Strength of Materials- Thermal Stresses- Hani Aziz Ameen

Strength of Materials Handout No.4

Thermal StressesAsst. 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.4- Thermal Stresses - Dr. Hani Aziz Ameen

4-1 Introduction Consider a bar of length L , in a material with a coefficient of

linear expansion . When the temperature of the bar is raised by T the bar extends by an amount ( )T L , if this extension is prevented by constraining the ends of the bar, a compressive internal stress is induced within the bar which is called thermal stresses. 4-2 Derivation Formula for Thermal Stress

It is well known, that the body is extended or contracted due to the effect of temp. change . The elongation or contraction due to temp. can be calculated by the formula : T.L.temp where : temp = extension or contraction due to temp. = thermal coefficient of linear expansion L = body length T= change in temp. If this deformation occurs freely , there is no external load acting on the body , but if there is a certain resistance to the expansion or contraction , the external load will show so,

-a- -b- There is no thermal stress There is thermal stress but thermal strain but no thermal strain Fig(4-1) To solve the problems of this type ( i.e. to find the thermal stress) , we must follow these steps :

1- Let the body expand or contract freely to calculate the deformation due to temp. difference , by removing all or part of constraints . In this case remove wall (B) in Fig(4-1b) we have :


2- Imagine there is a sufficient external force to restore the body to its

initial position , give the relation between geometric deformation due to temp. and external force Ptemp

AEPLT L ,

AP

Thus ,

E

T temp

T Etemp -1) In a similar manner , constraining the free expansine of a compound bar assembly will result in thermally induced stress and strain . The following cases arise 4-2-1- Bars between Fixed Ends Fig.(4-2) shows two bars of different material 1 and 2 connected end to end between fixed platens

Fig(4-2)


With a temperature rise of T , the free displacements in each bar sum to give a total unconstrained displacement of : )LL(T 221121 -2) when thin displacement is prevented , the force inducd in each bar is the same : 2211 AA -3) The compressive strecses 1 and 2 must depend upon eq.(4-2) in the following form to ensure zero axial displacement:

)44.........(....................0)LL(TE

LEL

22112

22

1

11

Combining eqs. (4-3) and (4-4) , gives

22

21

EALA

11

22111

)E/L()LL(T

4-2-2 Tubes or composite Bar with Constraive Ends

Fig(4-3) shows respectively concentric tubes and balanced strips of different materials A and B held between rigid platens .

Fig(4-3) Assuming ,BA the free expansion caused by T in These assemblies is shown in Fig.(4-3c) . When the ends are constrained the displacements in A and B equalise to a value intermediate to their free expansions ).( AB Consequently, A will be put in tension and B in compression . The stresses so induced in each material must satisfy two conditions

(i) Equilibrium the net load is zero , which gives : )5.4.......(........................................0AA BBAA

(ii) Compatibility the strains are equal , giving :


B

BB

A

AA E

LTLE

LTL -6)

Then from eqs.(4-5) & (4-6) : )EAEA/()EEA)((T ABBBBABABA -7) When there is a difference in the initial lengths of A and B then L in eq(4-6) cannot be cancelled. 4-2-3 Combined Mechanical and Thermel Effects The solutions to problems of this nature are found from superimposing the separate effects of mechanical and thermel straining . 4-3 Examples

The following examples explain the different concepts of the thermal stress problems .

Example (4-1) Fig (4-4) shows a compound bar made up by connecting a steel member and a copper member rigidly fixed at their ends. The cross sectional area of the copper member is A m2 while that of steel member is 2A m2 for the other half of the length. The coefficient of expansion of steel and copper are and 1.3 respectively while the elastic moduli are E and 0.5E respectively . Find the stress induced in the members due to rise of T degrees in temperature .

