Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
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Textbook Question
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Chapter 15, Problem 7P

In straight-bevel gearing, there are some analogs to Eqs. (14-44) and (14-45) pp. 766 and 767, respectively. If we have a pinion core with a hardness of (HB)11 and we try equal power ratings, the transmitted load Wt can be made equal in all four cases. It is possible to find these relations:

  Core Case
Pinion (HB)11 (HB)12
Gear (HB)21 (HB)22
  1. (a)     For carburized case-hardened gear steel with core AGMA bending strength (sat)G and pinion core strength (sat)P, show that the relationship is

( s a t ) G = ( s a t ) P J p J G m G 0.0323

This allows (HB)21 to be related to (HB)11.

  1. (b)     Show that the AGMA contact strength of the gear case (sac)G can be related to the AGMA core bending strength of the pinion core (sat)P by

( s a c ) G = C p ( C L ) G C H S H 2 ( s a t ) P ( K L ) P K x J P K T C s C x c N P I K s

If factors of safety are applied to the transmitted load Wt, then SH = S F and S H 2 / S F is unity. The result allows (HB)22 to be related to (HB)11.

  1. (c)     Show that the AGMA contact strength of the gear (sac)G is related to the contact strength of the pinion (sac)P by

( s a c ) P = ( s a c ) G m G 0.0602 C H

(a)

Expert Solution
Check Mark
To determine

The relation between AGMA bending strength of gear with AGMA bending strength of pinion.

Answer to Problem 7P

The relation between AGMA bending strength of gear with AGMA bending strength of pinion is satG=satP×(mG)0.0323×JPJG.

Explanation of Solution

Write the expression for gear ratio of gear set.

mG=Nn (I)

Here, the gear ratio is mG.

Write the critical expression for stress cycle factor of pinion for 3×106NLP1010.

KLP=1.683(NLP)0.0323 (II)

Here, the stress cycle factor for pinion is KLP and the life goal in revolution for pinion is NLP.

Write the critical expression for stress cycle factor of gear for 3×106NLP1010.

KLG=1.683(NLPmG)0.0323 (III)

Here, the stress cycle factor for gear is KLG.

Write the expression for permissible bending stress for pinion.

swtP=satPKLPSFKTKRSF=satPKLPswtPKTKR (IV)

Here, the permissible bending stress for pinion is swtP, the allowable bending stress number for pinion is satP, the reliability factor is KR, the bending safety factor is SF and the temperature factor is KT.

Write the expression for permissible bending stress for gear.

swtG=satGKLGSFKTKRSF=satGKLGswtGKTKR (V)

Here, the permissible bending stress for gear is swtG and the allowable bending stress number for gear is satG.

Substitute satKLGswtGKTKR for SF in Equation (IV).

satGKLGswtGKTKR=satPKLPswtPKTKR (VI)

Write the expression for bending stress for pinion.

stP=WtFPdKoKvKSKmKxJP (VII)

Here, the bending stress in pinion is stP, the transmitted load is Wt, the overload factor is Ko, length wise curvature factor is Kx and geometry factor for bending in pinion is JP.

Write the expression for bending stress for gear.

stG=WtFPdKoKvKSKmKxJG (VIII)

Here, the bending stress in gear is stG, the transmitted load is Wt and geometry factor for bending in gear is JG.

For limiting condition bending stress in pinion is equal to allowable bending stress in pinion.

Substitute stP for swtP in Equation (VI).

satGKLGswtGKTKR=satPKLPstPKTKR (IX)

For limiting condition bending stress in gear is equal to allowable bending stress in gear.

Substitute stG for swtG in Equation (IX).

satGKLGstGKTKR=satPKLPstPKTKR (X)

Substitute WtFPdKoKvKSKmKxJP for stP in Equation (X).

satGKLGstGKTKR=satPKLP(WtFPdKoKvKSKmKxJP)KTKRsatGKLGstG=satPKLP(WtFPdKoKvKSKmKxJP) (XI)

Substitute WtFPdKoKvKSKmKxJG for stP in Equation (XI).

satGKLGWtFPdKoKvKSKmKxJG=satPKLP(WtFPdKoKvKSKmKxJP)satG=satPKLPKLGJPJG (XII)

Substitute 1.683(NLP)0.0323 for KLP and 1.683(NLPmG)0.0323 for KLG in Equation (XII).

satG=satP{1.683(NLP)0.0323}{1.683(NLPmG)0.0323}JPJGsatG=satP×(mG)0.0323×JPJG (XIII)

Conclusion:

Thus, the relation between AGMA bending strength of gear with  AGMA bending strength of pinion is satG=satP×(mG)0.0323×JPJG.

