Fundamental Vehicle Loads & Their Estimationarahim/Lecture 5 smc4133.pdf · 2008-07-21 ·...
Transcript of Fundamental Vehicle Loads & Their Estimationarahim/Lecture 5 smc4133.pdf · 2008-07-21 ·...
Fundamental Vehicle Loads & Their Estimation
• The simplified loads can only be applied in the preliminary design stagewhen the absence of test or simulation data
• They should always be qualified and updated as more informationbecomes available
VEHICLE OPERATING CONDITIONS & PROVING GROUND TESTS• The significant proving ground events can be divided into two types:a) Instantaneous overloads (large pot holes, kerb bump, large bump,
panic braking, high g cornering, service loads)b) Fatigue damage ( medium size pot holes, Belgium block road,
twist course, Cobblestone track, service loads)
Example of service loads: towing, jacking, hoisting
LOADS CASES & LOAD FACTORS
• The vehicle designer needs to know the worst or most damaging loads in order to:a) ensure the structure does not fail in servicestructure does not fail in service due to instantaneous overload
b) ensure a satisfactory fatigue lifesatisfactory fatigue life
• Common assumption at early design stage:If the structure can resist the worst possible loads then it is likely to havesufficient fatigue strength
• For early design calculation, the actual dynamic loading is often replacedby a factored static loading:
Pdyn = Pst x M
Or sometimes an extra FOS is used; Pdyn = Pst x M x FOS
SYMMETRIC VERTICAL LOADS
( ) (kg)nrczszs GGmP −=
Pzs = the vertical force (kg)mzs = 2.5 for private cars,
2.5 for busses,3.0 for lorries
Gc = total weight of the car (kg)Gnr = weight of unsprung mass (kg)
ASYMMETRIC VERTICAL LOADS
( ) (kg)nrcznszn GGmP −=
Pzn = the asymmetric vertical force (kg)mzns = 1.3 for private cars,
1.3 for busses,1.5 for lorries
Gc = total weight of the car (kg)Gnr = weight of unsprung mass (kg)
ASYMMETRIC VERTICAL LOADS (cont.)
( ) (kg.m)2 ⎟⎟⎠
⎞⎜⎜⎝
⎛−= p
plppznss
rRRmM
Ms = torque (kg.m)mzns = 1.3 for private cars,
1.3 for busses,1.5 for lorries
(Rpp – Rpl) = difference of forces on the front wheels (kg)
fog = deflection of the tyres (mm)fr = deflection of the springs (mm)r = the track width (mm)z = the width between suspension
attachments (mm)
(mm)h1 ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛+=
t
prt
t
pog
p
prpog z
rf
rr
fzr
ff
For single bump:
ASYMMETRIC VERTICAL LOADS (cont.)
fog = deflection of the tyres (mm)fr = deflection of the springs (mm)r = the track width (mm)z = the width between suspension
attachments (mm)
(mm)h2 ⎟⎠⎞
⎜⎝⎛+=
zrff rog
For double bump:
ASYMMETRIC VERTICAL LOADS (cont.)
Maximum heights of actual road surface bumps
Type of vehicleType of vehicleBump Bump HeightHeight
±300±300±250±250±200±200HHrr, mm, mm
LorriesLorriesBusesBusesCarsCars
ASYMMETRIC VERTICAL LOADS (cont.)
If h1,2 < Hr, then one wheel may leave the road surface. The force acting on each wheel and the torque acting on the car body are given as:
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ +=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
=
=
2
2
2
0
ppznss
t
ppptznstp
t
ppptznstl
pznspp
pl
rRmM
rrRR
mR
rrRR
mR
RmR
R
ASYMMETRIC VERTICAL LOADS (cont.)
If h1,2 > Hr, then the force acting on each wheel and the torque acting on the car body are given as:
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎠⎞
⎜⎝⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟
⎠⎞
⎜⎝⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
2,1
2,1
2,1
2,1
2,1
2
22
22
12
12
hHr
RmM
rr
hHRRmR
rr
hHRRmR
hHR
mR
hHR
mR
rppznss
t
prptznstp
t
prptznstl
rpznspp
rpznspl
ASYMMETRIC VERTICAL LOADS (cont.)
