Strength of Materials- Deformation Due to Axial Load- Hani Aziz Ameen

Strength of Materials Handout No.3

Deformation due to Axial Load

Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Dies and Tools Eng. Dept. E-mail:[email protected]

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Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen

3-1 Introduction In machine design and structure design , the strain is another

important factor which can be considered . Strain is defined as the proportional changes in dimensions . Hence,

dimension originalndeformatioStrain

or

L

,

where ........... deformation

L .............. original length ............. strain

The kinds of strain are classified according to the types of stresses ; thus

1. Tensile Strain If the original length of the bar is L and under effect of the force P , it extends a distance . It is shown in Fig(3-1)

Thus L

Fig(3-1)

2. Compressive Strain If the original length of the bar is L and under effect of the force P , it compresses a distance . It is shown in Fig(3-2)

Thus L

Fig(3-2)


3. Shear Strain ( ) If the deformation in the direction of P is , as shown in Fig(3-3)

then Fig(3-3)

Lant

for small it can be deduced that L

3-2 Stress Strain Diagram The stress strain diagram can be obtained from the tensile test , as shown in Fig.(3-4)

Fig(3-4) The initial part of the tension curve which is recoverable immediately after unloading is termed as elastic range , and the rest of the curve which represents the manner in which the solid undergoes plastic deformation is termed as plastic range.


3- the deformation is directly proportional to the load producing it , since the stress is proportional to the load and the stain is proportional to the deformation , it follows that the stress is proportional to the strain i.e. the ratio stress / strain is a constant for any given material . i.e. or *.const This constant is known as the Moduland is denoted by E .

E

i.e. E

For shear this constant is known as the modulus of rigidity and is denoted by G ,

G

e :-

1- The material must be homogenous in any cross sectianal area. 2- The force applied is axial. 3- The stress and strain remain within elastic limit.

3-4 Determination of Strain in Two Dimension Consider the body subjected to the load as shown in Fig.(3-5)

Fig(3-5)

Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen To simplify this case , let us discuss the body in two dimensions as shown in Fig.(3-5b)

is deduced from tensile test and states that :

where

strain Lateral.......strain ngitudinal........Lo

ratio sPossion'........

Now from Fig(3-5b) , 41xx

y4

x1 ,

thus ; yxx1

Similarly y = 3 + 2

xyy1

3-5 Factor of Safety The stresses which are present in a component under normal working conditions are called the working stresses w ( allowable stress a ) , the ratio of ultimate stress u [ or yield stress y ] , to the working stress is the factor of safety , n , Hence ;

stress allowablestress ultimateor yieldn

w

y

w

u

where : 1< n < 10


3-6 Determination of Bodies Deformation 3-6-1 Rod or Bar with Constant Cross-Sectional Area When an axial load P is applied to a bar as shown in Fig(3-6) ,

E to it , we can find the deformation .

Fig(3-6) where:

L &

AP

(P/A) = E ( /L)

or AEPL

3-6-2 The Body with Variable Cross- Sectional Area

Case 1 If we have a body as in Fig.(3-7) ,

Fig(3-7)

Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen then the total deformation of all parts is equal to algebraic summation of individual i.e.

44

44

33

33

22

22

11

11n

1i ii

iiT EA

LPEALP

EALP

EALP

EALP

Case 2 We have a body as in Fig(3-8) ,

Fig(3-8)

law ,

EAPdx

)x()dx( ,

21)x( y2d

4A

L

0 21

L

0 )x(T

E]y2d[4

PdxEA

Pdx ........................... (3-1)

for similarity of triangles

L

2/)dd(xy 12 ..................... (3-2)

from eq.(3-1) and eq(3-2) , get

EddPL4

21T -3)

Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen 3-6-3 Deformation Due to the Weight of Bar If we have a bar as shown in Fig(3-9 a ) . If we apply the equilibrium condition on an element from a bar( Fig.(3-9 b))

- a - - b Fig(3-9) Thus,

0A]d[AdxA )x()x()x(

dxd x

Cdxd x

Cx)x( .............. (3-4) at 0CL Lx (L) LC , sub into eq.(3-4), yields

Lx)x( )xL()x(

)xL(E1

Ex

x

L

0

L

0xx dx)xL(

E1dx.

dx

E2L2

Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen if A is constant , hence

EAL AL

21 , ALW , hence

2EAL W

3-6-4 Deformation due to Weight of Bar & the External Force

If the bar with its weight is as shown in Fig.(3-10) , then we can prove that the total deformation will be

= (WL / 2AE) + (PL / AE) for constant cross section , and = ( WL / 2AL) + (4 PL / d1d2E) for variable cross section

Fig(3-10)

3-7 Statically Indeterminate Problems

The statically determinate calculation of the forces and stresses in members is possible considering only the static equation of equilibrium system i.e.

