Set-3 Final by Jahangir (27-36)R.p65 - SIA GROUP

Design of Machine Elements-II (April/May-2012, Set-3) JNTU-Anantapur
get,
(ii)
Substituting this value in the above equation, we
θ = 180 – 2 × 6.7
= 166.600°
π
⇒ θ = 166.600 × 180
θ = 2.908 rad
Using the following relation,
2
σmax
− mv ′
2
= e µ
σs
− mv ′
Where,
S 1
= Maximum allowable stress
= 2.5 MPa
S 2
= Stress in slack side of belt
m' = Mass per metre of unit cross-section
i.e., 1000 × 1 × 1 2 = 1000 kg
πDN π× 0 .3×
720
Belt velocity, V = =
= 11.310 m/sec
60 60
Now, substituting all the values in the above equation,
we get,
(iii)
2.5×
10
σ
s
6
−1000(11.310)
−1000(11.310)
2
2
= 0 .3×
2. 908
2372083.900
= 2.393
S 2 −127916.100
2.393 σ s
– 127916.100 × 2.393 = 2372083.900
2.393 σ s
– 306103.227 = 2372083.900
2.393 σ s
= 2372083.900
+ 306103.227
2.393 σ s
= 2678187.127
2678187.127
σ s
=
2.393
= 1.119 × 10 6 N/m 2
∴σ
s
=1.119 ~ 1.12 MPa
Power transmitted is given by,
P = (T 1
– T 2
) V
10×
10
T 1
– T 2
=
11.310
T
1 −T2
=
3
884.173N
e
(d) Let, b = Width of the belt in mm
t = Thickness of the belt in mm
Area
b = Thickness
1⎛
T
= ⎟ ⎞
⎜
1 −T2
t ⎝ σmax
− σs
⎠
= 6
S.33
B.Tech. III-Year II-Sem. ( JNTU-Anantapur )
3
10 ⎛ 884.173 ⎞ 1
⎜ ⎟×
9.5 ⎝ 2.5 −1.12
⎠ 10
b = 67 mm ~ 70 mm
= 0.067 m
∴ The width of belt is 70 mm.
Q6. A vertical screw with single start square
threads of 50 mm mean diameter and 12.5 mm
pitch is raised against a load of 10 kN by
means of a hand wheel, the boss of which is
threaded to act as a nut. The axial load is
taken up by a thrust collar which supports
the wheel boss and has a mean diameter of
60 mm. The coefficient of friction is 0.15 for
the screw and 0.18 for the collar. If the
tangential force applied by each hand to the
wheel is 100 N, find suitable diameter of the
hand wheel.
Answer :
April/May-12, Set-3, Q6
Given that,
Mean diameter of single start square threads,
d = 50 mm
Pitch of threads, P = 12.5 mm
Load, W = 10 kN = 10 × 10 3 N
Mean diameter of boss, D = 60 mm
Coefficient of friction for screw, µ = 0.15
Coefficient of friction for collar, µ c
= 0.18
Tangential force, P 1
= 100 N
Diameter of Hand Wheel
We have, tan α =
P
π d
= 12.5
π× 50
tan α = 0.08
Tangential force required at circumference of screw
is given as,
P = W tan(α + φ)
⎛ tan α + tan φ
= ⎟ ⎞
W
⎜
⎝1−
tan α.tan
φ ⎠
3⎛
0.08 + 0.15
= ⎟ ⎞
10×
10
⎜
⎝1−
(0.08×
0.15) ⎠
P = 2327.935 N
⎛Q
µ = tan φ
⎟ ⎞
⎜
⎝ = 0.15 ⎠