FE1073 Lab Report C1

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

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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 𝜀𝜀 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 𝜀𝜀. This showed that the truss member had undergone linear elastic

deformation, which meant the truss member obeyed Hooke’s law (𝜎𝜎 = 𝐸𝐸𝐸𝐸). 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.

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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 𝐹𝐹⃗ makes an angle 𝜃𝜃 with the horizontal axis (Direction), with a length λ (Magnitude).

Fig. 1 shows a vector 𝐹𝐹⃗ with length λ and makes an angle 𝜃𝜃 with the horizontal axis.

Vector forces acting on the same point of the object are called concurrent forces. The resultant forces 𝑭𝑭𝑟𝑟 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 𝑭𝑭𝑟𝑟 as the result of 𝑭𝑭1 + 𝑭𝑭2 . It also shows, 𝑭𝑭𝑒𝑒 , the equilibrant force which has the same magnitude as 𝑭𝑭𝑟𝑟 but is in the opposite direction.

Another vector 𝑭𝑭𝑒𝑒 shown in Figure 2 is the equilibrant of 𝑭𝑭1 and 𝑭𝑭2 . 𝑭𝑭𝑒𝑒 is the force needed to exactly offset the combined effect of 𝑭𝑭1 and 𝑭𝑭2 , which is 𝑭𝑭𝑟𝑟 . 𝑭𝑭𝑒𝑒 has the same magnitude as 𝑭𝑭𝑟𝑟 , but is in the opposite direction.

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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 𝐹𝐹𝐺𝐺𝐺𝐺 in GB and Compression force 𝐹𝐹𝐷𝐷𝐷𝐷 along DB) and

we can write two equilibrium equations to solve for 𝐹𝐹𝐺𝐺𝐺𝐺 and 𝐹𝐹𝐷𝐷𝐷𝐷 . However, we will be interested only in 𝐹𝐹𝐺𝐺𝐺𝐺 . 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

ℎ

𝐹𝐹𝐺𝐺𝐺𝐺 =

𝑃𝑃

sin 𝜃𝜃

𝑃𝑃

= ℎ /𝑙𝑙 =

𝑃𝑃𝑃𝑃

Equation (1)

ℎ

𝐹𝐹𝐷𝐷𝐷𝐷

𝑙𝑙

𝐹𝐹𝐺𝐺𝐺𝐺

𝜃𝜃

B

P 𝜃𝜃

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

𝜎𝜎 =

𝐹𝐹𝐺𝐺𝐺𝐺 𝑃𝑃𝑃𝑃 = 𝐴𝐴 𝐴𝐴ℎ

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 𝐹𝐹𝐺𝐺𝐺𝐺 and 𝐴𝐴 that are needed to calculate the stress 𝜎𝜎.

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

millimetre (mm), the stress 𝜎𝜎 will be in 𝑵𝑵/𝒎𝒎𝒎𝒎𝟐𝟐 or 𝑴𝑴𝑴𝑴/𝒎𝒎𝟐𝟐

or 𝑴𝑴𝑴𝑴𝑴𝑴.

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

𝜀𝜀 =

𝛿𝛿 𝐺𝐺𝐺𝐺

Equation (3)

𝑙𝑙

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

𝑙𝑙

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

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

𝜎𝜎 = 𝐸𝐸𝐸𝐸

Equation (4)

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

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

Substitute Equation (2) and Equation (4),

𝐸𝐸 =

𝜎𝜎 𝜀𝜀

Note: (a) The unit for E is 𝒌𝒌𝒌𝒌/𝒎𝒎𝒎𝒎𝟐𝟐 or 𝑮𝑮𝑮𝑮𝑮𝑮.

=

𝑃𝑃 𝑙𝑙

𝜀𝜀 𝐴𝐴ℎ

Equation (5)

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

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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 𝛿𝛿𝐺𝐺𝐺𝐺 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 𝛿𝛿𝐺𝐺𝐺𝐺 and deflection 𝛿𝛿 of B. Therefore, 𝛿𝛿

G

𝑙𝑙𝑙𝑙

𝛿𝛿 = sin𝐺𝐺𝐺𝐺𝜃𝜃 = ℎ /𝑙𝑙 =

𝑙𝑙 2 𝜀𝜀

Equation (6)

ℎ

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

𝜃𝜃

D

B 𝛿𝛿

B’

𝛿𝛿𝐺𝐺𝐺𝐺

L

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

𝑘𝑘 =

𝑃𝑃 𝛿𝛿

=

𝑃𝑃 ℎ

𝜀𝜀 𝑙𝑙 2

Equation (7)

By plotting the relationship between P and the measured strain 𝜀𝜀 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).

