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by: Clifford Bednar


Clifford Bednar
GPA 3.53

Thomas Burks

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Thomas Burks
Class Notes
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This 13 page Class Notes was uploaded by Clifford Bednar on Friday September 18, 2015. The Class Notes belongs to ABE 4171C at University of Florida taught by Thomas Burks in Fall. Since its upload, it has received 10 views. For similar materials see /class/206932/abe-4171c-university-of-florida in Agriculture Education at University of Florida.




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Date Created: 09/18/15
mm H v dummy a um m n mmmuu 1m Element Representing Stresses on a Stress l One main goals of stress analysis is to determine the point within a loadcarrying member that is subjected to the highest stress level I The orientation of the stress element is critical and it must be aligned wi specified axes on the member usually called X y and z Representing Stresses on a Stress Element con t l Tensile and compressive stresses called normal stresses act perpendicular to the opposite faces of the stress element l Tensile stresses pull on the element l Compressive stresses crush the element Stress Elements for 3 Types of i Stresses y a Direct tension 17 Direct compression 5 Pure shear Mon Machine Flemeni in Mechanical De hm 20m Representing Stresses on a Stress 7 Element con t gaggaga I Shear stresses are created by direct shear vertical shear in beams or torsion 7 I There is a tendency to cut the element by exerting a stress downward on one face while simultaneously exerting a stress 7 upward on the opposite parallel face Sign Convention for Shear i Stresses 7 l Positive shear ssses rotate the element in a clockwise direction 7 l Negative shear stresses rotate the element in a 7 counterclockwise direction l A double subscript notation is used to denote shear stresses in a plane For example drawn for the yz plane the pair of shear stresses Cyz indicates a shear stress acting on the element face that is perpendicular to the yaxis and parallel to the zaxis I So I acts on the face that is perpendicular to the y axis and parallel to the xaxis In this example Cyz is positive and CW is negative Direct Stresses Tension and Compression I Stress the internal resistance offered by a unit area of a material to an externally applied load Normal stresses 6 can be tensile positive or compressive negative I The magnitude of the stress can be calculated from the direct stress formula I G force area FA I US Units lb in2 psi SI Units N m2 pascal Pa Example Problem 34 A Iensxle force of 9500 N AS upphcd m a 12mmdiameter round bar as shown in Figure 376 ornpum the direct 1mm mass m nu 1m l 1 Sms element A w d i stat new rm section omecuve Compute lh tensile xtmss m m round bar Gwen ma F 9500 N diamcler o 12 mm Analysls 39 L 39 I 39 2 39 area fmrnA mm 7 Results 39A uni4 1112mm24 113 mm2 a FA 9500N113 mm 40 Nmm1 340 MP Comment V 39 39 M II 39 Theuubefonnof the element is as shown In Figure 375 a M ms Direct Stresses Tension and Compression con t I The conditions on the use of the previous equation are as follows 1 The loadcarrying member must be straight 2 The line of action of the load must pass through the centroid of the crosssection of the member 3 The member must be of uniform cross section near where the stress is being computed 4 The material must be homogeneous and isotropic 5 In the case of compression members the member must be short to prevent buckling Deformation Under Direct Axial 7 Loading i The following formula computes the stretch due to a direct axial tensile load or the short ening due to a direct axial compressive load i 5 FLEA 3 2 i where 5 total deformation of the member carrying the axial load F direct axial load i L original total length of the member E modulus of elasticity of the material A crosssectional area of the member 7 Noting that o FA we can also compute the deformation from oLE 7 3 3 Deformation Under Direct Axial 7 Loading 7 Stretch due to axial tension or shortening due to axial compression F L a 7 E A 7 Where F total deformation 7 F direct axial load 7 L Original length E modulus of elasticity A crosssectional area Noting that stress F FA we get F L F Direct Shear Stress 4 7 I Direct shear stress occurs when the applied force cuts through the member like scissors or when a punch and a die are used to punch a slug of material from a sheet I The method of computing direct shear stress is similar to method used for computing direct tensile stress Both methods assume the applied force is uniformly distributed across the cross section of the part that is resisting the force I 1 shearing force area in shear F AS 7 Relationship Among Torque 7 Power and Rotational Speed 7 I The relationship among the