Fettuccine Truss Bridge (Core Studies) by Cheah Ee Von

ARC2523: Building Structures Project 1: Fettuccine Truss Bridge 0311622 Gennieve Lee 0315210 Celine Tee 0308719 Cheah Ee Von 0316146 Choong Wan Xin 0311397 Chiang Kah Wai 0317197 Joash Lim YunAn 7 May 2015

Table of Contents Chapter 1: Introduction Introduction to the Project Aim and Objective Scope Limitations Equipment and Materials Used Testing of Fettuccine Testing of Adhesive Strength Methodology Schedule of Work

2 3 4 5 10 11 12

Chapter 2: Precedent Study Deep River Camelback Truss Bridge

Chapter 3: Experimentation and Progress 3.1 Bridge #1 3.2 Bridge #2 3.3 Bridge #3 3.4 Bridge #4

17 17 20 23 25

Chapter 4: The Final Bridge Design Process Members of the Bridge Connections in the Bridge Forces in the Bridge Calculation of Forces in the Bridge

28 29 30 32 36 39

References

Individual Case Studies Case Study 1 Case Study 2 Case Study 3 Case Study 4 Case Study 5 Case Study 6

Chapter 1: Introduction This project is commissioned by ARC2523: Building Structures. In a group of six, students are assigned to build a bridge using fettuccine as the materials for the bridge truss members. The weight of the bridge must not be more than 200g, but is required to carry a much larger weight for an extended period of time. The clear span of the bridge must be at least 750mm and students are to achieve a high efficiency with their bridge. As an unconventional building material, the fettucine represents a new material that could possibly be introduced into the industry in the future. With many unknown variables, including its compression and tension capabilities, students are asked to experiment and determine the best design for a bridge. Students will learn to explore truss members using different arrangements to achieve the best performance and how to build the perfect truss. One of the ways of determining the strengths and weaknesses of the fettuccine was simply to test its tension and compression capabilities. Through the case studies, students will apply the knowledge learned in class to calculate the moment force, reaction force, internal force, and force distribution of a truss. And thus, students will then be able to identify which members of the truss system must be strengthened in its tension or compression force. The purpose of the precedent studies was to used as a reference on how to properly design an efficient truss bridge, which is mandatory for the group of students. This report contains information regarding the students’ analysis and documentation of the experimentation with several fettuccine truss bridge designs. Individual case studies are also in this report, along with insight and suggestions for improvement.

Aim and objective The aim of this project is to develop students’ understanding of force distribution in a truss. It also aims to teach that design and construction methods can alter the efficiency in withstanding loads. The objective of this project is to discover the most successfully efficient bridge design specifically for a material like fettuccine (in regards to its tension and compression capabilities). Students are also to learn the importance of the forces present in the bridge design, especially considering a relatively weak material like fettuccine. Also, the project aims to teach students to build a perfect truss, which has positive aesthetic value, and minimizes construction material to the sufficient amount.

Scope The scope of this project was limited to using fettuccine as the construction material only. The bridge constructed had to have a clear span of 750 mm. Precedent studies, material testing, model making and load testing process were conducted to achieved the highest load possible withstand by the bridge. Structural analysis of the bridge was also conducted through calculations as well as observation of the load test from recorded videos.

Limitations Our main limitation was that some of the fettucine strands were not up to standards, as they were twisted, and not in uniform. This caused a loss in efficiency, as there needed to be a quality check to make sure that all the fettucine pieces used were up to par in quality. Consequently, this also created unnecessary waste. 4

Equipment and Materials Used 1. San Remo Fettuccine : The most common brand of fettucine was San Remo, as it was most easily available to us, and more cost effective than the other imported brands.

Figure 1.1 San Remo Fettuccine 2. Pen Knife : Pen knife was used to cut the fettuccine to pieces during the making of the bridge truss. Each group member had their own pen knife, so the design of the knife may have differed.

Figure 1.2 Pen Knife 3. UHU glue : Used during the first attempt when sticking the truss members together. However, it was not very successful as the glue was not strong enough to hold the joint.

Figure 1.3 UHU Glue 4. 3second glue : Used during the second attempt when glueing the bridge joint together. However, after discussing with our lecturer, we found out that the bridge truss would be brittle if the bridge is left after 2 days. Hence, this might affect the strength and efficiency of the bridge.

Figure 1.4 VTech 3 Second Glue 5. E6000 glue : This glue was used for several of our test bridges, as we realized that 3 second glue became too brittle too quickly, industrial glue seemed to fit as an equally strong glue

Figure 1.5 E600 Industrial Glue 6. Cutting mat : Used to protect the table surface when cutting the fettuccine. Each group member had their own cutting mat, so it may have differed in size. However the function remains the same.

Figure 1.6 Cutting mat 6

7. Ruler : Used to measure the length and marking on each fettuccine before cutting it. Metal rulers are more durable and therefore more ideal when slicing with the pen knife.

Figure 1.7 Metal ruler 8. Sandpaper : Used to smoothen the edges of the fettuccine after cutting and gluing.

Figure 1.8 Sandpaper 9. 500 ml water bottle : Used as load during the load testing of the bridge. We used a local brand of bottled water because it was more accessible and much cheaper.

Figure 1.9 500 mL water bottle 7

10. Lanyard : Used to replace the Shook during the load testing of the bridge truss.

Figure 1.10 Lanyard 11. Laptop : Used for researching precedent studies, references, and also for the analysing of the report.

Figure 1.11 Laptop: Macbook Pro 15” 12. DSLR Camera : Used to capture and record the load testing process. The model we used was the Canon 60D.

Figure 1.12 DSLR Camera 8

13. Shook : Used to hook on one point of the bridge to measure the load capacity of the bridge. The one we used was red and plastic.

Figure 1.13 Shook 14. Electronic Balance : To measure the weight of the bridge truss and the weight of the load applied on the bridge.

Figure 1.14 Electronic Balance 15. Bucket : Used as a load by filling it up with water during the load testing.

