Machine Design Tutorials - Week 10

 

Machine Design Tutorials

Spring/Summer 2021 July 16th - Week 11 DGD 10

 

Problem 1 The figure below shows a fully opened trap door covering a stairwell. The door weighs 60 lb, with center of gravity 2 ft from the hinge. hi nge. A torsion bar spring, extending along the hinge axis, serves as a counterbalance. a) Det Determ ermine ine tthe he le lengt ngth h and d diam iamet eter er of a ssoli olid d steel torsion bar that would counter-balance 80 percent of the door weight when closed and provide a 6-lb.ft torque holding the door against the stop shown. Use a maximum allowable torsional stress of 50 ksi. b) Make a gr graph aph showi showing ng gr gravi avity ty torque torque,, spr spring ing torque, and net torque all plotted against the door-open angle.

 

Assumptions 1. The bar is stra straight, ight, and the torque is applie applied d about the longitudinal axis. 2. The materia materiall is homogeneous and perfectly perfectly elastic within the stress rrange ange in involv volved. ed. 3. The cross section cons considered idered is sufficiently rremote emote fr from om points of load ap application plication and from stress raisers.

Analysis Adapting Eq 4.4 from J&M txtbk, we can find f ind the diameter of a solid round rod: =

16  

=

 

16/

 

Schematic

When the door is closed, the bar will counterbalance 80% of the door weight.   = 0.80(6 0.80(60 0 lb)( lb)(24 24 in) in)   = 1152 lb in

Given: a maximum allowable torsional stress of 50 ksi   = 50 ks ksii = 50 00 000 0 psi psi

So, the diameter of the torsion rod: =

 

16(1152)/(50 16(115 2)/(50 000)

 = 0.49 0.49 in. in.

ANS.

 

.., but what is L? We know tthat hat  = .   From Tbl 5.2 in J&M txtbk, ffor or a torsional case (Case 2):

=

  

and  

=

Δ Δ

 =

  

=

  (Δ/Δ)

Polar moment of inertia, J: π(0.49) π    = 0.00566 in  =   = 32 32

So, the length of the torsion rod: =

  (Δ/Δ)

 = 115.7 115.7 in. in.

 =

(0.0056 (0.0 0566)(1 6)(11.5 1.5 × 10 )

ANS.

1080/1.92

Δ =   −    = 1152 lb in − 6 lb ft ∗ 12 Δ = 1080 1080 lb in  rad Δ = 110° 180°  = 11. 11.5 5 × 10 psi

= 1.92 1.92 ra rad d

in ft 

 

b) Make a graph showing gravity torque, spring torque, torque, and net torque all plotted against the door-open angle Gravity Torque:   = (6 (60 0 lb)(2 lb)(24 4 in) in)     = 1440 

Spring Torque:   = 115 1152 2 lb in

@  = 0°

  = 6 lb ft = 72 72 lb in  =   +    →

  =

 

@  

= 110°

11 1152 52 − 72    − 1152 1152 → 110

  = 9.82 9.82 − 1152 1152

Net Torque:   =  + 

 

  = 1440   + 9.82 9.82 − 1152 1152

 

b) Make a graph showing gravity torque, spring torque, torque, and net torque all plotted against the door-open angle Gravity Torque:   = 1440 

Spring Torque:   = 9.82 9.82 − 1152 1152

Net Torque:   = 1440   + 9.82 9.82 − 1152 1152

 

Problem 2 The bar in Problem 1 is objectionably long and is to be replaced with a torsion spring made of steel having the same physical properties. Space is available for a spring of outside diameter up to 5 in. a) Determine Determine o one ne appropriat appropriate e combination combination of of values values of d, d, D, and and N. b) What is the over overall all leng length th of the coil coiled ed section section of the sprin spring g pro proposed posed?? c) How does does the w weight eight of of the torsion torsion spring spring compar compare e with that that of the torsion torsion bar?

Assumption: The friction between adjacent spring coils is negligible.

 

Schematic

Known:   = 1152 lb in   = 50 50 ksi ksi

Since according to the DE theory: We a assume: ssume:

  = 0.5 0.58 8 

ksi   =   = 50  = 86 ksi 0.58 0.58

  = 0.5 0.58 8 

 

There are many possible answers to this problem. We will assume that   = 1.1, 1.1, but any values on Fig 12.30 can be assumed. From Eq. (12.16):

=

32     



 

=

32 



=

  32(1152)(1.1) (86 000)

 

 = 0.531 in. ANS.