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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

𝜃𝜃1

Force Ring

𝑭𝑭1

𝑀𝑀1

𝑭𝑭2

𝜃𝜃2

Degree Scale

Holding Pin

Pulley 2

𝑭𝑭𝑒𝑒

𝜃𝜃𝑒𝑒

Zero-degree line

𝑀𝑀2

Experiment Board

Pulley 3

Figure 8 shows experiment 1 setup.

1) Use pulleys and hanging masses (𝑀𝑀1 = 55𝑔𝑔, 𝑀𝑀2 = 105𝑔𝑔) to setup the equipment as

shown in Figure 8, so that two known forces, 𝑭𝑭1 (=𝑀𝑀1 𝑔𝑔) and 𝑭𝑭2 (=𝑀𝑀2 𝑔𝑔), are pulling the force ring.

2) Use holding pin to prevent the ring from moving. The holding pin provides a force, 𝑭𝑭𝑒𝑒 , that is exactly opposite to the resultant of 𝑭𝑭1 and 𝑭𝑭2 .

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3) Adjust the spring balance to determine the magnitude of 𝑭𝑭𝑒𝑒 . 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 (𝑀𝑀1 = 55𝑔𝑔, 𝑀𝑀2 = 105𝑔𝑔); the magnitude of 𝑭𝑭1 , 𝑭𝑭2 , 𝑭𝑭𝑒𝑒 (in Newton); the angles 𝜃𝜃1 , 𝜃𝜃2 and 𝜃𝜃𝑒𝑒 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 𝑀𝑀1 = 135𝑔𝑔, 𝑀𝑀2 = 205𝑔𝑔 and repeat step (1) to (4) one more time.

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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 𝑙𝑙 (The length of truss member GB), and ℎ (The vertical length from joint G to the member DB).

Figure 9 shows how to measure 𝑙𝑙 and ℎ for experiment 2 setup.

Line that cuts through both joint point D and B

Note: The measurement, 𝑙𝑙, 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 (𝜀𝜀 ′ = 0). Then load a weight of 10N (which means P = 10N). Wait for the reading to

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

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

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 𝜀𝜀 ′ .

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7) Repeat step (6) until P = 0N. The results are shown on page 23 of the report. 8) Compute the values of strain 𝜀𝜀 for both loading and unloading. Strain 𝜀𝜀 = strain increment 𝜀𝜀 ′ × 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 𝑭𝑭1 , 𝑭𝑭2 , 𝑭𝑭𝑒𝑒 , 𝜃𝜃1 ,

𝜃𝜃2 and 𝜃𝜃𝑒𝑒 for the case where 𝑀𝑀1 = 55𝑔𝑔, 𝑀𝑀2 = 105𝑔𝑔. Table 2 shows all the recorded values

of 𝑭𝑭1 , 𝑭𝑭2 , 𝑭𝑭𝑒𝑒 , 𝜃𝜃1 , 𝜃𝜃2 and 𝜃𝜃𝑒𝑒 for the case where 𝑀𝑀1 = 135𝑔𝑔, 𝑀𝑀2 = 205𝑔𝑔. For each case, the

recorded values are used to construct 𝑭𝑭1 , 𝑭𝑭2 , 𝑭𝑭𝑒𝑒 with an appropriate scale (For 1st case:

10cm/Newton, For 2nd case: 5cm/Newton). 𝑭𝑭𝑟𝑟 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 𝑭𝑭𝑟𝑟 = 6.9cm ∴ Magnitude of 𝑭𝑭𝑟𝑟 =

6.9 10

= 0.69𝑁𝑁

Case 2: Length of 𝑭𝑭𝑟𝑟 = 6.3cm ∴ Magnitude of 𝑭𝑭𝑟𝑟 =

6.3 5

= 1.26𝑁𝑁

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

resultant vector, 𝑭𝑭𝑟𝑟 , for each case.

