Question

Find the maximum pressure P in the cylindrical steel vessel, using Tresca to yield a 50% safety margin.

Then find the axial stress σx and hoop stress σθ that results from P.

Calculate the shear stress on the inclined plane.

Calculate the force F needed for the axial stress to equal zero and the change in vessel length due to F alone.

diameter = 20 cm, thickness =1 cm, σf = 220 MPa, E = 200 GPa, ν = 0.3, and α = 12x10^-6

1 m ㅔ 0.2 m F P F 1300

Answer 1

Р F 130 20cm 200 mm Diameter Da thickness t= 10m 1 cm. lomm. failure 220 Mpa. E 200 Gpa. 8=0.3. 2-12 x 10-6 hoop stres to at longitudinal stress = PD ut OF-PD at PP PO TE ut ut T = PP 26. Since there is no shear stress due to loading, there are Principal streues. max in plane shear strey de-a Tmax= 2 OCO X2

Iman T - T 2 (Tresca theory PD PD ut Tman PP st Tmax is equal to equal to 50% failure stress. Tmax = 0.5 x 220 110 Mpa. 10 = Px 200 8x10 Internal Pressure IP 44 mpail axial (on longitudinal stress Jo ut 44x200 4x10 언 220 MPa Circumferential en hoop stress 05- PD at 440 MPa. 44 x 200 2 x 10 V POCO X2 = 440 MPa

440 MPa 60 + 220 MPa O 200 mpai 30 A 440 MPa. Since now, Draus the mohr's Circle. there are Principle stresser, Coordinates (o,,o) & (02, 0) (220,0) (440,0) are 110 て Tmax = Mpa. ABE て AB, CD 120 AD, BC (220,0) (440,0) 0 = 60° (from plane AB), E 20= 120° (from plane AB). E. (in mohr's Circle. from diadigram, To = 110x sin (60) = 95.26 MPa. shear stress on inclined plane POCO X2 95.26 MPa.

* force f needed for axial stress to be zero. F 11 A न 0 T-E A A = (tot). E T = F 220 = Tot 220 = TTX 200 xlo Z = 1382300 1382.3 KN * change in versel length due to f alone. FL Sl= АС 1382.3 xlo 103 x 1000 75X200 x10 x 200 x103 S1 = 1.1 mm change in verset length due to F alone is . 1.1 mm co
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