OHL06 - Supplementary material - OHLbasic mech calculation.pdf

Basic Mechanical Calculation 1

Overhead Line System

BASIC MECHANICAL CALCULATION


CONTENTS

PART 11.1 Relationship between Sag and Tension for a Contact Wire

or a Flexible Wire

PART 22.1 Stagger2.2 Blow-off2.3 Pantograph Sway2.4 Load on OCS Support

PART 33.1 Super-elevation of Track 3.2 Curve for Erection of Weights


1.1 Relationship between Sag and Tension for a Contact Wire or a Flexible Wire

1.1.1 Supports at a Horizontal Plane

1.1.2 Supports at Different Heights

1.1.2.1 Sag of a Wire from Supports at Different Heights

1.1.2.2 Inclined Sag

1.1.2.3 Tension of a Wire from Supports at Different

Heights

1.1.2.4 The Upward Force onto the Lower Support by

a Suspended Wire

Part 1


1.1 Relationship between Sag and Tension for a Contact Wire or a Flexible Wire


Suppose:* the sag is small in comparison with the span.* the contact wire, the messenger wire and the

additional conductors are regarded as flexible, the elasticity rigidity of which is neglected as approximately zero

* the wire densities of contact wire, messenger wire and the additional conductors are regarded as equal along the span although they are actually equal along the wires.


1.1.1 Supports at a Horizontal Plane


Assume:

L = span length mg= weight per meter of wire NFa, Fb = vertical force exerted onto the wire by supports A & B NTa, Tb = horizontal force exerted onto the wire by supports A & B N


According to the principle of force equilibrium, Σ Fx=0, so

Ta - Tb = 0or Ta = Tb = T


Σ Fy = 0

Fa + Fb - gL = 0or Fa = Fb = gL/2


Take a length of wire OA and by taking moments about point O

Ta Y - Fa X + gX* X/2 = 0 1.1.1.1

Substitute Fa = gL/2, and Ta = T into Equation 1.1.1.1, thenT Y = gL/2 * X -g X2 /2

whence Y = gX (L-X)/( 2T) 1.1.1.2


if X =L/2 and let Y= Ymax = F

F= g L2 /(8T) 1.1.1.3T = g L2 /(8F) 1.1.1.4


Substitute Equation 1.1.1.4 for Equation 1.1.1.2,

Y = 4 F X ( L - X )/ L2 1.1.1.5

which is an equation of a suspended wire and an equation of a parabola.


The tension in the suspended wire is not a constant and the tension is in the direction of its tangent at a specified point of the wire. But due to the small sag, R and T are approximately equal.,


1.1.2 Supports at Different Heights


1.1.2.1 Sag of a Wire from Supports at Different Heights


Ta = T, Tb = T

Fa = gL1 , Fb = gL2


Take moments around O for both parts

TF1 + gL1 2 /2 - gL1

2 =0and TF2 + gL2

2 /2 -gL2 2 =0

whence,F1 = gL1

2 /(2T) = g(2L1 )2 /(8T) 1.1.2.1.1F2 = g L2

2 /(2T) = g(2L2 ) 2 /(8T) 1.1.2.1.2


h = F2 - F1 = g ( L2 -L1 )/ (2T)= g (L2 +L1 )(L2 -L1 )/(2T)

and L2 + L1 = L 1.1.2.1.3L2 - L1 = 2Th/(g L) 1.1.2.1.4


Solve the above equation group

L2 =L/2 +Th/(gL) 1.1.2.1.5

L1 =L/2 - Th/ (gL) 1.1.2.1.6


Substitute Equation 1.1.2.1.5 & 1.1.2.1.6 into 1.1.2.1.1 & 1.1.2.1.2

F2 = g (L + 2Th/(gL)) 2 / 8T

= gL2 /(8T) + h/2 + Th2 /(2gL2 )


Substitute Equation 1.1.1.3 into the above

F2 =F +h/2 + h2 /(16F)

= F(1+h/(2F) + h2 /(16F2 ))


Sag at Supports

F2 =F(1+h/(4F)) 2 1.1.2.1.7

F1 =F ( 1 - h/(4F)) 2 1.1.2.1.8


1.1.2.2 Inclined Sag


F’=F

If the span length, the wire density and the tension in the left figure are the same as those in the right figure, the above relationship holds.The above relationship is very important and has been widely used in engineering calculations.


