CMats Lect4-Compatibility & Shear Stress and Strain

8/3/2019 CMats Lect4-Compatibility & Shear Stress and Strain

1/28

Civil Engineering Materials 267Stresses in Materials

Lecture 4: Compatibility

Shear Stress and Strain

Kerri Bland


2/28

Lecture 4 2

Civil Engineering Materials 267 - Stresses

References

P.P. Benham & R.J. Crawford, Mechanics of Engineering Materials, 1987,Longman Scientific & Technical

R.C. Hibbeler, Mechanics of Materials, 6th Ed., 2005, Prentice Hall/Pearson.

E.P. Popov, Engineering Mechanics of Solids, 2nd Ed., 1991, Prentice Hall.

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3/28

Lecture 4 3


Principal of Compatibility

Strains must be compatible with any

internal or external restraints

Force applied

resulting in xx

y

FF x=F/A

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=Determine x, y and z,using general strainequations:

Effect of Restraining Mechanical Strain:



4/28

Lecture 4 4


Strains must be compatible with any internal or externalrestraints

Force appliedresulting in xx

y

FFx=F/A

tEEE

tEEE

tEEE

yxz

z

zxy

y

zyxx

+=

+=

+=

If there is no restraint against strain in the y and z

direction (ie: it expands as it wants due to Poissons

ratio), then y and Z = 0, so:

E

E

E

xz

xy

xx

=

=

=

x

yy=0



5/28

Lecture 4 5



Force appliedresulting in x

x

y

FFx=F/A

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=

If there is restraint against strain in the y direction then y = 0,

y 0 (ie: in order to prevent the strain that wants to occur(due to ) a restraining force or stress is applied to the elementby the restraint) (Assume no restraint

in z direction so Z

= 0)

xy

xy

y 0EE

=

==

x

y=0y0

x primary or applied stress

y secondary or derivative stress



6/28

Lecture 4 6



Force appliedresulting in xx

y

FFx=F/A

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=

Also, if there is restraint against strain in the y direction then

x x/E(Assume no restraint in z direction so Z = 0)

( )2xx

xy

yxx

1E

and

EE

=

=

=

x

y=0y0



7/28

Lecture 4 7



x

y

FFx=F/A

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=

( )2xx

xy

yxx

1E

and

EE

=

=

=x

y=0y0

steelfor1.1E

)1(

E

)1(E

EModulussYoung'Apparent

2

2x

x

x

x'x

=

=

=

So, for 100% restraint in the y or z direction (not both) we get:



8/28

Lecture 4 8



x

y

FFx=F/A

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=

x

y=0y0

)1(

EEModulussYoung'Apparent

2

'x

=

For a reduced proportion of restraint (where 1.0> >0)in the y or z direction (not both) we get:

( )2xx

xy

yxx

1E

and

EE

=

=

=



9/28

Lecture 4 9



x

y

FFx=F/A

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=

x

y=0y0

)21(

)1EEModulussYoung'Apparent

2

'x

=

For 100% restraint in the y and z directions (both) we get:



10/28

Lecture 4 10


Strains must be compatible with any internal or external restraints

x

y

Know: l, t, , A, EFind x, induced axial force:

Need 2 equations to solve:

Effect of Restraining Thermal Strain:

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=

Heat t

(1)Compatibility:

x = thermal + mechanical strain = 0

=

000 === lll


(2) Axial force (AF) = xA

= -tEA

tEE

t

x

x

=

=+

0


11/28

Lecture 4 11


Principal of Compatibility Strains must be compatible with any internal or external restraints

x

y

Find st, br, induced axial force:

Need 3 equations to solve:

Effect of Restraining Thermal Strain:

tEEE

tEEE

tEEE

yxzz

zxy

y

zyxx

+=

+=

+=

Heat t

(2) & (3) Axial force (AF) = (xA)st=(xA)br

Solve forx.stand x.br, then AF

steel brass

lst

lbr

(1)Compatibility:

( ) ( )

0..

0:

00

=

++

+

=+

=

=

=+=

br

br

xst

st

x

brasssteel

brasssteeltotal

Et

Et

so

and

ll

ll

lll

l

lll


12/28

Lecture 4 12



Strains must be compatible with any internal orexternal restraints

Restraint problems are always statically

indeterminate and require knowledge of material

characteristics and member values (ie: crosssectional area, E, )


13/28

Lecture 4 13


Thermal strain example Concrete mass after initial set

Outer layers heat of hydration able to dissipate

cooled and set in fixed positionrelative to ambient temperature

Inner layers heat of hydration not able to

dissipate

heat builds up

wants to expand (=t) restrained due to outer fixed layers stresses induced

Compatibility

x

y

0

0

=+=

=+=

tEE

tEE

xy

y

yxx

cold

cold coldcold

cold

cold

coldcold

cold

coldheat

x

y

Two equations and two unknowns:

solve to find x

and y

.

Assuming no

restraint inthe z direction


14/28

Lecture 4 14


CompatibilityWorked exampleLong reinforced concrete wallGiven: Econcrete = 14 GPa

= 0.2 = 10-5/C

y= -2.5 MPax=?

y=?

x=?

t=100C

(a) (b)

(a) The compressive stress in an element of concrete at the bottom of areinforced concrete wall due to the self-weight of the concrete has been

calculated as -2.5 MPa.

