Question

a pressure vessel of 250-mm inner diameter and 6-mm wall thickness is fabricated from a 1.2-m section of spirally welded pipe AB and is equipped with two rigid end plates. the gage pressure inside the vessel is 2 MPa, and 45-kN centric axial forces P and P' are applied to the end plates. Determine a) the normal stress perpendicular to the weld b) the shearing stress parallel to the weld.

Answer 1

Missing Diagram is attached.

Answer:

a) 21.43 MPa

b) -14.2 MPa

Step-by-step explanation:

Given:

`di = 250 mm = 0.25 m`

`L = 1.2 m`

`t = 6 mm = 0.006m`

`d_o = d_i + 2t = 0.25+(2*0.006) => 0.262 m`

`P_g = 2 mpa`

`P = 45 KN`

`Area, A = (pi*(250+2(6)^2-(250)^2))/(4) = 4825.47 mm^2`

Calculating the stresses:

`Hoop stress = (Pd)/(2t) = (2*250)/(2(6))`

= 41.667 MPa

`Longitudinal stress, T_L= (Pd)/(4t) = \frca{2*250}{4*6}`

= 20.833 MPa

`Axial stress, T_a= (P)/(A) = (-45*10^3)/(4825.47)`

-9.325 MPa

For stress along x axis:

Tx = Tl - Ta

= 20.83 - 9.325

= 11.5 MPa

`Tavg = (Tx+Ty)/(2) => \frac{11.5+41.66}`

= 26.58 MPa

`Tmax = (Tx - Ty)/(2) => (11.5 - 41.66)/(2)`

= -15.08 MPa

Mohr's circle angle = 2∅

= 2 * 35°

= 70°

a) for normal stress perpendicular to weld, we have:

Tx' = Tavg + Tmax (cos70°}

= 26.590 - 15.08(cos70°)

= 21.43 MPa

b) shearing stress parallel to weld:

= Tmax (sin70°}

= -15.08 sin70°

= -14.2 MPa