6.20 A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain.
1 answer:
Answer:
A) Level D ( before development )
B) Level E ( After development )
Explanation:
A) Calculate the level of service before the development
First we calculate the base free flow speed ( BFFS )
= posted speed limit + 5 mi/hr.
= 50 + 5 = 55mi/hr.
Next determine free flow speed ( FFS )
= BFFS -
= 55 - 6.6 - 0.65 - 0 - 2.25 = 45 mi/hr.
The remaining part of the solution is attached below
The level of service = D
given that:
Vp = 1537.096 pc/hr/In
Density = 34.157 pc/mi/In
FFS = 45 mi/hr
B ) Calculate the level of service after the development
attached below is the detailed solution
The level of service = E
given that
Vp = 1816.568 pc/hr/In
Density = 40.368 pc/mi/In
FFS = 45 mi/hr
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Answer:
equivalent stiffness is 136906.78 N/m
damping constant is 718.96 N.s/m
Explanation:
given data
mass = 120 kg
amplitude = 120 N
frequency = 320 r/min
displacement = 5 mm
to find out
equivalent stiffness and damping
solution
we will apply here frequency formula that is
frequency ω = ω(n) √(1-∈ ²) ......................1
here ω(n) is natural frequency i.e = √(k/m)
so from equation 1
320×2π/60 = √(k/120) × √(1-2∈²)
k × ( 1 - 2∈²) = 33.51² ×120
k × ( 1 - 2∈²) = 134752.99 .....................2
and here amplitude ( max ) of displacement is express as
displacement = force / k × ( )
put here value
0.005 = 120/k × ( )
k ×∈ × √(1-2∈²) = 1200 ......................3
so by equation 3 and 2
solve it and we get
∈ = 1.00396
and
∈ = 0.08869
here small value we will consider so
by equation 2 we get
k × ( 1 - 2(0.08869)²) = 134752.99
k = 136906.78 N/m
so equivalent stiffness is 136906.78 N/m
and
damping is express as
damping = 2∈ √(mk)
put here all value
damping = 2(0.08869) √(120×136906.78)
so damping constant is 718.96 N.s/m
is the volume of the sample when the water content is 10%.
<u>Explanation:</u>
Given Data:
First has a natural water content of 25% = = 0.25
Shrinkage limit,
We need to determine the volume of the sample when the water content is 10% (0.10). As we know,
------> eq 1
The above equation is at ,
Applying the given values, we get
Shrinkage limit is lowest water content
Applying the given values, we get
Applying the found values in eq 1, we get
