Torsional and Bending Stresses in Machine Parts

Transcription

1 10 n A Textbook of Machine Design C 5 H A P T E R Torsional and Bending Stresses in Machine Parts 1. Introduction.. Torsional Shear Stress.. Shafts in Series and Parallel.. Bending Stress in Straight Beams. 5. Bending Stress in Curved Beams.. Principal Stresses and Principal Planes. 7. Determination of Principal Stresses for a Member Subjected to Biaxial Stress. 8. Application of Principal Stresses in Designing Machine Members. 9. Theories of Failure under Static Load. 10. Maximum Principal or Normal Stress Theory (Rankine s Theory). 11. Maximum Shear Stress Theory (Guest s or Tresca s Theory). 1. Maximum Principal Strain Theory (Saint Venant s Theory). 1. Maximum Strain Energy Theory (Haigh s Theory). 1. Maximum Distortion Energy Theory (Hencky and Von Mises Theory). 15. Eccentric Loading Direct and Bending Stresses Combined. 1. Shear Stresses in Beams. 5.1 Introduction Sometimes machine parts are subjected to pure torsion or bending or combination of both torsion and bending stresses. We shall now discuss these stresses in detail in the following pages. 5. Tor orsional Shear Stress When a machine member is subjected to the action of two equal and opposite couples acting in parallel planes (or torque or twisting moment), then the machine member is said to be subjected to torsion. The stress set up by torsion is known as torsional shear stress. It is zero at the centroidal axis and maximum at the outer surface. Consider a shaft fixed at one end and subjected to a torque (T) at the other end as shown in Fig As a result of this torque, every cross-section of the shaft is subjected to torsional shear stress. We have discussed above that the 10

2 Torsional and Bending Stresses in Machine Parts n 11 torsional shear stress is zero at the centroidal axis and maximum at the outer surface. The maximum torsional shear stress at the outer surface of the shaft may be obtained from the following equation: τ T C. θ...(i) r J l where τ Torsional shear stress induced at the outer surface of the shaft or maximum shear stress, r Radius of the shaft, T Torque or twisting moment, J Second moment of area of the section about its polar axis or polar moment of inertia, C Modulus of rigidity for the shaft material, l Length of the shaft, and θ Angle of twist in radians on a length l. Fig Torsional shear stress. The equation (i) is known as torsion equation. It is based on the following assumptions: 1. The material of the shaft is uniform throughout.. The twist along the length of the shaft is uniform.. The normal cross-sections of the shaft, which were plane and circular before twist, remain plane and circular after twist.. All diameters of the normal cross-section which were straight before twist, remain straight with their magnitude unchanged, after twist. 5. The maximum shear stress induced in the shaft due to the twisting moment does not exceed its elastic limit value. Notes : 1. Since the torsional shear stress on any cross-section normal to the axis is directly proportional to the distance from the centre of the axis, therefore the torsional shear stress at a distance x from the centre of the shaft is given by τx τ x r. From equation (i), we know that T τ J or T τ J r r For a solid shaft of diameter (d), the polar moment of inertia, π π π J I XX + I YY d + d d π π T τ d τ d d 1

3 1 n A Textbook of Machine Design In case of a hollow shaft with external diameter (d o ) and internal diameter (d i ), the polar moment of inertia, π d J [(do ) (d i ) o ] and r π π ( do) ( d ) i T τ [( do) ( d) ] τ do 1 do π τ ( do) (1 k ) 1... Substituting, d k i d o. The expression (C J) is called torsional rigidity of the shaft.. The strength of the shaft means the maximum torque transmitted by it. Therefore, in order to design a shaft for strength, the above equations are used. The power transmitted by the shaft (in watts) is given by P π N. T π N T. ω... ω 0 0 where T Torque transmitted in N-m, and ω Angular speed in rad/s. Example 5.1. A shaft is transmitting 100 kw at 10 r.p.m. Find a suitable diameter for the shaft, if the maximum torque transmitted exceeds the mean by 5%. Take maximum allowable shear stress as 70 MPa. Solution. Given : P 100 kw W; N 10 r.p.m ; T max 1.5 T mean ; τ 70 MPa 70 N/mm Let T mean Mean torque transmitted by the shaft in N-m, and d Diameter of the shaft in mm. We know that the power transmitted (P), π N. Tmean π 10 Tmean 1.7 T 0 0 mean T mean / N-m A Helicopter propeller shaft has to bear torsional, tensile, as well as bending stresses. Note : This picture is given as additional information and is not a direct example of the current chapter.

4 Torsional and Bending Stresses in Machine Parts n 1 and maximum torque transmitted, T max N-m N-mm We know that maximum torque (T max ), π τ d 1 π 70 d 1.75 d d / or d 81.5 mm Ans. Example 5.. A steel shaft 5 mm in diameter and 1. m long held rigidly at one end has a hand wheel 500 mm in diameter keyed to the other end. The modulus of rigidity of steel is 80 GPa. 1. What load applied to tangent to the rim of the wheel produce a torsional shear of 0 MPa?. How many degrees will the wheel turn when this load is applied? Solution. Given : d 5 mm or r 17.5 mm ; l 1. m 100 mm ; D 500 mm or R 50 mm ; C 80 GPa 80 kn/mm N/mm ; τ 0 MPa 0 N/mm 1. Load applied to the tangent to the rim of the wheel Let W Load applied (in newton) to tangent to the rim of the wheel. We know that torque applied to the hand wheel, T W.R W W N-mm and polar moment of inertia of the shaft, J π d π (5) mm We know that T τ J r 50 W or W 00 N Ans Number of degrees which the wheel will turn when load W 00 N is applied Let θ Required number of degrees. We know that T C. θ J l θ Tl Ans. CJ Example 5.. A shaft is transmitting 97.5 kw at 180 r.p.m. If the allowable shear stress in the material is 0 MPa, find the suitable diameter for the shaft. The shaft is not to twist more that 1 in a length of metres. Take C 80 GPa. Solution. Given : P 97.5 kw W; N 180 r.p.m. ; τ 0 MPa 0 N/mm ; θ 1 π / rad ; l m 000 mm ; C 80 GPa N/m N/mm Let T Torque transmitted by the shaft in N-m, and d Diameter of the shaft in mm. We know that the power transmitted by the shaft (P), π NT. π 180 T T 0 0 T / N-m N-mm Now let us find the diameter of the shaft based on the strength and stiffness.

5 1 n A Textbook of Machine Design 1. Considering strength of the shaft We know that the torque transmitted (T), π τ d 1 π 0 d d d / or d 7 mm...(i). Considering stiffness of the shaft Polar moment of inertia of the shaft, J π d d We know that T C. θ J l or d 000 d d / or d 10 mm...(ii) Taking larger of the two values, we shall provide d 10 say 105 mm Ans. Example 5.. A hollow shaft is required to transmit 00 kw at 110 r.p.m., the maximum torque being 0% greater than the mean. The shear stress is not to exceed MPa and twist in a length of metres not to exceed 1. degrees. Find the external diameter of the shaft, if the internal diameter to the external diameter is /8. Take modulus of rigidity as 8 GPa. Solution. Given : P 00 kw W; N 110 r.p.m. ; T max 1. T mean ; τ MPa N/mm ; l m 000 mm ; θ 1. π / rad ; k d i / d o /8 ; C 8 GPa N/m 8 10 N/mm Let T mean Mean torque transmitted by the shaft, d o External diameter of the shaft, and d i Internal diameter of the shaft. Cutting head roller Cutting teeth made fo tungsten carbide Archimedean screw lifts soil onto conveyer belt Powerful hydraulic rams push cutting head forward Control cab houses operator Conveyor belt carries soil away A tunnel-boring machine can cut through rock at up to one kilometre a month. Powerful hydraulic rams force the machine s cutting head fowards as the rock is cut away. Note : This picture is given as additional information and is not a direct example of the current chapter.