(b)

Expert Solution
Check Mark
To determine

The AGMA contact strength of the gear case relation with the AGMA core bending strength of the pinion core.

Answer to Problem 7P

The AGMA contact strength of the gear case is related to the AGMA core bending strength of the pinion core by sac=CPCLG{(SH2satPSF)(KLPKT(CHG)2)(KxJPKS)(CSCxcnI)}.

Explanation of Solution

Write the expression for pitch diameter of pinion.

dP=nPddPPd=n (XIV)

Here, the pitch diameter of pinion is dP, the number of teeth on pinion is n and the diametral pitch is Pd.

Write the expression for permissible bending stress for pinion.

swtP=satPKLPSFKTKR (XV)

Here, the permissible bending stress for pinion is swtP, the allowable bending stress number for pinion is satP, stress cycle factor for pinion is KLP, the reliability factor is KR, the bending safety factor is SF and the temperature factor is KT.

Write the expression for bending stress for pinion.

stP=WtFPdKoKvKSKmKxJP (XVI)

Here, the bending stress in pinion is stP, dynamic factor is Kv, size factor is KS, load distribution factor is Km, the transmitted load is Wt, the face width is F, the overload factor is Ko, length wise curvature factor is Kx and geometry factor for bending in pinion is JP.

For limiting condition bending stress in pinion is equal to allowable bending stress in pinion.

Substitute stP for swtP in Equation (II).

stP=satPKLPSFKTKR (XVII)

Substitute WtFPdKoKvKSKmKxJP for stP in Equation (IV).

WtFPdKoKvKSKmKxJP=satPKLPSFKTKRWt=satPKLPFKxJPPdKoKvKSKmSFKTKRWt=(satPSF)(KLPKTKR)(FKxJPPdKoKvKSKm) (XVIII)

Write the expression for reliability factor for pitting.

CR=KR (XIX)

Here, the reliability factor for pitting is CR and the reliability factor is KR.

Write the expression for permissible contact stress number for gear.

swcG=sacGCLGCHGSHKTCR (XX)

Here, the permissible contact stress for gear is swcG, the allowable bending stress number is sac, the contact safety factor is SH, the hardness ratio factor for pitting resistance for gear is CHG, the stress cycle factor for pitting resistance for gear is CLG.

Write the expression for contact stress for gear.

scG=Cp(WtFdPIKoKvCSKmCxc)1/2 (XXI)

Here, the contact stress in gear is scG size factor for pitting resistance is CS, the crowning factor for pitting is Cxc, Contact geometry factor is I and the Elastic Coefficient is CP.

For limiting condition contact stress in gear is equal to permissible contact stress in gear.

Substitute scG for swcG in Equation (XX).

scG=sacCLGCHGSHKTCR (XXII)

Substitute Cp(WtFdPIKoKvCSKmCxc)1/2 for scG in Equation (XXII).

Cp(WtFdPIKoKvCSKmCxc)1/2=sacCLGCHGSHKTCR(WtFdPIKoKvCSKmCxc)1/2=sacCLGCHGSHKTCRCpWt=(sacCLGCHGSHKTCRCp)2FdPIKoKvCSKmCxc (XXIII)

Substitute (sacCLGCHGSHKTCRCp)2FdPIKoKvCSKmCxc for Wt in Equation (XVIII).

(sacCLGCHGSHKTCRCp)2FdPIKoKvCSKmCxc={(satPSF)(KLPKTKR)(FKxJPPdKoKvKSKm)}(sacCLGCHGSHKTCRCp)2=(satPSF)(KLPKTKR)(KxJPPdKS)(CSCxcdPI)(sacCLGCHGSHKTCRCp)={(satPSF)(KLPKTKR)(KxJPPdKS)(CSCxcdPI)}sac={(satPSF)(KLPKTKR)(KxJPPdKS)(CSCxcdPI)}×SHKTCRCpCLGCHG (XXIV)

Substitute KR for CR and n for PddP in Equation (XXIV).

sac={(satPSF)(KLPKTKR)(KxJPKS)(CSCxcnI)}×SHKTKRCpCLGCHGsac=CPCLG{(SH2satPSF)(KLPKT(CHG)2)(KxJPKS)(CSCxcnI)} (XXV)

For equal amount of transmitted load SH2SF as 1.

