Topic 1-Stress Strain

8/10/2019 Topic 1-Stress Strain

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AVERAGE NORMAL STRESS

Will the total shear force over the anchor length be equal tothe total tensile force tensile A in the bar?

P =

Copyright 2011 Pearson Education South Asia Pte Ltd

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EXAMPLE 1

The bar in Fig. 116 a has a constant width of 35 mm and athickness of 10 mm. Determine the maximum average

shown.


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EXAMPLE 1 (cont)

By inspection, different sections have different internal forces.Solutions

, .


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EXAMPLE 1 (cont)

By inspection, the largest loading is in region BC ,Solutions

kN30= BC

P

Since the cross-sectional area of the bar is constant , the largestaverage normal stress is

( ) (Ans)MPa7.8501.0035.0

1030 3 === P BC BC


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DESIGN OF SIMPLE CONNECTION

P

allow =

or s ear orce requ rement

V

allow =


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. Double shear Single shear

8Beer FP, Johnston ER, Jr., DeWolf J.T, Mazurek DF. Mechanics of Materials, 5 th Edition, McGraw Hill, New York, 2009.


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EXAMPLE 2 (cont)

The allowable stresses areSolutions

( ) ( )

MPa3402

680

..

===S F

fail st allow st

( )

900

MPa352..

===S F

fail

fail al allowal

2..===

S F allow

ere are ree un nowns an we app y e equa ons o equ r um,

( ) ( ) (2) 075.02 ;0 .

==+ P F M B A AC B


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EXAMPLE 2 (cont)

For pin A or C,Solutions

( ) ( ) kN5.114009.010450 26 ==== A F V allow AC

( )( )25.114 ==. ,25.1

When P reaches its smallest value (168 kN), it develops the allowablenormal stress in the aluminium block. Hence,

(Ans)kN168= P


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Chapter Objectives

Understand the concept of normal and shear strain

types of problems


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NORMAL STRAIN

s s ' savg

=

s s '

sn A B along

( ) s s + 1'


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SHEAR STRAIN

' 2 along along t AC n A Bnt


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CARTESIAN STRAIN

e approx ma e eng s o e s esof the parallelepiped are

The approximate angles between sides, again originally

) ) z y x z y x +++ 1 1 1

defined by the sides x, y and z are

xz yz xy

2

2

2

of rectangular element, whereas the shear strain cause a

change in shape


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CARTESIAN STRAIN (cont)


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EXAMPLE 3

The slender rod creates a normal strain in the rod ofwhere z is in meters. Determine (a) displacement of end B

( ) 2/131040 z z =

ue o e empera ure ncrease, an e average normastrain in the rod.


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EXAMPLE 3 (cont)

Part (a)

Solutions

,deformed length of

dz z dz 2/13

10401'

+=

The sum along the axis yields the deformed length of the rod is

( )[ ] m20239.010401' 2.00

2/13 =+= dz z z

The displacement of the end of the rod is therefore

(Ans)mm39.2m00239.02.020239.0 === B


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EXAMPLE 3 (cont)

Part (b)Solutions

ssumes e ro as an or g na eng o mm an a c ange nlength of 2.39 mm. Hence,

Ansmm/mm0119.039.2' === s s

s


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EXAMPLE 4

Due to a loading, the plate is deformed into the dashed shapeshown in Fig. 26 a . Determine (a) the average normal straina ong e s e , an e average s ear s ra n n e p a eat A relative to the and y axes.


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EXAMPLE 4 (cont)

Part (a)

Solutions

, , ,the length of this line is

' 22 .

'( ) 3. 7.93 10 mm/mm (Ans)250 AB avg AB

= = =

The negative sign indicates the strain causes a contraction of AB.


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EXAMPLE 4 (cont)

Part (b)Part (b) Solutions

, ,referenced from the x, y axes, changes to due to the displacement of Bto B.

Since then is the angle shown in the figure.'2 =

xy xy

Thus,

(Ans)rad 121.02250

3tan 1 =

=

xy


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Chapter Objectives

Understand how to measure the stress and strainthrough experiments

Correlate the behavior of some engineering materialsto the stress-strain diagram.


