Strength of Materials- Thermal Stresses- Hani Aziz Ameen

Strength of Materials Handout No.4

Thermal Stresses Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Ba ghdad Dies and Tools Eng. Dept. E-mail:[email protected] www.mediafire.com/haniazizameen

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 bar exte xtends nds by an amount unt ( T ) L , if this extensio extension n is preven prevented ted 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 .L. T formula : temp where : temp

= extension or contraction due to temp.

= thermal coefficient of linear expansion L = body length T = change in temp. temp. If this deformation occurs freely , there is no external load acting on the body , but if there is a certain resistance resistance to the expansion expansion or contraction , the external load will show so,

-aThere is no thermal stress but thermal strain

-bThere is thermal stress 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 i ts initial position , give the relation between geometric deformation due to temp. and external force temp

L T

P

PL AE

,

P A

Thus , T

temp

temp

E -1)

E T

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 : T( 1L1 -2) 2L2 ) when thin displacement is prevented , the force inducd in each bar is the same : -3) 1A1 2A2 1

2

The compressive strecses 1 and 2 must depend upon eq.(4-2) in the following form to ensure zero axial axi al displacement: 1L1 2L2 T( 1L1 2 L 2 ) 0.......... .......... .........( 4 4) E1 E2 Combining eqs. (4-3) and (4-4) , gives T( 1L1 2L2 ) 1

(L1 / E1 )

A1L 2 A2E 2

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 .

(i) (ii)

Fig(4-3) Assuming T in These A B , the free expansion caused by 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 ( B Consequently, A will be put in tension and B A ). in compression . The stresses so induced in each material must satisfy two conditions Equilibrium the net load is zero , which gives : 0.......... .......... .......... .......... .......( 4.5) AAA BA B Compatibility the strains are equal , giving :

Strength of Materials- Handout No.4- Thermal Stresses - Dr. Hani Aziz Ameen AL

AL T

BL T

EA Then Then from from eqs.( eqs.(4-5 4-5)) & (4(4-6) 6) : A

T(

B

BL

-6)

EB

A )(A BE A E B ) /(A BE B

A BE A )

-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 Mechanical and Thermel Effects 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 th eir ends. The cross sectional area 2 2 of the copper member is A m while that of steel member is 2A m for the other half of the length. The coefficient of expansion of steel and copper are and 1.3 respect respective ively ly while while the elastic elastic moduli 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 th e 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 s

*A

c

s

s

* 2A

2

s

*A

c

If La is the actual length of the compound bar due to rise in temperature , then for copper rod : L(1

c T)

c

La

Ec

*L

For steel rod : L

's

Es 2 Adding eq(i) and eq(ii) we get

Es

La

(

L(1

s )T

c

c

s

s'

s

2

c

*

s

Ec

2E s c

*

L 2

s

2E s c

c

0.5E E 0.0109E T

4E

)T

(1.3

s

s T)

11

c

4E

0.0545E T

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 4. 5 m . Find the increase in i n 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 p er 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 ten sion 2 6 Take E=200 GN/m and = 12*10 per C °

Fig(4-5) Solution 2

55

A1

4 1000 75

A2

0.002376m 2

2

0.004148m 2

4 1000 When the turn buckle is turned by one quarter of a turn, 1 extension o =0.00125 m 0.2 * 1000 1

=0.001666 m 0.15 * 1000 Total extension of the two rods = 0.00125+0.001666 = 0.002916 m Total extension in the two bars PL1

PL 2

A1E

A2E

P * 4.5

1

1

4.5P

200 * 109 A1

A2

1

200 * 109 0.002376

1 0.004418

= 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 (T 2 T1) be T C °

TL T

0.002916 0.002916

12 * 10

6

* 4.5

12 *10 54 C

°

6

* T * 4.5

0.002916


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 mm respectivly. 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 temperature of the bar is is then raised by 50 C 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 Find the increase in the distance between the stops if the 2 force exerted between them is 60 kN , take E s = 200 GN/m , 2 5 5 E b = 90 GN/m , s = 1.2*10 per C = 1.2*10 per C °

i) ii)

°

° °

Solution

i)(a) When the distance between the stops remains constant stress in i n steel, 5 9 2 s= s.T.Es = 1.2*10 *50*200*10 = 120 MN/m 5 9 stress in brass, b= b t Eb = 2.1*10 *200*10 = 94.5 MPa (b) When the distance between the stops is increased by 0.3 mm exp exp ansion prevented s T L (0.2 / 1000) strain in steel , s original length 0.5

= stress in steel . strain in brass .