0Fx , 0Fy and 0M From these equilibrium equations ,one can find three unknown forces only .


But for statically indeterminate problems , the equations of statics are not sufficient to solve the unknown . We need another relation from elastic deformation to solve the problems and to solve similar problems :

1- Draw the free body diagram for the problems and apply the equilibrium equations .

0Fx , 0Fy and 0M

2- Find the geometric relation between bodies deformation )etc,........,( 4321 3- Solve the resulting equations in points (1) & (2) to get the unknowns . 3-8 Examples The following examples explains the different concepts of deformation due to axial load . Example (3-1) Fig.(3-11) shows the member AC is supported by a round structural steel tie rod BD and a pin at A , neglect the weight of member AC and assume that the ultimate strength ( u ) of structural steel rod is 490 MPa in tension and that the ultimate shear stress ( u ) of the pin is 315 MPa. Using a factor of safety of 3.5 for both tension and shear, find the minimum required diameters of tie rod & the pin .

Fig(3-11)


Solution

From equilibrium condition

0Rx 0MA

kN120S03*P2*S

kN402

80R01*P2*R0M AAB

hence, the allowable stress will be

MPa1405.3

490n

490nu

w

and

wAS

140*10*120*44S d

d4

S 3

wrw

2r

mm33dr

the working shear stress ( or allowable shear stress) will be

MPa905.3

315nu

w

)d)4/(*2/(R 2pAw

mm1790*10*40*2R2dp

3

w

A


Example(3-2) The following observations were made during a tensile test on a mild steel specimen 40 mm in diameter and 200 mm long . Elongation is with 40 kN load ( with in limit of proportionality )

mm0304.0 , yield load = 161 kN , Maximum load = 242 kN Length of specimen at fracture = 249mm Find :

1) 2) Yield point stress 3) Ultimate stress 4) Percentage elongation

Solution 1) nd & , hence ,

24

2m/kN10*18.3

)04.0(4

40AP

& 000152.02000304.0

L

then, 28

4m/kN 10*09.2

000152.010*3.18E E

2) Yield point stress = area

loadpoint yield

= 24

2kN/m 10*8.12

)04.0(*4

161

3) Ultimate stress = 24

2m/kN10*2.19

)04.0(*4

242area

load imummax

4) Percentage elongation =

%5.24245.0200

200249length original

length original fractureat specimen of lenght

Example (3-3) A square steel rod 20mm x 20 mm in section is to carry an axial load ( compressive) of 100 kN . Find the shortening in a length of 50 mm, E=2.14*108 kN/m2 Solution Area , A= 0.02 * 0.02 =0.0004 m2 Hence the shortening of the rod can be obtained :


2kN/m 2500000004.0100

AP

810*14.2250000

E , where

L

thus , 810*14.2250000

L

hence,

m0000584.005.0*10*14.2

2500008

Example (3-4) A hollow cast-iron cylinder 4m long , with 300mm outer diameter , and thickness of metal 50 mm is subjected to a central load on the top when standing straight . The stress produced is 75000 kN/m2

modulus for cast iron as 1.5*108 kN/m2 . Find 1- magnitude of the load , 2- longitudinal strain produced , and 3- total decrease in length .

Solution Inner diameter of the cylinder , d = D 2t = 0.3 2 * 0.05 = 0.2 m

1) Magnitude of the load P :

using the relation , AP

or )dD(4

*75000A*P 22

= 75000 * kN2.2945)2.03.0(4

22

2) Longitudinal strain produced , using the relation

= E

= 75000/(1.5* 108 )= 0.0005

3) Total decrease in length ( )

Strain = Llength Original

lengthin Change

0.0005= mm 2m 002.04

Hence decrease in length = 2 mm


Example(3-5) Fig.(3-12) shows a steel bar 900 mm long ; its two ends are 40 mm and 30 mm in diameter, the length of each rod is 200 mm . The middle portion of the bar is 15 mm in diameter and 500 mm long . If the bar is subjected to an axial tensile load of 15 KN , find its total extension, take E = 200 GPa

Fig(3-12) Solution Areas

A1= 2404

=1256.6 mm2

A2= 2154

=176.7 mm2

A3= 2304

= 706.8 mm2

Length : L1=200 mm=0.2 m ; L2=500 mm=0.5 m & L3= 200 mm=0.2 m Let 321 and , be the extensions for the parts 1,2 and 3 of the steel bar respectively .

