Question

Consider the pickup truck with the following specs: Vehicle dimensions and weight: Curb Weight Manual (lb.)- 8290 Front/Rear Axle Weights (lb.)- 4890/3400 Center of Mass Height (in)- 49 Wheelbase (in.) 172.40 Track Front (in.) 88.50 Track Rear (in.) -84.50 6 Speed Manual Transmission: st 5.79 2nd 3.30 3rd 2.10 4th .30 5th 1.00 Tires 225/70R20 Maximum Payload (lb.) 6120 Standard Towing (lb.) - 6000 Maximum Towing (Ib.) - 24,500 Max. trailer tongue load (lb.) 2500 6th 0.72 Final drive 4.88 nr0.9 Suspension Engine: Displacement: 6.4 liters V8 Diesel engine Power: 328 HP@ 3000 RPM Torque: 600 ft-lb @ 2000 RPM Idle RPM: 500 Independent front, solid axle rear Front Roll Stiffness (ft-lb/deg) 850 Rear Roll Stiffness (ft-lb/deg) 1550 Max RPM: 3000 Service Brakes: Description: Four wheel hydraulic actuated Master cylinder - Bore/Stroke (in): 1.375 1.55 Drum - inner diameter (in): 15.2 Line pressure at 100 lb. pedal load (psi): 2350 Pedal are ratio: 4.2:1 Brake Type Front/Rear: Disc/Drum Valving Type: Proportion (600/0.36) Rotor- Outer/inner diameter (in): 14.53/9.5 Wheel cylinder bore (in): 3.20 Drum brake cylinder bore (in): 1.5 Drum shoe width (in): 1.75 Dynamic friction coefficient for the lining material (F/R): 0.42/0.36 a) The fully loaded vehicle travels at 60 mph in the 5th gear on a level surface. The driver suddenly applies an 80 lb force on the brake pedal. Determine the total braking force, deceleration, and stopping distances for the second effectiveness tests. Does the vehicle pass the test? Notes: Neglect acrodynamic and rolling resistance forces. For drum brakes first cal culate the actuation force from the line pressure (use Faum instead of P, not to confuse it with the line pressure). Use the brake factors 4.2 and 2.5 for leading and trailing shoes respectively. b) Plot the maximum braking force on the front and rear axle, for the second effectiveness (consider that the payload distributes 30/70 % front/rear axle), on a dry road (Hp= 0.81). Plot the proportioning line on the same graph. Comment on the result c) Find a solutions (not involving the brake pads and/or shoes) to obtain a better braking characteristic. Try to make the proportioning line pass as close as possible to the point of simultaneous lockup (PSL)

Answer 1

Stoffen cas Vehicle travells at 60 mph . 5 th gear on level Surface force of so lb Total Break!! al Breaking force, deceleration and for the ness I distance Second effective tests. The deceleration produced by maximum breaking force , - - 64x64 (3.6) &51.8 serie dosa ou x 64 x 25.9 Gol misr. levels Brakes are applied to four wheel on Ma ointer = 0622.

Juch that, of 9 Ml Coso 5-Mh sono a ano lau o = 1 / 2 = 0.05 - Sino Cose :0.9999 = 1 {= 2.65-1.27 = 1.38 m Therefore, fr.0.622 x 1. 38 x 9.81 -0.05x9.81 2.65-0.622x0.76 = 3.87-0.49 = 3.38 mismo Stopping distance, S9WSG *2*3.38 (3-2) 2x3-4 2) x 2 x 3.36 = 46.8 mi. Distribution of loads Kfc I Cos 138 XI. (64] 2x 3.38 2 As bleh Repetto X133416 hoad is thin, = 8457.2 N RR = (b-l-Mh) Cose - bath 2.177-3.18 x 13341.6 2.177 988 4. 41884. N

b) maximum breaking force of front es rear axle. {fa wa .. fbt fw=for & fbrtfow sha Rr. 108.5" - Casun fost 398.4 lbs) Ro + Rf = 4t+393.7 fb = 2558.36 lbs. (1387.68) (24.6%) + Run Clogis) = (5213)(398) - Rr = 2096lbs . Maximum Breaking force is, =2870.69 Abso c) Better breaking characlisties- The designer load is greater than of 5490N . found earlier K = 0.288760.012) = 0.003464 in. c=1.2 Y = 4:03. - 297.3 Since, (Yk) (i) use Euler Equ Per= 0.025-10-012) PRY" (20-1) 109 (294.33" 0-003469 - 8321N.

- 2.65-1.27 - 1.38 m - 0.622 x 1.38 x 9.81 -0.05x9.81 2.65-0.622x0.76 $ 3.87-0.49 : 3.38 mis Slopping distance, . (66) af *2x 3.38 (3.23*2*3.33 = 46.8 mi. Distribution of loads Rf = lcoso w = 1.38 x 2x1334166 1 blh 2.147 : 8457.2 N Load is thin, RR = (b-l-Mh) Cosoi I both : 2.171-3:18 13341.6 2.197 488.4.4 N