Fig(4-4)


Solution

Since the coefficient of copper is more than that of steel , the copper member shall expand more than the steel member and as such the copper member shall be subjected to compressive force and the steel member to tensile forces when the temperature rises . The compound bar shall stabilize at an intermediate expansion between that of copper and steel , due to restraint at the ends , Compressive stress set up in copper bar = c Tensile stress set up in steel bar of cross-sectional area A= s Tensile stress set up in steel bar of cross-sectional area 2A= 's Also , total tension (pull) in steel bar = compression (push) in copper bar

ssc

css

2A*A2*A*

If La is the actual length of the compound bar due to rise in temperature , then for copper rod :

L*E

L)T1(Lc

cac

For steel rod :

2L*

E'

2L*

E)T1(LL

s

s

s

ssa

Adding eq(i) and eq(ii) we get

s

s

s

s

c

csc E2E2E

T)(

(1.3E4

11E4EE5.0

T) cccc

TE0109.0sc

TE0545.02

' ss

Example (4-2) Fig(4-5) shows two steel rods , one of 75mm diameter and the other of 55mm diameter are joined end to end by means of a turn buckle the other end of each rod is rigidly fixed and there is initially a small tension in the rods . If the effective length of each rod is 4.5 m . Find the increase in this


tension when the turn buckle is turned by one quarter of a turn . On one end of the bigger diameter rod there are 0.15 threads per mm length, while there are 0.2 threads per mm length on the other rod. Neglect the extension of the turn buckle . Find also what rise in temperature would nullify the increase in tension Take E=200 GN/m2 and = 12*10 6 per °C

Fig(4-5) Solution

22

1 m002376.01000

554

A

22

2 m004148.01000

754

A

When the turn buckle is turned by one quarter of a turn,

extension o1000*2.01 =0.00125 m

1000*15.01 =0.001666 m

Total extension of the two rods = 0.00125+0.001666 = 0.002916 m

Total extension in the two bars

004418.01

002376.01

10*200P5.4

A1

A1

10*2005.4*P

EAPL

EAPL

921

92

2

1

1

= 0.002916 m Solving we get P = 200241 N In order to nullify this tension by rise of temperature, the total expansion of the two rods must be equal to 0.002916 m Let the rise of temperature (T2 T1) be T °C

002916.05.4*T*10*12 002916.0L T 6

545.4*10*12

002916.0T 6 C °


Example(4-3) A steel rod with 40 mm in diameter is fixed concentrically in a brass tube having outside and inside diameters of 60 mm and 50 mm respectivly. Both the rod and the tube are 0.5 m long and their ends are level. The compound rod is held between two stops which are exactly 0.5 m apart and the temperature of the bar is then raised by 50 °C

i) Find the stresses in the rod and the tube if the distance between the stops (a) remains constant (b) is increased by 0.2 mm

ii) Find the increase in the distance between the stops if the force exerted between them is 60 kN , take Es = 200 GN/m2 , Eb = 90 GN/m2 , s = 1.2*10 5 per °C = 1.2*10 5 per C° Solution i)(a) When the distance between the stops remains constant stress in steel, s= s.T.Es = 1.2*10 5*50*200*109= 120 MN/m2 stress in brass, b= b t Eb = 2.1*10 5*200*109= 94.5 MPa (b) When the distance between the stops is increased by 0.3 mm

strain in steel , 5.0

)1000/2.0(L Tlength originalprevented ansionexp s

s

= 0002.05.0

0002.05.0*50*10*2.1 5

stress in steel . 4010*200*0002.0E 9sss MN/m2

strain in brass . 5.0

0002.0TLlength originalprevented ansionexp b

b

= 00065.05.0

0002.05.0*50*10*1.2 5

stress in brass , 69

bbb 10*5.5810*90*00065.0E N/m2 (compressive)


ii) When the force exerted between the stops is 60kN. let meter be the expansion of the composite member

strain in steel ,5.0

5.0*50*10*2.1LL T 5

ss

20006.0 stress in steel , 9

sss 10*200*)20006.0(E

similarly strain in brass,5.0

5.0*50*10*1.2L

TL 5b

b

)200105.0( stress in brass, 9

bbb 10*90*)200105.0(E

Area of steel rod , 262

s m10*6.12561000

40*4/A

Area of brass tube ,

622

b 10*9.8631000

501000

60*4/A

Load on steel +load on brass =total load between the stops PAA bbss

or 6969 10*9.863*10*90*200105.010*6.1256*10*200*20006.0

= 60000 or 6000010*7775.0*200105.010*513.2*20006.0 88 or 6000010*555.1026.510*164.810*078.15 844 or 484 10*66000010*581.610*242.23

mm262.0mm10*10*581.6

10*6242.23 38

4


Example(4-4) Fig(4-6) shows a copper bar 50mm in diameter is placed within a steel tube 75mm external diameter and 50 mm internal diameter of exactly the same length ,the two pieces are rigidly fixed together by two pins 18mm in diameter . one at each end passing through the bar and tube , find the stress induced in the copper bar and the steel tube and pins if the temperature of the combination is raised by 50°C take

