ANALYSIS & DESIGN OF MACHINE ELEMENTS I MET 301 SUMMARY OF TOPICS & FORMULAE

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1 AALYI & DEIG OF MACHIE ELEMET I MET 0 UMMARY OF TOPIC & FORMULAE Prepared b Prof. A. K. engupta Department of Engineering Technolog JIT Januar 8, 008

2 TABLE OF COTET. Resolving a force in two orthogonal directions.... Moment of a force about a given point.... tatic equilibrium in two dimensions Aial tension and compression: tress & strain & Hook s law Transverse loading: internal shear force & bending moment hear force & bending moment diagrams Bending stress (s) eutral ais (A) Moment of inertia (I) Transverse shear stress (t) in beams...6. Torsion of circular section & shear stress...7. Effect of combined stress in D: Mohr circle...8. tress manipulation in D train in two dimensions Theories of failure tress concentration Design for cclic loading Design for finite life hear stress in a shaft for combined bending & torsion Design of shaft with cclic load.... haft with bending loads in two planes.... Design of kes, kewas & couplings...4. Design of slender compression members or columns Deflection b double integration Deflection b strain energ method haft on three supports R, R & R Critical speed of a rotating shaft...7 of 7

3 . REOLVIG A FORCE I TWO ORTHOGOAL DIRECTIO F Y F.sin(θ) Y θ F X F.cos(θ) F ` If we know the magnitude of the force F, and its angle (θ) with -ais, then the component of the force in X and Y directions are: F F.cosθ,. MOMET OF A FORCE ABOUT A GIVE POIT ign convention: Clockwise (CW) positive Counter clockwise (CCW) negative. TATIC EQUILIBRIUM I TWO DIMEIO When several forces (and moments) are acting on a bod, the bod will maintain static equilibrium if the following three equalities are satisfied simultaneousl: (i) um of all forces acting on the bod in X direction 0; i.e., F 0 (ii) um of all forces acting on the bod in Y direction 0; i.e., F Y 0 (iii)um of moments about an point on the bod due to all forces acting on the bod 0; i.e., M 0 These equations are frequentl used to find unknown reaction forces and moments due to eternall applied known forces and moments. of 7

4 4. AXIAL TEIO AD COMPREIO: TRE, TRAI & HOOK LAW P A When a member is aiall loaded with a force P, the force acting on an cross-section across the length of the component P. The normal stress at an cross section can be P obtained b, where A cross-sectional area. A ign conventions: Tensile stress is positive, compressive stress is negative. P A taticall indeterminate problems: ome times, for aiall loaded members, the load shared b each member cannot be determined from static analsis of forces. In those cases, additional deformation relationships can be used to find the load shared. P ormal stress:, P Aial load, A cross sectional area and, A δ ormal strain: ε, δ elon`gation, L original length. l Hooks law: tress () is proportional to strain (ε) E.ε, E Elastic Modulus or Young s Modulus P δ Thus, E. A L Rearranging: δ PL AE of 7

5 5. TRAVERE LOADIG: ITERAL HEAR FORCE & BEDIG MOMET Transversel loaded members are often called beams. Beams can be supported as a cantilever beam or as a simple supported beam, as shown below. Cantilever te support generates a vertical reaction force R, and a reaction moment M R to support the beam. impl support generates onl vertical reaction forces, R A & R B w M Cantilever beam support P M R R w M Free-bod diagram P The magnitudes and locations of applied loads and moments are known. w P w P Tes of loads: P Concentrated applied load w Distributed applied load M Applied moment M imple beam support M Free-bod diagram For these two tes of supports, the unknown reaction force and reaction mome nt at the support can be readil determined b using the static equilibrium conditions. R A R B R B 6. HEAR FORCE & BEDIG MOMET DIAGRAM Due to transverse loading, shear forces and bending moments are generated internall within beam. The internal shear force (V) at an section of the beam: V sum of all eternal forces (including the reaction force at support) either to the left or right from the section. (Both will have same magnitude but opposite direction as the beam is in static equilibrium) ign convention of V positive, if the force is upward to the left of the section. To find the maimum V, often Vs along the length of the beam are determined and a diagram of variation of V is drawn, which is known as shear force diagram. The internal Bending Moment (M) at an section of the beam: M sum of moments about that section of all eternal forces (including the reaction force at support) either to the left or to the right from the section. (Both will have same magnitude but opposite direction as the beam is in static equilibrium) ign convention of M positive, if the moment of a force causes compression in the upper laer of the beam. To find the maimum M, often Ms along the length of the beam are determined and a diagram of variation of M is drawn, which is known as bending moment diagram. of 7