6 We know that power transmitted by the shaft (P), Torsional and Bending Stresses in Machine Parts n π NT. mean π 110 Tmean 11.5 T 0 0 mean T mean / N-m 5 10 N-mm and maximum torque transmitted by the shaft, T max 1. T mean N-mm Now let us find the diameter of the shaft considering strength and stiffness. 1. Considering strength of the shaft We know that maximum torque transmitted by the shaft, T max 1 π τ (do ) (1 k ) π. 10 ( do) ( do) 1 8 (d o ). 10 / or d o 17.7 mm...(i). Considering stiffness of the shaft We know that polar moment of inertia of a hollow circular section, We also know that π π di J ( do) ( di) ( do) 1 d o π π ( do) (1 k ) ( do) ( do) 8 T C. θ J l or ( d o ) 000 ( d o ) (d o ) / or d o 17. mm...(ii) Taking larger of the two values, we shall provide d o 17. say 180 mm Ans. 5. Shafts in Series and Parallel When two shafts of different diameters are connected together to form one shaft, it is then known as composite shaft. If the driving torque is applied at one end and the resisting torque at the other end, then the shafts are said to be connected in series as shown in Fig. 5. (a). In such cases, each shaft transmits the same torque and the total angle of twist is equal to the sum of the angle of twists of the two shafts. Mathematically, total angle of twist, Tl. 1 Tl. θ θ 1 + θ + C1 J1 C J If the shafts are made of the same material, then C 1 C C. θ T. l T. l T l l + CJ CJ C + J J

7 1 n A Textbook of Machine Design Fig. 5.. Shafts in series and parallel. When the driving torque (T) is applied at the junction of the two shafts, and the resisting torques T 1 and T at the other ends of the shafts, then the shafts are said to be connected in parallel, as shown in Fig. 5. (b). In such cases, the angle of twist is same for both the shafts, i.e. θ 1 θ Tl 1 1 Tl T1 l C1 J1 or or CJ 1 1 C J T l1 C J and T T 1 + T If the shafts are made of the same material, then C 1 C. T1 l J1 T l1 J Example 5.5. A steel shaft ABCD having a total length of.5 m consists of three lengths having different sections as follows: AB is hollow having outside and inside diameters of 100 mm and.5 mm respectively, and BC and CD are solid. BC has a diameter of 100 mm and CD has a diameter of 87.5 mm. If the angle of twist is the same for each section, determine the length of each section. Find the value of the applied torque and the total angle of twist, if the maximum shear stress in the hollow portion is 7.5 MPa and shear modulus, C 8.5 GPa. Solution. Given: L.5 m ; d o 100 mm ; d i.5 mm ; d 100 mm ; d 87.5 mm ; τ 7.5 MPa 7.5 N/mm ; C 8.5 GPa N/mm The shaft ABCD is shown in Fig. 5.. Fig. 5. Length of each section Let l 1, l and l Length of sections AB, BC and CD respectively. We know that polar moment of inertia of the hollow shaft AB, π π J 1 [(do ) (d i ) ] [(100) (.5) ] mm Polar moment of inertia of the solid shaft BC, J π (d ) π (100) mm

8 and polar moment of inertia of the solid shaft CD, Torsional and Bending Stresses in Machine Parts n 17 J π (d ) π (87.5) mm We also know that angle of twist, θ T. l / C. J Assuming the torque T and shear modulus C to be same for all the sections, we have Angle of twist for hollow shaft AB, θ 1 T. l 1 / C. J 1 Similarly, angle of twist for solid shaft BC, θ T. l / C. J and angle of twist for solid shaft CD, θ T. l / C. J Since the angle of twist is same for each section, therefore θ 1 θ l l l T. C. 1 J 1 T. C. J or l J J Machine part of a jet engine. Note : This picture is given as additional information and is not a direct example of the current chapter....(i) Also θ 1 θ T. l1 T. l l1 J or 1.7 C. J 1 C. J l J We know that l 1 + l + l L.5 m 500 mm l l l l1 l l l or l / mm Ans. From equation (i), l l 1 / / mm Ans. and from equation (ii), l l 1 / / mm Ans. Value of the applied torque We know that the maximum shear stress in the hollow portion, τ 7.5 MPa 7.5 N/mm For a hollow shaft, the applied torque,...(ii) T π ( do) ( d ) i π (100) (.5) τ do N-mm 7900 N-m Ans. Total angle of twist When the shafts are connected in series, the total angle of twist is equal to the sum of angle of twists of the individual shafts. Mathematically, the total angle of twist, θ θ 1 + θ + θ

9 18 n A Textbook of Machine Design T. l1 T. l T. l T l1 l l + + C. J1 C. J C. J C + + J1 J J [ ] 0.0 rad / π.0 Ans. 5. Bending Stress ess in Straight Beams In engineering practice, the machine parts of structural members may be subjected to static or dynamic loads which cause bending stress in the sections besides other types of stresses such as tensile, compressive and shearing stresses. Consider a straight beam subjected to a bending moment M as shown in Fig. 5.. The following assumptions are usually made while deriving the bending formula. 1. The material of the beam is perfectly homogeneous (i.e. of the same material throughout) and isotropic (i.e. of equal elastic properties in all directions).. The material of the beam obeys Hooke s law.. The transverse sections (i.e. BC or GH) which were plane before bending, remain plane after bending also.. Each layer of the beam is free to expand or contract, independently, of the layer, above or below it. 5. The Young s modulus (E) is the same in tension and compression.. The loads are applied in the plane of bending. Fig. 5.. Bending stress in straight beams. A little consideration will show that when a beam is subjected to the bending moment, the fibres on the upper side of the beam will be shortened due to compression and those on the lower side will be elongated due to tension. It may be seen that somewhere between the top and bottom fibres there is a surface at which the fibres are neither shortened nor lengthened. Such a surface is called neutral surface. The intersection of the neutral surface with any normal cross-section of the beam is known as neutral axis. The stress distribution of a beam is shown in Fig. 5.. The bending equation is given by M σ E I y R where M Bending moment acting at the given section, σ Bending stress,

10 Torsional and Bending Stresses in Machine Parts n 19 I Moment of inertia of the cross-section about the neutral axis, y Distance from the neutral axis to the extreme fibre, E Young s modulus of the material of the beam, and R Radius of curvature of the beam. From the above equation, the bending stress is given by σ E y R Since E and R are constant, therefore within elastic limit, the stress at any point is directly proportional to y, i.e. the distance of the point from the neutral axis. Also from the above equation, the bending stress, σ M M M y I I / y Z The ratio I/y is known as section modulus and is denoted by Z. Notes : 1. The neutral axis of a section always passes through its centroid.. In case of symmetrical sections such as circular, square or rectangular, the neutral axis passes through its geometrical centre and the distance of extreme fibre from the neutral axis is y d /, where d is the diameter in case of circular section or depth in case of square or rectangular section.. In case of unsymmetrical sections such as L-section or T- section, the neutral axis does not pass through its geometrical centre. In such cases, first of all the centroid of the section is calculated and then the distance of the extreme fibres for Parts in a machine. both lower and upper side of the section is obtained. Out of these two values, the bigger value is used in bending equation. Table 5.1 (from pages 10 to 1) shows the properties of some common cross-sections. Hammer strikes cartridge to make it explode Revolving chamber holds bullets Barrel Blade foresight Vulcanized Trigger rubber handle This is the first revolver produced in a production line using interchangeable parts. Note : This picture is given as additional information and is not a direct example of the current chapter.

11 10 n A Textbook of Machine Design ble 5.1. Pr Table 5.1. Proper operties of commonly used cross-sections oss-sections. Section Area Moment of inertia *Distance from the Section modulus Radius of gyration neutral axis to the I (A) (I) Z I extreme fibre (y) y k A 1. Rectangle I xx bh. 1 h bh. xx Z k xx 0.89 h bh I yy hb. 1 b hb. yy Z k yy 0.89 b. Square b b Ixx Iyy 1 b b xx yy Z Z k xx k yy 0.89 b. Triangle bh bh. I xx h bh xx 1 Z k xx 0.58 h * The distances from the neutral axis to the bottom extreme fibre is taken into consideration.

12 Torsional and Bending Stresses in Machine Parts n 11 Section (A) (I) (y) Z I. Hollow rectangle y b (h h 1 ) b Ixx ( h h1 ) 1 h Z xx b h h 1 h 5. Hollow square b h b h Ixx Iyy 1 b b h Zxx Zyy b. Trapezoidal a + b h I xx h ( a + ab + b ) ( a + b) a + b ( a + b) h Z xx a + ab + b 1 ( a + b) k I A k xx 0.89 h h h h b + h 0. a+ b h ( a + ab+ b )

13 1 n A Textbook of Machine Design Section (A) (I) (y) Z I y k I A 7. Circle π d π d Ixx Iyy d π d Zxx Zyy kxx kyy d 8. Hollow circle π π d ( d d 1 ) I xx I yy (d d 1 ) Z xx Z yy π d d 1 d k xx k yy 1 d + d 9. Elliptical πab π Ixx a b a π Iyy ab b π Zxx a b k xx 0.5a π Zyy ab k yy 0.5b

14 Torsional and Bending Stresses in Machine Parts n Hollow elliptical Section (A) (I) (y) Z I y I xx π (ba b 1 a 1 ) a π Z xx a (ba b 1 a 1 ) π (ab a 1 b 1 ) I yy π (ab a 1 b 1 ) b π Z yy b (ab a 1 b 1 ) 11. I-section bh b 1 h 1 I xx bh b 1h 1 1 h Z xx bh b 11 h h 1. T-section Bt + (H t) a l xx Bh b ( h t ) + ah 1 h H h 1 ah + bt ( ah + bt ) Z xx Ixx ( ah + bt) ah + bt k I A k xx 1 ba b a ab a b k yy 1 ab a b ab a b k xx 0.89 bh b h bh b h k xx I xx Bt + ( H t ) a