“Space is available for a spring of outside diameter up to 5 in.”  = 5 in − 

= 5 − 0.531 = 4.47 in

 = D/d = 4.47/0.531 4.47/0.531 = 8.4

  ≈ 1.09

 = 4.47 4.47 in. in. ANS.

 

From Problem 1: Δ 10 1080 80 lb lb in   = 563 lb in/rad in/rad Δ 1.92 1.9 2 ra rad d   = 563

From Tbl 5.1 in J&M txtbk, for bending with angular deflection deflection (Case 3):

=

 

or

Δ   = Δ 

Moment of inertia, I:   I = π   = π(0.531)  = 0.0039 in 64 64

 = 30 × 10 psi

Length of torsion spring, L: =

 Δ/Δ

=

(30 × 10 )(0.0039)   563

= 207. 207.8 8 in in..

(Appendix C-1)

 

Number of active turns, N: 

207.8

 =    = (4.47)   →

 = 14.8 14.8 ac acti tiv ve turn turnss ANS.

b) What is the overall length of the coiled section of the spring proposed? Length of coiled section,  :    ≈  + 1  = 14.8 + 1 0.531

→   = 8.4 8.4 in in.. ANS.

c) How does the weight of the torsion spring compare with that of the torsion bar? Torsion bar active volume,  : 

  =

 

   0.49 (115.7)   =   = 21 21.8 .82 2 in 4 4

Torsion spring active volume, :  0.531    (207.8)      = 46 46.0 .01 1 in   =   = 4 4

46.01 21.82

 = 2.1

The volume of the spring is ≈ two times greater than the bar, and since they are both made of steel, the spring weighs ≈ two times as much as the bar. ANS.

 

Problem 3 An automotive engine requires a valve spring to control the motion of a valve subjected to the accelerations shown in the figure below below.. •

•

•

Note: A spring is required to hold the follower in contact with the cam during negative accelerations only.   The critical point for the spring is the “acceler “acceleration ation reversal point,” corresponding in this case to a valve lift of 0.201 in. A larger spring force is required at maximum valv valve e lift (0.384 in.), but this force is easily obtained because the spring is further compressed. In fact, the problem will be to give the spring a high enough natural frequency without making it so stiff that the spring force at full valv valve e lift causes objectionably high contact stresses when the engine is running slowly.

 

Problem 3 – co con nt’d. The valve spring is to satisfy the following specifications. 1.

Spr Spring ing lleng ength th w when hen v val alve ve iiss clo closed sed:: not o ove verr 1.5 1.50 0 in. (b (beca ecause use o off spa space ce li limit mitati ations ons). ).

2.

Sp Spri ring ng fo forc rce ew whe hen nv val alve ve is cl clos osed ed:: at at llea east st 45 lb lb..

3.   Spring fforce orce w when hen v valve alve lift is 0.201 in. (“re (“reversal versal point”): at least 70 lb. 4. Spr Spring ing fo force rce at m maxi aximum mum va valv lve e li lift ft of 0.38 0.384 4 in in.: .: a att lleas eastt 8 86 6 lb but not ov over er 90 lb (to prevent excessive cam nose contact stresses). 5.

Spr Spring ing out outsid side ed diam iamete eter: r: not ov over er 1 1.65 .65 in. (be (becau cause se o off sspac pace e li limit mitati ations ons). ).

6. 7.

Clash allowance: 0.09 .094 in. Nat Natur ural al ffreq requen uency: cy: a att le least ast a ass hig high h as th the e thi thirte rteent enth h har harmon monic ic at 18 1800 00 ca camsh mshaft aft rpm (i.e., (i.e., at least 390 Hz).

High-quality valve valve spring wire is to be used, and full advantage take taken n of both shot-peening and presetting. Because of this, you can assume that fatigue failure will not occur if the calculated stress with spring solid is limited to 800 MPa. Ends are to be closed and ground.

Determine a suitable combination of d, D, N, and L f .  

Assumptions 1. Both en end d plate platess are in con contact tact w with ith a nearl nearly y full tu turn rn of wir wire. e. 2. The end p plate late lloads oads ccoinci oincide de wit with h the sp spring ring ax axis. is.

