MSQ1 - 1 Fiber? Check. My Shapes Matrix? Check. My...

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1 Fiber? Check. Matrix? Check. My Shapes My Cross-sectional Areas (For Normal) , = , ; , = AreaCross Circle πdf circle24 df circle diameter , = , * , = , ; , = AreaCross Square df square df square df square2 df square length of side , = , ; = AreaCross KeyedCircle 56πdf kcircle24 56 because 60o 16 of 360o the , , = circle df kcircle diameter , = * + ; = , = AreaCross Plus 4t l t2 t thickness l length of one side of plus My Surface Areas(For Shear) , = * = ; = , = AreaSurface Circle 2πr dx πdfdx df diameter dx length along longitudinal axis circumference * Length , = ; = , = AreaSurface Square 4dfdx df length of side dx length along longitudian axis * Sides Length , = - + ; = AreaSurface KeyedCircle πdf πdf60360 dfdx dx length along longitudian axis [Circumference – (Arc Length) + (Radius + Radius)]* Length , =( + ) AreaSurface Plus 4t 8l dx + * Thicknesses Lengths Length Determining Dimensions In order to determine the values for the diameters, side lengths, thicknesses, etc (it is all arbitrary anyway), I must set all cross sectional areas equal to each other (to have fair comparisons). I will choose an easy number as the cross sectional area I want to have (10 units2 ). Also because I have more unknowns than equations, I will make = t l4 (for plus shape). With this data made up, I can now solve for the dimensions:
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, = * = . ( df circle 10 4π 3 568 units units really don't matter in this problem * *) waves hand , = = . df square 10 3 162 units , = * * = . df kcircle 10 6 45π 3 909 units = + = . l 101 116 3 068 units = . =. t 3 0684 767 units These values will be used in the comparisons and discussions throughout the rest of this problem. o A] Developing Solutions for Normal Stress in Fiber as Function of Fiber Position (x) Static Equilibrium (Conservation) in X-Direction Circle = * = F Stress Area 0 Assumption: = ; = <= σ0 0 τx τyield rigid plastic + , - , - , = σf dσfAreaCross Circle σfAreaCross Circle τyAreaSurface Circle 0 ( + ) , - , - , = σf dσf πdf circle24 σfπdf circle24 τyπdf circledx 0 = , => => = ( , ) => , dσf 4dxdf circleτy Integrate σoσfdσf 0x 4dxdf circle τy σf = , ( ) Circle 4df circle τyx Square = * = F Stress Area 0 Assumption: = ; = <= σ0 0 τx τyield rigid plastic
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+ ( , )- , - ( , ) σf dσf AreaCross Square σfAreaCross Square τy AreaSurface Square ( + ) , - , - , = σf dσf df square2 σfdf square2 4τydf squaredx 0 = , => => = ( , ) => , dσf 4dxdf squareτy Integrate σoσfdσf 0x 4dxdf square τy σf = , ( ) Square 4df square τyx Keyway Circle = * = F Stress Area 0 Assumption: = ; = <= σ0 0 τx τyield rigid plastic + ( , )- , - ( , σf dσf AreaCross KeyedCircle σfAreaCross KeyedCircle τ AreaSurface ) KeyedCircle ( + )( )( , )- ( , )- ( , - , + , )= σf dσf 56 πdf kcir24 σf56 πdf kcir24 τy πdf kcir πdf kcir60360 df kcirdx 0 , =[[ , -( , )+ , ] ] 5πdf kcircle224dσf
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This note was uploaded on 03/25/2010 for the course ABBE BE430 taught by Professor Weiner during the Spring '10 term at Rose-Hulman.

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MSQ1 - 1 Fiber? Check. My Shapes Matrix? Check. My...

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