Then EA

PL , EA

PL , EA

PL

3

33

2

22

1

11 where P = 15 kN

Total extension of the Bar , 321

3

3

2

2

1

1

3

3

2

2

1

1AL

AL

AL

EP

EAPL

EAPL

EAPL

0007068.0

2.00001767.0

5.0001256.0

2.010200

10159

3= 0.0002454 m


Example(3-6) Fig(3-13a) shows a steel rod with variable cross-section subjected to a central force . Find the total elongation of the rod if GPa207 and the area of each cross section is as indicated on the Fig.

-a- - b Fig(3-13) The force acting on each portion of the rod is as indicated on the freebody diagrams of Fig(3-13 b) Hence, the total deformation is

3

33

2

22

1

11ALP

ALP

ALP

E1

AEPL

6

3

6

3

6

3

9 10*3.16161.0*10*9.8

10*6.32252.1*10*6.35

10*5.80613.2*10*6.88

10*2071

mm 1.2


Example (3-7) Fig(3-14) shows the bar subjected to a tensile load of 50 kN find : (i) the diameter of the middle portion if the stress is limited to 130MN/m2 (ii) the length of the middle portion if the total elongation of the bar is 0.15 mm , take E=200 GN/m2 .

Fig(3-14) Solution

i) The diameter of the middle portion , d :

Now , stress in the middle portion , 610*130d4/

1000*50AP

2

m 221.010*130*)4/(

1000*50d2/1

6

ii) Length of the middle portion : Let the length of the middle portion = x meter

Stress in the end portion E

x25.0*

also, elongation of the end portions + extension of the middle portion = 0.15*10-3

39

6

9

610*15.0

10*200x*10*130

10*200x25.0*10*79.39

3966 10*15.0*10*200x10*130x25.0*10*79.39 Dividing both sides by 39.79* 106 , we get

0.222m x 0.754 3.267x x -0.25


Example (3-8) Fig(3-15) shows a steel tie rod 50 mm in diameter and 2.5 m long subjected to a pull of 100 kN to what length should the rod be bored centrally so that the total extension will increase by percent under the same pull . The bore is 25 mm in diameter. Take E=200 GN/m2

Fig(3-15) Solution i)Length of the bore x :

Stress in the solid rod, 262 m/N10*92.50

05.0*)4/(1000*100

AP

Elongation of the solid rod, m 000636.010*200

5.2*10*92.50EL

9

6

Elongation after the rod is bored =1.15*0.636 = 0.731 mm Area at the reduced section = 222 m 001472.0025.005.0)4/(

Stress in the reduced section, 26 m/N10*93.67001472.0

1000*100

Elongation of the rod 310*731.0Ex

Ex5.2

39

6

8

610*731.0

10x200x10*9.67

10*200x5.210*92.50

3996 10*731.0*10*200x10*9.67x5.210*92.50 87.2x33.1x5.2 x = 1.12 m


Example(3-9) Fig(3-16) shows a brass bar having cross sectional area of 1000 mm2 subjected to axial force . Find the total elongation of the bar the

modulus of elasticity of brass = 100 GN/m2 Fig(3-16) Solution From equilibirum conditions

i) Total elongation of the bar :

Let 321 and , be the change in length LM ,MN and NP respectively

Then increase..............AE

LP 111

decrease..............AE

LP 222

decrease..............AE

LP 333

Net change in length 321

332211332211 LPLPLP

AE1

AELP

AELP

AELP

12303010

12.1*101*306.0*5010*100*10*1000

10596

3

m 00012.0


Example (3-10) Fig.( 3-17) shows a member LMNP is subjected to point load Find:

i) Force P necessary for equilibrium ii) Total elongation of the bar

Take E = 210 GN/m2

Fig(3-17) Solution i) Force P necessary for equilibrium Resolving the force on the rod along its axis, we get 50+500=P+200 P=350 KN ii)Total elongation of the bar : Let 321 & , be the change in lengths LM,MN and NP respectively

then m 10 3.97 1021010600

1100050 EA

LP 4-96

1

111 ...... increase (+)