Analysis For maximum natural frequency  frequency  , we must select the highest allowable stiffness: stiffness: ∆ when valve lift lift is 0.201 in. (“reversal point”): at least 70 lb Spec 3. Spring force when  = Spec 4. Spring force force at maximum valv valve e lift of 0.384 in.: at least 86 lb but not over over 90 lb 

=

90 − 70   = 109.29 lb/in 0.384 − 0.201 0.384 0.201

When the valve is closed,    =     −       = 70 − 109.2 09.29 9 0.2 .201 01 = 48.0 48.03 3 lb

Spec 2. Spring force when valve is closed: at least 45 lb 48.03 48.0 3 llb b > 45 lb

 

Spring rate requirement: (Eq. 12.8)

=

  8 

 = 11 11.5 × 10 psi ( (   − 1) =

 8  

 =

d (1 (11. 1.5 5 × 10 ) 8(109.29) 

Substituting in  = /  = 13 13 153 153  

Spring length requirement:

Spec 1. Spring length when valve is closed: not over 1.50 in.

Spec 6. Clash allowance: 0.094 in. Spring length when valve is open =    +  +   < 1.50  + 0.384 0.384 + 0. 0.09 094 4 < 1.50 (From fig. 12.8 for ground ends and eq. 12.10)

N + 2 d + 0.3 0.38 84 + 0.0 .09 94 < 1.5 1.50 0

     = N + 2 d N + 2 d < 1.022

   

Spring diameter requirement: Spec 5. Spring outside diameter: not over 1.65 in.  +  ≤ 1.65 1.65

Spring stress requirement: From the problem: “you can assume that fatigue failure will not occur if the calculated stress with spring solid is limited to 800 MPa”   =  +      = 90 + 0.094 .094 109. 109.29 29 = 100. 100.27 27 lb Corrected stress for fatigue (including the Wahl factor):   = 

8  

 

116 000  =

=

(Eq. 12.5)

8(100.2 8(1 00.27 7 )     

 454.74

800 MPa = 116 ksi

 

Spring natural frequency requirement: Spec 7. Natural frequency: at least as high as the thirteenth harmonic at 1800 camshaft rpm (i.e., at least 390 Hz).    =

13 900 900  

(Eq. 12.11)

Substituting in  = /    =

13 900   ≥ 390 390 Hz 

Solving equations 1-5 (Solution v1): For maximum nominal wire stress (minimum wire mass and max  max  ), try to satisfy with the smallest  . So, we take  = 9 → =

 454.74

  = 1.16



 

Solving equations 1-5 (Solution v1) – con cont’ t’d: d:

→ =

 454.74

=

9(1.16) 454.74

 →  = 0.152 in  

1.36 365 5 in   =   = 9(0.152)   →  = 1. 

→  +  ≤ 1.65 1. 1.36 365 5 + 0. 0.15 152 2 ≤ 1. 1.65 65 →  = 13 153 153

→

1.5 .52 2 in < 1.6 .65 5 in



0.152



9

 = 13 13 153 153 

  →  = 2. 2.74 74  

 

Solving equations 1-5 (Solution v1) – con cont’ t’d: d: → N + 2 d < 1.022 2.74 2.7 4 + 2 (0 (0.1 .152) 52) < 1.02 1.022 2 →

→

  =

0.7 .72 2 in < 1. 1.0 022 in

13 900 390 Hz    ≥ 390 13 900 900

(2.74)(1.365)(9)

 ≥ 390 Hz   → 413 Hz > 390 Hz

Free Free le lengt ngth h (  ):   =  +  +  /    =  + 2 

  =  + 2  +  +  /    = 2. 2.74 74 + 2 (0.1 (0.152 52)) + 0. 0.09 094 4 + 90/1 90/109 09.2 .29 9   = 1.638 in



 

Check for buckling:   = 1.638 in  = 1.36 1.365 5 in

   = 1.2

Figure 12.10 shows no buckling concern.

1.2

 

Solving equations 1-5 (Solution v2): Assume  = 7   →   = 1.21 → =

 454.74

=

7(1.21) 454.74

  →  = 0.137 0.137 in  

0.956 in   =    = 7(0.137)   →  = 0.956 

→  +  ≤ 1.65 .65 0.95 0.956 6 + 0.13 0.137 7 ≤ 1. 1.6 65

→  = 13 153 153

 

→

 = 13 13 153 153

1. 1.09 093 3 in < 1. 1.65 65 in

0.137 7

  →  = 5.25 5.25  

 

Solving equations 1-5 (Solution v2) – con cont’ t’d: d: → N + 2 d < 1.022 5.25 5.2 5 + 2 (0 (0.1 .137) 37) < 1.02 1.022 2 →

