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Chapter 3 Statically Determinate and Indeterminate Systems 3.1 Principles of Solid Mechanics For a stable system, the following requirements must be met: (1) Equilibrium of forces and moments, which directly relates to stresses. (2) Compatibility of strains or displacements, which relates to deformation of geometry. (3) Stress ~ strain relations of materials, which are linked by the physical properties of a

material. 3.2 Statically Determinate Systems 3.2.1 Definition of statically determinate systems A statically determinate system is one that can be solved by force equilibrium only, that is, the number of unknowns in the system, e.g., reactions and internal forces, equals the number of equations available from force equilibrium. Example 3-1 A stepped cylinder carries axial loads, as shown in the figure below. The diameters of bar B and bar C are 25 mm and 50 mm, respectively. Find stresses in the two bars. Solution: For bar B

50=Bf kN

102

225

1050

225 2

3

2 =

=

=

BBf MPa (Tensile)

50 kN

p = 5 MPa

C B

60 mm

2

For bar C

322

1050225

260 =

+ pfC

38320225

26051050

223 =

= Cf N

5.19

250

38320

250 22

=

=

=

CCf

MPa (Tensile)

3.2.2 Typical thin-walled structures (1) Thin-walled pressurized sphere In this structure, stress can be assumed to be uniform through the wall thickness. Radial

stress r is assumed to be zero.

For force equilibrium, RtRp 22 = , then

tpR2

= (Everywhere in the wall, in each direction)

t

p

2R

t

3

(2) Close-ended thin-walled pressurized cylinder In this structure, hoop stress and axial stress l are assumed to be uniform through the wall thickness. Radial stress r is assumed to be zero. Consider force equilibrium in horizontal direction,

RtRp l 22 = , tRp

l 2= (Longitudinal stress)

Consider force equilibrium in vertical direction,

2R

t

p

l l

t

4

ltlRp = 22 , tRp = (Hoop stress or circumferential stress)

l 2= (3) Thin rotating ring Hoop stress is assumed to be uniform through the wall thickness and radial stress r is zero. Consider force equilibrium of a differential element,

=

2sin2 AFr , A Cross-sectional area

The centrifugal force Fr is evaluated as

( ) 2 = RRAFr , Material density

Since is small, 22

sin

. Then 22 R=

R

t

r

r

t

5

3.3 Statically Indeterminate Systems A statically indeterminate system is one that cannot be solved by force equilibrium only, that is, the number of unknowns in the system, e.g., reactions and internal forces, is greater than the number of equations available from force equilibrium. In this case, deformation of geometry has to be involved. Example 3-2 A stepped bar is constrained between two walls with a total deformation , as shown below. Determine the reaction force. Solution: The internal forces of bar a and bar b are equal, that is,

Rff ba ==

ba += bbaa ll +=

bb

ba

a

a lE

lE

+=

b

b

b

b

a

a

a

a

Af

El

Af

El +=

bb

b

aa

a

AR

El

AR

El +=

+=

bb

b

aa

a

AEl

AElR

aabbba

baba

AElAElAAEER +=

R R

la lb

Aa Ab

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Example 3-3 A load P is applied on a rigid beam which is supported by two rods a and b, as shown in the figure below. Find stresses in the rods. Solution:

lflflP ab 25.1 += Then ba ffP += 25.1 fa and fb cannot be solved by force equilibrium only and deformation of the rods has to be considered.

= lb and = la 2 , therefore ba 2=

la

a = and

lb

b =

a

aa A

f= and b

bb A

f=

aaa E= and bbb E=

b a

l l A B C

l

A B C

a b

P l l

l/2

7

Integrating the above relations, the following expression can be obtained.

bb

b

aa

a

EAf

EAf

2= , Aa and Ab are the cross-sectional areas of rod a and rod b, respectively.

Thus bbb

aa fEAEAP

+= 145.1

bbaa

bbb EAEA

EPAf += 45.1

and bbaa

aaa EAEA

EPAf += 43

bbaa

a

a

aa EAEA

PEAf

+== 43 and

bbaa

b

b

bb EAEA

PEAf

+== 45.1

Example 3-4 Cylinder a and rod b are made of different materials. They are fitted into rigid ends, as shown in the figure below. When temperature changes by T, determine stresses in the cylinder and the rod, respectively. Solution: The length changes of the cylinder and the rod will be different due to the different thermal expansion of the materials. Thus one will be in tension and the other will be in compression.

