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] www.mediafire.com/haniazizameen

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, deformation Strain original dimension 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) tan

for small

3-2 Stress

L it can be deduced that

L

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.


3the 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 Modul and 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

123-

e :The material must be homogenous in any cross sectianal area. The force applied is axial. 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)


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 ........ Possion' s ratio ........Lo ngitudinal strain ....... Lateral strain Now from Fig(3-5b) , x

1

,

x

x

1

4

y 4

1

thus ;

x

x

y

Similarly y

=

3

+

2

1 y

y

x

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 ; n

u

y

w

w

where : 1< n < 10

yield or ultimate stress allowable stress


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: P & A

L

(P/A) = E ( /L)

or 3-6-2

PL AE The Body with Variable Cross- Sectional Area

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

Fig(3-7)


then the total deformation of all parts is equal to algebraic summation of individual i.e. n T i

Pi L i 1 AiEi

P1L1 A1E1

P2 L 2 A2E2

P3L 3 A3E 3

P4 L 4 A4E 4

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

Fig(3-8) law , Pdx , A(x)E

(dx )

A(x)

4 L

T

0

d1

2y

Pdx A (x) E

2

Pdx

L 0

4

[d1

........................... (3-1) 2

2 y] E

for similarity of triangles y ( d 2 d1 ) / 2 ..................... (3-2) x L from eq.(3-1) and eq(3-2) , get T

4PL d1d 2 E

-3)


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,

A d

(x)

dx

x

x L C L

0

C .............. (3-4)

L C 0 , sub into eq.(3-4), yields x L (L x)

(x )

x

E

1 (L x ) E L

x

( x ) ]A

(L)

( x)

x

d

x C

(x )

at

( x)

dx

x

d

Adx [

dx L2 2E

L

x .dx 0

1 ( L x )dx E0


if A is constant , hence 1 AL L 2 EA

, W

AL , hence

WL 2EA 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.

Fx

0 ,

Fy

0 and

M 0

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 . Fx 0 , Fy 0 and M 0

2-

Find the geometric relation between bodies deformation 1 2 ( 3 , 4 ,........ etc) 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 Rx

0

MA 0 S* 2 P * 3 0 MB

0

S 120kN

R A * 2 P *1 0

RA

80 2

40kN

hence, the allowable stress will be 490 490 u 140MPa w n n 3 .5 and S w A S w

4

d 2r

dr

dr

4S

4 * 120 * 10 * 140

w

3

33mm

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

dp

315 90MPa n 3.5 R A /(2 * ( / 4)d p 2 ) u

2R A w

2 * 40 * 10 * 90

3

17mm


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 ) 0.0304mm , yield load = 161 kN , Maximum load = 242 kN Length of specimen at fracture = 249mm Find : 1) 2) 3) 4)

Yield point stress Ultimate stress Percentage elongation Solution

1)

nd P A &

L

40

& , hence , 3.18 * 104 kN / m 2

(0.04) 2

4 0.0304 0.000152 200

then, 3.18 * 104 E E 2.09 * 108 kN / m 2 0.000152 yield point load 2) Yield point stress = area 161 12.8 * 104 kN/m 2 = * (0.04) 2 4 max imum load 242 3) Ultimate stress = 19.2 * 104 kN / m 2 area * (0.04) 2 4 4) Percentage elongation = lenght of specimen at fracture original length 249 200 0.245 24.5% original length 200 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 :


P A

100 250000 kN/m 2 0.0004 250000

E thus ,

, where 2.14 * 108 250000 L 2.14 * 108

L

hence, 250000 * 0.05 0.0000584m 2.14 * 108

123-

1)

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 magnitude of the load , longitudinal strain produced , and total decrease in length . Solution Inner diameter of the cylinder , d = D 2t = 0.3 2 * 0.05 = 0.2 m Magnitude of the load P : P using the relation , A or

P

* A 75000 * ( D 2 4 = 75000 *

2)

0.2 2 ) 2945.2kN

Longitudinal strain produced , using the relation =

3)

4

(0.32

d2 )

