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View Full DocumentExcerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.1 A stainless steel alloy bar 25 mm wide by 16 mm thick is subjected to an axial load of P = 145 kN. Using the stress-strain diagram given in Fig. P4.1, determine: (a) the factor of safety with respect to the yield strength defined by the 0.20% offset method. (b) the factor of safety with respect to the ultimate strength. Fig. P4.1 Solution (a) From the stress-strain curve, the yield strength defined by the 0.20% offset method is 550 MPa Y = The normal stress in the bar is (145 kN)(1,000 N/kN) 362.5 MPa (25 mm)(16 mm) F A = = = Therefore, the factor of safety with respect to yield is actual 550 MPa FS 1.517 362.5 MPa Y = = = Ans. (b) From the stress-strain curve, the ultimate strength is ult 1,100 MPa = Therefore, the factor of safety with respect to the ultimate strength is ult actual 1,100 MPa FS 3.03 362.5 MPa = = = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.2 Six bolts are used in the connection between the axial member and the support, as shown in Fig. P4.2. The ultimate shear strength of the bolts is 300 MPa, and a factor of safety of 4.0 is required with respect to fracture. Determine the minimum allowable bolt diameter D required to support an applied load of P = 350 kN. Fig. P4.2 Solution The allowable shear stress for the bolts is ult allow 300 MPa 75 MPa FS 4 = = = To support a load of P = 350 kN, the total area subjected to shear stress must equal or exceed: 2 2 allow 350,000 N 4,666.67 mm 75 N/mm V P A = = There are six bolts, and each bolt acts in double shear; therefore, the total area subjected to shear stress is 2 bolt (6 bolts)(2 surfaces per bolt) 4 V A D = The minimum bolt diameter can be found be equating the two expressions for A V : 2 2 bolt bolt (12 surfaces) 4,666.67 mm 4 22.252 mm 22.3 mm D D = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.... View Full Document
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