Figure 1.15 Plastic bucket

Testing of Fettuccine

Figure 1.16 Measuring the length Figure 1.17 15 cm of Fettuccine

Figure 1.18 Testing load

Figure 1.19 During Testing

Testing of the Adhesive strength Types of Adhesive

Analysis

UHU Glue

● ● ●

Slowest solidify time duration (30 Seconds) Average strength efficiency Weak bond efficiency

3Second Glue (VTech)

● ● ● ●

Fastest solidify time duration (35 seconds) High strength Efficiency High bond efficiency Causes bridge to be brittle after leaving it for two days or more

E6000 Industrial Glue

● ● ● ●

Average Solidify time duration (1015 seconds) High strength efficiency Medium to High bond efficiency Strength of the bond or joint increases after leaving for a period of time (24 hours or more)

Methodology In a group of six, students were told to design and build an efficient truss bridge using only fettuccine. A few precedent studies were conducted to introduce different truss bridges, such as the Deep River Camelback Truss Bridge. The next step was to test the strength and weakness of the fettuccine bridge, as well as experimenting with different design bridges to build. Step 1: To test the tension and compression strength of the fettuccine, we simply tried to pull the ends (tension) and push them together (compression). We found that the compressive strength of fettuccine is weak, while it has great tensile strength. Step 2: From references from external examples, including the precedent study, we sketched two plausible designs for an efficient perfect truss bridge. We had to be sure to design it for a clear span of 750mm (leaving 30mm on the sides to hoist up on the table). Step 3: Proceed to build the first bridge. We used pen knives to slice the fettuccine members to the correct length. To join them together laterally, we sliced the ends diagonally, and made sure the joints did not line up, so as to avoid breakage. Using the E6000 industrial glue, the members are joint together. The threesecond glue was used to glue the joints together during the first few attempts. Alignment is also crucial in this process, as is craftsmanship. Joints must fit together perfectly, without unnecessary gaps between them. Step 4: After waiting for the industrial glue to dry properly (at least 24 hours), we are ready for testing! To test the bridge, we set two tables (of equal height) exactly 750mm apart. and set the bridge in the middle. In preparation for the bridge testing, we must first weight the bridge to see how far above or below the 200 g mark we are. After noting the weight, the S hook and bucket are weight to calculate the total weight of the load that will be hung from the bridge. Step 5: Place the S hook on the midpoint or center point of the bridge, leaving the other side of the hook hanging down. So that the bucket is not too far from the ground. A piece of string is then used to connect the S hook and the bucket handle. Step 6: Using a 500ml water bottle (filled to the very top), water is slowly poured into the bucket. As load is being added, the bridge is checked for any deformities. As the fettuccine begins to deform, the points where the bridge are the weakest will then be noted down. We then continued pouring until the bridge broke. Using the 500ml bottle as a reference, we calculated how much water was poured in (millileters is equal to grams). Step 7: The second bridge design is built, and step three to six is repeated. Using the data collected, the stronger bridge is then selected and the design is refined according to the bridge testing data. The parts that deformed quickly are strengthen, where the bridge failed.

Schedule of Work (2015)

3 April 2015

Research and preliminary sketches, discussion of possible design ideas, possible glues to use and other possible techniques.

6 April 2015

First experimentation with the fettuccine. Testing of its tensile and compressive strength, as well as its ability to connect using different glues. Deciding on the design of the first bridge to be built (Bridge #1).

9 April 2015

Building of Bridge #1.

11 April 2015

Continued building of Bridge #1.

13 April 2015

Testing of Bridge #1. Discussion of a different design that may be stronger than Bridge #1.

14 April 2015

Building of Bridge #2.

15 April 2015

Testing of Bridge #2. Discussion of how to strengthen the bridge design, possibly by shortening the height and/or thickening the members.

18 April 2015

Building of Bridge #3.

21 April 2015

Testing of Bridge #3. Discussion on how to strengthen the areas that were faulty, especially the center piece that directly holds the weight of the bucket of water. Notation of craftsmanship. Discussion of possible change in glue used.

24 April 2015

Building of Bridge #4.

25 April 2015

Testing of Bridge #4. Finalization of the design to be tested for assessment.

26 April 2015

Building of Bridge #5.

27 April 2015

Final testing of the bridge (Bridge #5) for assessment.

Chapter 2: Precedent Study Deep River Camelback Truss Bridge

Figure 2.1 Deep River Camelback Truss Bridge History Built in 1910, the Deep River Camelback Truss Bridge spans the deep river of North Carolina; USA, providing access between Cumnock and the Gulf community in southern Chatham County. It underwent many reconstructions and renovations until 1992. For instance, the wooden deck was paved over with asphalt and modern guardrails were also added in later years. The guardrails were supported by their respective beams as opposed to being bolted onto the bridge's structural members. A few of the diagonals had been reinforced with parallel th cables as well. On June 9 1995, the Truss Bridge was listed in the National Register of Historic Places. The highway was rerouted, and the bridge is now open to pedestrian traffic only.

Figure 2.2 Side View of the bridge Structure The Truss Bridge is a steel camelback truss , with 360 ft. long in total and divided into 8panels, with the two trusses joined by a system of portal, lateral, and sway braces. The elevation of the bridge is 210 ft. above sea level. The camelback configuration owes its origins to the Pratt truss, and the two designs are very similar. With the exception of having two extra panels, the order of the diagonal structural members on this bridge are identical to that of the smaller Rocky River bridge. The two center panels, for example, have double diagonals that cross one another to form an "X." The most prominent feature that makes this bridge a camelback is the top chord, the uppermost part of the bridge that runs parallel to the deck. The Deep River Truss Bridge's top chord have three sections to it. The center section is straight, while the left and right ones angle down to the front posts. Thus, the top chord has four angles to it. The semicurve to the top chord allows the bridge to span a greater distance using less material. 14

Figure 2.3 Joint connection of Truss to Figure 2.4 Top view of the Joint Connection Base

Figure 2.5 Joint connection of Main Frame Load Analysis The truss system is pin joint. The trusses have diepunched eyes on the bottom chords as well as forgewelded double clevises on all tie bars. The bridge is supported by coarse stone and concrete piers, with a macadam covered plank road surface. The remains of the earthen base and timber support for the central pier of an earlier bridge may be seen below the bridge.

Figure 2.6 Base of Bridge The bottom chord is comprised of a series of eye bars, similar to the diagonals, as opposed to a single beam. The vertical members of the bridge used are referred to as "builtup" beams, meaning that the builders took two lengths of steel and "stitched" them together to form one. This is accomplished with short pieces of steel riveted together across the beam in a "zigzag" pattern. The top chord and front posts are builtup as well, using steel plates called batons.

Figure 2.7 Camelback Truss Conclusion The crossed ‘X’ braces in the center section of the bridge reinforced most of the members in tension. Hence, the Camelback truss design makes the bridge more efficient since fettuccine is good in tension but poor in compression. 16

Chapter 3: Experimentation and Progress Experimenting on different trusses and altering them to become stronger each time.

3.1 Bridge #1 As this was our first bridge, we started with just our instincts and tried to keep the design as simple as possible. Below is the diagram of our first bridge design. From our own research, we established that the triangle bridge would be the most efficient. We started with following the style of the Waddell “A” Truss and tried to incorporate elements of the Fink truss for more strength. Due to our concern for the strength of the joints, we also added gusset plates to the joints for extra support. 3Second glue was used for the entirety of the bridge, both for joining members together and for joint connection.

