Machine Design Databook 2010 Part 16 ppt

p pressure, MPa psiq load per unit length, kN/m lbf/in Qx, Qy shearing forces parallel to z-axis per unit length of sections of a plate perpendicular to x and y axis, N/m lbf/in Nr, N ra

p pressure, MPa (psi)

q load per unit length, kN/m (lbf/in)

Qx, Qy shearing forces parallel to z-axis per unit length of sections of a

plate perpendicular to x and y axis, N/m (lbf/in)

Nr, N radial and tangential shearing forces, N (lbf )

tension of a membrane, kN/m (lbf/in)

Mtxy twist of surface

u, v, w components or displacements, m (in)

x, y, z rectangular co-ordinates, m (in)

X, Y, Z body forces in x; y; z directions, N (lbf )

Z section modulus in bending, cm3(in3)

 density, kN/m3(lbf/in3)

x,y,z normal components of stress parallel to x, y, and z axis, MPa

(psi)

r, radial and tangential stress, MPa (psi)

r, z normal stress components in cylindrical co-ordinates, MPa

(psi)

 shearing stress, MPa (psi)

xy,yz,zx shearing stress components in rectangular co-ordinates, MPa

(psi)

" unit elongation, m/m (in/in)

"x,"y,"z unit elongation in x, y, and z direction, m/m (in/in)

"r," radial and tangential unit elongation in polar co-ordinates

xy,yz,zx shearing strain components in rectangular co-ordinate

r,z shearing strain in polar co-ordinate

r,z,rz shearing stress components in cylindrical co-ordinates, MPa

the design which will be used and observed throughout this Machine Design Data

Handbook

STRESS AT A POINT (Fig 27-1)

The stress at a point due to force
F acting normal to

an area dA (Fig 27-1b)

For stresses acting on the part II of solid body cut out

from main body in x, y and z directions, Fig 27-1b

Similarly the stress components in xy and xz planes

can be written and the nine stress components at the

point O in case of solid body made of homogeneous

and isotropic material

F ¼ force acting normal to the area
A

A ¼ an infinitesimal area of the body under theaction of F

dz a

(a) A solid body subject to action

of external forces (b) An infineticimal area ∆A of Part II of a solid body under the action of force

∆F at 0

(c) Stresses acting on the faces of a

small cube element cut out from the solid body

Summing moments about x, y and z axes, it can be

proved that the cross shears are equal

All nine components of stresses can be expressed by a

single equation

The FNx, FNy, and FNzunknown components of the

resultant stress on the plane KLM of elemental

tetra-hedron passing through point O (Fig 27-2)

The unknown components of resultant stress FNx, FNy

and FNz in terms of direction cosines l, m and n

FNx¼ xcos N; x þ xycos N; y þ xzcos N; z

FNy¼ yxcos N; x þ ycos N; y þ yzcos N; z

FNz¼ zxcos N; x þ zycos N; y þ zcos N; z ð27-6Þ

FNx¼ xl þ xym þ zxn

FNy¼ yzl þ ym þ yxn

FNz¼ zxl þ zym þ zn ð27-7Þwhere the direct cosines are

l ¼ cos  ¼ cos N; x; m ¼ cos ¼ cos N; y,

n ¼ cos  ¼ cos N; z,

lsþ m2þ n2¼ ðlÞ02þ ðm0Þ2þ ðn0Þ2¼ 1

τxy

Surface area KLM = A (normal to KLM)

x L

F z

z

y

K N

T N = stress vector in N direction

F bx , F by , F bz = Body forces in x, y and z - direction

FIGURE 27-2 The state of stress at O of an elemental

tet-rahedron.

x

y y’

σz τx’y’

+

+

+ + +

z

dx

dz

dy dy

dz

dx dz

The resultant stress FNon the plane KLM

The normal stress which acts on the plane under

consideration

The shear stress which acts on the plane under

consideration

Equations (27-1), (27-2) and (27-7) to (27-8) can be

expressed in terms of resultant stress vector as follows

(Fig 27-2)

The resultant stress vector at a point

The resultant stress vector components in x, y and z

directions

The resultant stress vector

The normal stress which acts on the plane under

consideration

The shear stress which acts on the plane under

consideration

The angle between the resultant stress vector TN and

the normal to the plane N

cos ¼ l ¼ angle between x axis and Normal Ncos ¼ m ¼ angle between y axis and Normal Ncos ¼ n ¼ angle between z axis and Normal N

