic range, i.e., the stress ๐๐ is linearly proportional to the strain ๐๐,

๐๐ = ๐ธ๐ธ๐ธ๐ธ

Equation (4)

This relation is known as Hookeโs law. Figure 6 shows a graph plotted ๐๐ against ๐๐. The coefficient E (Gradient of the graph) is called the modulus of elasticity (or Youngโs modulus). It is a measurement of stiffness of the material involved.

Fig. 6 shows graph plotted ๐๐ against ๐๐. The gradient of the graph is called the modulus of elasticity.

Substitute Equation (2) and Equation (4),

๐ธ๐ธ =

๐๐ ๐๐

Note: (a) The unit for E is ๐๐๐๐/๐๐๐๐๐๐ or ๐ฎ๐ฎ๐ฎ๐ฎ๐ฎ๐ฎ.

=

๐๐ ๐๐

๐๐ ๐ด๐ดโ

Equation (5)

(b) The strain ๐๐ equals to the measured strain increment ๐๐ โฒ multiplied by a conversion factor due to the instrumentation.

Page 6 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 Bโ due to the extension ๐ฟ๐ฟ๐บ๐บ๐บ๐บ 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 BBโ and LBโ due to rotation of DB and GL respectively, can be approximated to straight lines. It follows that the right-angled triangle BLBโ, shown in Figure 7, can represent the relationship between ๐ฟ๐ฟ๐บ๐บ๐บ๐บ and deflection ๐ฟ๐ฟ of B. Therefore, ๐ฟ๐ฟ

G

๐๐๐๐

๐ฟ๐ฟ = sin๐บ๐บ๐บ๐บ๐๐ = โ /๐๐ =

๐๐ 2 ๐๐

Equation (6)

โ

Fig. 7 shows deflection at joint B where force P is applied.

๐๐

D

B ๐ฟ๐ฟ

Bโ

๐ฟ๐ฟ๐บ๐บ๐บ๐บ

L

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

๐๐ =

๐๐ ๐ฟ๐ฟ

=

๐๐ โ

๐๐ ๐๐ 2

Equation (7)

By plotting the relationship between P and the measured strain ๐๐ for a range of applied P

values, the Youngโs modulus E and the stiffness coefficient k can be determined from the experiment using Equations (5) and (7).

Page 7 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:

Spring Balance

Pulley 1

๐๐1

Force Ring

๐ญ๐ญ1

๐๐1

๐ญ๐ญ2

๐๐2

Degree Scale

Holding Pin

Pulley 2

๐ญ๐ญ๐๐

๐๐๐๐

Zero-degree line

๐๐2

Experiment Board

Pulley 3

Figure 8 shows experiment 1 setup.

1) Use pulleys and hanging masses (๐๐1 = 55๐๐, ๐๐2 = 105๐๐) to setup the equipment as

shown in Figure 8, so that two known forces, ๐ญ๐ญ1 (=๐๐1 ๐๐) and ๐ญ๐ญ2 (=๐๐2 ๐๐), are pulling the force ring.

2) Use holding pin to prevent the ring from moving. The holding pin provides a force, ๐ญ๐ญ๐๐ , that is exactly opposite to the resultant of ๐ญ๐ญ1 and ๐ญ๐ญ2 .

Page 8 of 24

3) Adjust the spring balance to determine the magnitude of ๐ญ๐ญ๐๐ . 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 (๐๐1 = 55๐๐, ๐๐2 = 105๐๐); the magnitude of ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ (in Newton); the angles ๐๐1 , ๐๐2 and ๐๐๐๐ 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 ๐๐1 = 135๐๐, ๐๐2 = 205๐๐ and repeat step (1) to (4) one more time.

Page 9 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 ๐๐ (The length of truss member GB), and โ (The vertical length from joint G to the member DB).

Figure 9 shows how to measure ๐๐ and โ for experiment 2 setup.

Line that cuts through both joint point D and B

Note: The measurement, ๐๐, should be measured from the joint point G to the other one, B,

whereas the measurement, โ, 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 (๐๐ โฒ = 0). Then load a weight of 10N (which means P = 10N). Wait for the reading to

be stabilised. Record the value of ๐๐ โฒ , read off from the strain monitoring machine.

4) Load another 10N weight (which means P = 20N) and record the value of ๐๐ โฒ .

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 ๐๐ โฒ .

Page 10 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 ๐๐ for both loading and unloading. Strain ๐๐ = strain increment ๐๐ โฒ ร 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 ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ , ๐๐1 ,

๐๐2 and ๐๐๐๐ for the case where ๐๐1 = 55๐๐, ๐๐2 = 105๐๐. Table 2 shows all the recorded values

of ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ , ๐๐1 , ๐๐2 and ๐๐๐๐ for the case where ๐๐1 = 135๐๐, ๐๐2 = 205๐๐. For each case, the

recorded values are used to construct ๐ญ๐ญ1 , ๐ญ๐ญ2 , ๐ญ๐ญ๐๐ with an appropriate scale (For 1st case:

10cm/Newton, For 2nd case: 5cm/Newton). ๐ญ๐ญ๐๐ 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 ๐ญ๐ญ๐๐ = 6.9cm โด Magnitude of ๐ญ๐ญ๐๐ =

6.9 10

= 0.69๐๐

Case 2: Length of ๐ญ๐ญ๐๐ = 6.3cm โด Magnitude of ๐ญ๐ญ๐๐ =

6.3 5

= 1.26๐๐

From the above calculations, the equilibrant force vector, ๐ญ๐ญ๐๐ , does not exactly balance the

resultant vector, ๐ญ๐ญ๐๐ , for each case.

