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# Mechanics of Deformable Bodies 057 019

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This 41 page Class Notes was uploaded by Augustine Schmitt on Friday October 23, 2015. The Class Notes belongs to 057 019 at University of Iowa taught by Colby Swan in Fall. Since its upload, it has received 59 views. For similar materials see /class/228005/057-019-university-of-iowa in General Engineering at University of Iowa.

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Date Created: 10/23/15

057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Period 23 Spring Semester 2008 Textbook Sections 94 96 Topic Mohr s Circle Plane Stress Objectives 1 Review the plane stress transformation equations 2 Introduce Mohr s Circle and how to use it 3 Demonstrate the concepts with some examples 1 Review of planestress transformations At state of stress can be described in terms of components 6x try and Exy relatedto a Cartesian reference frame Xy The same state of stress can also be described in terms of components 0quot or 1W related to an alternative basis X y which corresponds to the original basis rotated ccw through an angle 9 039X 039 039X 7 039 0quot cos 28 1W s1n 28 Ky 039 7039 139 7s1n28r cos28 2 KY 039X 039y 039X 7 039y I 039y T 7 T cos 2877y s1n 28 not explrcrtly der1ved here 2 Mohr s Circle All combinations of normal and shear stress speci ed by the transformation equations above plot on a circle Mohr s circle in 613 space The radiusR of the circle and the center 0 on the Gaxis are given by the following relations Z 039 7039 039 039 y 15 C039Wg 2 y b R 057019BBB Mechanics ofDeformable Bodies College ofEngineeIing Instmctor CC Swan University of Iowa Proce dure 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 1 9776 Determine a the principal stress and b lt the maximum inplane shear stress and average normal stress Specify the orientation ofthe element in each case 5 ksi 5ksi 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 2 978 600 PM 3 9 Draw Mohr s circle that describes each a 10 MP ofthe following states of stress D College of Engineering 057019BBB Mechanics ofDeformable Bodies Instructor CC Swan University of Iowa Example 3 987 The bent rod has a diameter of 15mm 3900 mm Al 3900 mm and is subjected to the force of 600N Determine the principal stresses and the maximum inplane shear stress that are developed at point A and point B Shown the results on elements located at 600 N 600 N these points E 50 min 057019BBB Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa 057019BBB Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Period 11 Spring Semester 2009 Sections 55 8 56 of Textbook Topic Torsion of Statically Indeterminate Members Objectives Review Statically Indeterminate TorqueLoaded Members Torque of NonCircular CrossSections Example Problems PWN 1 Review of Torsion o The torsion Formula Tmax T T p J J o Shear Strain d p 7 P dx C 7max 0 Twist 7mm 2 2Imax ix dx 0 G0 JG JG 0 Sign Convention 0 Internal torques and their associated angle change are positive when the resultant vector points away from the cut face on which it acts The righthand rule is used 2 Statically Indeterminate TorqueLoaded Members In members where there are more support reactions than there are equations of static equilibrium additional kinematic constraints can be brought into play The principle of superposition can also be used 057019BBB Mechanics ofDeformable Bodies College ofEngineeIing Instructor CC Swan University of Iowa From statics TATBTc Additional constraint om0 LAC C AB 0 ix N jTAC Jrjf dx yB q A JG CJG TALAC To TALBC JG JG A TALAC LCBTCLBC TB TA TCLBC T8 TCLAC LAB LAB Solution of problems with statically indeterminate torqueloaded members is analogous to solution of problems with axially loaded indeterminate members 3 Torsion of Noncircular Members I So far we have considered torsion in axial members that have axi symmetric crosssections For such members the shear stress and shear strain vary linearly with distance from the axis For noncircular members the distribution of shear stress and strain over the crosssection is much more complicated and this complicated shear distribution leads to warping of the cross section See Figs 527 and 528 of the textbook Circular members are most efficient for carrying torsion Other crosssectional shapes are less efficient See Table 51 for effective polar moments of inertia of non