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TENSION AND COMPRESSION TEST


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STRESS STRAIN DIAGRAM

o e e cr ca s a us or s reng spec ca onproportional limite as c m

yield stressu t mate stressfracture stress


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EXAMPLE 5

The stressstrain diagram for an aluminum alloy that is usedfor making aircraft parts is shown in Fig. 319. If a specimeno s ma er a s s resse o a, e erm ne epermanent strain that remains in the specimen when the load

. ,and after the load application.


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EXAMPLE 5 (cont)

When the specimen is subjected to the load, the strain is approximately0.023 mm/mm.

Solutions

The slope of line OA is the modulus of elasticity,

GPa0.75006.0

450 == E

From triangle CBD,

( )100.7510600 9===CDCD

BD E

.


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EXAMPLE 5 (cont)

This strain represents the amount of recovered elastic strain .Solutions

The permanent strain is

Computing the modulus of resilience,

(Ans)mm/mm0150.0008.0023.0 ==OC

( ) ( )( ) (Ans) MJ/m35.1006.045011 3=== pl pl initial r u

( ) ( )( ) (Ans) MJ/m40.2008.06002

1

2

1 3=== pl pl final r u

Note that the SI system of units is measured in joules, where 1 J = 1 N


.

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POISSONs RATIO

long

v

=


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EXAMPLE 6

A bar made of A-36 steel has the dimensions shown in Fig.322. If an axial force of P = 80kN is applied to the bar,

e erm ne e c ange n s eng an e c ange n edimensions of its cross section after applying the load. The

.

Copyright 2011 Pearson Education South Asia Pte Ltd32


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EXAMPLE 6 (cont)

The normal stress in the bar isSolutions

( )( )( )

( )Pa100.1605.01.0

1080 63

=== A

P z

From the table for A-36 steel, E st = 200 GPa

( )mm/mm1080

100.16 66

=== z

10200 st E

( ) (Ans) m1205.11080 6z === L



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EXAMPLE 6 (cont)

The contraction strains in both the x and y directions areSolutions

m/m6.25108032.0 6 ==== z st y x v

The changes in the dimensions of the cross section are

( )( )( )( )[ ] (Ans) m28.105.0106.25 (Ans) m56.21.0106.25 66

===

===

y y y

x x x

L L



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SHEAR STRESS-STRAIN DIAGRAM

reng parame er ear mo u us o e as c y or emodules of rigidity

ratio v.

G=

( )v+=

12



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EXAMPLE 7

A specimen of titanium alloy is tested in torsion and theshear stress strain diagram is shown in Fig. 325 a .

e erm ne e s ear mo u us , e propor ona m , anthe ultimate shear stress. Also, determine the maximum

,Fig. 325 b, could be displaced horizontally if the materialbehaves elastically when acted upon by a shear force V.

What is the magnitude of V necessary to cause thisdisplacement?



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EXAMPLE 7 (cont)

By inspection, the graph ceases to be linear at point A. Thus, theSolutions

(Ans)MPa360= pl This value represents the maximum shear stress, point B. Thus the

ultimate stress is

Since the an le is small, the to of

=u

the will be displaced horizontally by

( ) mm4.0mm50

008.0rad 008.0tan == d


EXAMPLE 7 ( )


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EXAMPLE 7 (cont)

The shear force V needed to cause the displacement isSolutions

(Ans)kN270010075

MPa360 ; === V V V avg



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EXAMPLE 8


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EXAMPLE 8

The assembly shown in Fig. 47 a consists of an aluminumtube AB having a cross-sectional area of 400 mm 2. A steelrod havin a diameter of 10 mm is attached to a ri id collarand passes through the tube. If a tensile load of 80 kN isapplied to the rod, determine the displacement of the end Cof the rod. Take E = 200 GPa E = 70 GPa.


EXAMPLE 8 (cont )


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EXAMPLE 8 (cont.)

Find the displacement of end C with respect to end B.Solutions

( )[ ]( )( ) ( )[ ]

+=+== m003056.010200005.0

6.010809

3

/

AE

PL BC

Displacement of end B with respect to the fixed end A,

( )[ ] ( )[ ] ==

== m001143.0001143.01070104004.01080

96 AE PL

B

Since both displacements are to the right,==+= mm20.4m0042.0/ BC C C


PRINCIPLE OF SUPERPOSITION


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PRINCIPLE OF SUPERPOSITION

can e use o s mp y pro ems av ng comp ca eloadings. This is done by dividing the loading into

, .

and the deformation is small.