FN¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiF2

Nxþ F2

Nyþ F2 Nz

q

ð27-9eÞwhere the direction cosines are

cosðTN; xÞ ¼ TNx=jTNj, cosðTN; yÞ ¼ TNy=jTNj,cosðTN; zÞ ¼ TNz=jTNj

q

ð27-10bÞcosðTN; NÞ ¼ cosðTN; xÞ cosðN; xÞ

þ cosðTN; yÞ cosðN; yÞ

þ cosðTN; zÞ cosðN; zÞ ð27-10cÞ

EQUATIONS OF EQUILIBRIUM

The equations of equilibrium in Cartesian

coordi-nates which includes body forces in three dimensions

(Fig 27-3)

Stress equations of equilibrium in two dimensions

TRANSFORMATION OF STRESS

The vector form of equations for resultant-stress

vectors TN and TN0 for two different planes and the

outer normals N and N0in two different planes

The projections of the resultant-stress vector TNonto

the outer normals N and N0

Substituting Eqs (27-9b), (27-9c), (27-9d) and (27-9e)

in Eqs (27-13), equations for TN, N and TN, N0

The relation between TN, N0and TN0, N

By coinciding outer normal N with x0, N with y0,

and N with z0 individually respectively and using

Eqs (27-14a) to (27-14b), x0, y0 and z0 can be

x0¼ Tx0x0¼ xcos2ðx0; xÞ þ ycos2ðx0; yÞ

þ zcos2ðx0; zÞ þ 2xycosðx0; xÞ cosðx0; yÞ

þ 2yzcosðx0; yÞ cosðx0; zÞ

þ 2zxcosðx0; zÞ cosðx0; xÞ ð27-15aÞ

By selecting a plane having an outer normal N

co-incident with the x0 and a second plane having an

outer normal N0coincident with the y0and utilizing

Eq (27-14b) which was developed for determining

the magnitude of the projection of a resultant stress

vector on to an arbitrary normal can be used to

determine x0 y0 Following this procedure and by

selecting N and N0coincident with the y0and z0, and

z0 and x0 axes, the expression for y0 z0 and z0 x0 can

be obtained The expressions forx0y0,y0z0andz0x0are

y0¼ Ty0y0¼ ycos2ðy0; yÞ þ zcos2ðy0; zÞ

þ xcos2ðy0; xÞ þ 2yzcosðy0; yÞ cosðy0; zÞ

þ 2zxcosðy0; zÞ cosðz0; xÞ

þ 2xycosðy0; xÞ cosðy0; yÞ ð27-15bÞ

z0¼ Tz0z0¼ zcos2ðz0; zÞ þ xcos2ðz0; xÞ

þ ycos2ðz0; yÞ þ 2zxcosðz0; zÞ cosðz0; xÞ

þ 2xycosðz0; xÞ cosðz0; yÞ

þ 2yzcosðz0; yÞ cosðz0; zÞ ð27-15cÞ

x0y0¼ Tx0y0¼ xcosðx0; xÞ cosðy0; xÞ

þ ycosðx0; yÞ cosðy0; yÞ þ zcosðx0; zÞ cosðy0; zÞ

þ xy½cosðx0; xÞ cosðy0; yÞ þ cosðx0; yÞ cosðy0; xÞ

þ yz½cosðx0; yÞ cosðy0; zÞ þ cosðx0; zÞ cosðy0; yÞ

þ zx½cosðx0; zÞ cosðy0; xÞ þ cosðx0; xÞ cosðy0; zÞ

ð27-16aÞ

y0 z0¼ Ty0z0¼ ycosðy0; yÞ cosðz0; yÞ

þ zcosðy0; zÞ cosðz0; zÞ þ xcosðy0; xÞ cosðz0; xÞ

þ yz½cosðy0; yÞ cosðz0; zÞ þ cosðy0; zÞ cosðz0; yÞ

þ zx½cosðy0; zÞ cosðz0; xÞ þ cosðy0; xÞ cosðz0; zÞ

þ xy½cosðy0; xÞ cosðz0; yÞ þ cosðy0; yÞ cosðz0; xÞ

ð27-16bÞ

z0 x0¼ Tz0x0¼ zcosðz0; zÞ cosðx0; zÞ

þ xcosðz0; xÞ cosðx0; xÞ þ ycosðz0; yÞ cosðx0; yÞ

þ zx½cosðz0; zÞ cosðx0; xÞ þ cosðz0; xÞ cosðx0; zÞ

þ xy½cosðz0; xÞ cosðx0; yÞ þ cosðz0; yÞ cosðx0; xÞ

þ yz½cosðz0; yÞ cosðx0; zÞ þ cosðz0; zÞ cosðx0; yÞ

ð27-16cÞEquations (27-15a) to (27-15c) and Eqs (27-16a) to(27-16c) can be used to determine the six Cartesiancomponents of stress relative to the Oxyz coordinatesystem to be transformed into a different set of sixCartesian components of stress relative to an Ox0y0z0coordinate system

