Concentric Tension or Compression. Problems

8/3/2019 Concentric Tension or Compression. Problems

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Resolved Problems:

2.2.1 A steel bar with constant cross-section is loaded by two concentrated

forces NPNP 525

1 105,102 , as shown in Fig. 2.5. Determine the

necessary cross-sectional area for the bar, neglecting its own weight.

The cross-section is conceived in several variants: circular solid section,

rectangular section, circular hollow section, universal beam rolled shape, section

made of welded steel plates. The allowable strength in tension ( at ) or compression

(ac

) are equal to 1600 2/cmdaN .

Solution:

Fig. 2.5

A

B

C

1P

2P

12 PP

l2,0

l8,0

x

1P

1P

2P


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In order to determine the most loaded section, the axial force diagram must

be plotted.

At any section of the domain AB the axial force is:

constNPxN 51 102)(

The axial force is also constant along the domain BC and equals:

NPPNBC55

21 10310)52(

Along the domain AB the bar is subjected to concentric tension, while

domain BC is subjected to concentric compression.

As the material has the same strength in tension and compression, the most

loaded section is that of maximum normal stress in absolute value that is, any

section along the portion BC.

By applying relation (2.14), we obtain:

24

0

max 75,18

1600

103cm

NAnec

For this necessary area we shall determine the cross-sectional dimensions

for each shape of the section.

a) Circular solid section (Fig. 2.5.a)cmd

Ad

dA

nec

nec

necnec

88,475,1844

4

2

Fig. 2.5.a

d


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b) Rectangular section (Fig. 2.5.b)

cmacmbAbbAba necnecnec

necnec 63275,18

222 2

Fig 2.5.b

c) Circular hollow section (Fig 2.5.c)

cmDd

cmA

D

DA

D

d

necnec

nec

nec

nec

50,68,0

15,8)8,01(14,3

75,184

)1(

4

14

8,0

22

22

Fig. 2.5.c

a

b

d D


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d) Universal beam rolled shape (Fig. 2.5.d)From Appendix no. 13 we choose 16I with the area

28,22 cmA .

Fig. 2.5.d

e)

Section made of welded steel plates (Fig. 2.5.e)The section area is:

cmA

ttA

tttA

nec

necnec

t

43,0100

75,18

100100

100302202

2

222

Fig. 2.5e

Remark:

The material quantity does not depend on the cross-sectional shape because

the area is the same. The fibers are equally loaded because the normal stresses

are uniformly distributed over the section.

2.2.2. A member made of wood, with rectangular section, b=20cm and

h=2cm, is weakened by a circular hole, 20mm in diameter. The member is

loaded by two concentrated forces NPNP 424

1 105,5,103 (Fig. 2.6).

cmh 16

t20

t

t

t

t30

t


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It is required to check the strength and stiffness conditions for the considered

bar, knowing that the strengths of the material are:

- compression2

80cmdaN

oc ; tension 2100cmdaN

ot ,

the permissible elongation200

lla and the length l=20m.

Fig. 2.6

Solution:

The member is subjected to concentric tension along the domain AB by a

constant axial force: NPNAB4

1 103

11

1P

2P

A

d

b

2P

1P

C

l2,0

l8,0

N

1P

12 PP

B

h


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and to concentric compression along the domainBC, by a constant axial force:

NPPNBC44

21 105,210)5,53(

As the material has different strengths in tension and compression, the strength

requirement must be checked up for each domain.

- along portionAB (the most loaded section is the weakened one, 1-1):

2

22

3

3622220

)100(3,8336

103

cmAAA

cm

daN

cm

daN

A

N

holegrossnet

ot

net

AB

x

- along portionBC, the sections are equally loaded:

2

23

80/5,62220

105,2

cm

daNcmdaN

A

Noc

BC

x

The displacement of the member free end is the algebraic sum of the length

changes of the two component portions:

cmEA

lNl ABABAB 26

4

1012220102008,0103

cmEA

lNl BCBCBC

2

6

4

105,222010

2002,0105,2

cml

llcmml aa 1200

200

200095,010)5,212( 2

Remark:

In the evaluation of deformations, the bar is considered to have a constant

cross-section because the hole has no important influence.

On the contrary, in case of member design according to the strength

requirement, the most loaded section along domainAB is the weakened one.


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2.2.3 The elementAB belonging to a faade scaffold is supported by usingtwo steel bars AC and BC (Fig. 2.7). Determine the necessary cross-sectional

dimensions for the load-bearing elements of the scaffold and of the rod through

which the weight G is suspended. Compute the pressure exerted by the element AB

on the vertical wall.

