Answer:
Explanation:
Given that:
The Inside pressure (p) = 1402 kPa
= 1.402 × 10³ Pa
Force (F) = 13 kN
= 13 × 10³ N
Thickness (t) = 18 mm
= 18 × 10⁻³ m
Radius (r) = 306 mm
= 306 × 10⁻³ m
Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)
Then;
the state of the plane stress can be expressed as follows:

Since d = 2r
Then:







When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.
Thus;

Answer:
P = 18035.25 N
Explanation:
Given
D = 10.4 mm
ΔD = 3.2 ×10⁻³ mm
E = 207 GPa
ν = 0.30
If
σ = P/A
A = 0.25*π*D²
σ = E*εx
ν = - εz / εx
εz = ΔD / D
We can get εx as follows
εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴
Now we find εx
ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³
then
σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa
we have to obtain A:
A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²
Finally we apply the following equation in order o get P
σ = P/A ⇒ P = σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N
Answer:
2.275 %
Explanation:
see the attached picture for detailed answer.
Answer:
The minimum length of the vertical curve that can be used is 416.63 m
Explanation:
g₁ = -1.2% , g₂ = 0.8%
Station (Sta) = 75 + 00
Elevation (Ele) = 50.90 m
x₀ = Sta(VPC) = Sta(VPI) - L/2 ⇒ 7500 - L/2
y₀ = Ele(VPC) = Ele(VPI) - g₁L/2 == 50.90 - (-0.012)L/2 ⇒ 50.90 + 0.012L/2
at Sta 75+40, x = 7540 -x₀ = 40 + L/2
at Ele, y = 51.10 + C = 51.10 + 0.80 ⇒ y = 51.90 m
Equation of vertical curve is

Substituting the values for x and y



Solving : L = 416.63 m