Problem 11.1. Air supported tent

Inner cylindrical part of an air-supported tent is shown in the Figure (at the ends the tent is spherical). Radius of the cylinder is ρ = 7 m, the span is D = 12 m, thickness of the tent is t = 0.7 mm and the design inside air pressure is p_d = 1500 N/m². Considering the inner cylindrical part

a) calculate the design stress in the tent,

b) calculate the load on the edge beam,

c) verify whether the reinforced concrete edge beam is safe against sliding.

The cross-section of the beam is 800 x 800 mm, the friction coefficient between the beam and the soil is μ = 0.5, weight density is γ_c= 25kN/m³.

Solve Problem

SolveHide solve

Problem a)

Stress in the tent, σ₁ [MPa]=

Problem b)

Vertical load on the edge beam, A_y [kN/m]=

Horizontal load on the edge beam, A_x [kN/m]=

Problem c)

Friction force, F_f [kN/m]=

Do you need help?

Steps

Step by stepHide steps

Problem a)

Step 1. Determine the tensile force in the tent. Apply the pressure vessel formula.

Check tent's forceHide tent's force

According to the pressure vessel formula the hoop force in the cylindrical part of the tent is:

Eq(11-8)

$N_{1} = p ρ = 1500 \times 7.0 = 10500 \frac{N}{m} = 10.5 \frac{N}{mm}$

Step 2. Give the design stress

Check stressHide stress

The above force is the resultant of the uniformly distributed stress through the thickness of the tent.

$σ_{1} = \frac{N_{1}}{t} = \frac{10.5}{0.7} = 15 \frac{N}{{mm}^{2}}$

Problem b)

Step 1. Draw the direction of the edge beam’s load.

Check figureHide figure

The edge beam is loaded by the opposite of the tent’s force.

Step 2. Calculate the components of the edge load.

Check edge loadHide edge load

$A_{y} = A \sin α = A \frac{R}{ρ} = 10.5 \frac{6.0}{7.0} = 9.0 kN / m A_{x} = A \cos α = A \frac{\sqrt{ρ^{2} - R^{2}}}{ρ} = 10.5 \frac{\sqrt{7 . 0^{2} - 6 . 0^{2}}}{7.0} = 5.408 kN / m$

where α is given in the above Figure.

Problem c)

Step 1. Calculate the self-weight of the edge beam.

Check self-weightHide self-weight

$G = b h γ_{c} = 0.8 \times 0.8 \times 25 = 16.0 kN / m$

Step 2. Determine the friction force.

Check friction forceHide friction force

$F_{f} = μ (G - A_{y}) = 0.5 (16.0 - 9.0) = 3.50 kN / m$

Step 3. Check the beam for sliding

Check verificationHide verification

$F_{f} = 3.50 kN / m < A_{x} = 5.408 kN / m$

thus the edge beam is not safe for sliding.

Results

Worked out solutionHide solution

Problem a)

According to the pressure vessel formula the hoop force in the cylindrical part of the tent is:

Eq(11-8)

$N_{1} = p ρ = 1500 \times 7.0 = 10500 \frac{N}{m} = 10.5 \frac{N}{mm}$

The above force is the resultant of the uniformly distributed stress through the thickness of the tent.

$σ_{1} = \frac{N_{1}}{t} = \frac{10.5}{0.7} = 15 \frac{N}{{mm}^{2}}$

Problem b)

The edge beam is loaded by the opposite of the tent’s force.

Its vertical and horizontal components are

$A_{y} = A \sin α = A \frac{R}{ρ} = 10.5 \frac{6.0}{7.0} = 9.0 kN / m A_{x} = A \cos α = A \frac{\sqrt{ρ^{2} - R^{2}}}{ρ} = 10.5 \frac{\sqrt{7 . 0^{2} - 6 . 0^{2}}}{7.0} = 5.408 kN / m$

where α is given in the above Figure.

Problem c)

The self-weight of the edge beam is

$G = b h γ_{c} = 0.8 \times 0.8 \times 25 = 16.0 kN / m$

The friction force is determined from the vertical forces:

$F_{f} = μ (G - A_{y}) = 0.5 (16.0 - 9.0) = 3.50 kN / m$

$F_{f} = 3.50 kN / m < A_{x} = 5.408 kN / m$

thus the edge beam is not safe for sliding.