PROBLEM 1.29 P' 150 mm P Two wooden members of uniform rectangular cross section are joined 11 kN, by the simple glued scarf splice shown. Knowing that P determine the normal and shearing stresses in the glued splice. 45!! 45 75 mm SOLUTION 90 P A0 45 11 kN 45 11 103 N (150)(75) 11.25 103 mm 2 11.25 10 3 m 2 P cos 2 A0 (11 103 ) cos 2 45 11.25 10 3 489 103 Pa 489 kPa P sin 2 2 A0 (11 103 )(sin 90 ) (2)(11.25 10 3 ) 489 103 Pa 489 kPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 33 A PROBLEM 1.42 600 lb/ft Link AB is to be made of a steel for which the ultimate normal stress is 65 ksi. Determine the cross-sectional area of AB for which the factor of safety will be 3.20. Assume that the link will be adequately reinforced around the pins at A and B. 35! B C D E 5 kips 1.4 ft 1.4 ft 1.4 ft SOLUTION P MD (4.2)(0.6) 0: 2.52 kips (2.8)( FAB sin 35 ) (0.7)(2.52) FAB AB AAB (1.4)(5) 0 5.4570 kips FAB AAB ult F. S . ( F. S .) FAB ult (3.20)(5.4570 kips) 65 ksi 0.26854 in 2 AAB 0.268 in 2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 48 PROBLEM 1.69 2.4 kips The two portions of member AB are glued together along a plane forming an angle with the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine (a) the value of for which the factor of safety of the member is maximum, (b) the corresponding value of the factor of safety. (Hint: Equate the expressions obtained for the factors of safety with respect to the normal and shearing stresses.) A ! B 1.25 in. 2.0 in. SOLUTION A0 At the optimum angle, ( F. S.) ( F. S.) PU , U A0 2 P cos 2 A0 Normal stress: cos P sin A0 cos Equating, Solving, (b) PU U A0 2 P cos sin cos P P cos U A0 PU , sin cos U A0 PU , ( F. S.) U A0 2 PU , ( F. S .) Shearing stress: 2.50 in 2 (2.0)(1.25) P sin P cos U A0 P sin tan U A0 2 cos cos U U (12.5)(2.50) cos 2 27.5 F. S. 1.3 2.5 0.520 (a) opt 27.5 7.94 kips PU P 7.94 2.4 F. S. 3.31 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 76
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