HSC ENGINEERING STUDIES ASSESSMENT TASK 1 CIVIL STRUCTURES

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Adelia Townsend
6 years ago
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1 HSC ENGINEERING STUDIES ASSESSMENT TASK 1 CIVIL STRUCTURES ANDREW HARVEY

2 CONTENTS CONTENTS... REPORT BODY... Truss Design:... Analysis of orces in all Members:... Breaking Stress in irst Member to ail:... Comparative Ratio:... 4 Conclusion:... 4 Recommendations:...4 APPENDIX A... 6 Design 1 (Warren Truss):... 6 APPENDIX B Design (Warren Truss): APPENDIX C JOINTS... 1 HSC Engineering Studies Assessment Task 1 of 14

3 REPORT BODY Truss Design: We used the first truss sign, (See Appendix A & B). We chose this one because it h the lowest maximum force in any member, and therore would take more lo than sign number two. Analysis of orces in all Members: gb gh bh ch = = = = = gb = bh = gh = ch = = = Breaking Stress in irst Member to ail: The member with the most force in it is. So we predicted that it would fail first. The actual weight that caused it to fail was 5kg. So the tensile strength of 6mm square balsa is 96N. However this would not be true if the first member to fail was in compression, or if the truss failed first at a joint. The first member to fail was unclear. Many members broke, and we couldn t see which members failed first. The position of breakages is shown below. HSC Engineering Studies Assessment Task 1 of 14

4 In total the truss failed in 8 different places. However many of these breakages resulted from one member initially failing. I suspect from the analysis of forces that the first member to break was the one in red. Stress in the member that is thought to have broke first. orce Stress = = = MPa Area Comparative Ratio: lo (kg) 5 Comparative Ratio = = = mass (kg) Conclusion: Member orce at point of Nature failure (N) Compression Compression Tension Compression Compression gb Compression bh Compression gh Tension ch 99.0 Tension 99.0 Tension Tension was the first member to break. It was in tension and it h a force of 96.06N acting on it. This stress in this member was 11.00GPa. Recommendations: Although it is not likely that the break at joint BGH was the first breakage, that joint could have been manufactured better. This was the only break that was directly at the joint, as all the other breakages were either in the middle of a member or just outsi where the gusset was attached. The gusset at joint BGH was the first one ma, and it was not as good as the rest. This gusset didn t optimise the surface area available for gluing. Also we could have applied more glue to the joints to make them stronger, and in places such as is shown in the figure below. We could have d glue here, extending no further than the gusset. To strengthen the joint. However this may not have been allowed. HSC Engineering Studies Assessment Task 1 4 of 14

5 Alternate Height: We also consired changing the height of the truss. By use of method of section the lower middle member h a force of 1.8N. 5 kg A 171.5N As shown by the section line, we ignore the parts on the lt. M By taking moments about point A, we have external forces acting on the truss. With only one unknown of the cental lower member, M. + M = 0 = ( ) + ( 87.0 M ) A M = 1.8N ( Tension) This is a less force compared with our actual sign having 96.06N. There was no height limit, so we could have ma the truss higher, this would mean that it could have held a greater lo bore failure. HSC Engineering Studies Assessment Task 1 5 of 14

6 Design 1 (Warren Truss): ree Body Diagram: APPENDIX A R L and R R can be calculated now. By looking at it you can see that R L = and RR =. However for the sake of accuracy, these forces can be calculated, as shown below. R R M + R L = 0 = 0. = = 0.6 ( 0.6 R ) + ( 0.) R R L M + R R = = 0 = ( 0.6 R ) ( 0.) L The internal stresses can also be calculated. ADC: Close up of Joint ADC 0 The R L force is known in both magnitu and direction, so it is drawn in first. The forces in members and are known in direction but not magnitu or sense. However, the sense can be HSC Engineering Studies Assessment Task 1 6 of 14

7 obtained from the above force diagram, along with the magnitus. = = cos0 = tan 0 = ADE: We know the forces in member, so the forces in members and can be calculated. Close up of Joint ADE The forces in member are known in both magnitu and direction, so it is drawn in first. The forces in members and are known in direction but not magnitu or sense. However, the sense can be obtained from the above force diagram, along with the magnitus. and are easily resolved. It is an equilateral triangle because all the internal angles are, so = =. So far the forces in the truss are as follows: CDE: There are 4 members at joint CDE. two of which are unresolved. HSC Engineering Studies Assessment Task 1 7 of 14

8 Close up of Joint CDE The forces in members and are known in both magnitu and direction, so they is drawn in first. The forces in members and are known in direction but not magnitu or sense. However, the sense can be obtained from the above force diagram, along with the magnitus. To resolve and, I have split this qurilateral into two triangles. One of them is shown below. 0 n = = cos60 It can be seen that as the polygon is symmetrical =. So now the forces in the truss are as follows: inal Analysis As the truss is symmetrical, and the force is applied centrally, then we can assume the same forces for the other si of the truss. gb gh bh ch (All triangles are equilateral ( angles), and all members are 00mm.) Where: (as previously calculated) (magnitus only) HSC Engineering Studies Assessment Task 1 8 of 14

9 = = = = = gb = bh = gh = ch = = = So with these results, I know which members are in tension and which are in compression (from the above diagram), and I can also work out the forces in each member given different masses applied at force. Now I could see how the truss would behave. I tested it with a 50kg mass applied centrally at the top. The forces I calculated are shown on the diagram below. (50kg) 490N 8.9N 8.9 N 8.9 N 8.9 N 8.9 N 8.9 N 8.9 N 8.9 N N 45N 45N As there are 11 members, each being 0.m long. The total amount of balsa used in this sign is.m. m of balsa weights 100g, so the.m that was used in our truss sign weighed 75g. (100kg) 980N 565.8N N N N N N N N 8.9 N N 8.9 N 490N 490 N (5kg) 45N 70.7 N 8.9 N 70.7 N 1.5 N 1.5 N HSC Engineering Studies Assessment Task 1 9 of 14

10 APPENDIX B Design (Warren Truss): A B E G D H I J K L 10 Where all triangles are equilateral. C The forces in members,,,, and have been calculated in sign 1, and they are the same for this truss sign. So the following forces are known. AEG There are 4 members at joint AEG. two of which are unresolved. ag 10 ag 10 ree Body Diagram Close up of Joint AEG. orce Diagram = ag = CGH There are 4 members at joint CGH. two of which are unresolved. HSC Engineering Studies Assessment Task 1 10 of 14

11 gh ch ch gh ree Body Diagram orce Diagram gh = ch = As the truss is symmetrical I can fill in the rest of it. inal Analysis ag bi bk gh hi ij jk kl bl ch cj cl = = = = = = gh = hi = ij = jk = kl = bl = bk = = cl = = ag = bi = cj = ch Case 1 (5kg) 141N 71N 8N 44N HSC Engineering Studies Assessment Task 1 11 of 14

12 APPENDIX C JOINTS The joints are shown here. When we cid upon the joints we took into account whether the members were in tension or compression. We gave members in tension the greater surface area of gluing. ADC CDE fc CGH gh fc hc HSC Engineering Studies Assessment Task 1 1 of 14

13 CHB hb ch BGH bg gh bh AEGB bg HSC Engineering Studies Assessment Task 1 1 of 14

14 ADE HSC Engineering Studies Assessment Task 1 14 of 14

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