Determine the normal stress in the x-direction due to the confinement of

the concrete in the longitudinal direction of the wall.

y= -2.5 MPa

x=?

x= 0 MPaEE

ityCompatibil

yx

yxx

5.02.0*5.2

00

:

===

+==

As a wall is relatively thin

(cf width and height)

assume that z = 0


15/28

Lecture 4 15


CompatibilityWorked example

(2)tEE

(1)tEE

ityCompatibil

xy

y

yxx

+==

+==

0

0

:

Long reinforced concrete wallGiven: Ecocnrete = 14 GPa

= 0.2 = 10-5/C

y= -2.5 MPax=?

y=?

x=?

t=100C

(a) (b)

(b) A small part of the wall is being heated such that the temperature rise is100C.

Determine the normal stresses in the x and y-directions due to the

confinement of the concrete in the longitudinal direction of the wall.

y= ?

x=?

x= 0

As a wall is relatively thin (cf width andheight) assume thatz= 0

y= 0

)5.175.17

)2.01(

100*10*10*14

)1(

0)1()1(

53

2

MPa(andMPa

tE

tE

(2)intosubstituteandby(1)Multiply

x

y

y

==

=

=

=++


16/28

Lecture 4 16


CompatibilityWorked exampleLong reinforced concrete wallGiven: Ecocnrete = 14 GPa

= 0.2 = 10-5/C

y= -2.5 MPax=?

y=?

x=?

t=100C

(a) (b)

(c) If parts a and b occur simultaneously, determine the normal stresses inthe x and y-directions

y= ?

x=?

x= 0

As a wall is relatively thin (cf width andheight) assume thatz= 0

y= 0

By superposition:

x = -0.5+(-17.5) = -18 MPa

y = -2.5+(-17.5) = -20 MPa


17/28

Lecture 4 17


CompatibilityExample:Steel rod

Brass tube

airHeat t

Find st, br, xCompatibility of strains:

(1)tE

tE

tube

br

br

br

rod

st

st

stx

+=

+=

( ) ( ) (2)0AA

0F

brbrstst

x

=+=

Equilibrium:

Solve (1) and (2) to find st and br, then use to find x.

tuberodassembly

tuberodassembly

lll

lll

==

==

x,assembly=x,rod=x,tube


18/28

Lecture 4 18


External Shear ForceExternal Shear Force

External Axial Tension Force

Internal Normal Stress Response

External Shear ForceExternal Shear Force

Internal ShearStress Response

(AverageNormal Stress)A

P=

A

Vav =

(Average

Shear Stress)

Similarly:

Shear Stress


19/28

Lecture 4 19


Examples of shear in bolts:

Top bolt (a, b, c, d) shows single shear plane

Bottom bolt (e, f, g, h) shows double shear planes


20/28

Lecture 4 20


Class examples:

1. Determine the average shear stress in each bolt in the following detail.

Assume M24 bolts are being used. (ie bolt diameter = 24mm)

A

A

30

0kN

a) Section AA

b) Section AA


21/28

Lecture 4 21


Class examples:

2. If we have an applied load of 540kN, what bolt size would be required

for the three bolt arrangement shown below? Assume that the

maximum possible shear stress in the bolts is 400MPa. (Standard bolt

sizes: M12, M16, M20, M24, M30, M36).

A

A

540kN

a) Section AA

b) Section AA


22/28

Lecture 4 22


xxusually denoted asx

Stress direction and type (ie normal/shear) changes depending on direction and

type of externally applied loads. There needs to be a consistent method of

describing the stresses so that the type and direction of stress can

be easily recognised.

x

z

y Typically adopt an orthogonal set of axes:

In this unit the x-axis will correspond with

the longitudinal axis of the member inquestion.

Warning: In later design units axes changeand the x and z axes swap so that thelongitudinal axis becomes the z axis.

Stress expressed as with two subscripts.1

st

Subscript: direction normal to the plane on which stress acts2nd Subscript: actual direction of the stress

xy xz

Normal & Shear Stresses


23/28

Lecture 4 23


Normal & Shear Stresses 3D In a 3D elemental cube, the normal and shear

stresses can be designated as follows:-

1st subscript = direction

normal to

the planeon which

stress acts

2nd subscript = actualdirection of

the stress


24/28

Lecture 4 24


The element is in equilibrium:-MA = 0

yx= xy Complementary or conjugate shearstresses are paired and in equilibrium. For conjugate shear stresses in 3D:-

xy=yx yz=zy zx=xz

Conjugate Shear Stresses

x

z

y

xyxyxy

yx

yx(Initial action)(Reaction)

(Reaction)

(Reaction)

Considering the x-y plane only (ie: in 2D):

A


25/28

Lecture 4 25


Shear Strain Consider a small

element subjectedto shear:-

Shear stress = = V/A (MPa)

Shear strain = = l/l = angular strain = change of an angle that

was formally 90

Shear strain = l/l = tan for small values.


26/28

Lecture 4 26


Visualising shear straindeformation in 2D

subscript of (shear strain)= subscript of stresses causing distortion


27/28

Lecture 4 27


For a ductile material, we can draw a -relationship, which is analogous to -

relationship.

This can be idealised:-

Slope = G = = shear modulusor modulus of rigidity

Shear Modulus G


28/28

Lecture 4 28


SummaryShear stress and Strain

Average Shear stress :

Shear strain (angular strain):

Shear modulus (modulus of rigidity):

(a measure of the shear stiffness of

a linear elastic material)

AVav =

l

l =

=G

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CMats Lect4-Compatibility & Shear Stress and Strain

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