Numerical data:

)5,2(105,2,30,1600 40

2

tfNGcm

daNocot

Fig. 2.7

Solution:

The axial forces in the bars of the system are determined by isolating node

A.

B

C

1A

2A

3A

G

ABN

ACN

G

G

11A

diagrambodyFree

1

1


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)44,1(1044,13

105,2

)9,2(109,210

2

3

5,2

cos

44

44

tfNtgGN

tfNG

N

AB

AC

The selection of the necessary cross-sectional area of the rod used for weight

G suspension (stretched member):

2

1 56,11600

2500cm

GA

ot

nec

cmA

dnec

41,114,3

56,144 11

The selection of the necessary cross-sectional area of the stretched barAC:

2

2 81,11600

2900cm

NA

ot

AC

nec

cmA

dnec

52,114,3

81,144 22

The selection of the necessary cross-sectional area of the compressed bar AB

(considering that its bucking is not possible):

2

3 9,01600

1440cm

NA

oc

AB

nec

Two equal legs angle shapes 2L 50x50x5 are adopted, their area being

2x4,8=9,62

cm .

The pressure exerted on the wall is:

2

3

1506,9

1044,1

cm

daN

A

Np

AB

AB


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2.2.4 A rigid bar AB is supported by the vertical rods, AC made of steeland BD made of copper. The bar is acted by a concentrated force NP 4108 (Fig.

2.8). Knowing the allowable strength and the longitudinal modulus of elasticity for

steel 262 /101,2;/1600 cmdaNEcmdaN STSTo and for copper

262 /107,1;/1000 cmdaNEcmdaN COCOo , as well as the original

length of the rods l=2m, it is required to determine:

a) the cross-sectional areas for the rods AC and BD, so that the rigid bar toremain in the horizontal position under the action of load P

b) the displacement of the point where force P appliesc) the work done by force Pd) the strain energy stored by the vertical rods.

Fig.2.8

A B

C D

P

BDN

a04a06

l

P

ACN


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Solution:

The axial forces in the rods are obtained from the moment equilibrium

equations:

PNaNaPM

PNaNaPM

ACACB

BDBDA

4,004,00

6,006,00

The values of the two axial forces check up the equilibrium equation of

projection along the vertical direction:

-0,4P+P-0,6P=0

In the design of the rods, three requirements are involved: two strength

requirements and the requirement of elongations equality. The procedure that

should be followed is:

- the cross-sectional area of the rod BD is determined from its strength

requirement:

2

3

3

8,410

1086,0cm

NA

COo

BD

BD

- the second rod, AC, area is determined by imposing the equality of rods

elongations:

BDAC ll

26,28,41,2

7,1

6,0

4,0cmA

E

EA

N

NA

AE

lN

AE

lN

AC

ST

CO

BD

BD

AC

AC

BDCO

BD

ACST

AC

- the strength requirement for rodACis checked up:ST

AC

Ac

STefx

cm

daN

cm

N

A

N022

44

12301023,1

6,2

1084,0


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When this requirement is satisfied, the design of the rods is finished,

otherwise, the design should be started from the second rod,AC..

The displacement of load P point of application equals the elongation of the

rods:

cmAE

lNll

ACST

AC

BDAC 0117,06,2101,2

2001084,07

4

The work done by force P is:

)(68,41017,12

108

2

44

jouliNm

P

Lext

The strain energy stored in the rods is:

)(68,4

8,4107,1

108,4

6,2101,2

102,3200

2

1

2

1

2

1

7

24

7

24

22

jouliNmU

AE

lN

AE

lNU

COCO

BD

STST

AC

2.2.5 The truss structure from the roof of an industrial building is made ofequal legs angle shapes. Its free body diagram is shown in Fig. 2.9a.

Determine the necessary cross-sectional areas for the bars 1-2 and 5-5, considering

that their connection at the nodes is conceived in two variants:

-

riveted connection with rivets having the diameter d = 20mm.- welded connection;

Numerical data:

2

5 16002,1102cm

daNmaNP o


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Fig. 2.9

Solution:

a) Internal forces evaluationNode 1 (Fig. 2.9b) is isolated and from the equilibrium equation of

projection along the vertical direction, the axial force in the bar 1-2 is

obtained:

PPP

N

PN

11,1'4026sin

5,0

sin

5,0

05,0sin

012

12

The axial force in the bar 5-5 can be easily obtained by performing section

s-s.

