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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Engineering 36. Lab-05 Angle Problems. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. Problem. Cables AB & AC attached to Tree Trunk and fastened to Stakes in the Ground Given Cable Tension T AC = 3.6 kN Find

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Engineering 36 Lab-05AngleProblems Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. Problem • Cables AB & AC attachedto Tree Trunkand fastenedto Stakes in the Ground • Given Cable Tension TAC = 3.6 kN • Find • Components of Force exerted by cable AC on the Tree • The Space Angles θx, θy, θz for Cable AC

  3. Tree Prob by MATLAB >> TAC = 3.6; >> Fyv = -TAC*sind(45) Fyv= -2.5456 >> Fh = TAC*cosd(45) Fh= 2.5456 >> Fzv = Fh*cosd(25) Fzv= 2.3071 >> Fxv = -Fh*sind(25) Fxv= -1.0758 >> TACv = [FxvFyvFzv] TACv= -1.0758 -2.5456 2.3071 >> uAC = TACv/TAC uAC= -0.2988 -0.7071 0.6409 >> Qxyz = acosd(uAC) Qxyz= 107.3877 135.0000 50.1443

  4. Problem • Given theGeometryof the SteelFrameWork as Shown • Given • EF & EG are Cables • Pt-E is at MidPt of BC • Tension in Cable EF is 330N • Find • Angle Between EF and BC • Projection on BC of the force exerted by Cable EF at Pt-E

  5. ûEF ûBC ûEF

  6. I-Beam Prob by MATLAB >> B = [0 8.25 0]; F = [1 0 0]; C = [16 3.75, -12]; >> E = [16/2 (3.75+8.25)/2 -6] E = 8 6 -6 >> EFv = F-E EFv= -7 -6 6 >> BCv = C - B BCv= 16.0000 -4.5000 -12.0000 >> EFm = norm(EFv) EFm= 11 >> BCm = norm(BCv) BCm= 20.5000 >> uEF = EFv/EFm uEF= -0.6364 -0.5455 0.5455 >> uBC = BCv/BCm uBC= 0.7805 -0.2195 -0.5854 >> Qceg = acosd(dot(uEF,uBC)) Qceg= 134.1254

  7. Problem • Given • Geometryas Shown • BungiCordPC with Tension of 30N • Distance OP = 120 mm • Find • Angle Between PC and OA • Projection on OA of the force exerted by the BungiCord PC at Pt-P

  8. =

  9. ûOA ûOA |û|

  10. BungiProb by MATLAB >> OAm = norm([240 240 -120]) OAm= 360 >> OPm = 120; >> Ratio = OPm/OAm Ratio = 0.3333 >> A = [240 240 -120] A = 240 240 -120 >> P = Ratio*A P = 80 80 -40 >> C = [180 300 240] C = 180 300 240 >> PCv = C-P PCv= 100 220 280 >> OAv = A; >> PCm = norm(PCv) PCm= 369.8648 >> OAm = norm(OAv) OAm= 360 >> Qcpa = acosd(dot(PCv,OAv)/(PCm*OAm)) Qcpa= 71.0682 >> uPC = PCv/PCm uPC= 0.2704 0.5948 0.7570 >> Tpcv = 30*uPC Tpcv= 8.1111 17.8444 22.7110 >> FOAproj = 30*cosd(Qcpa) FOAproj= 9.7333

  11. TBC = 5.3 kN Problem • Given • Geometry as Shown • Angle OAB is 90° • Tension in Cable BC is 5.3 kN

  12. TBC = 5.3 kN Problem • Find • The Angle, θ, Between the Cable and Pipe BA • The Components of FBC (cable force acting at Pt-B) that are || and  to Pipe BA

  13. −

  14. Pipe Prob by MATLAB >> TBC = 5.3; % in kN >> A = [2400 0 0]; C = [0 1050 1800]; >> B = [2400 -1200*sind(30) 1200*cosd(30)] B = 1.0e+03 * 2.4000 -0.6000 1.0392 >> BAv = A-B BAv= 1.0e+03 * 0 0.6000 -1.0392 >> BCv = C-B BCv= 1.0e+03 * -2.4000 1.6500 0.7608 >> BAm = norm(BAv) BAm= 1200 >> BCm = norm(BCv) BCm= 3.0102e+03 >> Qabc = acosd(dot(BAv,BCv)/(BAm*BCm)) Qabc= 86.8358 >> Fpar = TBC*cosd(Qabc) Fpar= 0.2925 >> Fper = TBC*sind(Qabc) Fper= 5.2919

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