1ENEC20002: Structural Design IIProject Title: Masonry DesignNature of the Work: Individual WorkSubmission Due Date: 24th May 2019 (5:00 PM AEST)Weight: 35%1. IntroductionObjective and assessment criteria of this design project is explained in the unit profile. Allstudents are encouraged to read and understand them before start of this project work.2. Project DescriptionAll the masonry walls of the single storied industrial building shown in Figure 1, is to bedesigned. The details of the walls are shown in Figure 2 – 6.Figure 1: Building planA steel truss roof system will be installed on top of the building which will be supportedby walls BE, FG, HA, HH*, CC* and DD*, and the two steel columns. Furthermore, theA BCEDFHGC*H*Steel ColumnD*2structural design assumes that no vertical pre-compression loading from the roof truss willbe directly applied on walls AB and EF.Figure 2: Elevation Wall ABFigure 3: Elevation Wall EFA BSteelColumnSteel BeamSteel BeamOpeningOpeningX1,AB X2,ABY1,ABY2,ABY3,ABYAB= Y1,AB+ Y2,ABXAB= X1,AB+ X2,AB =0.25MPa =0.1 MPaF ESteelColumnOpening OpeningX1,FE = X1,AB X2,FE = X2,ABY1,FEY2,FE = Y3,ABYFE = YAB3Figure 4: Elevation (a) Wall HA, (b) Wall FGFigure 5: Elevation Wall HH*/CC*/DD*HXHH*YHH* = YABH* HXHAYHA = YABA F XFGYFG = YABG(a) (b)4Figure 6: Elevation Wall BEDesign the component walls of the building using suitable masonry units (you may referto the CMAA/TBA website to find suitable units). Consider the cyclonic wind pressure as3kPa and suction as 2kPa. Dimension of the walls are given in Table 1.Assume any reasonable value for the missing data as per the relevant Australian standard.The unconfined characteristics compressive strength of unit and the steel yield strength ofsteel can be obtained from the supplier’s brochure or through laboratory testing.As part of the continuous design process, the building was analysed for lateral andgravitational forces as per AS1170. The in-plane force acting on the component walls ofbuilding is shown in Table 2. The vertical pre-compression load from the roof truss onwalls BE, FG, HA, HH*, CC* and DD* were 0.5MPa (calculated considering 110mm thicksolid brick).BXBEXC D EY1,BEOpeningY2,BEYBE = Y1,BE + Y2,BE5Table 1: Wall dimension based on student IDTable 2: In-plane loading on the masonry wallsStudentIDLoad on the following walls in kNAB EF HA FG BE HH*/CC*/DD*12088571 250 255 277.5 277.5 50 3012091465 275 280 302.5 302.5 50 3012090286 300 305 327.5 327.5 50 3012093080 200 205 227.5 227.5 50 3012088963 225 230 252.5 252.5 50 3012080837 250 255 277.5 277.5 50 3012060601 100 105 115 115 25 3012087181 120 125 135 135 25 3012096930 135 140 150 150 25 3012087018 100 105 115 115 25 3012062456 100 105 112.5 112.5 20 2512093836 125 130 137.5 137.5 20 2512074708 300 305 325 325 45 25StudentID/WallNameAB EF HA FG HH* BE CDX1,AB X2,AB Y1, AB Y2,AB Y3,AB Y1, EF Y2,EF XHA XFG XHH* XBE Y1,BE Y2,BE XCD12088571 4 4 4 3 1.5 5.5 1.5 3.5 3.5 1 8 5 2 412091465 4.5 4.5 4 3 1.5 5.5 1.5 3.5 3.5 1 8 5.25 1.75 412090286 5 5 4 3 1.5 5.5 1.5 3.5 3.5 1 8 5.5 1.5 412093080 4 4 4 2 1 4.5 1.5 4 4 1 9 4.75 1.25 512088963 4.5 4.5 4 2 1 4.5 1.5 4 4 1 9 4.5 1.5 512080837 5 5 4 2 1 4.5 1.5 4 4 1 9 4.25 1.75 512060601 3 3 4 2 1 4.5 1.5 4 4 1 9 4 2 512087181 3.5 3.5 4 2 1 4.5 1.5 4 4 1 9 4.25 1.75 512096930 3.8 3.8 4 2.5 1.5 5.5 1 3.5 3.5 1 8 5 1.5 412087018 4 4 4 3 1.25 4.75 2.25 3 3 1 7 5.5 1.5 312062456 4.5 4.5 4 3 1.25 4.75 2.25 3 3 1 7 5.25 1.75 312093836 5 5 4 3 1.25 4.75 2.25 3 3 1 7 5 2 312074708 4 4 4 2 1.5 4.25 1.75 3 3 1 7 5.5 0.5 312081265 4.5 4.5 4 2 1.5 4.25 1.75 3 3 1 7 5.25 0.75 312080609 5 5 4 2 1.5 4.25 1.75 3.5 3.5 1 8 5 1 412089402 3 3 4 2 1.5 5 1 3.5 3.5 1 8 4.75 1.25 412092461 3.5 3.5 4 2 1.5 5 1 3.5 3.5 1 8 4.5 1.5 412064843 3.8 3.8 4 2.5 2 5 1.5 4 4 1 9 5 1.5 512079240 5 5 4 2.5 2 5.5 1 4 4 1 9 5.5 1 5612081265 325 330 350 350 45 2512080609 375 380 400 400 45 2512089402 200 205 225 225 45 2512092461 220 225 245 245 45 2512064843 230 235 255 255 45 2512079240 200 205 225 225 45 257Your TasksDesign all the masonry walls of the building against lateral and gravitational loading. Thefollowing steps may help you the in the structural design:(1) It is a masonry building, identify the walls, check all details, design the control joints(for URM) and determine the necessary material parameters.(2) Design for in-plane and out-of-plane shear: The demand in-plane shear is given in Table2. Calculate the capacity of your wall. If capacity > demand, the wall is safe otherwiseconsider altering your design. Calculate the out-of-plane shear demand and capacity andensure the capacity > demand. Start your design considering URM.(3) Design for out-of-plane bending: Take the cyclonic wind pressure as 3kPa and suctionas 2kPa. Design the walls for the maximum out-of-plane pressure load. Start your designconsidering URM. If not safe, choose RM.(4) Design for vertical compression: calculate the compression load acting on the walls andcompare with the compression load capacity of the component walls. Start your designconsidering URM. If not safe, choose RM.(5) Draw the designed walls using appropriate CAD tool.Note:? Develop am excel design spreadsheet (Mandatory), it will assist you to designmultiple walls and include necessary adjustments, quickly.? The lintel on the door opening in BE is a steel UB.? Use suitable CAD software for drawing.? Submit a reflection report on CMAA/TBA guest lecture.Report:Your report should contain all of the following,1. Report title page2. Calculations for all the walls in excel file3. Final design drawing of the walls and the building (CAD) – DXF file preferred4. Reflection on CMAA/TBA guest lecture.Important: submit your report through Moodle.8Reflection on Concrete Masonry Association Australia/Think BrickAustralia Guest Lecture (Mandatory Component to Meet LO2)Reflection Writing Guideline:1. Write about the guest lecture/seminar (date/time/location/title), name and role ofthe presenter, details about the industry/organisation conducting the presentation.(one paragraph)2. Write about the topic presented in the guest lecture/seminar. (one paragraph)3. Write how this guest lecture/seminar helped to enhance your understanding on thedesign of masonry structures. (one paragraph)4. Write how this guest lecture/seminar helped to enhance your understanding onprofessional level of communication and leadership skills. (one paragraph)5. Write whether this guest lecture/seminar helped to enhance yourunderstanding/interest on masonry structures. (one paragraph)