Figure 3.1.1 Combination of Waddell “A” truss and Fink truss Design

Figure 3.1.2 Model of Bridge #1 Clear Span: 750 mm Total Length: 820 mm Height of the bridge : 200 mm Weight of the bridge : 186.1 g Total Load withstand : 2440 g Efficiency : 31.99% Test results : The results were fair for this trial, as it was our first. Considering our unconventional bridge design, we did not know what to expect. At the suspected fault of both craftsmanship and also a lack of support at the higher end of bridge. We felt the gusset plates played very well in strengthening the joints of the bridge. But due to its height, we were concerned that each member was not thick enough to support the forces.

Figure 3.1.3 Breakage Point of Bridge #1 18

Figure 3.1.3 shows the whole bridge design of Bridge #1. Colored in red are the members that did not have sufficient internal force to withstand the external force. The member at the top was already susceptible to breakage, but the lower left ends were unexpected, and therefore our group suspected that it was due to poor craftsmanship as it was our first experiment.

Figure 3.1.4 After testing of Bridge #1

3.2 Bridge #2 For the second bridge, we had to change the design of the truss as the analysis calculation for the first bridge showed that it was redundant. Hence, we redesign the second bridge truss by combining the Waddell “A” Truss and Camelback Truss instead. For the second bridge, we added more vertical and diagonal members in order to increase the compressive and tensile strength of the trusses. The thickness of the fettuccine members were also the same as Bridge #1. The height of the bridge was also similar to the first bridge, which was 200 mm. After researching about the strength of the glue, we found out that the 3second glue could actually cause the fettuccine bridge to be more brittle if leaving the bridge to dry more than 24 hours. Hence, we decided to limit the usage of the 3second glue to the joint connections only, while using the E6000 industrial glue for the joining of the fettuccine members.

Figure 3.2.1 Combination of Waddell “A” Truss and Camelback Truss with height of 200mm

Figure 3.2.2 Model of Bridge #2 20

Clear Span: 750 mm Total Length: 820 mm Height of the bridge : 200mm Weight of Bridge: 220 g Total Load Withstand: 2170 g Efficiency : 21.40% Test result: The total load withstand was lower compared to the first bridge due to the exceeding weight of the bridge and poor craftsmanship at the joint connection of the truss members. Based on the efficiency analysis, the higher the weight of the bridge, the lower the efficiency of the bridge. The second bridge broke at the center part, where the load is withstand at the highest. Although the load withstand was lower, the overall result and recording showed that the bridge performs better in terms of tensile and compressive strength compare to the first one as it was near to a perfect truss design. Besides, the used of industrial glue also might have affected the strength of the bridge when compare to the 3seconds glue. In terms of strength, the 3second glue was faster and stronger in holding the bridge together. However, we still continued using the industrial glue as we had to leave the bridge overnight before we could test it the following day.

Figure 3.2.3 Breakage point of Bridge #2 Shown in Figure 3.2.3 is the whole bridge design. Colored in red are the members that did not have sufficient internal force to withstand the external force. Similarly to our first bridge testing, we found that our top member was at fault for the breakage. However, since it broke on only one side, we were unsure whether it was a craftsmanship issue or a member issue.

Figure 3.2.4 First type of Figure 3.2.5 Second type of center piece of center piece

Figure 3.2.6 Second type of middle piece snapped in the middle

3.3 Bridge #3 After reviewing the results from the last testing, we believed that if we shortened the total height of the bridge, then it would be stronger. Since fettuccine has better tensile strength than compressive strength, the shorter each member is, the stronger it becomes. Also, by shortening the total height of the bridge, we could thicken members of the bridge we believed were weak. One of the things we were concerned about was the base of the bridge, because of its length. However, our only solution was to thicken the base for strength reinforcement. We thought little of the centerpiece of the bridge, that would directly hold the weight of the bucket and water. In hindsight, it was quite foolish of us to so easily overlook this aspect, considering the direct force would be placed on this piece before being distributed to the rest of the bridge.

Figure 3.3.1 Bridge #3 with height of 150 mm

Figure 3.3.2 Adding load to the Bridge Clear Span: 750 mm Total Length: 820 mm Height of the bridge : 150 mm Weight of the bridge : 186.6 g Total Load withstand : 2570 g 23

Efficiency : 35.39% Test results : For the third bridge, the load withstand was increased by a few hundred grams. After analysing the second bridge, we continued using the same truss design as we find it to be the most efficient one. However, we changed a few of the measurements on the truss members so as to reduce the center of gravity of the bridge as well as increasing the tensile strength in hopes that it would increase the load withstand by the bridge. For instance, we decreased the height of the bridge by 50 mm from 200 mm to 150 mm. As a result, our center piece that directly held the load of the Shook and therefore the bucket struggled quite a bit, and at first, broke under the weight of 770 grams. However, after rebuilding the center piece another two times, the final load it could withstand was 2570 grams.

Figure 3.3.3 Breakage point of Bridge #3 Shown in Figure 3.3.3 is the whole bridge design. Colored in red are the members that did not have sufficient internal force to withstand the external force. The first to break was the base, and it could either be a craftsmanship issue (regarding the connections between the fettuccine members) or simply that the base wasn’t thick and supported well enough.

Figure 3.3.4 After testing

3.4 Bridge #4 Using the same truss design from the second and third bridge for this fourth one, we were able to achieve a higher efficiency due to the change in measurements, the types of glue used and the improvement of craftsmanship especially at the joint areas. As a result, the load withstand by the fourth bridge was also increase up to 4590 grams, which was by far our highest achievement. After testing and analysing the previous bridges, we reduced the height of the bridge truss as well as increased the thickness of the base and diagonal members of the bridge by glueing one to two pieces more of fettuccine. The thickness of the base consisted of 5 layers of fettuccine while the top main diagonal member was 4 layers thick. The height was decreased by another 50 mm again, which makes the bridge 100 mm in total height. The load testing had to be tested over and over again due to the breakage at the center member of the fettucine bridge, that was hooked directly with the Shook and bucket of water. Hence, we realised that the solution to the problem was to instantly test the load withstand by the bridge after glueing the middle piece of the fettuccine member to the bridge using the 3seconds glue as it causes the fettuccine members to be brittle after leaving it to dry overnight.

