Chapter 7- DOZER PRODUCTIONA track-type dozer equipped with a power shift can push an average blade load of 6 lcy. Thematerial being pushed is silty sand. The average push distance is 88 ft. What production, inloose cubic yards (Icy), can be expected?Push time: 3 mph average speed (sandy material):distance (ft)Push time =X 60min/hr = .min5280 ft/mispeed (mph)Return time: Second gear because less than 100 ft.Maximum speed 5 mph:distance (ft)T.Return time =X 50min/hr =min5280 ft/mispeed (mph)The dozer must accelerate to attain the maximum velocity. Therefore, when using such speeddata, it is always necessary to make an allowance for acceleration time. Because, in thisexample, the change in speed is very small, an allowance of 0.05 min is made for accelerationtime.Return time =-0.05 =minManeuver time 0.05 minUsing Equation [7.4], pg.195 determine the production60 min x blade load[Eq. 7.4] =push time (min+return rimeimin+manexar rime imin

Given Data: Blade load=6 lcy It is asked to find the expected production in loose cubic yards

A track-type dozer equipped with a power shift can push an average blade load of 6 lcy. The material being pushed is silty sand. The average push distance is 88 ft. What production, in loose cubic yards (Icy), can be expected?

A track-type dozer equipped with a power shift can push an average blade load of 6 lcy. The material being pushed is silty sand. The average push distance is 88 ft. What production, in loose cubic yards (Icy), can be expected?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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