Page 332 - Marks Calculation for Machine Design

P1: Shashi
January 4, 2005
Brown˙C07
Brown.cls
314
Step 5. Multiply the alternating normal stress (σ a ) by the reduced stress concentration
factor (K f ) to give 15:4 STRENGTH OF MACHINES
σ a = (1.15)(20 kpsi) = 23 kpsi
Step 6. There are no alternating axial stresses, so proceed to step 7.
Step 7. Calculate the two sets of principal stresses using Eqs. (7.36) and (7.37); one set
for the mean normal and shear stresses and one set for the alternating normal and shear
stresses.

2 (10 kpsi) 10 kpsi 2
m
2
σ ,σ m = σ m ± σ m + τ = ± + (8 kpsi) 2
1 2 m
2 2 2 2

2 2
= (5 kpsi) ± (25 + 64) kpsi = (5 kpsi) ± (89) kpsi
= (5 kpsi) ± (9.4 kpsi) = 14.4 kpsi, −4.4 kpsi

2
a a σ a σ a 2 2 (23 kpsi) 23 kpsi 2
σ ,σ = ± + τ = ± + (0 kpsi)
1 2 2 2 a 2 2

2
= (11.5 kpsi) ± (11.5 kpsi) = (11.5 kpsi) ± (11.5 kpsi)
= 23 kpsi, 0 kpsi
Step 8. Using Eqs. (7.38) and (7.39) calculate the effective mean stress and the effective
alternating stress.

eff m 2 m 2 m m 2 2 2

σ m = σ 1 + σ 2 − σ 1 σ 2 = (14.4) + (−4.4) − (14.4)(−4.4) kpsi

2 2
= (207.36) + (19.36) + (63.36) kpsi = (290.08) kpsi
= 17 kpsi

eff a 2 a 2 a a 2 2 2

σ a = σ 1 + σ 2 − σ 1 σ 2 = (23) + (0) − (23)(0) kpsi
= 23 kpsi
Step 9A. Using Eq. (7.40) calculate the factor-of-safety (n) as
σ a eff σ m eff 1 23 kpsi 17 kpsi
+ = = + = (0.799) + (0.227) = 1.026
S e S ut n 28.8 kpsi 75 kpsi
1
n = = 0.975 (unsafe!)
1.026
which as the factor-of-safety (n) is less than 1 means the design is unsafe.
Step 9B. Plot the mean effective stress (σ eff ) and alternating effective stress (σ eff ) from
m a
step 8, the given ultimate shear strength (S ut ), and the endurance limit (S e ) calculated in
step 1F in a Goodman diagram like that shown in Fig. 7.31.
Notice that the point (σ m eff ,σ a eff ) is just outside the Goodman line, confirming the calcu-
lation in step 9A that the design is unsafe.