Fig(4-6) Solution Area of copper bar , 2

c )1000/50(*4/A =1963.49*10 6 m2

Area of steel tube , 2622

m10*37.24541000

501000

754/As

Difference in free expansion L*50*10*5.1110*17TL 66

sc =275*10-6L A Compressive Force P exerted by the steel tube on the copper rod opposes the extra expansion of the copper rod , and in return , the copper rod exerts an equal tensile force P to pull the steel tube so that the consequent combined effect of reduction in the length of copper rod and the increase in length of steel tube equalize the difference in free expansions of the two reductions in the length

of copper red due to force P 96cc 10*105*10*49.1963

PLEA

PL

increase in length of steel tube due to force P

96ss 10*210*10*37.2454

PLEA

PL

but , 9696 10*210*10*37.2454PL

10*105*10*49.1963PL

= 27.5*10 6 L 4.85*10 9 P +1.94*10 9 P=275*10 6 P=40500 N

Cper10*17,Cper10*5.11,m/GN105E,m/GN210E 6c

6s

2c

2s


stress in copper rod, MPa 36.2010*10*49.1963

40500AP 6

6c

c

stress in steel tube, Mpa 5.1610*10*37.2454

40500AP 6

6s

s

shears stress in pins 57.7910*

100018*

4*2

40500A*2P 6

2pin

pin MPa

Example(4-5) Fig(4-7) shows a steel rod of 10 mm diameter passes centrally through a copper tube of external diameter 40 mm and internal diameter 30 mm and steel length 2m. The tube is closed at each end by 20 mm thick steel plates which are screwed by the nuts. The nuts are tightened until the copper tube is reduced to length to 1.9996m .Find the stresses in the rod and the tube. If the whole assembly is heated through 60°C what are then the stresses in the rod and tube , assuming that the thickness of the plates remains unchanged ? Take Es=210 GN/m2 , Ec=100 GN/m2 C per 10*12 6

s and C per 10*5.17 6c

Fig(4-7) Solution

Area of steel rod , 262

s m10*54.78100010

4A

Area of copper tube, Ac = 2622

m10*78.5491000

301000

404

i) Stress due to tightening of the nuts : When the nuts are tightened the steel rod will be subjected to tensile stress and the copper tube will be subjected to compressive stress . Let

s = tensile stress in steel rod, c= compressive stress in copper tube,


As there is no external load , push on the copper tube = pull in the steel rod

7or 01.0*4

*03.004.04

* cs2

s22

c

Strain in the copper tube, 0002.029996.12

c

Stress in the copper tube MPa 2010*10*100*0002.0E* 69

cccc and stress in the steel rod

MPa 14020*7 7 sssc ii) Stresses due to rise of temperature :

If the two members were free to expand, free expansion of steel = ss L T Free expansion of copper = cc L T

Since sc an greater th is , the free expansion of copper is greater than the free expansion of steel , but since the ends are provided with washers and nuts, the members are not free to expand fully. Final expansion of each of the members will be the same. Let this final expansion be , the free expansion of copper is greater than , while the free expansion of steel is less than . Hence the steel rod will be subjected to a tensile stress while the copper tube will be subjected to a compressive stress. Let

cs and be the stresses in the steel and copper respectively equilibrium of the whole system .

Total tension (pull) in steel = total compression (push) in copper.

cs

ccsccss 7

AAor AA

Final expansion of steel = final expansion of copper

cc

cccs

s

sss L

EL T L

EL T

but T = 60 °C Ls = 2+2*0.02 = 2.04 m & Lc = 2m

12*10 6 *60*2.04+ 2*10*100

2*60*10*5.1704.2*10*210

79

c69

c

12*10 6*60*2.04*210*109+7 c *2.04 = 17.5*10 6*60*2*210*109 2.1*2 c c= 0.0717*108 N/m2 s = 7 c = 50.19 MPa Stress in Copper = cc = 20 + 7.17 = 27.17 MPa Stress in steel = ss = 140 + 50.19 =190.19 MPa


