Page 37 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
12:26
Brown.cls
Brown˙C01
FUNDAMENTAL LOADINGS
Substituting for the respective polar moments of inertia from Eqs. (1.22) and (1.23), and
performing some simple algebra, gives 19
1 4 4 4 4 4 4 4
T hollow 2 π R − R i R − R i R o R i R i
o
o
= 1 = 4 = 4 − 4 = 1 − (1.26)
T solid π R 4 R o R o R o R o
2 o
For Example 2, the ratio of the inside radius to the outside radius is one-half. So
Eq. (1.26) becomes
4 4

T hollow R i 1 1 15
= 1 − = 1 − = 1 − = = 0.9375 = 93.75% (1.27)
T solid R o 2 16 16
So what is remarkable is that removing such a large portion of the shaft only reduces
the torque carrying capacity by just a little over 6 percent. Note that Eq. (1.27) is a general
relationship and can be used for any ratio of inside and outside diameters.
Strain. As a consequence of the torsional loading on the circular shaft, there is a twisting
of the shaft along its geometric axis. This produces a shear strain (γ ) which is given in
Eq. (1.28), without providing the details,
rφ
γ = 0 ≤ r ≤ R (1.28)
L
where (φ) is the angle of twist of the shaft, measured in radians.
Deformation. The angle of twist (φ) is given by Eq. (1.29).
TL
φ = (1.29)
GJ
and shown graphically in Fig. 1.21.
Note that Eq. (1.29) is valid only in the region up to the proportional limit as it derives
from Hooke’s law for shear, Eq. (1.15). The shear stress (τ) is substituted from Eq. (1.21)
and the shear strain (γ ) is substituted from Eq. (1.28), then rearranged to give the angle of
twist (φ) given in Eq. (1.29). This algebraic process is shown in Eq. (1.30).
Tr rφ TL
τ = Gγ → = G → φ = (1.30)
J L GJ

f

T
R
0

FIGURE 1.21 Angle of twist.