For two-dimensional stress fields, the Eqs (27-15a) to

(27-15c) and (27-16a) to (27-16c) reduce to, since

z¼ zx¼ yz¼ 0 z0 coincide with z and  is the

angle between x and x0, Eqs (27-15a) to (27-15c)

and Eqs (27-16a) to (27-16c)

FIGURE 27-5 The stress vector T N

PRINCIPAL STRESSES

By referring to Fig 27-5, where TN coincides with

outer normal N, it can be shown that the resultant

stress components of TN in x, y and z directions

Substituting Eqs (27-9b) to (27-9d) into (27-18), the

following equations are obtained

Eq (27-19) can be written as

From Eq (27-20), direction cosine (N, x) is obtained

and putting this in determinant form

Putting the determinator of determinant into zero,

the non-trivial solution for direction cosines of the

principal plane is

x0¼ xcos2 þ ysin2 þ 2xysin cos 

¼ xþ y

2 þx
y

2 cos 2 þ xysin 2 ð27-17aÞ

y0¼ ycos2 þ xsin2
2xysin cos 

Expanding the determinant after making use of Eqs.

(27-4) which gives three roots They are principal

stresses

For two-dimensional stress system the coordinating

system coinciding with the principal directions,

Eq (27-23) becomes

The three principal stresses from Eq (27-23a) are

The directions of the principal stresses can be found

from

From Eq (27-15)

From definition of principal stress

Substituting the values of TN1and TN2in Eq (27-15)

2

þ 2 xy

¼ x0y0þ y0z0þ z0x0
2

x0y0
2

y0z0
2

z0x0ð27-28bÞ

For the coordinating system coinciding with the

principal direction, the expression for invariants

from Eq (27-28)

STRAIN (Fig 27-6)

The normal strain or longitudinal strain by Hooke’s

law (Fig 27-6) in x-direction

The lateral strains in y and z-direction

The normal strains caused byyandz

THREE-DIMENSIONAL STRESS-STRAIN

SYSTEM

The general stress-strain relationships for a linear,

homogeneous and isotropic material when an element

subject tox,yandzstresses simultaneously

where I1¼ first invariant, I2¼ second invariantand I3¼ third invariant of stress

z

dx’

dy’

dx dz

K’ K k n

m I L L’

The expressions forx,yandzstresses in case of

three-dimensional stress system from Eqs (27-33)

BIAXIAL STRESS-STRAIN SYSTEM

The normal strain equations, when z¼ 0 from

Eq (27-33)

The normal stress equation, when z¼ 0 from

Eq (27.34)

SHEAR STRAINS

For a homogeneous, isotropic material subject to

shear force, the shear strain which is related to shear

stress as in case of normal strain

It has been proved that the shear modulus (G) is

related to Young’s modulus (E) and Poisson’s ratio

as

From Eqs (27-37), shear strain in terms of E and

STRAIN AND DISPLACEMENT

(Figs 27-8 and 27-9)

The normal strain in x-direction

The normal strain in y and z-directions

The shear strains xy, yz and zx planes

N’

w

v K

Unstrained element

FIGURE 27-8 Deformation of a cube element in a solid

body subject to loads.

dy K

K’

N u

X y

ν

Y

M’ L’