Example (4-6) Fig (4-8) shows a brass rod of 6mm diameter and 1 m long joined at one end to a rod of steel 6 mm diameter and 1.3m long . The compound rod is placed in a vertical position with the steel rod at the top and connected top and bottom are connected to rigid fixing in such a way that it is carrying a tensile load of 3.5 kN . An attachment is fixed at the junction of the two rods and to this a vertical axial load of 1.3 kN is applied downwards . Find the stresses in steel and brass . The temperature is then raised through 30°C . What are the final stresses in steel and brass ? Take Es =200 GPa , Eb = 85 GPa , Cper 10*19 , Cper 10*12 -6

b6

s

Fig(4-8) Solution

262

bs m10*27.281000

6*4

AA

In the first case , when the composite bar LMN is held in fixtures both steel and brass portions of the rod carry the tensile load of 3.5 kN . Thus the initial tensile stresses in each of the metals will be equal , as the areas of their cross-sections are equal .

MPa8.12310*10*27.28

1000*5.3 661b1s

When the load of 1.3 kN is applied at junction M, it will induce tensile stress in steel and compressive stress in brass . The magnitudes of the loads Ps and Pb due to the action of 1.3 kN load M , will be such that the stretch in steel will be equal in magnitude to the compression in brass


Ps + Pb=1.3*1000 = 1300 N

)A(A P LL

EEP

EALP

EALP

bsbs

b

b

ss

bb

bb

ss

ssbs

= bb9

9P81.1P*

3.11*

10*8510*200

1.81 Pb+ Pb=1300 Pb= 462.6 N and Ps =1300 462.6 = 837.4 N Thus the corresponding additional stresses in steel and brass , due to the application of 1300 N load are

66

s

s2s 10*

10*27.284.837

AP

= 29.62 MPa (tensile)

(comp.) MPa 36.1610*10*27.286.462

AP 6

6b

b2b

The resulting stress )tensile(MPa44.1036.168.1232b1bbr )tensile(MPa42.15362.298.1232s1sst In the absence of fixtures the rod LN would have expanded freely due to temperature LN , but the pressure of fixture effectively compresses the rod LN by an amount LL which is equal to ( TLTL bbss ) thus if is the compressive stress in steel and brass , then we have

TLTLLE

LE bbssb

bs

s

99 10*851

10*2003.1 30 ( 12* 10 6*1.3+19*10 6*1)

e)compressiv ( MPa 10*0117.00065.0

10*10*038.1

10*038.10117.00065.010

696

69

Hence the final stresses in steel and brass are :

(tensile) MPa41.5003.5744.107)tensile( MPa39.9603.5742.153

bbf

ssf


Example(4-7) Fig(4-9) shows a circular section tapered bar is rigidly fixed at both ends . If the temperature is raised by 40°C . Find the stress in the bar , take E=200 GN/m2 , 610*12 per °C

Fig(4-9) Solution When the temperature is raised by T ° C a compressive force P is induced since this force is the same for all cross sections , a maximum stress will be at section MM . If the bar were free to expand its expansion would be :

cm 000576.040*2.1*10*12)TT(L 612

The force induced will be a force P which is required to prevent a free expansion of 0.000576 m For an element dx the expansion is

EA

Pdx

x

Total extension L

0 x EAPdx

But 22

121x kxd

4x

Lddd

4A

where k = ( d2 d1)/L

)ii.......(..........dEd

PL4d1

ddd1

)dE(d4PL

kEP4

)kxd(dx

EP4

21112112

1

0

L

02

1 kxd1

1


From eq(i) and eq(ii) we get

21

12 dEdPL4)TT(L

From which , 4

dd)TT(EP 2112

Maximum stress induced is given by d4

*dd)TT(E4

21

2112

1

212 d

d)TT(E

MPa 1441.0

15.0*40*10*12*10*200 69

Example(4-8) Fig(4-10) shows a steel cylinder is placed centrally on a bronze cylinder and their ends are secured between two rigid supports. If the stress in the cylinder is zero at a temp. of 30 °C find the stress in each cylinder when the temperature rises to 60 °C . or the steel E= 207 GPa , 610*5.6 m/m/°C L=203.2mm , A = 968 mm2 For bronze , E = 83 MPa 610*5.10 m/m/°C L= 152.4 mm , A = 1290mm2