6 7. BEDIG TRE (s) Bending stress due to bending moment M will be zero at the neutral ais (A). Bending stress is tensile in one side of the A, and compressive on the other side of A. At an point in the loaded beam, bending stress () can be calculated from the following formula: Where, M Internal bending moment at the point stress is being calculated, M I Moment of Inertia of the beam cross section about the neutral v ais (A), I v distance of the point from A where the stress is determined. Ma bending stress will occur at the outermost laer of the beam (vmaimum), furthest awa from the A. ma Also, M c I ma M Z If the radius of curvature of the beam r, due to bending is known, bending stress () can be found from the following alternative formula: E r v Where, c distance of the furthest outer laer of the beam from the neutral ais (A) Z I/c known as section modulus Where, E Elastic Modulus r Radius of curvature due to bending at that section v distance of the point from the neutral ais (A) where the stress is determined 4 of 7

7 8. EUTRAL AXI (A) A passes through the Center of gravit (CG) of the beam cross section. For rectangular or circular cross-section of the beam, CG is at the geometric center of the section. For a composite section, the location of the CG can be determined b the following formula, A + A + A +... A + A + A MOMET OF IERTIA (I) I, for rectangular and circular sections about their A can be found using following formulae: I for I-sections, Bo sections and channel sections can be found using following formulae: Transfer of ais for Moment of Inertia This formula is used to find MI of a T or other sections, whose A or CG is not located at the geometric smmetricall. I I 0 + A 5 of 7

8 0. TRAVERE HEAR TRE (t) I BEAM (i) The transverse shear stress at a section BB is given b the following formula: V va Ib a VQ Ib Where, V Internal shear force, I MI about A of the beam section b Width of the section v Distance of CG of the section BB CC from A A a Area of the section BB CC Q v Aa Area moment of the section BB CC (ii) Ma. transverse shear stress, ma alwas occurs at A (because Q is ma at A) (iii) When finding transverse shear stress () for a composite section, the following formula can V Ib be used: va ( v A + v A ) a V Ib a a (iv) Ma Transverse shear stresses (t ma ): olid rectangular cross-section olid circular cross-section Circular cross section with thin wall * I cross-section V ma A 4 V ma A V ma A V ma A... A area of the cross-section b.d A area of the cross-section A area of the cross-section π ( do d i ) 4 A t.d t thickness of the web, and d total depth of the I-beam π 4 d * A more eact analsis gives the values of.8 V/A and. V/A for the transverse shear stress at the center and ends, respectivel, of the neutral ais. 6 of 7

9 . TORIO OF CIRCULAR ECTIO & HEAR TRE When a torque T is applied to a circular section, the elastic deformation produces an angular twist φ, and shear stress, within the member. At an radial distance r, the shear stress, can be obtained from: T J r Where, J polar moment of inertia of the circular section I The shear stress 0 at the ais, as r 0, and the shear stress ma, at r r, the outermost laer. Thus, ma T J r 4 4 π d π r For, solid circular section: J For, hollow circular section: π ( d o di ) π ( ro ri ) J Putting the values of J, in the above equation: For solid circular section: T T d 6T ma r J 4 ( π d /) π d For hollow circular section: ma T r J ( π ( d d where, λ d i o 4 0 T d 4 i do ) / ) 6T 4 π ( d d d ) ratio of inner to outer diameter The shear stress t and the angle of twist f is related b φgr 0 4 i o 6T 4 π d ( λ ) 0, where G shear modulus of elasticit, l length of the shaft and r radius. l Relationship between power (kw/hp), torque (T) and rpm (n) T T 6,05HP in lb n 9,550,000 kw mm n 7 of 7

