Design of threaded jopints and mechanical spring

for the purpose to learn easy the subject of mechanical engineering

1. Machine Design I Unit5: Design of Threaded Joints and Mechanical Springs

2. Unit5: Design of Threaded Joints and Mechanical Springs Stresses in screw fasteners, bolted joints under tension, torque requirement for bolt tightening, preloading of bolt under static loading, eccentrically loaded bolted joints. Power Screws : Forms of threads used for power screw and their applications, torque analysis for square and trapezoidal threads, efficiency of screw, collar friction, overall efficiency, self-locking in power screws, stresses in the power screw, design of screw and nut, differential and compound screw, re-circulating ball screw. Welded Joints : Type of welded joints, stresses in butt and fillet welds, strength of welded joints subjected to bending moments. Mechanical Springs: Stress deflection equation for helical spring, Wahl’s factor, style of ends, design of helical compression, tension and torsional spring under static loads, construction and design consideration in leaf springs, nipping, strain energy in helical spring, shotpeening .

3. Power Screws Forms of threads used for power screw and their applications, torque analysis for square and trapezoidal threads, efficiency of screw, collar friction, overall efficiency, self-locking in power screws, stresses in the power screw, design of screw and nut, differential and compound screw, re-circulating ball screw.

4. Power Screws A power screw is a mechanical device used for converting rotary motion into linear motion and transmitting power. The main applications of power screws are as follows: ( i ) to raise the load, e.g., screw-jack; (ii) to obtain accurate motion in machining operations, e.g., lead-screw of lathe; (iii) to clamp a work piece, e.g., a vice; and (iv) to load a specimen, e.g., universal testing machine. There are three essential parts of the power screw, viz., screw, nut and a part to hold either the screw or the nut in its place. Examples: The lead screw of the lathe A screw jack and machine vice Advantages: ( i ) A power screw has large load carrying capacity. (ii) The overall dimensions of the power screw are small, resulting in compact construction (iii) A power screw is simple to design. (iv) The manufacturing of a power screw is easy (v) A power screw provides large mechanical advantage. (vi) A power screw provides precisely controlled and highly accurate linear motion (vii) A power screw gives smooth and noiseless service without any maintenance. (viii) There are few parts in a power screw. (ix) A power screw can be designed with self locking property. Disadvantages: A power screw has very poor efficiency, (ii) High friction in threads causes rapid wear of the screw or the nut. The efficiency of a power screw is increased if sliding friction is replaced by rolling friction. This principle is used in recirculating ball screw.

5. FORMS OF THREADS There are two popular types of threads used for power screws, viz., square and ISO metric trapezoidal, as shown in Fig. Advantages: 1. The efficiency is more 2. There is no radial pressure or side thrust on the nut. Disadvantages: Square threads are difficult to manufacture. The strength of a screw and load carrying capacity is less It is not possible to compensate for wear in square threads. Advantages: Threads are economical to manufacture. More strength and load carrying capacity The axial wear on the surface of trapezoidal threads can be compensated by means of a split-type of nut. Disadvantages: The efficiency is less Threads result in side thrust or radial pressure on the nut. Combines the advantages of square and trapezoidal threads, can transmit power and motion only in one direction. A special type of trapezoidal thread called acme thread.

6. Proportions of square and ISO metric trapezoidal threads Proportions of ISO metric trapezoidal threads Proportions of square threads (normal series) Sq 30 X 6 : Square thread, Single start 30 mm nominal diameter and 6 mm pitch Tr 40 X 7 : single-start trapezoidal threads with 40 mm nominal diameter and 7 mm pitch. Tr 40 X 14 ( P 7) : Double-start trapezoidal threads with 40 mm nominal diameter and 7 mm pitch. MULTIPLE THREADED SCREWS: large axial motion per revolution, higher efficiency, Low Mechanical advantage, loss of self locking property, used in high speed actuators and sluice valves

7. TERMINOLOGY OF POWER SCREW Pitch: (p) Lead: (l) Nominal Diameter: (d) Core Diameter: (d c ) Mean Diameter: ( d m ) Helix Angle: It is also called lead angle. The helix angle is denoted by alpha. 2. Development of Thread 1. Terminology of Power Screw Fig. 2 is used as basis for determining the torque required to raise or lower the load. The following conclusions can be drawn on the basis of development of thread: Screw is considered as inclined plane with angle of inclination alpha, load W ( acts always vertically downwards) is raised, it moves up the inclined plane. When the load W is lowered, it moves down the inclined plane. The load W is raised or lowered by means of an imaginary force P acting at the mean radius of the screw. The force P multiplied by the mean radius ( d m /2) gives the torque required to raise or lower the load. Remember P is perpendicular to the load W.

