Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
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Chapter 13, Problem 26P

(a)

To determine

The reason for greater acceleration in all wheel drive concept.

(b)

To determine

Whether the center or rear differential, or both provides greater traction for a certain road condition during the locked state.

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See the answer A flat belt drive system is to be designed for an application in which the input shaft speed (driving pulley) is 1725 rpm, the driven shaft speed is to be approximately 960 rpm, and the power to be transmitted has been estimated as 3.0 horsepower. The driven machine has been evaluated and found to have characteristics of moderate shock loading during operation. The desired center distance between driving and driven pulleys is approximately 18 inches.
480 ft-lbf First Gear (2.66) Final Drive (3.42) Figure 2 - Corvette Z06 and its transmission schematic. You start by measuring the 60-0 mph braking performance, which is accomplished along 30.56 m. The test weight (including the driver) was found to be 1510.5 kgf. Peak torque (@4800 rpm) 1st gear reduction Axle Torque Thrust Force After the braking test, you are asked to determine the maximum acceleration in the first gear. Consulting the manufacturer, you find out the following technical data: Final drive (differential) reduction Rolling radius of the rear tyres % weight on the rear tyres Table 1 - Technical data for the Corvette Z06 480 lbf-ft 2.66 3.42 0.327 m 65% 1) Assuming the maximum traction capacity of the tyres has been determined in the braking test, CALCULATE the maximum traction generated by the rear tyres in first gear. CALCULATE the available thrust force (i.e. traction) from the torque on the wheels.
3)Design a V – Belt drive to the following specifications Power transmitted = 75kw Speed of driving wheel = 1440rpm Speed of driven wheel = 400rpm Diameter of driving wheel = 300mm Center distance = 2500mm Service = 16hrs/day

Chapter 13 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 13 - Prob. 11PCh. 13 - Prob. 12PCh. 13 - Prob. 13PCh. 13 - Prob. 14PCh. 13 - A parallel-shaft gearset consists of an 18-tooth...Ch. 13 - The double-reduction helical gearset shown in the...Ch. 13 - Shaft a in the figure rotates at 600 rev/min in...Ch. 13 - The mechanism train shown consists of an...Ch. 13 - The figure shows a gear train consisting of a pair...Ch. 13 - A compound reverted gear trains are to be designed...Ch. 13 - Prob. 21PCh. 13 - Prob. 22PCh. 13 - Prob. 23PCh. 13 - A gearbox is to be designed with a compound...Ch. 13 - The tooth numbers for the automotive differential...Ch. 13 - Prob. 26PCh. 13 - In the reverted planetary train illustrated, find...Ch. 13 - Prob. 28PCh. 13 - Tooth numbers for the gear train shown in the...Ch. 13 - The tooth numbers for the gear train illustrated...Ch. 13 - Shaft a in the figure has a power input of 75 kW...Ch. 13 - The 24T 6-pitch 20 pinion 2 shown in the figure...Ch. 13 - The gears shown in the figure have a module of 12...Ch. 13 - The figure shows a pair of shaft-mounted spur...Ch. 13 - Prob. 35PCh. 13 - Prob. 36PCh. 13 - A speed-reducer gearbox containing a compound...Ch. 13 - For the countershaft in Prob. 3-72, p. 152, assume...Ch. 13 - Prob. 39PCh. 13 - Prob. 40PCh. 13 - Prob. 41PCh. 13 - Prob. 42PCh. 13 - The figure shows a 16T 20 straight bevel pinion...Ch. 13 - The figure shows a 10 diametral pitch 18-tooth 20...Ch. 13 - Prob. 45PCh. 13 - The gears shown in the figure have a normal...Ch. 13 - Prob. 47PCh. 13 - Prob. 48PCh. 13 - Prob. 49PCh. 13 - The figure shows a double-reduction helical...Ch. 13 - A right-hand single-tooth hardened-steel (hardness...Ch. 13 - The hub diameter and projection for the gear of...Ch. 13 - A 2-tooth left-hand worm transmits 34 hp at 600...
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