From the moment equilibrium equation expressed with respect to node 3

(Fig. 2.9c), it results:

PN

aNaPPaP

'55

'55 025,15,1)5,2(5,25,0

a

1

2

3

4 5

P

'1

'2

'4'5

a75,0

P5,0

P2

'40260

12N

14

N

a5,0

PV 5,2

'55N

a aa

P

a5,0

a5,0

P5,0

3

2

14

P5,0

2

)b

)a)c

P s

s


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b) Selection of the necessary cross-sectional areas for the considered bars incase of riveted connections

The areas will be determined by considering a coefficient 1,1wK , which

takes into account that the sections are weakened by holes. After that, the strength

requirements will be checked by using the net cross-sectional areas.

Fig. 2.9.a

- bar 1-2:

24

0

1212 2,15

1600

10211,11,1 cm

NKA wnec

There are adopted 2L: 70x70x6 with the area:

23,1615,82 cmAgross

The net area of the bar is:

29,136,0223,16 cmAAA holegrossnet

Bar strength requirement checking up:

5

25N

45N '55N

53N


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14

02

412 1600

9,13

10211,1

cm

daN

A

N

net

efx

- bar 5-5:

66060275,131600

1021,1 2

4

0

'55'55 xxLcm

NKA w

2

2

42,116,02282,13

82,1391,62

cmA

cmA

net

gross

Bar strength requirement checking up:

02

4'55 1750

42,11

102

cm

daN

A

N

net

efx

This condition is not satisfied and therefore the bar cross-sectional area must

be increased. There are chosen 2L 70x70x6. The stresses in the bar are now :

02

4'55 1440

9,13

102

cm

daN

A

N

net

efx

The area is overestimated, but in this case there are no better possibilities.

c) Selection of the necessary cross-sectional areas for the considered bars incase of welded connections

In this case the design of the bars can be directly performed:

)12,13(7505025,121600

102

)82,13(660602875,131600

10211,1

224

0

'55'55

224

0

1212

cmAxxLcmN

A

cmAxxLcmN

A

nec

nec


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Remarks:

1. The riveted connections lead to greater cross-sectional areas that implygreater material quantities.

2. When the section of the bar is a rolled shape, its area cannot be the exactlyneeded one, so that it is generally overestimated.

2.2.6 A column made of cast iron is supported on a concrete foundation byusing a steel plate. The column is loaded by a concentric compressive force

NP 5105 (Fig. 2.10). Determine the necessary cross-sectional areas for the

column, steel plate and foundation, knowing:

mc

N

mc

N

D

dmhmh

cm

N

cm

N

cm

N

cicf

soilaconcreteaironcasta

44

22

2

2

4

102,7;108,1;8,0;80,0,1

20;104;10

Solution:

a) The column cross-sectional area is:2,710

105

1010102,710

1054

5

2644

5

h

PA

ciaci

nec

The product hci can be neglected in comparison with aci , so that the own

weight influence can be ignored in case of structural elements with small length and

high strength.


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Fig. 2.10

So:

cmd

cmA

D

DA

cmA

nec

nec

nec

nec

nec

7,103,138,0

3,13)8,01(14,3

504

)1(

4

)1(4

5010

105

22

22

2

4

5

fb

pb

fh

aa

a

D

d

a

P

h


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b) The steel plate area results from the condition that the pressures on the

foundation block mustnt exceed the concrete allowable strength. The own weight

of the column and of the steel plate are ignored.

2

2

5

1250104

105cm

PA

acnecp

If the plate has a square shape, its side is:

cmbp 4,351250

c) The dimensions of the foundation block cross-section result from the

condition that the pressures on the foundation foot mustnt exceed the soil

allowable strength.

cmb

cmh

PA

f

fcsanecf

16427000

2700044,120

105

108,010108,120

105 25

264

5

The own weight of the foundation block cannot be ignored.

2.2.7 The structural element shown in Fig. 2.11 is subjected to tension byan axial force NN 41050 .

Compute the stresses on a section inclined at an angle of 030 with respect to

the element longitudinal axis and plot the trajectories of first and second kind.

Discuss the element failure, that depends on the material nature.


