Page 321 - Marks Calculation for Machine Design

P1: Shashi
January 4, 2005
Brown˙C07
Brown.cls
(s )
329.0 a S e 15:4 FATIGUE AND DYNAMIC DESIGN 303
Scale: 17.5 MPa × 17.5 MPa
280
Calculated stresses
210
Goodman line
140 s m
124.3
70 s a
S ut
0 (s )
0 140 198.9 280 420 560 700 770 m
735
FIGURE 7.23 Goodman diagram for Example 2 (SI/metric).
Step 11. Plot the mean axial stress (σ m ) and alternating axial stress (σ a ) from step 10, the
given ultimate tensile strength (S ut ), and the endurance limit (S e ) calculated in step 6 in a
Goodman diagram like that shown in Fig. 7.23.
Step 12. To answer question (a), calculate the factor-of-safety (n) using Eq. (7.25), which
represents the distance (d) in Fig. 7.12.
1 σ a σ m 124.3MPa 198.9MPa
= + = + = (0.378) + (0.271) = 0.649
n S e S ut 329.0MPa 735 MPa
1
n = = 1.54
0.649
Step 13. To answer question (b), calculate the factor-of-safety (n m ) using Eq. (7.27), which
represents the distance (d m ) in Fig. 7.13.

σ m 198.9MPa
S e 1 − (329.0MPa) 1 −
σ a | σ m S ut 735 MPa (329.0MPa)(0.729)
n m = = = =
σ a σ a 124.3MPa 124.3MPa
239.84 MPa
= = 1.93
124.3MPa
) was substituted from Eq. (7.28).
where the alternating stress (σ a | σ m
Step 14. Multiply the factor-of-safety (n m ) found in step 13 with the alternating axial force
(F a ) to give a maximum alternating axial force (F a max ) as
F max = n m F a = (1.93)(2,250 N) = 4,343 N
a
Step 15. Use the maximum alternating axial force (F a max ) found in step 14 to deter-
mine the limiting values of the maximum axial force (F max ) and the minimum axial force
(F min ).

lim
F max = F m + F a max = (3,600 N) + (4,343 N) = 7,943 N
lim max
F = F m − F = (3,600 N) − (4,343 N) =−743 N
min a