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4 m 3 m HA P EI (KL)2 Per = πRET Iy.y = bh³ B The steel bar AB of the frame is assumed to be pin connected at its ends for y-y axis buckling. If P=19kN, σ = 300 MPa = 200 GPa Determine: Esteel 50 mm x x 50 mm 50 mm- 6 m bh3 50 100 12 3 The factor of safety with respect to buckling about the y-y axis due to the applied loading. Pcr= SF = 50.100 = 1.04166 12 Per = π² (200) (1.04166) = 57.1154 110612 SF = 57.1154 = 3.006073 19 πT²EI SF=3.006073 im not sure (KL)2 Pcr P what im doing Wrong

4 m 3 m HA P EI (KL)2 Per = πRET Iy.y = bh³ B The steel bar AB of the frame is assumed to be pin connected at its ends for y-y axis buckling. If P=19kN, σ = 300 MPa = 200 GPa Determine: Esteel 50 mm x x 50 mm 50 mm- 6 m bh3 50 100 12 3 The factor of safety with respect to buckling about the y-y axis due to the applied loading. Pcr= SF = 50.100 = 1.04166 12 Per = π² (200) (1.04166) = 57.1154 110612 SF = 57.1154 = 3.006073 19 πT²EI SF=3.006073 im not sure (KL)2 Pcr P what im doing Wrong

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter2: Axially Loaded Members
Section: Chapter Questions
Problem 2.8.3P
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please help i am not sure what i am doing wrong

4 m 3 m HA P EI (KL)2 Per = πRET Iy.y = bh³ B The steel bar AB of the frame is assumed to be pin connected at its ends for y-y axis buckling. If P=19kN, σ = 300 MPa = 200 GPa Determine: Esteel 50 mm x x 50 mm 50 mm- 6 m bh3 50 100 12 3 The factor of safety with respect to buckling about the y-y axis due to the applied loading. Pcr= SF = 50.100 = 1.04166 12 Per = π² (200) (1.04166) = 57.1154 110612 SF = 57.1154 = 3.006073 19 πT²EI SF=3.006073 im not sure (KL)2 Pcr P what im doing Wrong
Transcribed Image Text:4 m 3 m HA P EI (KL)2 Per = πRET Iy.y = bh³ B The steel bar AB of the frame is assumed to be pin connected at its ends for y-y axis buckling. If P=19kN, σ = 300 MPa = 200 GPa Determine: Esteel 50 mm x x 50 mm 50 mm- 6 m bh3 50 100 12 3 The factor of safety with respect to buckling about the y-y axis due to the applied loading. Pcr= SF = 50.100 = 1.04166 12 Per = π² (200) (1.04166) = 57.1154 110612 SF = 57.1154 = 3.006073 19 πT²EI SF=3.006073 im not sure (KL)2 Pcr P what im doing Wrong
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