888_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 888_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 11Ch11.qxd 9/27/08 882 3:39 PM Page 882 CHAPTER 11 Columns Solution 11.6-4 Steel pipe column Pinned ends. DATA 2.1 m E d1 60 mm 0.06 m P 10 kN 10,000 N r 210 * 109 N/m2 210 GPa d2 68 mm e 30 mm 0.068 m p2 (d 2 4 p4 (d 64 2 22.671 * 10 ec 0.03 m 1.9845 r2 6 L P 2r A EA 3 m m2 d2 2 c 0.034 m 0.35638 Substitute into Eq. (1): TUBULAR CROSS SECTION I 513.99 * 10 Units: Newtons and meters L A I A r2 d 2) 1 804.25 * 10 6 4 d1 ) 413.38 * 10 9 38.8 * 106 N/m2 smax m2 m4 38.8 MPa ; (b) MAXIMUM PERMISSIBLE LENGTH sallow 50 MPa P(ec/r 2) EI arccos c d AP smaxA P Solve Eq. (1) for the length L: (a) MAXIMUM COMPRESSIVE STRESS ec P P L c 1 + 2 sec a bd A 2r A EA r L Secant formula (Eq. 11-59): smax P A 6 (1) (2) 2 Substitute numerical values: L max 5.03 m ; 2 12.434 * 10 N/m Problem 11.6-5 A pinned-end strut of length L 5.2 ft is constructed of steel pipe (E 30 * 103 ksi) having inside diameter d1 2.0 in. and outside diameter d2 2.2 in. (see figure). A compressive load P 2.0 k is applied with eccentricity e 1.0 in. (a) What is the maximum compressive stress smax in the strut? (b) What is the allowable load Pallow if a factor of safety n 2 with respect to yielding is required? (Assume that the yield stress sY of the steel is 42 ksi.) Solution 11.6-5 Pinned-end strut Steel pipe. DATA L d1 P (a) MAXIMUM COMPRESSIVE STRESS 5.2 ft 62.4 in. E 2.0 in. d2 e 30 * 103 ksi 2.2 in. 1.0 in. 2.0 k I p2 (d 42 p4 (d 64 2 smax P A r2 TUBULAR CROSS SECTION A P ec L P c 1 + 2 sec a bd A 2r A EA r d2 3.0315 ksi c 1.1 in. 2 Secant formula (Eq. 11-59): Units: kips and inches d 2) 1 0.65973 in.2 4 d 1) 4 0.36450 in. r I A L P 2r A EA 0.55250 in.2 0.74330 in. ec r2 1.9910 0.42195 (1) ...
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