8_17 - x = My (270 lb-ft)(1.4747 in.)(12 in./ft) = =...

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4 (270 lb-ft)(1.4747 in.)(12 in./ft) 4,69 4,700 psi 8.164 psi 1.016999 in. (C) x z My I σ =− = Ans.
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8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17 a ) having the cross section shown in Fig. P8.17 b . Determine the maximum tension and compression bending stresses produced in segment BC of the beam. Fig. P8.17 a Fig. P8.17 b Solution Centroid location in y direction: Shape Area A i y i (from bottom) y i A i (mm 2 ) (mm) (mm 3 ) top flange 3,000.0 167.5 502,500.0 stem 1,440.0 80.0 115,200.0 4,440 mm 2 617,700 mm 3 3 2 617,700 mm 139.1216 mm 4,440 mm ii i yA y A Σ == = Σ (measured upward from bottom edge of stem) Moment of inertia about the z axis: Shape I C d = y i y d²A I C + d²A (mm 4 ) (mm) (mm 4 ) (mm 4 ) top flange 56,250.00 28.38 2,415,997.08 2,472,247.08 stem 3,072,000.00 59.12 5,033,327.25 8,105,327.25 Moment of inertia about the z axis (mm 4 ) = 10,577,574.32 Comment [TAP1]: max tension:
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This note was uploaded on 02/09/2012 for the course MECHANICAL EM217 taught by Professor Joyce during the Spring '11 term at Naval Academy.

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8_17 - x = My (270 lb-ft)(1.4747 in.)(12 in./ft) = =...

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