Axial Stress mech sol

7/31/2019 Axial Stress mech sol

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Strain and Axial Loading

Calculation of stress is essential to ensure that a mechanical or structuralcomponent does not fail. Displacement calculations are also necessary, in order

to avoid excessive deformations/displacements during the operation of amachine, or the use of a structure. Furthermore, the relationship between the

forces (or moments) and displacements (or rotations) is needed in the stressanalysis of statically indeterminate structures.

Normal Strain

Let us consider the axial displacement of a bar due to a given set of centric axial

loads:

Let u(x) be the axial

displacement of apoint (or section)

originally (at theunloaded state) located

at distance x from theleft hand support.

By the definition, the

normal strain is given by= ..(1)

The strain may be regarded as a rate of change of axial displacement withrespect to axial co-ordinate x.

From Hooke's Law, in the linear-elastic range, = ..(2)

But since stress is the intensity of force, = ..(3)

Therefore =dx

du..(4)

where F is the induced force in the bar, and A is its cross sectional area.

By

u = ..(5)

= 12 uu ..(6)

If F, A and E remain constant for x1 < x


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2

= 12 uu ..(7)

Equation (7) may be used for calculating displacements in a bar or a bar

segment subject to a constant axial force, provided the material is homogeneous(constant E) and uniform (constant A). In all other cases, equation (6) should

be used.

Sign Convention: Tensile strain is positive.

Other notations in use:

i.e. u1 = u(x1).

1 may also be used instead of u1. The relative displacement between twopoints x1 and x2 is sometimes written as u21 or 21. Thus an alternative form of

equation (6) would be = 21

x

x21dx

AE

F

For a uniform bar segment under constant induced axial force equation (7) mayalso be written as:

21 =)A)(E(

)L(F

2121

212112 = (7a)

where the subscript 2 and 1 refer to the points on the bar segment 12.

The use of subscript numbers may cause confusion when there are several barswhose displacements need to be distinguished by using a subscript number

identifying a bar. In such cases it may be best to denote the displacements using

letters as the subscripts. For example uB or B would indicate displacement atpoint B.

Quiz:In calculating axial displacements, it is important to know when to use

equations (6) and (7). For each of the following problems indicate which of thetwo equations is most appropriate or applicable: It should be noted here that

although equation (6) is applicable for all cases, use of equation (7) would bemore convenient and simpler for some of the problems, while others can only be

solved using equation (6).


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3

P1L

P2L2

E1,A1 E2,A2B C

Two bars in series under two concentratedloads

BC

L

P

Tapered Bar subject to an end load

Vertical bar underself-weight

P

Uniform, homogeneousbar, subject to an end load

L

B C

Rotating bar under centripetal force


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4

Case1: Uniform, homogeneous Bar subject to an end load.

Determine the displacement of the loaded

end of a uniform bar of length L, andcross-sectional area A, subject to a load P

at its unrestrained end as shown in thefigure. The bar is made of a

homogeneous material having an elasticmodulus E.

x1=0, x2 =

Using equation (7), u2 - u1=

F =

u1 =u2 = u(L) = cThe end displacement c = PL/(AE).

As expected the end displacement is proportional to load and the length, andinversely proportional to the cross-sectional area and the elastic modulus.

Case 2: Two bars in series under two concentrated loads

Problem: Find the displacement of D.

Let us apply equation (7) to the first bar:x1 = 0, x2 =

F =

u1(0) =

Let u1(L1) = C

P1L1

P2

L2

E1,A1 E2,A2

B C D

P

x1=0 x2 =

L

B


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Using equation (7) to Bar 1

C = ..(8)

Similarly applying equation (7) to Bar 2 gives:

D - C = ..(9)

Adding equations (8) and (9) gives:

D = .....(10)

Case 3: Assembly of bars under two several concentrated loads

Determine the relative end displacement of the right end (D) of the following

bar assembly with respect to its left end (A). i.e. D - A = ? The elasticmodulus of the bars is 200 GPa.