15 1 n A Textbook of Machine Design 1. Channel Section Section (A) (I) (y) Z I y Bt + (H t) a I xx Bh b ( h t ) + ah 1 h H h 1 ah + bt ( ah + bt ) Ixx ( ah + bt) Z xx ah + bt 1. H-Section BH + bh I xx BH + bh 1 H Z xx BH + bh H 15. Cross-section BH + bh I xx Bh + bh 1 H Z xx BH + bh H k I A k xx I xx Bt + ( H t ) a k xx 0.89 BH + bh BH + bh k xx 0.89 BH + bh BH + bh

16 Torsional and Bending Stresses in Machine Parts n 15 Example 5.. A pump lever rocking shaft is shown in Fig The pump lever exerts forces of 5 kn and 5 kn concentrated at 150 mm and 00 mm from the left and right hand bearing respectively. Find the diameter of the central portion of the shaft, if the stress is not to exceed 100 MPa. Fig. 5.5 Solution. Given : σ b 100 MPa 100 N/mm Let R A and R B Reactions at A and B respectively. Taking moments about A, we have R B R B / kn N and R A (5 + 5) kn N Bending moment at C R A N-mm and bending moment at D R B N-mm We see that the maximum bending moment is at D, therefore maximum bending moment, M.1 10 N-mm. Let Section modulus, d Diameter of the shaft. Z π d d We know that bending stress (σ b ), 100 M Z The picture shows a method where sensors are used to measure torsion Note : This picture is given as additional information and is not a direct example of the current chapter d d d. 10 / or d 8. say 90 mm Ans. Example 5.7. An axle 1 metre long supported in bearings at its ends carries a fly wheel weighing 0 kn at the centre. If the stress (bending) is not to exceed 0 MPa, find the diameter of the axle. Solution. Given : L 1 m 1000 mm ; W 0 kn 0 10 N ; σ b 0 MPa 0 N/mm The axle with a flywheel is shown in Fig. 5.. Let d Diameter of the axle in mm.

17 1 n A Textbook of Machine Design Section modulus, Z π d d Maximum bending moment at the centre of the axle, W. L M N-mm We know that bending stress (σ b ), M Z d d d / or d 108. say 110 mm Ans. Example 5.8. A beam of uniform rectangular cross-section is fixed at one end and carries an electric motor weighing 00 N at a distance of 00 mm from the fixed end. The maximum bending stress in the beam is 0 MPa. Find the width and depth of the beam, if depth is twice that of width. Solution. Given: W 00 N ; L 00 mm ; σ b 0 MPa 0 N/mm ; h b The beam is shown in Fig Fig. 5.7 Let b Width of the beam in mm, and h Depth of the beam in mm. Section modulus, b. h b ( b) b Z mm Maximum bending moment (at the fixed end), M W.L N-mm We know that bending stress (σ b ), Fig. 5. M Z b b b / or b 1.5 mm Ans. and h b 1.5 mm Ans. Example 5.9. A cast iron pulley transmits 10 kw at 00 r.p.m. The diameter of the pulley is 1. metre and it has four straight arms of elliptical cross-section, in which the major axis is twice the minor axis. Determine the dimensions of the arm if the allowable bending stress is 15 MPa. Solution. Given : P 10 kw W ; N 00 r.p.m ; D 1. m 100 mm or R 00 mm ; σ b 15 MPa 15 N/mm Let T Torque transmitted by the pulley. We know that the power transmitted by the pulley (P), π N. T π 00 T T 0 0 T / 8 N-m 8 10 N-mm

18 Torsional and Bending Stresses in Machine Parts n 17 Since the torque transmitted is the product of the tangential load and the radius of the pulley, therefore tangential load acting on the pulley T N R 00 Since the pulley has four arms, therefore tangential load on each arm, W 9.7/ 99. N and maximum bending moment on the arm, M W R N-mm Let b Minor axis in mm, and a Major axis in mm b b Section modulus for an elliptical cross-section, Z π a b π (b) b π b mm We know that bending stress (σ b ), M Z π b b or b 18 9/15 1 or b 10.8 mm Minor axis, b mm Ans. and major axis, a b mm Ans. 5.5 Bending Stress in Curved Beams...(Given) We have seen in the previous article that for the straight beams, the neutral axis of the section coincides with its centroidal axis and the stress distribution in the beam is linear. But in case of curved beams, the neutral axis of the cross-section is shifted towards the centre of curvature of the beam causing a non-linear (hyperbolic) distribution of stress, as shown in Fig It may be noted that the neutral axis lies between the centroidal axis and the centre of curvature and always occurs within the curved beams. The application of curved beam principle is used in crane hooks, chain links and frames of punches, presses, planers etc. Fig Bending stress in a curved beam. Consider a curved beam subjected to a bending moment M, as shown in Fig In finding the bending stress in curved beams, the same assumptions are used as for straight beams. The general expression for the bending stress (σ b ) in a curved beam at any fibre at a distance y from the neutral

19 18 n A Textbook of Machine Design axis, is given by M y σ b A. e Rn y where M Bending moment acting at the given section about the centroidal axis, A Area of cross-section, e Distance from the centroidal axis to the neutral axis R R n, R Radius of curvature of the centroidal axis, R n Radius of curvature of the neutral axis, and y Distance from the neutral axis to the fibre under consideration. It is positive for the distances towards the centre of curvature and negative for the distances away from the centre of curvature. Notes : 1. The bending stress in the curved beam is zero at a point other than at the centroidal axis.. If the section is symmetrical such as a circle, rectangle, I-beam with equal flanges, then the maximum bending stress will always occur at the inside fibre.. If the section is unsymmetrical, then the maximum bending stress may occur at either the inside fibre or the outside fibre. The maximum bending stress at the inside fibre is given by M. yi σ bi A. e. Ri where y i Distance from the neutral axis to the inside fibre R n R i, and R i Radius of curvature of the inside fibre. The maximum bending stress at the outside fibre is given by M. yo σ bo A. e. Ro where y o Distance from the neutral axis to the outside fibre R o R n, and R o Radius of curvature of the outside fibre. It may be noted that the bending stress at the inside fibre is tensile while the bending stress at the outside fibre is compressive.. If the section has an axial load in addition to bending, then the axial or direct stress (σ d ) must be added algebraically to the bending stress, in order to obtain the resultant stress on the section. In other words, Resultant stress, σ σ d ± σ b The following table shows the values of R n and R for various commonly used cross-sections in curved beams. Table 5.. Values of R n and R for var arious commonly used cross-section in curved ed beams. Section Values of R n and R h Rn R log o Ri h R Ri +

20 Torsional and Bending Stresses in Machine Parts n 19 Section Values of R n and R R n Ro + Ri R R + i d bi + bo h R n br i o bori R log o e ( bi bo) h Ri h ( b + ) + i b R R o i ( bi + bo) Rn br h h R Ri + 1 bi h R log e bi Ri i o o ( b t)( ti + to) + t. h Rn R + log i ti R + log o R +.log o t b o e e t e + Ri Ro to Ri ti h. t + ti ( b t) + ( b t) to( h to) R R + i ht. + ( b t)( ti + to)

21 10 n A Textbook of Machine Design Section Values of R n and R t ( ) +. i bi t t h Rn R + ( ) log i ti R b +.log o i t e t e Ri Ri R Ri h t + ti ( bi t) ht. + ti( bi t) t ( ) + ( ) +. i bi t to bo t th Rn R + log i ti R + log o to R b + log o i e t e bo e + Ri Ri ti Ro to R Ri ht+ ti ( bi t) + ( bo t) to( h to) ti( bi t) + to( bo t) + t. h Example The frame of a punch press is shown in Fig Find the stresses at the inner and outer surface at section X-X of the frame, if W 5000 N. Solution. Given : W 5000 N ; b i 18 mm ; b o mm ; h 0 mm ; R i 5 mm ; R o mm We know that area of section at X-X, A 1 (18 + ) 0 80 mm The various distances are shown in Fig We know that radius of curvature of the neutral axis, bi + bo h R n br i o bo Ri Ro log e ( bi bo) h Ri log e (18 ) mm ( ) 1 Fig. 5.9