Page 11 of 24

Percentage error:

๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ โ๐๐โ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ ๐๐โ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ๐ฃ

Case 1: Percentage error =

0.70โ0.69

Case 2: Percentage error =

1.20โ1.26

0.69 1.26

ร 100%

ร 100% = 1.45% ร 100% = โ4.76%

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 81.7๐๐๐๐2 . 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 ๐๐ , while loading and unloading. The table also includes computed strain ๐๐, which can be found by Strain ๐๐ = strain increment ๐๐ โฒ ร conversion factor (CF)

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

A graph of load P versus strain increment ๐๐ 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 ๐๐. Slope of the straight line graph =

๐๐ ๐๐

60

= 12ร10 โ6 = 5 ร 106 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 =

Truss stiffness k =

๐๐ ๐๐

๐๐ ๐ด๐ดโ

๐๐ โ ๐๐

๐๐ 2

450

= 5 ร 106 ร 81.7ร427 = 64.5 ๐๐๐๐/๐๐๐๐2 427

= 5 ร 106 ร 450 2 = 10500 ๐๐/๐๐๐๐

Page 12 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, ๐ญ๐ญ๐๐ , does not exactly balance the resultant vector, ๐ญ๐ญ๐๐ , 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 ๐๐1 or ๐๐2 (See Figure 10). Hence, the spring balance may be

slanted at an angle, which results in an inaccurate reading of ๐ญ๐ญ๐๐ .

Fig. 10 shows the portion of string that connects the spring balance is not parallel to the portion of the string that hangs the ๐๐1 or ๐๐2

Page 13 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 ๐ญ๐ญ๐๐ .

Fig. 11 shows spring balance is not upright, hence results in an inaccurate reading of ๐ญ๐ญ๐๐ .

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 ๐ญ๐ญ๐๐ . Accuracy in

this experiment is important because the force to be measured, ๐ญ๐ญ๐๐ , is very small (< 2N).

Further Discussion: For the given masses ๐๐1 , ๐๐2 and the measured angles ๐๐1 , ๐๐2 , calculate the equilibrant ๐ญ๐ญ๐๐

and its direction ๐๐๐๐ using the equilibrium conditions. Compare the calculated results with the

measured values. Case 1:

๐ญ๐ญ1 : X-component = 0.54 cos 168 = โ0.5282๐๐ Y-component = 0.54 sin 168 = 0.1123๐๐

Page 14 of 24

๐ญ๐ญ2 : X-component = 1.03 cos 26 = 0.9258๐๐

Y-component = 1.03 sin 26 = 0.4515๐๐

๐ญ๐ญ๐๐ : X-component = 0.9258 โ 0.5282 = 0.3976N Y-component = 0.4515 + 0.1123 = 0.5638N Under equilibrium conditions, |๐ญ๐ญ๐๐ | = |๐ญ๐ญ๐๐ | = 0.690๐๐

๐๐๐๐ = 180ยฐ โ ๐๐๐๐ = 180 โ tanโ1

0.5638

0.3976

= 125ยฐ

Comparison:

Calculated Measured

๐ญ๐ญ๐๐ (๐๐) 0.69 0.70

๐๐๐๐ (๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท) 125 126

Case 2: ๐ญ๐ญ1 : X-component = 1.32 cos 168 = โ1.291๐๐ Y-component = 1.32 sin 168 = 0.2744๐๐

๐ญ๐ญ2 : X-component = 2.01 cos 26 = 1.807๐๐

Y-component = 2.01 sin 26 = 0.8811๐๐

๐ญ๐ญ๐๐ : X-component = 1.807โ 1.291 = 0.516N

Y-component = 0.2744 + 0.8811 = 1.156N

Under equilibrium conditions, |๐ญ๐ญ๐๐ | = |๐ญ๐ญ๐๐ | = 1.26๐๐

๐๐๐๐ = 180ยฐ โ ๐๐๐๐ = 180 โ tanโ1

1.156

0.516

= 114ยฐ

Page 15 of 24

Comparison: ๐ญ๐ญ๐๐ (๐๐) 1.26 1.20

Calculated Measured

๐๐๐๐ (๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท๐ท) 114 117

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.

M e 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).

3) Instead of using a spring balance, use a high tech digital force gauge to obtain a much accurate reading of ๐ญ๐ญ๐๐ . This will minimize error (4) stated in Page 14 of the report. Page 16 of 24

Experiment 2: Discussion: What is the significance of observed strains fitting into a straight line in the plot of P vs.๐๐? โข

Fitting all the points into a straight line signifies load P is linearly proportional to the strain increment ๐๐. This shows that the truss member had undergone linear elastic deformation, which means the truss member obeys Hookeโs law (๐๐ = ๐ธ๐ธ๐ธ๐ธ).