Let us find the axial force in the bar segments by applying the method of

sections:

-FCD +50 = 0Therefore FCD = 50 kN. (Tensile)

Note that we could have summed the forces to the

left.

i.e. FCD 50 = 0 also gives FCD = 50kN

Now taking a cut between B and C andconsidering the right hand side free-

body gives:

. FBC + 20050 = 0FBC = -150 kN (Compressive)

4 m

50 kN

3 m

A=0.001

D200250

A =0.002

R

2 m

A=0.001

A

50 kND

FCD

50 kND

200

CFBC


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Similarly for FAB we get:

FAB 250 + 20050 = 0giving FAB =

Induced Axial Forces:FAB = 100 kN; FBC

= -150 kN; FCD = 50 kNUsing equation (6a) for

segment AB, we have

)001.0)(10200(

)4)(10100(

)A)(E(

)L(F9

3

ABAB

ABAB

AB

== m ..(12a)Note that all units used are standard form of S.I. units and the resultingdisplacement would be in meters.

Similarly for segment BC we get,

)002.0)(10200(

)2)(10150(9

3

BC = m ..(12b)

)001.0)(10200(

)3)(1050(9

3

CD

= m ..(12c)

Adding equations (12a,b and c) gives:

D - A = 0.002 m = 2 mm. ..(13)

This is the relative displacement of D with respect to A. Since it is positive, Dwould move right in relation to A, and there will be a net increase in the total

length of the bars.

50 kNB D200250FAB

A B C D

100

50 kN

-150

Axial Force Diagram


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Case 4: Displacement caused by self weight of a vertical bar

A uniform column of height H, weight

(total) W, and cross sectional area A isshown in the figure. Let the origin of the

axial co-ordinate x be at the base, wherethe displacement B = 0. Thedisplacement of the tip C is required.

To find the axial force in the member weneed to isolate a free-body as shown in the diagram. This free-body is subjectto the gravity force P and the induced axial force at the cut FBC.

P =

Fy = 0 gives: P + FBC = 0

FBC = (13)Hence the axial force diagram is triangular.

Using equation (6),

=

== H

0

H

0

BC12 dx

AE

Fuu

This is equal to half the displacement caused by a force W applied at the top of

the column. The negative sign indicates that the displacement is opposite to thedirection of x. (i.e. top moves down, as expected).

C

FBC

P

x

H

-W

Axial force Diagram


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Case 5: Displacement caused by centripetal acceleration of a rotating bar

Let us now revisit

the rotating barwhose axial force

distribution wasobtained in Chapter

1. Let us considerthe stresses and

displacements in auniform bar of

length L, cross-sectional area A,

which is fixed to a

hub of radius r.The hub and thebar are rotating at a

constant angular speed of rad/s. Let the density and elastic modulus of thematerial the bar is made of be and E respectively. From Chapter 1, the resultfor the axial force in the bar is given by:

( )2222222

)xr()Lr(2

A

2

)xr(A

2

)Lr(AF ++

=

+

+= (14)

Substituting the r.h.s of equation (14) for axial force in equation (6) gives:

)

++=

L

0

222

dxAE2

)xr()Lr(A)0(u)L(u

=

[ ]3322

r)Lr(L)Lr(3E6

+++

=

Taking the hub as rigid, u(0)=0 which gives

[ ]3322

r)Lr(L)Lr(3E6

)L(u +++

= (15)

This is the radial displacement of the tip of the rotating bar.

Dividing equation (14) by A gives an expression for the stress as

( )222

)xr()Lr(2

++

=

The maximum stress in the bar (at the hub) may be obtained by substituting x =

0 into this equation, giving

Plan

rad/s

Lr

x x

rad/s

Elevation F+FF

2(r+x)

Ax

Infinitesimal

Element


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( ) )Lr2L(2

r)Lr(2

22

222

max+

=+

= (16)

Note thatshould be given in radians per second.

Case 6: Displacement of a tapered bar

Consider the case of a tapered bar of length L which is

subject to an axial force P at the free end as shown inthe figure. The cross sectional area A may be assumed

to decrease linearly from A0

at the left end to A1

at the

free end. The material has an elastic modulus E. Find

the displacement of the loaded end C.

