Page 314 - Marks Calculation for Machine Design

P1: Shashi
January 4, 2005
15:4
Brown.cls
Brown˙C07
STRENGTH OF MACHINES
296
solution
Step 1. Using Eq. (7.8) and values for the coefficient (a) and exponent (b) from Table 7.1,
calculate the surface finish factor (k a ) as
b −0.085
k a = aS ut = (1.58 MPa)(595 MPa) = (1.58)(0.5810) = 0.92
Step 2. Using Eq. (7.12) calculate the effective diameter (d e ) as

2
d e = 0.808 (bh) 1/2 = 0.808 (0.16 cm)(1.25 cm) = (0.808) (0.20 cm )
= (0.808)(0.4472 cm) = 0.361 cm = 3.61 mm
Step 3. Using Eq. (7.10) calculate the size factor (k b ) as
−0.1133 −0.1133

d e 3.61 −0.1133
∼
k b = = = (0.474) = 1.09 = 1
7.62 7.62
Step 4. As the beam is in bending the load type factor (k c ) from Eq. (7.14) is
k c = 1
Step 5. As the beam is operating at room temperature, the temperature factor (k d ) from
Eq. (7.15) and Table 7.2 is
k d = 1
Step 6. Using the given reduced stress concentration factor (K f ), calculate the miscella-
neous effect factor (k e ) using Eq. (7.16) as
1 1
k e = = = 0.83
K f 1.2
Step 7. Using the given ultimate tensile stress (S ut ) and the guidelines in Eq. (7.1), calculate
the test specimen endurance limit (S ) as

e

S = 0.504 S ut = (0.504)(595 MPa) = 300 MPa
e
Step 8. Using the test specimen endurance limit (S ) found in Step 7 and the modifying

e
factors found in Steps 1 through 6, calculate the endurance limit (S e ) for the cantilevered
beam using the Marin equation in Eq. (7.7) as

S e = k a k b k c k d k e S = (0.92)(1)(1)(1)(0.83)(300 MPa)
e
= (0.764)(300 MPa) = 229.1MPa
Step 9. Calculate the mean force (F m ) and the alternating force (F a ) as
F max + F min (25.2N) + (10.8N) 36 N
F m = = = = 18 N
2 2 2
F max − F min (25.2N) − (10.8N) 14.4N
F a = = = = 7.2N
2 2 2
Step 10. Calculate the mean bending moment (M m ) and the alternating bending moment
(M a ) as
M m = F m L = (18 N)(6cm) = 108 N · cm = 1.08 N · m
M a = F a L = (7.2N)(6cm) = 43.2N · cm = 0.43 N · m