E

= 75000/(1.5* 108 )= 0.0005

Total decrease in length ( ) Strain = 0.0005=

Change in length Original length

4 Hence decrease in length = 2 mm

L

0.002 m 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=

402 =1256.6 mm2

4

A2=

152 =176.7 mm2

4

302 = 706.8 mm2

A3=

4 Length : L1=200 mm=0.2 m ; L2=500 mm=0.5 m & L3= 200 mm=0.2 m Let 1, 2 and respectively . Then

1

3 be

the extensions for the parts 1,2 and 3 of the steel bar

PL1 , A1 E

2

PL 2 , A 2E

3

PL3 where P = 15 kN A 3E

Total extension of the Bar , 1

2

PL1 A1 E

3

PL 2 A 2E

15 103 200 109

PL 3 A3E

0.2 0.001256

P L1 E A1

L2 A2

0.5 0.0001767

L3 A3 0.2 = 0.0002454 m 0.0007068


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 207GPa 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 free body diagrams of Fig(3-13 b) Hence, the total deformation is PL AE

1 P1L1 E A1

P2 L 2 A2

1 88.6 * 103 * 2.13 207 * 109 806.5 * 10 6

2.1 mm

P3L 3 A3 35.6 * 103 * 1.52 322.6 * 10 6

8.9 * 103 * 0.61 161.3 * 10 6


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) i)

Solution The diameter of the middle portion , d : P 50 * 1000 Now , stress in the middle portion , 130 * 106 2 A /4 d 1/ 2

ii)

50 * 1000 d 0.221 m ( / 4) * 130 * 106 Length of the middle portion : Let the length of the middle portion = x meter 0.25 x Stress in the end portion * E also, elongation of the end portions + extension of the middle portion = 0.15*10-3 39.79 *106 * 0.25 x 9

130 *106 * x 9

0.15 *10

3

200 *10 200 *10 6 39.79 *10 * 0.25 x 130 *106 x 200 *109 * 0.15 *10 Dividing both sides by 39.79* 106 , we get 0.25 - x 3.267x 0.754 x 0.222m

3


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,

P A

100 *1000 ( / 4) * 0.05

2

50.92 *106 N / m2

L 50.92 * 106 * 2.5 Elongation of the solid rod, 0.000636 m E 200 * 109 Elongation after the rod is bored =1.15*0.636 = 0.731 mm Area at the reduced section = ( / 4) 0.052 0.0252 0.001472 m 2 100 * 1000 67.93 * 106 N / m 2 Stress in the reduced section, 0.001472 2.5 x x Elongation of the rod 0.731* 10 3 E E 6 50.92 * 10 2.5 x 67.9 * 106 x 0.731* 10 3 8 9 200 * 10 200x10 6 9 50.92 *10 2.5 x 67.9 * 10 x 200 *109 * 0.731*10 3 2.5 x 1.33x 2.87 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 : 1 , 2 and 3 be the change in length LM ,MN and NP respectively P1L1 Then 1 .......... ....increase AE Let

2

P2 L 2 .......... ....decrease AE

P3 L 3 .......... ....decrease AE Net change in length 3

1

P1L1 AE

2

P2 L 2 AE 103

3

P3L 3 AE

1000 * 10 6 * 100 * 109 0.00012 m

1 P1L1 AE

P2 L 2

P3L 3

50 * 0.6 30 * 1 10 * 1.2

1 105

30 30 12


i) ii)

Example (3-10) Fig.( 3-17) shows a member LMNP is subjected to point load Find: Force P necessary for equilibrium 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 then

1, 2 1

2

3

& 3 be the change in lengths LM,MN and NP respectively P1L1 50 1000 1 3.97 10- 4 m ...... increase (+) 6 9 A1E 600 10 210 10 P2 L 2 300 1000 1 = 5.95 10-4 m ..... decrease (-) 6 9 A 2 E 2400 10 210 10 P3L 3 200 1000 0.6 4.76 10- 4 m........increase (+) 6 9 A 3 E 1200 10 210 10 -4 5.95*10-4 +4.76*10-4 = 2.78*10-4 m 1 2 3 = 3.97*10