The amount of counterclockwise rotation of a line

segment located at R in xy, yz and zx planes

The strain"z and first strain invariant J1in case of

plane stress

The strains components"x0,"y0 and"z0, along x0, y0

and z0 axes line segments with reference to the

O0x0y0z0system

The shearing strain components (due to angular

changes) x0y0, y0z0 and z0x0 with reference to the

O0x0y0z0system

xy¼12

"x0¼ "xcos2ðx; x0Þ þ "ycos2ðy; x0Þ

þ "zcos2ðz; x0Þ þ xycosðx; x0Þ cosðy; x0Þ

þ yzcosðy; x0Þ cosðz; x0Þ þ zxcosðz; x0Þ cosðx; x0Þ

ð27-42aÞ

"y0¼ "ycos2ðy; y0Þ þ "zcos2ðz; y0Þ

þ "xcos2ðx; y0Þ þ yxcosðy; y0Þ cosðz; y0Þ

þ zxcosðz; y0Þ cosðx; y0Þ þ xycosðx; y0Þ cosðy; y0Þ

ð27-42bÞ

"z0¼ "zcos2ðz; z0Þ þ "xcos2ðx; z0Þ

þ "ycos2ðy; z0Þ þ zxcosðz; z0Þ cosðx; z0Þ

þ xycosðx; z0Þ cosðy; z0Þ þ yzcosðy; z0Þ cosðz; z0Þ

ð27-42cÞ

x0y0¼ 2"xcosðx; x0Þ cosðx; y0Þ þ 2"ycosðy; x0Þ cosðy; y0Þ

þ 2"zcosðz; x0Þ cosðz; y0Þ

þ xy½cosðx; x0Þ cosðy; y0Þ þ cosðx; y0Þ cosðy; x0Þ

þ yz½cosðy; x0Þ cosðz; y0Þ þ cosðy; y0Þ cosðz; x0Þ

þ zx½cosðz; x0Þ cosðx; y0Þ þ cosðz; y0Þ cosðx; x0Þ

ð27-43aÞ

For the case of two-dimensional state of stress when z0

coincides with z and zx¼ yz¼ 0, the angle between

x and x0coordinates

The cubic equation for principal strains whose three

roots give the distinct principal strains associated

with three principal directions, is

The three strain invariants analogous to the three

stress invariants

y0z0¼ 2"ycosðy; y0Þ cosðy; z0Þ þ 2"zcosðz; y0Þ cosðz; z0Þ

þ 2"xcosðx; y0Þ cosðx; z0Þ

þ yz½cosðy; y0Þ cosðz; z0Þ þ cosðy; z0Þ cosðz; y0Þ

þ zx½cosðz; y0Þ cosðx; z0Þ þ cosðz; z0Þ cosðx; y0Þ

þ xy½cosðx; y0Þ cosðy; z0Þ þ cosðx; z0Þ cosðy; y0Þ

ð27-43bÞ

z0 x0¼ 2"zcosðz; z0Þ cosðz; x0Þ þ 2"xcosðx; z0Þ cosðx; x0Þ

þ 2"ycosðy; z0Þ cosðy; x0Þ

þ zx½cosðz; z0Þ cosðx; x0Þ þ cosðz; x0Þ cosðx; z0Þ

þ xy½cosðx; z0Þ cosðy; x0Þ þ cosðx; x0Þ cosðy; z0Þ

þ yz½cosðy; z0Þ cosðz; x0Þ þ cosðy; x0Þ cosðz; z0Þ



"x"yþ "y"zþ "z"x


2 xy

4


2 yz

4
2zx4

4
"y zx2

4
"z

2 xy

4 þxyyzzx4



¼ 0ð27-45Þ

J1¼ "xþ "yþ "z¼ first invariant of strain

ð27-45aÞ

BOUNDARY CONDITIONS

The components of the surface forces Fsfxand Fsfyper

unit area of a small triangular prism pqr so that the

side qr coincides with the boundary of the plate ds

4


2 yz

4
zx24

¼ second invariant of strain ð27-45bÞ

y r

O

N ds

The volume dilatation of rectangular parallelopiped

element subject to hydrostatic pressure whose sides

are l1, l2and l3

The final dimensions of the element after straining

Substituting the values of l1f, l2f, l3f, l1, l2, l3 in

Eq (27-48) and after neglecting higher order terms

of strain

If hydrostatic pressure (0) or uniform compression

is applied from all sides of an element such thatx¼

y¼ z¼
0¼ 1¼ 2¼ 3,xy¼ yz¼ zx¼ 0,

Eq (27-48) becomes

The bulk modulus of elasticity

GENERAL HOOKE’S LAW

The general equation for strain in x-direction

accord-ing to general Hooke’s law in case of anisotropic or

non-homogeneous and non-isotropic materials such

as laminate, wood and fiber-filled-epoxy materials as

a linear function of each stress

For relationships between the elastic constants

The three-dimensional stress-strain state in anisotropic

or non-homogeneous and non-isotropic material such

as laminates, fiber filled epoxy material by using

generalized Hooke’s law which is useful in designing

machine elements made of composite material

(Fig 27-1c)

Note:½S matrix is the compliance matrix which gives

the strain-stress relations for the material The inverse

of the compliance matrix is the stiffness matrix and

the stress-strain relations If no symmetry is assumed,

there are 92¼ 81 independent elastic constants

pre-sent in the compliance matrix [Eq (27-55)]

ð1þ 2þ 3Þ ð27-50bÞ

3 7 7 7 7 7 7 7 7 7 7

3 7 7 7 7 7 7 7 7 7 7 5

3 7 7 7 7 7 7 7 7 7 7 5

ð27-55Þ

Equation (27-55) can be written as given here under

Eq (27-56) with the following use of change of

nota-tions and principle of symmetrical matrix in case of

3 7 7 7 7 7 7

3 7 7 7 7 7 7

3 7 7 7 7 7 7

3 7 7 7 7 7 7 ð27-56Þ

a Courtesy: Extracted from Ashton, J E., J C Halpin, and

P H Petit, Primer on Composite Materials—Analysis, Technomic Publishing Co., Inc., 750 Summer Street, Stamford, Conn 1969.