If P = P1 + P2 and d d1 d2, then the deflection atlocation x is sum of two cases, = +


COMPATIBILITY CONDITIONS


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COMPATIBILITY CONDITIONS

en e orce equ r um con on a one cannodetermine the solution, the structural member is called

.

,locations shall be used to obtain the solution. For example,the stresses and elongations in the 3 steel wires are

different, but their displacement at the common joint A mustbe the same.


EXAMPLE 9


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EXAMPLE 9

The bolt is made of 2014-T6 aluminum alloy and is tightenedso it compresses a cylindrical tube made of Am 1004-T61magnes um a oy. e u e as an ou er ra us o mm,and both the inner radius of the tube and the radius of the bolt

.considered to be rigid and have a negligible thickness. Initiallythe nut is hand-tightened slightly; then, using a wrench, the

nut is further tightened one-half turn. If the bolt has 25threads per mm, determine the stress in the bolt.


EXAMPLE 9 (cont )


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EXAMPLE 9 (cont.)

Equilibrium requiresSolutions

(1) 0 ;0 ==+ t b y F F F

en e nu s g ene on e o , e u e w s or en.

( bt =+ 5.0


EXAMPLE 9 (cont )


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EXAMPLE 9 (cont.)

Taking the 2 modulus of elasticity,Solutions

( )

[ ] ( )[ ]

( )

[ ] ( )[ ]10755

605.0

104551060

32322bt F F =

Solving Eqs. 1 and 2 simultaneously, we get

(2) 911251255 bt =

kN56.3131556 === t b F F

The stresses in the bolt and tube are therefore

(Ans)MPa8.401 N/mm8.40131556 2

==== b

b

F

(Ans)MPa9.133 N/mm9.13331556 2

22 ====

t s

b

F


t

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EXAMPLE 10


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EXAMPLE 10

The A-36 steel rod shown in Fig. 417 a has a diameter of 10mm. It is fixed to the wall at A, and before it is loaded there isa gap e ween e wa a an e ro o . mm.Determine the reactions at A and Neglect the size of the

. .



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EXAMPLE 10 (cont.)


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( )

From the free-body diagram,Solutions

( )

005.420

0

=+

=+ x F

F

(Ans) kN0.16 = A F



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Chapter Objectives (Section 4.6 ~ 4.9)

Deal with thermal stress problems

Deal with inelastic deformation problems


THERMAL STRESS


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Ordinarily, the expansion or contraction T is linearlyrelated to the temperature increase or decrease T thatoccurs.

T TL =

= linear coefficient of thermal expansion , property of the material= algebraic change in temperature of the member =

T

= algebraic change in length of the member T

If the change in temperature varies throughout the length ofthe member, i.e. T = T (x), or if varies along the length,

dxT T =


EXAMPLE 11


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The rigid bar is fixed to the top of the three posts made of A-36 steel and 2014-T6 aluminum. The posts each have alen th of 250 mm when no load is a lied to the bar and thetemperature is T1 = 20 C. Determine the force supportedby each post if the bar is subjected to a uniform distributedload of 150 kN/m and the tem erature is raised to T2 =80 C.


EXAMPLE 11 (cont.)


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From the free-body diagram we haveSolutions

(1) 010902 ;0 3 =+=+ al st y F F F

The top of each post is displaced by an equal amount and hence,

Final osition of the to of each ost is e ual to its dis lacement caused(2) al st

=+

by the temperature increase and internal axial compressive force.

( ) ( ) ( ) F al T al al F st T st st

+=++=+


EXAMPLE 11 (cont.)


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Applying Eq. 2 givesSolutions

( ) ( ) ( ) ( ) F al T st F st T st +=+

With reference from the material properties, we have

( )[ ]( )( )( ) ( )[ ]

( )[ ]( )( )( ) ( )[ ]( ) (3) 109.165216.1

101.7303.0

.25.020801023

1020002.0

.25.020801012

3

926

926

=

+=+

al st

al st

F F

Solving Eqs. 1 and 3 simultaneously yields

(Ans) kN123 and kN4.16 == al st F F


STRESS CONCENTRATION


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The stress concentration factor K is a ratio of themaximum stress to the average stress acting at thesmallest cross section; i.e.

maxavg


STRESS CONCENTRATION (cont.)


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K is independent of the material properties K de ends onl on the s ecimens eometr and the t e

of discontinuity


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Topic 1-Stress Strain