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TORSION IN CIRCULAR SHAFT
Torque is a moment that tends to twist a member about its longitudinal axis. When a member is subjected to a moment about its longitudinal axis , then torque is said to be applied and the member is said to be under torsion. 2
- In workshops and factories, rotating shafts used to transmit energy are subjected to torsion.
- Its effect is of primary concern in the design of axles or drive shafts used in vehicles and machinery. 3
A’
A
o
Angle of Twist
When a shaft is subjected to torsion, point A o surface of shaft comes to A’. Then angle AOA’ subtended at center is termed as angle of twist.
- A beam bends due to bending moment, a shaft twists due to torsion.
- A member is said to be under pure torsion when it is subjected to a torque only, without being associated with any bending moment or axial force. When a shaft is under the action of pure torsion, its cross-section are under pure shear stresses.
- Moment is applied in a vertical plane perpendicular to longitudinal axis of the beam it will be subjected to torsion. 4
- When section is subjected to axial force tensile or compressive , the stresses induced in section are tensile stresses or compressive stresses also known as direct stresses. The stresses are uniform throughout the section.
- When section is subjected to bending moment , the stresses induced in section are bending stresses. The stresses are maximum at extreme fiber in a plane of bending of section and zero at neutral axis.
Stresses due to different loading
- When section is subjected to shear force, the stresses induced in section are shearing stresses. The shearing stresses are maximum at N.A. in most of the sections and zero at extreme fibers for all sections.
- When circular section is subjected to twisting moment , the stresses induced in section are shearing stresses. The shearing stresses are maximum at circumference and zero at center of circular section.
Stresses due to different loading
1 Bending Stress (due to B.M.) 2 Shear Stress (due to S.F.) 3 Direct Stress (due to A.F.) 4 Shear Stress (due to torsion)^7 1 Bending Stress (due to B.M.) 2 Shear Stress (due to S.F.) 3 Direct Stress (due to A.F.)
So, Shear Strain = fs/N = Shear stress/Modulus of rigidity Also = AA’/CA’ = OA* /CA = R /L (Length of arc AA’ = R )
C
A
A’ A’
A
o
L
T
T
T
T
Hence = fs/N = R /L or fs/R = N /L Similarly layer such as DB. Distance ‘r’ from axis of shaft will be distorted to DB’, through an angle 1. Q / N = 1 = BB 1 /DB = r / L or q/r = N /L fs/R = N /L = q/r Relation between angle of twist and shear stress
A
A’
R
r
Q q
Note: The intensity of shear stress at any point in the C.S. of the shaft subjected to pure torsion is proportional to distance from the center. The intensity is zero at the center and increases to fs at the surface varying linearly. Resisting Torque: From equilibrium condition Mx = 0 i.e. External torque T = Resisting Torque. = Moment of tangential shearing stresses acting on any transverse section. 15 Consider any concentric ring of radius ’r’ and thickness ‘r’ Shear Force = fs * a Turning Moment = fs* a *r Area of strip a = 2 r r r
R
r
Further,
Net torsional moment, T = ∫ ோ (fs) a r= ∫ ோ (fs/R) r^2 a= ∫ ோ(N /L) r^2 a now r^2 a , represents polar moment of inertia J about longitudinal axis of shaft. Hence T= (N /L) *J or T /J= (N /L) T /J= (fs/R) = N /L ----------(Equation of Pure Torsion) 17 Compare: Tr / J = fs/ R = N / L with M / I = f / y = E / R The expression J/R corresponds to I/y = I/(d/2) and is called Torsional Section Modulus. The expression EI = Flexural Rigidity The expression AE = Axial Rigidity The expression NJ = Torsional Rigidity of shaft and is the relative stiffness of two shafts measured by the inverse ratio of the angles of twist in equal lengths of shaft when subjected to equal torques. 18
Polar Moment of Inertia
J = I (^) zz=Ixx +Iyy For a circular section Ixx =Iyy= (/64)D^4 Polar Moment of Inertia = J = 2(/64)D^4 = (/32)D^4 Torsional section Modulus = J/R = (/32)D^4 /(D/2) = (/16)D^3 For hollow circular shaft J= (/32)*(D^4 -d^4 ) 19
DESIGN OF SHAFTS
In S.I. System power (P) is measured in Watts(W). 1 W = 1 Joule/s = 1 Nm/s External Work done in transmitting P kilo-watts. = 1 x P x 10^3 Nm/s = P x 60 x 10^3 Nm/min = 60 P kNm/min If T = torque in kNm, Internal work done = Torque x angle turned through one minute. = T 2n (Where n=no. of revolution per min.) 60 P = 2 n T P = 2 n T/60 kW 20 So, T = 60 P/ 2n kNm Note: 1 Horse-power(metric) = 0.75 kW