circular crosssections 057019BBB Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa 4 Example Problems Example 1679 The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core If it is fixed to a rigid support at A and a torque of T50bft is applied to it at C determine the angle of twist that occurs at C and compute the maximum shear and maximum shear strain in the brass and steel Take GS 115103ksi Gb 56103ksi 057019BBB Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 2 580 The two 3ftIong shafts are made of 2014T6 aluminum Each has a diameter of 15in and they are connected using the gears fixed to their ends Their other ends are attached to fixed supports at A and B They are also supported by bearings at C and D which allow free rotation of the shafts about their axes If a torque of 600 Ibft is applied to the top gear as shown determine the maximum shear stress in each shaft Prob 5 80 057019BBB Mechanics of Deformable Bodies Instructor C C Swan College of Engineering University of Iowa Example 3 583 The A36 steel shaft is made from two segmentsAC has a diameter of 05in and CB has a diameter of 1 in If the shaft is fixed at its endsA and B and subjected to a uniform distributed torque of 60binin along segment CB determine the absolute maximum shear stress in the shaft Fab 6 WLEMlDQr Ag TAT3 web lain m 39 Zoo 154 kin gum was V Mu t 39 shs B Ii 5 T711 39 er z A C 6 6 P 3015 O 155 TALEB 6091a minm C 3 1 jun Imam R39 05 in 20 1 1 L 4 JTA bv 45 T4975 915 New 9 3 4 a ml 012 L3 2 TE3 5 regfdlxs an 4L6 ls 3016 Evian arty TR L13 TLB z 1 3103quot 51 m Tn 3 L93 30 L43 161ml 1 Biz MkiLw The H I ax miv 4 04 2175qu 291 i204 Ta molly1n 4 IB jlxia grim 153 T LC 39 Upbin Iqi gt 39 am Ac T QUEEN 957951 21 7 955L27PSZ aI m 057019BBB Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Period 38 Spring Semester 2008 Textbook Sections 1413 Topic Strain Energy and Conservation of Energy Objectives External work and internal energy Strain energy in linear elastic systems Conservation of energy Example problems hMNt t H External Work and Internal Energy 0 When external forces act on mechanical systems the systems deform Fig lb 0 The general expression for the resulting work done by a force is W IF 615 Fig 1c 0 For a linear elastic system the resulting work is W F 5 Fig 1d Unloaded undeformed Loaded Deforme system system 1a 1b 1 a Nonlinear system Linear system gt11 gt11 1c 5 1d a 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa The energy associated with the external work is stored in the strain energy U of the material that comprises the system Conservation of energy requires equality between external work and stored energy and is expressed W U Over the next two periods we will explore how the relationship W and U can be used to compute the deformations of linear elastic mechanical systems 2 Strain energy in linear elastic systems At a given point in a linear elastic solid the strain energy per unit volume is given by the following relation l 01681 O39y8y 0282 Tag7x TyZVyz 72 M 2 Utilizing Hooke s Law for a linear isotropic elastic material this can be rewritten as i 2 2 2 i i 2 2 2 u 2E 039x 039y O39Z EO39XO39y O39yO39Z 0Z0 261W ryz rzx The total strain energy in a structural system is obtained by integrating the strain energy density over the material volume y 2 U Iu dV JOXZ 0 2 0 2oXo39y o yo z O39ZO39XTWZ L39yz2 er2dV Next specific forms of strain energy for systems undergoing axial loading bending and torsion 41 Axial Loading 016 For members undergoing pure axial loading the strain energy density would be u 2E Integrating over the entire volume of the axially loaded member 2 2 UIudVI EdVJ dx N2 JN2 Adx Adx JZAZE ZAE 01 2E b Ben ding For members undergoing pure bending neglecting shear deformation the strain energy 2 039 a density at a point would be u 2E Using the exure formula and integrating over the entire volume of the beam 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa M 2 MR 2 MR UIudVIdVI jy dAdxjTExdx 0 Transverse Sh ear 2 13916 For a beam the strain energy density due to transverse shear is u 22 Using the transverse shear formula and integrating over the volume of the beam gives T 2 V2 2 V2 2 V2 fslz szs UjuarVj arVjAabcjEjdAdxjE A dxj dx 2GIt Here is the formfactor of a crosssection For rectangular sections 65 d Torsional Deformation For circularcrosssectioned shafts undergoing torsional loadings the strain energy 2 r dens1ty 1s u Applying the tors1on formula