Building Construction Handbook, Eighth Edition

Output and Cycle Times ~ all items of plant have optimum output

and cycle times which can be used as a basis for estimating

anticipated productivity taking into account the task involved,

task efficiency of the machine, operator's efficiency and in the case

of excavators the type of soil. Data for the factors to be taken

into consideration can be obtained from timed observations,

feedback information or published tables contained in

manufacturer's literature or reliable textbooks.

Typical Example ~

Backacter with 1 m^3 capacity bucket engaged in normal trench

excavation in a clayey soil and discharging directly into an

attendant haulage vehicle.

An allowance should be made for the bulking or swell of the solid

material due to the introduction of air or voids during the

excavation process

\Net output allowing for a 30% swell = 36 † (36 ¾ 0„3)

= say 25 m^3 per hr.

If the Bill of Quantities gives a total net excavation of 950 m^3

time required =^95025 =38 hours

or assuming an 8 hour day---1/2 hour maintenance time in

days =7„5^38 = say5 days

Haulage vehicles required = 1+round trip time of vehicle

loading time of vehicle

If round trip time = 30 minutes and loading time = 10 mins.

number of haulage vehicles required = 1+^30

10 =4

This gives a vehicle waiting overlap ensuring excavator is fully

utilised which is economically desirable.

Optimum output = 60 bucket loads per hour

Task efficiency factor = 0„8 (from tables)

Operator efficiency factor = 75% (typical figure)

\Anticipated output = 60 ¾ 0„8¾ 0„75

= 36 bucket loads per hour

=36¾ 1 = 36 m^3 per hour

Builders Plant