96-2

222 10210102400

11000300 EA

LP = 5.95 10-4 m ..... decrease (-)

4-96-

3

333 104.76

102101012000.61000200

EALP m........increase (+)

321 = 3.97*10-4 5.95*10-4 +4.76*10-4 = 2.78*10-4 m


Example (3-11) Fig(3-18 a) shows a steel plate 6.35mm thick and having the dimensions shown in its Fig . Find the total elongation of the plate when subjected to a central force of 40kN taking E as 207 GPa

Fig(3-18a) Solution

Fig(3-18b) From the free body diagram of Fig(3-18 b) , the elongation of the uniform part of plate is

9331 10*207*10*35.6*6.101*10914.0*1000*40

mm27.01 The elongation of the tapered part of the plate is computed as discussed before in article (3-6-2) . If one refers to the free body diagram of Fig(3-18b) , the elongation is

1.22

0 (y)2 EA

Pdy

from Fig (3-18 b) A(y)= (0.0508+2x)*0.00635 from the similar triangles of Fig(3-18b)

y208.0 x yx

22.110*4.25 3


from above A(y) = (0.0508+2*(0.208y))*0.00635

hence , 1.22

092 dy)y00264.00003225.0(

10*2071000*40

mm507.02 the total elongation of the plate = 21 = 0.27+ 0.507= 0.777 Example(3-12) Fig.(3-19) shows a central force of 89 kN is applied to a steel bar at a distance of 14 m from its free end . The total elongation due to the central force and the weight of the bar itself was found as 1.6mm find the total length of the bar and the elongation of the bar due to its own weight. Also , find the max . normal stress caused by the central force and the weight of the bar .Density of the material is 76000 N/m3 and the area of the cross section of the bar is constant at 1290mm2 take E as 207GPa Solution 1 = elongation of the bar due to its own weight

1 = AE2

WL .......................... (i)

W = AL =1290*10 4*L*76000 = 9804L The length L is assumed in meter , substituting into eq .(i) for W Fig(3-19)

96

2

1 10*207*10*1290*2L*9804 .................................. (ii)

The central force causes an elongation of 2 in the part (L 14) of the bar . Hence 2 =P(L 14)/AE ........................... (iii) The total elongation of the bar is the sum of eq . (ii) & (iii)

= 1 + 2 = 1.6*10 3

39696

210*6.1

10*207*10*1290)14L(*1000*89

10*207*10*1290*2L*9804

L = 18.6 mm Using eq.(ii) , the elongation of the bar under its own weight is :


m10*0635.010*207*10*1290*2

)6.18(*9804 376

2

2

The max . normal stress occur at the fixed end of the bar

A

PW

70.4MPa 10*1290

10*896.18*98046

3

Example (3-13) (A) If the strain in the y-direction due to the simultaneous action of

the uniform forces Px, Py, & Pz acting on the steel block shown in Fig(3-20) was found as 1.75 * 10 4 m/m , what was the magnitude of the applied uniform force Pz ? E=207 GPa , =0.25

(B) What single uniform force must be applied to the block in the Z- direction only in order to produce the strain of 1.75*10 4 m/m the y-direction ? Solution E/)( zxyy The stress acting on the block is Fig(3-20)

MPa58.2710*8.50*4.25

10*6.36AP

6

3x

x

MPa16.5510*2.76*8.50

10*5.2136

3

y

6zz

z 10*5.1935P

2.76*4.25P ( assumed as positive)

Substituting the results in the expression for y gives

)10*207/())10*5.1935/P(*25.058.27*)4/3(10*16.55(10*75.1 96z

64

Pz =200 kN (tensile)


We have axial strain in the direction of Pz is E

z and the lateral strain in

the direction of y is *y axial strain , E

zy

or 49

zy 10*75.1

10*207*

41

from which 76.2*25.4

P(comp.) MPa8.144 zz

or Pz= 280 kN (comp) Example (3-14) Fig.(3-21) shows two parallel steel wires 6m long 10 mm diameter are hung vertically 70 mm , apart and support a horizontal bar at their lower ends. When a load of 9 kN is attached to one of the wires it is observed

Fig(3-21) Solution Let the inclination of the bar after the application of the load be The extension in the length of steel wire ST will be

0.00293mmm 933.20419.0*70 2.4tan *70 tan70

strain in the wire, 000488.06

00293.0L

and stress in the wire


272 N/m 10*46.11

100010*4/

9000AP

297

N/m 10*235000488.0

10*46.11

Example(3-15) Fig.(3-22 a) shows a steel wire 1 mm diameter is stretched horizontally between two fixed points apart. A vertical load applied at the mid span of the wire causes a vertical displacement of 45 of the point of application of the load applied. What will be the stress induced in the wire and load applied? Neglect the weight of the wire. Take E for the wire material as 200 GN/m2 .

