3_torsion.ppt

CHAPTER

3

MECHANICS OF SOLIDS Torsion

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MECHANICS OF SOLIDS Contents Introduction

Statically Indeterminate Shafts

Torsional Loads on Circular Shafts

Sample Problem 3.4

Net Torque Due to Internal Stresses

Design of Transmission Shafts

Axial Shear Components Shaft Deformations Shearing Strain Stresses in Elastic Range

Normal Stresses Torsional Failure Modes Sample Problem 3.1 Angle of Twist in Elastic Range

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MECHANICS OF SOLIDS Torsional Loads on Circular Shafts • Interested in stresses and strains of circular shafts subjected to twisting couples or torques • Turbine exerts torque T on the shaft • Shaft transmits the torque to the generator • Generator creates an equal and opposite torque T’

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MECHANICS OF SOLIDS Net Torque Due to Internal Stresses • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, T    dF     dA

• Although the net torque due to the shearing stresses is known, the distribution of the stresses is not • Distribution of shearing stresses is statically indeterminate – must consider shaft deformations • Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Axial Shear Components • Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis. • Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft • The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats.

The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

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MECHANICS OF SOLIDS Shaft Deformations • From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length.  T L

• When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted. • Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric. • Cross-sections of noncircular (nonaxisymmetric) shafts are distorted when subjected to torsion. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Shearing Strain • Consider an interior section of the shaft. As a torsional load is applied, an element on the interior cylinder deforms into a rhombus. • Since the ends of the element remain planar, the shear strain is equal to angle of twist. • It follows that L   or  

 L

• Shear strain is proportional to twist and radius  max 

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c  and    max L c

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MECHANICS OF SOLIDS Stresses in Elastic Range • Multiplying the previous equation by the shear modulus, G 

 c

G max

From Hooke’s Law,   G , so 

 c

 max

The shearing stress varies linearly with the radial position in the section.

J  12  c 4

• Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section,   T    dA  max   2 dA  max J c c



J  12  c24  c14



• The results are known as the elastic torsion formulas,

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 max 

Tc T and   J J 3-8

MECHANICS OF SOLIDS Normal Stresses • Elements with faces parallel and perpendicular to the shaft axis are subjected to shear stresses only. Normal stresses, shearing stresses or a combination of both may be found for other orientations. • Consider an element at 45o to the shaft axis, F  2 max A0 cos 45   max A0 2

 45o 

F  max A0 2    max A A0 2

• Element a is in pure shear. • Element c is subjected to a tensile stress on two faces and compressive stress on the other two. • Note that all stresses for elements a and c have the same magnitude GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Torsional Failure Modes • Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear. • When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis. • When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Sample Problem 3.1 SOLUTION: • Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings

Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown, determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD if the allowable shearing stress in these shafts is 65 MPa. GIK Institute of Engineering Sciences and Technology

• Apply elastic torsion formulas to find minimum and maximum stress on shaft BC • Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter

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MECHANICS OF SOLIDS Sample SOLUTION:Problem 3.1 • Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings

 M x  0  6 kN  m   TAB

 M x  0  6 kN  m   14 kN  m   TBC

TAB  6 kN  m  TCD

TBC  20 kN  m

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MECHANICS OF SOLIDS Sample Problem 3.1 • Apply elastic torsion formulas to find minimum and maximum stress on shaft BC

J

  c24  c14   0.060 4  0.045 4  2 2



 13.92 10

 max   2 

6

m

4

TBC c2 20 kN  m 0.060 m   J 13.92 10 6 m 4

• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter

 max 

Tc Tc  J  c4 2

65 MPa 

6 kN  m  c3 2

c  38.9 10 3 m

d  2c  77.8 mm

 86.2 MPa

 min c1   max c2

 min 86.2 MPa

 min  64.7 MPa



45 mm 60 mm

 max  86.2 MPa  min  64.7 MPa

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS Angle of Twist in Elastic Range • Recall that the angle of twist and maximum shearing strain are related,  max 

c L

• In the elastic range, the shearing strain and shear are related by Hooke’s Law,  max 

 max G



Tc JG

• Equating the expressions for shearing strain and solving for the angle of twist, 

TL JG

• If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations Ti Li i J i Gi

 

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS Statically Indeterminate Shafts • Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B. • From a free-body analysis of the shaft, TA  TB  90 lb  ft

which is not sufficient to find the end torques. The problem is statically indeterminate. • Divide the shaft into two components which must have compatible deformations,   1  2 

TA L1 TB L2  0 J1G J 2G

TB 

L1 J 2 TA L2 J1

• Substitute into the original equilibrium equation, LJ TA  1 2 TA  90 lb  ft L2 J1

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MECHANICS OF SOLIDS Sample Problem 3.4 SOLUTION: • Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0

• Apply a kinematic analysis to relate the angular rotations of the gears • Find the maximum allowable torque on each shaft – choose the smallest

Two solid steel shafts are connected by gears. Knowing that for each shaft • Find the corresponding angle of twist G = 11.2 x 106 psi and that the for each shaft and the net angular allowable shearing stress is 8 ksi, rotation of end A determine (a) the largest torque T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which end A of shaft AB rotates. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Sample Problem 3.4 SOLUTION: • Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0

 M B  0  F 0.875 in.  T0

• Apply a kinematic analysis to relate the angular rotations of the gears

rB B  rCC rC 2.45 in. C  C rB 0.875 in.

 M C  0  F 2.45 in.  TCD

B 

TCD  2.8 T0

 B  2.8C

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MECHANICS OF SOLIDS Sample Problem 3.4 • Find the T0 for the maximum • Find the corresponding angle of twist for each allowable torque on each shaft – shaft and the net angular rotation of end A choose the smallest

A / B 

 max 

T 0.375 in. TAB c 8000 psi  0  0.375 in.4 J AB 2

T0  663 lb  in.

 max 

TCDc 2.8 T0 0.5 in. 8000 psi   0.5 in.4 J CD 2

T0  561 lb  in.

T0  561 lb  in

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561lb  in.24in. TAB L  J AB G  0.375 in.4 11.2  106 psi 2





 0.387 rad  2.22o

C / D 

TCD L 2.8 561lb  in.24in.  J CD G  0.5 in.4 11.2  106 psi 2



 0.514 rad  2.95o







 B  2.8C  2.8 2.95o  8.26o  A   B   A / B  8.26o  2.22o

 A  10.48o 3 - 20

MECHANICS OF SOLIDS Design of Transmission Shafts • Principal transmission shaft performance specifications are: - power - speed • Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable shearing stress.

• Determine torque applied to shaft at specified power and speed, P  T  2fT T

P





P 2f

• Find shaft cross-section which will not exceed the maximum allowable shearing stress,  max 

Tc J

J  3 T  c  c 2  max



solid shafts



J  4 4 T  c2  c1  c2 2c2  max

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hollow shafts

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS Practice Problems

3.5, 3.8, 3.11, 3.18, 3.23, 3.24, 3.27, 3.28, 3.32, 3.36, 3.38, 3.40, 3.42, 3.43, 3.44, 3.49, 3.53, 3.57, 3.64, 3.68, 3.72, 3.74, 3.76 GIK Institute of Engineering Sciences and Technology

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