Question

A spherical tank for storing gas under pressure is 25 m in diameter and is made of steel 15 mm thick. The yield point of the material is 240 MPa. A factor of safety of 2.5 is desired. The maximum permissible internal pressure is most nearly: 90 kPa 230 kPa 430 kPa D. 570 kPa csauteol psotolem here Pcr 8. A structural steel tube with a 203 mm x 203 mm square cross section has an average wall thickness of 6.35 mm. The tube resists a torque of 8 N m. The average shear flow is most nearly
A. 100 N/m
B. 200 N/m
C. 400 N/m
D. 800 N/m

Answer 1

Answer:

a) (option B) 230 kPa

b) (option A) 100 N/m

Explanation:

Given:

Diameter, d = 25 m

Thicknesses, t = 15 mm

Yield point = 240 MPa

Factor of safety = 2.5

a) To find the maximum internal pressure, let's use the formula:

$\sigma l = \frac{\sigma y}{FOS} = \frac{PD}{4t}$

$\frac{\sigma y}{FOS} = \frac{PD}{4t}$

Solving for P, we have:

$P = \frac{\sigma y * 4t}{FOS * D}$

$P = \frac{240 * 4 * 15}{2.5 * 25}$

P = 230.4 kPa

≈ 230 kPa

The maximum permissible internal pressure is nearly 230kPa

b) Given:

Thickness, t = 6.35 mm

L = 203 mm

Torque, T = 8 N m

Let's find the mean Area,

mA = (l - t)²

= (203 - 6.5)²

= 38671.22mm²

≈ 0.03867 m² (converted to meters)

To find the average shear flow, let's use the formula:

$q = \frac{T}{2* mA}$

$= \frac{8}{2 * 0.03867}$

q = 103.4 N/m approximately 100N/m

The average shear force flow is most nearly 100 N/m

Answer 2

Answer:

1) 2304 kPa

2) B. 200 N/m

Explanation:

The internal pressure of the of the tank  can be found from the following relations;

Resisting wall force F = p×(1/4·π·D²)

σ×A = p×(1/4·π·D²)

Where:

σ = Allowable stress of the tank

A = Area of the wall of the tank = π·D·t

t = Thickness of the tank = 15 mm. = 0.015 m

D = Diameter of the tank = 25 m

p = Maximum permissible internal pressure pressure

∴ σ×π·D·t = p×(1/4·π·D²)

p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa

With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa

2) The formula for average shear flow is given as follows;

$q = \dfrac{T}{2 \times A_m}$

Where:

q = Average shear flow

T = Torque = 8 N·m

$A_m$ = Average area enclosed within tube

t = Thickness of tube = 6.35 mm = 0.00635 m

Side length of the square cross sectioned tube, s = 203 mm = 0.203 m

Average area enclosed within tube, $A_m$ = (s - t)² = (0.203 - 0.00635)² = 0.039 m²

$\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m$

Hence the average shear flow is most nearly 200 N/m.