An expression for the area A is needed. Since itdecreases linearly and is equal to A0 at x = 0, and A1 at x = L, it will be of the

formA = A0 -

For this case, the induced axial force is constant.FBC = P

Therefore, using equation (6)

= L

010

0

BC dx

L

x)AA(AE

P

Defining a new variableLA

AAr

0

10 = for convenience, this may be written

( ) ( )( ) =

=

= L0

1

0

L

00

L

00

BC dxxr1EA

Pdx

xr1

1

EA

Pdx

xr1EA

P

=

=

=

1

0e

101

0e

0 A

Alog

)AA(E

PL

A

Alog

rEA

P(17)

BC

L

P

Tapered Bar


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Statically Indeterminate Problems

In all the above cases, the axial force distribution in the bar(s) was calculated by

applying equations of equilibrium. As explained in Chapter 1, since this is notsufficient for Statically Indeterminate Problems, we need to obtain

compatibility conditions too. Static indeterminacy in bars may occur in caseswhere any of the following is encountered:

additional constraints, temperature changes in bars restrained at more than one point, compounding two or more bars in parallel, or turn-buckles

Some examples from the above categories are presented in this section, to

explain the strategy for solving these problems by applying:

Statics (equation(s) of equilibrium) Compatibility condition(s) Constitutive equation(s)

Case 7: A bar restrained at both ends and subject to a point load

A uniform bar of length L, cross sectional area A is

restrained at both ends. Calculate the maximum stressin the bar due to a point load P acting at distance a

from one support as shown in the diagram. The bar ismade of a material having an elastic modulus E.

Equilibrium:

Let the unknown reactions at B and D be RB andRD respectively. The directions were chosen

arbitrarily.

By inspection, P= (18)

Compatibility:

The net is zero.This may be written as

= 0 (19)

B C D

L-aa

L-a

RB RD

B C D

Overall Free-body


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From equation (21a)EAL

a)aL(P

)A)(E(

)L(F

BCBC

BCBCBC

==

Since B = 0,EAL

a)aL(PC

=

The same result would be obtained, if equation (21b) were used.

If a = L/2, thenEA4

LPC =

Alternative Approach (Flexibility Method):

This problem may be solved using a slightly different procedure, in three steps.In the first step, one of the restraints is initially removed from the bar to render

the structure statically determinate. The resulting bar (which is referred to as

the released structure) can be statically analysed, and its displacement at thepoint where the restraint was released could be calculated. In the second step,the restraint may be reapplied so as to bring the bar back to where it would be if

the restraints had acted all along. The magnitude of the restraining forcerequired is determined by calculating the displacement caused by it andequating it to the negative of the displacement due to the loading on the released

structure. The final step involves adding the induced forces (and/or stresses anddisplacements if necessary) obtained in the first two steps. This procedure is

based on the Principle of Superposition as described in Chapter 1, and isapplicable only for linear elastic structures undergoing small, deformations.

Although it may appear to be cumbersome, it is computationally easier to applyfor complex problems, and illustrates some problems (for example, the

calculation of temperature induced stresses) more clearly.

Step 1: Analysis of the released bar:

In this case, any one of the two end reactions could be moved, and the barwould still be stable, although the resulting stress distribution would bedifferent and may cause failure. Any additional constraints that are not needed

for maintaining the stability of a structure are referred to as redundants and thefirst step in the flexibility method is to remove all redundants. For the current

problem, only one of the two reactions could be removed. The choice isarbitrary since the bar would be stable without either one. Let us choose the

reaction at D as the redundant and remove it. The resulting structure is shownbelow:

Let the displacement of the bar at D due to

the loading on the released structure be D,L,

where the subscript D refers to the location of B C D

P

L-aa D,L

Loadin on the Released Bar


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the bar where the displacement takes place, and the subscript L stands for theLoading.

The induced forces in the segments BC and CD for this case areFBC,L = P ...(24a)

FCD,L = 0 ...(24b)Substituting these into equation (6a) we get:

0L,CL,D = and ( ) EAa)P(

L,BL,C =

Adding the above equations gives

( )EA

PaL,BL,D =

But B,L = 0 since it is a fixed end.