The general stress-strain equations under linear

The matrix expression from Eq (27-55) for

ortho-tropic material in a three-dimensional state of stress

The two-dimensional or a plane stress state matrix

expression after putting 3¼ 23¼ 13¼ 0 and

23¼ 13¼ 0 and "3¼ S131þ S232 in Eq (27-59)

for orthotropic material

The stress-strain relationship for homogenous

iso-tropic laminae of a laminated composite in the

matrix form, which is assumed to be in state of

plane stress

Alternatively Eqs (27-61) can be written for the nth

layer of laminated composite, which is assumed to

be in a state of plane stress

377777

377777

377777

ð27-59Þwhere there are 9 independent constants in theabove compliance matrix which is inverse ofstiffness matrix

"1

"2

12

26

37

3

7 12

12

26

37n

37n

"1

"2

12

26

37n

Substituting strain-displacement, Eqs (27-40) and

(27-41) into stress-strain Eqs (27-33) and (27-37) or

(27-39), displacement stress equation are obtained

with from 15 unknowns to 9 unknowns

Combining stress equation of equilibrium from Eqs

(27-11) with stress displacement Eqs (27-63) (from 9

to 3 unknowns)

Six stress equations of compatibility are obtained by

making use of stress strain relations of Eqs (27-33)

and (27-39), the stress equations of equilibrium

Eq (27-11) and the strain compatibility Eq (27-47)

in three dimension in Cartesian system of coordinates

AIRY’S STRESS FUNCTION

Differential equations of equilibrium for

two-dimensional problems taking only gravitational

force as body force

The stress components in terms of stress function

and body force

Substituting Eqs (27-66c) for stress components into

Eq (27-66b) that the stress function must satisfy the

equation

The stress compatibility equation for the case of plane

strain

If components of body forces in plane strain are

Substituting Eqs (27-68) into Eqs (27-11d), (27-11e)

and Eq (27-67) and taking2 ¼ 1

1
v

By assuming that the stress can be represented by a

stress function  such that x¼@2

@z þ

@Fbz

@y

ð27-65eÞ

@z þ

@Fbz

@x

ð27-65fÞ

Stresses for plane-stress can be obtained by letting

v

1
v! v in Eq (27-70) and it becomes

If body forces are zero or constant then Eq (27-70)

becomes

The biharmonic Eq (27-71a) can be written in

expanded form as

CYLINDRICAL COORDINATES SYSTEM

General equations of equilibrium in r,  and z

coordinates (cylindrical coordinates) taking into

consideration body force (Figs 27-13 to 27-15)

@r

@r þ

1r

Equations of equilibrium for axial symmetry Eqs.

(27-73) reduce to Eqs (27-74) when there are body

forces acting on the body

Equations of equilibrium in two dimension in r and 

coordinates (polar coordinates) taking into

considera-tion body force components

Equations of equilibrium for an axially symmetrical

stress distribution in a solid of revolution when there

are no body forces acting on the body (Fig 27-13),

since the stress components are independent of

STRAIN COMPONENTS (Fig 27-14)

The strain components in r,  and z

coordinates system

The strain in the radial direction

The strain in the tangential direction

@r

@ þ

1

rðr
Þ þ FbR¼ 0 ð27-75aÞ1

∂ν

∂

ν + d

v B B’

r

θ

d O

∂u

∂ dθθ FIGURE 27-14 Strain components in polar co-ordinates.

The strain in the axial direction

The shear strains

The rotation of the element in the counter clock-wise

direction in the r, z and zr planes

AIRY’S STRESS FUNCTION IN POLAR

COORDINATES

When components of body force Fbrand Fbare zero,

Eqs (27-74a) and (27-74b) are satisfied by assuming

stress function forr,andr

The stress equation of compatibility Eq (27-72) in

terms of Airy’s stress function referred to Cartesian

coordinates x and y, has to be transferred to Airy’s

stress function referred to polar coordinates r and

 system In this transformation from x and y

coordinates transform to r and  coordinates

Eq (27-72) can be written as

Using Eqs (27-79) and (27-80) and transforming

Eq (27-72) into stress equation of compatibility in

polar coordinates r and  system

r¼1r

1r

Stress equation of compatibility in terms of Airy’s

stress function in polar coordinate r and  is obtained

by substituting (Eq (27-83) to Eq (27-82))

