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Exam II solutions W10

# Exam II solutions W10 - ME 382 Winter 2010 Mid-term 2...

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ME 382 Winter 2010 Mid-term 2 Monday, March 29, 2010 This is a closed book exam; students are allowed one unmarked copy of the ME382 Data Book. Answer all questions. Show all calculations for full credit. If you need more paper, staple it to the back of this exam. There are a total of 10 pages to this exam. Use a ruler / straight edge to read from graphs! Name (PRINTED): Circle Your Section: Prof. Garikipati Prof. Jones Honor Pledge:__________________________________________________________ _______________________________________________________________________ Signed: ____________________________ Problem Score 1 _______ 2 _______ 3 _______ 4 _______ 5 _______

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Name: 2 Problem 1 . A thin–walled steel cylinder ( R = 2.5 m, t = 10 mm) has a circumferential crack. The steel has a yield strength σ Y = 500 MPa and critical stress intensity factor K Ic = 40 MPa√m. The cylinder was initially used to store acetylene at 2 MPa. Was this safe? Why? Subsequently the same cylinder was used to store nitrogen. Up to what pressure can the nitrogen be stored safely? Stresses due to internal pressure: ° ±± = ²? ? = 500 ³²´ , ° ?? = ²? 2 ? = 250 ³²´ , ° µµ = 0 ³²´ Tresca effective stress = 500MPa. Since this is equal to σ o , this pressure vessel is not safe against yield while in service. σ zz is perpendicular to the circumferential crack and will therefore be the stress acting to cause fracture. To find the size of the crack at which fracture will occur (critical crack length): ·¸ = ° ?? ¹ c c (assuming F(α) =1) c c = 8.15mm < t Since the critical crack length is less than the thickness, the design does not satisfy the leak-before-break criterion and is therefore unsafe. In order to safely store nitrogen, the minimum critical crack length should be at least equal to the thickness (or better still, larger, with a factor of safety incorporated). For c c = t = 10mm, ° ?? = ·¸ º¹ t = 225.7 ³²´ With this, the max shear stress (Tresca) is 451 MPa < σ o and the maximum allowable pressure is ² = 2 ?? ? = 1.81 ³²´
Name: 3 Problem 2 . Atoms of element A are 30% larger than those of element B. A dissolves in B to form a dilute solid substitutional solution (an alloy). At concentration > 0.5% A (by mass), a precipitate is formed with formula A 2 B 3 . The alloy is thereafter a mixture of A 2 B 3 precipitates and pure B. Explain the strengthening mechanisms and the progression (any changes) in these mechanisms for this alloy if it is subjected to the following sequence of treatments: 1. The alloy is initially formed by cooling slowly from the molten state with 0.1% A by mass. Given that size(A) = 1.3 size(B)

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Exam II solutions W10 - ME 382 Winter 2010 Mid-term 2...

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