8. TORQUE REQUIREMENT Force Diagram for Lifting Load Considering equilibrium of horizontal forces, Considering equilibrium of vertical forces, Dividing expression and simplifying Similarly torque required to lower the load can be found out by :

9. SELF-LOCKING SCREW Neglecting collar friction, the rule for a self locking screw is as follows, “ A screw will be self-locking if the coefficient of friction is equal to or greater than the tangent of the helix angle ”. The following conclusions are drawn by examination of above equation: Self-locking of screw is not possible when the coefficient of friction is low. Self-locking property of the screw is lost when the lead is large. Self-locking condition is essential in applications like screw jacks. Efficiency of a self-locking square threaded power screw is less than 50%. EFFICIENCY OF SQUARE THREADED SCREW We know that, Maximum efficiency of a square threaded power screw is given by,

10. COLLAR FRICTION TORQUE Collar Friction According to the uniform pressure theory, According to the uniform wear theory, Where, In certain applications, the collar between the cup and the screw is replaced by thrust ball bearing to reduce friction. The advantage of using thrust ball bearing at the collar is that the sliding friction is replaced by rolling friction. The collar friction torque becomes almost negligible in these cases. OVERALL EFFICIENCY The total external torque required to raise the load, The overall efficiency,

11. DESIGN OF SCREW AND NUT COEFFICIENT OF FRICTION An average value of 0.15 can be taken for the coefficient of friction at the thread surface, when the screw is lubricated with mineral oil. The values of the coefficient of friction for the thrust collar with sliding contact are given in Table There are three basic components of a power screw, viz., screw, nut and frame. Screw material should have sufficient strength, good wear resistance and machinability. Screw s are made of plain carbon steel such as 30C8, 40C8 and 45C8 or alloy steels like 40Cr1. The screws are case hardened, e.g., the lead screw of a lathe is case hardened by the nitriding process. Bronze is the ideal material for nut . Mainly tin bronze and phosphor bronze are used for nut. The frame of a power screw is usually made of grey cast iron of grade FG 200. Cast iron is cheap, can be given any complex shape and possesses high compressive strength

12. DESIGN OF SCREW AND NUT The torsional shear stress is given by, The principal shear stress is given by, The transverse shear stresses in the nut

13. DESIGN OF SCREW AND NUT The bearing pressure between the contacting surfaces of the screw and the nut is an important consideration in design. The bearing area between the screw and the nut for one thread is Therefore, or Where, The permissible bearing pressure depends upon the materials of the screw and the nut and the rubbing velocity. The permissible values of the unit bearing pressures are given in Table

14. Numerical: Power Screw The nominal diameter of a triple threaded square screw is 50 mm, while the pitch is 8 mm. It is used with a collar having an outer diameter of 100 mm and inner diameter as 65 mm. The coefficient of friction at the thread surface as well as at the collar surface can be taken as 0.15. The screw is used to raise a load of 15 kN. Using the uniform wear theory for collar friction, calculate: ( i ) torque required to raise the load; (ii) torque required to lower the load; and (iii) the force required to raise the load, if applied at a radius of 500 mm. Torque required to raise the load

15. Numerical: Power Screw Torque required to lower the load The negative sign indicates that the screw alone is not self-locking. However, due to the restraining torque of collar friction, the screw is self-locking. Force required to raise the load The force P i at the radius of 500 mm is given by, Practice Problem: Solved 6.3 and 6.4 TW Problem :DESIGN OF SCREW JACK : 6.15, page 206; Solved Problem. Write all design calculations on Journal Pages and draw assembly drawing and detailed part drawing on two separate sheets. ( for all batches.) Assignment Note book : S. Notes:1. Differential and compound screw, 2. Recirculating Ball screw.

