A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is
1 answer:
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Answer:
Explanation:
Using the kinematics equation to determine the velocity of car B.
where;
initial velocity
= constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
Then:
The distance traveled by car B in the given time (t) is expressed as:
For car A, the needed time (t) to come to rest is:
Also, the distance traveled by car A in the given time (t) is expressed as:
Relating both velocities:
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
Rubber block is not shown. I have attached an image of it.
Answer:
A) ε_x = 0.0075
B) ε_y = 0.00375
C) γ_xy = 0.0122 rad
Explanation:
We are given;
δ = 0.03 in
L = 4 in
ν_r = 0.5
θ = 89.3° = 89.3π/180 rad
Let's calculate ε_x in the direction of axis x
Thus, ε_x = δ/L = 0.03/4 = 0.0075
Let's calculate ε_y in the direction of axis y;
ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375
Now, shear strain is angle between π/2 rad surfaces at that point.
Thus,
γ_xy = π/2 - θ = π/2 - 89.3π/180
γ_xy = π(0.003889) = 0.0122 rad