Cylinder a

Rod b

Cylinder a

Rod b

fa

fa

fb

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ba ff = fa represents the total internal force in the cylinder and fb represents the total internal force in the rod.

bbaa AA = Since ba ll = and ba = , then ba =

TE aa

aa +=

and TE bb

bb +=

Therefore TE

TE bb

ba

a

a +=+

TAEAT

E bbbaa

aa

a +=+

( )aabb

bbaaba AEAE

AEET +=

( )aabb

abaabb AEAE

AEET +=

If a and b are negative, the cylinder expands more than the rod, then the former is in compression and the latter in tension. Example 3-5 A symmetrical frame consisting of three pin-connected steel bars (E = 200 GPa) is loaded by a force P at the joint, see the figure below. The middle bar is 2 m long and its axial strain is measured to be 0.008. The angle between the inclined bars and the horizontal is = 50. Determine stress in the inclined bars.

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Solution:

8.261040cos

200040cos

===B

Cll mm

2.167840tan200040tan' === Bll mm ( ) ( ) 1.26232000008.020002.1678008.0 2222' =++=++= BB llll mm

3.128.26101.2623 === Clll mm

0047.08.2610

3.12 ===Cll

940102000047.0 3 === E MPa Example 3-6 A trimetallic bar is uniformly compressed by an axial force P = 2 kN applied through a rigid end plate, as shown in the figure below. The bar consists of a circular steel core surrounded by a brass tube and a copper tube. The steel core has diameter 10 mm, the brass tube has outer diameter 12 mm, and the copper tube has outer diameter 15 mm, with the corresponding elastic constant Es = 200 GPa, Eb = 100 GPa and Ec= 120 GPa. Calculate stresses in the steel, brass and copper, respectively, due to the force P.

50

40 l

l

10

Solution:

cbs fffP ++= , llll cbs === and cbs ==

lEA

flE

lss

s

s

sss ===

, lEA

f

bb

bb = and lEA

f

cc

cc = . Then

ss

bbsb AE

AEff = and ss

ccsc AE

AEff = , ss

ccs

ss

bbss AE

AEfAEAEffP ++=

5.782

10 2 =

= sA mm2

54.342

102

12 22 =

= bA mm2

59.632

122

15 22 =

= cA mm2

+

+=

5.781020059.6310120

5.781020054.341010012000 3

3

3

3

sf , 1172=sf N

84.2575.7810200

54.34101001172 33

==bf N

6.5695.7810200

59.63101201172 33

==cf N

93.145.78

1172 ===s

ss A

f MPa

46.754.3484.257 ===

b

bb A

f MPa

96.859.636.569 ===

c

cc A

f MPa

Example 3-7 A rigid triangular frame is pivoted at C and held by two identical horizontal wires at point A and B, see the figure below. Each wire has axial rigidity 120=EA klb and coefficient of thermal expansion = 6105.12 /F. (1) If a vertical load P = 500 lb acts at point D, what are the tensile forces TA and TB in the wires

at A and B, respectively? (2) If, while the load P is acting, both wires have their temperatures raised by 180F, what are

the forces TA and TB? (3) What further increase in temperature will cause the wire at B to become slack?

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Solution: 022 = bTbTbP BA PTT BA 22 =+

b

bA

B

2=

, BA 2=

P

TA

TB

Rx

Ry

A

B

12

(1) AE

lTlE

l AAAA === ,

AElTl

El BBBB ===

Therefore AE

lTAE

lT BA 2= , BA TT 2=

( ) PTT BB 222 =+ , 20050052

52 === PTB lb

40020022 === BA TT lb

(2) lTAE

lTAA += and lTAE

lTBB +=

lTAE

lTlTAE

lT BA +=+ 22 TAETT BA = 2 ( ) PTTAET BB 222 =++ ( ) ( ) 92180105.1210120500

52

52 63 === TAEPTB lb

454180105.12101209222 63 =+=+= TAETT BA lb

(3) Set 0=BT , then ( ) 052 == TAEPTB , TAEP = .

Therefore 33.333105.1210120

50063 =

== AEPT F

Further increase in temperature = 333.33F 180F = 153.33F Example 3-8 The shaft assembly shown in the figure below consists of a steel rod A with Youngs modulus of 210 GPa, cross- sectional area of 150 mm2, and coefficient of thermal expansion of 6106 /C, a rigid bearing plate C that is securely fastened to bar A, and a bronze bar B with Youngs modulus of 110 GPa, cross-sectional area of 250 mm2, and coefficient of thermal expansion of 6109 /C, A clearance of 0.5 mm exists between the bearing plate C and bar B before the assembly is loaded. (1) Determine the value of the applied force P to the bearing plate which just closes the gap

between the bearing plate C and bar B, and compute the stresses in bar A and bar B, respectively.