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Fig. 2.11

Fig. 2.12

22

x

x 1

21

x

245

x 02

2

02

1

45

4545

1

2S

1S

cmt 2

cmb 20N N

N


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Solution:

On the element cross-section the stresses are:

0

0

1250202

500002

xy

y

xcm

daNAN

On a section inclined at 030 , the normal stress is obtained by using relation

(2.2):

2

0

30 9352

112

1250302cos22 cm

daNxx

and the shear stress with relation (2.3):

2

0

30 5402

3

2

1250302sin

2 cm

daNx

As there are no shear stresses on the bar cross-section, the normal stresses

are quite the principal stresses:

x 1 and 02

The trajectories of first kind are lines parallel to the element longitudinal

axis, while the trajectories of second kind are lines perpendicular to the longitudinal

axis.

On the planes inclined at

0

45 with respect to the bar longitudinal axis (thebisectors of the principal planes) the extreme shear stresses occur:

22,1625

2 cm

daNx

(a)

and the normal stresses are:

245625

2 cm

daNx

(b)


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Let us analyze the strength requirement on two sections :

- on the cross-section, where only normal stresses exist:

ax that is a1250 (c)

- on the sections inclined at 045 , where both normal and shear stresses exist:

a 45 (d)

a 2,1 (e)

aech 21

242 3 (f)

By taking into account relations (a) and (b), it results:

a

x

2

(d)

axa

x

22

(e)

axxxxech

22

232 (f)

From the four strength requirements (c,d,e,f), the important ones are (c) and

(e), so that:

- for the materials characterized by2

a

a

, the most dangerous section is

the inclined one at

0

45 with respect to the longitudinal axis and the failure occursalong this direction (wood case);

- for the materials characterized by2

a

a

, the most dangerous section is

the cross-section of the bar and the failure occurs in this section (steel case).


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2.2.8 A brick wall is subjected to loads NP 41 1011 and NP4

2 1015

transmitted by two floor slabs (Fig. 2.13).

Select the necessary width of the wall in three different variants:

a) constant cross-section over the whole height;

b) constant cross-section over the height of each storey;

c) element of constant strength.

Select the necessary width of the foundation for each case.

Numerical data:

mc

N

mc

N

cm

daNmh

mhmhcm

daN

cbsoilf

b

44

20

2120

102,2108,131

5,2215

Fig. 2.13

)a )b )c

a

1P

2P

)(2 xA

1P

2P

a

)(1 xA

cfa

bfa

afa

2h

1h

fhmb 1

1a

2a

x

ee

ee

x

m1


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Solution:

The loads and the section being constant along the wall, the desigh is

performed for the unit length (1m)see Fig. 2.13.

a) Wall of constant cross-section over the whole height (Fig. 2.13a)The most loaded section is the lower section of the wall. The necessary area

is:

2

263

321max 3680

105,410108,115

10)1511(2)(2cm

h

PP

h

NA

bobbob

nec

cmanec 37100

36810

b) Stepped wall (Fig. 2.13b)The most loaded section of each portion is the lower one:

cma

cmh

P

h

NA

nec

bobbob

151001500

150021010108,115

101122

1

2

263

3

1

1

1

11

cmacm

h

GPP

h

NA

nec

bobbob

36100

36003600

105,210108,115

108,110200150010)1511(222

2

2

264

363

2

121

2

22

c) Wall of constant strength (Fig. 2.13c)For the upper portion, by applying relation (2.25), it results:

2101215

10108,131

1

5

63

147015

101122)( cmeee

PxA

xxx

ob

ob

b

The concentrated force at the free end of the lower portion is

121 22 GPP , so that:


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23

x

ob

b ob

b

eGPP

xA

1212

)(2)(

The own weight of the upper portion is:

151

51

0

1012

5

0

1012

0

111012

114701470)(

hx

b

h

b

x

b

h

b edxedxxAG

NeG 5340)1(10220510200123

1

By coming back to the relation for the area selection:

210121510108,13

2

5

63

3500350015

53410)1511(2)( cmeeexA xx

ab

b

The foundations:

- for variant a)wall with constant cross-sectional area:

2

63

363

0

max 1977710010102,23

108,110450368010)1511(2cm

h

GNA

fcts

b

fnec

cmaaf

198100

19777

- for variant b)stepped wall:

daNG

daNG

cmh

GGNA

b

b

fcos

bb

fnec

1620108,1102503600

540108,1102001500

1948010010102,23

162054010)1511(2

36

2

36

1

2

63

321max

cmabf

195100

19480

- for variant c)wall of constant strength:

2

63

321max 19470

10010102,23

160053410)1511(2cm

h

GGNA

fcos

bb

fnec


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daNG b 5341 , previously determined

)1(108,110123500

108,13500)(

25010123

50

31012

022

52

52

edxedxxAG

h

x

h

bb

daN1600)10305,1(10525 2

cmacf

195100

19470

Remarks:

1. By comparing the widths of the wall in the three analyzed variants, thegreatest width has been obtained for the wall with constant cross-section,

while the smallest one has been obtained for the wall of constant strength.