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 40 * 1000 * 0.914 1 10 3 * 101.6 * 6.35 * 10 3 * 207 * 109 1 0.27mm 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 Pdy 2 A (y) E 0 from Fig (3-18 b) A(y)= (0.0508+2x)*0.00635 from the similar triangles of Fig(3-18b) 25.4 * 10 3 x x 0.208y 1.22 y


from above A(y) = (0.0508+2*(0.208y))*0.00635 1.22 40 * 1000 (0.0003225 0.00264y)dy hence , 2 207 * 109 0 2 0.507mm the total elongation of the plate = 1 2 = 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 = elongation of the bar due to its own weight WL .......................... (i) 1 = 2AE W = AL =1290*10 4*L*76000 = 9804L The length L is assumed in meter , substituting into eq .(i) for W 1

Fig(3-19) 2

9804 * L

.................................. (ii) 2 * 1290 * 10 6 * 207 * 109 The central force causes an elongation of 2 in the part (L 14) of the bar . Hence ........................... (iii) 2 =P(L 14)/AE The total elongation of the bar is the sum of eq . (ii) & (iii) = 1 + 2 = 1.6*10 3 1

9804 * L2 6

89 *1000 * (L 14) 9

6

9

1.6 *10

3

2 *1290 *10 * 207 *10 1290 *10 * 207 * 10 L = 18.6 mm Using eq.(ii) , the elongation of the bar under its own weight is :


9804 * (18.6) 2 2

6

7

0.0635*10 3 m

2 *1290 *10 * 207 *10 The max . normal stress occur at the fixed end of the bar W P A 9804 * 18.6 89 * 103 70.4MPa 1290 * 10 6 Example (3-13) (A) If the strain in the y-direction due to the simultaneous action of the uniform forces P x, P y, & P z 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 P z ? E= 207 GPa , = 0.25 (B) What single uniform force must be applied to the block in the Zdirection only in order to produce the strain of 1.75*10 4 m/m the ydirection ?

Solution

( y x The stress acting on the block is y

z)/ E

Fig(3-20) x

Px A

3

36.6 * 10

25.4 * 50.8 * 10 213.5 * 103

y

50.8 * 76.2 * 10

6

6

27.58MPa

55.16MPa

Pz Pz ( assumed as positive) 25.4 * 76.2 1935.5 * 10 6 Substituting the results in the expression for y gives z

1.75 * 10 4 (55.16 * 106 Pz =200 kN (tensile)

(3 / 4) * 27.58 0.25 * ( Pz / 1935.5 * 10 6 )) /(207 * 109 )


z

We have axial strain in the direction of Pz is the direction of y is or

y

1 z * 4 207 * 109

from which or

z

y

* axial strain , 1.75 * 10

E

and the lateral strain in

y

z

E

4

144.8MPa (comp.)

Pz 25.4 * 76.2

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 70 tan 70 * tan 2.4 70 * 0.0419 2.933 mm 0.00293m 0.00293 strain in the wire, 0.000488 L 6 and stress in the wire


P A

9000 /4*

10 1000

2

11.46 * 107 N/m 2

11.46 * 107 0.000488

235 * 109 N/m 2

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, P P P 2T sin , T [since is very small, 2sin 2 tan sin tan ] P T 11.11 P 0.045 2* 1


stress in wire and change in length

11.11P 2

14.14 *106 P N/m2

/4 * 0.001 in each half portion of the wire

0.045 2 12

1 1.001 1 0.001m 0.001 0.001 and strain = L 1 stress 14.14 * 106 * P 9 But : 200 * 10 P 14.14N strain 0.001 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 1 tensile stress in LQ 2 compressive stress in QM 3 compressiv e stress in NP * 0.2 m Elongation OF QM = 2 70 * 109 * 0.5 LQ = 1 m 70 * 109 * 0.25 NP = 3 m 210 * 109 But force in QM =force in NP 2

* 1250 * 10

6 3

* 2500 * 10

6 3

2

2

Fig (3-23)


2

Contractio n of NP

* 0.25

2 * 210 * 109 1 * 0.5 2 * 0.2 Net slongation 9 70 * 10 70 * 109 This must be equal to 0.0003 m 1 * 0.5 2 * 0.2 2 * 0.25