SOLUTION OF ELASTICITY PROBLEMS

USING AIRY’S STRESS FUNCTION

Any Airy’s stress function  either in Cartesian

coordinates or polar coordinates used in solving any

two-dimensional problems must satisfy Eqs (27-66)

and (27-72) in Cartesian coordinates and Eqs

(27-79) and (27-84) in polar coordinates and

bound-ary conditions (27-46)

Cartesian coordinates

Solutions of many two-dimensional problems can be

found by assuming Airy’s stress function in terms of

polynomial and Fourier series, which are

r4 ¼



@2

@r2þ1r

ð2l

0  dx

an¼1l

ð2l

0  cosn x

l dx if n 6¼ 0

Polar coordinates

r4 ¼ 0 is a fourth order biharmonic partial

differen-tial equation The fourth order differendifferen-tial equation

can be obtained by using a function  in r4 ¼ 0

which in term gives four different stress functions

One of the stress function  for solving many

problems in polar coordinates

The second order stress function2

The third order stress function3

The fourth order stress function

The general expression for the stress function 

which satisfy boundary conditions and compatibility

Eq (27-84)

bn¼1l



A1r þB1

r þ C1r3þ D1r ln r

sin

þ



A01r þB

0 1

r þ C0

1r3þ D0

1r ln r

cos

In a general case the loading can be represented by the

trigonometric series

The stress function can also be represented by

APPLICATION OF STRESS FUNCTION

Thick cylinder

Stress function used in this case, Eq (27-93)

Boundary conditions are

Equation of equilibrium used in this problem

The expression for radial stress in thickness wall of

thick cylinder under external pressure (po) and

inter-nal pressure (pi) at any radius r

The expression for tangential stress in thickness wall

of thick cylinder under pressures po and pi at any

radius r

The shear stress

Expression for displacement of an element in the

thickness wall of cylinder at any radius r in radial

direction and tangential direction respectively

Curved bar under pure bending (Fig 27-15)

Stress function used in this problem Eq (27-93)

 ¼ ðA eyþ B e
yþ Cy eyþ Dy e
yÞ sin x

ð27-99Þ

 ¼ A ln r þ Br2

ln r þ Cr2þ D ð27-93ÞðrÞr ¼ di=2¼
pi and ðrÞr ¼ d0=2¼
po ð27-100aÞ

@r

@r þ

r


Since it is a case of problem of symmetry with respect

to axis of cylinder and no body force acting on it

r¼pidi2
podo2

d2

o
d2 i

di2do2ðpi
poÞ4r2ðd2

þdi2do2ðpi
poÞ4r2ðd2

o
d2

iÞ ð27-101bÞ

u ¼ 1E

FIGURE 27-15

Equation of equilibrium used in this problem of

sym-metry with respect to the xy-plane perpendicular to

axis of the bar

The expression for the radial stress component in the

bar at r radius

The expression for the tangential stress component in

the bar at r radius

The expression for shear stress component

STRESS DISTRIBUTION IN A FLAT PLATE

WITH HOLES OR CUTOUTS UNDER

DIFFERENT TYPES OF LOADS

An infinite flat plate with centrally located circular

cutout or hole subject to uniform uniaxial tension

ðrÞr ¼ b¼  cos2 ¼1

2ð1 þ cos 2Þ ð27-107aÞ

Boundary conditions are

The radial stress in an infinite plate with a centrally

located circular hole (cut-out) subject to uniform

uniaxial tension at infinity (Fig 27-16)

The tangential stress in an infinite plate with centrally

located circular hole (cutout) under uniform uniaxial

tension at infinity

The shear stress in an infinite flat plate with a centrally

located circular cutout (hole) subject to uniform

uniaxial tension at infinity

The tangential stress at hole boundary at ¼ =2 or

3 =2

The stress concentration factor

For distribution of tangential stressaround circle

of hole under uniform uniaxial tension

For superposition of stresses in a flat plate with a

centrally located circular hole subject to tension,

compression and uniform pressure

The shear stress around hole at ¼ =2 or 3 =2

ðrÞr ¼ b¼1 sin 2 ð27-107bÞ

r¼2



1
a2

r2

þ

ð27-108Þ

¼2

ð27-109Þ

r¼
2

FIGURE 27-16 A large flat plate with a

centrally located circular hole under

uni-form uniaxial stress at infinity.

