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Please solve part b only within 30 minutes please help me

Please solve part b only within 30 minutes please help me

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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Please solve part b only within 30 minutes please help me
(a) Determine the load carrying capacity of a 6-m tall (L = 6 m) steel column which has a Pin- Roller boundary condition as shown in the following figure. The column is made of steel (E = Modulus of elasticity = 200 GPa and oy = Yield stress= 300 MPa). The cross section of the column is 200 mm x 200 mm that is b = 200 mm and h = 200 mm. (5) X Elevation b Cross-section x-x Column with Pin-Roller boundary condition (b) Is this column safe if the boundary condition is change to Pin-Fixed (not shown in the above figure) and the column is subjected to 15000 kN? (5)
Transcribed Image Text:(a) Determine the load carrying capacity of a 6-m tall (L = 6 m) steel column which has a Pin- Roller boundary condition as shown in the following figure. The column is made of steel (E = Modulus of elasticity = 200 GPa and oy = Yield stress= 300 MPa). The cross section of the column is 200 mm x 200 mm that is b = 200 mm and h = 200 mm. (5) X Elevation b Cross-section x-x Column with Pin-Roller boundary condition (b) Is this column safe if the boundary condition is change to Pin-Fixed (not shown in the above figure) and the column is subjected to 15000 kN? (5)
Formulas/ Help: You may not need to use all of them Oy - 0x sin 20 + Try cos 20 Ox + 0y Oav = Per (Lefs)? Ox + 0y Ox Or = + cos 20 + Try sin 20 2 Ox Ox + Oy cos 20 – Txy sin 20 01,2 = Oav ± + Tảy of - 0,02 + o = 03 |(0, – 02): = Oy = 2ty 2Txy tan 20p Ox - Oy Oy - ơx tan 20, %3D 2Txy bh3 I of a rectangular section = 12 Py = A. Oy
Transcribed Image Text:Formulas/ Help: You may not need to use all of them Oy - 0x sin 20 + Try cos 20 Ox + 0y Oav = Per (Lefs)? Ox + 0y Ox Or = + cos 20 + Try sin 20 2 Ox Ox + Oy cos 20 – Txy sin 20 01,2 = Oav ± + Tảy of - 0,02 + o = 03 |(0, – 02): = Oy = 2ty 2Txy tan 20p Ox - Oy Oy - ơx tan 20, %3D 2Txy bh3 I of a rectangular section = 12 Py = A. Oy
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