and 1ntegrat1ng over the volume of the shaftyields 12 Tlpl T2 2 T2 U dV dV Adx dAdx d Iquot I2C 2G12 Izerlp I2G1 x 3 Conservation of Energy a Truss Systems Consider a truss structure subjected to a loading P as shown at right The external work done by the loading is W PA and the internal energy stored in the truss members is expressed 9 2 e a U N Ll 11 KZAE 1 quot By conservation of energy 2U 2 9 A NZL P P11 ZAE amp D 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa b Beams Similarly the de ection of a beam loaded by a point load P can be calculated by P conservation of energy as follows i L 2 kg M dx g f 7 in P P 0 2E A The rotation of a beam loaded by a concentrated moment can also be computed by conservation of energy as follows L 2 Mo 62U 2 JM M0 MOOZEI 39 fry 93 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa 4 Example Problems Example 1 14 8 Determine the total aXial and bending strain energy in the A 1395 km 36 steel beam A2300 mmz l l l l 195106mm4 91 l5 kN amp IO m 4 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 2 14 26 Determine the vertical displacement of joint D AB is constant B 05701900B Mechanics ofDeformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 3 1430 Use the method of work and energy and lt3 B C Mo determine the slope of the beam at point E E1 is constant 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Period 28 Spring Semester 2008 Textbook Section 106 Topic More on Hooke s Law Objectives 1 Revisit Hooke s Law 2 Solve Relevant Example Problems 1 MaterialProperty Relationships Last period we established the following form of Hooke s Law valid for linear isotropic elastic materials x l V V 0 0 0 o y V l V 0 0 0 039 Z 1 V V 1 0 0 0 0392 7yz E 0 0 21 v 0 0 ryz 72x 0 0 0 0 21 v 0 rm 7W 0 0 0 0 21 V TV The inverse form of this relation can be written 113v lVZV 1 0 0 0 0x 112ij lVZV x O O O 039 U 8 y V V l V 0 0 0 y Oz E 1 2V 1 2V 1 2V gz T W V 0 0 0 l 0 0 7 Ta 2 7 xy 0 0 0 0 l 0 7w 2 l 0 0 0 0 0 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 1 1053 The principal stresses at a point are shown in the gure If the material is aluminum for which E3110103ksi and 1033 l 26 ksi determine the principal strains l ksl lSksi 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 2 10 54 A thinwalled cylindrical pressure vessel has an inner radius r thickness t and length L If it is subjected to an internal pressure p show that the increase in its inner radius is dr r81 pr2l Et and the increase in its length is AL erQ v Et Using these results show that the change in internal volume becomes dV 72721 81 2 1 82 L n39rzL Since 81 and 82 are small quantities show further that the change in volume per unit volume called volumetric strain can be written as dVV pr25 2v Et 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 3 10 58 The smooth rigidbody caVity is lled with liquid 606lT6 aluminum When cooled it is 0012 in from the top of the cavity If the top of the caVity is not covered and the temperature is increased by 200 F determine the strain components 8x 8y 82 in the aluminum Hint Use Eqs 1018 with the additional strain term of OLAT Eq 44 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa Period 27 Spring Semester 2008 Textbook Sections 105 106 Topic Strain Rosettes and Hooke s Law Objectives 1 Introduce Rosettes for measuring of strains 2 Revisit Hooke s Law 3 Solve Relevant Example Problems 1 Strain Rosettes In experimental mechanics measurement of strain states is very common One way to measure strains on the surface of a mechanical specimen is to bond an electrical strain gage to the surface The gage shown below responds only to extension or contraction along its axis To measure the complete state of strain ax 8y yxy at a point on the surface of a material a cluster of three gages called a rosette is typically used A single strain gage A W sumn rosette If the extensional strain can be measured along the three axes with orientations 9 9b SC then the in inplane strain components ax 8y yxy can be determined by solving the system of transformation equations a cosZ 8 sin2 8 sin 8 cos 8 a 5 cosZ 85 sin2 85 sing cosHb 5y 2 39 Z 39 5 cos 8 sm 8 sm 8 cos 8 yxy Two speci c examples ofstrain rosettes are the 45 rosette with 9 9b SC being 0 45 and 90 respectively and the 60 rosette which has 9 9b SC being 0 60 and 120 respectively 05701900B Mechanics of Deformable Bodies Instructor CC Swan C 45 b I 45 strain rosette W College of