Page 17 - Mechanical Engineer's Data Handbook

6 MECHANICAL ENGINEER'S DATA HANDBOOK

Maximum principal strain theory (used for nD37,,,
special cases) Torque capacity T=- 16
FS = smallest of u,J(ul - vu2 -vu,), Power capacity P=- n2ND3
u,J(u2-vuI -vu,) and o~,/(u,-v~~ -vu1) 8
where: N = the number of revolutions per second.
Example
Angle of twist e = rad
In a three-dimensional stress system, the stresses nGD4
are a,=40MNm-2, ~,=20MNm-~ and u3= where: G =shear modulus, L = length
-10MNm-2. ~,,=200MNm-~ and v=0.3. Cal-
culate the factors of safety for each theory. T
Answer: (a) 5.0; (b) 4.0; (c) 4.5; (d) 4.6; (e) 5.4.

I. I .6 Strain energy (Resilience)

Strain energy U is the energy stored in the material of a
component due to the application of a load. Resilience
u is the strain energy per unit volume of material.
Tension and compression

Fx u2AL Hollow circular shaft
Strain energy u = - = -
2 2E
16TD n(D4-d4)
5, =
02
Resilience U = - n( D4 - d4); T= 160 5m
2E
where: D =outer diameter, d=inner diameter.
Shear n2N(D4 - d4)5,,,, 32 TL
P= , %=
8D nG(D4 - d4)
22
Resilience U = -
2G
The units for U and u are joules and joules per cubic
metre.
I. I .7 Torsion of various sections

Formulae are given for stress and angle of twist for a
solid or hollow circular shaft, a rectangular bar, a thin
tubular section, and a thin open section. The hollow
shaft size equivalent in strength to a solid shaft is given
for various ratios of bore to outside diameter. Rectangular section bar
Solid circular shafi For d>b:

16T 5, = (1'8b+3d)T (at middle of side d)
Maximum shear stress t,=- b2d2
nD3 7TL(b2 +d2)
where: D=diameter, T= torque. %= 2~b3d3