Figure 3.4.1 Bridge #4 with height 100 mm

Figure 3.4.2 Model of Bridge #4

Figure 3.4.3 Perspective view of the bridge

Clear Span: 750 mm Total Length: 820 mm Height of the bridge : 100 mm Weight of the bridge : 187.6 g Total Load withstand : First try 1670 g Second try 2240 g third try 3470 g Final test 4590 g Efficiency (final test) : 112.30% Test result: Overall, the total load withstand by the fourth bridge was 4590 g, which by far was our highest load. The efficiency calculation of the bridge also exceeded more than a hundred. From the test result, we realised that decreasing the height as well as increasing the thickness of the bridge members could actually affect so much of the final load withstand as it not only increase the compressive but also the tensile strength. The weight of the bridge was also within the requirement of the brief, which was under 200 g.

Figure 3.4.4 Breakage point of Bridge #4 Figure 3.4.4 is the fourth bridge design. Colored in red are the members that did not have sufficient internal force to withstand the external force. Although fettuccine is great in tensile strength, the vertical member in the middle is seen to be faulty. This may be due to the joint connection with the base and top members. As also shown in other images below, the bridge seems to have snapped in half. This leads us to believe that the top members are much weaker than we anticipated. The closer to the middle vertical, the higher the top member, and therefore, the longer each lattice member is. As a conclusion, the results prove that fettuccine is weak under compressive strength, and even weaker when the members are too long.

Figure 3.4.5 Adding load to the bridge

Figure 3.4.6 After testing

Figure 3.4.7 Bridge broke in half after testing the load

Chapter 4: Final Bridge 4.1 Design of Truss

Figure 4.1.1 3D model of the final bridge For the fifth and final bridge, we continued using the same truss design and measurements from bridge #4. Hence, the efficiency calculation was expected to be the same as the previous testing. However, it was not deem so as the expected load withstand by this bridge was not even close to the actual aim and goal which was more than 4000 grams. The top main diagonal members and base were the thickest members of the entire bridge as there were 5 layers of fettucine glued together. Another layer of fettuccine was added to the main diagonal member and base member from bridge #4, which originally consist of 4 layers only. For the base, it was glued similar to an Ibeam design, while the top main diagonal members were layered on top of one another instead. Based on our analysis, we could increase the compressive strength of the bridge by increasing the thickness of the main frame members as they are the main support member of the entire bridge. The next support members are the diagonal and vertical trusses on the bridge. Adding more diagonal members would actually help in increasing the tensile strength of the bridge, which is to the advantage of the fettucine as it works best under tension due to its flexibility to bend. Besides, adding vertical trusses to the bridge can also help to increase the compressive strength as fettucine are weak in compression. The vertical and diagonal trusses only consist of 28

two layers of fettuccine, whereas compared to the main diagonal and horizontal frame of the bridge that has more layers. For the middle piece of the fettuccine bridge that directly holds the shook to withstand the load, the thickness of it consist of 8 layers and is the thickest members out of all the bridge members because it directly supports the heaviest load.

Figure 4.1.2 Final model of the bridge Figure 4.1.3 Side view of the bridge model

4.2 Design Process

Figure 4.2.1 The design process of different bridges Figure 4.2.1 shows the different changes that were made to the bridge over the period of time when testing the different bridge design for its compressive and tensile strength. The first diagram actually showed the first bridge design that we came up with, which ended up to be redundant based on the calculation. The following bridge designs were a combination of the Waddle “A” truss and Camelback truss. The only difference between the last three bridges was the change in height, which they were decreased by every 50 mm each time they were tested. By reducing the height, we believe that it would reduce the compression as fettuccine works best under tensile strength.

Members of the Bridge

Figure 4.2.2 Main Frame of the bridge

Figure 4.2.3 Vertical truss members of the bridge to support the compressive strength

Figure 4.2.4 Diagonal truss members of the bridge to support the tensile strength

Figure 4.2.5 Horizontal bracing joining two sides of the bridge

Figure 4.2.6 Center Piece of the bridge that directly holds the Shook and Load

4.3 Connections of the Bridge

Figure 4.3.1 Fettuccine members of Final Bridge The above figure 4.1.10 shows the different colored fettuccine members representing the different type of members as well as difference in the number of layers on each member. The pink colored fettuccine represent the horizontal beams that connects the two sides of the frame and is made of one layer. The grey colored fettuccines are the diagonal and vertical trusses, which are made of two layers of fettuccine. As for the Blue colored ones, they are the main diagonal beams that hold the bracing together and are made up of 4 layers. Finally, the yellow members, which consist of 8 layers is the center component of the bridge that directly holds the Shook and carry the most load. 32

1. Beams

Figure 4.3.2 Diagonal joints of the bridge truss The horizontal beams or known as the base of the final bridge are made up of 5 layers of fettuccine. The fettuccine were layered over one another using the 3seconds glue with a minimum total length of 820 mm, as the clear span requires at least 750 mm. The end of each fettuccine piece were cut diagonally in precision before glueing them together to achieve the total length required. Precision and good workmanship is vital as it determines whether the ends of each fettuccine member can fit each other perfectly when connecting them to form a long, straight single beam. This will help to strengthen the stability of the bridge. 2. Diagonal Beam Connection

Figure 4.3.3 Diagonal Beam Joints 33

Figure 4.3.3 shows the connection between the top main diagonal beams of the bridge. The ends of each fettuccine piece is cut diagonally at an angle to achieve an accurate and perfect beam to beam connection. A direct and precise contact of the end of each beam surfaces is crucial as it allows the adhesive to bond the beams evenly and securely. Thus, creating a durable and strong joint. The length of each main diagonal beam is 450 mm long and made up of 5 layers of fettuccine as well to increase the compressive strength. 3. Vertical and Diagonal members (Bracing)

Figure 4.3.4 Vertical and diagonal members The connections that support the top two diagonal beams are crucial in distributing the load to the two ends as well as vertically supporting the long 450mm members. When constructing this part of the bridge, good craftsmanship and precision is vital to achieve an efficient and strong bridge. Each fettuccine member, which is two layers thick, must be cut exactly to the measurement to fit perfectly the corner of each fettuccine member when connecting them. There must not be any gap or extruding parts at the joint area as it will affect the strength and stability of the bridge. The excess parts must be removed, either by a pen knife or by sandpaper, and molded to fit the corner between the vertical members and base, as well as between the vertical member and the top member. The height of the highest vertical member of the bridge is 100 mm, while the gap distance between each vertical member is around 50 mm. 4. Horizontal members, Center Component 34

Figure 4.3.5 Horizontal members and center component of Final Bridge The horizontal members of the bridge strengthen the connection between the two bridge frames together to form a bridge truss. The member is placed with horizontal facing, which has a larger surface area for easy joint connection. The total length of each horizontal member is 60 mm long. The top horizontal members, which are the blue colored ones based on figure 4.3.5, are made up of one layer of fettuccine each, as they do not resist much load except tensile force. In addition, the top horizontal layer of fettuccine member is connected directly on top of the diagonal beam using the 3seconds glue. Similar to the blue members, the yellow colored bottom horizontal members of the final bridge from figure 4.3.5, are also made up of only one layer of fettuccine each. They are attached on top of the base or bottom beam of the bridge using the 3seconds glue also. Referring back to Figure 4.3.5, the purple colored members on the bridge represent the center component of the bridge, which is made up of 8 layers of fettuccine. It bears the most weight because the Shook, carrying the load is directly hooked on to it. It is crucial for the center component to be very thick in order to withstand more loads before the structure collapse. The length of the center component of the fettuccine bridge has the same length as the horizontal members, which is 60 mm long.

