Suppose we need to design a light weight and strong tubular beam as an intramedullary bone...

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Suppose we need to design a light weight and strong tubular beam as an intramedullary bone fixation nail, shown in the upper right figure, which is mounted by screws on its ends. The beam has a fixed length L, and a fixed thickness t, while the outer radius r is adjustable to fit the design requirement, and t<<r. The function of the beam is to resist bending moments. We know the moment of inertia of a tubular (or ring) cross section is Ic = Tr³t, and the bending moment Mb where E is elasticity modulus of the material. Since the thickness t is much smaller than beam outer radius r, we can assume the tubular (or ring) cross section area s = 2πrt. ICE = r 2 1000 Modulus of Elasticity, E (GPa) 100 10 0.1 0.1 Elp=C (a) If our objective is to maximize the bending moment Mb that the beam can afford and minimize the mass m of the beam, please derive the material performance index in terms of elasticity modulus E and density p of the beam material. (b) Based on the derived material performance index, if the titanium alloy and stainless steel are the two candidate materials for the tubular beam, draw the lines with the corresponding slopes of material performance index across those two materials points in the lower right Ashby charts (if you draw the schematic Ashby charts manually on a paper, you can simply just include the points of titanium alloy and stainless steel, the two axis labels, and denote the correct slope in terms of the material performance index), and determine which one is the better choice for the tubular beam. EV/2/p=C Hydroxyapatite E1/3/p=C Graphite/Carbon Poly propylene Fiber UHMW-PE Titanium- Alloys Glass PMMA Alumina Polyethylene 1 Density, p (g/cm³) Nylon (-0.002) Silicone (<0.01) Pr Titanium Cortical Bone Stainless Steel Co-Cr PTFE (Teflon) 10 Suppose we need to design a light weight and strong tubular beam as an intramedullary bone fixation nail, shown in the upper right figure, which is mounted by screws on its ends. The beam has a fixed length L, and a fixed thickness t, while the outer radius r is adjustable to fit the design requirement, and t<<r. The function of the beam is to resist bending moments. We know the moment of inertia of a tubular (or ring) cross section is Ic = Tr³t, and the bending moment Mb where E is elasticity modulus of the material. Since the thickness t is much smaller than beam outer radius r, we can assume the tubular (or ring) cross section area s = 2πrt. ICE = r 2 1000 Modulus of Elasticity, E (GPa) 100 10 0.1 0.1 Elp=C (a) If our objective is to maximize the bending moment Mb that the beam can afford and minimize the mass m of the beam, please derive the material performance index in terms of elasticity modulus E and density p of the beam material. (b) Based on the derived material performance index, if the titanium alloy and stainless steel are the two candidate materials for the tubular beam, draw the lines with the corresponding slopes of material performance index across those two materials points in the lower right Ashby charts (if you draw the schematic Ashby charts manually on a paper, you can simply just include the points of titanium alloy and stainless steel, the two axis labels, and denote the correct slope in terms of the material performance index), and determine which one is the better choice for the tubular beam. EV/2/p=C Hydroxyapatite E1/3/p=C Graphite/Carbon Poly propylene Fiber UHMW-PE Titanium- Alloys Glass PMMA Alumina Polyethylene 1 Density, p (g/cm³) Nylon (-0.002) Silicone (<0.01) Pr Titanium Cortical Bone Stainless Steel Co-Cr PTFE (Teflon) 10

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The image contains text describing a design scenario for creating a lightweight and strong tubular beam to be used as an intramedullary bone fixation nail The beam must have a fixed length and thickne... View the full answer