16. Design of threaded Joints Stresses in screw fasteners, bolted joints under tension, torque requirement for bolt tightening, preloading of bolt under static loading, eccentrically loaded bolted joints.

17. Terminology of Threads Profile of Internal and External Threads

18. Basic dimensions for ISO metric screw threads (coarse series) Table 7.2 Basic dimensions for ISO metric screw threads (fi ne series)

19. BOLTED JOINT—SIMPLE ANALYSIS Bolt in Tension The strength of the bolt in tension is given by, The height of the standard hexagonal nut is (0.8 d ). Hence, the threads of the bolt in the standard nut will not fail by shear. Or

20. Numerical An electric motor weighing 10 kN is lifted by means of an eye bolt as shown in Fig. 7.13. The eye bolt is screwed into the frame of the motor. The eye bolt has coarse threads. It is made of plain carbon steel 30C8 ( Syt = 400 N/mm2) and the factor of safety is 6. Determine the size of the bolt. Size of bolt

21. Numerical Two plates are fastened by means of two bolts as shown in Fig. The bolts are made of plain carbon steel 30C8 ( S yt = 400 N/mm 2 ) and the factor of safety is 5. Determine the size of the bolts if, P = 5 kN Permissible shear stress The shank portion of the bolts is subjected to direct shear stress. Let the diameter of the shank be denoted by d .

22. ECCENTRICALLY LOADED BOLTED JOINTS IN SHEAR The eccentricity of the external force P is e from the centre of gravity. This eccentric force can be considered as equivalent to an imaginary force P (results in primary shear forces P′ 1 , P′ 2 , ,..., etc.,) at the centre of gravity and a moment (P x e) about the same point. The moment (P x e) about the centre of gravity results in secondary shear forces P’ ’1, P’′ 2 , , ..., etc. If r 1 , r 2 , ..., etc., are the radial distances of the bolt centres from the centre of gravity, Primary and Secondary Shear Forces and so on. The primary and secondary shear forces are added by vector addition method to get the resultant shear forces P 1 , P 2 , P 3 , and P 4 .

23. Numerical The structural connection shown in Fig. is subjected to an eccentric force P of 10 kN with an eccentricity of 500 mm from the CG of the bolts. The centre distance between bolts 1 and 2 is 200 mm, and the centre distance between bolts 1 and 3 is 150 mm. All the bolts are identical. The bolts are made from plain carbon steel 30C8 ( Syt = 400 N/mm2) and the factor of safety is 2.5. Determine the size of the bolts. Vector Addition of Shear Forces Similarly, it can be proved that Resultant shear force

24. Numerical Bolts 2 and 4 are subjected to maximum shear forces. Size of bolts From Table, the standard size of the bolts is M 20. Example 7.4

25. TORQUE REQUIREMENT FOR BOLT TIGHTENING The bolts are tightened with a specific magnitude of pre-load P i . It is necessary to determine the magnitude of the torque which will induce this pre-tension. The torque required to tighten the bolt consists of the following two factors: torque required to overcome thread friction and induce the pre-load, i.e., (M t ); and torque required to overcome collar friction between the nut and the washer (M t ) c . The equations derived for trapezoidal threads are suitably modified for ISO metric screw threads. The torque required to overcome thread friction is given by, Where, According to uniform wear theory, the collar friction torque ( M t ) c is given by, Where, For ISO metric screw threads the total torque ( M t ) t required to tighten the bolts is given by, or to determine the wrench torque ( M t ) t required to create the required pre-load P i .

26. Welded Joints Type of welded joints, stresses in butt and fillet welds, strength of welded joints subjected to bending moments.

