Page 360 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C08
Brown.cls
342
U.S. Customary 15:14 APPLICATION TO MACHINES SI/Metric
A T E steel A T E steel
k T = k T =
L T L T
2
9
6
2
2
2
(2.26 × 10 −1 in )(30 × 10 lb/in ) (1.57 × 10 −4 m )(207 × 10 N/m )
= =
(1.0625 in) (0.027 m)
9
6
= 6.38 × 10 lb/in = 1.20 × 10 N/m
Step 5. As the thicknesses (t 1 ) and (t 2 ) are not Step 5. As the thicknesses (t 1 ) and (t 2 ) are not
equal, a third thickness (t middle ) is found from equal, a third thickness (t middle ) is found from
Eq. (8.64) as Eq. (8.64) as

L grip L grip
t middle = − h t middle = − h
2 2
1.0625 in 0.027 m

= − (0.3125 in) = − (0.008 m)
2 2
= 0.219 in = 0.0055 m
Step 6. Using the third thickness (t middle ) Step 6. Using the third thickness (t middle )
found in step 5 in Eq. (8.66), calculate the found in step 5 in Eq. (8.66), calculate the
special diameter (D*) as special diameter (D*) as
∗
∗
D = 1.5 d + 2 h tan 30 ◦ D = 1.5 d + 2 h tan 30 ◦
= (1.5)(0.625 in) = (1.5)(0.016 m)
+ (2)(0.3125 in)(0.577) + (2)(0.008 m)(0.577)
= (0.9375 in) + (0.625 in) (0.577) = (0.024 m) + (0.016 m) (0.577)
= 1.30 in = 0.033 m
Step 7. Substitute the special diameter (D*) Step 7. Substitute the special diameter (D*)
found in step 6, the third thickness (t middle ) found in step 6, the third thickness (t middle )
found in step 5, and the given cone angle (α) found in step 5, and the given cone angle (α)
and modulus of elasticity (E steel ), in Eq. (8.20) and modulus of elasticity (E steel ), in Eq. (8.20)
to find the stiffness (k middle ) as to find the stiffness (k middle ) as
(π tan α) Ed (π tan α) Ed
k middle = k middle =
(2 t tan α + D − d)(D + d) (2 t tan α + D − d)(D + d)
ln ln
(2 t tan α + D + d)(D − d) (2 t tan α + D + d)(D − d)
◦ ◦
(π tan 30 ) E steel d bolt (π tan 30 ) E steel d bolt
= =
C 1 C 1
ln ln
C 2 C 2
where where
C 1 = (2 t tan α + D − d)(D + d) C 1 = (2 t tan α + D − d)(D + d)
= (2 t middle tan 30 + D − d bolt ) = (2 t middle tan 30 + D − d bolt )
∗
∗
◦
◦
∗
∗
×(D + d bolt ) ×(D + d bolt )

(2)(0.219 in)(0.577) (2)(0.0055 m)(0.577)
= =
+(1.30 in) − (0.625 in) +(0.033 m) − (0.016 m)
× ((1.30 in) + (0.625 in)) × ((0.033 m) + (0.016 m))
= (0.928 in)(1.925 in) = (0.023 m)(0.049 m)
= 1.79 in 2 = 0.00113 m 2