10 . EFFECT OF COMBIED TRE I D: MOHR CIRCLE The two dimensional general state of stress at a point is shown, where, normal stress in X direction, normal stress in Y direction, and shear stress. The corresponding Mohr circle can be drawn b plotting the X(, ), and Y (, ) points in the plane. The circle drawn, using the XY line as diameter, is the Mohr circle. + The center of the circle avg Radius of the circle R The angle, θ tan + ma ο, and the angle, φ 90 θ Y Y(,) X (avg,ma) avg Y ais φ θ (avg,-ma) X ais X (,) All Mohr circle angles are double the actual angles Principal normal stresses will occur along the diameter at an angle θ from the original X direction. avg + R avg R + Maimum shear stress will occur along the vertical diameter at an angle φ from the original X direction. ma R + tresses in an arbitrar direction u-v which is at an angle f with respect to the - aes sstem, can be readil found if s avg, R & q are known: u avg +Rcos[(θ+φ)], v avg - Rcos[(θ+φ)], and uv Rsin[(θ+φ)] min avg Y avg avg ma avg θ φ X 8 of 7

11 tress relationship for a rotation of ais sstem b 45 o (a) When the XY coordinate sstem is rotated b 45 o counter-clockwise to X Y sstem + + Then, ' ', ' avg and ' avg+, where avg (b) When the X Y coordinate sstem is rotated b 45 o clockwise to XY sstem ' ' ' + ' + Then,, avg+ '' and avg '', where avg ' ' X'(','') (-)/ X(,) ('-')/ Y(,-) (-)/ ('-')/ Y'(',-'') 9 of 7

12 . TRE MAIPULATIO I -D For a general state of stress in D, shown in (a), normal and shear stresses ma be present in orthogonal directions. It can be shown that at a certain orientation, three principal normal stresses, orthogonal to each other, are equivalent to the stress condition at (a)., and are the three principal normal stresses, which are three roots of the following cubic equation: a. + b. c 0, where, a + + z b + z + z z z c z + z z z z z The maimum shear stress ma Ma of,, 0 of 7

13 4. TRAI I TWO DIMEIO If the & are normal stresses, then normal strains ε & ε can be determined from: ε ε ( E ( E µ µ ), and ) Where µ Poisson s ratio, which is a material propert olving the above two equations, we can find the normal stresses in two directions. E µ E µ ( ε ( ε + µε + µε ), and ) of 7

14 5. THEORIE OF FAILURE Uniaial Tensile tress Testing During uniaial stress testing of ductile materials, the first mechanical failure occurs b ielding, at the material s ield strength. The maimum stress that the material can withstand before breakage occurs at the ultimate tensile strength, u, and u >. For ductile materials, and u values are same in tension and compression. During uniaial stress testing of brittle materials, the first mechanical failure occurs b fracture, at the material s ultimate tensile strength u. For brittle materials, is greater than u and thus is non-eistent. Also, for brittle materials, fracture strength in compression uc is higher than fracture strength in tension ut. General D state of stress For these tes of stresses, predicting failure is not as straight forward as in case of uniaial stresses. The following theories of failure are developed to predict failure in such general state of stress. To appl these theories, first the principal normal stresses, and are computed, and then the theories are applied, with a factor of safet. For D stresses, one of the principal normal stresses 0. A positive value of principal normal stress means the principal stress is tensile, and a negative value means that the principal stress is compressive. uc ut Maimum ormal tress theor Applicable for brittle materials Maimum hear tress theor Applicable for ductile materials Maimum train Energ: Applicable for ductile materials Maimum Distortion Energ theor: Applicable for ductile materials (Also known as von Mises-Henck theor) µ uc uc ut + + ( + + ) ut + + of 7