Substitute 1 for SH2SF in in equation (XXV).

sac=CPCLG{(1×satP)(KLPKT(CHG)2)(KxJPKS)(CSCxcnI)}sac=CPCLGCHG{(1×satP)(KLPKT)(KxJPKS)(CSCxcnI)} (XXVI)

Conclusion:

Thus, the AGMA contact strength of the gear case is related to the AGMA core bending strength of the pinion core by sac=CPCLG{(SH2satPSF)(KLPKT(CHG)2)(KxJPKS)(CSCxcnI)}.

(c)

Expert Solution
Check Mark
To determine

The AGMA contact strength of gear relation with contact strength of pinion.

Answer to Problem 7P

The AGMA contact strength of gear is related to contact strength of pinion by sacP=sacGCHG(mG)0.0602.

Explanation of Solution

Write the expression for gear ratio of gear set.

mG=Nn (XXVII)

Write the expression for stress cycle factor for pitting resistance for pinion.

CLP=3.4822(NLP)0.0602 for 104NLP1010 (XXVIII)

Here, the stress cycle factor for pitting resistance for pinion is CLP and the life goal in revolution for pinion is NLP.

Write the critical expression for stress cycle factor for pitting resistance for gear.

CLG=3.4822(NLPmG)0.0602 for 3×106NLP1010 (XXIX)

Here, the stress cycle factor for pitting resistance for gear is CLG.

Write the expression for permissible contact stress number for pinion.

swcP=sacPCLPSHKTCRSH=sacPCLPswcPKTCR (XXX)

Here, the permissible contact stress number for pinion is swcP, the contact safety factor is SH and the temperature factor is KT.

Write the expression for permissible contact stress number for gear.

swcG=sacGCLGCHGSHKTCRSH=sacGCLGCHGswcGKTCR (XXXI)

Substitute sacGCLGCHGswcGKTCR for SH in Equation (XXX).

sacGCLGCHGswcGKTCR=sacPCLPswcPKTCR (XXXII)

Write the expression for contact stress for pinion.

scP=Cp(WtFdPIKoKvCSKmCxc)1/2 (XXXIII)

Here, the contact stress in pinion is sCP, the transmitted load is Wt, the overload factor is Ko, crowning factor for pitting resistance is Cxc and geometry factor for pitting resistance is I.

Write the expression for contact stress for gear.

scG=Cp(WtFdPIKoKvCSKmCxc)1/2 (XXXIV)

Here, the contact stress in gear is scG and the transmitted load is Wt.

For limiting condition contact stress on gear and pinion is equal to permissible contact stress on gear and pinion.

Substitute scG for swtG and scP for swtP in Equation (XXXII).

sacGCLGCHGscGKTCR=sacPCLPscPKTCR (XXXV)

Substitute Cp(WtFdPIKoKvCSKmCxc)1/2 for scG in Equation (XXXV).

sacGCLGCHG{Cp(WtFdPIKoKvCSKmCxc)1/2}KTCR=sacPCLPscPKTCR (XXXVI)

Substitute Cp(WtFdPIKoKvCSKmCxc)1/2 for scP in Equation (XXXVI).

sacGCLGCHG[{Cp(WtFdPIKoKvCSKmCxc)1/2}KTCR]=sacPCLP[{Cp(WtFdPIKoKvCSKmCxc)1/2}KTCR]sacPCLP=sacGCLGCHG (XXXVII)

Substitute 3.4822(NLP)0.0602 for CLP and 3.4822(NLPmG)0.0602 for CLG in Equation (XXXVII).

sacP(3.4822(NLP)0.0602)=sacG(3.4822(NLPmG)0.0602)CHGsacP=sacGCHG(mG)0.0602sacP=sacGCHG(mG)0.0602

Conclusion:

Thus, AGMA contact strength of gear is related to contact strength of pinion by sacP=sacGCHG(mG)0.0602.

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