Brake problems

1. A motor whose inertia is 0.3 kg.m² drives the rope drum of a hoist through a 5:1 gear reduction. The average diameter, radial thickness and face width of the larger gear's rim are   360, 16 and 50 mm. The mass of the rope drum is 120 kg, its radius of gyration is 110 mm, and it is equipped with grooves of 250 mm pitch diameter, on which is wrapped the hoisting rope whose mass is 0.5 kg/m. The maximum extended length of the rope is 60 m.
   A brake is incorporated into the motor shaft. Determine the brake torque and average power over the braking period when stopping within 1 m, a load of 1 t (1 tonne) being lowered at 3 m/s.         [ 435 Nm, 26.1 kW ]
    
2. Derive equations ( 13).
    
3. ( Problems 4-6 are similar ) For each brake, determine the sensitivity, the hinge and drum shaft reactions, and the parameters which are not defined in the duty statements. The brakes are symmetric, except for the mechanism of Problems 5 & 6; the geometry of Problem 6 is identical to that of Problem 5, however the drum rotation is clockwise in 5, counterclockwise in 6.
   Duty statements are   [ psi ≡ lbf/in²; 1000 lbf/kip ] :

   	problem	3	4	5	6
   	friction coefficient	0.32	0.24	0.3	0.3
   	braking torque	-	-	-	25 kip.in
   	actuation force	1.2 kN	-	400 lbf	-
   	lining mean pressure limit	-	1.0 MPa	150 psi	-
   	lining width	28 mm	75 mm	-	1.7 in
   Solutions
   	braking torque	231 Nm	2646 Nm	2150 lbf.ft	-
   	actuation force	-	8604 N	-	404 lbf
   	lining mean pressure	543 kPa	-	-	125 psi
   	lining width	-	-	1.7 in	-
   	sensitivity	1.337	1.117	1.143	1.101
   	hinge reaction - left	462 N	8900 N	6540 lbf	4720 lbf
   	hinge reaction - right	2716 N	8900 N	4130 lbf	5925 lbf
   	drum shaft reaction	2323 N	0	2510 lbf	1290 lbf

   
    

7. Design an external rigid shoe brake for the hoist of Problem 1, given that braking occurs twice a minute, that the lining is expected to reach a temperature of 300^oC, and that the required lining life is 5 khr. A design factor of 1.2 should be applied to the system energy loss of Problem 1.
    
8. A skip hoist for lifting bulk material, consists of two identical buckets or   "skips" connected by a wire rope which passes around a motor- driven head pulley. The loaded skip is partially counterbalanced by the empty skip. The head pulley is equipped with the brake illustrated, which is so arranged that in normal operation the brake is disengaged by an electric solenoid. In the event of an electrical power failure however, the brake is automatically engaged by a steel compression spring.
   Determine the initial compression of the spring necessary to arrest the loaded skip, falling at 2 m/s, within 2 s of brake engagement.
   The inertia of the pulley/drum is 0.5 t.m², the coefficient of friction for the brake may be taken as 0.3, and the spring, which is made from 10 mm diameter stock, has 10 active turns of 60 mm mean coil diameter.       [ 27 mm]
    
9. At first glance, a sprag type of over-running, uni-directional clutch looks rather like a cylindrical roller bearing in that it consists of two concentric circular rings. However instead of having cylinders between these, there is a series of closely spaced   "sprags" or cams, similar to the one sketched. These fit loosely for one direction of relative ring rotation to allow "free-wheeling". Light springs keep the sprags in touch with the rings. A reversal of relative rotation causes a rocking of the sprags and a tightening-up so that a high torque may be carried.
   The two lines drawn from the centre O of the rings to the centres of curvature O₁ and O₂ of the sprag surfaces at the contact points, make a small angle   α with one another.
   Draw the free body of a sprag, perhaps exaggerating the angle α for clarity, and obtain graphically the relative magnitudes and directions of the two contact forces with their normal and tangential components.
   Derive approximate equations for the angles betwen the forces and their normal components as functions of the angle α and the ring contact radii r₁ and r₂, and find the maximum value of α in terms of the coefficient of friction μ if slipping is not to occur.       [ α_max ≅ μ ( 1 - r₁/r₂ ) ]
    
10. The centre of mass of a vehicle lies at   c_F = h = 1/3
    ( a)   Plot the vehicle braking characteristic along with representative loci of constant adhesion coefficient and of constant decelerations;
    ( b)   The vehicle is equipped with proportional braking of 1:4 (rear:front). What is the maximum deceleration that can be achieved safely, and the corresponding necessary adhesion coefficient ?       [ 0.4]
    ( c)   Repeat (b) if the normalised rear braking force is limited to 0.06.       [ 0.765]
     
11. The centre of mass of a 1.2 t vehicle lies midway between front and rear axles, at a height above road level of one quarter of the wheelbase (the distance between front and rear axles). The vehicle is equipped with hydraulically operated, symmetric brakes as shown, the front and rear sets being identical except for lining width and hydraulic cylinder diameter.

    Further details are :-	Front	Rear
    hydraulic cylinder diameter, mm	29	20.5
    maximum hydraulic pressure, MPa	5.5	4

    The friction coefficient between lining and drum is 0.4 and the tyres are 640 mm diameter.
    ( a)   Criticise the safety of this braking arrangement.
    ( b)   Determine the maximum vehicle deceleration that may be obtained, and the tyre-road adhesion coefficient necessary to achieve it.       [ 7.3 m/s², 0.79]
    ( c)   Calculate the maximum value of the average lining pressure if the front and rear lining widths are 60 and 40 mm respectively.       [ 1.34 MPa]
     
12. Repeat Problem 5 with the shoes pivoted to the posts at   b = 11 in, θ_G = 80^o.
    The sensitivity expression for a pivoted shoe is more complex than that of a rigid shoe - the   m & n parameters are no longer appropriate. One way to proceed is to denote   ' ≡ d/dμ, whereupon   α' and   β' follow from ( 6c) and ( 11). Call the scalar ratio between the pressure components   ν and evaluate it from ( 11) as   ν = N_c /N_s = p_c /p_s = - β J_s /β J_c .
    Thereafter form the vector   J_o ≡ J_s + ν J_c and hence   ν' = - δΔ.β' J_o /β J_c . It follows that   η = μr J_o[3] / αJ_o . Differentiating this last expression and inserting into ( vii) leads to the expression for sensitivity.
    [   T = 2160 lbf.ft;   w = 1.67 in;   S = 1.148;   R_HL = 3610 lbf;   R_HR = 1400 lbf;   R_O = 2475 lbf ]