Worked Examples 2010 - Vehicle Testing - Mechanical System

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Worked Examples

Vehicle Acceptance and Testing(Mechanical Systems)


Emergency Brake Rate

Sample Test Result

Time (s)

Instantaneous Brake Rate a_i

Trai

n S

peed

(km

/h)

Brake Cylinder Pressure (bar)

Average Brake Rate a_e

t_0 t_1



Acceptance Criteria

1. Emergency brake rate >= 1.4 m/s2

Question :Train speed = 35 km/h at t = 3 sec, and the train stopped at t = 9.5 sec. Find the emergency brake rate.

Answer :

Measured emergency brake rate = (35/3.6) / (9.5 – 3) = 1.496 m/s2

Therefore the required emergency brake rate >= 1.4m/s2 can be satisfied.



Acceptance Criteria

2. Maximum braking distance <= calculated braking distance (based on 1.4 m/s2 - 14%)

Question :As read from graph, t_0 = 0.4 sec and t_1 = 1.4 sec. If the initial train speed = 40 km/h. Check the maximum allowable braking distance for minimum instantaneous emergency brake rate 1.4 m/s2 -14%.

Answer :The maximum allowable emergency braking distance can be

calculated by:S_i = Uo*(to+0.5*t1) + Uo^2/(2*a_i) or S_e = Uo^2/(2*a_e)



=Uo^2/(2*a_e) or= Uo*(to+0.5*t1) +

Uo^2/(2*a_i)m63.5S_e or S_i

Maximum Allowable Braking Distance

=a_i / (1 + ( 2*t_0 + t_1 )*a_i / Uo)m/s20.972a_eAverage brake rate

Design requirement specificationm/s21.4*0.86=

1.204a_iEmergency brake rate (instantaneous)

Read from graphsec1.4t_1Brake force build up time (90%)

Read from graphsec0.4t_0Dead time

= Speed in km/h / 3.6m/s40/3.6=11.11UoVelocity

Read from graphkm/h40SpeedInitial speed

FormulaUnitValueAbbr.Parameter


Vehicle Weighing

10,20710,303 10,30810,11310,106Axle Load, AL

5,1045,3134,9905,2685,0404,9805,1335,0455,061Wheel Load, WL

Average5,2405,0605,3604,9454,9755,1355,0605,050Fourth

5,3604,9455,2055,1005,0105,1055,0255,080Third

5,3154,9855,2905,0204,9505,1605,0705,035Second

5,3354,9705,2155,0954,9855,1305,0255,080First

Wheel Load Measure

ment

W4BW4AW3BW3AW2BW2AW1BW1AWheel No

Axle 4Axle 3Axle 2Axle1Axle No

Question: Check for axle load balance and wheel load balance on axle 1 from vehicle weighing results.


Vehicle Weighing

0.9%<=4%

*Calculate the total WL for each sideFor Side A, 5061+5133+5040+4990 = 20224For Side B, 5045+4980+5268+5313 = 20606Average load on both sides of vehicle = (20224+20606)/2 = 20415=Max(20224-20415,20606-20415)/(20415)

= Maximum of Abs ( total WL on one side - average of both sides WL) /(average of both sides WL)

b. Difference in load from one side of the vehicle to the average load on both sides of the vehicle

0.2%<=4%

For Axle 1,*Calculate AL/2 = 10106/2 = 5053

=Max(5061-5053,5045-5053)/(5053)

= Maximum of Abs (WL -AL/2) / (AL/2)

a. Difference in load from the average load per wheel for a given axle

2. Wheel load balance :

1.0%<=2%

*Calculate the average of AL = 10207

For Axle 1,=Abs (10106-10207)/(10207)

= Abs (AL - average of AL) /( average of AL)

1. Axle load balance(for power car)


Wheel Off-loading Calculation

Question: Given twisted track condition as above, calculate track twists at bogie centre distance 15.6m and bogie wheelbase distance 2.5m.


Wheel Off-loading Calculation

Answer:

For bogie centre distance 15.6m, 1:260 cant gradient will give track twist of 15.6*1000*1/260 = 60mm.

For bogie wheelbase distance 2.5m, 1:230 cant gradient will give track twist of 2.5*1000*1/230 = 10.9mm.

Note: Cant gradient is a measure of rate of change in cant.


for 2 bogie vehicle (8 wheels)

250kNW

Total car weight

6000KN m/rad+Bogie torsional stiffness

8000KN m/rad*Body torsional stiffness

5000KN/mkAnti roll bar stiffness per bogie

7000KN/m2ksSecondary vert stiffness per bogie

1800KN/m2kpPrimary vertical stiffness per wheelset

1.88m2bsSecondary suspension spacing

1.94m2bpPrimary suspension spacing

1.505m2b

Wheel spacing

Question: Calculate dQ/Q based on the following parameters together with track twist 60mm over bogie centre distance and 10.9mm overbogie wheelbase distance.


