Worked Examples 2010 - Vehicle Testing - Mechanical System
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Transcript of Worked Examples 2010 - Vehicle Testing - Mechanical System
1MTRMTR--ASSUREASSURETMTM ©2006 MTR Corporation Limited. All rights reserved.
Worked Examples
Vehicle Acceptance and Testing(Mechanical Systems)
2MTRMTR--ASSUREASSURETMTM ©2006 MTR Corporation Limited. All rights reserved.
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
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Emergency Brake Rate
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.
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Emergency Brake Rate
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)
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Emergency Brake Rate
=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
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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.
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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)
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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.
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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.
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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.
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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:
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“+" 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.
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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.
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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
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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
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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
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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
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Vehicle Overturning
c,
b
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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.