Page 260 - Marks Calculation for Machine Design

P1: Shibu
14:56
January 4, 2005
Brown.cls
Brown˙C06
STRENGTH OF MACHINES
242
Example 2 (Sec. 5.3):
1/2
2 2 2 2 1/2

σ + σ − σ 1 σ 2 1 ((12) + (0) − (12)(0))
1 2
= =
S y n 12
1 (144 + 0 + 0) 1/2 (144) 1/2 12
= = = = 1.0
n 12 12 12
1
n = = 1.0 (okay, but marginal)
1.0
Example 3 (Sec. 5.3):
1/2
2 2 2 2 1/2

σ + σ − σ 1 σ 2 1 ((16) + (8) − (16)(8))
1 2
= =
S y n 12
1 (256 + 64 − 128) 1/2 (192) 1/2 13.86
= = = = 1.15
n 12 12 12
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 ((10) + (−10) − (10)(−10))
1 2
= =
S y n 12
1 (100 + 100 + 100) 1/2 (300) 1/2 17.32
= = = = 1.44
n 12 12 12
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 (Sec. 5.3) and the maximum-shear-stress theory for Examples 5
(Sec. 5.2) and 4 (Sec. 5.3).
As stated earlier, the combination for Example 2 (Sec. 5.3) falls directly on the boundary
and where all three theories coincide. As both the principal stresses in Example 3 (Sec. 5.3)
are positive, use the first expression in Eq. (6.2) to give
Example 3 (Sec. 5.3):
σ 1 1 16
= = = 1.33
S y n 12
1
n = = 0.75 (unsafe)
1.33
where the factor-of-safety is smaller than obtained with the distortion-energy theory.
For Examples 5 (Sec. 5.2) and 4 (Sec. 5.3), use the expression in Eq. (6.6) to give
Example 5 (Sec. 5.2):
σ 1 − σ 2 1 11 − (−4) 15
= = = = 1.25
S y n 12 12
1
n = = 0.80 (unsafe)
1.25