power P the rotational speed n and the torque T in a shaft is described by the equation 7 I T P n SI or T 63000 P n English I In SI units power is expressed in i I watt W or its equivalent I newton meter per second Nms 7 I Rotational speed is in radians per second rads Torsional Shear Stress I A torque causes a rotation of one part of the member relative to another which leads to 7 shear stress I In torsional shear the distribution of stress is not uniform across the cross section 7 I Most common case a round circular sha transmitting power E Torsional Shear Stress Formula I When the outer surface of a solid round shaft is subjected to torque it experiences 7 the greatest shearing strain and the largest 7 torsional shear stress I tmax Tc J 7 I Where 0 radius of the sha to its ouwide 7 surface I J polar moment ofinertia Stress Distribution in a Solid 7 Shaft a 320 Mon Machine Flement m Morhnnirnlne I n 2003 7 Example Problem 3 6 Compute the maximum torsional shear stress in a shaft having a diameter of 10 min when it carries a torque of 4 10 N m olution Objective Compute the torsional shear stress in the shaft Given Torque T 410 Nm shaft diameter D I 10 mm c radius ofthe shaft D2 50 mm 7 w Analysis 7 Use Equation 3 7 to compute the torsional shear stress me TcJ 7 I J is the polar moment of inertia for the shaft J nD432 see Appendix 1 4 Results A J qu432 2 n101nm432 982 mm j 410 Nm50 mm 103 mm W 2 2 Tmux 982 mm4 m 209 Nmm 209 MPa Comment 7 The maximum torsional shear stress occurs at the outside surface of the shaft around its en tire circumference 39 Mon Machine Flement m Morhnnirnlne I n 2003 General Formula I To calculate the torsional shear stress at some point inside the shaft the formula used is I 1 Tr J I Where r radial distance from the center of the shaft to the point of interest 7 I In hollow shafts the maximum shear stress occurs 7 at the outer surface I The hollow shaft is more ef cient because the entire cross section carries a relatively high stress level Stress Distribution in a Hollow 7 Shaft 7EEEEEEEE Stress distribution 71 I Polar Section Modulus 7 A design simpli cation uses the polar section modulus 7 I Zp J c i I The equation for the maximum torsional shear stress is i I Tmax T Zp 7 I When one crosssection is rotated relative to other cross 7 sections due to torque the angle of twist is computed from I 9 TL GJ I Where 9 angle of twist radians I L length of the shaft over which the angle is being computed I G modulus of elasticity of the shaft material in shear Vertical Shearing Stress 7 l A beam carrying loads transverse to its axis will experience shearing forces V 7 I To analyze the beams you compute the variation in 7 shearing force across the entire length of the beam and 7 draw the shearing force diagram I The resulting vertical shearing stress 17 is computed by r VQ It I Where I rectangular moment of inertia of the cross section i of the beam I t thickness of the section at the place where the shearing stress is to be computed I Q first moment with respect to the overall centroidal axis Vertical Shearing Stress con t 39Q I To calculate the value of Q Apy I Where Ap that part of the area of the section above the place Where the stress is to be computed ly distance from the neutral aXis of the section to the centroid of the area Ap 7 Vertical Shearing Stress con t I For most section shapes the maximum vertical shearing stress occurs at the centroidal axis i x 1 1 1 rev 7 7 4L i fllrl Rf Fill l l 4 FIGURE 3 14 200h 600 H A 2m m Starter A 0 1 7 m 3 mg 1 51 l h sness mrbeam R quotmm R2 mm J i um m V i lb 0 E 7 n 03 X psi 7 son sum dxsuibuuon snumuu Fume umguuu Problem 3 10 Obleclwe Gwen Analysls Figure 3714 show u mupiy mppmred beam currymg mo concenmued mm The 39 u r r We q on m h m dlsmbmion is puruhuuu mm m maximum sums up ur mg u xhc nunnll ux Use Fquunon 3 167 m compmc hc muxunnm shearing 1er m the beam Compute ma muxrmum heunng sues 7 m hc beam m Fuguuc 314 The hemquot shape is ecmnguhr h xxx m4 200 in 7 V Use Fquntmn Kr 11 0 compmc r Vaml 1 an gmquot From Appendix 1 I r luv 12 The value 01 m mu mnmunl of KM um Q be cumpmed rrum Izqumiun 3717 For the rectangular cross secnon shown m Figure 343m A IUzZ uml 3 14 Then Q AV mzjum mlx nn Results 71 th312 20 in80in312 85317114 Q A 1128 20in80 inf8 2160in3 Then the maximum shearing stress is VQ 1000 lb 161 in 2 I I 853 1410 in 9381btn 938 p51 Comments The maximum shearing stress of 938 psi occurs at the neutral axis of the rectangular 7 section as shown in Figure 3 4 The stress distribution within the cross section is gen erally parabolic ending with zero shearing stress at the top and bottom surfaces This is the nature of the shearing stress everywhere between the lcft support at A and the point of application of the 120041 load at B Thc maximum shearing stress at any other point in the 7 beam is