Attachments:

Masonry-Proje….pdf

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1a) In a study of the gas phase decomposition of dinitrogen pentoxide at 335 K N2O5(g)2 NO2(g) + ½ O2(g) the concentration of N2O5 was followed as a function of time. It was found that a graph of ln[N2O5] versus time in seconds gave a straight line with a slope of -6.72×10-3 s-1 and a y-intercept of -1.67 . Based on this plot, the reaction is [ ] order in N2O5 and the rate constant for the reaction is 1b) In a study of the rearrangement of cyclopropane to propene at 500 °C (CH2)3CH3CH=CH2 the concentration of (CH2)3 was followed as a function of time. It was found that a graph of ln[(CH2)3] versus time in seconds gave a straight line with a slope of -1.03×10-3 s-1 and a y-intercept of -2.83 . Based on this plot, the reaction is [ ] order in (CH2)3 and the half life for the reaction is

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18. Code generation for programs. Translate the following program:

intresult;?intfibo(intn) {

intresult;?if(nreturn-1;switch(n) {

case0 :return0;break;?case1 :return1;break;?default:returnfibo(n-1)+fibo(n-2);}

}?intmain() {

intn;?n?5;?result? fibo(n);return0;

}

19. C MAinterpreters.?Implement an interpreter for the C-Machine in the programming language of your choice.

• Choose a suitable data type for instructions. Take into account that some instructions have arguments, but some have not.?

• Implement the datastructuresCandS.?

• Test your interpreter with the factorial function and the main function:?

intmain() { returnfac(9);}?

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16. Engineering personnel of companies have major influences in selection of A. product components B.supplier selection C.buyer selection D.trial supplier?s 17. Individuals who have formal authority in selection of supplier?s are classified as A.approvers B.deciders C.buyers D.evaluators 18. Buying situations in business markets include A.straight rebuy B.new task C.modified rebuy D.all of the above 19.Type of contract in which single supplier provides buyer with all requirements is classified as A.systems buying B.management supplies C.systems contracting D.system selling 20. With passage of time, new-buy situations become A.straight rebuy B.straight buy C.new rebuy D.modified rebuy

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