Answer:
The curve length (<em>L</em>) will be = 1218 ft
The elevations and stations for PVC and PVI
a. station of PVC = 103 + 91.00
b. station of PVI = 116 + 09.00
c. elevation of PVC = 432.18ft
d. elevation of PVI = 426.09ft
Explanation:
First calculate for the length (<em>L</em>)
To calculate the length, use the formula of "elevation at any point".
where, elevation at any point = 424.5.
and ∴ PVC Elevation = (420 + 0.01L)
Then, calculate for Station of PVC and PVI and elevation of PVC and PVI
This an example solved please follow up with they photo I sent ok
Answer:
L= 312.75 mm
Explanation:
given data
elastic modulus E = 106 GPa
cross-sectional diameter d = 3.9 mm
tensile load F = 1660 N
maximum allowable elongation ΔL = 0.41 mm
to find out
maximum length of the specimen before deformation
solution
we will apply here allowable elongation equation that is express as
ΔL = 
....................1
put here value and we get L
L = 
solve it we get
L = 0.312752 m
L= 312.75 mm
Answer:
676 ft
Explanation:
Minimum sight distance, d_min
d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time
v_max is given as 50 mi/h
t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds
Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft
For practical purposes, this distance is taken as 676 ft
The answer is E to increase the shear strength of the beam beyond that provided by the concrete