1.2 * 10 sEs

s

5

* 50 * 0.5

0.0002

0.5 9 0.0002 * 200 * 10

exp exp ansion prevented b

=

0.0002 2

40 MN/m 0.0002 b TL

original length 2.1 *10

5

0.5

* 50 * 0.5 0.0002 0.5

0.00065

stress in brass , b

bE b

0.00065 * 90 * 10 9

58.5 * 10 6 N/m (compressive) 2

Strength of Materials- Handout No.4- Thermal Stresses Stresses - Dr. Hani Hani Aziz Ameen Ameen

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

sT

s

L

5

1.2 * 10

* 50 * 0.5

L 0.5 0.0006 2 (0.0006 2 ) * 200 * 109 s sEs

stress in steel ,

similarly strain in brass,

stress in brass,

b

Area of steel rod ,

b TL

b

2.1 * 10

L (0.00105 2 )

5

* 50 * 0.5 0.5

(0.00105 2 ) * 90 * 10 9

bEb

As

/ 4 *

2

40 1000

1256.6 *10

6

863.9 *10

6

m

2

Area of brass tube , Ab

/ 4 *

2

60 1000

2

50 1000

Load on steel +load on brass =total load between the stops P s As b Ab or

0.0006 2 * 200 *109 *1256.6 *10 or

0.0006 2 * 2.513*108

or 15.078*104

8.164 *104

or 23.242 *104

6.581*10

8

23.242

0.00105 2 * 90 *109 * 863.9 * 10 = 60000

0.00105 2 * 0.7775 *108 5.026 1.555 *108 4

60000 6 *10

6 * 10

6.581 * 108

6

4

* 103 mm

0.262mm

60000

60000

6


Example(4-4) Fig(4-6) shows a copper bar 50mm in diameter diameter is placed placed within a steel tube 75mm external external diameter and 50 mm mm internal diameter diameter of exactly the same length ,the two pieces pieces are rigidly fixed fixed together together by two 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 °

Es

210GN / m 2 , E c

105GN / m 2 ,

s

11.5 * 10

6

per per C,

17 * 10

c

6

per per C

Fig(4-6) Solution

Area of copper bar , A c Area of steel tube , As

/ 4 * (50 / 1000) 2 =1963.49*10

/ 4

75 1000

2

50 1000

6

2

m

2

2454.37 *10

6

m2

Difference in free expansion TL 17 * 10 6 11.5 * 10 6 * 50 * L =275*10 L A Compressive Compressive Force P exerted by the steel tube on the copper rod opposes the extra expansion of of the copper copper rod , and and in return , the copper rod exerts exerts an equal tensile force P to pull the steel tube so so that the consequent combined effect 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 PL PL of copper red due to force P A c E c 1963.49 * 10 6 * 105 * 109 increase in length of steel tube due to force fo rce P PL PL -6

c

AsEs but ,

s

2454.37 * 10 PL

6

* 210 * 109 PL

1963.49 *10 6 * 105 *10 9 2454.37 *10 6 * 210 *10 9 6 = 27.5*10 L 9 9 6 4.85*10 P +1.94*10 P=275*10 P=40500 N


P

stress in copper rod, stress in steel tube,

c

40500

Ac

1963.49 * 10 40500

P s

shears stress in pins

6

As

pin pin

6

6

* 10

* 10

6

16.5 Mpa

2454.37 * 10 P 40500

2 * A pin pin

2*

4

*

18

20.36 MPa

2

* 10

6

79.57 MPa

1000

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 ? 2 2 Take E s=210 GN/m , E c=100 GN/m °

s

12 * 10

6

and

per C

c

17.5 * 10

6

per C

Fig(4-7) Solution

Area of steel rod , A s

4 1000

Area of copper tube, Ac = i)

2

10

40 4

1000

78.54 * 10

6

2

2

30 1000

m2 549.78 * 10

6

m2

Stress due to tightening 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 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 c

*

0.04 2

4

0.032

Strain in the copper tube,

s

*

* 0.012 or

4 2 1.9996

c

2

s

7

c

0.0002 000 2

Stress in the copper tube * Ec 0.0002 * 100 * 10 9 * 10 6 20 MPa c and stress in the steel rod 7 s 7 * 20 c s s 140 MPa Stresses due to rise of temperature : If the two members were free to expand, free expansion of steel = s T Ls Free expansion of copper = cT Lc c

ii)

c

Since c is greater th an s , 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 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 and c be the stresses in the steel and copper respectively equilibrium s and of the whole system . Total tension (pull) in steel = total compression (push) in copper. Ac 7 c sAs c A c or s c As Final expansion of steel = final expansion of copper s

