Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:

where;
R= radius
= coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
= 0.20
So;



R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
Answer:
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Explanation:
Answer:
total weight of aggregate = 5627528 lbs = 2814 tons
Explanation:
given data
dry density = 119.7 lb/cu ft
area = 2000 ft × 48 ft × 6 in
aggregate = 3.1%
required compaction = 95%
solution
we get here volume of space to be filled with aggregate that is
volume = 2000 × 48 × 0.5 = 48000 ft³
when here space fill with aggregate of density is
density = 0.95 × 119.7 = 113.72 lb/ft³
and
dry weight of this aggregate will be is
dry weight = 48000 × 113.72 = 5458320 lbs
and
we consider take percent moisture by weigh so that there weight of moisture in aggregate is express as
weight of moisture = 0.031 × 5458320 = 169208 lbs
and
total weight of aggregate will be
total weight of aggregate = 5458320 + 169208
total weight of aggregate = 5627528 lbs = 2814 tons
Width * Length * Thickness * Density = Weight.
48″ * 96″ * . 1875″ * 0.284 lb/in3 = 245 lb.