Engineering University of Iowa 60 strain rosette C a 5 For the 45 rosette the transformation equations yield X a g 5 y C 7v 255 5agc 5X 5 Similarly for the 60 rosette the transformation equations give 5y 3255 25 5a 2 7v 55 gc 2 MaterialProperty Relationships Here we will look at the multiaxial form of Hooke s Law for a linear elastic and isotropic material A material point will generally be experiencing multiple different stress and strain components simultaneously as shown be 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa If the material is experiencing a triaxial state of stress With components 6x 6y 6 we can consider the resulting strains and then add them up using superposition x 0 v0 Under the stress 6x2 9 i X 715 7 i E y Z X 039y vay Under the stress 6y 5y 39 5X gz 715 E y E 039 v0 Under the stress oz 52 z 39 5X a V z E y z Summing the effects gives 1 X E 0X 7 1039y 02 5 039y VO39X 02 2 1 0392 VO39X ow Under pure shear stresses the only deformation response of the material is the corresponding shear stress For example when a shear stress rxy is applied to the material the only resulting strain is yxy yxy 7 rxy Similar results occur With application of shear stress ryz yyz l r G yz E 1 J and ZX yxzi ru G 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa For a linear isotropic elastic material there are only two independent elastic parameters Knowing any of these two the others can be determined 0 If E and v are known the shear modulus can be determined by G V o If E and v are known the bulk modulus can be determined by K 31 E2 1 Under a triaXial state of stress 6x 6y oz the bulk modulus relates the mean normal stress 6m to the volumetric strain 6 of the material The mean normal stress 6m of the material is the mean of 6x 6y oz or 1 039quot UI 039y 02 The volumetric strain of a material in small deformation theory is e 8 8y 8z OE m3 6v Finally Ee 039quot 31 2v Ke 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 1 1032 The 45 strain rosette is mounted on a steel sha The following readings are obtained from each gauge 5a 80010396 5b 52010 6 ac 745010 6 Determine a the inplane principal strains and their orientation 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 2 10 36 A bar of copper alloy is loaded in a tension machine and it is determined that 8x94010396 and 6xl4kSl 6y0 620 Determine the modulus of elasticity Ecu and the dilatation ecu of the copper The Poisson s ratio vcu035 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 3 1048 The spherical pressure vessel has an inner diameter of 2m and a thickness of 10m A strain gauge having a length of 20mm is attached to it and it is observed to increase in length by 001me When the vessel is pressurized Determine the pressure causing this deformation and nd the maximum in plane shear stress and the absolute maximum shear stress at a point on the outer surface of the vessel The material is steel for Which EfZOOGPa and vst03 7 10 mm 05701900B Mechanics of Deformable Bodies Instructor CC Swan College of Engineering University of Iowa Period 26 Spring Semester 2008 Textbook Sections 103 104 Topic Mohr s Circle for Plane Strain Transformations Objectives p A Asap Review ofthe plane strain transformation equations Show that transformed plane strain states plot on a Mohr s circle Absolute Maximum shear strains Apply the Mohr s circle ofstrain transformation in some problems 1 Review of Plane Strain amp the Transformation Equations When a state of strain ex is imposed it creates strains axv ayv and A state of plane strain occurs when material deformation occurs within and only within a speci c plane For example if a state of strain were X 0 y 0 yxy 0 520 yyz0 y 0 it would represent a state of plane strain in the Xy plane Plane strain and plane stress are usuall not the same thing Under plane stress for example the material deforms in the outofplane direction which violates the de nition of plane strain Plnnc stress 6 slnlin in he does not cause plmic 39 x 0 pluneslllcc e VX y Similarly when a state of strain 8y is imposed it also creates strains axv ayv and yxvyv Finally when a state of strain yxy is imposed it also creates strains axv ayv and yxvyv Superposition of all these effects yields the following transformation equations 5 5Xgygxigy cos28yTxysin28 axigy cos 287sin28 sin28yTxycos 28 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa 3 Mohr39s Circle of Plane Strain All transformed states of strain