strength of materials- deformation due to axial load- hani aziz ameen
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strength of materials, deformation due to axial load, Hani Aziz AmeenTRANSCRIPT
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,
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)
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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.
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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 .
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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)
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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 :
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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 :
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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)
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
ii) Elongation of left portion.
= m 0000882.010*200*10*25025.0*10*647.17
E*A25.0*R
96
3L
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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
Strength of Materials- Handout No.3- Deformation due to Axial Load- Dr. Hani Aziz Ameen
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