4.4 Forces in the Bridge 35

Figure 4.4.1 Strength of Bridge The forces in the bridge are a mixture of compression and tension. The weakest point of the bridge is highlighted in yellow while the strongest point is highlighted in red, as seen from figure 4.4.1. As fettuccine performs best under tension, we increased the thickness of the main diagonal frame and base member by another layer of fettuccine. Thus, this was able to also increase the tensile strength of the bridge. In Figure 4.4.2 The black solids at both ends of the truss represent the table that the bridge was set on. The clear span between the tables is 750 mm apart. As the yellow arrows demonstrate, there is an upward force from the table, which is holding up the bridge. The base is in tension, which is good because fettuccine has good tensile strength. However, the the vertical and top members are in compression, which is a weak aspect of fettuccine, so the diagonal lattice members are there to support them with tensile strength. The way that the diagonal lattice members are cut also aid in resisting the shear forces. Figure 4.4.2 Forces acting on the bridge

Clear Span: 750 mm Total Length: 820 mm Height of the bridge : 100 mm Weight of the bridge : 182 g Total Load withstand : 2297 g Efficiency : 28.99% Test result: On the day of presentation and final testing of the final bridge, the load withstand by the bridge was half of the original load that was tested with bride #4, which was 2297 g only. Based on our analysis, the factors that affected the load carried by the bridge was due to the poor craftsmanship as well as the time difference used by each group member to complete glueing all the layers of fettuccine to form the bridge. During the process of building the bridge truss, the task was segregated differently to different group members compared to the fourth testing. Thus, affecting the workmanship. Besides, the time span for allowing the glue to dry was also different and longer compared to the fourth bridge as we had to redo some part of the members due to the uneven layers of fettuccine. Besides, we also started constructing the final bridge earlier than the fourth bridge. Hence, it was to our conclusion that the fettucine was left to dry for too long. As a result, the fettuccine members became more brittle during testing, causing them to break easily when load was added to it.

Figure 4.4.3 Final testing, the bridge broke into half 37

Group photo on the day of final testing

References Bridgehunter.com,. 'Bridgehunter.Com | Deep River Bridge'. N.p., 2015. Web. 12 Apr. 2015. Sites.google.com,. 'Deep River Camelback Truss Bridge Matthew B Ridpath'. N.p., 2015. Web. 12 Apr. 2015. Cridlebaugh, B. (2008, June 3). Bridge Basics A Spotter's Guide to Bridge Design. Retrieved May 6, 2015, from http://pghbridges.com/basics.htm Lamb, Robert, and Michael Morrissey. "How Bridges Work" 01 April 2000. HowStuffWorks.com. <http://science.howstuffworks.com/engineering/civil/bridge.htm> 06 May 2015.

Individual Case Studies Case Study 1: Cheah Ee Von Case Study 2: Chiang Kah Wai Case Study 3: Gennieve Lee Case Study 4: Celine Tee Case Study 5: Joash Lim Case Study 6: Choong Wan Xin

Case Study 1 0308719 Cheah Ee Von

Case Study 2 0311397 Chiang Kah Wai (Natalie)

Determination of truss: 2J = M + 3 2(9) = 15 + 3 18 = 18 (PERFECT)

Calculation of External Forces (F): ∑M @ J = 0 100(2) + 150(2) + 60(6) + 50(8) – 20(2.5) + FAx (3.5) = 0 FAx = -345.71 kN ∑Fx = 0 FJx + 50 + 80 – 20 – 345.71 = 0 FJx = 235.71 kN ∑Fy = 0 -100 – 150 – 60 – 50 + FJy = 0 FJy = 360 kN

Calculation of Internal Forces (F): 1. @ J: ∑Fy = 0 FAJ = 360 kN (C) ∑Fx = 0 FHJ = 235.71 kN (C)

2. @ A: ∑Fy = 0 3.5 360 = FAH (4.03) FAH = 414.51 kN (T) ∑Fx = 0 -345.71 + 414.51 (

2 ) 4.03

+ FAB = 0 FAB = 140 kN (T)

3. @ H: ∑Fy = 0 3.5 -FBH + 4140.51(4.03) – 150 = 0 FBH = 210 kN (C) ∑Fx = 0 2 235.71 – 414.51(4.03) – FHG = 0 FHG = 30 kN (C)

4. @ B: ∑Fy = 0 3.5 210 – 100 – FBG(5.315) = 0 FBG = 167.04 (T) ∑Fx = 0 FBC + 167.04(

4 ) 5.315

– 140 = 0 FBC = 14.287 (C)

5. @ G: ∑Fy = 0 3.5 -FCG – 60 + 167.04(5.315) = 0 FCG = 50kN (C) ∑Fx = 0 4 80 + 30 + FGF – 167.04(5.315) = 0 FGF = 15.712kN (T)

6. @ C: ∑Fy = 0 3.5 -FCF( ) + 50 + FCD(

1 ) 2.236

4.03

∑Fx = 0 2 -FCD( ) + FCF(

2 )– 4.03

2.236

From Eq.1: FCF =

50 + 𝐹𝐶𝐷 (0.447) 0.868

= 0 ── Eq.1

14.287 = 0 ── Eq.2

── Eq.3

Eq.3 sub into Eq.2: -14.287 + 28.57 + FCD (0.255 – 0.894) = 0 FCD = 22.35 kN (C) FCD sub into Eq.3: FCF = 69.11 kN (T)

7. @ D: ∑Fy = 0 1 FDF – 50 – 22.35(2.236) = 0 FDF = 60kN (C) ∑Fx = 0 -20 + 22.35(

2 ) 2.236

2.5 ) 3.5355

+ FDE (

=0 FDE = 0 (N)

8. @ E: ∑Fx = 0 FEF = 50 kN (T)

9. @ F: check for EQUILIBRIUM ∑Fy = 0 3.5 -60 + 69.11(4.03) = 0 ∑Fx = 0 2 50 – 15.712 – 69.11(4.03) = 0

