Page 387 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
15:28
Brown.cls
Brown˙C09
MACHINE ENERGY
stresses into a single shear stress, with no normal stresses present, is actually the maximum
shear stress (τ max ) and given by the two terms in Eq. (9.3) as 369
τ max = τ + τ torsion (9.3)
direct
shear
From Eq. (1.12) in Chap. 1, and using Eq. (9.1), the shear stress due to direct shear is
given by Eq. (9.4) as
V F
τ = = (9.4)
direct A A
shear
where (A) is the cross-sectional area of the wire.
From Eq. (1.21) in Chap. 1, and using Eq. (9.2), the shear stress due to torsion is given
by Eq. (9.5) as

D
F × r
Tr 2 FDr
τ torsion = = = (9.5)
J J 2J
where (r) is the outside radius and (J) is the polar moment of inertia of the wire.
For a circular cross section with a diameter (d), the area (A), the outside radius (r), and
the polar moment of inertia (J) are given by the following equations
π 2
A = d (9.6)
4
d
r = (9.7)
2
4
1 4 1 d π 4

J = πr = π = d (9.8)
2 2 2 32
Substituting these expressions for the area (A), the outside radius (r), and the polar
moment of inertia (J) in Eqs. (9.4) and (9.5) gives
F F 4F
τ = = = (9.9)
direct A π 2 πd 2
shear d
4
d

FD
FDr 2 8FD
τ torsion = = π = (9.10)
2J 2 d 4 πd 3
32
Substituting the expression for the shear stress due to direct shear from Eq. (9.9) and the
expression for the shear stress due to torsion from Eq. (9.10) in Eq. (9.3) gives Eq. (9.11).
4F 8FD d 8FD
τ max = τ + τ torsion = + = + 1 (9.11)
direct πd 2 πd 3 2D πd 3
shear
If a spring index (C) is defined as
D
C = where 6 ≤ C ≤ 12 (9.12)
d