Answer:
=>> 167.3 kpa.
=>> 60° from horizontal face.
Explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;
=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "
=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."
The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).
magnitude of the stresses on the failure plane = 167.3 kpa.
The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.
y = 60 sin 60° = 30√3 = sheer stress.
the orientation of this plane with respect to the major principle stress plane.
Theta = 45 + 15 = 60°.
Answer:
design angle ∅ = 4.9968 ≈ 5⁰
Explanation:
First calculate the force Fac :
Fac = 
= 
= 708.72 Ib
using the sine law to determine the design angle
hence ∅ = 
= 
= 4.9968 ≈ 5⁰
Answer:
v₀ = 2,562 m / s = 9.2 km/h
Explanation:
To solve this problem let's use Newton's second law
F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v
F dx = m v dv
We replace and integrate
-β ∫ x³ dx = m ∫ v dv
β x⁴/ 4 = m v² / 2
We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max
-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)
x_max⁴ = 2 m /β v₀²
Let's look for the speed that the train can have for maximum compression
x_max = 20 cm = 0.20 m
v₀ =√(β/2m) x_max²
Let's calculate
v₀ = √(640 106/2 7.8 104) 0.20²
v₀ = 64.05 0.04
v₀ = 2,562 m / s
v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)
v₀ = 9.2 km / h