Fig(3-22) Solution i) Fig.(3-22 b) has been drawn by taking line AB (representing load P) of any length and then from the points A and B, the lines AC and BC have been drawn parallel to LN and MN to represent tension T produced in each half portion of the wire now form ABC,

tan2P

2sinPT ,sinT2P [since is very small,

tansin ]

P 11.11

1045.0*2

PT


262 N/m P 10*14.14

0.001*/4P 11.11 in wire stress

and change in length in each half portion of the wire

m001.01001.11 1045.0 22

and strain = 001.01001.0

L

But : N14.14P001.0

P*10*14.1410*200strainstress 6

9

Stress in the wire =14.14*106 *14.14=199.9 MN/m2 Example(3-16) Fig(3-23) shows a 700 mm length of aluminum alloy bar is suspended from the ceiling such away to provide a clearance of 0.3 mm between it and a 250 mm length of steel bar as shown in figure. Aal = 1250 mm2 Eal =70 GN/m2 ; As = 2500 mm2 , Es = 210 GN/m2 . Find the stress in the aluminum and in the steel due to a 300 kN load applied 500 mm from the ceiling. Solution On application of load of 300 kN at Q, the portion LQ will move forward and come in contact with N so that QM and NP will both be under compression. LQ will elongate while QM and NP will contract and the net elongation will be equal to gap of 0.3 mm between M and N. Let LQin stress tensile1 QMin stress ecompressiv2 NPin stress ecompressiv3

Elongation OF QM = m10*70

2.0*9

2 Fig (3-23)

LQ = m10*70

5.0*9

1

NP = m10*21025.0*

93

But force in QM =force in NP

210*2500*10*1250* 2

36

36

2


10*210*225.0* NP of nContractio 9

2

92

92

91

10*210*225.0*

10*702.0*

10*705.0* slongationNet

This must be equal to 0.0003 m

0003.010*210*225.0*

10*702.0*

10*705.0*

92

92

91

0003.0*10*210*225.02.13 9221

Tensile force in LQ + Compressive force in QM =300000 28

2126

16 N/m 10*4.2300000*10*1250*10*1250

solving (i) and (ii) we get 282

328

228

1 N/m 10*667.02

: N/m 10*335.1 : N/m 10*065.1

Example (3-17) Fig.(3-24) shows a steel bar of cross sectional area 250 mm2 held firmly by the end supports and loaded by an axial force of 25 kN. Find :

i) reactions at L and M ii) extension of the left portion, E=200 GN/m2

Fig(3-24) Solution

i) reaction at L and M , from Equilibrium condition ; RL

+RM =25 kN .......... ( i ) Also, since total length of the bar remains unchanged extension in LN = contraction in MN

MLLML R 4.2

25.06.0*RMR6.0*RM25.0*R

E*A6.0*R

E*A25.0*R

Substituting the value of RL in (i) we get 2.4 RM + RM= 25

From which RM =7.353 kN kN 647.17353.725RL


ii) Elongation of left portion.

= m 0000882.010*200*10*25025.0*10*647.17

E*A25.0*R

96

3L


Example (3-18) Fig.(3-25) shows a bar . Find the reaction produced by the lower support on the bar. Take E = 200GN/m2 . Find also the stresses in the bars.

Fig(3-25) Solution Let R1 = reaction at the upper support R2 = reaction at the lower support when the bar touches it. If the bar MN finally rests on the lower support , we have

R1+R2=55000 N For bar LM, the total force R1 = 55000 R2 (tensile) For bar MN , the total force = R2 ( compression)

1= extension of LM = m10*200*10*1102.1*R55000

962

2 =Contraction of MN = 962

10*200*10*2204.2*R

In order that N rests on the lower support, we have from compatibility equation m0012.01000/2.121

