Exam 1 Review Concepts - Midterm 1 Review Topics Vb F...

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Unformatted text preview: Midterm 1 - Review Topics Vb . . F Average Normal Stress In an AXIal Member (7an = X Axial Strain = Ax1a1_-Deformatlon e: = £ = E Or1gma1_Length L0 L Axial Strain Due to Temperature Change 5T = a-AT Axial Deformation Due to Temperature Change 5T = L-sT 0' = E-€(normal stress) E n G = ——- HOOke S Law T 2 Gay (shear stress) 2(1 + U) "elateral ‘Tetransverse- POISSO S Ra '0 slongitudinal eaxial Ufailure Allowable stress being the reduced stress Factor Of Safety FS = ——-—- we allow the specimen to safely reach with 0' allowable the factor of safety. If given a stress limit on a material, that would be a failure stress . . 0f '1 V (I. e. 'uIt/mate stress 9, and would then be _ 31 “re. reduced to an allowed level to obtain a Allowable Stress “allowable T FS factor of safety. 11 F L- . . FL 1 1 Axral Deformation ,6 = —— = 8-0' or 5 = A-E A1 Ei = 1 . . . F-L Axual Deformation With Temperature 6 = E + mAT-L Stress-Strain Curve for Steel Rupture. | | l l l | l | l l l l l I I l l : l | l [-e——>-:<—-——-——-———-—>-‘ -lv<—-—-—>ll iYield: Strain-hardening Necking I ' I l 0.02 0.2 0.25 0.0012- Slope of linear-elastic region = modulus of elasticity (Hooke's Law) 2/6 (Shear Force) divided by - _ 1 (Cross-Sectional Area Parallel Average DlreCt Shear Stress Tavg — A to Shearing Force) _ V (Transverse Force) divided by - _ [(Perimeter of Punching Member) * PunChmg Shear Stress T _ A (Thickness of Punched Member)] V F , (Force) divided by (Area of Contact O-b _ X1; Between 2 Components) [i. e. (Diameter of Bolt) * (Plate Thickness)] Rigid Bar: Represents an object that undergbes no deformations. Depending upon how it is supported, it may translate (i.e. shift) position, or rotate, but it will not deform/elongate or bend. (An acceptable assumption for significantly stronger members attached to weaker members. Shear Stress and Strain —-—————————:-)'C . 5 > Can ignore the tan”() term ifa small - ’Y = arcrtan[— : angle approximation is applicable Shear Stram 'h V Shear Stress T = — AV. Bolts and Pins Single Shear 3/6 Indeterminate Axial Members, general approach 1) Equilibrium: Express static equations in terms of unknown axials forces (use a FBD) 2) Force-Deformation Relationship: Relate INTERNAL member forces to corresponding deformation 3) Geometry of Deformation: Evaluate the geometry of the system to determine how the deformations of the axial members are related 4) Compatibility Equation: Substitute the force-deformation relationships into the geometry of deformation relationship to obtain an equation expressed in unknown axial forces based on the system's geometry 5) Solve the Equations: Simultaneously solve the equilibrium and compatibility equations 61: Common Geometry Deformation Relationships 1. Coaxial or parallel axial members . No Gap Gap 51:52 61+Gap=52 2. Axial Members Connected End-to-End in Series No‘Gag 51+ 52 = 0 ”/6 $9 61+ 52 = Gap Gag ,(51 + Gap) 52. 5/6 Torsion Torque is a moment twisting a shaft about its longitudinal axis T = Torque (Force*Perpendicular Distance) 1’1 ¢= Angle of Twist [Radians] 4’: 43:21:11 SOIid Cigcle 4 Hollow Circular Tube J — Polar Moment of IntertIa J : 7T : 77 J = :(R4 _ r4) = 1(D4 _ d4) 2 32 2 32 . D" c = Distance to outermost surface of shaft 0 = R = — (Often maximum radius) 2 p = radial coordinate from zero to'c' L = Shaft Length G = Shear Modulus, Modulus of Rigidity Stress and Strain vary ’ p I i / p. linearly with radial distance T = —T = — from longitudinal axis P C max WP C Wmax (I) T c Tmax L T L ’lmax‘T TIM—T = cc 2% Equations egg/y to shaft that is: *Homogeneous (is. constant G) *Prismatic (i.e. Uniform diameter) *has Constant internal torque If a shaft IS subjected to external torques at intermediate points (not only the end), or if it consists of various diameters or materials, it must be divided into segments to satisfy the above 3 requirements. (Make a torque diagram) Torsion Of Thin-Walled Tubes T Am = All area enclosed by Not necessarily circular T = 2_ A . t median line of tube thickness In t = tube thickness Closed sections only Shear flow is constant and tangential to wall of tube Statics Truss Member Axial Force F1 = (F1x)2 + (Fly)2 Trusses: Utilize either the Method of Joints, or the Method of Sections. Either way, equilibrium of forces must be satisfied FBD: Remember to satisfy equilibrium for the Joint or Section you are inspecting by including all applicable External Forces, Reaction Forces, and internal Forces ...
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