Page 104 - Analysis and Design of Machine Elements

Analysis and Design of Machine Elements
82
Steps Computation Results Units
(3) Compute the force on the bolts to resist the
bending moment from
ML max 530400 × 80
F max = z = 2 2 N = 3315N
∑ 2 80 + 80
L
i=1 i
F = 5083 N
The total axial working force on the left bolt
F = F + F = 1768 + 3315 = 5083N
a max
(4) The connection has tendency of sliding relative
to each other under the load F .Toprevent such
Σh
situation, the following condition should be met.
( )
k m
f zQ − F Σv ≥ K F
p
s Σh
k + k m
b
k
Select b = 0.2, so
k + k m
b
k k
m = 1 − b = 0.8, select antiskid factor
k + k k + k
b m b m
K = 1.2, The preload of each bolt is Q = 14 674 N
s
p
( )
1 K F k m
s Σh
Q ≥ + F Σv =
p
z f k + k m
b
1 ( 1.2 × 3536 )
× + 0.8 × 3536 N = 14674N
2 0.16
(5) The total force the left bolt subjected to is
k b
Q = Q + F = 14674 + 0.2 × 5083 =
p
k + k m
b
15691N Q = 15 691 N
3. Determine the Select the bolt property class as 4.6. From
diameter of the Table3.2,the yieldstrengthofmaterialis
bolts = 240 MPa.
s
Assuming the safety factor is S = 1.5. The allowable
stress of bolt material is
240
[ ]= s = = 160MPa
S 1.5
The required minor diameter of the bolt will be
M16 mm
√ √
4 × 1.3Q 4 × 1.3 × 15691
d ≥ = = 12.7mm
1 [ ] 3.14 × 160
From Table 3.1, select major diameter d = 16 mm
(d = 13.6 > 12.7 mm).
1