DAF 95XF, Maintenance Manual

Page: 95 / 188

95 XF
series Inspection and adjustment
EXPLANATORY NOTES ON THE MAINTENANCE ACTIVITIES
3-35
3.26 INSPECTION AND ADJUSTMENT, LOAD-DEPENDENT CONTROL VALVE,
AIR SUSPENSION
Explanatory notes on instruction plate
The information contained on the instruction
plate relates to the axle loads, the output
pressures and bellows pressures, in accordance
with the order of axles beneath the vehicle.
So “1” is the front axle, etc.
To check the load-dependent control valve, the
details on the instruction plate relating to the
“driven axle” are therefore vital.
L1 = Effective length of unloaded spring
between thrust piece and adjustable
plug. Spring length in mm.
L2 = bolt length up to counter nut in mm.
1. Check whether the correct valve is fitted
(see instruction plate).
2. Connect pressure gauge (4) to the test
connection of the load-dependent valve
(input pressure).
3. Connect pressure gauge (2) to the test
connection on the rear axle brake cylinder
(output pressure).
4. Connect pressure gauge (43) with a reducer
valve to the simulation connection of the
load-dependent valve (simulated adjustable
bellows pressure).
5. Make sure that the reservoir pressure is
higher than 6.5 bar throughout the
measurement.
MEEREGELVENTIEL EMPTY-LOAD VALVE LAST-LEER VENTIEL VALVE CHARGE-VIDE VALVOLA VUOTO-CARICO VALVULA VIDA-CARGA
ASLAST AXLE LOAD ACHSLAST CHARGE SOUS ESSIEU CARICO ASSE CARGA EJE
UITGESTUURDE DRUK DELIVERY PRESSURE AUSGESTEUERTER DRUCK PRESSION DELIVREE PRESSIONE USCITA PRESION DE SALIDA
G
p
2
BALGDRUK PRESSURE BELLOWS BALG DRUCK PRESSION COUSSIN PRESS.CUSCINI ARIA PRESION FUELLES
p3
bar
bar
x 10
4
N
AUTOM. LASTAFHANKELIJKE REMKRACHTREGELING
AUTOM. LOAD SENSING DEVICE
AUTOM LASTABHANGIGE BREMSKRAFT REGELINR.
DISPOSITIF DE CORRECTION AUTOM DE FREINAGE
REGOLATORE AUTOM. DELLA FORZA FRENANTE
REGULADOR AUTOM. DEL ESFUERZO DE FRENADA
1263639
TYPE - TIPO : FA
W 475 711 071 0 p1= > 6.5 bar p4= 6.0 bar
L1 = 114.0 MM L2 = 4.0 MM
2
1
2.0 2.5 3.0 4.0
10.0 11.5 12.0 13.0
2
1
1.4 1.5 1.6 2.0 4.6 5.3 5.5 5.9
4.3 4.4 4.4 4.5 5.8 5.9 5.9 6.0
2
1
0.3 0.5 0.7 1.2 4.1 4.8 5.1 5.6
0.2
0.4
L2
p1
p4
p3
L1
i = 1: 1.5
M6046
42
1
2
4
2
4
41
43
M6102
5
 200424