7. In class, we derived material indices for stiffness-limited designs. In stiffness-limited designs, it is elastic...

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7. In class, we derived material indices for stiffness-limited designs. In stiffness-limited designs, it is elastic deflection that is the active constraint. However, in strength-limited design, deflection is acceptable given that the component does not fail, so strength here is the active constraint. Derive the material index for selecting materials for a beam of length (L), specified strength, and minimum weight. Assume that the beam has a solid square cross section (t x t). Equations for the failure load of a beam and the second moment of the cross-section are given below. Here, ym is the distance between the neutral axis of the beam and its outer filament. 10000 1000 Strength, of (MPa) 100 10 (a) List the design requirements, e.g., function, constraints, objective, and free variables. (b) Derive the material index for this strength-limited design for minimum mass. 0.1 8. Suppose that the beam in Q7 must support a design load of 750 kN and it has dimensions of 1 m in length and 0.25 m in width and height. By looking at the strength vs. density chart given below, which material or materials would be your choice? 0.01- Strength Density Metals and polymers: yield strength Ceramics and glasses: MOR Elastomers: tensile tear strength Composites: tensile failure 0.01 Foams Rigid polymer foams Natural materials 0.1 Polymers and elastomers Free Composites. Cork Wood, PMMA Flexible polymer foams CFRP Mg alloys, GFRP PEEK Butyl rubber Ceramics SigN4 Ti alloys Steels SIC AlO3 Al alloys Silicone elastomers Ff 1 = Density, p (Mg/m) Zinc alloys Concrete = 12 Iof YmL 10 A 12 Ni alloys/ Metals Tungsten alloys Lead alloys Tungsten carbide Copper alloys Guidelines for minimum mass design 7. In class, we derived material indices for stiffness-limited designs. In stiffness-limited designs, it is elastic deflection that is the active constraint. However, in strength-limited design, deflection is acceptable given that the component does not fail, so strength here is the active constraint. Derive the material index for selecting materials for a beam of length (L), specified strength, and minimum weight. Assume that the beam has a solid square cross section (t x t). Equations for the failure load of a beam and the second moment of the cross-section are given below. Here, ym is the distance between the neutral axis of the beam and its outer filament. 10000 1000 Strength, of (MPa) 100 10 (a) List the design requirements, e.g., function, constraints, objective, and free variables. (b) Derive the material index for this strength-limited design for minimum mass. 0.1 8. Suppose that the beam in Q7 must support a design load of 750 kN and it has dimensions of 1 m in length and 0.25 m in width and height. By looking at the strength vs. density chart given below, which material or materials would be your choice? 0.01- Strength Density Metals and polymers: yield strength Ceramics and glasses: MOR Elastomers: tensile tear strength Composites: tensile failure 0.01 Foams Rigid polymer foams Natural materials 0.1 Polymers and elastomers Free Composites. Cork Wood, PMMA Flexible polymer foams CFRP Mg alloys, GFRP PEEK Butyl rubber Ceramics SigN4 Ti alloys Steels SIC AlO3 Al alloys Silicone elastomers Ff 1 = Density, p (Mg/m) Zinc alloys Concrete = 12 Iof YmL 10 A 12 Ni alloys/ Metals Tungsten alloys Lead alloys Tungsten carbide Copper alloys Guidelines for minimum mass design

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The question provided relates to material selection for a strengthlimited design of a beam with the goal of minimizing mass The image attached shows a StrengthDensity plot with different materials and ... View the full answer