Answer:
The provided length of the vertical curve is satisfactory for the reconstruction design speed of 60 mi/h
Explanation:
The explanation is shown on the first uploaded image
Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ
= displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ
------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
Under no circumstances
Explanation:
I'm not 100% sure why, but I remember hearing that you're not suposed to go over the speed limit no matter what
Answer:
w = 10.437 kips
deflection at 1/4 span 20.83\E ft
at mid span = 1.23\E ft
shear stress 7.3629 psi
Explanation:
area of cross section = 18*76
length of span = 32 ft
moment = 334 kips-ft
we know that
moment = load *eccentricity
334 = w * 32
w = 10.437 kips
deflection at 1/4 span



= 20.83\E ft
at mid span



shear stress