22 and radius of curvature of the centroidal axis, Torsional and Bending Stresses in Machine Parts n 11 h ( bi + bo) 0 (18 + ) R Ri mm ( bi + bo) (18+ ) mm Distance between the centroidal axis and neutral axis, e R R n mm and the distance between the load and centroidal axis, x R mm Bending moment about the centroidal axis, M W.x N-mm The section at X-X is subjected to a direct tensile load of W 5000 N and a bending moment of M N-mm. We know that direct tensile stress at section X-X, σ t W N/mm 10. MPa A 80 Fig Distance from the neutral axis to the inner surface, y i R n R i mm Distance from the neutral axis to the outer surface, y o R o R n mm We know that maximum bending stress at the inner surface, σ bi 87. MPa (tensile) and maximum bending stress at the outer surface, σ b0 M. yi N/mm A. e. R i M. yo N/mm A. e. R o 09. MPa (compressive)

23 1 n A Textbook of Machine Design Resultant stress on the inner surface σ t + σ bi MPa (tensile) Ans. and resultant stress on the outer surface, σ t σ bo MPa MPa (compressive) Ans. A big crane hook Example The crane hook carries a load of 0 kn as shown in Fig The section at X-X is rectangular whose horizontal side is 100 mm. Find the stresses in the inner and outer fibres at the given section. Solution. Given : W 0 kn 0 10 N; R i 50 mm ; R o 150 mm ; h 100 mm ; b 0 mm We know that area of section at X-X, A b.h mm The various distances are shown in Fig We know that radius of curvature of the neutral axis, h R n mm Ro loge loge R i 50 and radius of curvature of the centroidal axis, h 100 R R i mm Distance between the centroidal axis and neutral axis, e R R n mm and distance between the load and the centroidal axis, x R 100 mm Bending moment about the centroidal axis, M W x N-mm

24 Torsional and Bending Stresses in Machine Parts n 1 The section at X-X is subjected to a direct tensile load of W 0 10 N and a bending moment of M 10 N-mm. We know that direct tensile stress at section X-X, σ t W A N/mm 10 MPa 000 Fig Fig. 5.1 We know that the distance from the neutral axis to the inside fibre, y i R n R i mm and distance from the neutral axis to outside fibre, y o R o R n mm Maximum bending stress at the inside fibre, M. yi σ bi 9 N/mm A. e. Ri MPa (tensile) and maximum bending stress at the outside fibre, Resultant stress at the inside fibre and resultant stress at the outside fibre M. yo σ bo N/mm A. e. Ro MPa (compressive) σ t + σ bi MPa (tensile) Ans. σ t σ bo 10 MPa MPa (compressive) Ans. Example 5.1. A C-clamp is subjected to a maximum load of W, as shown in Fig If the maximum tensile stress in the clamp is limited to 10 MPa, find the value of load W. Solution. Given : σ t(max) 10 MPa 10 N/mm ; R i 5 mm ; R o mm ; b i 19 mm ; t i mm ; t mm ; h 5 mm We know that area of section at X-X, A mm

25 1 n A Textbook of Machine Design The various distances are shown in Fig We know that radius of curvature of the neutral axis, R n ti ( bi t) + t. h Ri + ti Ro ( bi t) loge t loge R + i R i (19 ) (19 ) loge + loge mm and radius of curvature of the centroidal axis, i h. t t ( bi t) R Ri + h. t + ti ( bi t) (19 ) (19 ) mm Distance between the centroidal axis and neutral axis, e R R n mm and distance between the load W and the centroidal axis, x 50 + R mm Bending moment about the centroidal axis, M W.x W W N-mm Fig. 5.1 Fig. 5.1 The section at X-X is subjected to a direct tensile load of W and a bending moment of 8. W. The maximum tensile stress will occur at point P (i.e. at the inner fibre of the section). Distance from the neutral axis to the point P, y i R n R i mm

26 Direct tensile stress at section X-X, W W σ t W N/mm A 1 and maximum bending stress at point P, Torsional and Bending Stresses in Machine Parts n 15 M. yi 8. W. σ bi W N/mm A. e. Ri We know that the maximum tensile stress σ t(max), 10 σ t + σ bi W W 0.1 W W 10/ N Ans. Note : We know that distance from the neutral axis to the outer fibre, y o R o R n mm Maximum bending stress at the outer fibre, M. yo 8. W 18. σ bo 0.1 W A. e. Ro and maximum stress at the outer fibre, σ t σ bo W 0.1 W 0.15 W N/mm 0.15 W N/mm (compressive) From above we see that stress at the outer fibre is larger in this case than at the inner fibre, but this stress at outer fibre is compressive. 5. Principal Stresses and Principal Planes In the previous chapter, we have discussed about the direct tensile and compressive stress as well as simple shear. Also we have always referred the stress in a plane which is at right angles to the line of action of the force. But it has been observed that at any point in a strained material, there are three planes, mutually perpendicular to each other which carry direct stresses only and no shear stress. It may be noted that out of these three direct stresses, one will be maximum and the other will be minimum. These Field structure (magnet) perpendicular planes which have no shear stress are known as principal planes and the direct stresses along these planes are known as principal stresses. The planes on which the maximum shear stress act are known as planes of maximum shear. Big electric generators undergo high torsional stresses. Armature containing several coils The ends of the coils are arranged round the shaft

27 1 n A Textbook of Machine Design 5.7 Determina mination of Principal Stresses for a Member Subjected to Bi-axial Stress When a member is subjected to bi-axial stress (i.e. direct stress in two mutually perpendicular planes accompanied by a simple shear stress), then the normal and shear stresses are obtained as discussed below: Consider a rectangular body ABCD of uniform cross-sectional area and unit thickness subjected to normal stresses σ 1 and σ as shown in Fig (a). In addition to these normal stresses, a shear stress τ also acts. It has been shown in books on Strength of Materials that the normal stress across any oblique section such as EF inclined at an angle θ with the direction of σ, as shown in Fig (a), is given by σ 1 + σ σ 1 + σ σ t + cos θ + τ sin θ...(i) and tangential stress (i.e. shear stress) across the section EF, τ 1 1 (σ 1 σ ) sin θ τ cos θ...(ii) Since the planes of maximum and minimum normal stress (i.e. principal planes) have no shear stress, therefore the inclination of principal planes is obtained by equating τ 1 0 in the above equation (ii), i.e. 1 (σ 1 σ ) sin θ τ cos θ 0 τ tan θ...(iii) σ σ 1 (a) Direct stress in two mutually prependicular planes accompanied by a simple shear stress. (b) Direct stress in one plane accompanied by a simple shear stress. Fig Principal stresses for a member subjected to bi-axial stress. We know that there are two principal planes at right angles to each other. Let θ 1 and θ be the inclinations of these planes with the normal cross-section. From Fig. 5.1, we find that sin θ ± τ 1 ( σ σ ) + τ

28 Torsional and Bending Stresses in Machine Parts n 17 sin θ 1 and sin θ Also cos θ + ± τ 1 ( σ σ ) + τ τ 1 ( σ σ ) + τ σ σ 1 1 ( σ σ ) + τ σ1 σ cos θ 1 + ( σ1 σ ) + τ Fig. 5.1 σ1 σ and cos θ ( σ1 σ ) + τ The maximum and minimum principal stresses may now be obtained by substituting the values of sin θ and cos θ in equation (i). Maximum principal (or normal) stress, 1 σ t1 and minimum principal (or normal) stress, σ+σ 1 + ( σ 1 σ ) + τ...(iv) σ 1+ σ 1 σ t ( σ1 σ ) + τ...(v) The planes of maximum shear stress are at right angles to each other and are inclined at 5 to the principal planes. The maximum shear stress is given by one-half the algebraic difference between the principal stresses, i.e. 1 τ max σ σ 1 ( σ 1 σ ) + τ...(vi) A Boring mill. Note : This picture is given as additional information and is not a direct example of the current chapter.

29 18 n A Textbook of Machine Design Notes: 1. When a member is subjected to direct stress in one plane accompanied by a simple shear stress as shown in Fig (b), then the principal stresses are obtained by substituting σ 0 in equation (iv), (v) and (vi). σ t1 σ ( σ 1) + τ σ 1 1 ( σ 1) + τ σ t 1 and τ max ( σ 1) + τ 1. In the above expression of σ t, the value of ( 1) σ 1 σ + τ is more than. Therefore the nature of σ t will be opposite to that of σ t1, i.e. if σ t1 is tensile then σ t will be compressive and vice-versa. 5.8 Application of Principal Stresses in Designing Machine Members There are many cases in practice, in which machine members are subjected to combined stresses due to simultaneous action of either tensile or compressive stresses combined with shear stresses. In many shafts such as propeller shafts, C-frames etc., there are direct tensile or compressive stresses due to the external force and shear stress due to torsion, which acts normal to direct tensile or compressive stresses. The shafts like crank shafts, are subjected simultaneously to torsion and bending. In such cases, the maximum principal stresses, due to the combination of tensile or compressive stresses with shear stresses may be obtained. The results obtained in the previous article may be written as follows: 1. Maximum tensile stress, σt 1 σ t(max) + ( σ t ) + τ. Maximum compressive stress, σc 1 σ c(max) + ( σ c ) + τ. Maximum shear stress, 1 τ max ( σ t ) + τ where σ t Tensile stress due to direct load and bending, σ c Compressive stress, and τ Shear stress due to torsion. Notes : 1. When τ 0 as in the case of thin cylindrical shell subjected in internal fluid pressure, then σ t (max) σ t. When the shaft is subjected to an axial load (P) in addition to bending and twisting moments as in the propeller shafts of ship and shafts for driving worm gears, then the stress due to axial load must be added to the bending stress (σ b ). This will give the resultant tensile stress or compressive stress (σ t or σ c ) depending upon the type of axial load (i.e. pull or push). Example 5.1. A hollow shaft of 0 mm outer diameter and 5 mm inner diameter is subjected to a twisting moment of 10 N-m, simultaneously, it is subjected to an axial thrust of 10 kn and a bending moment of 80 N-m. Calculate the maximum compressive and shear stresses. Solution. Given: d o 0 mm ; d i 5 mm ; T 10 N-m N-mm ; P 10 kn N; M 80 N-m N-mm We know that cross-sectional area of the shaft, π π A ( do) ( d ) i (0) (5) 7 mm