Example 3 A solid shaft has to transmit 90 kW power at 160 rpm. Find suitable diameter for the shaft if maximum torque transmitted in each revolution exceeds the mean by 25%.Take fs=70MPa. Solution: Tav = 60 P/2n = 6090/(2 160) = 5.37 kNm T = 1.25 Tav = 1.255.37 = 6.71 kNm T = (fs/R) J = 70 / R * (/32D^4 ) From which, D = 78.74 mm; say 80 mm 25 Example 4 450kW power has to be transmitted at 100 rpm. Find (i) the necessary diameter of a solid circular shaft.(ii) the necessary diameter of a circular section, the inside diameter being 3/4 of external diameter. Allowable shear stress =75MPa.Density of material 77 kN/m^3. Which is more economical? Solution: T = 60 P/2 n = 60450/(2100) = 42.9710^6 Nmm (i) T/J =fs/R T = fs[ d^3 /16] d = [16T/(fs)]1/ = [1642.9710^6 /(75)] 1/ = 142.89 mm, say 143 mm 26 (ii) for hollow circular section d = ଷ ସ D 42.9710^6 =75(/16)[D^4 -d^4 ]/D From which, D=162.2 mm, d= 121.67mm wt. of solid shaft = ALdensity wt. of hollow shaft = ALdensity So, wt. of hollow shaft / wt. of solid shaft = Ah/As = 1. 27 Example 5 Select a suitable diameter of solid shaft of circular section to transmit 112.5 kW of power at 200 rpm.If the allowable fs =75 MPa and allowable twist is 1^0 in a length of 3 meter. Take N=0.08210^6 MPa. Solution: T = 60 P/2n = 60112.5/(2 200) = 5.37210^6 Nmm
- Strength consideration T/J =fs/R T =fs[ d^3 /16] d = [16T/(fs)]1/ = [165.37210^6 /(*75)] 1/3^ = 71.45 mm 28
- Stiffness consideration T/J = N /L 32 T/d^4 =N /L ; (here take1^0 =0.0174rad) d = [32 TL/ N ]1/ d = 103.56 mm hence stiffness criteria is critical for selection. Adopt 103.56 mm, say 104 mm. 29 Example 6 The angle of twist of a 5 meter length of shaft whose diameter is 80 mm is observed to be 0.06 radian when the shaft is revolving at 4 Hz. If N=80 GPa, find the power transmitted and maximum shear stress induced. Solution: T/J = N /L T = N J /L J = d^4 /32 = 80^4 /32 = 4.02210^6 mm^4 T = 8010^3 0.06 4.02210^6 / = 3861120 Nmm = 3861.12 Nm 30 Angular frequency ω =2 f= 2 4=25.13 rad/sec. Power =Torqueangular frequency P = T* ω =3861.1225. = 97030 W = 97.03 kW Shear Stress fs=T (R/J) = T[(d/2)/ d^4 /32] = 16 T/ d^3 = ଵ∗ଷ଼ଵ .ଵଶ∗ଵଷ ଼ య fs=38.4 N/mm^2 31 A key is a piece of metal used to connect a rotating machine element to the shaft. A key prevents a relative rotation between the two parts, and may enable torque transmission to occur. They are subjected to twisting moment and consequently shear stresses develop in it. Let qk= shear stress in the key r = distance of key from center of shaft. L = length of key B = width of key Moment of resistance of key =(qkbL)r Torque transmitted = T = d^3 /16*fs equating these, size of key could be worked out. b key 32
Keys
Example 7 A shaft of 100 mm diameter transmits 112.5 kw power at 180 rpm. A flanged coupling is keyed to the shaft by means of a key 100 mm ling and 30 mm wide. The coupling has 6 bolts of 20 mm diameter symmetrically arranged along the bolt circle of 300 mm diameter. Calculate shearing stresses in the shaft, the key and the bolts of the coupling. Solution : 60P = 2 nT so T = 60P/2n=60112.5/(2180) =5.97 kN-m=5.9710^6 N-mm Now, T = (d^3 /16)fs so fs = 16T/d^3 = 165.9710^6 / 100^3 =30.4MPa 37 Stress in key qk (bL)r =T so qk=T/bLr = 5.9710^6 /3010050=39.8MPa stress in bolts qB T = n(qB/4 dB^2 )RB so, qB= 4T/ (n dB^2 RB) = 45.9710^6 /(6 20^2 *150) = 21.11 N/mm^2 38
Closely Coiled Helical Spring
L = length of the wire d = diameter of the spring wire R = mean radius of the coil n = no. of turns or coils = Angle of twist δ = deflection of the spring due to axial load fs = maximum shear stress induced in the wire due to twisting moment N = modulus of rigidity of spring material
Total length of wire, L = Length of one coil x No. of coils = 2 R * n From: T/J = N/L 𝜃 =
𝑁𝐽 =^
ସ∗ோమ ே∗ௗర 40
The free end will be twisted through an angle . Axial movement of the free end = R . Deflection of the spring, = R = ோ∗ସ∗ோ మ ே∗ௗర
R
The Deflection , can also be obtained by equating torsional strain energy to work done by axial force and deflection. Strain energy stored in spring = work done (1/2) * P = (1/2)* T = (1/2)(PR)(64PR^2 n/Nd^4 ) 41
Example 8 In a close coiled helical spring, the diameter of each coil is to be 10 times that of wire of the spring and the maximum shear stress is not to exceed 60 N/mm^2. Maximum permissible deflection under a load of 400 N is 10 cm. Taking the shear modulus as 9 x 10^4 N/mm^2 , determine the number of coils, the diameter of the coil and energy stored in the coil. 42 Solution dia. of coil D=10d where, d = dia. of wire radius R=5d max. Shear stress, fs = 60 N/mm^2 Max. deflection, = 10cm = 100mm Load, W = 400 N Modulus of Rigidity, N = 9 * 10^4 N/mm^2 Diameter of wire:- let d= dia. of the wire. W * R = (/16)fsd^3 Diameter of wire:- let d= dia. of the wire. W * R = (/16)fsd^3 400 * 5d = (/16)60d^3 d = 13.02 mm Dia. of the coil, D = 10d = 130.2 mm Radius of the coil, R = 130.2/2 = 65.1 mm Number of coils = n From relation = (64WR^3 n)/(Nd^4 ) 100 = (6440065.1^3 n)/(9 * 10^4 13.02^4 ) n = 36.61 or say 37 Energy Stored, U = (1/2)W = (1/2)400*100 = 20000 N.mm 43 44
Questions?