1.1.2.3 Tension of a Wire from Supports at Different Heights


Around Point D

FbX - gX2 /2- Hb Y cos a + Hb X sin a = 0

Y = hX/L + ( gLX/2 - gX2 /2)/T 1.1.2.3.1


Substitute Equation 1.1.1.3 into the above

Y = hX/L + 4 FX(L -X )/L2 1.1.2.3.2

R=T ( 1+(h/L+ 4F/L - 8 FX/L2 )) 1/2 1.1.2.3.3


Substituting X=L and X=0

Ra = T ( 1 + (h/L-4F/L) 2 ) 1/2

and Rb = T ( 1 + (h/L+4F/L) 2 ) 1/2 1.1.2.3.4


Va = Fa - Ha sin a = gL/2-T tg a = T (4F/L-h/L)

and Vb = Fb -Hb sin a = gL/2 + T tg a= T ( 4F/L + h /L) 1.1.2.3.5


from Equation 1.1.2.3.5Va = Fa - Ha sin a

= gL2 - Th/L= g(L/2 - Th/(gL) )= g L1

and Vb = Fb - Hb sin a= gL/2 - Th/L= g(L/2 + Th/(gL))= gL2


Also

Ra =(T +Va ) =( T +T (4F/L-h/L) ) = T ( 1+ (4F/L-h/L) )

Rb =(T + Vb )= ( T + T (4F/L+h/L) ) =T (1+(4F/L+h/L) )


1.1.2.4 The Upward Force onto the Lower Support by a Suspended Wire


L =L2 -L1 1.1.2.4.1

h = gL2 2 /(2T)-gL1

2 /(2T) = g (L2 + L1 )(L2 - L1 )/(2T)

Thus, L2 +L1 =2Th/(gL) 1.1.2.4.2


From Equation 1.1.2.1.6

L1 = L/2 - Th/(gL) 1.1.2.4.3 Substitute Equation 1.1.1.3 into Equation 1.1.2.4.3 and

If L1 > 0, it gives:

4F < h 1.1.2.4.4 In the case that Equation 1.2.4.4. holds, the0 lowest

point of the wire is located outside the span.


Based on Equation 1.1.2.3.5

If 4F=H Va=0

If 4F>H Va>0

If 4F<H Va<0


2.1 Stagger

2.2 Blow Off

2.3 Pantograph Sway

Part 2


2.1 Stagger

The position of the contact wire relatively to the track should be such that the contact surface of the pantograph will be worn uniformly throughout its width. To obtain this result, it is necessary to ‘stagger’ the contact wire with respect to the center line of the track, a stagger of 200-400 mm, on each side of the center line being the usual allowance


2.2 Blow-offB / y = P / qvB = y (P / qv) 2.2.1From Sag Equationy = g X (L-X) /2T 2.2.2

Put g = qvy = qv X(L-X)/2 T 2.2.3

Put 2.2.3 into 2.2.1

B = P L (L-X)/2T

B

y

P

qvqj


B = PX (L-X)/( 2T) 2.2.4

Thus, if X =L/2, we have

Bmax= P L2 /(8T) 2.2.5


Equal and Opposite Staggers

X

Bmax

L

y1

y2


Equal and Opposite Staggersy1 = P X (L- X)/ 2Ty2 = a (L-2X)/L

B = y1 + y2 = P X (1-X)/2T + a (L-2X)/L

set d B / d X = 0, ⇒

Xmax = L/2 – 2aT/(PL)

⇒

Bmax= P L2 /(8T)+ 2a2 T/(PL2 )


Unequal and Opposite Staggers

y1B max

y2

L

(a1 + a2)/2a1

a2

y3


Unequal and Opposite Staggers

Calculate y1, y2 and y3.

B = y1 + y2 + y3

Bmax= P L2 /(8T)+ (a1 +a2 )2 T/(2PL2) +(a1 -a2 )/2


2.3 Pantograph Sway

- Pantograph lateral movement


Pantograph Sway and Blow off

Bmax < D/2 – d

In which

D = length of carbon strips on pantographd = Pantograph Sway


Pantograph on Car Roof


The loads exerted on the mast by OCS equipment

compose of the following items.