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Percentage error:

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 −𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑇𝑇ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣

Case 1: Percentage error =

0.70−0.69

Case 2: Percentage error =

1.20−1.26

0.69 1.26

× 100%

× 100% = 1.45% × 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 81.7𝑚𝑚𝑚𝑚2 . 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 𝜀𝜀 , while loading and unloading. The table also includes computed strain 𝜀𝜀, which can be found by Strain 𝜀𝜀 = strain increment 𝜀𝜀 ′ × conversion factor (CF)

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

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

𝑃𝑃 𝜀𝜀

60

= 12×10 −6 = 5 × 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 =

𝑃𝑃 𝑙𝑙

𝜀𝜀 𝐴𝐴ℎ

𝑃𝑃 ℎ 𝜀𝜀

𝑙𝑙 2

450

= 5 × 106 × 81.7×427 = 64.5 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚2 427

= 5 × 106 × 450 2 = 10500 𝑁𝑁/𝑚𝑚𝑚𝑚

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Discussion Experiment 1: From the diagrams drawn in Appendix A2 and the calculations on Page 11 of the report, the equilibrant force vector, 𝑭𝑭𝑒𝑒 , does not exactly balance the resultant vector, 𝑭𝑭𝑟𝑟 , 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 𝑀𝑀1 or 𝑀𝑀2 (See Figure 10). Hence, the spring balance may be

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

Fig. 10 shows the portion of string that connects the spring balance is not parallel to the portion of the string that hangs the 𝑀𝑀1 or 𝑀𝑀2

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 𝑭𝑭𝑒𝑒 .

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

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 𝑭𝑭𝑒𝑒 . Accuracy in

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

Further Discussion: For the given masses 𝑀𝑀1 , 𝑀𝑀2 and the measured angles 𝜃𝜃1 , 𝜃𝜃2 , calculate the equilibrant 𝑭𝑭𝑒𝑒

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

measured values. Case 1:

𝑭𝑭1 : X-component = 0.54 cos 168 = −0.5282𝑁𝑁 Y-component = 0.54 sin 168 = 0.1123𝑁𝑁

Page 14 of 24

𝑭𝑭2 : X-component = 1.03 cos 26 = 0.9258𝑁𝑁

Y-component = 1.03 sin 26 = 0.4515𝑁𝑁

𝑭𝑭𝑟𝑟 : X-component = 0.9258 – 0.5282 = 0.3976N Y-component = 0.4515 + 0.1123 = 0.5638N Under equilibrium conditions, |𝑭𝑭𝑒𝑒 | = |𝑭𝑭𝑟𝑟 | = 0.690𝑁𝑁

𝜃𝜃𝑒𝑒 = 180° − 𝜃𝜃𝑟𝑟 = 180 − tan−1

0.5638

0.3976

= 125°

Comparison:

Calculated Measured

𝑭𝑭𝑒𝑒 (𝑁𝑁) 0.69 0.70

𝜃𝜃𝑒𝑒 (𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷) 125 126

Case 2: 𝑭𝑭1 : X-component = 1.32 cos 168 = −1.291𝑁𝑁 Y-component = 1.32 sin 168 = 0.2744𝑁𝑁

𝑭𝑭2 : X-component = 2.01 cos 26 = 1.807𝑁𝑁

Y-component = 2.01 sin 26 = 0.8811𝑁𝑁

𝑭𝑭𝑟𝑟 : X-component = 1.807– 1.291 = 0.516N

Y-component = 0.2744 + 0.8811 = 1.156N

Under equilibrium conditions, |𝑭𝑭𝑒𝑒 | = |𝑭𝑭𝑟𝑟 | = 1.26𝑁𝑁

𝜃𝜃𝑒𝑒 = 180° − 𝜃𝜃𝑟𝑟 = 180 − tan−1

1.156

0.516

= 114°

Page 15 of 24

Comparison: 𝑭𝑭𝑒𝑒 (𝑁𝑁) 1.26 1.20

Calculated Measured

𝜃𝜃𝑒𝑒 (𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷) 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 𝑭𝑭𝑒𝑒 . 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.𝜀𝜀? •

Fitting all the points into a straight line signifies load P is linearly proportional to the strain increment 𝜀𝜀. This shows that the truss member had undergone linear elastic deformation, which means the truss member obeys Hooke’s law (𝜎𝜎 = 𝐸𝐸𝐸𝐸).

