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A 200-Kip 1oad P is applied to the pin-ended rectongular lumber column Shown below. The column is braced in the weak direction Shown below. The lumber used has an allowable stress in compraSSIon Gau=1500 psi and E= 450x 10? psi find the smablest cross secton that cam be uscd ; d-ab factor of safety of 2.5 P. 7 ft 8 ft B.

A 200-Kip 1oad P is applied to the pin-ended rectongular lumber column Shown below. The column is braced in the weak direction Shown below. The lumber used has an allowable stress in compraSSIon Gau=1500 psi and E= 450x 10? psi find the smablest cross secton that cam be uscd ; d-ab factor of safety of 2.5 P. 7 ft 8 ft B.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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A 200-Kip l0ad P is applied to the Pin-ended rectongular lumber colomn Shown below. The column is braced in the weak direction Shown below. The lumber used has an allowable Stress in compresston Gal=150u psi and Ez 450 x 10° psi find the smablest cross section that can. be used ; factor of safety of 2.5 d-ab 7 ft 8 ft
Transcribed Image Text:A 200-Kip l0ad P is applied to the Pin-ended rectongular lumber colomn Shown below. The column is braced in the weak direction Shown below. The lumber used has an allowable Stress in compresston Gal=150u psi and Ez 450 x 10° psi find the smablest cross section that can. be used ; factor of safety of 2.5 d-ab 7 ft 8 ft
Мy = I Mc 17. 18. Omax I M 1 M 19. Отах 20. %3D S EI 1= Σ( + Ad?) E(yA) y 21. 22. ΣΑ 23. bh3 I : 24. bh2 S = 12 2EI 25. Per 26. Ocr = L2 (L/r)² VQ T = It VQ 27. 28. I 3V V 29. Ттах 30. Ттах 2A dtw kN 1 GPa = 106 m2 103 тт kN 1 MPа — 103 m2 1 т %3 1° -rad 180 lb k ksi = in? psi in? lb.ft = 550 S 1 hр :
Transcribed Image Text:Мy = I Mc 17. 18. Omax I M 1 M 19. Отах 20. %3D S EI 1= Σ( + Ad?) E(yA) y 21. 22. ΣΑ 23. bh3 I : 24. bh2 S = 12 2EI 25. Per 26. Ocr = L2 (L/r)² VQ T = It VQ 27. 28. I 3V V 29. Ттах 30. Ттах 2A dtw kN 1 GPa = 106 m2 103 тт kN 1 MPа — 103 m2 1 т %3 1° -rad 180 lb k ksi = in? psi in? lb.ft = 550 S 1 hр :
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