• Vertical Load

• Static Radial Load

• Wind Pressure

2.4 Load on OCS Support


g = vertical force of catenary system (N/m)H = Height difference in support (m)T = Tension in Messenger Wire (N)L = Span Length (m)

Fa = (g L)/2 - (H T) / LFb = (g L)/2 + (H T) / L

Refer Equation 1.1.2.1.5 & 1.1.2.1.6

gFb

H

Fa

L

Vertical Loads of Catenary System


So = Stagger at registration (m)S = Stagger at adjacent registration T = Tension of Conductor (N)L = Span Length (m)

sin θ

z tan θ= (So + S) / L

The radial load (one side):

R = T sin θ

z T (So + S) / L

Radial Loads by Stagger

θ


T = Tension in Conductor (N)L = Span LengthRAD = Radius of Track (m)V = versine of track

The radial load by angle (one side)

sin θ

z tan θ= L/ 2 RAD

R = Tsin θ

= T L / 2 RAD( = 4V T / L)

Given : V = L2 / 8 RAD

Radial Loads on Curve Track


• Operational (44.4m/s)

It is the base wind speed condition that the train service is

still maintained.

• Survival (72.2m/s, Non-operational) - Typhoon

The strength of all structures and components must be

designed to be survived under this wind speed but some

operational parameters can be exceeded. (blow-off,

mast deflection)

Base Wind Speeds (Open Route)


Wind Pressure EquationP = 0.613 x K x (V x Kd)2 x S

Kd : wind factors consists of surface roughness factor,

topographical exposure factor and height factor.

K: Drag coefficient

V: Base Wind Speed

S: Unit section of an object

Wind Pressure


Total Vertical Load

Total Radial Load (Along Track)

Turning Moment

Total Radial Load (Across

Track)

Load on Mast Base


3.1 Super-elevation of Track

3.2 Curve for Erection of Weights

Part 3


In the case the train is running o the curved track, the train tends to tilt toward the outside of the track curve by a centrifugal force. The superelevation of the track is necessary in order to produce a centripetal force and to avoid any accident caused by the train tilt.

3.1 Superelevation of Track


Superelevation of Track


F = GV2 /(gR) 3.1.1

in whichF = the centrifugal force of the train NG = the weight of the train Ng = gravity acceleration m/s2

V = train average speed km/hR = radius of track m


Now we have P = G sin θ = G h/S 3.1.2

In which P = the centripetal force caused by superelevation

GV2 /(gR) =Gh/S h = SV2 /(gR) S = 1.5 mg =9.81 m/sh = 1500 (1/3.6)2 V2 /(9.81R)

= 11.8 V2/R 3.1.3


3.2 Curve for Erection of Weights

a = amin + nL α

(tx -tmin ) 3.2.1

b = bmin + nL α

(tmax -tx ) 3.2.2


In which,α

= coefficient of linear expansion 1/degC

a = distance between the top of weights and the pulley mmamin = min. allowable distance between the top weight and the pulley mmb = distance between the bottom of weights and ground mmbmin = min. allowable distance between the bottom weight and the ground mmL = length of the half tension section, or the distance between the mid-point anchor

and the auto-tensioning device. Mmtmin = min. temperature (design) degCtmax = max. temperature (design) degCtx = temperature at which the erection work is done degCn = the ratio of the balance wheel


Example 1 : Wheel Turn Ratio 1:2


Example 2 : Wheel Turn Ratio 1:3


AssignmentsA1 Derive the formula of the relationship between sag and tension for a

flexible wire and the unit of each item, in the case of support at horizontal.

A2 Assume that for the simple suspension (supports at the same horizontal plane), the span length is 40m and the working tension of the 120 mm2 Cu CW (1070kg/km) is 12000N. Find the sag of the contact wire.

A3 Assume that in Question A2 above, the supports are at different elevation that is the support A is 300mm lower than support B.

Find a. The distance between the lowest point of the contact wire and the two supports.b. The sag relative to the two support respectivelyc. Inclined sag at the mid point of the span lengthd. Vertical force at support


AssignmentsB1 Derive the formula of the maximum blow-off under the following conditions

a. On tangent track, stagger is zerob. On tangent track, staggers are “a” but in opposite directionsc. On tangent track, staggers are in opposite direction and

respectively “a1” and “a2”

B2 Given the wind loading on contact wire is 5.678 N/m, span length is 60m, and the working tension of the wire is 12000N, calculate the max. blow off:a. On tangent track, stagger is zerob. On tangent track, staggers are 200mm but in opposite directionsc. On tangent track, staggers are 100mm and 300mm in opposite

directionsB3 Given the length of carbon strip on pantograph (D) = 1500mm, find the max pantograph

sway (d) for the case “c” in Question B2

B4 Given the coefficient of linear expansion for contact wire is 17 E-6 /degC. The distance between a mid point anchor is 600m. The operation temperature are in the range 0 – 60 degC. Calculate the wire variation in length.

OHL06 - Supplementary material - OHLbasic mech calculation.pdf

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