Civil Engineering - Highway Engineering - Discussion

Discussion Forum : Highway Engineering - Section 1 (Q.No. 1)
1.
A district road with a bituminous pavement has a horizontal curve of 1000 m for a design speed of 75 km ph. The super-elevation is
1 in 40
1 in 50
1 in 60
1 in 70
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
95 comments Page 5 of 10.

Anwar zaib said:   7 years ago
To determine super elevation we need only f- value, design speed and radius of the curve. So, if it is not mentioned then directly take the value with the unit of distance i.e meter.
the formula for mix traffic.

e + f = V^2/225R
HERE
f = 0 R=1000m and V =75kmph
so
e= 0.025 (1/40).

Dev said:   7 years ago
Here it's v^2/127r.

Anmol said:   7 years ago
It's we have already given the design speed so we will use v2/127r.

Haroon said:   8 years ago
e+f=v^2/gr, v in m/s.
e+f=v^2/127r, v in km/h.

For 75% of design speed f=0,as per irc.
e=(0.75v)^2/127r =v^2/225r.
e=75*75/(225*1000)=0.025=1:40.

DJ.thakur said:   8 years ago
For design of superelevation on road:-

(1)- design superelevation for 75% of design speed. (Hence s.e. resists 75% of centrifugal force rest is balanced by road friction)

MANISH CHANDEL said:   8 years ago
e=v^2/225*R is correct.

Deven negi said:   8 years ago
Radius are not mentioned.

Himanshu kumar said:   8 years ago
But there speed in kilometer and horigental curve in meter so how it is possible? Please explain.

Janardhan said:   8 years ago
e=V2/225R when V in KMPH,
e=v2/127R-f ,if f is given,
and v in m/s,other wise e=v2/127R if f is not given.
(1)

Azanaw said:   8 years ago
Why 75% is taken for design speed?


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