(2) Under the applied load P, the assembly is heated from room temperature (20C) to 100C. Determine the stresses in bar A and bar B, respectively.

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Solution:

(1) 41025.6800

5.0 ===A

AA l

25.131102101025.6 34 === AAA E MPa 310688.1915025.131 === AA AP N = 19.688 kN 0=B (2) 0= PRR AB AA Rf = and BB Rf = , thus Pff AB =

,TE AA

AA +=

,TE BB

BB +=

A

AA l

= , B

BB l

=

RA

RB

P/2 P/2

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Since 0=+ BA , 0=+ BBAA ll

0=

++

+ BB

B

BAA

A

A lTE

lTE

0=

++

+ BBBB

BAA

AA

A lTAE

flTAE

f

02008010910110250

8008010610210150

63

63 =

+

+

+

BA ff

26400364.01268.0 =+ BA ff

Since 19688== Pff AB N, then 11785=Af N = 11.785 kN and 31468=Bf N = 31.468 kN

53.78150

11785 ===A

AA A

f MPa (Compressive stress)

88.125250

31468 ===B

BB A

f MPa (Compressive stress)

Example 3-9 A copper tube in the figure below is sealed by two rigid washers. After the nut is tightened by 1/8th of a turn of the thread, the components are in contact with each other. Find the stresses in the bolt and in the copper tube. 207=sE GPa and 120=cE GPa.

200 mm

180 mm

Pitch of thread 1 mm

Rigid washer

Steel bolt 12 mm

Copper tube Do = 25 mm Di = 15 mm

Hamza Mahmood

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Solution:

sc ff = sscc AA =

Assume that the deformation of copper tube is (compressive), and given that the axial deformation of steel bolt is 125.01

81 = mm (tensile), then the net increment in length of the

bolt is 125.0 .

( ) 310207200

125.0 === s

sssss l

EE

310120180

=== c

ccccc l

EE

( ) 232

1210207200

125.0

=

223

215

22510120

180 , therefore

0448.0= mm

( ) 8310207200

0448.0125.0 3 ==s MPa

30101201800448.0 3 ==c MPa

Example 3-10 In the figure below, a steel pipe (1) is attached to an aluminum pipe (2) at flange B. Both the pipes are attached to rigid supports at A and C, respectively. Pipe (1) has a cross-sectional area of A1 = 3600 mm2, an elastic modulus of E1 = 200 GPa, and an allowable normal stress of 160 MPa. Pipe (2) has a cross-sectional area of A2 = 2000 mm2, an elastic modulus of E2 = 70 GPa, and an allowable normal stress of 120 MPa. Determine the maximum load P that can be applied to flange B without exceeding either allowable stress.

Cylinder a

Rod b

fc

fc

fs

16

Solution:

021 =+ Pff

11

1

1

11 EA

fE

== and 22

2

2

22 EA

fE

==

11

11111 EA

lfl == and 22

22222 EA

lfl ==

Since 021 =+ , 022

22

11

11 =+EAlf

EAlf , then

112

22112 EAl

EAlff = .

Therefore, PEAlEAlff =+

112

22111 or PEAl

EAlf =

+1

112

2211

111 Af = and 222 Af = Let 1601 = MPa,

=

+1

112

2211 EAl

EAlf 53333

112

22111 102.71102003600104.1

10702000108.136001601 =

+

=

+

EAlEAlA N

= 720 kN 720P kN, in this case, 5111 1076.53600160 === Af N = 576 kN,

=== 57672012 fPf 144 kN

722000

1044.1 5

2

22 =

==Af MPa < 120 MPa

221

11221 EAl

EAlff = , then PfEAlEAlf =+ 2

221

1122 or PEAl

EAlf =

+1

221

1122

P/2

P/2

f1 f2

Hamza Mahmood

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Let 1202 = MPa, then

=

+1

221

1122 EAl

EAlf 63333

221

11222 102.1110702000108.1

102003600104.120001201 =

+

=

+

EAlEAlA N

= 1200 kN 1200P kN, in this case, 5222 104.22000120 === Af N = 240 kN,

960240120021 === fPf kN

2673600

106.9 5

1

11 =

==Af MPa > 160 MPa

Therefore, the maximum load P that can be applied to flange B without exceeding either allowable stress is 720 kN.