9 torsion 2015 bach khoa

Đây là tài liệu của các bạn sinh viện hiện tại đang học tại Đại học Bách Khoa TP HCM. Đồng thời cũng là giáo án của giảng viên tại Đại học Bách Khoa. Nó sẽ rất hữu ích cho công việc học tập của các Bạn. Chúc Bạn thành công.

9.0 Introduction

9.1 Torsional loads on circular shafts

9.2 Torsion of noncircular members

9.3 Helical spring under axial load

A bar subjected to loadings has internal torsion moment only on cross sections: M z

• Interested in stresses and strains of circular shafts subjected to twisting

NET TORQUE DUE TO INTERNAL STRESSES

• Although the net torque due to the shearing stresses is known, the distribution of the stresses

is not

• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform

• Distribution of shearing stresses is statically indeterminate – must consider shaft

deformations

AXIAL SHEAR COMPONENTS

• Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis

• 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

• Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft

• When subjected to torsion, every cross-section

of a circular shaft remains plane and undistorted

• Cross-sections of noncircular axisymmetric) shafts are distorted when subjected to torsion

(non-• Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric

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

• Shear strain is proportional to twist and radius

max max  and  



c L

• It follows that

• Since the ends of the element remain planar, the shear strain is equal to angle of twist

STRESSES IN ELASTIC RANGE

J c

dA c

4 2

1 c c

and max

J

T J

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

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

max 0

0 max 45

0 max 0

max

2 2

2 45

cos 2

F

A A

F

• Consider an element at 45o to the shaft axis,

• Element a is in pure shear

• Note that all stresses for elements a and c have

the same magnitude

• Element c is subjected to a tensile stress on

two faces and compressive stress on the other two

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

EXAMPLES 9.01

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

• Apply elastic torsion formulas to find minimum and maximum

stress on shaft BC

EXAMPLES 9.01

SOLUTION:

• Cut sections through shafts AB and BC

and perform static equilibrium analysis

to find torque loadings

CD AB

AB x

T T

T M

m kN 6

m kN 20

m kN 14 m

kN 6 0

T

T M

EXAMPLES 9.01

• Apply elastic torsion formulas to

find minimum and maximum

stress on shaft BC

       

4 6

4 4

4 1

4

2

m 10 92

.

13

045 0 060

0 2 2

J

MPa 2

86

m 10 92 13

m 060 0 m kN 20

4 6

2 2

64

mm 60

mm 45 MPa

2 86 min

min 2

1 max

64

MPa 2

86

m 10 9 38

m kN 6 65

3

3 2

4 2 max

Tc J

2 

 c d

• Recall that the angle of twist and maximum shearing strain are related,





i i i

i i G J

L T



SUMMARY

4 2

The relative angle of twist (góc xoắn tỷ đối),

• 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,

which is not sufficient to find the end torques The problem is statically indeterminate

ft lb

90 



 B

A T T

ft lb 90 1

J L T

• Substitute into the original equilibrium equation,

A B

B

J L

J L T

G J

L T G J

L T

1 2

2 1 2

2 1

1 2

EXAMPLE 9.02

Two solid steel shafts are connected

by gears Knowing that for each shaft

G = 11.2 x 106 psi and that the

allowable shearing stress is 8 ksi,

determine (a) the largest torque T 0

that may be applied to the end of shaft

AB, (b) the corresponding angle

through which end A of shaft AB

• Find the maximum allowable torque

on each shaft – choose the smallest

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

SOLUTION:

• Apply a static equilibrium analysis on

the two shafts to find a relationship

in.

45 2 0

in.

875 0 0

T T

T F

M

T F

M

CD

CD C

C C

B

C B

C C B

B

r r

r r

in.

875 0

in.

45 2









EXAMPLE 9.02

• Find the T 0 for the maximum

allowable torque on each shaft –

choose the smallest

in.

5 0

in.

5 0 8 2 8000

in.

lb 663

in.

375 0

in.

375 0 8000

0

4 2

0 max

0

4 2

0 max

J

c T T

T psi J

c T

CD CD

AB AB

0  

T

• Find the corresponding angle of twist for each

shaft and the net angular rotation of end A

/

o o

o

6 4

2 /

o

6 4

2 /

2.22 26

8

26 8 95

2 8 2 8

2

95 2 rad 514 0

psi 10 2 11 in.

5 0

24 in.

lb 561 8 2

2.22 rad

387 0

psi 10 2 11 in.

375 0

24 in.

lb 561

A

C B

CD

CD D

C

AB

AB B

A

in G

J

L T

in G

J

L T



A



• Principal transmission shaft

performance specifications are:

T

fT T

2

max

4 1

4 2 2 2

max 3

c c c

J

T c

c J

J Tc

• Designer must select shaft

material and cross-section to

meet performance specifications

without exceeding allowable

shearing stress

STRESS CONCENTRATIONS

• The derivation of the torsion formula,

assumed a circular shaft with uniform cross-section loaded through rigid end plates

• At large values of a/b, the maximum

shear stress and angle of twist for other open sections are the same as a

rectangular bar

G ab c

TL ab

c

T

3 2

2 1



• For uniform rectangular cross-sections,

• Previous torsion formulas are valid for axisymmetric or circular shafts

• Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly

SUMMARY FOR TORSION OF RECTANGULAR MEMBERS

• Summing forces in the x-direction on AB,

shear stress varies inversely with thickness

flow shear

t

x t x

t F

B B A A

B B A

A x

qA dA

q dM

T

dA q pds

q ds t p dF p dM

2

2 2

2 0

• Compute the shaft torque from the integral

of the moments due to shear stress





t

ds G A

of 0.160 in and wall thicknesses of (b) 0.120

in on AB and CD and 0.200 in on CD and

1 in.

986 8 2

in.

kip 24 2

-in.

986 8 in.

34 2 in.

84 3

A

• Find the corresponding shearing stress with each wall thickness with a uniform wall thickness,

in.

160 0

in.

kip 335 1

in.

kip 335 1

in.

kip 335 1



 CD

BC 



INTRODUCTION

INTRODUCTION

τ : Torsional corrected stress

K : Application correction factor

BASIC FORMULAS FOR SPRING

3 4



1

25 0

D K