Page 380 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
15:14
Brown.cls
Brown˙C08
362
L
1
L –D
D o APPLICATION TO MACHINES
o
1
A
a
t
r o t shear torsion
L /2
2
a
b = 180∞ - a t max
O
FIGURE 8.20 Maximum shear stress diagram.
The shear stress (τ shear ) acts downward on all the welds; however, the shear stress (τ torsion )
acts perpendicular to the radial distance (r o ), an angle (α) from horizontal.
Using the dimensions in Fig. 8.20, the angle (α) is calculated as
α = tan −1 L 1 − D o (8.87)
L 2
2
and the angle (β), which is the supplement of the angle (α),isgivenby
◦
β = 180 − α (8.88)
Therefore, using the law of cosines on the resulting scalene triangle formed by the three
shear stresses in Fig. 8.20, the maximum shear stress (τ max ) is determined from Eq. (8.78).
To find the required weld throat (H), divide the maximum shear stress (τ max ), which will
have units of (stress–width), by the weld strength (S weld ), which will have units of (stress),
that is,
τ max (stress − width)
(weld throat) H = = = (width) (8.89)
S weld (stress)
The required weld size (t) is then determined from the weld throat (H) as
H
(weld size) t = (8.90)
cos 45 ◦
The weld strength in shear is specified by the particular code governing the design of the
machine element. For the AWS code, the weld strength in shear is taken as 30 percent of
the ultimate tensile strength (S ut ) of the electrode material, that is

S weld = (0.30) S ut (8.91)
For example, E60xx electrode material has an ultimate tensile stress of 60 kpsi or
420 MPa; therefore the weld strength (S weld ) in shear would be

(0.30)(60 kpsi) = 18.0 kpsi
S weld = (0.30) S ut = (8.92)
(0.30)(420 MPa) = 126.0MPa