CHAPTER 9.ppt

Prof (Dr) B Prof (Dr) B DayalDayal

SUSPENSION/SPRINGING SYSTEM OF TANK

It is the combination of the parts connecting the tank hull with the axles of the road wheels.

Purpose Total realization of the engine power to combat Ensure fighting efficiency of the crew Proper observation of the battle field Accuracy of delivery of fire on move.The parameters to be considered Periods Amplitude Speeds The acceleration of vertical and longitudinal angular

oscillations of the tank hull.


the main parts springs. – they cushion the jerks and shocks transmitted

to the hull dampers / shock absorbers. – they absorb hull oscillations

caused due to external actions during tank motion.

types of tracked suspension bogie suspension. consists of system of links, arms and

springs interconnected in a manner to permit two or more road wheels to function in tandem. was used extensively on earlier tracked vehicles. the suspension provides an effect that devides the load equally

between a pair of tandem wheels.

BOGIE WHEEL SUSPENSION


when one of the wheels is displaced or subjected to a vertical track force, an equal force is reflected on the wheel of the bogie unit.

the suspension behaves in this manner only for speeds upto 10mph and above it the effect is lost.

independent suspension. it employs separate elastic element for each road wheel. the vertical displacement of one road wheel does not change the spring force of the remaining suspension springs except through an influence on the sprung mass.

INDEPENDENT TRAILING ARM SUSPENSION


Independet suspension Bogie suspension

1. The displacement of one or several road wheels does not change the displacement of remaining road wheels.

1. The maximum displacement of single road wheel is greater than the maximum simultaneous displacement of all the wheels of the entire bogie unit.

Changing the loading of one road wheel does not change the loading of others on a stationary hull.

2. Changes the loading of other road wheels of bogie unit.

3. On terrain irregularities the impact received by one road wheel

3. All wheels of particular bogie unit.

4. Dismantling one road wheel does not effect function of other wheels

4.Impairs functioning of entire bogie unit.


Independet suspension Bogie suspension

Large diameter road wheels are usually employed

Numerous small diameter road wheels are usually used.

Mechanically less complex 5. More complex.

6. Easy to maintain 3. More malfunctioning and maintenance problems.

4. More protected from damage by enemy action and terrain debris.

4. More exposed to ballistic attack

Softer suspension with high reserve of energy absorption and road wheel deflection.

Harder suspension leading to harsh jolting ride.

Used for high speeds over 10 mph Inferior for high speeds.


terrain induced dynamic loads

three basic approaches are used:to write the equation of motion of vehicular

system, and then assuming a deterministic terrain function, which is representative of real configuration of the terrain.

similar to above but actual terrain profile is used as input.

using empirical methods from structural and experimental technique of actual vehicles.


THE FIRST APPROACH: Classical deterministic approach. Assume a terrain profile in terms of a series of periodic

functions such as in sine waves, with any numbers of frequency components.

Differential equation to be made for such idealised system and solved.

IDEALISED SYSTEM OF WHEEL – TYRE - SPRINGING


the periodic displacement will be:x1 = x1 sin ωt (1)

where,x1 – vertical displacement of tyre at any time in mx1 – amplitude of the wave in mω – frequency of ground wave in rad/sect – time in sec

the differential equation of the system, assuming viscous damping is:

mx2” + cx2’ + (k + k0) x2 = k0x1 sin ωt (2)where, m – unsprung mass of the vehicle

x2 – vertical displacement of the mass in mk – spring rate of the suspension spring, kg/mk0 – spring rate for tyre kg/m


c – DAMPING CONSTANTTHE DYNAMIC FORCE Fdy1 (kg) developed at the point of tyre

and ground contact is:Fdy1 = k0 (x1 – x2) (3)

The second approach: Geomewtric terrain profile is used as input to the equation

of motion. Profilometer is used to measure the ground profile. In profilometer the horizontal component ‘x’ and vertical

component ‘y’ are to be determined assuming the following relationship:

y – y0 = ∫0s sin θ ds

x = ∫0s cos θ ds

Where, y0 – elevation at the beginning of the runy – instantaneous elevation at any point alongthe run.


θ – slope angle relative to the horizontal reference at distance s from the origin.

In order to solve above equations, it is necessary to make a continuous measurement of θ and s.

The third approach:In order to evaluate most complex dynamic

system of road loads, a semi empirical system does exists.

It is based on the data collected from actual vehicle by means of oil pressure readings, strain gauge and accelerometer measurements taken under motion.

