HJD Institute Of

Technical

Education & Research

>Thakkar Brije

sh

>Vora Nitin

>Bhudiya Dinesh

>Mara Mustaq

>Gorsiya

Jetendra

1.TORSION

Torsion Of Circular Shafts1.Torque Or Turning Moment Or Twisting Moment:-

In factories and workshops,shaft is use to transmit energy from one end to the other end.to transmit the energy a turning forces is applied either to the rim of pulley,key to the shaft or to any other suitable point at some distance from the axis of shaft. The moment of couple acting on the shaft is called torque or turning moment or twisting moment.

Torque=turning force*dia.of shaftT=F*2R

Where T=torqueF=turning forceR=radius of shaft

2.Angle Of Twist () :-

When a shaft is subjected to torue(T),Point A on the surface of the shaft comes to A’ position. The angle AOA’ aat the center of shft is called the angle of twist().

AOA’=

Angle of twist () is measured in radiance.

3.Shear Stress in shaft ():-When a shaft is subjected to equal and opposite end couples, whose axis coincide with the axis of the shaft, the shaft is said it be in pure torsion and at any point in the section of the shaft stress will be induced. That stress is called shear stress in shaft.

4.Strenth of the shafts:-Maximum torque or power the shaft can transmit from one pulley to another, is called strength of shaft.

(A)For solid circular shaft: (B)For hollow circular shafts:

T=*()*()T= T=*()*{(-)/D}

Where WhereD=Dia. Of shaft

D=outer dia. Of the shaft ()=Shear stress in shaftd=inner dia. Of the shaft

5.Polar moment of the inertia(J):-The moment of inertia of a planner area , with respect to an axis perpendicular to the plane of the fig. is called polar moment of inertia.

Izz = Ixx+Iyy = J = *+ * J= *

For solid circular section……Ixx = Iyy = *

For hollow circular shaft…..J=(- )

Polar section Modulus (Zp) :-

Polar section Modulus = (polar M.I.)/(distance of extreme fibre from c.g.)

Zp = J/Y = J/R

For solid circular shaft : Zp = =

For Hollow circular Shaft : Zp = = =

6.Assumption in the theory of Torsion:

1. The material of the shaft is uniform throughout the length.2. The twist along the shaft is uniform.3. The shaft is of uniform circular section throughout the length.4. Cross sections of the shaft , which are plane before twist remain

plane after twist.5. All radii which are straight before twist remain straight after twist.

7.Theory of torsion and torsion Equation:Consider a shaft fixed at one end and subjected to torque at other end.

Let, T= Torque l=length of the shaft R=Radius of the shaft

At a result of torque every cross-section of the shaft will be subjected to shear stresses.

Line CA on the surface of the shaft will be deformed to CA’ and OA to OA’ ,as shown on figure.

Let,ACA’= Ø in degree = shear strain AOA’= in degree = angle of twist

C= modulus of rigidity

Now,Shear strain = AA’/l = tan Ø

= Ø (.: Ø is very small )Length of arc AA’= R.

Ø = AA’/l = R./l-----------------(1)

Now, Modulus of rigidity =:. C = Ø:. Ø = /C-------------------(2)

Equating (1) and (2)

:. ------------------(3)We know that for solid circular shaft, T= >>>>>>>>

Substitute the value of in equation (3) ,

(.: R=D/2)

T/J=(C* /l------------------(4)

From equation 3 and 4

this is equation of torsion.

8.Power Transmitted by a shaft :-

(a)Power in horse power(h.p):

P=2NT/4500 h.p Where N=R.P.M T=Torque in kg.m

(b) Power in watts :P=2NT/60 Watt

9.Torsional Rigidity :-

Let twisting moment T , produce a twist radians in a length l.

So that T/J = C./l

So that T.l/CJThe quantity CJ in the above equ . Is known as Torsional rigidity.

The quantity CJ/l is known as Torsional stiffness.The quantity l/CJ is known as Torsional flexibility.

10. Shaft Couplings:-When length of shaft required is very large, due to non-availability of a single shaft of required length, it becomes necessary to connect two shafts together. This is usually done means of flanged coupling as shown below.

* The flanges of two shafts are joined together by bolts and nuts or rivets and the torque is then transferred from one shaft to another through the couplings.* As the torque is transferred through the bolts , bolts will be subjected to shear stress. As the diameter of bolts is small ,as compared to be uniform in the bolts.

(A)Design of bolts.

*()*=n*8*d* b*D

(B) Design of keys.

*()*=(l*b* k)d/2

Design of keys :-

11.Closed coil helical springs :-

Stiffness of a spring : The load required to produced a unit deflection in a spring is called stiffness of a spring.

Uses of springs : a spring is used to absorb energy due to resilience, which may be restored as and when required. Followings are uses of springs.

1. Spring in which for storage of energy.2. Spring in railways, automobile etc. to absorb shocks.3. Spring in brakes and clutches to apply required force.4. Spring in spring balance in which deflection of spring gives

weight or mass.5. Spring in seat of cycles , scooter , sofa set etc.

Types of springs:-

Bending springs : a spring, which is subjected to bending only , and the resilience is also due to bending is called bending spring.Leaf springs or laminated springs are bending spring.

Torsion spring : a spring , which is subjected to torsion or twisting moment only and resilience is also due to it is called torsion spring.Helical springs are torsion springs.

Closed coil helical springs Subjected to an axial load :-

Consider a closed coil helical spring subjected to axial load as shown in figure.Let,d=dia. Of spring wireR= radius of the spring coiln= no. of coils or turnsW= axial load on spring maximum shear stress in spring wire due to twist C= modulus of rigidity of spring material.=angle of twist in the spring wire=deflection of spring.Torque produced by axial load w,

T= W.R.--------------(1)

But, we know that twisting moment,

T=--------------------------------------(2)

So that W.R=

Length of wire,

l=No.of coils*Length of one coil

=n*2

We, already know the torsion formula T/J=C

So that T=W.R J=

Deflection of the spring,

---------------------------------deflection of spring.

Stiffness of a spring,S=

Thank you….!!!