A single cylinder air compressor head sees forces that range

A single cylinder air compressor head sees forces that range from 0 to 18.5 kN at each cycle. The compressor head is 100 mm thick 10CM aluminum, and the compressor block is cast iron. To fix the compressor head on the block it is proposed to use M 6 times 1 cap screws of rolled threads, class 5.8. (a) Assume that the load is static with a value of 18.5 kN and find the number of screws of N needed to give a load factor of n_L = 1.2. (b) Based on the number of screws calculated in part (a), find the proof factor of safety (n_p), load factor (n_L) and joint separation factor (n_0). (c) Now find the fatigue factor of safety (n_f) using the Goodman criterion. Consider the following while solving this problem: The effective grip length is l = 160 mm. The connection is permanent. The fully corrected endurance limit for the fatigue analysis is S_e = 127 MPa.

Solution

solution:

1)for given cap screw assembly we have member alluminium and cast iron is series and cap screw is parallel to member and stiffness of block memeber is given by below empirical relation

Km=E*d*Aexp^(B*d/l)

where for alluminium

l=100 mm,A=.71670 and B=.63816 and E=71000 MPa

for screw of d=6 mm and pitch as 1 mm

we have value for Km1=442627.67 N/mm2

for cast iron we have value as

l=60mm,A=.77871,B=.61616,E=100000 MPa

Km2=496920.016 N/mm2

hence resultant member 1/Km=1/Km1+1/Km2

Km=234102.59 N/mm2

for bolt we have steel as material we get

Kb=A*E/l=pi*.25*6^2*207000/160=36579.91 N/mm2

where value of C as load taken by single bolt is given by

C=Kb/(Kb+Km)=.135139

where for load factor is given by

nl=Sp*A-Fi/C*(P/N)

where for steel of class 5-8 as

Sp=380 MPa,Sut=520 MPa,Syt=420 MPa

Fi=.9*Sp*A- for permanent connection

so we get number of screw for given load factor of nl=1.2

N=2.79227 approx to 3

so number of screw is N=3

2)for same condition

proof factor of safety is=nf=Sp/Sb

Sb=[C(P/N)+.9Sp*A]/A=371.47

Sp=380

so nf=380/371.47=1.0229

where load factor of safety is

nl=Sp*A-Fi/C*(P/N)

Fi=.9*Sp*A

nl=380*pi*.25*6^2(1-.9)/[.135139*(18500/3)]

nl=1.2892

where factor of joint separation is given by

no=Fi/[(P/N)(1-C)]=1.8130

3)where for fatigue loading we have that

Pmin=0

Pmax=18.5 KN

Fbmin=C(Pmin/N)+Fi=9669.82 N

Fbmax=C(Pmax/N)+Fi=10503.17 N

where amplitude stress is

Sa=Fbmax-Fbmin/2*A=14.73 MPa

mean stress

Sm=Fbmax+Fbmin/2*A=356.73 MPa

where initial stress is

Si=Fi/A=Sm-Sa=341.99 MPa

here factor of fatigue is

nfatigue=Se(Sut-Si)/[Sut*Sa+Se(Sut-Si)]=2.3718

4)in this way all value are calculated for capscrew assembly and corresponding factor of safety is calculated as above.