Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 36)
36.
In a braking test, a vehicle travelling at 36 km ph was stopped at a braking distance of 8.0 m. The average value of the vehicle's skid resistance (friction coefficient) is
Discussion:
16 comments Page 1 of 2.
Kshitiz said:
8 years ago
Friction coefficient value never exceed 0.4
Longitudinal friction coefficient range 0.3 to o.4
Lateral friction coefficient value generally 0.15
You can use the formula for finding friction coefficient but it exceeds limiting value. So Ans is 0.16 correct ie it took lateral friction value.
Longitudinal friction coefficient range 0.3 to o.4
Lateral friction coefficient value generally 0.15
You can use the formula for finding friction coefficient but it exceeds limiting value. So Ans is 0.16 correct ie it took lateral friction value.
Shubham gour said:
5 years ago
Option c is correct.
Given:- SSD :- 8 M,
V=36kmph then convert.
V= 36 * 5/18= 10 t=2.5 sec by IRC.
SSD = (V*T)+(V^2/2gf),
8 = (10*2.5)+(10^2/2*9.18*f),
8 = 25+100/18.36fm
8 = 30.44f,
f = 8/30.44,
f = 0.26.
It is the nearest value options C.
So, options C is correct.
Given:- SSD :- 8 M,
V=36kmph then convert.
V= 36 * 5/18= 10 t=2.5 sec by IRC.
SSD = (V*T)+(V^2/2gf),
8 = (10*2.5)+(10^2/2*9.18*f),
8 = 25+100/18.36fm
8 = 30.44f,
f = 8/30.44,
f = 0.26.
It is the nearest value options C.
So, options C is correct.
Suman said:
7 years ago
It seems to be option D is correct as 'f' in the breaking distance formula signify longitudinal friction, which varies from 0.4 to 0.35 depending on the speed, while the lateral friction 'f' comes to design the superelevation instead. So option D be the correct answer.
(1)
Shubham gour said:
5 years ago
Options C is correct.
We know that;
SSD= (V*T)+(V^2/2*g*f).
First off ..V= 36*5/18=10 and t=2.5 by IRC.
8 = (10*2.5) +(10^2/2*9.81*f)
8 = 25+100/19.62*f
8 = 25+30.09*f
f = 8/30.09
f = 0.26.
So, take the value is nearest in options is 0.16.
We know that;
SSD= (V*T)+(V^2/2*g*f).
First off ..V= 36*5/18=10 and t=2.5 by IRC.
8 = (10*2.5) +(10^2/2*9.81*f)
8 = 25+100/19.62*f
8 = 25+30.09*f
f = 8/30.09
f = 0.26.
So, take the value is nearest in options is 0.16.
(1)
Swain lipan said:
5 years ago
Option c is correct. Because it asking in skid resistance coefficient.
Hence.
S=1/2*at^2,
Hence a= g*f.
S is given 8.
t = 2.5 sec. As per IRC.
g = 9.81,
f = ?.
If we put this formula answer comes 0.26 which is nearly equal to option C.
Hence.
S=1/2*at^2,
Hence a= g*f.
S is given 8.
t = 2.5 sec. As per IRC.
g = 9.81,
f = ?.
If we put this formula answer comes 0.26 which is nearly equal to option C.
Gubendhiran said:
8 years ago
SSD=lag distance + braking distance.
SSD=Vt +[(V^2)÷(2g(f+-g))],
Breaking distance=(v^2)÷(254f+-g),
We don't have gradients so leave it.
8=(36^2)÷(254f),
f=0.63779=0.64,
So option A is correct.
SSD=Vt +[(V^2)÷(2g(f+-g))],
Breaking distance=(v^2)÷(254f+-g),
We don't have gradients so leave it.
8=(36^2)÷(254f),
f=0.63779=0.64,
So option A is correct.
(1)
Salim said:
5 years ago
f = v2/254L = 36*36/254*8 = 0.64 but the value of friction factor lies in between 0.35 to 0.40. So, the correct Ans will be (D) none of the above.
(4)
Akhila said:
6 years ago
Yes, in sight distance its always Longitudinal friction not lateral. So, Option D as it exceeds the range (0.35-0.40).
Raj said:
8 years ago
Yes, absolutely correct @Kshitiz.
The Limiting value is 0.4. The correct option A is correct.
The Limiting value is 0.4. The correct option A is correct.
ARUN KUMAR said:
9 years ago
f' = V*2 / 2gL
= (36/3.6)**2/ 2 * 9.81 * 8.0 v = V /3.6
= 36/3.6
= 10m/sec
= 0.64m (say)
= (36/3.6)**2/ 2 * 9.81 * 8.0 v = V /3.6
= 36/3.6
= 10m/sec
= 0.64m (say)
(1)
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