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

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

From equation (1), 𝐹𝐹𝐺𝐺𝐺𝐺 =

𝑃𝑃 𝑃𝑃 𝑃𝑃𝑃𝑃 = = sin 𝜃𝜃 ℎ/𝑙𝑙 ℎ

𝐹𝐹𝐺𝐺𝐺𝐺 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 𝐹𝐹𝐺𝐺𝐺𝐺 will be the same under the

same load with different cross sectional area. •

From equation (6), 𝛿𝛿 =

𝛿𝛿𝐺𝐺𝐺𝐺 𝑙𝑙𝑙𝑙 𝑙𝑙 2 𝜀𝜀 = = sin 𝜃𝜃 ℎ/𝑙𝑙 ℎ

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

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

𝐸𝐸 =

𝜎𝜎 𝜀𝜀

=

𝑃𝑃 𝑙𝑙

𝜀𝜀 𝐴𝐴ℎ

Page 17 of 24

Rearranging equation (5), 𝜀𝜀 =

𝜎𝜎 𝑃𝑃 𝑙𝑙 = 𝐸𝐸 𝐸𝐸 𝐴𝐴ℎ

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 𝜀𝜀 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 𝐹𝐹𝐷𝐷𝐷𝐷 .

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 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? •

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 19 of 24

Conclusion Experiment 1: In theory, the equilibrant vector force 𝑭𝑭𝑒𝑒 exactly offsets the combined effect of 𝑭𝑭1 and 𝑭𝑭2 , which is the resultant force 𝑭𝑭𝑟𝑟 . However, in this experiment we observed that 𝑭𝑭𝑒𝑒 does not

exactly balance out 𝑭𝑭𝑟𝑟 . 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 𝜀𝜀. This showed that the truss member had undergone linear elastic deformation,

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

𝑃𝑃

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

𝑃𝑃 𝑙𝑙

𝜀𝜀 𝐴𝐴ℎ

𝑃𝑃 ℎ

𝜀𝜀 𝑙𝑙 2

𝜀𝜀

60

12×10−6

= 5 × 106 ×

= 5 × 106 ×

450

= 5 × 106 N

81.7×427

427

450 2

= 64.5 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚2

= 10500 𝑁𝑁/𝑚𝑚𝑚𝑚

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 20 of 24

Appendix A1

Appendices

Table 1: (Case 1) 𝑀𝑀1 = 55𝑔𝑔, 𝑀𝑀2 = 105𝑔𝑔 𝑀𝑀1 (g) 55

𝑀𝑀2 (g)

105

𝜃𝜃1 (degree) 168

𝜃𝜃2 (degree) 26

Table 2: (Case 2) 𝑀𝑀1 = 135𝑔𝑔, 𝑀𝑀2 = 205𝑔𝑔 𝑀𝑀1 (g)

135

𝑀𝑀2 (g)

205

𝜃𝜃1 (degree) 168

𝜃𝜃2 (degree) 26

𝑭𝑭1 = 𝑀𝑀1 𝑔𝑔 (N)

𝑭𝑭2 = 𝑀𝑀2 𝑔𝑔 (N)

𝑭𝑭1 = 𝑀𝑀1 𝑔𝑔 (N)

0.54

1.32

1.03

𝑭𝑭𝑒𝑒 (N)

0.70

𝜃𝜃𝑒𝑒 (degree)

𝑭𝑭2 = 𝑀𝑀2 𝑔𝑔 (N)

𝑭𝑭𝑒𝑒 (N)

𝜃𝜃𝑒𝑒 (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 (𝑚𝑚𝑚𝑚2 )

81.6

82.2

81.3

Average Area, A =

Other measurements: h = 427mm

81.6+82.2+81.3 3

= 81.7𝑚𝑚𝑚𝑚2

l = 450mm

Table 4:

Strain Increment P Loading (N) 0

𝜀𝜀 , (× 10−6 ) 0

Unloading

𝜀𝜀 (× 10−6 ) 0

𝜀𝜀 , (× 10−6 )

𝜀𝜀 (× 10−6 )

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