BMayerChabotCollegeedu Engineering 36 Lab05 Angle Problems Problem Cables AB amp AC attached to Tree Trunk and fastened to Stakes in the Ground Given Cable Tension T AC 36 kN ID: 759121
Download Presentation The PPT/PDF document "Bruce Mayer, PE Licensed Electrical &..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Bruce Mayer, PELicensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu
Engineering 36
Lab-05
AngleProblems
Slide2Problem
Cables AB & AC attachedto Tree Trunkand fastenedto Stakes in the GroundGiven Cable Tension TAC = 3.6 kNFindComponents of Force exerted by cable AC on the TreeThe Space Angles θx, θy, θz for Cable AC
Slide3Slide4Slide5Slide6Slide7Tree 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
= [
Fxv
Fyv
Fzv
]
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
Slide8Problem
Given theGeometryof the SteelFrameWork as ShownGivenEF & EG are CablesPt-E is at MidPt of BCTension in Cable EF is 330NFindAngle Between EF and BCProjection on BC of the force exerted by Cable EF at Pt-E
Slide9û
EF
û
BC
û
EF
Slide10Slide11Slide12I-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
Slide13Problem
GivenGeometryas ShownBungiCordPC with Tension of 30NDistance OP = 120 mmFindAngle Between PC and OAProjection on OA of the force exerted by the BungiCord PC at Pt-P
Slide14Slide15=
Slide16û
OA
|û|
û
OA
Slide17Bungi Prob 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
Slide18Problem
GivenGeometry as ShownAngle OAB is 90°Tension in Cable BC is 5.3 kN
T
BC
= 5.3 kN
Slide19Problem
T
BC
= 5.3 kN
Find
The Angle,
θ
, Between the Cable and Pipe BA
The Components of
F
BC
(cable force acting at Pt-B) that are || and
to Pipe BA
Slide20Slide21−
Slide22Slide23Slide24Pipe 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