P Q

q = σ

q = σ

Pure shear

An infinite flat plate with centrally located circular

cutout or hole subject to uniform uniaxial tension

and compression (i.e., pure shear) (Fig 27-17)

The expression for stress function

Boundary conditions are

The tangential stress in an infinite plate with a centrally

located circular hole subject to uniform tensile and

compressive stresses as shown in Fig 27-17

The radial stress

The shear stress

The tangential stress

The maximum tangential stress at  ¼ =2 or 3 =2

i.e., at P and Q (Fig 27-17)

2a 2a

σ 2

σ 2

σ 2

σ 2

FIGURE 27-19 Principle of superposition.

The maximum tangential stress at ¼ 0 or  ¼ , i.e.,

at R and S (Fig 27-17)

The stress concentration factor

Bi-axial tension (Fig 27-20)

An infinite flat plate with centrally located circular hole

(cutout) under biaxial uniform tension (Fig 27-20)

The radial stress at hole boundary

The shear stress at hole boundary

The tangential stress in an infinite flat plate with a

centrally located circular hole subject to uniform

biaxial tensile stress at infinity

The stress concentration factor

Finite plate (Fig 27-21)

Uniaxial tension (Fig 27-21)

h w

σmax

0 0.1 0.2 2.0

2.2 2.4 2.6 2.8 3.0

0.3 0.4 0.5 0.6 0.7

Ratio of d/w

s q

FIGURE 27-21 Stress concentration factor for a plate of finite width with a circular hole (cutout) in tension (Howland).

The stress distribution in a flat plate of finite width w

with a centrally located small circular hole according

to Howland11, which can be expressed in terms of

stress concentration around hole

The tangential stress at the points of q and s when

w ¼ 2d according to Howland

The tangential stress at the point P according to

Howland11

It can be seen from Fig 27-21 that maximum stress

which is at the hole boundary, decreases very rapidly

and approach the value of average stress at the edge of

the infinite plate

ROTATING SOLID DISK WITH UNIFORM

THICKNESS (Fig 27-22)

From Eqs (27-75a) and (27-75b), which can be

made use of for a rotating disk of uniform thickness

with z-axis perpendicular to the xy-plane and stress

components do not depend on  Hence Eqs

(27-75a) taking a body force equal to inertia force

i.e., FbR¼ !2

r, becomes

Equation of force equilibrium from Eqs (27-122a)

after substituting value of FbRbecomes

Equation of compatibility

Using compatibility equation (27-123) and Hooke’s

law after simplification, the expression for force



ð27-121bÞwhere

The general solution of Eq (27-124a), when r ¼ eis

substituted in it, becomes

Boundary conditions

(a) The radial stress at outer boundary of rotating

disc of radius b

(b) stress at center of rotating disc

The expression for radial stress at any radius r

The expression for tangential stress as any radius r

ROTATING DISK WITH A CENTRAL

CIRCULAR HOLE OF UNIFORM

THICKNESS, Fig 27-23

Boundary conditions

Using force equilibrium Eq (27-124b) and boundary

conditions Eqs (27-128) the tangential and radial

stresses at any radius r are

x

b r

The expression for maximum radial stresses which

Rotating disk as a three-dimensional problem

The differential equations of equilibrium from Eqs

(27-76) when body force which is an inertia force

(cen-trifugal force) is included, becomes

After substituting the body forces Fbx¼ !2

x,

Fby¼ !2

y, Fbz¼ 0 in Eqs (27-65) and the last

three equations containing shearing stress

compo-nents remain the same as in Eqs (27-65), and the

first three equations in polar coordinates become

The solution of Eq (27-132) consists of a particular

solution and complementary function The particular

solution call be obtained by assuming

The complementary solution is obtained by assuming

a stress function, which has to satisfy boundary

conditions, compatibility equations and having a

form of a polynomial of the fifth degree

The particular solution

b h

FIGURE 27-24 Rotating disc of variable thickness.

The complementary function obtained from assuming

stress function Eq (b)

r max¼



3þ v8

@I

@r¼

2!2

1
vð27-133bÞ

r2zþ 1

1þ v

@2I

@z2¼
2v!2

The equations of shearing stress components in Eqs.(27-65) remain the same without any change evenwhen the body forces are acting

An expression for uniform radial tension on the disk

which is superposed on the resultant stresses to express

the resultant radial compression along the boundary

The final expressions for stress components

For disk of uniform strength rotating at! rad/s

For solid cylinder rotating at! rad/s, hollow cylinder

rotating at ! rad/s and solid thin uniform disk

rotating at! rad/s under external pressure

For asymmetrically reinforced circular holes/cutouts

in a flat plate subject to uniform uniaxial tensile

force/stress

ROTATING DISK OF VARIABLE

THICKNESS (Fig 27-24)

