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?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![Chapter 7- DOZER PRODUCTION
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?
Push time: 3 mph average speed (sandy material):
distance (ft)
Push time =
X 60min/hr = .
min
5280 ft/mi
speed (mph)
Return time: Second gear because less than 100 ft.
Maximum speed 5 mph:
distance (ft)
T.
Return time =
X 50min/hr =
min
5280 ft/mi
speed (mph)
The dozer must accelerate to attain the maximum velocity. Therefore, when using such speed
data, it is always necessary to make an allowance for acceleration time. Because, in this
example, the change in speed is very small, an allowance of 0.05 min is made for acceleration
time.
Return time =
-0.05 =
min
Maneuver time 0.05 min
Using Equation [7.4], pg.195 determine the production
60 min x blade load
[Eq. 7.4] =
push time (min
+return rimeimin
+manexar rime imin](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01c298ef-cb75-4403-b7a0-b52192ebb929%2F4e06a4bd-b006-4f8e-8d96-cb8f73603b50%2F838uv9m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Chapter 7- DOZER PRODUCTION
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?
Push time: 3 mph average speed (sandy material):
distance (ft)
Push time =
X 60min/hr = .
min
5280 ft/mi
speed (mph)
Return time: Second gear because less than 100 ft.
Maximum speed 5 mph:
distance (ft)
T.
Return time =
X 50min/hr =
min
5280 ft/mi
speed (mph)
The dozer must accelerate to attain the maximum velocity. Therefore, when using such speed
data, it is always necessary to make an allowance for acceleration time. Because, in this
example, the change in speed is very small, an allowance of 0.05 min is made for acceleration
time.
Return time =
-0.05 =
min
Maneuver time 0.05 min
Using Equation [7.4], pg.195 determine the production
60 min x blade load
[Eq. 7.4] =
push time (min
+return rimeimin
+manexar rime imin
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