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The best choice of material for a light strong column depends on its aspect ratio: the ratio of its height H to its diameter D. This is because short, fat columns fail by crushing; tall slender columns buckle instead. Derive two performance equations for the material cost of a column of solid circular section and specified height H, designed to support a load F large compared to its selfload, the first using the constraints that the column must not crush, the second that it must not buckle. The table summarizes the needs. Function Constraints Objective Free variables Column Must not fail by compressive crushing Must not buckle Height H and compressive load F specified. Minimize material cost C Diameter D Choice of material Force F (a) Proceed as follows (1) Write an expression for the material cost of the column - its mass times its cost per unit mass, Cm. M₁ Cross-section area A DR4 (2) Express the two constraints as equations, and use them to substitute for the free variable, D, to find the cost of the column that will just support the load without failing by either mechanism. (3) Identify the material indices MI and M2 that enter the two equations for the mass, showing that they are: CmP LE¹/2 1-(mp) and M₂ where C is the material cost per kg. p the material density, & its crushing strength and E its modulus. Height H (b) Data for six possible candidates for the column are listed in the table below. Use these to identify candidate materials when F-10³ N and H=3m. Ceramics are admissible here, because they have high strength in compression.

The best choice of material for a light strong column depends on its aspect ratio: the ratio of its height H to its diameter D. This is because short, fat columns fail by crushing; tall slender columns buckle instead. Derive two performance equations for the material cost of a column of solid circular section and specified height H, designed to support a load F large compared to its selfload, the first using the constraints that the column must not crush, the second that it must not buckle. The table summarizes the needs. Function Constraints Objective Free variables Column Must not fail by compressive crushing Must not buckle Height H and compressive load F specified. Minimize material cost C Diameter D Choice of material Force F (a) Proceed as follows (1) Write an expression for the material cost of the column - its mass times its cost per unit mass, Cm. M₁ Cross-section area A DR4 (2) Express the two constraints as equations, and use them to substitute for the free variable, D, to find the cost of the column that will just support the load without failing by either mechanism. (3) Identify the material indices MI and M2 that enter the two equations for the mass, showing that they are: CmP LE¹/2 1-(mp) and M₂ where C is the material cost per kg. p the material density, & its crushing strength and E its modulus. Height H (b) Data for six possible candidates for the column are listed in the table below. Use these to identify candidate materials when F-10³ N and H=3m. Ceramics are admissible here, because they have high strength in compression.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Data for candidate materials for the column Material Density p Cost/kg (kg/m³) Cm ($/kg) Wood (spruce) Brick Granite Concrete Cast iron Structural steel Al-alloy 6061 700 2100 2600 2300 7150 7850 2700 0.5 0.35 0.6 0.08 0.25 0.4 1.2 Modulus E (MPa) 10,000 22,000 20,000 20,000 130,000 210,000 69,000 Compression strength σc (MPa) 25 95 150 13 200 300 150
Transcribed Image Text:Data for candidate materials for the column Material Density p Cost/kg (kg/m³) Cm ($/kg) Wood (spruce) Brick Granite Concrete Cast iron Structural steel Al-alloy 6061 700 2100 2600 2300 7150 7850 2700 0.5 0.35 0.6 0.08 0.25 0.4 1.2 Modulus E (MPa) 10,000 22,000 20,000 20,000 130,000 210,000 69,000 Compression strength σc (MPa) 25 95 150 13 200 300 150
Problem E8.2. Multiple constraints: a cheap column that must not buckle or crush (pg. 314). The best choice of material for a light strong column depends on its aspect ratio: the ratio of its height H to its diameter D. This is because short, fat columns fail by crushing; tall slender columns buckle instead. Derive two performance equations for the material cost of a column of solid circular section and specified height H, designed to support a load F large compared to its selfload, the first using the constraints that the column must not crush, the second that it must not buckle. The table summarizes the needs. Function Constraints Objective Free variables Column Must not fail by compressive crushing Must not buckle Height H and compressive load F specified. . Minimize material cost C Diameter D Choice of material (a) Proceed as follows (1) Write an expression for the material cost of the column - its mass times its cost per unit mass, Cm. Force F Cmp Cross-section area AD14 (2) Express the two constraints as equations, and use them to substitute for the free variable, D, to find the cost of the column that will just support the load without failing by either mechanism. and M₂ (3) Identify the material indices MI and M2 that enter the two equations for the mass, showing that they are: CmP LE ¹/2 Height H M₁ where C is the material cost per kg. p the material density, o its crushing strength and E its modulus. (b) Data for six possible candidates for the column are listed in the table below. Use these to identify candidate materials when F= 10³ N and H=3m. Ceramics are admissible here, because they have high strength in compression.
Transcribed Image Text:Problem E8.2. Multiple constraints: a cheap column that must not buckle or crush (pg. 314). The best choice of material for a light strong column depends on its aspect ratio: the ratio of its height H to its diameter D. This is because short, fat columns fail by crushing; tall slender columns buckle instead. Derive two performance equations for the material cost of a column of solid circular section and specified height H, designed to support a load F large compared to its selfload, the first using the constraints that the column must not crush, the second that it must not buckle. The table summarizes the needs. Function Constraints Objective Free variables Column Must not fail by compressive crushing Must not buckle Height H and compressive load F specified. . Minimize material cost C Diameter D Choice of material (a) Proceed as follows (1) Write an expression for the material cost of the column - its mass times its cost per unit mass, Cm. Force F Cmp Cross-section area AD14 (2) Express the two constraints as equations, and use them to substitute for the free variable, D, to find the cost of the column that will just support the load without failing by either mechanism. and M₂ (3) Identify the material indices MI and M2 that enter the two equations for the mass, showing that they are: CmP LE ¹/2 Height H M₁ where C is the material cost per kg. p the material density, o its crushing strength and E its modulus. (b) Data for six possible candidates for the column are listed in the table below. Use these to identify candidate materials when F= 10³ N and H=3m. Ceramics are admissible here, because they have high strength in compression.
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