Question

if a thin-walled cylindrical pressure vessel with a radius of 14 cm and a wall thickness of 1.6 mm has an internal pressure of 35 atm, along with a material with a yield strength of 200 mpa, by what factor do we need to multiply thickness to get a safety factor of 5?

Answer 1

The wall thickness of the cylindrical pressure vessel needs to be multiplied by a factor of 2.07 to achieve a safety factor of 5.

Step-by-step explanation:

The correct answer is that we need to multiply the thickness by a factor of 2.07 to achieve a safety factor of 5. To determine this factor, we can use the formula for hoop stress in a cylindrical pressure vessel: σ = (P * r) / t, where P is the internal pressure, r is the radius, and t is the wall thickness.

We know the yield strength (σ_y) of the material is 200 MPa, and we need to include the safety factor (SF) of 5, so our target is to keep the actual hoop stress below σ_y / SF.

The existing internal pressure is 35 atm, which we convert to Pa (1 atm = 101325 Pa), giving P = 35 * 101325 Pa. The original wall thickness (t) is 1.6 mm or 0.0016 m. Solving for hoop stress gives σ = (35 * 101325 Pa * 0.14 m) / 0.0016 m = 386703125 Pa or 386.7 MPa, which exceeds σ_y / SF = 200 MPa / 5 = 40 MPa.

To reduce hoop stress to this level, we multiply thickness by σ / (σ_y / SF) giving a multiplication factor t_new = t * σ / (40 MPa) = 2.07. Thus, the wall thickness needs to be multiplied by approximately 2.07 to maintain the desired safety factor.