If member GB is replaced by another member of the same length but different cross sectional area, will the tensile force ๐น๐น๐บ๐บ๐บ๐บ be different under the same load P (assuming small

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

From equation (1), ๐น๐น๐บ๐บ๐บ๐บ =

๐๐ ๐๐ ๐๐๐๐ = = sin ๐๐ โ/๐๐ โ

๐น๐น๐บ๐บ๐บ๐บ 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 ๐น๐น๐บ๐บ๐บ๐บ will be the same under the

same load with different cross sectional area. โข

From equation (6), ๐ฟ๐ฟ =

๐ฟ๐ฟ๐บ๐บ๐บ๐บ ๐๐๐๐ ๐๐ 2 ๐๐ = = sin ๐๐ โ/๐๐ โ

๐ฟ๐ฟ depends on 3 factors, ๐๐, l, and h. l, and h are not related to the breadth and the thickness

of member GB. However, ๐๐ is dependant on cross sectional area. Based on equation (5),

๐ธ๐ธ =

๐๐ ๐๐

=

๐๐ ๐๐

๐๐ ๐ด๐ดโ

Page 17 of 24

Rearranging equation (5), ๐๐ =

๐๐ ๐๐ ๐๐ = ๐ธ๐ธ ๐ธ๐ธ ๐ด๐ดโ

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 ๐๐ 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? โข

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 โ 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 ๐น๐น๐ท๐ท๐ท๐ท .

Hence, Equation (6) (on Page 7 of the report) is not valid because circular arcs BBโ and LBโ due to rotation of DB and GL respectively, cannot be approximated to straight lines. Therefore, making our calculation much more difficult.

Page 18 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? โข

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.

โข

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.

โข

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 19 of 24

Conclusion Experiment 1: In theory, the equilibrant vector force ๐ญ๐ญ๐๐ exactly offsets the combined effect of ๐ญ๐ญ1 and ๐ญ๐ญ2 , which is the resultant force ๐ญ๐ญ๐๐ . However, in this experiment we observed that ๐ญ๐ญ๐๐ does not

exactly balance out ๐ญ๐ญ๐๐ . 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 ๐๐. This showed that the truss member had undergone linear elastic deformation,

which meant the truss member obeyed Hookeโs law (๐๐ = ๐ธ๐ธ๐ธ๐ธ). The slope of the graph was determined,

๐๐

Slope of the straight line graph = = And hence, the elastic modulus Elastic modulus E = and truss stiffness. Truss stiffness k =

๐๐ ๐๐

๐๐ ๐ด๐ดโ

๐๐ โ

๐๐ ๐๐ 2

๐๐

60

12ร10โ6

= 5 ร 106 ร

= 5 ร 106 ร

450

= 5 ร 106 N

81.7ร427

427

450 2

= 64.5 ๐๐๐๐/๐๐๐๐2

= 10500 ๐๐/๐๐๐๐

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, โMaterials Science and Engineering, An Introductionโ, 7th Edition, John Wiley & Sons, Inc. 4) A/P Tan Ming Jen, 2009, FE1005 Materials Science Lecture Notes. Page 20 of 24

Appendix A1

Appendices

Table 1: (Case 1) ๐๐1 = 55๐๐, ๐๐2 = 105๐๐ ๐๐1 (g) 55

๐๐2 (g)

105

๐๐1 (degree) 168

๐๐2 (degree) 26

Table 2: (Case 2) ๐๐1 = 135๐๐, ๐๐2 = 205๐๐ ๐๐1 (g)

135

๐๐2 (g)

205

๐๐1 (degree) 168

๐๐2 (degree) 26

๐ญ๐ญ1 = ๐๐1 ๐๐ (N)

๐ญ๐ญ2 = ๐๐2 ๐๐ (N)

๐ญ๐ญ1 = ๐๐1 ๐๐ (N)

0.54

1.32

1.03

๐ญ๐ญ๐๐ (N)

0.70

๐๐๐๐ (degree)

๐ญ๐ญ2 = ๐๐2 ๐๐ (N)

๐ญ๐ญ๐๐ (N)

๐๐๐๐ (degree)

2.01

1.20

126

117

Page 21 of 24

Appendix A2

Appendices

Page 22 of 24

Appendix B1

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 (๐๐๐๐2 )

81.6

82.2

81.3

Average Area, A =

Other measurements: h = 427mm

81.6+82.2+81.3 3

= 81.7๐๐๐๐2

l = 450mm

Table 4:

Strain Increment P Loading (N) 0

๐๐ , (ร 10โ6 ) 0

Unloading

๐๐ (ร 10โ6 ) 0

๐๐ , (ร 10โ6 )

๐๐ (ร 10โ6 )

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

Page 23 of 24

Appendix B2

Appendices

Page 24 of 24

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