2. The same order has been obtain for the quantity of material used to carry outthe foundations.

3. There are small differences between variants b) and c) ,which means that thestepped wall (variant b)) practically coincides to the ideal shape (variant c).

2.2.8 A stepped bar made of steel is fixed at end A and simply supportedat end B (Fig. 2.14). Compute the stresses in the bar and the pressure

exerted on the two walls, when the temperature increasing is

Ct

00

50

.Numerical data:

mlmlcmAcmA 25,22540 122

22

1

cmcm

daNECt 1,0101,2102,1 2

6105


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Fig. 2.14

Solution:

The elongation of the stepped bar produced by the axial N, is:

2

2

1

1

EA

Nl

EA

Nll

and the elongation produced by the temperature change:

0tll tt

By equating them, it results :

2

2

1

1

0

A

l

A

l

EtlN t

(a)

For a bar with n portions, the axial force relation becomes:

n

i i

i

t

A

l

EtlN

1

0

In our case, the global free elongation caused by the temperature change is:

1A 2A

N NA B

1l 2l


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cmtll tt 27,050450102,150

In the bar axial forces occur only when its elongation exceed cm1,0 , that

is, corresponding to a temperature change:

0

1

00

1 5,3127,0

17,050

l

ltt t

By applying relation (a), the pressure exerted on the wall is:

daNN4

65

1038,2

25

250

40

200

101,25,314501021

and the stresses in the two component portions are:

24

2

2

24

1

1

/95025

1038,2

/59540

1038,2

cmdaNA

N

cmdaNA

N

x

x

2.2.9 A column made of reinforced concrete is acted by a concentriccompressive load P (Fig. 2.15). Select the necessary dimensions for

the column cross-section by using the Allowable Strength Method

and the Failure Method. Compare the obtained results.

Numerical data:

8,0

25008,1701600

4015%1104

222

2

4

D

d

cm

daNc

cm

daNR

cm

daN

cm

daN

E

E

nA

A

daNP

yscra

ac

c

r

c

r

Solution:

1. The Allowable Strength Method


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27

24

870)01,0151(40

104

)1(cm

n

PA

ac

necc

Fig. 2.15

cmA

DD

A necc

necnecc5,55

)8,01(14,3

8704

)1(

4)1(

4 222

2

cmDd necnec 5,445,558,08,0

270,887001,0 cmAA bnecr

D

d

P


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The reinforcement consists of four circular solid bars. The diameter1d of

these bars is:

cmA

dcmA

A rnec

r

r 66,114,3

18,24418,2

4

70,8

4

11

2

1

The strength requirement checking up for the reinforcement bars is not

necessary becausec

ac

r

ra

EE

2. The Failure MethodIt is considered that at the failure moment the stresses in the concrete are

uniformly distributed over the cross-section and equal toc

R while the stress in the

reinforcement bars reaches the yield stress. The equilibrium equation expressed for

the element section is:

ycccyrccs ARAARAPc

24

760250001,070

1048,1cm

R

PcA

yc

s

necc

cmdcmD necnec 5,41528,052)8,01(14,3

76042

cmdcmAcmAnecrnecr

56,19,14

6,76,776001,0 1

2

1

2

The Failure Method leads to smaller sections for concrete and

reinforcement, which means that the material carrying capacity is better used.

.2.11. The chord of a roof truss made of wood is subjected to concentric tension by

a load NP 41070 . Knowing that it is made of wood, stiffened by two channel

rolled shapes U12, symmetrically disposed and their connection is done by using


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bolts 16 , compute the stresses in wood and steel.

Fig. 2.16

Solution:

Numerical data:- area of wood section 23001520 cmAw

- area of steel section 24,282,142 cmAs

- Youngs modulus for wood 25 /10 cmdaNEw

- Youngs modulus for steel 26 /101,2 cmdaNEs

- coefficient 6,104,28

300 s

w

AA

- coefficient 0477,021

1

s

w

E

En

The axial force in steel is: Nn

PNs

44

105,460477,06,101

1070

1

cm20 cm12

cm15P


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30

and in wood: Nn

nPNw

44

105,230477,06,101

0477,06,101070

1

The equilibrium equation is satisfied lOL NNN .