Therefore EA

PaL,D = ...(25)

Step 2: Analysis of the bar subject to the redundant:

In this step we reapply the redundant RD tonullify the displacement caused by the loadon the released bar.At this stage, we consider the effect of the

reaction RD only. The bar is under an axialforce of - RD (negative because it is a

compressive force).

FBC,R = - RD ...(26a)FCD,R = - RD ...(26b)

In fact, since the entire bar is under a uniform axial force - RD the displacement

of the bar at D may be obtained by putting FBD = - RD and LBD = L into equation(6a) giving:

( ) EALR D

R,BR,D

= And since B is a fixed end B,R = 0 which when substituted into the previous

equation gives:EA

LR DR,D

= ...(27)

The minus sign indicates that this displacement is in the opposite direction topositive u and positive x, and D moves left.

Step 3: Superposition of the effects of the load and redundant:

B C D

R

L-aa D,R

Reapplying RD on the ReleasedBar


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From the Principle of superposition (see Chapter 1), the net result of P and RDacting simultaneously may be found by adding the results in Steps 1 and 2.Using equations (25) and (27), net displacement of D

EA

LRPa DR,DL,DD

=+= ...(28)

Compatibility Condition:

Compatibility is enforced, by equating the net displacement due to the load andthe redundant to zero. Since D cannot actually move RD must be such that the

above expression, equation (28) gives zero.Therefore P a RD L = 0

This gives RD = P where = a/L .....(29a)

There is another way to consider compatibility. The displacement due to theredundant (given by the right-hand side of equation (27)) must be equal andopposite to the displacement due to the load (given by the right-hand side of

equation (25)).

Substituting equation (29a) into equations (26a,b) and adding the results toequations (24a,b) results in the following expressions for the net induced forces:

FBC = FBC,L + FBC,R =P(1- ) ...(29b)

and FCD =FCD,L + FCD,R = P ...(29c)

The stresses in the bars can be found by dividing the induced forces by the cross

sectional area A.

BC = P(1- )/A ...(30a)and CD = P/A ...(30b)


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Case 8: A restrained bar subject to temperature change

A uniform bar of length L, cross sectional area A is

restrained at both ends. The bar is then subject to a

temperature increase of TC. Find the stress induced inthe bar if it is made of a material having an elastic

modulus E and the coefficient of linear thermal expansion /C.

This is done using the flexibility method. One end of the bar may be firstreleased, thus allowing free elongation to take place.

Step 1: Analysis of the released bar:

Let us remove the reaction at D and consider the

effect of the temperature increase. At this stagethe bar is free to elongate and is therefore under

no internal forces or stresses. Denoting theeffects associated with the temperature change

with a subscript T, we haveFBD,T =

.(31)The displacement of the end D is caused by the temperature and is given by

D,T =.....(32)

Step 2: Analysis of the bar subject to the redundant:

This is similar to Case 7, and the application of the reaction RD would result in a

displacement at D given by equation (27). i.e.EA

LR DR,D

=

Step 3: Superposition of the effects of the load and redundant:

Using equations (32) and (27), D = D,T + D,R = LT +EA

LR D ,

.....(33)


Since D is actually restrained, D = 0.

Putting equation (33) into this gives, LT +EA

LR D = 0

giving RD = TEA.....(34)

B D

L

TC

B D

TC

D,T


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The force in the bar is caused by the reaction only, and is equal to RD.

i.e. FBD = TEA.Therefore the stress in the bar is BD = FBD/A = TE

.....(35)As expected an increase in temperature causes compressive (negative) stress.

Expansion Joints:

In designing bridges, floor-slabs and railway tracks, a clearance is provided to

allow for the free movement of rails due to temperature changes. If the feedue to temperature increase is greater than the clearance, then

compressive stresses will develop. They can cause buckling type failure. These

thermal stresses can be calculated using a modified compatibility condition.