0.9 .99 93 in < 1. 1.02 022 2 in

13 900 →   =    ≥ 390 390 Hz 13 900 900 (5.25)(0.956)(7)

 ≥ 390 Hz   → 396 Hz > 390 Hz

Free Free le lengt ngth h ( ):  

  =  +  +  /    =  + 2 

  =  + 2  +  +  /    = 5. 5.25 25 + 2 (0.1 (0.137 37)) + 0. 0.09 094 4 + 90/1 90/109 09.2 .29 9   = 1.911 in



Ok, but not as good as 413 Hz with C=9

 

Check for buckling:   = 1.911 in  = 0.95 0.956 6 in

   = 2.0

Figure 12.10 shows no buckling concern.

 

Problem 4 ASTM A229 oil-tempered oil-tempered carbon steel is used for a helical coil spring. The spring is wound with  = 25 mm,  = 5.0 mm, and a pitch (distance between corresponding points of adjacent coils) of 7 mm. If the spring is compressed solid, would the spring return to its it s original free length when the force is removed? removed?

ASTM A229 Oil-tempered carbon steel

p = 7 mm d = 5 mm

D = 25 mm

 

Assumptions 1. There are are no unfavorab unfavorable le resi residual dual str stresses esses 2. Both end plates plates are are in contac contactt with near nearly ly a ful fulll turn of wire (pressure applied by plates is uniformly distributed) 3. The end plate plate loads loads coincide coincide with with the spring spring axi axiss (i.e external loads are applied along axis of helix) helix)

Fig 12.2 J&M txtbk 

 

From Fro m J&M txtbk, txtbk, Section 12.4 12.4 pg 506: “To limit long -term -term set in compression coil springs to less than 2%, shear stresses calculated from Eq 12.6 (normally with force F corresponding to spring “solid”) should be:”

Where equation 12.6 for shear stress is:

 

Adapting Eq 12.7 from J&M txtbk, we can find force, F F,, to compress spring solid (all coils touching): =

8    

=

   8  

If spring is compressed solid, the total deflection of the spring would be the pitch minus d multiplied by the number of active coils in the spring (…use figure for visual… we are looking for deflection required for all coils to be touching) ASTM A229 Oil-tempered carbon steel

 = ( (  − )  

p = 7 mm

 = 7 −5 = 2m mm m d = 5 mm

D = 25 mm

 

Shear modulus (also called modulus of rigidity) , G , of carbon steel:    = 79 

Known:

 = 5 

=

 

   8  

 = 25 

=

  

 = 79 



 = 2 

2  5     (79 × 10 ) 8(25 )

 =   

 

The corresponding stress can be calculated using Eq. 12.6 in J&M txtbk:   = 8     

First, find Ks : Method 1: Use Eq. 12.4

 :  =  

Method 2: Use Fig 12.4

 

  = 1 +

→

0.5 5

 = 25  = 5 5 =

 

 =

25  =5 5

  = . . 

  = . . 

 

Now calculate corresponding stress using Eq. 12.6: 8    =     

  = 8 7 79 90  25     5 

Known:

 = 5 

  = 1.1

 = 25  

 = 790 

1.1 = . 

Shear stresses calculated using Eq. 12.6 should satisfy Eq. 12.9 for no set-in compression :

 

Ferrous Ferr ous without pre-setting:   ≤ 0.45 Using Figure 12.7 in J&M txtbk … → ASTM A229 A229 oil temper tempered ed carbo carbon n steel steel → wire wire diamet diameter er =  = 5

  = 1300    ≤        = 442.6 442.6    Since 442.6 MPa < 585 MPa, no set should occur, therefore spring should return to original length

 

* Similar to sample problem 12.1D in J M textbook

 

Problem 5 A helical compression spring with squared and ground ends is to be made of steel, and presetting is to be used. The T he loading can be considered static. Maximum working force is 90 lb. lb. A force of 40 lb is required when the spring is 1.5 in. longer. Use recommended clash allowance of 10% of the maximum deflection and longer. a steel having   = 200 ksi. For a spring index of  of   = 8, determine appropriate values for D, d, N, and  . 40 

90   .5  11.5



Figure adapted from Fig 12.1 in J&M txtbk 

 

Assumptions 1. Both end plates plates are in contact contact with nearly nearly a full turn turn of wire. 2. The end plate plate loads loads coincide coincide with with the spring spring ax axis. is. 3. The clash clash allowance allowance is 10% 10% of the maximum maximum deflection. deflection. Clash allowance: allowance: Difference in spring length between the spring solid height and the lowest point of deflection (max load) the spring reaches in service.