* 0.25 2 * 210 * 109 2

0.0003 70 * 109 70 * 109 2 * 210 * 109 3 1 1.2 2 0.25 2 2 * 210 * 109 * 0.0003 Tensile force in LQ + Compressive force in QM =300000 1250 * 10 6 * 1 1250 * 10 6 * 2 300000 2.4 * 108 N/m 2 1 2 solving (i) and (ii) we get 1

i) ii)

1.065 * 108 N/m 2 :

2

1.335 * 108 N/m 2 :

3

2

2

0.667 * 108 N/m 2

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

Fig(3-24) i)

Solution 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 R L * 0.25 R M * 0.6 R L * 0.25 RM * 0.6 R L A*E A*E Substituting the value of RL in (i) we get 2.4 RM + RM= 25 From which RM =7.353 kN R L 25 7.353 17.647 kN

RM * 0.6 0.25

2.4 R M


ii)

Elongation of left portion. R L * 0.25 17.647 *103 * 0.25 = A*E 250 *106 * 200 *109

0.0000882 m


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) 55000 R 2 * 1.2 m 1 = extension of LM = 110 * 10 6 * 200 * 109 R 2 * 2.4 2 =Contraction of MN = 220 * 10 6 * 200 * 109 In order that N rests on the lower support, we have from compatibility equation 1 2 1.2 /1000 0.0012m 55000 R 2 *1.2 R 2 * 2.4 0.0012 110 *10 6 * 200 *109 220 *10 6 * 200 *109 or 2*(55000 R2)*1.2 2.4 R2=52800 R 2 16500 N R1 = 55 16.5=38.5 N Stress in LM =

R1 A1

38.5 110 * 10

6

0.35 * 106 kN/m 2


Stress in MN =

R2 A2

16.5 220 * 10

6

0.075 * 106 kN / m 2

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 b 2 b1 the width at the section, b x b1 x b1 kx L Where k

b2

b1

L Area = ( b1 + kx ) t Now extension of a short length x

P x b1 kx Et

Total extension of the bar is given by P b kL ln 1 ktE b1

P b ln 2 k t E b1

L 0

P x b1 kx Et

P 1 . ln b1 tE k

kx

L 0


Putting k

b2

b1

, we get

PL b ln 2 b1 E t b1

L b2 Substituting the numerical values we get 100 * 103 * 800 * 10 3 200 ln 0.0001352m 200 100 20 100 9 * 3 * 205 * 10 103 10 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


1

2

3

P1L1 P2 L 2 P3 L 3 A1E A 2 E A 3 E Substituting the value, we get simple stresses and strains R1 *103 * 0.5 50 R1 *103 * 0.75

* 200 *106 1500 *10 6 * 200 *109 2000 *10 50 R 1 * 0.75 R 2 or R1*0.5= 1.5 2 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 125 R2 83.33 kN and R 1 66.67 kN 1.5 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 : 1000 * 10

6

R 2 *103 *1 6

* 200 *109

P1L1 66.67 * 103 * 0.5 0.1666 * 10 3 m 1 6 9 A1E 1000 * 10 * 200 * 10 Displacement of point , N : Displacement of point N= 1 2 where :P2 L 2 16.67 * 103 * 0.75 4.17 * 10 5 m 2 6 9 A 2 E 1000 * 10 * 200 * 10 Displacement of point N=0.1666+0.0417=0.2083mm also, P3 L3 83.33 * 103 * 1 2.083 * 10 3 m 3 6 9 A 3 E 2000 * 10 * 200 * 10

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 Pcu L Pal L Pzn L A cu Ecu A zn E zn A al E al 375 *10 6 100 15 * * Pcu Pcu 6 130 13 250 * 10 A al E al 500 * 10 6 80 16 Pal = Pcu * * * Pcu Pcu 6 A cu E cu 250 * 10 130 13 Substituting the value of Pzn and Pal in eq. (i) we get 15 16 Pcu 125 Pcu 36.93 kN Pcu + Pcu 13 13 Pzn =42.61 kN Pal = 45.45 kN Stress induced in the bar cu , zn , al A E Pzn = Pcu * zn * zn A cu E zn

36.93 * 103

cu

Pcu A cu

42.61* 103

zn

Pzn A zn

al

Pal A al

6

250 * 10

6

375 * 10

45.45 * 103 500 * 10

6

147.72 MPa

113.63 MPa

90.9 MPa

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

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