30 Torsional and Bending Stresses in Machine Parts n 19 Direct compressive stress due to axial thrust, P σ o 1.05 N/mm 1.05 MPa A 7 Section modulus of the shaft, π ( do) ( d ) i π (0) (5) Z 55 mm do 0 Bending stress due to bending moment, M σ b 15.0 N/mm 15.0 MPa (compressive) Z 55 and resultant compressive stress, σ c σ b + σ o N/mm 8.07 MPa We know that twisting moment (T), π ( do) ( d ) i π (0) (5) τ τ τ 1 do 1 0 τ / N/mm 11.7 MPa Maximum compressive stress We know that maximum compressive stress, σc 1 + ( σ c ) + τ σ c(max) (8.07) + (11.7) MPa Ans. Maximum shear stress We know that maximum shear stress, τ max 1 ( ) 1 (8.07) (11.7) 18 MPa σ c + τ + Ans. Example 5.1. A shaft, as shown in Fig. 5.17, is subjected to a bending load of kn, pure torque of 1000 N-m and an axial pulling force of 15 kn. Calculate the stresses at A and B. Solution. Given : W kn 000 N ; T 1000 N-m 1 10 N-mm ; P 15 kn N; d 50 mm; x 50 mm We know that cross-sectional area of the shaft, A π d Fig π (50) 19 mm Tensile stress due to axial pulling at points A and B, P σ o 7. N/mm 7. MPa A 19 Bending moment at points A and B, M W.x N-mm

31 150 n A Textbook of Machine Design Section modulus for the shaft, Z π d π (50) mm Bending stress at points A and B, σ b M Z N/mm 1.1 MPa This bending stress is tensile at point A and compressive at point B. Resultant tensile stress at point A, This picture shows a machine component inside a σ A σ b + σ o crane 8.7 MPa Note : This picture is given as additional information and and resultant compressive stress at point B, is not a direct example of the current chapter. σ B σ b σ o MPa We know that the shear stress at points A and B due to the torque transmitted, 1 T τ 0.7 N/mm π 0.7 MPa... T τ d π d π (50) 1 Stresses at point A We know that maximum principal (or normal) stress at point A, A σ A(max) σ 1 + ( σ A ) + τ (8.7) + (0.7) MPa (tensile) Ans. Minimum principal (or normal) stress at point A, A σ A(min) σ 1 ( σ A) + τ MPa 18.9 MPa (compressive ) Ans. and maximum shear stress at point A, Stresses at point B τ A(max) 1 1 ( σ A ) + τ (8.7) + (0.7) 5. MPa Ans. We know that maximum principal (or normal) stress at point B, σ B(max) 1 σ B + ( σ B) + τ (5.) + (0.7) MPa (compressive) Ans.

32 Minimum principal (or normal) stress at point B, σb 1 σ B(min) ( σ B) + τ MPa MPa (tensile) Ans. and maximum shear stress at point B, Torsional and Bending Stresses in Machine Parts n 151 τ B(max) 1 1 ( B ) σ + τ (5.) + (0.7) 8.7 MPa Ans. Example An overhang crank with pin and shaft is shown in Fig A tangential load of 15 kn acts on the crank pin. Determine the maximum principal stress and the maximum shear stress at the centre of the crankshaft bearing. Fig Solution. Given : W 15 kn N; d 80 mm ; y 10 mm ; x 10 mm Bending moment at the centre of the crankshaft bearing, M W x N-mm and torque transmitted at the axis of the shaft, T W y N-mm We know that bending stress due to the bending moment, M M π σ b... Z d Z π d N/mm 5.8 MPa π (80) and shear stress due to the torque transmitted, 1 T τ 0.9 N/mm 0.9 MPa π d π (80) Maximum principal stress We know that maximum principal stress, σt 1 σ t(max) + ( σ t ) + τ (5.8) + (0.9)... (Substituting σ t σ b ) MPa Ans.

33 15 n A Textbook of Machine Design Maximum shear stress We know that maximum shear stress, τ max 1 ( ) 1 (5.8) (0.9) σ t + τ MPa Ans. 5.9 Theories ies of Failur ailure e Under Static tic Load It has already been discussed in the previous chapter that strength of machine members is based upon the mechanical properties of the materials used. Since these properties are usually determined from simple tension or compression tests, therefore, predicting failure in members subjected to uniaxial stress is both simple and straight-forward. But the problem of predicting the failure stresses for members subjected to bi-axial or tri-axial stresses is much more complicated. In fact, the problem is so complicated that a large number of different theories have been formulated. The principal theories of failure for a member subjected to bi-axial stress are as follows: 1. Maximum principal (or normal) stress theory (also known as Rankine s theory).. Maximum shear stress theory (also known as Guest s or Tresca s theory).. Maximum principal (or normal) strain theory (also known as Saint Venant theory).. Maximum strain energy theory (also known as Haigh s theory). 5. Maximum distortion energy theory (also known as Hencky and Von Mises theory). Since ductile materials usually fail by yielding i.e. when permanent deformations occur in the material and brittle materials fail by fracture, therefore the limiting strength for these two classes of materials is normally measured by different mechanical properties. For ductile materials, the limiting strength is the stress at yield point as determined from simple tension test and it is, assumed to be equal in tension or compression. For brittle materials, the limiting strength is the ultimate stress in tension or compression Maximum Principal or Normal Stress Theory (Rankine s Theory) According to this theory, the failure or yielding occurs at a point in a member when the maximum principal or normal stress in a bi-axial stress system reaches the limiting strength of the material in a simple tension test. Since the limiting strength for ductile materials is yield point stress and for brittle materials (which do not have well defined yield point) the limiting strength is ultimate stress, therefore according Limestone Coke Waste gases are removed Hot air blasted into furnace Iron ore Pig iron and scrap steel are poured into converter Mixed raw maerials Oxygen is blown into molten metal The molten steel can then be tapped off. Converter pours out molten steel Molten slag removed Iron Ladle Molten pig iron Molten steel fluid can be poured into moulds or cast while fuild Oxygen burns off carbon to turn the pig iron into steel Pig iron is made from iron ore in a blast furnace. It is a brittle form of iron that contains -5 per cent carbon. Note : This picture is given as additional information and is not a direct example of the current chapter.

34 Torsional and Bending Stresses in Machine Parts n 15 to the above theory, taking factor of safety (F.S.) into consideration, the maximum principal or normal stress (σ t1 ) in a bi-axial stress system is given by σ yt σ t1, for ductile materials FS.. σ u, for brittle materials FS.. where σ yt Yield point stress in tension as determined from simple tension test, and σ u Ultimate stress. Since the maximum principal or normal stress theory is based on failure in tension or compression and ignores the possibility of failure due to shearing stress, therefore it is not used for ductile materials. However, for brittle materials which are relatively strong in shear but weak in tension or compression, this theory is generally used. Note : The value of maximum principal stress (σ t1 ) for a member subjected to bi-axial stress system may be determined as discussed in Art Maximum Shear Stress ess Theory y (Guest s s or Tresca esca s Theory) y) According to this theory, the failure or yielding occurs at a point in a member when the maximum shear stress in a bi-axial stress system reaches a value equal to the shear stress at yield point in a simple tension test. Mathematically, τ max τ yt /F.S....(i) where τ max Maximum shear stress in a bi-axial stress system, τ yt Shear stress at yield point as determined from simple tension test, and F.S. Factor of safety. Since the shear stress at yield point in a simple tension test is equal to one-half the yield stress in tension, therefore the equation (i) may be written as σ yt τ max FS.. This theory is mostly used for designing members of ductile materials. Note: The value of maximum shear stress in a bi-axial stress system (τ max ) may be determined as discussed in Art Maximum Principal Strain Theory (Saint Venant enant s Theory) According to this theory, the failure or yielding occurs at a point in a member when the maximum principal (or normal) strain in a bi-axial stress system reaches the limiting value of strain (i.e. strain at yield point) as determined from a simple tensile test. The maximum principal (or normal) strain in a bi-axial stress system is given by σt1 σt ε max E m. E According to the above theory, σt1 σ σ t yt ε max ε...(i) E m. E E F. S. where σ t1 and σ t Maximum and minimum principal stresses in a bi-axial stress system, ε Strain at yield point as determined from simple tension test, 1/m Poisson s ratio, E Young s modulus, and F.S. Factor of safety.

