week3-3 - Assignment Assignment#3 Forsubmission 6.18 6.62...

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ssignment #3 Assignment #3 For submission .18 .62 .86 6.18 6.62 6.86 For practice 6.21 6.55 6.90 Td Fb 3 t 12 Due Tuesday, February 3 rd at 12 noon
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rque ist relationship Torque twist relationship = i i L T φ i pi i I G For compound members, divide shaft into sections that are omogeneous onstant G) • homogeneous (constant G) • uniform cross section (constant I p ) • constant internal torque T
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atically Determinate Problems Statically Determinate Problems olvable using the principles of statics Solvable using the principles of statics 0 = x F 0 = A M Determine reactions at supports and internal rques through the use of free body diagrams torques through the use of free body diagrams and the solving of equilibrium equations
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atically Indeterminate Problems Statically Indeterminate Problems The number of unknowns exceed the number f static equilibrium equations of static equilibrium equations. Reactions and internal forces cannot be determined by equilibrium equations alone. The problems are solved by additionally considering deformations and the deformed ometry geometry.
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rategy for Solving Strategy for Solving uilibrium Equations 1. Equilibrium Equations o expressed in terms of unknown internal rques torques 2. Torque twist relationships TL/ I o φ = TL/GI p 3. Geometry of deformations o Determine how the deformations of the members are related to each other
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rategy for Solving Strategy for Solving ompatibility equation 4. Compatibility equation o Substitute the torque twist relationships to the geometry of deformation into the geometry of deformation equation lve the equations 5. Solve the equations o Simultaneous solving of compatibility d equilibrium equations and equilibrium equations
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pes d to End Types End to End End to End Members he sum of the angle of The sum of the angle of twists for the members will be zero 0 2 1 = φ
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pes d to End with misfit Types End to End with misfit End to End Members with misfit he sum of the angle of twists for tt The sum of the angle of twists for the members will equal a specified angular rotation constant = + 2 1 φ
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pes oaxial Types Coaxial Coaxial Members ngle of twist for both Angle of twist for both members will be the same 2 1 φ φ=
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ample Example Member 1, solid Aluminum Member 2, solid Bronze D = 3.00 in G = 4,000 ksi D = 2.00 in G = 6,500 ksi a) What is the maximum shear stress in each member?
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This note was uploaded on 12/01/2009 for the course CIVE 207 taught by Professor Shao during the Winter '09 term at McGill.

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week3-3 - Assignment Assignment#3 Forsubmission 6.18 6.62...

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