Page 361 - Marks Calculation for Machine Design

P1: Sanjay
15:14
January 4, 2005
Brown.cls
Brown˙C08
U.S. Customary MACHINE ASSEMBLY SI/Metric 343
and and
C 2 = (2 t tan α + D + d)(D − d) C 2 = (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))
= (2.18 in)(0.675 in) = (0.055 m)(0.017 m)
= 1.47 in 2 = 0.00094 m 2
Therefore, Therefore,
lb N
(1.814) 30 × 10 6 (0.625 in) (1.814) 2.07 × 10 11 (.016 m)
in 2 m 2
k middle = k middle = 2
1.79 in 2 0.00113 m
ln ln 2
1.47 in 2 0.00094 m
7
9
3.40 × 10 lb/in 6.00 × 10 N/m
k middle = k middle =
0.197 0.184
8
= 1.73 × 10 lb/in = 3.26 × 10 10 N/m
Step 8. The remaining thickness of the steel Step 8. The remaining thickness of the steel
member plus the thickness of the steel washer member plus the thickness of the steel washer
is half the grip length (L grip /2). Substitute this is half the grip length (L grip /2). Substitute
thickness (L grip /2), the given bolt diameter this thickness (L grip /2), the given bolt diame-
(d bolt ), and the modulus of elasticity (E steel ) in ter (d bolt ), and the modulus of elasticity (E steel )
Eq. (8.22) to find the stiffness (k 1 ) as in Eq. (8.22) to find the stiffness (k 1 ) as
(π tan 30 ) Ed (π tan 30 ) Ed
◦
◦
k 1 = k 1 =
◦
2 t tan 30 + 0.5 d 2 t tan 30 + 0.5 d
◦
ln 5 ln 5
2 t tan 30 + 2.5 d 2 t tan 30 + 2.5 d
◦
◦
◦ ◦
(π tan 30 ) E steel d bolt (π tan 30 ) E steel d bolt
= =
C 1 C 1
ln 5 ln 5
C 2 C 2
where where
◦
◦
C 1 = 2 t tan 30 + 0.5 d C 1 = 2 t tan 30 + 0.5 d

L grip L grip
= 2 tan 30 + 0.5 d bolt = 2 tan 30 + 0.5 d bolt
◦
◦
2 2
= 2 (1.0625 in/2) tan 30 ◦ = 2 (0.027 m/2) tan 30 ◦
+ 0.5 (0.625 in) + 0.5 (.016 m)
= 0.93 in = 0.024 m
and and
◦
◦
C 2 = 2 t tan 30 + 2.5 d C 2 = 2 t tan 30 + 2.5 d

L grip L grip
= 2 tan 30 + 2.5 d bolt = 2 tan 30 + 2.5 d bolt
◦
◦
2 2