35 15 n A Textbook of Machine Design From equation (i), we may write that σ σ t yt σ t1 m F. S. This theory is not used, in general, because it only gives reliable results in particular cases. 5.1 Maximum Strain Energy gy Theory y (Haigh s Theory) y) According to this theory, the failure or yielding occurs at a point in a member when the strain energy per unit volume in a bi-axial stress system reaches the limiting strain energy (i.e. strain energy at the yield point ) per unit volume as determined from simple tension test. This double-decker A 80 has a passenger capacity of 555. Its engines and parts should be robust which can bear high torsional and variable stresses. We know that strain energy per unit volume in a bi-axial stress system, 1 σ t1 σt U 1 ( t1) ( t) E σ + σ m and limiting strain energy per unit volume for yielding as determined from simple tension test, 1 σ yt U E F. S. According to the above theory, U 1 U. 1 σ t1 σt 1 σ yt ( t1) ( t) E σ + σ m E F. S. σ t1 σ σ t yt or (σ t1 ) + (σ t ) m F. S. This theory may be used for ductile materials. 5.1 Maximum Distortion tion Energy gy Theory (Hencky y and Von Mises Theory) y) According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy (also called shear strain energy) per unit volume in a bi-axial stress system reaches the limiting distortion energy (i.e. distortion energy at yield point) per unit volume as determined from a simple tension test. Mathematically, the maximum distortion energy theory for yielding is expressed as (σ t1 ) + (σ t ) σ yt σ t1 σ t FS.. This theory is mostly used for ductile materials in place of maximum strain energy theory. Note: The maximum distortion energy is the difference between the total strain energy and the strain energy due to uniform stress.

36 Torsional and Bending Stresses in Machine Parts n 155 Example 5.1. The load on a bolt consists of an axial pull of 10 kn together with a transverse shear force of 5 kn. Find the diameter of bolt required according to 1. Maximum principal stress theory;. Maximum shear stress theory;. Maximum principal strain theory;. Maximum strain energy theory; and 5. Maximum distortion energy theory. Take permissible tensile stress at elastic limit 100 MPa and poisson s ratio 0.. Solution. Given : P t1 10 kn ; P s 5 kn ; σ t(el) 100 MPa 100 N/mm ; 1/m 0. Let d Diameter of the bolt in mm. Cross-sectional area of the bolt, π A d d mm We know that axial tensile stress, P t σ 1 kn/mm A d d and transverse shear stress, P s 5.5 τ kn/mm A d d 1. According to maximum principal stress theory We know that maximum principal stress, 1 σ t1 σ+σ 1 + ( σ1 σ ) + τ 1 σ 1 + ( σ 1) + τ...( σ 0) d d d d d kn/mm N/mm d + + d d According to maximum principal stress theory, 15 5 σ t1 σ t(el) or 100 d d 15 5/ or d 1. mm Ans.. According to maximum shear stress theory We know that maximum shear stress, τ max...( σ 0) 1 1 ( σ 1 σ ) + τ ( σ 1) + τ d d d 9 kn/mm 9000 N/mm d d According to maximum shear stress theory, σ tel ( ) τ max or 50 d d 9000 / or d 1. mm Ans.

37 15 n A Textbook of Machine Design. According to maximum principal strain theory We know that maximum principal stress, 1 σ t1 σ ( 1) σ + τ...(as calculated before) d and minimum principal stress, 1 σ t σ 1 ( σ 1) + τ d d d d d kn/mm d d 5 N/mm Front view of a jet engine. The rotors undergo high torsional and bending stresses. d We know that according to maximum principal strain theory, σt1 σ σ t tel ( ) σt or σ t1 σt( el) E me E m or 100 d d d d 1 15 / or d 1.7 mm Ans.. According to maximum strain energy theory We know that according to maximum strain energy theory, t t (σ t1 ) + (σ t ) σ 1 σ [σ t(el) ] m (100) d + d d d d d d or 1 d d d d d 7 or d 1.78 mm Ans. 5. According to maximum distortion energy theory According to maximum distortion energy theory, (σ t1 ) + (σ t ) σ t1 σ t [σ t(el) ] (100) d d d d d d d or 1 d d d d d 91 or d 1. mm Ans.

38 Torsional and Bending Stresses in Machine Parts n 157 Example A cylindrical shaft made of steel of yield strength 700 MPa is subjected to static loads consisting of bending moment 10 kn-m and a torsional moment 0 kn-m. Determine the diameter of the shaft using two different theories of failure, and assuming a factor of safety of. Take E 10 GPa and poisson's ratio 0.5. Solution. Given : σ yt 700 MPa 700 N/mm ; M 10 kn-m N-mm ; T 0 kn-m 0 10 N-mm ; F.S. ; E 10 GPa N/mm ; 1/m 0.5 Let d Diameter of the shaft in mm. First of all, let us find the maximum and minimum principal stresses. We know that section modulus of the shaft π Z d d mm Bending (tensile) stress due to the bending moment, M σ 1 N/mm Z d d and shear stress due to torsional moment, 1 T τ N/mm π d π d d We know that maximum principal stress, σ t1 σ+σ ( σ1 σ ) + τ σ ( σ 1) + τ...( σ 0) d d d (101.8) (15.8) d + d N/mm d d d and minimum principal stress, σ t σ+σ 1 1 ( σ1 σ ) + τ σ 1 1 ( σ 1) + τ...( σ 0) N/mm d d d Let us now find out the diameter of shaft (d) by considering the maximum shear stress theory and maximum strain energy theory. 1. According to maximum shear stress theory We know that maximum shear stress, σ t1+ σ 1 t τ max + d d d We also know that according to maximum shear stress theory, σ yt τ max or 175 FS.. d d / or d 97. mm Ans.

39 158 n A Textbook of Machine Design Note: The value of maximum shear stress (τ max ) may also be obtained by using the relation, τ max 1 ( σ 1) + τ d d 1 10 (101.8) + (15.8) d N/mm...(Same as before) d d. According to maximum strain energy theory We know that according to maximum strain energy theory, 1 σ t1 σt 1 σ yt ( t1) ( t) E σ + σ m E F. S. or (σ t1 ) + (σ t ) σ t1 σ t σ yt m F. S d d d d or d d d d d / or d 90.8 mm Ans. Example A mild steel shaft of 50 mm diameter is subjected to a bending moment of 000 N-m and a torque T. If the yield point of the steel in tension is 00 MPa, find the maximum value of this torque without causing yielding of the shaft according to 1. the maximum principal stress;. the maximum shear stress; and. the maximum distortion strain energy theory of yielding. Solution. Given: d 50 mm ; M 000 N-m 10 N-mm ; σ yt 00 MPa 00 N/mm Let T Maximum torque without causing yielding of the shaft, in N-mm. 1. According to maximum principal stress theory We know that section modulus of the shaft, π π Z d (50) 1 7 mm Bending stress due to the bending moment, σ 1 M 10 Z N/mm and shear stress due to the torque, 1 T 1 T τ d T N/mm π π (50) π... T τ d 1 We know that maximum principal stress, σ t1 σ ( σ 1) + τ (1) + ( T )

40 Torsional and Bending Stresses in Machine Parts n 159 Minimum principal stress, and maximum shear stress, T N/mm 1 σ t σ 1 ( σ 1) + τ 1 1 (1) + ( T ) τ max T N/mm 1 1 ( σ 1) + τ (1) + ( T ) T N/mm We know that according to maximum principal stress theory, σ t1 σ yt...(taking F.S. 1) T T ( ) 1 0 T or T N-mm 118 N-m Ans.. According to maximum shear stress theory We know that according to maximum shear stress theory, τ max τ yt 9 00 σ yt T T (100) T T 1 10 N-mm 1 N-m Ans.. According to maximum distortion strain energy theory We know that according to maximum distortion strain energy theory (σ t1 ) + (σ t ) σ t1 σ t (σ yt ) T T T T (00) 9 9 (81.5) T (81.5) T (00) (81.5) T (00) T T T N-mm 17 N-m Ans.

