S.A. - I (130604)
Torsion
Civil Branch (Sem. – III)
Assignment - 2
01.08.2013
- : FORMULAE SUMMARY : 1.
Strength of Shaft
For Solid Circular shaft:
For Hollow Circular shaft:
Maximum Torque (T) is given by T =
2.
𝜋𝜏𝐷3
Polar Moment of Inertia : (J)
For Solid Circular shaft:
𝜋𝜏(𝐷4 − 𝑑 4 )
16𝐷
For Hollow Circular shaft:
Polar Moment of Inertia Izz = Ixx + Iyy = J =
3.
Maximum Torque (T) is given by T =
16
𝜋𝐷4
Polar Moment of Inertia Izz = Ixx + Iyy = J =
32
Section Modulus: (Zp)
For Solid Circular shaft:
𝜋(𝐷4 − 𝑑 4 )
32
For Hollow Circular shaft:
Section Modulus = Zp =
𝜋𝐷3
Section Modulus = Zp =
16
𝜋(𝐷4 − 𝑑 4 )
16𝐷
Equation of Torsion:
𝑇
𝜏
𝐶𝜃
=
=
𝐽
𝑅
𝑙
4.
5.
6.
Modulus of rigidity =
8.
9.
=> C =
𝜏
𝜙
Power Transmitted by a shaft
Power in horse power (h.p.)
2𝜋𝑁𝑇
𝑁 = 𝑅. 𝑃. 𝑀.
𝑃 =
𝑊ℎ𝑒𝑟𝑒:
4500
𝑇 = 𝑇𝑜𝑟𝑞𝑢𝑒 𝑖𝑛 𝑘𝑔. 𝑚
1kw = 10000watt
-
7.
𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑒𝑠𝑠
𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛
Degree = ( Degree X
𝜋
180
)radian
Two shaft in parallel:
Angle of twist for both shafts will be same.
Applied torque (T), will be shared by both the
shafts.
Design of shaft coupling
(a) Design of bolts:
𝜋𝜏𝑑 3
𝜋
= 𝑛 𝑑𝑏2 𝜏𝑏 𝐷
16
8
Where:d = Dia. Of shaft.
𝜏 = Shear Stress in shaft
𝜏𝑏 = Shear Stress in bolt
𝑑𝑏 = Dia. of bolt.
n = No. of bolt.
Closed coil helical springs subjected to an axial load.
Angle of twist in spring
-
Θ =
64 𝑊 𝑅 2 𝑛
𝐶 𝑑4
Stiffness of spring
-
01
02
03
S=
𝑇
𝐽
=
𝐶 𝑑4
64 𝑅 3 𝑛
Power in Watts
2𝜋𝑁𝑇
𝑁 = 𝑅. 𝑃. 𝑀.
𝑊ℎ𝑒𝑟𝑒:
60
𝑇 = 𝑇𝑜𝑟𝑞𝑢𝑒 𝑖𝑛 𝑁. 𝑚
1h.p. = 746watt
𝑃 =
-
𝑃𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑏𝑦 ℎ𝑜𝑙𝑙𝑜𝑤 𝑠ℎ𝑎𝑓𝑡
𝑃𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑏𝑦 𝑠𝑜𝑙𝑖𝑑 𝑠ℎ𝑎𝑓𝑡
=
𝑇𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑎𝑚𝑖𝑡𝑡𝑒𝑑 𝑏𝑦 ℎ𝑜𝑙𝑙𝑜𝑤 𝑠ℎ𝑎𝑓𝑡
𝑇𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑎𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑏𝑦 𝑠𝑜𝑙𝑖𝑑 𝑠ℎ𝑎𝑓𝑡
Two shaft in series:
Angle of twist will be the algebraic sum of the twist of
each shaft.
Applied torque (T), will be minimum of torque
capacity of two shaft & torque capacity obtained by
the angle of twist criteria.
(a) Design Key:
𝜋𝜏𝑑 3
𝑑
=
(𝑙 𝑏 𝜏𝑘 )
16
2
Where:d = Dia. Of shaft.
𝜏 = Shear Stress in shaft
𝜏𝑘 = Shear Stress in key
b = Width of key.
𝑙 = length of key.
Deflection in spring
δ =
-
64 𝑊 𝑅 3 𝑛
𝐶 𝑑4
Energy stored in the spring.
1
U = W.δ
2
A solid shaft of 80 mm diameter is to be replaced by a hollow shaft of external diameter
100m. Determine the internal diameter of the hollow shaft if the same power is to be
transmitted by both the shafts at the same angular velocity and shear stress.
A 60 mm diameter shaft transmits 80 kW at 100 r.p.m. The shaft is connected to machine
components by means of key, which is 20 mm wide and 100 mm long. Find the shear stress
developed in the shaft and key.
In which section the maximum torque will occur?
Department of Civil Engineering. | Darshan Institute Of Engineering & Technology.
Contact me for any quarry on Blog Address:- kckoradia.wordpress.com
Dec
2009
07
Dec
2009
07
March 02
1
S.A. - I (130604)
Torsion
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Civil Branch (Sem. – III)
Assignment - 2
01.08.2013
2010
Calculate the diameter of the shaft required to transmit 45 kW at 120rpm. The maximum March 02
torque is likely to exceed the mean by 30%, for a maximum permissible shear stress of 2010
55N/mm2. Calculate also the angle of twist for a length of 2 m. G = 80 x 103 N/mm2.
A composite shaft consisting of a solid steel core 80 mm diameter is enclosed in a closely March 02
fitting bronze sleeve. Find the outside diameter of the sleeve so that a pure torque applied 2010
to the composite shaft is shared equally by the two materials. If the torque is 16 kN.m
calculate maximum shear stress induced in each material. G steel = 80 x 103 N/mm2, G
3
2
bronze = 40 x 10 N/mm .
Dec
07
A curved beam has semicircular shape with radius of 3m and is loaded by a point load of
10kN at centre of arc. Calculate the maximum bending moment, shear force and torsion.
2010
Dec
07
For a structure as shown in the figure, draw bending moment, shear force and torsion
diagrams.
2010
A solid steel shaft has to transmit 120 kW at 600 r.p.m.Find the diameter of the shaft if
the shear stress is to be limited to 100 N/mm2. Estimate the possible % saving in the
material of the shaft if hollow shaft of internal diameter equals 0.75 times external
diameter is replaced against solid shaft.
A shaft has to transmit 105 kW power at 160rpm.If the shear stress is not to exceed
65N/mm2 & the twist in a length of 3.5m must not to exceed 1°. Find suitable diameter.
Take G=8 x 104 N/mm2.
A 100mm diameter shaft transmits 105kw power at 120rpm. A flanged coupling is keyed
to the shaft, the key being 25mm wide & 140mm long. Six bolts of 20mm dia. Are
symmetrically arranged along a bolt circle of 280mm dia. Find the shear stress induced in
the shaft, the key & bolts.
Analyse the grid shown in the Fig.2 and draw shear force, bending moment and twisting
moment diagrams.
Department of Civil Engineering. | Darshan Institute Of Engineering & Technology.
Contact me for any quarry on Blog Address:- kckoradia.wordpress.com
May
2011
07
Dec
2011
07
Dec
2011
07
May
2012
07
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