In the case of a bar that is fully restrained at oneend and is placed at distance e from a fixed wall

at the other end, equation (33) will need to bechanged to

LT +EA

LR D e where e is the clearance. If

the left-hand side is less than e then there would be no stress, and only if the

temperature increase is sufficiently high then equality would result.

i.e. if LT


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Case 9: A compound bar subject to end loads applied through rigid end plates

Consider the case of a compound bar which

consists of two identical copper bars and asteel bar, all having length L. The cross

sectional areas of the bars are A (for copper)and n(A) (for steel). The elastic moduli for

copper and steel are E and r(E) respectively.We may want to find the stresses in the bars

in terms of applied load P and the propertiesof the bars.

The problem with this case is that from equilibrium alone, we cannot obtain the

axial forces in each of the bars. Applying the method of sections for a segment

of the compound bar we get

Fx = 0 gives: -P = 0,P = .......(37)where FS and FC are the axial forces in the

steel bar and copper bar respectively.

This is the only equation of equilibrium.


Since the load is applied through rigid end plates, the elongation of the copper

bar =

Denoting the changes in length of the steel and copper bar by S and Crespectively, S = C .....(38)And from equation (7a) we have,

S = .....(39a)

C = .....(39b)

Substituting equations (39a,b) into equation (38) gives

This gives: FS = n.r.FC .....(40)

Substituting equation (40) into equation (37) gives: -P n.r. FC-2FC = 0

Therefore FC =-P/(2+n.r) ...(41a)

and substituting this back into equation (40) gives

L

P

Compound Bar

Copper Steel

P

FC

FC

FS

Free-body Diagram of a Segment


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FS = -n.r.P/(2+n.r) ...(41b)

The stresses in the bars are given by C =)r.n2(A

P

+

...(42a)

and S = )r.n2(A P.r+ (42b)

It is interesting to note that S/C = r = ES/EC ..(43)That is to say that the normal stress induced in the components of a compound

bar is proportional to their elastic modulus. A material that has a higher elasticmodulus (stiffer material) will undergo greater stress when subject to axial

loading under same constraints.

Case 10: Turnbuckle Problems

Turnbuckle is a simple device that can be used to adjust the tension instructures, particularly in cables. It consists of two rods having threads running

in opposite directions, and a threaded buckle. When the buckle is turned, itwould tend to move either in or out simultaneously on both rods. If the other

ends of the rod are constrained, say by a cable or directly on to a fixed wall,then the threaded rods and/or the cable that it is attached to, will undergo a

change in the axial force. As an example, let us consider the case of two rods

connected to a turnbuckle and rigid walls as shown in the following figure.

JJust for the fun of it, one can think of a possible similarity between thisbehaviour of materials and what happens to people under stress. Do people who

have a stiff/rigid attitude in life, undergo more mental stress than those who take a

flexible, compromising attitude?

Another point of observation: Why do women have a longer life-span? Is it

because they are more flexible (less stiff) than men?

Comments like these are given just to have a break from the s (stresses) of

learning this course, and should not be taken as the authors view on these matters.


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The steel rods having the same cross sectional area (A) are connected by aturnbuckle, and are clamped to rigid walls at the other ends. Calculate theincrease in stress caused by tightening the turnbuckle by a third of a turn. The

turnbuckle may be assumed to be rigid. For steel E = 207 GPa.

Equilibrium:

For equilibriumthe axial force

in both rodsmust be same

at any time.Let the axial

force due to turning the buckle be F and any force that was already present be

F0.


Let the elongations of the rods on the left and right hand sides be eL and eRrespectively. If the buckle is turned (tightened) by a third of the turn, the two

ends of the rods inside the buckle would have travelled by a distance of onethird of the pitch each.

i.e. eL + eR = .....(44)

Constitutive Equations:

If both rods were of the same length, the elongations would be the same.

Otherwise the longer rod would stretch more. If one rod were to be extremelyshort, then the elongation of the other rod would be 0.8 mm. In the presentcase, the elongations can be expresed in terms of the length, force etc, using

equation (7). This gives:eL = .....(45a)

and eR = .....(45b)

Substituting these into equation (44) gives: (1.8 F + 1.2 F)/(AE) = 0.8 10-3m

i.e. F = (AE) (0.8 10-3) /3.0The stress in the rods are given by, =

= 55.2 Mpa .....(46)

1.2 m1.8 m

Threads at 1.2 mm itch

Turnbuckle


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20

Multi-Axial Loading

Poissons Ratio

When a rubber block is squeezed it tends to bulge out sideways. When a piece

of rubber is stretched, it tends to shrink sideways.

