bruce mayer, pe licensed electrical & mechanical engineer [email protected]
DESCRIPTION
Engineering 25. Tutorial: Catenary Cables. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. Partial Catenary. The cable has a mass of 0.5 kg/m and is 25 m long. Determine the vertical and horizontal components of force it exerts on the top of the tower. - PowerPoint PPT PresentationTRANSCRIPT
ENGR36_H13_Tutorial_Catenary_Cables.pptx1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 25
Tutorial: Catenar
yCables
ENGR36_H13_Tutorial_Catenary_Cables.pptx2
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Partial Catenary
The cable has a mass of
0.5 kg/m and is 25 m long.
Determine the vertical and horizontal components of force it exerts on the top of the tower.
ENGR36_H13_Tutorial_Catenary_Cables.pptx3
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Eqns Used
0
0
sinh
sinh
Tµx
dxdy
Tµx
dxdy
p
xx p
00
0 sinhsinhTµx
Tµx
µTS PQ
00 cosh T
µxTT p
p0
sinhtanTµxp
p
ENGR36_H13_Tutorial_Catenary_Cables.pptx4
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ENGR36_H13_Tutorial_Catenary_Cables.pptx5
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ENGR36_H13_Tutorial_Catenary_Cables.pptx6
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ENGR36_H13_Tutorial_Catenary_Cables.pptx7
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ENGR36_H13_Tutorial_Catenary_Cables.pptx8
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
MATLAB Code% Bruce Mayer, PE% ENGR36 * 22Jul2% ENGR36_Tutorial_Partial_Catenary_H13e_P7_119_1207.m%K1 = tand(30)K2 = asinh(K1)S0 = @(z) (z/K2)*(sinh(K2*(z+15)/z)-K1) - 25xB = fzero(S0, 10)xA = xB+15u = 0.5*9.81TO = u*xB/K2TA = TO*cosh(u*xA/TO)QA = atand(sinh(u*xA/TO))WA = TO*tand(QA)
ENGR36_H13_Tutorial_Catenary_Cables.pptx9
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
MATLAB ResultsxB = 8.2804xA = 23.2804u = 4.9050TO = 73.9396TA = 181.0961QA = 65.9026WA = 165.3141
ENGR36_H13_Tutorial_Catenary_Cables.pptx10
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Chain Lift Problem The man picks up the 52-ft chain
and holds it just high enough so it is completely off the ground. The chain has points of attachment A and B that are 50 ft apart. If the chain has a weight of 3 lb/ft, and the man weighs 150 lb, determine the force he exerts on the ground. Also, how high h must he lift the chain? Hint: The slopes at A and B are zero.
ENGR36_H13_Tutorial_Catenary_Cables.pptx11
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Eqns Used
00
0 sinhsinhTµx
Tµx
µTS PQ
00 cosh T
µxTT p
p
0
sinhtanTµxp
p
1cosh
0
0
TxTy
ENGR36_H13_Tutorial_Catenary_Cables.pptx12
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ENGR36_H13_Tutorial_Catenary_Cables.pptx13
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ENGR36_H13_Tutorial_Catenary_Cables.pptx14
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ENGR36_H13_Tutorial_Catenary_Cables.pptx15
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
MATLAB Code% Bruce Mayer, PE% ENGR36 * 22Jul2% ENGR36_Tutorial_Chain_Lift_Catenary_H13e_P7_124_1207.m%Zf1 = @(q) (q/3)*sinh(75/q) - 26TO = fzero(Zf1, 150)h = (TO/3)*(cosh(75/TO) - 1)Q = atand(78/TO)Th = 3*h + TOTup = Th*sind(Q)
ENGR36_H13_Tutorial_Catenary_Cables.pptx16
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
MATLAB ResultsZf1 = @(q)(q/3)*sinh(75/q)-26
TO = 154.0033
h = 6.2088
Q = 26.8614
Th = 172.6297
Tup = 78