27. Type of welded joints Welding can be defined as a process of joining metallic parts by heating to a suitable temperature with or without the application of pressure. BUTT JOINTS : A butt joint can be defined as a joint between two components lying approximately in the same plane. FILLET JOINTS : A fillet joint, also called a lap joint, is a joint between two overlapping plates or components. Cross-section of Fillet Weld

28. STRENGTH OF BUTT WELDS Butt Weld in Tension A butt welded joint, subjected to tensile force P , is shown in Fig.. The average tensile stress in the weld is given by, where, Equating the throat of the weld h to the plate thickness t where, There are certain codes, like code for unfired pressure vessels, which suggest reduction in strength of a butt welded joint by a factor called efficiency of the joint where, Butt welded joint, when properly made, has equal or better strength than the plates and there is no need for determining the stresses in the weld or the size and the length of the weld. All that is required is to match the strength of the weld material to the strength of the plates.

29. Numerical A gas tank consists of a cylindrical shell of 2.5 m inner diameter. It is enclosed by hemispherical shells by means of butt welded joint as shown in Fig. 8.6. The thickness of the cylindrical shell as well as the hemispherical cover is 12 mm. Determine the allowable internal pressure to which the tank may be subjected, if the permissible tensile stress in the weld is 85 N/mm 2 . Assume efficiency of the welded joint as 0.85. Tensile force on plates The length of the welded joint is equal to the circumference of the cylindrical shell. Allowable internal pressure Corresponding pressure inside the tank is given by,

30. STRENGTH OF PARALLEL FILLET WELDS Parallel Fillet Weld in Shear The size of the weld is specified by the leg length h is equal to the plate thickness. The throat is the minimum cross-section of the weld located at 45° to the leg dimension. Therefore, or Failure of the fillet weld occurs due to shear along the minimum cross-section at the throat. The cross-sectional area at the throat is ( tl ) or (0.707 hl). The shear stress in the fillet weld is given by, or Usually, there are two welds of equal length on two sides of the vertical plate. In that case, or In determining the required length of the weld, 15 mm should be added to the length of each weld calculated by above equations to allow for starting and stopping of the weld run. In case of a static load, the permissible shear stress for the fillet welds is taken as 94 N/mm 2 as per the code of American Welding Society (AWS).

31. STRENGTH OF TRANSVERSE FILLET WELDS The tensile stress in the transverse fillet weld is given by, Failure of Fillet Weld the strength equation of the transverse fillet weld is written in the following form, Usually, there are two welds of equal length on two sides of the plate as shown in Fig. (a). In such cases, or The nature of stresses in the cross-section of the transverse fillet weld is complex. The weld is subjected to normal stress as well as shear stress. In addition, the throat is subjected to bending moment, which adds to the complications. Theoretically, it can be proved that for transverse fillet weld, the inclination of the plane, where maximum shear stress is induced, is 67.5° to the leg dimension as shown in Fig.(b). In order to simplify the design of fillet welds, many times shear failure is used as the failure criterion. It is assumed that the stress in the transverse fillet weld is shear stress on the throat area for any direction of applied load. With this assumption, equations derived for parallel fillet welds are also used for the transverse fillet welds.

32. AXIALLY LOADED UNSYMMETRICAL WELDED JOINTS (a) shows an angle section welded to a vertical beam by means of two parallel fillet welds 1 and 2. G is the centre of gravity of the angle section. The external force acting on the joint passes through G . Suppose P 1 and P 2 are the resisting forces set up in the welds 1 and 2 respectively. The free body diagram of forces acting on the angle section with two welds is shown in Fig.(b).

33. Numerical An ISA 200 x 100 x 10 angle is welded to a steel plate by means of fillet welds as shown in Fig. The angle is subjected to a static force of 150 kN and the permissible shear stress for the weld is 70 N/mm 2 . Determine the lengths of weld at the top and bottom. Step I Total length of weld Step II Weld lengths l 1 and l 2 Also,

34. Numerical How much length of a 10 mm fillet weld is required to weld the long side of an ISA angle 150 ¥ 75 ¥ 10 to a steel plate with side welds only? A static load of 125 kN acts through the centre of gravity of the angle section which is 53.2 mm from the short side. The allowable load per mm of the weld length is 665 N. Step I Total length of weld The welded joint is shown in Fig. 8.18. The total length ( l ) of two fillet welds is given by,

35. ECCENTRIC LOAD IN THE PLANE OF WELDS A bracket subjected to an eccentric force P and attached to the support by means of two fillet welds W 1 and W 2 is shown in fig.(a). Suppose G is the centre of gravity of two welds and e is the eccentricity between the centre of gravity and the line of action of force P . According to the principle of Applied Mechanics, the eccentric force P can be replaced by an equal and similarly directed force ( P ) acting through the centre of gravity G , along with a couple ( M = P x e ) lying in the same plane [Fig.(b)]. The effects of the force P and the couple M are treated separately as shown in Fig (c) and (d).

