A cylindrical tank is assembled by welding steel sections in a helical pattern with angle $\alpha = 50 ^ { \circ }$ . Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus $2.75 \mathrm { MPa }$ $E = 210 \mathrm { GPa }$ and Poisson's Ratio ν = 0.28. The normal stress acting perpendicular to the weld is approximately:

Question

A cylindrical tank is assembled by welding steel sections in a helical pattern with angle $\alpha = 50 ^ { \circ }$ . Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus $2.75 \mathrm { MPa }$ $E = 210 \mathrm { GPa }$  and Poisson's Ratio ν = 0.28. The normal stress acting perpendicular to the weld is approximately:

Quizplus · Accepted Answer

First, we need to find the hoop stress which can be calculated using the formula below:

$$\sigma_{hoop}=\frac{pD}{2t}$$

where:

p = internal pressure = 2.75 MPa
D = diameter = 1.6 m
t = thickness = 20 mm = 0.02m

$$\sigma_{hoop}=\frac{2.75 \times 1.6}{2 \times 0.02}=220 \mathrm{MPa}$$
Next, we need to find the axial stress which can be calculated using the formula below:

$$\sigma_{axial}=\frac{\sigma_{hoop}}{2}\left(1+\frac{\cos{\alpha}}{1-\nu}\right)$$

where:

ν = Poisson's Ratio = 0.28
α = angle between welds = 50°
$$\cos{\alpha}=\cos{50^{\circ}}=0.643$$
$$\sigma_{hoop}=220 \mathrm{MPa}$$

$$\sigma_{axial}=\frac{220}{2}\left(1+\frac{0.643}{1-0.28}\right)=78 \mathrm{MPa}$$

Therefore, the normal stress acting perpendicular to the weld is approximately 78 MPa, which corresponds to choice C.

A Cylindrical Tank Is Assembled by Welding Steel Sections in a Helical