SolutionsHW6

SolutionsHW6 - HW #6 Problem 8.2 a. To Find: Theoretical...

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HW #6 Problem 8.2 a. To Find: Theoretical fracture strength of the given material b. Given: o = 1200 MPa a = 0.25 mm t = 1.2 X 10 -3 mm c. Assumptions: (i) Surface crack; semi-elliptical crack (ii) Applied stress is perpendicular to the crack/ crack length (see Fig. 8.8 a) (iii) No other stress-raisers present in the sample (iv) Given values of o, a and t are correct d. Solution: From equation 8.1: m = 2 0 a t 1/ 2 = (2)(1200 MPa) 0.25 mm 1.2 10 3 mm 1/2 = 3.5 10 4 MPa = 35 GPa (5.1 10 6 psi ) Fracture strength = m = 35 GPa
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Problem 8.4 a. To Find: Maximum permissible surface crack length in the given polystyrene component b. Given: Applied stress, = 1.25 MPa S = 0.5 J/m 2 E = 3GPa c. Assumptions: (i) Given data is accurate (ii) There is no other flaw in the component (iii) The experiment is carried out at low temperature – other mechanisms of failure may set in at higher temperatures (iv) Applied stress is perpendicular to the crack/ crack length (see Fig. 8.8 a) d. Solution:    c Rearranging equation 8.3: a = 2 E s  c 2 = (2) ( 3 10 9 N/m 2 ) (0.50 N/m) ( ) ( 1.25 10 6 N/m 2 ) 2 = 6.1 10 -4 m = 0.61 mm (0.024 in.) 0.61 mm
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Problem 8.22
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This note was uploaded on 12/08/2011 for the course MSE 2090 taught by Professor Leoindzhiglei during the Fall '10 term at UVA.

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SolutionsHW6 - HW #6 Problem 8.2 a. To Find: Theoretical...

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