1. Two angles t* and t+ are calculated from the measured track twistsTwist* Twist+

t* = ------------ t+ = -----------2b 2b

2. The stiffness C* and C+ can be worked out from the vehicle parameters1

C* = ------------------------------------------------------[ 1/ * + 2/ (2ksbs2 + k ) + 2/ (4kpbp2) ]

1C+ = ------------------------------

[ 1/ + + 2/ (2kpbp2 ) ]3. Two values of wheel off-loading dQ* and dQ+ are calculated by

C* t* C+t+dQ* = ---------- dQ+ = ---------

2 x 2b 2b4. Then add these two values to give the total wheel off-loadingdQ = dQ* + dQ+5. The value of dQ/Q can be calculated.

Answer : Using the following formulae:


“+" for bogie wheelbase distance (2.5m)

“*" for bogie centre distance (15.6m)Notes :

=18.37/31.2558.9%%dQ/QDelta Q / Q

=250/831.25kNQNominal wheel load

=14.81+3.5618.37kNdQ = dQ* + dQ+

=(742.1*0.0072)/1.5053.56kNdQ+

=(1118.2*0.0399)/(2*1.505)14.81kNdQ*Wheel off-loading

=1/(1/6000+2/(1800*0.97^2))742.1KN m/radC+

=1/(1/8000+2/(7000*0.94^2+5000)+2/(2*1800*0.97^2))1118.2KN m/radC*Total stiffness

=(10.9/1000)/1.5050.0072radt+

=(60/1000)/1.5050.0399radt*Twist angle

=2500/ 23010.9mmTwist+

=15600/26060mmTwist*Twist height

The wheel off-loading factor is calculated to be 0.589 and is acceptable.


Wheel Off-loading Test

kN70.0754.3535.726.7420.0235.9152.963.18Weight:mm-0.7-0.100.50.69.960.270.9Height:

Jack to maximum height at wheel 1AStep 2

kN45.443.1744.0948.1144.9647.345.3941.69Weight:mm0.1-0.20.20-0.10.1-0.10.1Height:

Level TrackStep 1

4B3B2B1B4A3A2A1AWheel:

Question: Check for maximum dQ/Q from wheel off-loading test results.


Wheel Off-loading Test

Q0 = 44.96 kN, Min. Qt = 20.02 kNMax dQ/Q = (Q0 - Qt ) / Q0 = (44.96 – 20.02) / 44.96 = 55.5%

Acceptance limit :The wheel off-loading of any wheel shall not exceed 60% of the nominal static load value, i.e. dQ/Q <= 0.6.

Q0 – Qt

Q0

dQ =Q

Answer: Wheel Off-loading Factor :

where,Q0 = nominal wheel static load at level track conditionQt = wheel load at maximum track twist condition


2T

Bogie Rotational Resistance

Question : Given that : Vehicle weight = 42085 kg for a 4-axled vehicle

Bogie semi-wheel base = 1.25mCalculate the X-factor from bogie rotational resistance test result.

Measured T are :+18.52 kN

-18.59


Bogie Rotational Resistance

Answer :

• The expression for Bogie Rotational Resistance (X-factor) is :

X = T / (W * 2a)

where, T = Torque to rotate the bogieW= average axle load on the bogie being rotateda = bogie semi-wheel base

Axle load W = 42085 * 9.81 / (4 *1000 ) = 103.2 kNTotal torque, 2T = 18.52+18.59 = 37.11 KNX = (37.11 /2) / (103.2 * 2 * 1.25) = 0.072 < 0.1


Vehicle Overturning

Question : Assess the vehicle design against overturning for the

worst combination of wind loading and track cant.

m/s29.81gAcceleration due to gravity

N53538FBody lateral force

m1.505bWheel contact patch spacing

mm150cMaximum installed cant

m0.005dLateral offset of CG

m1.649h_xHeight of lateral load above rail

m1.483h_bodyHeight of body CG above rail

kg32000MCarbody mass


Vehicle Overturning

c,

b


Vehicle Overturning

Answer := arcsin (150 / 1505)

Q1 + Q2 = M/2 * gTake moment about A :Mg/2 * sinØ x h_body + Mg/2 * d + F/2 * h_x = (Q_2 - Q_1) x 1.505 / 2

Thus, Q1 = Mg/4 - Mg * d/3 - F * h_x/3 – Mg * sinØ * h_body/3

To prevent overturning, check if min. Q1 > 0.

Q1 = 32000*9.81/4 – 32000*9.81*0.005/3 – 53538*1.649/3 –32000*9.81*(150/1505)*1.483/3

= 33062 N > 0.

Worked Examples 2010 - Vehicle Testing - Mechanical System

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