Conclusion:

Case Study 3 0311622 Gennieve Lee

Case Study 4 0315210 Celine Tee

Determination of truss: 2J = M + 3

J = 9, M = 15

2(9) = 15 + 3 18 = 18 (Perfect truss) Calculation of External Forces (F): ∑M (J) = 0 100(2) + 150(2) + 60(6) + 50(8) – 20(2.5) + FAx (3.5) = 0 FAx = -345.71 kN ∑Fx = 0 FJx + 50 + 80 – 20 – 345.71 = 0 FJx = 235.71 kN ∑Fy = 0 -100 – 150 – 60 – 50 + FJy = 0 FJy = 360 kN

Calculation of Internal Forces (F): Joint A: ∑Fx = 0 -FAX + FAB = 0 FAB = 345.71 kN

∑Fx = 0 -FAJ = 0

Joint J: ∑Fy = 0

FJBy = FJB sin 60.15

FJy – FJBy = 0

FJBy = FJB cos 60.15

360 – 0.8682 x FJB= 0 FJB = 414.65 kN

∑Fx = 0 FJx – FJH – FJBx = 0 235.71 – FJH – 414.65 x 0.4962 = 0 FJH = 29.96 kN

Joint H: ∑Fy = 0 FBH – 150 = 0 FCH = 150 kN

∑Fx = 0 FJH – FHG = 0 FHG = 29.96 kN

Joint B: ∑Fy = 0

FBGy = FBG sin 41.19

-100 + FJBy - FBH - FBGy= 0

FBGx = FBG cos 41.19

-100 + 360 – 150 – FBG sin 41.19 = 0 FBG = 110 kN

∑Fx = 0 -FAB + FJBx + FBGx + FBC = 0 -345.71 + 205.7498 + 125.72 + FBC = 0 FBC = 14.25 kN

Joint C: ∑Fx = 0

FCDx = FCD sin 63.26

- FBC + FCD x = 0

FCDy = FCD cos 63.26

-14.25 + FCDx = 0 FCDx = 14.25 kN

∑Fy = 0 FCDy - FCG= 0 7.12 - FCG = 0 FCG = 7.12 kN

Joint G: ∑Fy = 0

FDGy = FDG sin 51.20

-FCG + FBGy -60 –FDGy = 0 FDGx = FDG cos 51.20 -7.12 + 110 – 60 – FDG sin 51.20 = 0 FDG = 42.88 kN

∑Fx = 0 -FBGx + FHG + 80 + FGF – FDGx = 0 -125.72 + 29.96 + 80 + FGF – 34.31 = 0 FGF = 50.07 kN

Joint F: ∑Fx = 0 - FGF + FFE = 0 FFE = 50.07 kN

∑Fx = 0 FDF = 0 kN

Joint D: ∑Fx = 0

FDEx = FDE sin 45

-FCDx + FGDx – FDRx – 20 = 0 FDEy = FDE cos 45 -14.25 + 34.31 – FDEx – 20 = 0 FDEx = 0.0652 kN

∑Fy = 0 -50 + FCDy+ FDEy - FDF + FDEy = 0 -50 + 7.12 + 42.88 – 0 + FDEy = 0 FDEy = 0 kN

Joint E: ∑Fx = 0 FFe – 50 + FDEx = 0 FDEx = 50.07 – 50 = 0.07 kN

∑Fy = 0 FDEy = 0

Result: Member AB is in highest tension force which is 345.71 kN and Member BJ is in highest compression which is 414.65 kN.

Case Study 5 0317197 Joash Lim Yun-An

tanϴ =

1) 2J = m+3, J = 9, m = 15 18 = 18 → perfect truss

3 , ϴ = 36.9ᵒ 4

sin 36.9ᵒ =

xy , xy = 60kN 100

cos 36.9ᵒ =

xz , xz = 80kN 100

2) ƩFy = 0 0 = RJ - Fxy - 150 - 50 - 100 RJy = 60 + 150 + 50 + 100 = 360kN ƩFx = 0 0 = RAx + RJx - 20 + Fzx + 50 RAx + RJx = - 110kN _____ (1) ƩM(J) = 0 RAx(3.5) + 100(2) + 150(2) + 80(0) + 60(6) + 50(8) +50(0) – 20(2.5) = 0 Sub RAx = -345.71 → (1) -345.71 + RJx = -110 RJx = 235.71kN

3) JOINT A

ƩFx = 0 -RAx + FAB = 0 345.71 = FAB ƩFy = 0 -FAJ = 0kN JOINT J

tanϴ =

2 , ϴ = 29.7ᵒ 3.5

ƩFy = 0 RJy + FJB cosϴ + FAJ = 0 360 + FJB cos29.7 + 0 = 0 FJB = -414.63kN ƩFx = 0 RJX - FJB sinϴ + FJH = 0 235.71 - 414.63 sin29.7 + FJH = 0 FJH = -30 kN

JOINT B

tanϴ =

2 , ϴ = 29.7ᵒ 3.5

ƩFx = 0 FBA - FBJ sinϴ + FBC = 0 -345.71 + 414.45 sin29.7 +FBC = 0 FBC = 140.08kN ƩFy = 0 -100 + FBH - FBJ cosϴ = 0 -100 + FBH + 414.63 cos 29.7 = 0 FBH = -260kN

JOINT H

tanϴ =

4 , ϴ = 48.8ᵒ 3.5

ƩFx = 0 -FHJ + FHG +FHC sinϴ = 0 +30 + FHG + FHC sin48.8 = 0_____ (1) ƩFy = 0 -150 - FHB +FHC cosϴ = 0 -150 + 260 + FHC cos 48.8= 0 Sub FHC = -167.05kN → (1) +30 + FHG - 167.05 sin48.8 = 0 FHG = 95.71kN

JOINT C

tanϴ =

1 , ϴ = 26.6ᵒ 2

tanα =

4 , α = 48.8ᵒ 3.5

ƩFx = 0 FBC + FCD cosϴ - FCH sinα = 0 -140.08 + FCD cos26.6 + 167.05sin48.8 = 0 FCD = 16.06kN ƩFy = 0 FCD sinϴ - FCH cosα + FCG = 0 16.06 sin26.6 + 167.05 cos 48.8 + FCG = 0 FCG = -117.18kN

JOINT G

tanϴ =

2.5 , ϴ = 51.34ᵒ 2

ƩFx = 0 FGH + FGF + 80 + GD cosϴ = 0 -95.71 + FGF + 80 + GD cos51.34 = 0_____ (1) ƩFy = 0 -FGC - 60 + GD sinϴ = 0 117.18 - 60 + GD sin51.34 = 0 Sub GD = -73.23kN → (1) -95.71 + FGF + 80 - 73.23 cos51.34= 0 FGF = 61.46kN