Note that the strain in the direction of loading is opposite to the strain in thedirection that is perpendicular to the loading. In the figure shown, when thebody is compressed vertically, it expands horizontally, and when it is stretched

vertically it shortens horizontally. It was Poisson who first discovered that theratio of the strain in the direction perpendicular to the loading and the strain in

the direction of the loading is a constant for many materials.

This ratio is called Poisson's ratio and is denoted by the Greek

alphabet (nu).

A rubber squeezed between the thumb and a finger has abarrel shape. The surfaces of rubber in contact with the

finger/thumb do not move due to friction, causing thisdistortion.

When a material is subject to a state of uni-axial stress, and it is free from anyrestraints in lateral directions, the strain in the lateral directions is proportional

to the strain in the axial direction, the constant of proportionality being -.

From Hooke's law, the strain in the direction of the loading is given by

=

Since we are going to deal with multi-axial loading, we need to assign

subscripts to the stress and strain to indicate the direction. The notation we use

is as follows: xx refer to the strain in the x-direction caused by the stress inthe x-direction (x ) The strains in y, z directions due to x (the stress in x-direction) are denoted by yx and zx respectively.


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Complete the expression for yx and zx in terms of x , and E:

yx = zx = x ( )

Stress - Strain Relationship

We now have the following relationship between the stress and strain:

xx = x/E, and yx = zx = x/EA material that has the same elastic properties in all directions is called anisotropic material. The above equation is applicable only for isotropic

materials. For anisotropic materials the Poisson's ratio and the elastic moduluswould depend on the orientation. For isotropic materials, we may write the

stress - strain relationship as follows:

xx

xy

xz

yx

yy

yz

zx

zy

zz

If an element is subject to stresses in all three directions, then how does onecalculate the resulting strains? For example when an element is under a state of

tri-axial stress (x, , y, , z)what is the net strain in the x direction?

x == ++ ++ .........(1).........(1)

Similarly,

y = (= (y ( ( x+ + z))/ .........(2)))/ .........(2)

and z = (= (z ( ( x + + y))/ .........(3)))/ .........(3)

These together with equations relating shear stress and shear strain are calledGeneralised Hooke's law or General constitutive equations for isotropic

materials.

Bulk Modulus and Dilatation


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The change in volume per unit volume is referred to as the dilatation of amaterial and is denoted by e. Dilatation may be regarded as volumetric strain.

Consider the change in volume of an infinitesimal element of dimensions dx,dy and dz. Let the new lengths of these sides be dx', dy' and dz'. Using

Hooke's law,dx'= dx(1+x), dy'= dy(1+y) and dz'= dz(1+z)Change in volume = New volume - old volume

=dx'dy'dz'-dxdydz =dxdydz((1+x)(1+y)(1+z)-1)= old volume ((1+x)(1+y)(1+z)-1)But dilatation e = change in volume/old volume

=old volume ((1+x)(1+y)(1+z)-1) /old volume=1+xyz+xy+xz+yz+x+y+z-1

=xyz+xy+xz+yz+x+y+zNeglecting the product terms this becomes,

e = x+y+z = sum of normal strains

Now we have e = x+y+z .......(4).......(4)

From Generalised Hooke's Law, we also have the stress strain relationship.

Substituting this relationship into the above equation yields:

e = x-((y+z)/ +)/ +y-((x+z)/+)/+z-((x+y)/)/

= (= (x+y+z)(12)/)(12)/ .......(5).......(5)

Special Case:

If an element is subject to a uniform pressure p

then x = x = x = - p

e = -3p(1-2)/E ......................(6) which may be written ase = -p/k ......................(7)

where k = E/(3(1--2)) ............(7a)k is called the bulk modulus or modulus of compression of the material.

From the above equation, one can deduce the practical limits of Poisson's ratiofor an isotropic material. When a body is subject to a uniform pressure can its

volume increase?

dy'

dx' dz'


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23

For common structural metals such as steel, is about 0.3. Rubber has a valueclose to 0.5. Did you see the Walt Disney movie "Absent minded Professor" ?