If h1,2 > Hr, then the force acting on each wheel and the torque acting on the car body are given as:
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎠⎞
⎜⎝⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎠⎞
⎜⎝⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
2,1
2,1
2,1
2,1
2,1
2
22
22
12
12
hHr
RmM
rr
hHRRmR
rr
hHRRmR
hHR
mR
hHR
mR
rppznss
t
prptznstp
t
prptznstl
rpznspp
rpznspl
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ +=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
=
=
2
2
2
0
ppznss
t
ppptznstp
t
ppptznstl
pznspp
pl
rRmM
rrRR
mR
rrRR
mR
RmR
R
If h1,2 < Hr, then one wheel may leave the road surface. The force acting on each wheel and the torque acting on the car body are given as:
fog = deflection of the tyres (mm)fr = deflection of the springs (mm)r = the track width (mm)z = the width between suspension
attachments (mm)
(mm)h1 ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛+=
t
prt
t
pog
p
prpog z
rf
rr
fzr
ff
For single bump:
Type of vehicleType of vehicleBump Bump HeightHeight
±300±300±250±250±200±200HHrr, mm, mm
LorriesLorriesBusesBusesCarsCars
These loads are caused by braking or accelerating the vehicle or by driving over a bump.
LONGITUDINAL LOADS
( ) (kg)nrcxx GGmP −±=
Px = the longitudinal force (kg)mx = 0.7 to 1.0Gc = total weight of the car (kg)Gnr = weight of unsprung mass (kg)
The case of a car driving into a bump or kerb:.
LONGITUDINAL LOADS (cont.)
(kg)tanθpzx RmP =
Px = the longitudinal force (kg)mz = 2.5 for carsRp = static load on the front axle (kg)
The angle θ is obtained as follows:
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−= −
d
r
rH1sin 1θ
rd is the dynamic radius of the wheel (from tyre catalogue data) and Hr is the actual height of road surface bump.
Side loads arise from driving on a curved track, or through a sideways collision with an obstruction.The maximum possible loading from the centrifugal force C is determined by the relationship between the wheel track and the height of the centre of gravity:
SIDE LOADS
czs
b
sc GmC
Zr
==2
tanγ
r = track width (mm)Zsc = height of centre of gravity (mm)Cb = lateral component of the
inertial force (kg)mzs = 2.5 for carsGc = total weight of the car (kg)
The loads at the suspension attachment points, front and rear, are
SIDE LOADS (cont.)
(kg)2
(kg)2
(kg)2
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛==
o
p
scczsty
o
t
scczspy
scczsyb
LL
ZrGmR
LL
ZrGmR
ZrGmRC
Lo = the wheelbase (mm)Lt = the distance of the centre of
gravity from the rear axle (mm)Lp = the distance of the centre of
gravity from the front axle (mm)
Vehicle structures are divided into three types:
(i) closed integral structures(ii) open integral structures(iii) flat or punt-type structures
STRESS ANALYSIS
The idealistion of a closed integral structure.
The idealistion of an open integral structure.
The idealistion of a flat or punt-type structure.
z
(i) Closed Integral Structures
STRESS ANALYSIS (cont.)
A structure is closed if the main design surfaces form a closed system and if shear forces occur between them when loaded in torsion.
Figure (b) shows the distribution of edge forces when this closed integral structure is under torsional loading.
The application of torque Ms onto the structural surface ABCD produces edge forces K1, K2, etc. in all surfaces and also produces the reaction Ms on the surface A’B’C’D’.
STRESS ANALYSIS (cont.)
Figure (c) shows the distribution of edge forces when this closed integral structure is under bending.
(i) Closed Integral Structures (cont.)
(ii) Open Integral Structures
A structure is open if it lacks the upper or the front and rear structural surfaces and if edge shear forces do not arise between all the surfaces.
Figure (a) shows the open integral structure open at the top.
Figure (b) shows the distribution of edge forces when the structure is under torsion.
STRESS ANALYSIS (cont.)
(ii) Open Integral Structures (cont.)
Figure (c) shows the distribution of edge forces when the structure is in bending.
STRESS ANALYSIS (cont.)
(ii) Open Integral Structures (cont.)
Figure (a) shows the idealisation of a vehicle structure open front and rear.
STRESS ANALYSIS (cont.)
Figure (b) shows the distribution of forces in torsion.
a)
(ii) Open Integral Structures (cont.)
Figure (a) shows the idealisation of a vehicle structure open front and rear.
STRESS ANALYSIS (cont.)
Figure (c) shows the distribution of forces in bending.
a)
Figure (a) shows the idealisation of a flat or punt type vehicle structure.
Figure (b) shows the distribution of forces in torsion.
STRESS ANALYSIS (cont.)
(iii) Flat or punt type structures