Fig(4-10) Solution

Total expansion of the cylinder = tstb ................ (i) due to temperature change Total contraction in the cylinder = sb ....................... (ii)


due to compressive force P The expansions in eq(i) & eq(ii) are numerically equal , Hence ,

sbtstb

steelbronzesteelbronze )EAPL)

EAPL)TL)TL

sbsobo )AEPL)

AEPL)TT(L)TT(L

10.5*10 6*152.4*10 3 *(60-30)+6.5*10 6*203.2*10 3 *(60 30)

= 66

3s

66

3b

10*207*10*96810*2.203*P

10*83*10*129010*4.152*P

Putting Pb = Ps = P , gives P = 96 kN

Stress in the steel cylinder , (comp.) MPa6.9810*968

960006s

Stress in the bronze cylinder , (comp.) MPa1.7410*1290

960006b

Example(4-9) Fig(4-11) shows the assembly is symmetrical about the y-axis and in which two aluminum bars and one brass bar support a load of 249 kN . Find the following :

(a) The normal stress in each bar (b) The change in temp. that ought to occur in order that each

aluminum bar to carries (1/4) of applied load For each aluminum bar E= 69 GPa , 610*8.12 m/m/°C , L=254 mm , A=968 mm2 For the brass bar ,

610*2.10 m/m/°C , E=103.5 GPa, L=203 mm , A=1613mm2

Fig(4-11)


Solution

(a) Due to load , the contraction is the same amount = ba

or 96

3b

96

3a

10*5.103*10*161310*203*P

10*69*10*96810*254*P

from which Pb=25Pa also , it can be deduced that 2Pa+Pb solving eq.(i) & (ii) gives Pa = 48.6 kN Pb= 152.12 kN

Stress in the aluminum bar , .)comp(MPa33.5010*968

46006a

Stress in the brass , .)comp(MPa5.9410*1613

1521206b

(b) From fig-(4-11b), the following relation is established

Fig(4-11 b) tbbtaa

or

T*10*203*10*2.1010*5.103*10*1613

10*203*PT*10*254*10*8.12

10*69*10*96810*254*P 36

96

3b36

96

3a

from static 2 Pa + Pb = 249000 since it is required that each aluminum bar is carry 0.25 of the applied load , then , Pa = 0.25 * 249000 = 62.3 kN

Pb = 249000 2 * 623000 = 124.4 kN Sub. into eq. (iii) yields T = 22.5 °C


Example(4-10) Fig(4-12) shows a brass ring having an inside diameter of 0.914m and a thickness of 9.525mm thickness if the length of each ring of 50.8 mm and the temperature at the time of assembly of both rings 149 ° C , find the pressure between the rings when they cool to 50 °C , Assume that the pressure between the rings is Zero at 150 °C for the brass ring

=10.2*10 6 m/m/°C , E=103.5 GPa , for steel ring, =6.5*10-6 m/m/C , E=207 GPa

Fig(4-12) Solution

The deformation ,due temp .and bearing pressure for both rings is btbsts

Or

93ss

s6

10*207*0508.0*10*525.9L*F50150L*10*5.6

= 10.2*10-6*Lb(150 50)+ 93bb

10*5.103*0508.0*10*7.12L̀F


Putting Ls = Lb = r =457.2mm & simplified , gives

Fs+3Fb=148118.4 ............(i) 2Fs = total pressure force

2Fs = Pressure * projected area projected area = diameter * length of the ring in the Z-direction =914.4*50.8=46451.52 mm2 , hence 2Fs = 46.451*10 3 P P10*22.23F 3

s similarly 2Fb=FP Fb = 23.22*10 3 P Sub into (i) 4.148116P10*22.23*3P10*22.323 33 P =1.6 MPa Stress in the steel ring is

(comp) MPa3.7650508.0*10*525.9

FareaF

3ss

s

Stress in the brass ring is

tensileMPa366.570508.0*10*7.12P10*22.23

3

3

b


4.4 Problems

4-1) A 12mm diameter steel rod passes centrally through a copper tube

48mm external and 36mm internal diameter and 2.5m long. The tube is closed at each end by 25mm thick steed plates which are secured by nuts , the nuts are tightened until the copper tube is reduced in length to 2.4995m and the whole assembly is then raised in temp . by 60°C. Find the stresses in the copper and steel before and after the rise in temp . Assuming the thickness of the plate remains unchanged Esteel =200 GN/m2 ,Ecopper =100GN/m2 and

coef .of expansion per °C for steel is 12*10-6 for copper 17.5*10-6

4-2) Fig.(4-13) shows a copper bar and steel bar each with the

same length and diameter are assembled . Find the stress in each material following temperature rise of 10°C . For steel C/10*12.5, GPa 207E -6 For copper E=105 GPa , =18*10-6/°C

