UNIT-05. Torsion And Buckling of columns Lecture Number-1 Mr. M.A.Mohite Mechanical Engineering S.I.T., Lonavala.

UNIT-05. Torsion And Buckling of columns

Lecture Number-1Mr. M.A.Mohite

Mechanical EngineeringS.I.T., Lonavala

Torsion of circular shafts

• Torsion is a moment that twists/deforms a member about its longitudinal axis

• By observation, if angle of rotation is small, length of shaft and its radius remain unchanged

• By definition, shear strain is

Let x dx and = d

BD = d = dx

= (/2) lim ’CA along CA

BA along BA

= ddx

Since d / dx = / = max /c

= maxc( )

• For solid shaft, shear stress varies from zero at shaft’s longitudinal axis to maximum value at its outer surface.

• Due to proportionality of triangles, or using Hooke’s law and Eqn 5-2,

= max

c( )

= maxc ∫A 2 dA

• The integral in the equation can be represented as the polar moment of inertia J, of shaft’s x-sectional area computed about its longitudinal axis

max =T cJ

max = max. shear stress in shaft, at the outer surfaceT = resultant internal torque acting at x-sectionJ = polar moment of inertia at x-sectional areac = outer radius of the shaft

Torque T tends to rotate the right-hand end of the bar counter clockwise

Therefore the shear stresses τ acting on the surface stress element will have the directions shown in fig 3-6b

G is shear modulus of elasticity

cL

G

max

TORSION FORMULA FOR A MEMBER OF CIRCULAR C/S

L

G

J

T

c

Solid shaft• J can be determined using area element in the form of a

differential ring or annulus having thickness d and circumference 2 .

• For this ring, dA = 2 d

• J is a geometric property of the circular area and is always positive. Common units used for its measurement are mm4 and m4.

J = c42

J = (co4 ci

4) 2

Hollow shaft

Ex.1) A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The shaft may be assumed to be made of mild steel with an allowable shear stress of 42 MPa. Determine the diameter of the shaft, neglecting the bending moment on the shaft.

Soln.) Given : N = 200 r.p.m. ; P = 20 kW = 20 × 103 W; τ = 42 MPa = 42 N/mm2

Let d = Diameter of the shaft.

We know that torque transmitted by the shaft,

NmN

PT 955

2002

601020

2

60 3

We also know that torque transmitted by the shaft ( T ),

333 421616

10955 dd

mmd 507.48

Ex. 2) Find the diameter of a solid steel shaft to transmit 20 kW at 200 r.p.m. The ultimate shear stress for the steel may be taken as 360 MPa and a factor of safety as 8.

If a hollow shaft is to be used in place of the solid shaft, find the inside and outside diameter when the ratio of inside to outside diameters is 0.5.

Soln. Given : P = 20 kW = 20 × 103 W ; N = 200 r.p.m. ; τu = 360 MPa = 360 N/mm2 ; F.S. = 8 ; k = di / do = 0.5

We know that the allowable shear stress,

245

8

360

. mm

N

SF

u

Diameter of the solid shaft:-Let d = Diameter of the solid shaft.We know that torque transmitted by the shaft,

NmN

PT 955

2002

601020

2

60 3

We know that torque transmitted by the shaft ( T ),

333 451616

10955 dd

mmd 506.47

Diameter of hollow shaft:-Let di = Inside diameter, and

do = Outside diameter.

We know torque transmitted by the hollow shaft ( T )

)1(16

10955 433 kdo

)1()(4516

43 kdo

do = 48.6 mm say 50mm.di = 0.5do = 0.5 x 50 = 25 mm