The equation of force equilibrium in case of rotating

disk of variable thickness from Eq (27-122b)

Using rhr¼ F and thickness variation as h ¼ Cr
,

Hooke’s law, r ¼ e and Eq (27-123), Eq (27-135)

become

r¼
!2

vð1 þ vÞ

Refer to Chapter 4, Figs 4-7(a), 4-7(b) and 4-7(c)

Boundary conditions

The expression for radial stress

The expression for tangential stress

For a symmetrically reinforced circular cutout in a

flat plate under uniform uniaxial tension according

to the analysis of Timoshenko

NEUTRAL HOLES (MANSFIELD THEORY)

Reinforced holes which do not affect the stress

distri-bution in a plate are said to be neutral

Resolving the forces in x-direction, and using stress

function for stresses as x¼@@y2z, y¼@2

@x2 and

xy¼@x@y@2 and after integrating, an expression is

obtained as

Resolving the forces in y-direction after performing

integration etc as done under Eq (a), another

expres-sion is obtained

From Eqs (a) and (b)

Eq (c) may be written as

rb

1þ
1

rb

1þ
1

1þ 3v

3þ v

rb

2

ð27-139aÞwhere

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 2

2ð1 þ v Þ

s

ð27-139bÞ

1and 2are roots of Eq (27-139b)Refer to Fig 27-25

There is negligible bonding on the reinforcement since

it is thin compared to the radius of the curvature ofthe hole, and shear across the section of reinforcement

Integrating Eq (c) an expression for  is obtained as

The term Ax þ By þ C can be included or excluded

from the stress function without changing the stress

distribution These terms do not affect the shape of

the neutral hole but determines the position of the

hole By omitting A; B and C the shape of the neutral

hole is given by stress function

Compatibility

Considering the displacements and strain of

triangu-lar element klm as shown in Fig 27-26 x and y

direc-tions due tox,yandxystresses and equating strain

in the plate equal to strain in the reinforcement, an

expression for area of reinforcement (AR) for neutral

σσ

d o d i

h H

KσB

KσB

KσB

Where Bend cross-section Hole cross-section (do-di) (H-h)

dih

di w

0.2 ANALYSIS OF TIMOSHENKO

Neutral hole in a thin walled cylinder

(Fig 27-27)

The stress function may be taken as

For neutral hole

Eq (27-144) becomes an ellipse whose minor axis is 2r

and ratio of major axis (2a) to minor axis (2b) is ffiffiffi

2

p: 1Substituting stress function from Eq (27-144a) in

Eq (27-142), the area of reinforcement

 ¼2



x2þy22



þ Ax þ By þ C ð27-143aÞ

 ¼2

k

Element Reinforcement

dx

dz

k F + dF dy

m m

(b)

σy

σ σ

σx

ψ

τxy

τxy

FIGURE 27-26 (a) A reinforced circular hole under tension,

(b) element at reinforcement under the action of normal

stres-ses  x and  y and shear stress  xy due to force F acting on the

cross-section of the reinforcement.

σ

σ σ

y

y 2a

2a 2b

COMPLEX VARIABLE METHOD APPLIED

The body force complex potentials and components

of body force complex potentials

The equations of equilibrium can be reduced to two

equations using the stress combinations of Eqs

(27-146) and variable complex number

When the body force is zero Eq (27-148) can be

written as

Stress strain relation

The complex displacement

Stress-displacement equations by using Eqs (27-146)

STRAIN COMBINATIONS

The strain combinations are

Strain transformation rules (Fig 27-28)

x x’

FIGURE 27-28 Strain transformation.

Stress transformation rules

 ¼ "xþ "y

 ¼ "x
"yþ ixy ð27-152Þ ¼ xzþ iyz; D ¼ u þ iv

PLANE STRAIN (Figs 27-29 and 24-30)

External forces have to be applied on both top and

bottom flat ends of the cylinder to prevent its

move-ment in order to meet the condition that w ¼ 0 The

Eq (27-151b) becomes

And Eq (27-151d) becomes

The expression for F

The stress combinations are

The displacement D

BOUNDARY CONDITIONS

Specified stresses

The expression for F

By using Eq (27-155), Eq (27-159) becomes

The solution of Eq (27-132) consists of a particular

solution and complementary function The particular

solution call be obtained by assuming

The...

located circular hole (cut-out) subject to uniform

uniaxial tension at infinity (Fig 27 -16)

The tangential stress in an infinite plate with centrally

located circular hole...

ẳ2

27-109ị

rẳ
2

FIGURE 27 -16 A large at plate with a

centrally located circular hole under