41 10 n A Textbook of Machine Design 5.15 Eccentric ic Loading - Direct and Bending Stresses Combined An external load, whose line of action is parallel but does not coincide with the centroidal axis of the machine component, is known as an eccentric load. The distance between the centroidal axis of the machine component and the eccentric load is called eccentricity and is generally denoted by e. The examples of eccentric loading, from the subject point of view, are C-clamps, punching machines, brackets, offset connecting links etc. Fig Eccentric loading. Consider a short prismatic bar subjected to a compressive load P acting at an eccentricity of e as shown in Fig (a). Let us introduce two forces P 1 and P along the centre line or neutral axis equal in magnitude to P, without altering the equilibrium of the bar as shown in Fig (b). A little consideration will show that the force P 1 will induce a direct compressive stress over the entire cross-section of the bar, as shown in Fig (c). The magnitude of this direct compressive stress is given by P σ o 1 A or P, where A is the cross-sectional area of the bar. A The forces P 1 and P will form a couple equal to P e which will cause bending stress. This bending stress is compressive at the edge AB and tensile at the edge CD, as shown in Fig (d). The magnitude of bending stress at the edge AB is given by P. e. yc σ b (compressive) I and bending stress at the edge CD, P. e. yt σ b (tensile) I

42 Torsional and Bending Stresses in Machine Parts n 11 where y c and y t Distances of the extreme fibres on the compressive and tensile sides, from the neutral axis respectively, and I Second moment of area of the section about the neutral axis i.e. Y-axis. According to the principle of superposition, the maximum or the resultant compressive stress at the edge AB, P. e. yc P * M P σ c + + σ b + σo I A Z A and the maximum or resultant tensile stress at the edge CD, P. e. yt P M P σ t σ b + σo I A Z A The resultant compressive and tensile stress diagram is shown in Fig (e). Air in Fuel injector Exhaust Turbines Turbine shaft Combustion chamber Spark plug Fuel line In a gas-turbine system, a compressor forces air into a combustion chamber. There, it mixes with fuel. The mixture is ignited by a spark. Hot gases are produced when the fuel burns. They expand and drive a series of fan blades called a turbine. Note : This picture is given as additional information and is not a direct example of the current chapter. Notes: 1. When the member is subjected to a tensile load, then the above equations may be used by interchanging the subscripts c and t.. When the direct stress σ o is greater than or equal to bending stress σ b, then the compressive stress shall be present all over the cross-section.. When the direct stress σ o is less than the bending stress σ b, then the tensile stress will occur in the left hand portion of the crosssection and compressive stress on the right hand portion of the crosssection. In Fig. 5.19, the stress diagrams are drawn by taking σ o less than σ b. In case the eccentric load acts with eccentricity about two axes, as shown in Fig. 5.0, then the total stress at the extreme fibre P P. e. P. ey. y x x ± ± A I I XX YY Compressor I I * We know that bending moment, M P.e and section modulus, Z y yc or yt Bending stress, σ b M / Z Fig Eccentric load with eccentricity about two axes.

43 1 n A Textbook of Machine Design Example A rectangular strut is 150 mm wide and 10 mm thick. It carries a load of 180 kn at an eccentricity of 10 mm in a plane bisecting the thickness as shown in Fig Find the maximum and minimum intensities of stress in the section. Solution. Given : b 150 mm ; d 10 mm ; P 180 kn N; e 10 mm We know that cross-sectional area of the strut, A b.d mm Direct compressive stress, P σ o A N/mm 10 MPa Section modulus for the strut, YY. /1. Z I d b d b y 10 (150) b/ mm Bending moment, M P.e N-mm M Bending stress, σ b Z Fig. 5.1 N/mm MPa Since σ o is greater than σ b, therefore the entire cross-section of the strut will be subjected to compressive stress. The maximum intensity of compressive stress will be at the edge AB and minimum at the edge CD. Maximum intensity of compressive stress at the edge AB σ o + σ b MPa Ans. and minimum intensity of compressive stress at the edge CD σ o σ b 10 MPa Ans. Example 5.0. A hollow circular column of external diameter 50 mm and internal diameter 00 mm, carries a projecting bracket on which a load of 0 kn rests, as shown in Fig. 5.. The centre of the load from the centre of the column is 500 mm. Find the stresses at the sides of the column. Solution. Given : D 50 mm ; d 00 mm ; P 0 kn 0 10 N; e 500 mm We know that cross-sectional area of column, A π (D d ) π [(50) (00) ] 17 7 mm Direct compressive stress, P 0 10 σ o 1.1 N/mm A MPa Fig. 5.

44 Section modulus for the column, π D d I Z y D/ mm Bending moment, M P.e N-mm Bending stress, σ b M Z Torsional and Bending Stresses in Machine Parts n N/mm 11.0 MPa Since σ o is less than σ b, therefore right hand side of the column will be subjected to compressive stress and the left hand side of the column will be subjected to tensile stress. Maximum compressive stress, σ c σ b + σ o MPa Ans. and maximum tensile stress, π (50) (00) 50/ Positioning gears Turbine head Control electronics adjust position of wind turbine head Internal ladders allow access to wind turbine head Transmission Turbine blade Vents for cooling air Wind turbine. Drive shaft Note : This picture is given as additional information and is not a direct example of the current chapter. σ t σ b σ o MPa Ans. Example 5.1. A masonry pier of width m and thickness m, supports a load of 0 kn as shown in Fig. 5.. Find the stresses developed at each corner of the pier. Solution. Given: b m ; d m ; P 0 kn ; e x 0.5 m ; e y 1 m We know that cross-sectional area of the pier, A b d 1 m Moment of inertia of the pier about X-axis, b. d I XX 9m 1 1 and moment of inertia of the pier about Y-axis, d. b I YY 1 m 1 1 Distance between X-axis and the corners A and B, x / 1.5 m Distance between Y-axis and the corners A and C, y / m Fig. 5. We know that stress at corner A, P P. e. P. ey. y x x σ A [ At A, both x and y are +ve] A I I XX YY

45 1 n A Textbook of Machine Design kn/m Ans. Similarly stress at corner B, P P. e.. x. x P ey y σ B +... [ At B, x is +ve and y is ve] A IXX IYY kn/m Ans. Stress at corner C, P P. e... x x P ey y σ C +... [At C, x is ve and y is +ve] A I I and stress at D, XX YY kn/m Ans. P P. e.. x. x P ey y σ D... [At D, both x and y are ve] A IXX IYY kn/m.75 kn/m (tensile) Ans. Example 5.. A mild steel link, as shown in Fig. 5. by full lines, transmits a pull of 80 kn. Find the dimensions b and t if b t. Assume the permissible tensile stress as 70 MPa. If the original link is replaced by an unsymmetrical one, as shown by dotted lines in Fig. 5., having the same thickness t, find the depth b 1, using the same permissible stress as before. Solution. Given : P 80 kn N; σ t 70 MPa 70 N/mm Fig. 5. When the link is in the position shown by full lines in Fig. 5., the area of cross-section, A b t t t t...( b t ) We know that tensile load (P), σ t A 70 t 10 t t / or t 19.5 say 0 mm Ans. and b t 0 0 mm Ans. When the link is in the position shown by dotted lines, it will be subjected to direct stress as well as bending stress. We know that area of cross-section, A 1 b 1 t

46 Direct tensile stress, σ o P P A b1 t and bending stress, σ b M P. e P. e Z Z t ( b1) Total stress due to eccentric loading Torsional and Bending Stresses in Machine Parts n 15 P. e P P e σ b + σ o ( 1) 1. t b b t t b1 b1 Since the permissible tensile stress is the same as 70 N/mm, therefore b1 t ( b1)... Z b b Eccentricity, e b b1 b / say 0 mm Ans. Example 5.. A cast-iron link, as shown in Fig. 5.5, is to carry a load of 0 kn. If the tensile and compressive stresses in the link are not to exceed 5 MPa and 80 MPa respectively, obtain the dimensions of the cross-section of the link at the middle of its length. Fig. 5.5 Solution. Given : P 0 kn 0 10 N; σ t(max) 5 MPa 5 N/mm ; σ c(max) 80 MPa 80 N/mm Since the link is subjected to eccentric loading, therefore there will be direct tensile stress as well as bending stress. The bending stress at the bottom of the link is tensile and in the upper portion is compressive. We know that cross-sectional area of the link, A a a + a a 5.7 a mm Direct tensile stress, Fig. 5. P σ o N/mm A 5.7 a a Now let us find the position of centre of gravity (or neutral axis) in order to find the bending stresses. Let y Distance of neutral axis (N.A.) from the bottom of the link as shown in Fig. 5.. a a a + a y 1. a mm 5.7 a

47 1 n A Textbook of Machine Design Moment of inertia about N.A., a ( a) a a a I + a (1. a 0.5 a) + + ( a 1. a) 1 1 (0.5 a a ) + (0.a a ). a mm Distance of N.A. from the bottom of the link, y t y 1. a mm Distance of N.A. from the top of the link, y c a 1. a 1.8 a mm Eccentricity of the load (i.e. distance of N.A. from the point of application of the load), e 1. a 0.5 a 0.7 a mm We know that bending moment exerted on the section, M P.e a 1 10 a N-mm Tensile stress in the bottom of the link, σ t M Z t M M. yt 1 10 a 1. a 907 I / y I. a a t and compressive stress in the top of the link, σ c M M M. yc 1 10 a 1.8 a 580 Z I/ y I. a a c c We know that maximum tensile stress [σ t (max) ], σ t + σ c + a a a a 977 / or a 19.7 mm...(i) and maximum compressive stress [σ c(max) ], 80 σ c σ a a a a 0 / or a 5. mm...(ii) We shall take the larger of the two values, i.e. a 19.7 mm Ans. Example 5.. A horizontal pull P 5 kn is exerted by the belting on one of the cast iron wall brackets which carry a factory shafting. At a point 75 mm from the wall, the bracket has a T-section as shown in Fig Calculate the maximum stresses in the flange and web of the bracket due to the pull.