36. ECCENTRIC LOAD IN THE PLANE OF WELDS Primary and Secondary Shear Stresses where A is the throat area of all welds. The couple M causes torsional shear stresses in the throat area of welds [Fig.(b)]. They are called secondary shear stresses and given by The resultant shear stress is obtained by vector addition of primary and secondary shear stresses [Fig.(c)]. Figure shows a weld of length l and throat t . G 1 is the centre of gravity of the weld, while G is the centre of gravity of a group of welds. The moment of inertia of this weld about its centre of gravity G1 is given by, The polar moment of inertia about an axis passing through G is determined by the parallel axis theorem. Thus, The above value of J is to be used in Eq. to determine secondary shear stresses.

37. Numerical A welded connection, as shown in Fig. is subjected to an eccentric force of 7.5 kN. Determine the size of welds if the permissible shear stress for the weld is 100 N/mm 2 . Assume static conditions. the primary shear stress is given by, The distance r of the farthest point in the weld from the centre of gravity is given by the polar moment of inertia J 1 of the weld W 1 about G is given by

38. Numerical Due to symmetry, the polar moment of inertia of the two welds ( J ) is given by the secondary shear stress is given by Step III Resultant shear stress Figure shows the primary and secondary shear stresses. The vertical and horizontal components of these shear stresses are added and the resultant shear stress is determined. Therefore, from the figure, Practice :Example 8.10/ page 287

39. WELDED JOINT SUBJECTED TO BENDING MOMENT A cantilever beam of rectangular cross-section is welded to a support by means of two fillet welds W 1 and W 2 as shown in Fig. the principle of Applied Mechanics, the eccentric force P can be replaced by an equal and similarly directed force P acting through the plane of welds, along with a couple ( M b = P x e ) as shown in Fig. The force P through the plane of welds causes the primary shear stress t 1 , which is given by where A is the throat area of all welds. The moment M b causes bending stresses in the welds. The bending stresses are given by, The resultant shear stress in the welds is given by, the moment of inertia of weld W 1 about the X -axis is given by Since there are two such symmetrical welds,

40. Numerical A bracket is welded to the vertical column by means of two fillet welds as shown in Fig. Determine the size of the welds, if the permissible shear stress in the weld is limited to 70 N/mm 2 .

41. Numerical Practice: Example 8.14, 8.15

42. Mechanical Springs Stress deflection equation for helical spring, Wahl's factor, style of ends, design of helical compression, tension and torsional spring under static loads, construction and design consideration in leaf springs, nipping, strain energy in helical spring, shotpeening .

43. SPRINGS A spring is defined as an elastic machine element, which deflects under the action of the load and returns to its original shape when the load is removed. F unctions and applications of springs are as follows: Springs are used to absorb shocks and vibrations , e.g., vehicle suspension springs, railway buffer springs, buffer springs in elevators and vibration mounts for machinery. Springs are used to store energy, e.g., springs used in clocks, toys, movie-cameras, circuit breakers and starters . Springs are used to measure force, e.g ., springs used in weighing balances and scales. Springs are used to apply force and control motion TYPES OF SPRINGS: Helical Springs: (a) Compression Spring (b ) Extension Spring (c )Torsion Spring Semi-elliptic Leaf Spring

44. TYPES OF SPRINGS:

45. TERMINOLOGY OF HELICAL SPRINGS The main dimensions of a helical spring are as follows: d = wire diameter of spring (mm ) Therefore, Di = inside diameter of spring coil (mm) Do = outside diameter of spring coil (mm) D = mean coil diameter (mm ) The spring index is defined as the ratio of mean coil diameter to wire diameter . A spring index from 4 to 12 is considered best from manufacturing considerations . Therefore, in practical applications, the spring index usually varies from 4 to 12. However , a spring index in the range of 6 to 9 is still preferred particularly for close tolerance springs and those subjected to cyclic loading. ( i ) Solid Length Solid length = N t d where, N t = total number of coils

46. Spring Length Terminology (c) Solid Length ( b ) Compressed Length Total gap = ( N t – 1) x Gap between adjacent coils ( a ) Free Length The stiffness of the spring ( k ) is defined as the force required to produce unit deflection . Therefore, Is also known as rate of spring , gradient of spring , scale of spring or simply spring constant .