proportional to the magnitude of the vertical shearing force at the point of interest Special Shearing Stress Formulas I The most common cross sections have unique formulas for finding the maximum i vertical shearing stress 7 Tmax 3V2A for rectangle 7 Tmax 4V3A for circle Stress Due to Bending m u a I A beam is a member that carries loads transverse to its axis producing bending moments in the beam Bending stresses are normal stresses tensile 0r compressive I The maximum bending stress in a beam cross section occurs in the part farthest from the neutral axis of the section i o Mc I i Where M magnitude of bending moment at section i I moment of inertia of section wrt neutral axis 0 distance from na to outer most ber Stress Due to Bending con t I The value varies from zero at the neutral axis to the maximum tensile stress on one side of the neutml axis to the maximum compressive stress on the other side I Positive bending only occurs when the de ected shape of the beam is concave upward resulting in compression on the upper part ofthe cross section and tension on the lower part I Negative bending causes the beam to be concave downward Stress Due to Bending con t I The exure formula was developed subject to the following conditions 1 The beam must be in pure bending Shearing stresses must be zero or negligible No axial loads present The beam must not twist or be subjected to a torsional load The material ofthe beam must obey Hooke s law N t The modulus of elasticity of the material must be the same in both tension and compression Stress Due to Bending con t 5 The beam is initially straight and has a constant cross section 0 Any plane cross section ofthe beam remains plane during bending No part ofthe beam shape fails because of local buckling or wrinkling u F Load due to pipe RI 7 E V 0 b 7 a 7 1 R2 Fba 7 r Mmax R1 M ra b 7 70 l a b Shear and bending moment diagrams 7 7 7 7 Compression 7 i 7 7 Neutral axis I s gt X X7 Beam 7 U 39C Tension cross section 7 c Stress distribution on beam section 7 7 1 Stress element in compression 79 Stress element in tension in top part of beam in bottom part of beam Mott 39 39 Mechanical wasquot 20 03 For design it is convenient to de ne the term section modulus S as 7 S 3 323 7 The flexure formula then becomes 7 7 039 MS 7 344 Since I and c are geometrical properties of the cross section of the beam S is also Then in39 7SMo has Mott 39 39 Mechanical Design 2003 design it is usual to de ne a design stress 01 and with the bending moment kirowiigsolv39 7 a For the beam shown in Figure 346 the load F due to39the pipe is 12 000 lb The distances are a 4 ft and b n 6 ft Determine the required section modulus for the beam to limit the stress due to bending to 30 000 psi the recommended design stress for atypical structural 7 steel in static bending 7 Objective 7 Compute the required section modulus S for the beam in Figure 3 16 Given The layout and the loading pattern are shown in Figure 3 16 Lengths Overall length L 10 ft a 4ft b 6 ft Load F 12 0001b Design stress 0 30 000 psi 7 Analysis Use Equation 3 25 to compute the required section modulus S Compute the maximum i bending moment that occurs at the point of application of the load using the formula shown in Part b of Figure 3 16 V 1 7 7 Results 7M Fba 12000 lb6ft4 ft m R1a a b 6ft Mm 288001bft M 28 8001bft 12m 7 L 3 S 01 30000 ibin2 ft 1 5 1 7 7 Comments 7 A beam section can now be selected from Tables A16 3 and A164 that has at least this value for the section modulus The lightest section typically preferred is the W8X15 widc 7 flange shape with S 118 in Mott 39 39 Mechanical Design 2003 11 Combined Normal Stresses When the same cross section of a load carrying member experiences both a direct normal stress and a bending stress the resulting normal stress can be computed by the method of superposition I UiMcIrFA I Where tensile stresses are positive and compressive stresses are negative Combined Normal Stresses con t An example of this situation involves a beam subjected to a load applied downward and to the right through a bracket below the beam The load s horizontal and vertical components show that the effect can be broken into three parts The vertical component places the beam in bending with tension on the top and compression in the bottom The horizontal component causes direct tensile stress across the entire cross section Combined Normal Stresses con t 1 2 Prepare the shearing force and bending moment diagrams Combine the effecm of the bending stress and the direct tensile stress at any point KA 8560 m M nxmu lb m V sum 0 o 7 asnmng um and hmdmg 51960 moment may mus M 7 lbm7 nxmo a 391 gt21 0psx n 5921 a 2 A a Nth 9232 pm C 725 if 2360pr C 6872 pg 1 Bending stress 2 Dimcl mam ms 4 Cumhmcd sum an m harimnul wmpoumu of load 13


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