I suspect the Professor's invention flubbermay have a greater than 0.5 :-) Itshould be noted here that some anisotropic materials may have Poisson's ratiogreater than 0.5 in some directions._

Summary:

Dilatation (volumetric strain) e = x+y+ z.......(3).......(3)

= (= (x+y+z)(12)/.......(4))(12)/.......(4)

Also e = -p/k ......................(7)

where Bulk modulus k = E/(3(1-2))............(7a)_

Example:Here is a chance to design your own problem. Choose the volume of a steel

block (in mm3) to be subject to hydrostatic pressure:Initial volume V0 =

For steel, take=0.3 and E=

From equations (7) and (7a) k =

The pressure that the block is going to be subject to is MPa

But change in volume = e Vo = (Hint: Use equation (7))


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Shear Stress and Shear Strain

We have seen that the effect of normal stress is normal strain which is definedas the change in length divided by the original length. Shearing stress does not

cause a change in the dimension of an element. It causes distortion, or a changeof shape. The shear strain is defined as the change in the angle of a corner of a

rectangular infinitesimal element as illustrated in the following diagram:

In cross-section, a rectangular element becomes a parallelogram under shearing

stresses. The angles at the corners change from /2 to /2-or /2+. Angle is defined as the shearing strain.

The relationship between the shear stress and shear strain is very similar to the

relationship between normal stress and strain. The linear part of the stress-straincurve is expressed

= = G

where G is called the shear modulus or modulus of rigidity. Since is non-

dimensional (radians) G has the same units as that of stress.


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Summary: Determination of axial force, stress and displacement in bars

Strategy:

Determine whether or not the problem is statically indeterminate. If it is not

possible to calculate the internal forces from statics then the problem isstatically indeterminate.

Statically determinate problems

1. Apply method of sections to calculate the internal axial force. This involves

sketching a free-body and applying either equations of equilibrium, orequations of motion (Newton's second law).

2. For a series of loads, make cuts at various points, or use singularity

functions, although this is not usually needed for axial loading problems.3. Axial force may vary as a continuous function of axial co-ordinate due to

body forces (such as force due to gravity or acceleration) or other externalforces (friction and drag forces in 'pile driving'). For uniform bar subject to

constant acceleration, finite free-body diagrams may be sketched, and eitherequations of equilibrium or equations of motion (depending on whether the

body is stationary or not) may be applied.4. If the body forces were to vary with distance (such as in rotating bars, or

accelerating tapered bars), then consider the motion of an infinitesimal

element, and apply Newton's second law of motion.5. Calculate the stress by using the formula = F/A where F is the induced

axial force. This does not include stress concentration effects and isapplicable only for centric uni-axial loading.

6. Use one of the following equations to determine the relative displacement

between two ends of a bar segment: dxAE

Fuu

x

x=2

112 ..(i) or

AE

)xx(Fuu 12

12

= ..(ii) depending on whether any of the parameters, F,A

andEvary withx or not.

Statically indeterminate problems

An equilibrium problem is said to be statically indeterminate if the number of

unknowns exceed the number of independent equations of equilibrium. Thedifference between the two numbers is referred to as the degree of static

indeterminacy. To solve these problems one needs to use:(a) Equation(s) of equilibrium (see steps 1-4 above)

(b) compatibility condition(s)

(c) constitutive equation(s) formed using formula (i) or (ii) above.


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Compatibility conditions are geometric conditions imposed by the supports orconnections to other structures, and give a relationship between displacements.Substituting the force-displacement equation(s) into compatibility condition will

result in additional equations in terms of unknown forces. In some problems, itis convenient to remove some of the constraints to make the system statically

determinate and stable. The actions thus removed are called redundants.Number of redundants = degree of static indeterminacy. The compatibility

condition is enforced by superimposing the displacements due to the loadingand redundant(s) on the released structure.

Examples of statically indeterminate problems include constrained bars, thermal

stresses, compound beams and stresses due to imposed displacements such as inturn-buckles.

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Axial Stress mech sol