T Ls

Ls

c

Lc Es Ec but T = 60 C Ls = 2+2*0.02 = 2.04 m & Lc = 2m 7 c 6 c 12*10 6 *60*2.04+ * 2 . 04 17 . 5 * 10 * 60 * 2 *2 210 *10 9 100 *10 9 6 9 12*10 *60*2.04*210*10 +7 c *2.04 = 17.5*10 6*60*2*210*109 2.1*2 c 8 2 c = 0.0717*10 N/m s

c

T Lc

c

= 20 + 7.17 = 27.17 MPa

°

= 7 c = 50.19 MPa Stress in Copper = c s

Stress in steel =

s

s

= 140 + 50.19 =190.19 MPa


Example (4-6) Fig (4-8) shows a brass rod of of 6mm diameter 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 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 E s =200 GPa , E b = 85 GPa , 12 * 10 6 per C , 19 * 10 -6 per C s b °

Fig(4-8) Solution

As

Ab

*

6

2

28.27 *10

6

m2

4 1000 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 . 3.5 *1000 *10 6 123.8MPa s1 b1 6 28.27 *10 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 compression in brass


s

Ps + Pb=1.3*1000 = 1300 N Ps Ls Pb L b Es Lb Ps Pb As E s A b E b Eb Ls

b

9

=

200*10

9

1

*

* Pb

(A s

Ab )

1.81Pb

1.3 85*10 1.81 Pb+ Pb=1300 Pb= 462.6 462.6 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 Ps

s2

837.4 6

A s 28.27 *10 = 29.62 MPa (tensile) Pb 462.6 b2

Ab

28.27 * 10

6

*10

6

* 10

6

16.36 MPa (comp.)

The resulting stress 123.8 16.36 10.44MPa ( tensile ) 123.8 29.62 153.42MPa ( tensile ) st s1 s2 In the absence of fixtures the rod LN would have expanded freely due to temperature NL , but the pressure of fixture effectively compresses the rod LN by an amount LL which is equal to ( s Ls T b L bT ) thus if is the compressive stress in steel and brass b rass , then we have br

b1

Es

10

Ls

b2

Eb

Lb

s Ls T

1.3

1

200 *10 9

85 *10 9

9

0.0065 0.0117 1.038 * 10

6

* 10

b L bT

30 ( 12* 10 6*1.3+19*10 6*1) 1.038 * 10

9

0.0065 0.0117

* 10

6

6

MPa ( compressiv e)

Hence the final stresses in steel and brass are : 153.42 57.03 96.39 MPa sf s bf

b

107.44 57.03

50.41MPa

( tensile ) (tensile)


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 , 2 take E=200 GN/m , 12 *10 6 per C °

°

Fig(4-9) Solution When the temperature is raised 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 : °

6

L (T2 T1 ) 12 * 10 * 1.2 * 40 0.000576 cm 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 Pdx

Ax E L

Total extension 0

But

Ax

where k = ( d2

4P

Pdx Ax E

d1

4 d1)/L

L

d1 L

dx

E 0 (d1

2

x

kE 1

d1 ) d1

4

d2

d1

d

kx

2

1

1

4P

kx ) 2

4PL E(d 2

d2

d1 kx

0

1

4 PL

d1

Ed1d 2

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


From eq(i) and eq(ii) we get 4PL L(T2 T1 ) Ed1d 2 From which ,

E (T2

P

T1 )d1d 2

4 Maximum stress induced is given by 4

E (T2

E (T2

T1 )d1d 2 *

T1 )

4 d12

d2 d1

200 * 10 9 * 12 * 10

6

* 40 *

0.15

144 MPa

0.1

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 , 6.5 *10 6 m/m/ C 2 L=203.2mm , A = 968 mm For bronze , E = 83 MPa °

°

° °

6

10.5 *10 m/m/ C 2 L= 152.4 mm , A = 1290mm ° °

Fig(4-10) Solution

Total expansion of the cylinder = due to temperature change Total contraction in the cylinder =

tb

b

ts

s

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

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


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

ts

L T) bronze

b

s

L T) steel

PL

PL

) steel EA PL L(T To ) b L (T T o ) s )b )s AE AE 6 3 6 3 10.5*10 *152.4*10 *(60-30)+6.5*10 *203.2*10 *(60 =

) bronze

EA PL

Pb * 152.4 * 10

3

30)

Ps * 203.2 * 10

3

1290 * 10 6 * 83 * 10 6 968 * 106 * 207 * 106 Putting Pb = Ps = P , gives P = 96 kN 96000 Stress in the steel cylinder , s 98.6MPa (comp.) (comp.) 968 * 10 6 96000 Stress in the bronze cylinder , b 74.1MPa (comp.) (comp.) 1290 * 10 6 Example(4-9) Fig(4-11) shows the assembly assembly is symmetrical about the y-axis and in which two aluminum bars bars and one 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 2 E= 69 GPa , 12.8 *10 6 m/m/ C , L=254 mm , A=968 mm For the brass bar , 2 10.2 *10 6 m/m/ C , E=103.5 GPa, L=203 mm , A=1613mm ° °

° °

Fig(4-11)