Fig 12.2 J&M txtbk 

 

40   90 



1. 1.5  1. 1.5 55 

Adapted from Fig 12.1 in J&M txtbk 

Ultimate Tensile Strength: Strength:   = 200 

Mean coil diameter:  = ?

Spring Index:  = 8

Number of active coils: coils:  = ?

Wire diameter:  = ?

F         ::   = ?



  

40  90  

Using Eq. 12.9 (ferrous with presetting):



1. 1.5  1. 1.5 55 

  ≤ 0.65   ≤ 0.65 200 ksi   ≤ 130 ksi Considering a 10

Adapted from Fig 12.1 in J&M txtbk 

clash allowanc allowance e of the max deflection, and using Eq 12.8 to find spring constant, k :

(90 0 − 40) 40) lb  ∆ (9   = . .  / /   =  =  = 1.5 1.5 in ∆ 

Force when when solid ((ie ie..

→ Clash Allowance Allowance = 0.1

90    = 0.1   =  .     33.33/

aximum force that must be resisted without ‘set’) is:

  = 90 + 33.33lb/in(0 33.33lb/in(0.27) .27) = 99 lb

 

Using Eq. 12.6 in i n J&M txtbk, the wire diameter diameter,, d , can be found: 8  =  

=

→

=

8   

Finding Ks :

8 99 99 lb (8)(1.06) 130 × 10 psi 

 = .   

Since  = /: /:  =   = 8 0. 0.12 128 8 in

 =8   = .

 = 1.02 1.02 in



 

Using Eq. 12.8 in J&M txtbk, the number of active coils, N , can be found:

=



From Appendix C-1, shear modulus/ modulus of rigidity of carbon steel is:

8   



0.12 128 8 in 11.5 × 10 psi N = 0. 8 1.02 in    (33.33lb/in)

 = 11.5 1.5 × 10 psi Found previously in analysis:  = 0.128 in

N = 10.92 

 = 1.02 1.02 in



 =8  = 33. 33.33 33 lb/i lb/in n

  

From problem statement: “A helical compression spring with squared and ground ends ”

Using Fig. 12.8 in J&M txtbk , the number of solid length, L s, can be found:   =     =        =  + 2 where, whe re, N = number number of active active coils coils   = ( + 2) 2)    = 10.92 + 2 0. 0.1 128 in = 1.65 in

  = 1. 1.65 65 in

  

Using Fig for problem illustration, the spring free length, Lf  , can be found:

  =  +  + Clash Clash Allowance Allowance

40  90   



1. 1.5  1. 1.5 55    =   = 90 lb = 2.70 2.70  lb  33.33 in

  = 1.6 1.65 in + 2.7 .70 0 in + 0.2 .27 7 in Adapted from Fig 12.1 in J&M txtbk 

  = 4. 4.62 62 in



* Similar to sample problem problem 12.4 in J M textbook textbook

 

Problem 6 A semi-elliptic leaf spring for use in a light l ight trailer is to be made of steel having   = 1200 MPa,    = 1080 MPa, and a fully corrected endurance limit of 550 MPa. The spring is 1.2m long and has five leaves leaves of 5 mm thickness and 100mm width.    = 1.4. When the trailer is fully loaded, the static load applied to the center of the spring is 3500 N. a)

b) c)

d)

The load alternat alternates es as as the trailer trailer is is driven driven over over a rough rough road. Estimate what alternating load, when superimposed onto the fully loaded spring, would verge on causing eventual failure from fatigue. What What will will the maximum maximum deflectio deflection n of the spring spring be when when loaded as determined in part (a)? How much much ener energy gy is abs absorb orbed ed by by the spri spring ng in goin going g from from minimum load to maximum load when loaded as determined determine d in part (a)? To what what value value could could the the altern alternatin ating g load load be increas increased ed if 4

only 10 cycles of life are required?

Image not intended for problem (Taken from J&M textbook pg 524 to help with visualization visualization))  

Assumptions 1. 2. 3. 4.

The end pivots pivots apply a uniform lload oad ov over er the width of the spring ends. Failure ailure does not not occur occur at the spring spring end end.. The central central force force is aligned aligned so as not to induce twis twisting ting in the sprin spring. g. The deflections deflections do not significantl significantly y change the geometry geometry..

