# hw5_sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 02139 2.002 MECHANICS AND MATERIALS II HOMEWORK SOLUTION NO. 5 Distributed : Wednesday, April 10, 2003 Due : Wednesday, April 16 , 2003 Problem 1 ( 30 points ) The critical stress intensity factor, K IC , is given by the inequality K IC πa (1) Solving for the critical crack length, a crit we get 2 1 ± K IC a a crit (2) π For an edge crack (whose length is less than or equal to 0.65 times the plate’s width, w ), the conﬁguration correction factor, Q , is 1 . 12 Q = ² 1 0 . 7 ³ a ´ 1 . 5 µ 3 . 25 (3) w a We are told that the the plate is large , which implies that w 1. It is also explicitly stated that Q = 1 . 12 (4) We are also told to use a design stress equal to 0 . 6 times the minimum yield strength. σ = 0 . 6 σ y min (5) Substituting equations 4 and 5 into equation 2 we get 1 ± K IC 2 a crit = (6) π 0 . 672 σ y min 1

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We are asked to compare the critical crack sizes for four diﬀerent grades of steel. The values of σ y min and K IC for these steels are given in the second and third columns of table 1. The fourth column of the table contains the critical crack length as computed by equation 6. Steel (ASTM Grade) σ y min [ ksi ] K IC at 60 o F, [ ksi in ] a crit [ in ] A36 36 60 1.958 A441 50 53 0.792 A572 50 100 2.82 A514 100 60 0.254 Table 1: Material properties and Critical crack length for available steels Based on table 1, the steel that would most likely be selected is A572. A572
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## This note was uploaded on 02/23/2012 for the course MECHANICAL 2.002 taught by Professor Davidparks during the Spring '04 term at MIT.

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hw5_sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT...

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