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PROBLEM (9.9) A turbine shaft of diameter d is under a thrust load P and a torque T (Fig. P9.9).
Calculate the largest permissible value of the load P if the allowable normal stress is not to exceed
Given:
σ all = 220 MPa
T = 2 kN ⋅ m,
d = 60 mm,
σ all .
SOLUTION
y
z
C
A
Maximum normal stress
occurs at a point A.
d
We have
16T 16(2 ×103 )
τ= 3=
= 47.16 MPa
πd
π (0.06)3
4P
4P
σ= 2=
= 353.7 P (P in MN)
πd
π (0.06)2
Hence
σ1 =
σ
σ
+ ( ) 2 + τ 2 = σ all
2
2
220 = 176.8 P + (176.8P) 2 + (47.16) 2
(220 − 176.8P) 2 = 31, 258 P 2 + 2, 224
or
48, 400 − 77, 792 P + 31, 258 P 2 = 31, 258 P 2 + 2, 224
Solving,
P = 0.594 MN = 594 kN
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