LOAD FACTORS FOR ESTIMATING THE ROAD LOAD


select the vehicle classification and operation category by selecting the appropriate cureve.

determine the load factor ‘n’ corresponding to the inter section of the selected curve with gross weight ‘ w’ of the vehicle. tis represents the maximum acceleration in g’s experienced by the front end of the tank. at successive stations it will decrease (n/2)g at the vehicle cg and remain constant at the stations behind.

horizontal and vertical design loads on forward sprocket will be: = 0.15n w in newtons

vertical load on no. 1 road wheel: = 0.1 n w kg vertical load on no. 2 and successive road wheels

= 0.05 n w kg vertical and horizontal load on rear sprocket = 0.07nw kg transverse load acting horizontally down below of road

wheel = w/2 kg.


Example:Assume that a tracked combat vehicle under

consideration, which is expected to have 5 road wheels per side and a gross weight 23000 kg.


Example:Assume that a tracked combat vehicle under

consideration, which is expected to have 5 road wheels per side and a gross weight 23000 kg.

SOLUTION:From the graph: n = 33Load on forward idler wheel – 0.15 x 33 x 23000 g =

113850g NewtonLoad on no.1 road wheel = 0.1 x 33 x 23000g =

75900g newtonLoad on no,2, 3, 4 and 5 road wheel = 0.05 x 33 x

23000g = 37950g newtonLoad on sprocket = 0.07 x 33 x 23000g = 53130gTransverse load on each wheel = 23000/2 = 11500g


Apparent ground contact area:Aa = 2. L. T

The actual ground contact area of tracked vehicle is a finction of track tension, suspension design and soil properties.

Apparent average ground pressure (NGP):

NGP = W/Aa

= W/2LTWhere, W – gross weight in newton

L – length of track bearing surface between centre lines of extreme road wheels.

T – over all track width

STATIC LOAD ON ROAD WHEELS

Assume a vehicle with three equidistance road wheels on each side with a rigid suspension and loose track.

W1 + w2 + w3 – w = 0W.L/2 - w2.L - w3. 2l = 0

An additional assumption of sinkage of wheels is considered. Sinkage will be such that the centres of the road wheels will remain in straight line.

The left wheel will sink more than the right wheelZ1 – z2 = z2 – z3

Sinkage for road wheel is given byZi = (wi/AK)1/n

Where, wi – loading on each wheelA – area of ground contact of each

wheel

STATIC LOAD ON ROAD WHEELSK – soil constant

Taking value of n = ½,

w32 = 2w2

2 – w12

The solution is

W1 = 0.556 w

W2 = 0.390 w

W3 = 0.054 wConsider the vehicle possessing 5 road wheels on each side.

The deflection of spring is f1, f2, f3, ---fn

W1 = cf1; w2 = cf2; w3 = cf3; -----wn = cfn

Where, w1, w2, ---wn – load acting on each wheelC – rate of individual springs (assumed equal)

kg/cm

F2 = f1 + l2 tan θ

STATIC LOAD ON ROAD WHEELS

F3 = f1 + l3 tan θ

- -

- -

Fn = f1 + ln tan θ

Finally two more equations can be written

W1 + w2 + w3 + ------ wn = w

W2l2 + w3l3 + ---------wnln = w l0

Since total equations are (2n + 1) and unknowns are also (2n + 1), i.E., N forces w, n deflections f, and angle θ. This permits the determination of unknowns by algebric manipulation.

STATIC LOADS ON ROAD WHEELS

LOADS ON ROAD WHEELS

Dynamic loads on the road wheelsThe configuration of torsion bar springs used in combat

vehicles are straight bars of circular cross section, splined at each end.

For a round bar in torsion, the relationships are as under:Wind up angle ø = 32 tl/ (π. D4. G) = 2sl/dg radTorsional rate t’ = t/ø = π. D4. G/ 32.L kg m/radShear stress s = 16t/π d3 = ødg/2l kg/m2

Stress rate = s/ø = dg/2l kg/m2/radDesign load p = π. D3. Sv/ 16 r kgResilience u = pf/2 = sv

2. V/4g kgmWhere, t – torque applied to the bar kgm l

– active length of the barD – diameter of the barG – shear modulus of elasticity kg/m2

R – length of moment arm in mF – deflection at the end of moment arm

CHAPTER 9.ppt

Documents

road wheels of bogie

road wheel deflection

road wheel3

pair of tandem wheels

vehicular system

system of links

tank hull

independent suspension