47. Spring Coils Active coils and inactive coils . Active coils are the coils in the spring which contribute to spring action , support the external force and deflect under the action of force . A portion of the end coils, which is in contact with the seat, does not contribute to spring action and are called inactive coils . These coils do not support the load and do not deflect under the action of an external force. The number of inactive coils is given by , inactive coils = N t – N, where, N = number of active coils. STYLES OF END End Styles of Helical Compression Springs End Styles of Helical Extension Springs

48. STRESS AND DEFLECTION EQUATIONS (a) Helical Spring (b) Helical Spring-unbent A helical spring made from the wire of circular cross-section is shown in Fig .(a ). D and d are the mean coil diameter and wire diameter respectively. The number of active coils in this spring is N . The spring is subjected to an axial force P . When the wire of the helical spring is uncoiled and straightened, it takes the shape of a bar as shown in Fig .(b ). The force P acting at the end of the bracket induces torsional shear stress in the bar. The torsional moment M t is given by, The torsional shear stress in the bar is given by, or

49. STRESS AND DEFLECTION EQUATIONS Stresses in Spring Wire: (a) Pure Torsional Stress (b) Direct Shear Stress (c) combined Torsional , Direct and Curvature Shear Stresses The effect of direct shear stress and stress concentration due to curvature effect , requires modification K S = factor to account for direct shear stress K c = factor to account for stress concentration due to curvature effect The combined effect of these two factors is given by, K = K S K C As shown in Fig ., the direct shear stress in the bar is given by, Superimposing the two stresses of expressions The shear stress correction factor ( K s ) is defined as , or Hence AM Wahl 1 derived the equation for resultant stress , The Wahl factor ‘K’ is given by, where C is the spring index. In normal applications, the spring is designed by using the Wahl factor. When the spring is subjected to fluctuating stresses, two factors K s and K c are separately used.

50. STRESS AND DEFLECTION EQUATIONS L oad-deflection equation: Axial deflection due to load P is given by, G = modulus of rigidity The rate of spring ( k ) is given by, or The load is linearly proportional to the deflection of the spring. The load-deflection curve for helical spring is shown in Fig . The area below the load-deflection line gives the strain energy stored in the spring. Assuming that the load is gradually applied , the energy stored in the spring is given by, E = area below load-deflection line Load-deflection Diagram where, E = strain energy stored in spring (N-mm) SERIES AND PARALLEL CONNECTIONS IN SERIES IN PARALLEL where k is the combined stiffness of all springs in the connection.

51. DESIGN OF HELICAL SPRINGS Design objectives: ( i ) It should possess sufficient strength to withstand the external load. (ii) It should have the required load-deflection characteristic. ( iii) It should not buckle under the external load . Load-stress equation: Factor of Safety : The factor of safety in the design of springs is usually 1.5 or less The Indian Standard 4454–1981 has recommended a much higher value for the permissible shear stress. According to this standard, Basic Steps in Design of Helical Springs: Estimate value of Load P and corresponding deflection required Select suitable material and calculate permissible shear stress by, Assume suitable value of C ( Between 8-10) Calculate the Wahl factor v) Determine wire diameter , mean coil diameter and , number of active turns.( G= 81 370 N/mm2 for steel) vi) Determine length, actual deflection and other spring specifications. The thumb rules for provision of guide are as follows:

52. Numerical : It is required to design a helical compression spring subjected to a maximum force of 1250 N. The deflection of the spring corresponding to the maximum force should be approximately 30 mm . The spring index can be taken as 6. The spring is made of patented and cold-drawn steel wire. The ultimate tensile strength and modulus of rigidity of the spring material are 1090 and 81 370 N/mm2 respectively . The permissible shear stress for the spring wire should be taken as 50% of the ultimate tensile strength. Design the spring and calculate : ( i ) wire diameter ; ( ii) mean coil diameter ; ( iii) number of active coils ;( iv) total number of coils ; Step III Number of active coils

53. The dimensions of the spring are shown in Fig. Refer solved problems: 10.2 to 10.4

54. Numerical : A helical tension spring is used in the spring balance to measure the weights. One end of the spring is attached to the rigid support while the other end, which is free, carries the weights to be measured. The maximum weight attached to the spring balance is 1500 N and the length of the scale should be approximately 100 mm. The spring index can be taken as 6. The spring is made of oil-hardened and tempered steel wire with ultimate tensile strength of 1360 N/mm2 and modulus of rigidity of 81 370 N/mm2. The permissible shear stress in the spring wire should be taken as 50% of the ultimate tensile strength . Design the spring and calculate: ( i ) wire diameter ; ( ii) mean coil diameter ; ( iii) number of active coils ; ( iv) required spring rate; and (v ) actual spring rate. Step I Wire Diameter The working principle of the spring balance is illustrated in Fig. As the load acting on the spring varies from 0 to 1500 N, the pointer attached to the free end of the spring moves over a scale between highest and lowest positions. The length of the scale between these two positions of the pointer is 100 mm. In other words, the spring deflection is 100 mm when the force is 1500 N.

55. The permissible shear stress for spring wire is given by, Step III : Number of active coils Assignment Problems: Unsolved 10.2, 10.3 Solved : 10.24 ( Multi leaf spring) Short note : Shot peening

56. MULTI-LEAF SPRING Used for the suspension of cars, trucks and railway wagons . consists of a series of flat plates, usually of semi-elliptical shape, as shown in Fig . The flat plates are called leaves of the spring. The leaves have graduated lengths. The leaf at the top has maximum length. The length gradually decreases from the top leaf to the bottom leaf. The longest leaf at the top is called master leaf . It is bent at both ends to from the spring eyes. Two bolts are inserted through these eyes to fix the leaf spring to the automobile body. The leaves are held together by means of two U-bolts and a center clip. Rebound clips are provided to keep the leaves in alignment and prevent lateral shifting of the leaves during operation . At the center, the leaf spring is supported on the axle. Multi-leaf springs are provided with one or two extra full length leaves in addition to master leaf. The extra full-length leaves are stacked between the master leaf and the graduated length leaves . The extra full-length leaves are provided to support the transverse shear force.

57. MULTI-LEAF SPRING n f = number of extra full-length leaves n g = number of graduated-length leaves including master leaf n = total number of leaves b = width of each leaf (mm) t = thickness of each leaf ( mm) L = length of the cantilever or half the length of semi-elliptic spring ( mm) P = force applied at the end of the spring ( N) P f = portion of P taken by the extra full-length leaves (N) P g = portion of P taken by the graduated-length leaves (N ) For the purpose of analysis, the leaves are divided into two groups namely, master leaf along with graduated-length leaves forming one group and extra full-length leaves forming the other. The following notations are used in the analysis: The bending stress in the plate at the support is given by, and The deflection at the end of the spring is determined from Equation.

58. NIPPING OF LEAF SPRINGS The stresses in extra full-length leaves are 50% more than the stresses in graduated-length leaves. One of the methods of equalising the stresses in different leaves is to pre-stress the spring by bending the leaves to different radii of curvature, before they are assembled with the center clip. As shown in Fig. , the full-length leaf is given a greater radius of curvature than the adjacent leaf. The radius of curvature decreases with shorter leaves. The initial gap C between the extra full-length leaf and the graduated-length leaf before the assembly, is called a ‘ nip ’. Such pre-stressing , achieved by a difference in radii of curvature, is known as ‘nipping’. Nipping is common in automobile suspension springs . Required value of C to achieve stress equalization is given by, The initial pre-load P i required to close the gap C is given by Since the stresses are equal in all leaves it is calculated by The deflection of the multi-leaf spring due to the external force P is given by,