0012.010*200*10*220

4.2*R10*200*10*1102.1*R55000

962

962

or 2*(55000 R2)*1.2 2.4 R2=52800 N 16500R2

R1 = 55 16.5=38.5 N

Stress in LM = 266

1

1 kN/m 10*35.010*1105.38

AR


Stress in MN = 2 66

2

2 m/kN10*075.010*2205.16

AR

Example (3-19) Fig.(3-26) shows a flat steel plate of trapezoidal form of uniform thickness of 20mm tapers uniformly from a width 100mm to 200mm in a length of 800mm. If an axial tensile force of 100 kN is applied at each end, find the elongation of the plate. Take E = 205 GN/m2

Fig(3-26) Solution Consider a small section of length x at a distance x from the width b1

the width at the section, kxbxL

bbbb 112

1x

Where

Lbbk 12

Area = ( b1 + kx ) t

Now extension of a short length Etkxb

xPx1

L01

L

0 1kxbln

k1.

tEP

EtkxbxPby given isbar theof extension Total

1

2

1

1bbln

E t kP

bkLbln

ktEP


Putting

1

2

12

12bbnl

t EbbPLget we,

Lbbk

Substituting the numerical values we get

m0001352.0100200ln

10*205*1020*

10100200

10*800*10*1009

33

33

Example (3-20) Fig.(3-27) shows a bar LMNP fixed at L and P is subjected to axial force. Find the force in each portion of the bar and the displacement of points M and N. Take E = 200 GN/m2 .

Fig(3-27) Solution

P2 = 50 R1 similarly P2 = R2 100 Or

R1+R2=100+50=150 kN ..............(i) (the above equation can also be obtained by considering the static equilibrium of the bar) Now, extension of LM = compression of MN and NP


321

EALP

EALP

EALP

3

33

2

22

1

11

Substituting the value, we get simple stresses and strains

96

32

96

31

66

31

10*200*10*20001*10*R

10*200*10*150075.0*10*R50

10*200*10*10005.0*10*R

or R1*0.5=2

R5.1

75.0*R50 21

or 1.5R1= 75 1.5R1 + 1.5R2 or 3R1 1.5R2 = 75 or R1 0.5R2 = 25 kN ............ (ii) From (i) and (ii) we get

kN 66.67R and kN 33.835.1

125R 12

Hence P1 = R1 =66.67 kN (tensile) P2 =50 R1 =50 66.67 = 16.67 kN =16.67 kN (tensile)

Displacement of point M , 1:

m10*1666.010*200*10*1000

5.0*10*67.66EA

LP 396

3

1

111

Displacement of point , N : Displacement of point N= 21 where :-

m10*17.410*200*10*1000

75.0*10*67.16EA

LP 596

3

2

222

Displacement of point N=0.1666+0.0417=0.2083mm also,

m10*083.210*200*10*2000

1*10*33.83EA

LP 396

3

3

333

Example (3-21) Fig(3-28) shows three bars made of copper , zinc and aluminum and of equal length rigidly connected at their ends, they have cross-sectional areas of 250 mm2 , 375 mm2 and 500 mm2 respectively. If the compound member is subjected to a longitudinal pull of 125 kN , find the proportion of load carried by each rod and the induced stresses. Take. E cn =130 GN/m2 : Ezn=100 GN/m2 : Eal=80 GN/m2


Fig(3-28) Solution Loads carried by each bar, Pcu, Pzn, Pal : Considering equilibrium of the bar, we have Pcu + Pzn + Pal = P = 125 kN ...............(i) Since all the bars are rigidly connected at their ends , their deformation will be equal

alal

al

znzn

zn

cu

cuEALP

EALP

Ecu ALP

Pzn = Pcu * cucu6

6

zn

zn

cu

zn P1315P*

130100*

10*25010*375

EE*

AA

Pal = Pcu * cucu6

6

cu

al

cu

al P1316P

13080*

10*25010*500

EE*

AA

Substituting the value of Pzn and Pal in eq. (i) we get

Pcu + Nk 93.36P125P1316P

1315

cucucu

Pzn =42.61 kN Pal = 45.45 kN Stress induced in the bar alzncu ,,

MPa 72.14710*250

10*93.36AP

6

3

cu

cucu

MPa 63.11310*375

10*61.42AP

6

3

zn

znzn

MPa 9.9010*500

10*45.45AP

6

3

al

alal

Strength of Materials- Deformation Due to Axial Load- Hani Aziz Ameen

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Transcript of Strength of Materials- Deformation Due to Axial Load- Hani Aziz Ameen