JOINT D

tanϴ =

2.5 , ϴ = 45ᵒ 2.5

tanα =

1 , α = 26.57ᵒ 2

ƩFx = 0 FDC cosα - FDG sin  FDG + FDE sinϴ - 20 = 0 -16.06 cos 26.57 + 73.23 sin38.7 + FDE sin 45 – 20 = 0 FDE = -16.1kN ƩFy = 0 -FDF + FDC sinα - FDG cos  FDG - FDE cosϴ - 50 = 0 -FDF - 16.06 sin26.57 + 73.23 sin38.7 - 16.1 cos45 - 50 = 0 FDF = -22.76kN

JOINT F

ƩFx = 0 FFG + FFE = 0 -61.46 + FFE = 0 FFE = 61.46kN

JOINT E

tanϴ =

2.5 , ϴ = 45ᵒ 2.5

ƩFx = 0 FFE + 50 + FDE cosϴ = 0 -61.46 + 50 +FDE cos45 = 0 FDE = 16.1kN INTERNAL REACTION FORCE DIAGRAM – Tension and Compression

Case Study 6 0316146 Choong Wan Xin

Determination of truss: 2J = M + 3 J = 9, M = 15 2(9) = 15 + 3 18 = 18 (Perfect truss) Calculation of External Forces (F): ∑M (J) = 0 100(2) + 150(2) + 60(6) + 50(8) – 20(2.5) + FAx (3.5) = 0 FAx = -345.71 kN ∑Fx = 0 FJx + 50 + 80 – 20 – 345.71 = 0 FJx = 235.71 kN ∑Fy = 0 -100 – 150 – 60 – 50 + FJy = 0 FJy = 360 kN Calculation of Internal Forces (F): Joint A: ∑Fx = 0 -FAX + FAB = 0 FAB = 345.71 kN ∑Fx = 0 -FAJ = 0

Joint J: ∑Fy = 0 FJy – FJBy + FAJ = 0 0 – FJB sin 60.26 + 0 = 0 FJB = 414.61 kN

FJBx = FJB cos 60.26 FJBy = FJB sin 60.26

∑Fx = 0 FJx – FJH – FJBx = 0 235.71 – FJH – 414.61 x cos 60.26 = 0 FJH = 30.04 kN Joint B: ∑Fx = 0 -FAB + FJBx + FBC = 0 -345.71 + 414.61 x cos 60.26 + FBC = 0 FBC = 140.04 kN ∑Fy = 0 -100 + FJBy + FBH = 0 -100 + 414.61 x sin 60.26 + FBH = 0 FBH = -260 kN Joint H: ∑Fy = 0 FCHx = FCH cos 41.19 FBH + FCHy – 150 = 0 FCHy = FCH sin 41 260 + FCH sin 41.19 – 150 = 0 FCH = -167.03 kN ∑Fx = 0 FJH – FCHx – FHG = 0 30.04 – 167.03 x cos 41.19 - FHG = 0 FHG = -95.66 kN

Joint G: ∑Fy = 0 FCG - FGy = 0 FCG - 60 = 0 FCG = 60 kN

FGx = 80 kN FGy = -60 kN

∑Fx = 0 FGx – FHG + FGF = 0 80 – 155.74 +FGF = 0 FGF = 75.74 Kn

Joint C: FCDx = FCD cos 26.57 FCDy = FCD sin 26.57

FCFx = FCF cos 60.26 FCFy = FCF sin 60.26

∑Fx = 0 - FBC + FCHx + FCD x – FCFx = 0 -140.04 + 125.70 + FCD cos 26.57 – FCF cos 60.26 =0 FCD cos 26.57 – FCF cos 60.26 = 14.34 ----- (1) ∑Fy = 0 FCFy – FCG + FCFy + FCDy = 0 110 – 60 + FCF sin 60.26 + FCD sin 26.57 = 0 FCF sin 60.26 + FCD sin 26.57 = -50 FCF = (-FCD sin 26.57 – 50) / sin 60.26 ----- (2), sub (2) into (1) FCD cos 26.57 – cos 60.26 x [(-FCD sin 26.57 – 50) / sin 60.26] = 14.34 0.8944 FCD + 0.1926 FCD + 21.54 = 14.34 FCD = -6.62 kN sub into (2) FCF = (-6.62 x sin 26.57 – 50) / sin 60.26 = -60.99k

Joint D: ∑Fx = 0 FDEx = FDE cos 45 FCDx – 20 - FDEx = 0 FDEy = FDE sin 45 6.62 x cos 26.57 – 20 – FDE cos 45 = 0 FDE = -19.91 kN ∑Fy = 0 -50 – FCDy + FDF + FDEy = 0 -50 – 6.62 x sin 60.26 + FDF + 19.91 x sin 45 = 0 FDF = -41.67 kN

Joint F: ∑Fx = 0 - FGF + FFE – FCFx = 0 -75.74 + FFE – 60.99 x cos 60.26 = 0 FFE = 106 kN

Result:

Member AB is in highest tension force which is 345.71 kN and Member BJ is in highest compression which is 414.65 kN.

Calculation of Forces in Final Truss

1) 2J = m + 3, J = 17, m =31 2(17) = 31 + 3 34 = 34 perfect truss. 2) ƩFy = 0 2R – Load (2.297 × 9.81) ÷ 2 = 0 R = 11.27N 3) JOINT A cosϴ =

5.275 10

ϴ = 58.16ᵒ α = 90ᵒ -‐ 58.16ᵒ = 31.84ᵒ ƩFx = 0 FAB cosα = 0 FAB = 0N ƩFy = 0 FAB sinα -‐ FAQ = 0 FAQ -‐ 0 = 0

JOINT Q

8.75 5.125

tan ϴ =

ϴ = 59.64ᵒ ƩFx = 0 FQB cosϴ + FQP = 0_____ (1) ƩFy = 0 FQB sinϴ + FAQ + Load = 0 FQB sin59.64 + 0 + 22.53 = 0 Sub FQB = -‐26.11 → (1) -‐26.11 cos59.64 + FQP = 0 FQP = 13.2N JOINT B tan α =

5.125 5.125 cos ϴ = 8.75 5.275

α = 30.36ᵒ ϴ = 13.7ᵒ ƩFx = 0 FBC cosϴ -‐ FBQ sinα + FBA cosϴ = 0 FBC cos13.7 + 26.11 sin30.36 + 0 = 0 FBC = -‐13.58N ƩFy = 0 FBC sinϴ -‐ FBQ cosα + FBA sinϴ + FBP = 0

