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Determine the maximum safe capacity of an axially loaded hinged ends column having an unsupported length of 4m. Fy = 248 MPa and E = 200 GPa. Use AISC Specifications. 25 CIVIL ENGINEERING: STEEL DESIGN 100 25 175

Determine the maximum safe capacity of an axially loaded hinged ends column having an unsupported length of 4m. Fy = 248 MPa and E = 200 GPa. Use AISC Specifications. 25 CIVIL ENGINEERING: STEEL DESIGN 100 25 175

Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter5: Beams
Section: Chapter Questions
Problem 5.11.5P
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Please show solutions on how to get the 1500 value. The circle green in the figure.

 

Civil engineering- steel design

 

Note: I thought the value of area 2 is 3750.

 

Area 2: 25x150= 3750 mm2. Feel free to correct me . What should put in circle in the figure.

 

 

 

Determine the maximum safe capacity of an axially loaded hinged ends column having an unsupported length of 4m. Fy = 248 MPa and E = 200 GPa. Use AISC Specifications. 25 CIVIL ENGINEERING: STEEL DESIGN 100 25 175
Transcribed Image Text:Determine the maximum safe capacity of an axially loaded hinged ends column having an unsupported length of 4m. Fy = 248 MPa and E = 200 GPa. Use AISC Specifications. 25 CIVIL ENGINEERING: STEEL DESIGN 100 25 175
A₂ = = area of section 2 = 150mm x 25mm = 3750 mm² X₁ = centroid of section 1 = 50 mm X₂ = centroid of section 2 = 100 - 12.5 = 87.5 mm (2500 x 50) + (3750 x 87.5) 2500+ 3750 => X = => X = 72.5 mm Now, A₁Y₁ + A₂Y2 A₁ + A₂ where, y₁ = centroid of section 1 = 175 - 12.5 = 162.5 mm y = => y => y = 110 mm Y₂ = centroid of section 2 = 75 mm (2500 x 162.5) + (3750 x 75) 2500+ 3750 = According to figure above, 3 3 1 1 2 - ₁. - ( D ₁0₂²³ + A₁ X ²) + (0₂0/₂²0 + A₂ V ² ) => x = Y2² 12 12 where, y₁= distance between centroid = 175 - 110 - 12.5 = 52.5 mm Y₂ = distance between centroid = 110 - 75 = 35 mm Putting values, => x= Now, 100 x 25³ 12 => x = 14.05 x 106 mm² = (0₁/ + 2500 (52.5)² + 52²) + 3 d₁ b₁³ 3 +A₁X²³) + (0²₂D₂2² + A₂X²³) 2 2 1 12 12 where, x₁ = distance between centroid = 72.5 - 50 = 22.5 mm X₂ = distance between centroid = 100 - 72.5 - 12.5 = 15 mm Putting values, 25 x 100³ => ly = 12 => y = 3.88 x 106 mm4 25 x 150³ 12 53²) + ( + 2500 (22.5)² + + 1500 (35)² 150 x 25³ 12 + 1500 (15)²
Transcribed Image Text:A₂ = = area of section 2 = 150mm x 25mm = 3750 mm² X₁ = centroid of section 1 = 50 mm X₂ = centroid of section 2 = 100 - 12.5 = 87.5 mm (2500 x 50) + (3750 x 87.5) 2500+ 3750 => X = => X = 72.5 mm Now, A₁Y₁ + A₂Y2 A₁ + A₂ where, y₁ = centroid of section 1 = 175 - 12.5 = 162.5 mm y = => y => y = 110 mm Y₂ = centroid of section 2 = 75 mm (2500 x 162.5) + (3750 x 75) 2500+ 3750 = According to figure above, 3 3 1 1 2 - ₁. - ( D ₁0₂²³ + A₁ X ²) + (0₂0/₂²0 + A₂ V ² ) => x = Y2² 12 12 where, y₁= distance between centroid = 175 - 110 - 12.5 = 52.5 mm Y₂ = distance between centroid = 110 - 75 = 35 mm Putting values, => x= Now, 100 x 25³ 12 => x = 14.05 x 106 mm² = (0₁/ + 2500 (52.5)² + 52²) + 3 d₁ b₁³ 3 +A₁X²³) + (0²₂D₂2² + A₂X²³) 2 2 1 12 12 where, x₁ = distance between centroid = 72.5 - 50 = 22.5 mm X₂ = distance between centroid = 100 - 72.5 - 12.5 = 15 mm Putting values, 25 x 100³ => ly = 12 => y = 3.88 x 106 mm4 25 x 150³ 12 53²) + ( + 2500 (22.5)² + + 1500 (35)² 150 x 25³ 12 + 1500 (15)²
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