Missing Diagram is attached.
Answer:
a) 21.43 MPa
b) -14.2 MPa
Explanation:
Given:
[tex]di = 250 mm = 0.25 m[/tex]
[tex]L = 1.2 m[/tex]
[tex]t = 6 mm = 0.006m[/tex]
[tex] d_o = d_i + 2t = 0.25+(2*0.006) => 0.262 m [/tex]
[tex]P_g = 2 mpa[/tex]
[tex]P = 45 KN [/tex]
[tex]Area, A = \frac{pi*(250+2(6)^2-(250)^2)}{4} = 4825.47 mm^2 [/tex]
Calculating the stresses:
[tex] Hoop stress = \frac{Pd}{2t} = \frac{2*250}{2(6)} [/tex]
= 41.667 MPa
[tex] Longitudinal stress, T_L= \frac{Pd}{4t} = \frca{2*250}{4*6}[/tex]
= 20.833 MPa
[tex] Axial stress, T_a= \frac{P}{A} = \frac{-45*10^3}{4825.47} [/tex]
-9.325 MPa
For stress along x axis:
Tx = Tl - Ta
= 20.83 - 9.325
= 11.5 MPa
[tex] Tavg = \frac{Tx+Ty}{2} => \frac{11.5+41.66}[/tex]
= 26.58 MPa
[tex]Tmax = \frac{Tx - Ty}{2} => \frac{11.5 - 41.66}{2}[/tex]
= -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