-‐13.58 sin13.7 + 26.11 cos30.36 + FBP + 0 = 0 FBP = -‐19.31N JOINT P tan ϴ =

7.5 5.125

ϴ = 55.65ᵒ ƩFx = 0 FPC cosϴ + FPO + FPQ = 0_____ (1) ƩFy = 0 FPC sinϴ + FPB = 0 FPC sin55.65 + 19.31 = 0 Sub FQB = -‐23.39 → (1) -‐23.39 cos55.65 + FPO -‐ 13.2 = 0 FPO = 26.4N JOINT C tan α =

5.125 5.125 cos ϴ = 7.5 5.275

α = 34.35ᵒ ϴ = 13.7ᵒ ƩFx = 0 FCB cosϴ + FCD cosϴ +FCP sinα = 0 13.58 cos13.7 + FCD cos13.7 + 23.39 sin34.35 = 0 FCD = -‐27.16N ƩFy = 0 FCB sinϴ + FCD sinϴ + FCP cosα + FCO = 0 13.58 sin 13.7 -‐ 27.16 sin13.7 + 23.39 cos34.35 + FCO = 0 40

FCO = -‐16.1N JOINT O tan ϴ =

6.25 5.125

ϴ = 50.65ᵒ ƩFx = 0 FOD cosϴ + FOP + FON = 0_____ (1) ƩFy = 0 FOD sinϴ + FOC = 0 FOD sin50.65 + 16.1 = 0 Sub FQB = -‐20.82 → (1) -‐20.82 cos50.65 -‐ 26.4 + FON= 0 FON = 39.6N JOINT D tan α =

5.125 5.125 cos ϴ = 6.25 5.275

α = 39.35ᵒ ϴ = 13.7ᵒ ƩFx = 0 FDC cosϴ + FDE cosϴ +FDO sinα = 0 27.16 cos13.7 + FDE cos13.7 + 20.82 sin39.35 = 0 FDE = -‐40.75N ƩFy = 0 FDC sinϴ + FDE sinϴ + FDO cosα + FDN = 0 27.16 sin 13.7 -‐ 40.75 sin13.7 + 20.82 cos39.35 + FDN = 0 FDN = -‐12.88N 41

JOINT N tan ϴ =

5 5.125

ϴ = 44.29ᵒ ƩFx = 0 FNE cosϴ + FNO + FNM = 0_____ (1) ƩFy = 0 FNE sinϴ + FND = 0 FNE sin44.29 + 12.88 = 0 Sub FQB = 18.45N → (1) 18.45 cos44.29 -‐ 39.6 + FNM= 0 FNM = 26.39N JOINT E tan α =

5.125 5.125 cos ϴ = 5 5.275

α = 45.71ᵒ ϴ = 13.7ᵒ ƩFx = 0 FED cosϴ + FEF cosϴ +FEN sinα = 0 40.75 cos13.7 + FEF cos13.7 -‐ 18.45 sin45.71= 0 FEF = -‐27.16N ƩFy = 0 FED sinϴ + FEF sinϴ + FEN cosα + FEM = 0 40.75 sin13.7 -‐ 27.16 sin13.7 -‐ 18.45 cos45.71 + FEM = 0 FEM = 9.67N 42

JOINT M tan ϴ =

3.75 5.125

ϴ = 36.19ᵒ ƩFx = 0 FMF cosϴ + FMN + FML = 0_____ (1) ƩFy = 0 FMF sinϴ + FME = 0 FMF sin36.19 -‐ 9.67 = 0 Sub FMF = 16.38N → (1) 16.38 cos36.19 -‐ 26.39 + FML= 0 FML = 13.16N JOINT F tan α =

5.125 5.125 cos ϴ = 3.75 5.275

α = 53.81ᵒ ϴ = 13.7ᵒ ƩFx = 0 FFE cosϴ + FFG cosϴ +FFM sinα = 0 27.16 cos13.7 + FFG cos13.7 -‐ 16.38 sin53.81= 0 FFG = -‐13.55N ƩFy = 0 FFE sinϴ + FFG sinϴ + FFM cosα + FFL = 0 27.16 sin13.7 -‐ 13.55 sin13.7 -‐ 16.38 cos53.81+ FFL = 0 FFL = 6.45N 43

JOINT L tan ϴ =

2.5 5.125

ϴ = 26ᵒ ƩFx = 0 FLG cosϴ + FLM + FLK = 0_____ (1) ƩFy = 0 FLG sinϴ + FLF = 0 FLG sin26 -‐ 6.45 = 0 Sub FLG = 14.71N → (1) 14.71 cos26 -‐ 13.16 + FLK= 0 FLK = 0.06N JOINT G tan α =

5.125 5.125 cos ϴ = 2.5 5.275

α = 64ᵒ ϴ = 13.7ᵒ ƩFx = 0 FGF cosϴ + FGH cosϴ +FGL sinα = 0 13.55 cos13.7 + FGH cos13.7 -‐ 14.71 sin64= 0 FGH = 0.058N ƩFy = 0 FGF sinϴ + FGH sinϴ + FGL cosα + FGK = 0 13.55 sin13.7 + 0.058 sin13.7 -‐ 14.71 cos64+ FGK = 0 FGK = 3.23N 44

JOINT K tan ϴ =

1.25 5.125

ϴ = 13.71ᵒ ƩFx = 0 FKH cosϴ + FKL + FKJ = 0_____ (1) ƩFy = 0 FKH sinϴ + FKG = 0 FKH sin13.71 -‐ 3.23 = 0 Sub FKH = 13.63 → (1) 13.63 cos13.71 -‐ 0.06 + FKJ= 0 FKJ = -‐13.18N JOINT H tan α =

5.125 5.125 cos ϴ = 1.25 5.275

α = 76.29ᵒ ϴ = 13.7ᵒ ƩFx = 0 FHG cosϴ + FHI cosϴ +FHK sinα = 0 -‐0.058 cos13.7 + FHI cos13.7 -‐ 13.63 sin76.29= 0 FHI = 13.69N 45

ƩFy = 0 FHG sinϴ + FHI sinϴ + FHK cosα + FHJ = 0 -‐0.058 sin13.7 + 13.69 sin13.7 -‐ 13.63 cos76.29+ FHJ = 0 FHJ = 0.0018N ≈ 0N JOINT J ƩFx = 0 FJK + FJI = 0 13.18 + FJI = 0 FJI = -‐13.18N ƩFy = 0 FJH = 0 JOINT I tan ϴ =

1.25 5.125

ϴ = 13.71ᵒ ƩFx = 0 FIH cosϴ + FIJ = 0 -‐13.69 cos13.71 + FIJ = 0 FIJ = 13.18N