plot on a circle are given by the following relations Y 2 g 53975Wg2 R2 where am X y R Given the state of strain de ned with respect to axes X and y Mohr s circle can be used to nd the shear and normal strains associated with a different reference basis rotated by angle 9 counterclockwise from that of the Xaxis Given ax 8y yxy compute 8an and R Plot the circle on axes labeled a and y2 Identify the reference states of strain ax ny 2 and 8y yyxZ To nd the state of strain for a ber oriented at an angle 9 ccw with respect to the Xaxis move 29 on the circle from the point ax ny 2 To nd the state of strain for a ber oriented at an angle 9 ccw with respect to the yaxis move 29 on the circle from the point 8y yyxZ 59 V39 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa The Mohr s circle makes clear that the principal inplane normal strains and the maximum inplane shear strain are 1gwgR 525WE7R 7 4 Absolute Maximum Shear Strain To determine the absolute maximum shear strain associated with a triaxial state of strain the maximum and minimum principal strains must be determined Then 1 7 2 abmlutermax 2 5min 5am 5m Yum um I 2 e 05701900B Mechanics of Deformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 1 1022 The state of strain at the point on the fan blade has components 25250006 ey4501039 yxy8251039 Use Mohr s circle to determine a the in plane principal strains and b the maximum inplane stmin and average normal strain In each case specify the orientation of the element and show how the stmin deforms the element in the xy plane 05701900B Mechanics ofDeformable Bodies Instructor CC Swan College ofEngineering University of Iowa Example 2 1023 The strain at point A on the bracket has components g 30010 6 8 55000 yxy 65010 6 82 0 Determine a the principal strains at A b the maximum shear strain in the Xy plane and c the absolute maximum shear strain 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 3 1027 The steel bar is subjected to the tensile load of 500 lb Ifit is 05 in thick determine the absolute maximum shear strain E29103 ksi v03 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Period 3 Spring 2009 Semester Section 15 of Textbook Topic Average Shear Stress Objectives 1 Review key ideas of stress Note differences between shear and normal stresses Show symmetry of the stress tensor Solve example problems hWN 1 Key Ideas of Stress Stress is often defined as W per unit Q When working with stresses we need to be very specific about the directionality ofthe forces and the orientation or directionality of the area on which the forces are acting This is done by representing stress as a rank 2 tensor as follows rDC rxy In 0 TV rx2 T Tquot Tyy Tyz Tquot 0y Tyz sz sz 122 TD sz 02 The diagonal entries in the stress tensor are said to be normal stresses because the associated force component acts normal to the plane The off diagonal entries in the stress tensor are called shear stresses because the associated force components act parallel to the plane 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa It is very straightforward to demonstrate that the stress tensor is symmetric or specifically that Tyz sz Ta TXZ Consider the case in which only the two conjugate shear stresses ryz ancl rzy act as shown below The moments about the x axis are as follows 2M 0 ryzmAzAy 2y AyXAxXAZ ryzmAzAy a WWW TT The same arguments can be made to show that rxy Ty and I 1x 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 1 1 49 The open square butt joint is used to transmit a force of 50 kip from one plate to the other Determine the average normal and average shear stress components that this loading creates on the face of the weld section AB Prob l 4 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University oroWa ExampleZI63 The railcar docklight is supported by the 18 h in diameter pin at A lfthe lamp weighs 4lb and the extension arm has a weight of 05lbft determine the average shear stress in the pin needed to supportthe lamp Hint The shearforce in the pin is caused bythe couple moment required for equilibrium atA L 125 in Prob 1 63 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 3 1 69 The frame is subjected to the load of 200 lb Determine the average shear stress in the bolt at A as a function of the bar angle The bolt has a diameter of 025in and is subjected to single shear quot l 2 r i iiis ft 200 lb Prob 1 69

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