Page 263 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
14:56
Brown.cls
Brown˙C06
STATIC DESIGN AND COLUMN BUCKLING
Example 3 (Sec. 5.3):
1/2 245
2 2 2 2 1/2

σ + σ − σ 1 σ 2 1 ((112) + (56) − (112)(56))
1 2
= =
S y n 84
1 (12,544 + 3,136 − 6,272) 1/2 (9,408) 1/2 96.99
= = = = 1.15
n 84 84 84
1
n = = 0.87 (unsafe)
1.15
Example 4 (Sec. 5.3):
1/2
2 2 2 2 1/2

σ + σ − σ 1 σ 2 1 ((70) + (−70) − (70)(−70))
1 2
= =
S y n 84
1 (4,900 + 4,900 + 4,900) 1/2 (14,700) 1 / 2 121.24
= = = = 1.44
n 84 84 84
1
n = = 0.69 (unsafe)
1.44
Step 4. Compare the factors-of-safety found in step 3 with the maximum-normal-stress
theory for Example 3 and the maximum-shear-stress theory for Examples 5 and 4.
As stated earlier, the combination for Example 2 falls directly on the boundary and where
all three theories coincide. As both of the principal stresses in Example 3 are positive, use
the first expression in Eq. (6.2) to give
Example 3 (Sec. 5.3):
σ 1 1 112
= = = 1.33
S y n 84
1
n = = 0.75 (unsafe)
1.33
where the factor-of-safety is smaller than obtained with the distortion-energy theory.
For Examples 5 and 4, use the expression in Eq. (6.6) to give
Example 5 (Sec. 5.2):
σ 1 − σ 2 1 83 − (−33) 116
= = = = 1.38
S y n 84 84
1
n = = 0.72 (unsafe)
1.38
Example 4 (Sec. 5.3):
σ 1 − σ 2 1 70 − (−70) 140
= = = = 1.67
S y n 84 84
1
n = = 0.60 (unsafe)
1.67