Springs Lecture 3. Extension Springs
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1 Springs Lecture 3 Extension Springs
2 Carry tensile loading Extension springs
3 Max. Normal stress at A due to Bending and Axial loading σ A K A 6D 4 F K A + 3 πd πd 4C 4C C ( C ) r d C Max. torsional stress at B τ B K B 8FD 3 πd Improved design K B 4C 4 C 4 r d C (K B for the tensile case not to be confused with BergstrasserK B)
4 The free length is given by L 0 + ( D d) + ( Nb + ) d (C Nb) d Where N b is the number of body coils and D is the mean coil diameter The equivalent number of active coils taking the bending of the loops into consideration is N a Nb + G E we need to account for this to calculate spring rate
5 F Fi + ky L ( D d) + ( Nb + ) d (C Nb) d 0 +
6 Initial tension in the spring is incorporated by twisting the wire while coiling the spring Uncorrected torsional stress is given by τ i e C ± C MPa Maximum Allowable stresses for static applications Table 0-7
7
8 EXAMPLE 0-6 (self study) Hard-Drawn steel wire extension spring has a wire diameter of 0.9 mm, an outside coil diameter of 6.3 mm, hook radii of r.7 mm and r.3 mm and an initial tension of 5N. The number of body turns is.7. From the given information: a) Determine the physical parameters of the spring b) Check the initial preload stress conditions. c) Find the factor of safety under a static 3 N load
9 Example 0-7 (Self study) The helical coil extension spring in the previous example is subjected to a dynamic loading from 6.5 to 0 N. Estimate the factor of safety using the Gerber failure criterion for a) Coil fatigue b) Coil yielding c) End hook bending fatigue at A d) End hook torsionalfatigue at B
10 Helical coil torsion springs
11 End position commercial Tolerance of springs Table 0-9
12 End position commercial Tolerance of springs Table 0-9
13 Bending Stress in torsional springs σ K Mc I Wahl stress correction factor 4 K C C 4C 4C( C + C K i o ) 4C( C ) iand o for inner and outer respectively For a round wire torsion spring σ K i 3Fr 3 πd
14 Deflection and spring rate of torsion spring Angular deflection in radians or revolutions prime notation indicate revolutions k' M θ The end deflection is given by ' M θ ' y Fl Fl 64 64Ml θ e 4 4 l 3EI 3Eπd 3πd E Tabel A-9- M M θ θ ' '
15 Strain energy in bending is For a torsion spring, MFlFr and integration over the entire length of the body coil. Applying Castigliano s theorem gives rθ For round wire U F θ U M dx EI F F r dx EI πdn b πdn b 0 64FrDN 4 d E b 64MDN 4 d E b 0 Fr dx EI
16 The total deflection is given by adding the two cantilever end deflections θ t 64MDNb 64Ml 64Ml 64MD l + l + + N b d E 3πd E 3πd E d E 3πD The equivalent number of active turns is l + l 3πD The spring rate in torque per radian is given by In torque per turn it is given by k N a Fr θ t N b + M θ t 4 d E 64DN a DN a a 4 4 d E d E k' π 64/pi DN 0.
17 Due to friction between arbor and the coil the constant 0. was found by testing to be closer to 0.8 and the angle The diametric clearance Δ is In terms of N b DN a E d k 0.8 ' D l l N E d MD b π θ ' p c b b p D d N D N D d D + ' ' θ p p c b D d D D d N + + ) ( ' θ
18 Static strength from Distortion Energy Theory S y Fatigue Strength Since the spring is in bending the Sine equation is not applicable. Zimmerli s data is for compression springs. Use Data from Table 0-0 with R0 For Gerber Allowable _ Torsional _ Stress _ Table _ S e S r S r Sut
19 Strength amplitude is given by Where the slope of the load line is rm a /M m The factor of safety guarding against fatigue failure is Or + + ut e e ut a rs S S S r S a a f S n σ + + a e ut m m ut e a f S S S S n σ σ σ σ
20 Example 0-8 (Self study, for interest only) A stock spring is shown. It is made from.8mm music wire and has 4.5 body turns with straight torsion ends. It works over a pin of 0mm diameter. The coil outside diameter is 5mm a) Find max torque and rotation b) Calculate coil diameter and clearance c) Estimate the fatigue factor of safety
21 Belleville Springs
22 Miscellaneous Springs Constant force spring
23 Volute spring Variable-spring scale Damp out vibrations if coil touches
24 Conical spring Only one wire high when solidly compressed
25 Flat stock Clock Springs Clips Power springs Good approximation can be found by applying Castigliano s Theorem
26 For the cantilever The width b M Fx b bo x l Ignoring transverse shear the deflection is given by y o Fl 3 b h E M ( M EI l xdx F) 3 6Fl b h E 0 3 o E Fx( x) ( bo x / l) h 3 dx
Chapter 10 1" 2 4" 1 = MPa mm. 201 = kpsi (0.105) S sy = 0.45(278.7) = kpsi D = = in. C = D d = = 10.
budynas_sm_ch10.qxd 01/29/2007 18:26 Page 261 Chapter 10 10-1 1" 4" 1" 2 1" 4" 1" 2 10-2 A = Sd m dim( A uscu ) = dim(s) dim(d m ) = kpsi in m dim( A SI ) = dim(s 1 ) dim ( d m ) 1 = MPa mm m A SI = MPa
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