 

Quarter-, Quart er-, Semi- and Full-E Full-Ellipt lliptic ic Leaf Springs

 

a)

The load alternates alternates as the trailer is driven over a rough road. road. Estimate Estimate what alternating load, when superimposed onto the fully loaded spring, would verge on causing eventual failure from fatigue.

Each half of a semielliptic spring acts as a cantilever carrying half of the total load… “When the trailer is fully loaded, the static load applied to the center of the spring is 3500 N” 

  = 

3500    = 1750 1750  2

Using Fig 12.22b, we can find the mean stress of the leaf spring, were  =  and  = 

6    =   ℎ  

  =

6 1750 N 60 600 0 mm

 = 600 600 mm    = 1750  |

   (1.4)

(1.2m/2 because we are analyzing half of leaf spring as cantilever)

→   =  ..    

b = 100 100 mm |

K   = 1.4  

|

ℎ = 5 mm

5 leav aves es 100 mm 5 mm

 

We can determine the corresponding alternating stress,  , using the Modified Goodman Eq with SF=n=1 (with n=1, spring is just barely reaching infinite life and is therefore on the verge of an eventual fatigue failure) •

 

+

 

=

 



70 705.6 5.6 MPa MPa 550 MPa + 1200 MPa   = 1  550 MPa

 (MPa)

Modified Goodman Criterion

  = 550

230

 = 1 − 0.41 0.412 2 

(MPa)

  =   

 

Use Eq from Fig 12.22b to solve for alternating load,  :

  =

6    ℎ  

230 MPa    =

 

→

  =

 ℎ 

  = 230 230   

K  = 1.4

b = 100 100 mm

ℎ = 5 mm

6

5 leaves 100 mm 5 mm    6 600 mm (1.4)

 = 600 600 mm (1.2m/2 because we are analyzing half of leaf spring as cantilever)

  =   

* Remember, Remember, we are analyzing half the semielliptic spring and treating it as a cantilever… cantilever … therefore the total alternating load required is:   = 2 57 570 0 N =    



ANS

 

(b)

What will the maximum deflection of the spring be when loaded a ass determined determined in part (a)?

Using Fig 12.22b, we can find the max deflection,  , of the   570 N and   = 175 1750 0N leaf spring when   = 57

=

*

E,  = 207 207 GPa GPa

 

δ=

6 57 570N + 1750N 600mm 207 × 10 MPa 5 leaves 100mm 5mm   

δ = 232. 232.4 4 mm ANS

6 ℎ 

 

c) How much much energy is absorbed absorbed by the spring in going from minimum minimum load to maximum load when loaded as determined in part (a)? Work is defined by  ×  and is a measure of the energy expended in apply applying ing a force to move an object

Averag Ave rage e for force ce = Mean Mean For Force ce =   = 3500 N    =        ∗    



Dis Distan tance = δ due to F (2F (2F  = 1140 N)

δ=

6 114 1140N 0N 60 600m 0mm m  6FL  = 114. 114.2 2 mm  = 207 20 7 × 10 MPa 5 leaves 100mm 5mm    Ebh

Ener En ergy gy = 35 3500 00 N 0.11 0.1142 42m m =     ANS



 

d) To w what hat value could the alternating load be increased if only 104 cycles of life are required? Construct S-N diagram (or use S-N formulas in Appendix I) to determine corresponding 104 strength Modified Goodman, Yield and 104 Life

S-N Diagram

 (MPa)   = 1080

0.9  = 1080 = 

  = 860   = 550

    a 860      P      M  ,      S     g     o      L

350 230

  = 550



(MPa) 10  

10

10

10

# cycl cycles es,, N

d) To w what hat value could the alternating load be increased if only 104 cycles of life are required?

Use Eq from Fig 12.22b to solv solve e for alternating load,  , when  = 350 MPa   =

  =

6    ℎ  

350 MPa

 

→

  =

 ℎ  6

5 leaves 100 mm 5 mm

  = 350  

K  = 1.4

b = 100 100 mm

ℎ = 5 mm

 

6 600 mm (1.4)

 = 600 600 mm (1.2m/2 because we are analyzing half of leaf spring as cantilever)

  =   

* Remember, Remember, we are analyzing half the semielliptic spring and treating it as a cantilever cantilever… … therefore the total alternating load required is:  =2 8 86 68 N =  

  

2 8 86 68 N

 

Good luck!

 

ANS