Page 335 - Marks Calculation for Machine Design

P1: Shashi
15:4
January 4, 2005
Brown.cls
Brown˙C07
317
FATIGUE AND DYNAMIC DESIGN
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
2 (70 MPa) 70 MPa

σ m

σ m
m
2
σ ,σ m = ± + τ = ± + (56 MPa) 2
1 2 m
2 2 2 2

2 2
= (25 MPa) ± (1,225 + 3,136) MPa = (35 MPa) ± (4,361) MPa
= (35 MPa) ± (66 MPa) = 101 MPa, −31 MPa

2
2

a a σ a σ a 2 (161 MPa) 161 MPa 2
σ ,σ = ± + τ = ± + (0MPa)
a
1 2 2 2 2 2

2
= (80.5MPa) ± (80.5MPa) = (80.5MPa) ± (80.5MPa)
= 161 MPa, 0MPa
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 = (101) + (−31) − (101)(−31) MPa

2 2
= (10,201) + (961) + (3,131) MPa = (14,293) MPa
= 120 MPa

eff a 2 a 2 a a 2 2 2

σ a = σ 1 + σ 2 − σ 1 σ 2 = (161) + (0) − (161)(0) MPa
= 161 MPa
Step 9A. Using Eq. (7.40) calculate the factor-of-safety (n) as
σ eff σ eff 1 161 MPa 120 MPa
a m
+ = = + = (0.797) + (0.229) = 1.026
S e S ut n 202 MPa 525 MPa
1
n = = 0.975 (unsafe!)
1.026
which means the design is unsafe as the factor-of-safety n is less than 1.
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 on a Goodman diagram like that shown in Fig. 7.33.
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.