48 Torsional and Bending Stresses in Machine Parts n 17 Fig. 5.7 Solution. Given : Horizontal pull, P 5 kn 5000 N Since the section is subjected to eccentric loading, therefore there will be direct tensile stress as well as bending stress. The bending stress at the flange is tensile and in the web is compressive. We know that cross-sectional area of the section, A (90 1) mm P 5000 Direct tensile stress, σ A 1.51 N/mm.51 MPa Now let us find the position of neutral axis in order to determine the bending stresses. The neutral axis passes through the centre of gravity of the section. Let y Distance of centre of gravity (i.e. neutral axis) from top of the flange y 8. mm Moment of inertia of the section about N.A., 0 (1) 9 (78) I + 70 (8. ) (51 8.) 1 1 ( ) + ( ) mm This picture shows a reconnoissance helicopter of air force. Its dark complexion absorbs light that falls on its surface. The flat and sharp edges deflect radar waves and they do not return back to the radar. These factors make it difficult to detect the helicopter. Note : This picture is given as additional information and is not a direct example of the current chapter.

49 18 n A Textbook of Machine Design Distance of N.A. from the top of the flange, y t y 8. mm Distance of N.A. from the bottom of the web, y c mm Distance of N.A. from the point of application of the load (i.e. eccentricity of the load), e mm We know that bending moment exerted on the section, M P e N-mm Tensile stress in the flange, M M M. yt σ t N/mm MPa and compressive stress in the web, Z I / y I t t M M M. yc σ c.8 N/mm Zc I/ yc I MPa We know that maximum tensile stress in the flange, σ t(max) σ b + σ o σ t + σ o MPa Ans. and maximum compressive stress in the flange, σ c(max) σ b σ o σ c σ o MPa Ans. Example 5.5. A mild steel bracket as shown in Fig. 5.8, is subjected to a pull of 000 N acting at 5 to its horizontal axis. The bracket has a rectangular section whose depth is twice the thickness. Find the cross-sectional dimensions of the bracket, if the permissible stress in the material of the bracket is limited to 0 MPa. Solution. Given : P 000 N ; θ 5 ; σ 0 MPa 0 N/mm Let t Thickness of the section in mm, and b Depth or width of the section t...(given) We know that area of cross-section, A b t t t t mm t b and section modulus, Z t ( t) t mm Horizontal component of the load, P H 000 cos N Fig. 5.8 Bending moment due to horizontal component of the load, M H P H N-mm

50 A little consideration will show that the bending moment due to the horizontal component of the load induces tensile stress on the upper surface of the bracket and compressive stress on the lower surface of the bracket. Maximum bending stress on the upper surface due to horizontal component, M H σ bh Z Torsional and Bending Stresses in Machine Parts n t Turbine shaft Generator rotor Water Curved blades Water Schematic of a hydel turbine. Note : This picture is given as additional information and is not a direct example of the current chapter. t Vertical component of the load, 77 5 N/mm (tensile) P V 000 sin N Direct stress due to vertical component, P σ ov V 11 N/mm (tensile) A t t Bending moment due to vertical component of the load, M V P V N-mm This bending moment induces tensile stress on the upper surface and compressive stress on the lower surface of the bracket. Maximum bending stress on the upper surface due to vertical component, M V σ bv N/mm (tensile) Z t t and total tensile stress on the upper surface of the bracket, σ t t t t t Since the permissible stress (σ) is 0 N/mm, therefore or + 1 t t t t t 8. mm Ans.... (By hit and trial) and b t mm Ans. Example 5.. A C-clamp as shown in Fig. 5.9, carries a load P 5 kn. The cross-section of the clamp at X-X is rectangular having width equal to twice thickness. Assuming that the clamp is made of steel casting with an allowable stress of 100 MPa, find its dimensions. Also determine the stresses at sections Y-Y and Z-Z. Solution. Given : P 5 kn 5 10 N; σ t(max) 100 MPa 100 N/mm Dimensions at X-X Let t Thickness of the section at X-X in mm, and b Width of the section at X-X in mm t...(given)

51 170 n A Textbook of Machine Design We know that cross-sectional area at X-X, A b t t t t mm Direct tensile stress at X-X, σ o P 5 10 A t N/mm t Bending moment at X-X due to the load P, Section modulus, Z M P e N-mm t. b t ( t) t mm...( b t) Fig. 5.9 Bending stress at X-X, M σ b N/mm (tensile) Z t t We know that the maximum tensile stress [σ t (max) ], σ o + σ b + t t or t t t 8.5 mm Ans....(By hit and trial) and b t mm Ans. Stresses at section Y-Y Since the cross-section of frame is uniform throughout, therefore cross-sectional area of the frame at section Y-Y, A b sec 5 t mm Component of the load perpendicular to the section P cos N This component of the load produces uniform tensile stress over the section. Uniform tensile stress over the section, σ / 19. N/mm. MPa Component of the load parallel to the section P sin N This component of the load produces uniform shear stress over the section. Uniform shear stress over the section, τ / 19. N/mm. MPa

52 We know that section modulus, Torsional and Bending Stresses in Machine Parts n 171 t ( b sec 5 ) 8.5 (77 1.1) Z 7 10 mm Bending moment due to load (P) over the section Y-Y, Bending stress over the section, M N-mm M.5 10 σ b N/mm MPa Z 7 10 Due to bending, maximum tensile stress at the inner corner and the maximum compressive stress at the outer corner is produced. Maximum tensile stress at the inner corner, σ t σ b + σ o MPa and maximum compressive stress at the outer corner, σ c σ b σ o. 1.8 MPa Since the shear stress acts at right angles to the tensile and compressive stresses, therefore maximum principal stress (tensile) on the section Y-Y at the inner corner σt ( ) t (50.) (.) MPa σ + τ MPa Ans. and maximum principal stress (compressive) on section Y-Y at outer corner σc ( ) c (1.8) (.) MPa σ + τ MPa Ans. Maximum shear stress 1 ( ) 1 σ t + τ (50.) + (.) 5. MPa Ans. Stresses at section Z-Z We know that bending moment at section Z-Z, N-mm t. b 8.5 (77) and section modulus, Z 8 10 mm Bending stress at section Z-Z, M 1 10 σ b. N/mm. MPa Ans. Z 8 10 The bending stress is tensile at the inner edge and compressive at the outer edge. The magnitude of both these stresses is. MPa. At the neutral axis, there is only transverse shear stress. The shear stress at the inner and outer edges will be zero. We know that *maximum transverse shear stress, P 5 10 τ max 1.5 Average shear stress bt N/mm 1.5 MPa Ans. * Refer Art. 5.1

53 17 n A Textbook of Machine Design Sluice gate Dam Spillway River Water from the reservoir passes through a gate The flow of water makes the turbine shaft turn Turbines drive generator to produce electricity Cables carry away the electricity for use Excess water flows over spillway General layout of a hydroelectric plant. Note : This picture is given as additional information and is not a direct example of the current chapter. 5.1 Shear Stresses in Beams In the previous article, we have assumed that no shear force is acting on the section. But, in actual practice, when a beam is loaded, the shear force at a section always comes into play along with the bending moment. It has been observed that the effect of the shear stress, as compared to the bending stress, is quite negligible and is of not much importance. But, sometimes, the shear stress at a section is of much importance in the design. It may be noted that the shear stress in a beam is not uniformly distributed over the cross-section but varies from zero at the outer fibres to a maximum at the neutral surface as shown in Fig. 5.0 and Fig Fig Shear stress in a rectangular beam. Fig Shear stress in a circular beam. The shear stress at any section acts in a plane at right angle to the plane of the bending stress and its value is given by F τ Ay. I. b