15 6. TRE COCETRATIO If there is a sudden change in geometr at a point, then the actual stress at that point is K times the calculated stress, where K>. The value of K, the stress concentration factor, for combinations of geometr and loading te, can be obtained from the graphs given in pages 9-45 in the tet book. When the material is ductile, loads are not cclic or are not applied suddenl or the application is not working in a low temperature condition, then the effect of stress concentration factors can be ignored, or K, even if there is a sudden change in geometr. For brittle materials all tes of applications, and for ductile materials when, loads are applied suddenl, or for low temperature applications, the actual stress K times the theoreticall calculated stress. For ductile materials, when the load is cclic, a stress concentration factor K f is used which is lesser than K. K f K f can be determined from the formula q. K Here, q is the notch sensitivit factor or notch sensitivit inde, which depends on the material and its surface condition. The value of q can be obtained from handbooks and 0<q<. If the value of q is unknown, then conservativel, it is assumed q, which gives K f K. 7. DEIG FOR CYCLIC LOADIG For pure cclic stress ( r ), that varies cclicall from 0 to r to 0 to r, e r K f, where e endurance limit of the material, & Factor of safet. If the stress changes cclicall between ma and min, then the equivalent stead stress ( avg ) and equivalent cclic stress ( r ) can be given b ma + min avg and ma min r. In such situations the design equations are as followings: oderberg s equation: avg + rk f e u u Goodman s equation: avg + rk f e u u Modified Goodman s equation: avg + rk f & e avg + r K f of 7

16 8. DEIG FOR FIITE LIFE (i) For a completel reversing cclic stress r, the - curve is represented b, B A, where number of stress reversals. r A & B are determined from: log( e ) log( 0.9u ) B, where e endurance limit, and u ultimate tensile strength. A e (6 ) 0 B Once, A & B are known, for a given completel reversing stress r, the number of stress reversals before failure can be found b: r A (ii) For a combined reversing ( r ) and stead ( avg ) stress situation, the equivalent completel K f r ult reversing stress R. For this te of loading, R should substitute r of the equation shown in (i) ult avg n n n (iii) Miner s equation: , where n, n, n, are actual number of reversals with R, R, R,.. equivalent completel reversing stress levels, and,,, are maimum number of reversals before failure with R, R, R,.. equivalent completel reversing stress levels. B 4 of 7

17 9. HEAR TRE I A HAFT FOR COMBIED BEDIG & TORIO For a shaft carring bending moment (M) and torque (T), the maimum shear stress ( ma ),: ma + M 6T 6 For solid shaft: and thus ma M + T πd πd πd For hollow shaft: M 6T 6 and thus M + T 4 4 ma 4 πd ( λ ) πd ( λ ) πd ( λ ) o where, λ d d i o o 0. DEIG OF HAFT WITH CYCLIC LOAD Based on maimum distortion energ theor, the design equation is: Where, av + K r fb e + av tead normal stress av + K r ft ma + min e o ma r Cclic normal stress min K fb Fatigue stress concentration factor in bending ma + av tead shear stress min ma r Cclic shear stress min K ft Fatigue stress concentration factor in torsion Yield stress e Endurance limit Factor of afet. HAFT WITH BEDIG LOAD I TWO PLAE (i) Resolve each bending load in vertical and horizontal direction. (ii) Determine the bending moment diagram separatel for horizontal and vertical loads (iii) The resultant bending moment at an point on the beam M M + M R v H 5 of 7