Fig(4-13) 4-3) Fig.(4-14) shows assembly of three materials I , II , III with the given data a) E1 = E2 = E3 = E = 19.6*1010 N/m2 a = 0.4m , A1=1200 mm2 b = 1.2m , A2 = 1400 mm2 c = 0.4m , A3 = 16cm2 1 = 45° 2 = 60° 3 = 30° and MPa 157 Find P & Pmax b) 1= 2 = 3 = = 12.5*10-6 and 40T Find IIIIII &, c) mm2.12 which is the amount by which bars II are manufactured shorter than the design length


Fig(4-14) 4- 4) Fig(4-15) shows the structure of copper & steel with Esteel=2*1010N/m2 710*125 Ecopper =9*1010 N/m2 7

copper 10*165 ,and the change of temperature is 10T Find the thermal stresses

Fig(4-15) 4-5) Fig(4-16) shows the assembly of the three materials, find the stresses due to the action of the forces and those due to the change in temp . take E= 19.6*1010 N/m2 , 610*12

Fig(4-16)


4-6) A mass of 150 kg is suspended by three vertical wires ,the two outer wires are of steel and the middle one of aluminum each of area 8mm2 , the lengths is adjusted so that each wire carries an equal share of the load .If the temperature is raised by 50 °C , find the stresses in the wires.

Find also what rise of temperature would just cause the aluminum wire to become slack , E Steel = 210 GN/m2 , E Aluminum=70 GN/m2

6Al

6steel 10*24 & 10*12

4-7) A copper tube of mean diameter 120mm , and 6.5 mm thick , has its open ends sealed by two rigid plates connected by two steel bolts of 25 mm diameter, initially tensioned to 20 kN at a temperature of 30 °C , thus forming a pressure vessel . Find the stresses in the copper and steel at freezing point , and the temp. at which the vessel would cease to be pressure tight Esteel = 200 GN/m2 , C/10*11 6

steel Ecopper = 100 GN/m2 , C/10*18 6

copper 4-8) Fig(4-17) shows the rigid bar CDE is attached to a pin support at E

and rests on the 30mm diameter brass cylinder BD . A 22 mm diameter steel rod AC passes through a hole in the bar and is secured by a nut which is snugly fitted when the temp. of the entire assembly is 20°C . The temp. of the brass cylinder is then raised to 50°C while the steel rod remains at 20°C . Assuming that no stresses were present before the temp. change , find the stress in the cylinder .

Rod AC , steel Cylinder BD , brass E = 200 GPa E = 105 GPa

C/10*12 6 C/10*6.18 6

Fig(4-17)


4-9) A steel tie rod 25 mm diameter is placed concentrically in a brass tube 3 mm thick and 60 mm mean diameter . Nuts and washers are fitted on the tie rod so that the ends of the tube are enclosed by the washers . The nuts are initially tightened to give a compressive stress of 30 MN/m2 in the tube , and a tensile load of 45 kN is then applied to the tie rod . Assuming the rod and the tube have the same effective length , find the resultant stresses in the tie-rod and tube , (i) when there is no change of temperature (ii) when the temperature increases by 60°C.

Esteel = 200 GN/m2 , C/10*1.1 5steel

Ebrass=80 GN/m2 , C/10*89.1 5brass

4-10) A steel rod of 320 mm2 cross-sectional area and a coaxial copper

tube of 800 mm2 cross sectional area , are rigidly bonded together at their ends . An axial compressive load of 40 kN is applied to the composite bar , and the temperature is then raised by 100° C . Find the stresses then existing in both steel and copper . The moduli of elasticity for steel and copper are 200 GN/m2 and 100 GPa and the coef. of linear expansion 12*10 6 /°C and 16*10 6 /°C respectively .

Strength of Materials- Thermal Stresses- Hani Aziz Ameen

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Transcript of Strength of Materials- Thermal Stresses- Hani Aziz Ameen