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

Pa * 254 * 10

or

3

Pb * 203 * 10

a

b

3

968 * 10 6 * 69 * 10 9 1613 * 10 6 * 103.5 * 109 from which Pb=25Pa also , it can be deduced that 2P a+Pb solving eq.(i) eq.(i) & (ii) gives gives Pa = 48.6 kN Pb= 152.12 kN 4600 Stress in the aluminum bar , a 50.33MPa (comp.) 6 968 * 10 152120 Stress in the brass , b 94.5MPa (comp.) 6 1613 * 10 (b) From fig-(4-11b), the following relation is established

Fig(4-11 b) a

ta

b

tb

or Pa * 254*10 968*10

6

3

* 69 *10

6

9

12.8 *10 * 254*10

3

*T

Pb * 203*10 1613*10

6

3

*103.5 *10

9

10.2 *10

6

* 203*10

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 into eq. eq. (iii) (iii) yields T = 22.5 C °

3

*T


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

°

°

°

Fig(4-12) Solution

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

s

tb

b

Or 6.5 * 10

6

* L s 150

-6

= 10.2*10 *Lb(150

50

Fs * L s 9.525 * 10

50)+

3

* 0.0508 * 207 * 10

9

Fb `L b 12.7 * 10

3

* 0.0508 * 103.5 * 109


Putting Ls = Lb = r =457.2mm & simplified , gives Fs+3Fb=148118.4 ............(i) 2Fs = total pressure force 2Fs = Pressure * projected area projected area = diameter diameter * length of the ring in the Z-direction 2 =914.4*50.8=46451.52 mm , hence 3

Fs

2Fs = 46.451*10 P similarly 2Fb=FP

23.22 * 10 3 P

3

Fb = 23.22*10 P Sub into (i) 323.22 *10 3 P 3 * 23.22 *10 3 P 148116.4 P =1.6 MPa Stress in the steel ring is Fs Fs 765.3MPa (comp) s area 9.525 * 10 3 * 0.0508 Stress in the brass ring 23.22 *10 3 P b

12.7 *10

3

* 0.0508

57.366MPa tensile

is


4.4 Problems 4-1) A 12mm diameter steel rod passes centrally centrally through a copper tube 48mm external 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 raised in temp . by 60 C. Find the stresses in the copper and steel before and after the rise in temp temp . Assuming the thickness of the plate remains unchanged 2 2 Esteel =200 GN/m ,Ecopper =100GN/m 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 E 207 GPa , 12.5 *10-6 / C -6 For copper E=105 GPa , =18*10 / 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*10 10 N/m2 2 a = 0.4m 0.4m , A1=1200 mm b = 1.2m , A2 = 1400 mm2 c = 0.4m 0.4m , A3 = 16cm2 1 = 45 2 = 60 3 = 30 and 157 MPa Find P & Pmax -6 b) 1= 2 = 3 = = 12.5*10 ° ° °

and T 40 Find I , II & III c) 2 1.2mm 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 10 2 Esteel=2*1010N/m2 125 * 10 7 Ecopper =9*10 N/m copper

165 * 10

7

,and the change of temperature is

T 10

Find the thermal stresses

Fig(4-15) 4-5) Fig(4-16) shows the assembly of the t he three materials, find the stresses due to the action of the forces and those due to the change c hange in temp . 10

2

take E= 19.6*10 N/m

,

12 *10

Fig(4-16)

6


4-6) A mass of 150 kg is suspended by three vertical wires ,the two outer 2 wires are of steel and the middle middle one of aluminum each of area 8mm 8mm , 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 °

12 * 10 6 & 24 * 10 6 Al 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 steel

°

6

2

cease to be pressure tight Esteel = 200 GN/m , Ecopper = 100 GN/m2 ,

copper

steel

11 * 10 / C

18 * 10 6 / C

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 . °

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Rod AC , steel E = 200 GPa

Cylinder BD , brass E = 105 GPa

12 *10 6 / C

18.6 *10 6 / C

Fig(4-17)


4-9) A steel tie rod 25 mm diameter is placed placed concentrically concentrically in a brass tube 3 mm thick and 60 mm mean diameter . Nuts and washers are fitted on 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 2 of 30 MN/m in the tube , and a tensile load of 45 kN is t hen applied to the tie rod . Assuming the rod and the tube tube have the same effective length , find the resultant stresses in the tie-rod and tube , (i) when there is no change of temperature temperature (ii) when the temperature temperature increases by 60 C. °

Esteel = 200 GN/m2 , 2

Ebrass=80 GN/m

,

1.1 * 10 5 / C

steel

5

brass

1.89 * 10 / C 2

4-10) A steel rod of 320 mm mm cross-sectional area and a coaxial copper 2 tube of 800 mm 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/m 2 and 100 GPa and the 6 6 coef. of linear expansion 12*10 / C and 16*10 / C respectively . °

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Strength of Materials- Thermal Stresses- Hani Aziz Ameen

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