18 . DEIG OF KEY, KEYWAY & COUPLIG From the HP or kw rating and rpm n, torque T can be determined using the following formula: 6,05HP 9,550,000 kw T in lb or, T mm n n Then, the tangential force F T/r, where r radius at which the tangential force is required. (i) For designing kes and kewas, tangential force transmitted b the ke is assumed to be acting at the outer diameter of the shaft, ie., r d o /, and F T/d o If, L length of the ke or the kewa a width of the ke or the kewa, and b depth of the ke Then, shear stress in the ke It is assumed, thus, F A s F a. L F a. L F F F F Bearing stress in the ke or, Ab b bl bl. L For square Ke, ab, and hence ke designed from either shearing or bearing stress, will result in same dimension, and square kes are equall strong from shearing and bearing. (ii) In rigid couplings, (a) Bolts ma fail due to shearing or bearing: Tangential force carried b each bolt pitch circle diameter of the bolts. hear stress in bolts F A s 4F π d b, T F, where, n number of bolts, & d p nd p F F Bearing stress in bolts, Ab tdb where, d b diameter of the bolt, and t flange thickness of the coupling (b) The coupling can shear from the hub/flange joint: T Tangential force F, where d h hub diameter d h hear stress F A h F π d h 6 of 7

19 . DEIG OF LEDER COMPREIO MEMBER OR COLUM For columns with an initial crookedness, the design load P can be determined from the following quadratic equation: ac P + + APcr P A + 0 Pcr i Where, P the design load ield strength A cross sectional area a initial crookedness c distance from the neutral ais to the edge of the cross section i I / A radius of gration P c r Critical load for a centrall loaded column factor of safet. Roots of quadratic equation b ± b a + b + c 0; a 4ac 7 of 7

20 4. DEFLECTIO & LOPE BY DOUBLE ITEGRATIO Epress bending moment (M) as a function of the longitudinal distance, of the shaft. i.e., M f(). If represents downward deflection, then for a shaft with constant EI, d EI M f ( )...() d Integrating both sides of (): d EI f ( ) + C d...() d Here represents the slope of the shaft, and C is the constant of integration. If the d value of the slope is known for an value of, then using those values, C can be determined from the above equation. Integrating both sides of (): EI f ( ) + C + C...() Here represents the downward deflection of the shaft, and C is another constant of integration. If the value of the deflection, is known for an value of, then using those values, C can be determined from the above equation. Usuall at the support 0. Once C and C are known, equation () and () are the slope and deflection equations for an valid value of. 5. DEFLECTIO & LOPE BY TRAI EERGY METHOD This method can take into account when diameter of the shaft is not uniform M pm f d Deflection E I M pm md and, slope θ E I Where M p Bending moment due to applied loads M f Bending moment due to a fictitious unit load applied at the point where deflection is needed. M m Bending moment due to a fictitious unit moment applied at the point where the slope is needed. The above integrations can be calculated graphicall. The integration is the volume enclosed b the M p and M f diagrams for deflection, or the volume I M p enclosed b the and M m diagrams for slope for the I entire length of the shaft. For a prismoidal solid shown, l Volume ( A 4Am A ) l For a pramid Volume A A l 8 of 7

21 6. HAFT O THREE UPPORT R, R & R (i) Remove R and find the downward deflection, of the shaft at R, due to the downward applied loads. (ii) ow, remove the applied loads and write an equation relating the upward force (R) and upward deflection () at the support point in the form RK. (iii)ow three tes of situations can arise: (a) If all three supports are in the same level, then the magnitude of the reaction force R K. (b) If R is ofet b δ from R & R, then R K(-δ) (c) If R is an elastic support with a spring constant K with no ofet at no load, then: (-RK)K R or, R K/(+K.K) (iv) When the reaction force R is known, then the problem becomes staticall determinate, and the R & R can be found b the application of static equilibrium conditions. 7. CRITICAL PEED OF A ROTATIG HAFT f g( W + W + W +...) W + W + W +... n 60 f r. p. m. ccles / sec Where, W, W, W are the vertical loads on the shaft, and,, are deflections at of the loads, W, W, W due to bending of the shaft. g acceleration due to gravit, * 86 in/sec, 9806 mm/sec 9 of 7