The stresses in the two component materials are:

23

/17777,06,124,28

105,46cmdaN

A

N

nets

s

xs

23

/856,115300

105,23cmdaN

A

N

netw

w

xw

2.2.12 Four steel bars 12mm in diameter are stretched by using a hydraulic

machine. The tensile axial force is NN 40 1045 . These pre-stressed bars are

used to carry out an element made of reinforced concrete with the cross-section

20x20cm. Knowing that this element is finally subjected to a tensile axial force

NN 41050 , compute the stresses in concrete and reinforcement bars

corresponding to the transfer and final stages.

Fig. 2.17

NN

0N 0N

cm20

cm20

122

122

stressing-prebars

transferstresses

stagefinal

cm20

cm20


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31

Solution:

We shall compute the stresses corresponding to each stage:

1. Reinforcement bars pre-stressing- in the reinforcement bars:

254

00, /10

5,4

1045cmN

A

N

r

r

222

5,42,114,34

4cm

dAr

- in concrete: 00, c

2. Stresses transfer (the released reinforcement bars compress the concrete)- in the reinforcement bars:

2

2

240

, /14350)1012,1151(5,4

1012,1151045

)1(cmN

nA

nN

r

Tr

1201,0400

5,415

c

r

c

r

A

A

E

En

- in concrete:

2

2

40

, /962)1012,1151(400

1045

)1(cmN

nA

N

c

Tc

3. Action of load N- in the reinforcement bars:

2

2

24

, /15900)1012,1151(5,4

1012,1151050

)1(cmN

nA

nN

r

Nr

- in concrete:

2

2

4

, /1070)1012,1151(400

1050

)1(cmN

nA

N

c

Nc

4. Final stage


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The final stresses are determined by superposing the stresses corresponding

to the three previously mentioned stages:

- in the reinforcement bars:2/1015501590014350100000 cmNr

- in concrete:

2/10810709620 cmNc

Remarks:

1. Before applying the load N, the stresses in the reinforcement bars andconcrete, respectively, are:

2

2

/962

/8565014350100000

cmN

cmN

c

r

The global axial force in the reinforcement bars is:

NAN rrr 3840005,485650

and in the concrete:

NAN ccc 384000962400

The axial forces in the reinforcement bars and concrete have the same

magnitude but with opposite sign, before applying the load (they are self-

balanced).

2. The initial stresses are useful for concrete but they increase the tensilestresses in the reinforcement bars. This remark leads to the idea of using

high strength steel.


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33

3. In case of an usual element made of reinforced concrete, the stresses in theconcrete would be acNc cmdaN

2

, /108 , that is , the element could

not resist load N.

2.2.13 In the system shown in Fig. 2.18 the central bar has a clearance

cml 03,0 . A load NP4103 applies upon the rigid bar AC when the system

has been assembled. There are known: 22

1 25,2 AcmA 26

1 101,2cm

daNE

ma 3

Fig. 2.18

Solution:

In order to resolve the problem, the superposition principle is applied, taking

into account that first off all the central bar has been stretched, then, connected to

the rigid bar and finally loaded by force P.

a a

A C

P

0l

a1512 2


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34

1. Axial forces and stresses corresponding to the system assemblingThe tensile load, which must be applied on bar 1 in order to obtain a final

length equal to that of bar 2, is determined by using the relation:

daNAEl

N 315030015

25,2101,23,0

1

620

0

In this stage, the stress in the bar is:

ax

cm

daN

A

N

21

0

01400

25,2

3150

At this moment, the bar tends to return to its original length, which is

equivalent to the action of a compressive load 0N , equally distributed to the three

deformable bars due to the system symmetry.

daNN

N 10503

3150

3

01

The axial forces in the bars that correspond to the assembling stage are

presented in Fig. 2.18a.

Fig. 2.18a Fig. 2.18b Fig. 2.18c Fig. 2.18d

2. Axial forces and stresses corresponding to load P action

3000P

1100

2050

daNP 3000

3100

1000

2050

2100 501050

501000

1000

P

1050


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35

Load P will be equally resisted by the three bars (Fig. 2.18b).

daNP

N3

2 10

3

3. The final axial forces are obtained by superposing the effects from the

previous mentioned stages (Fig. 2.18.c).

The stresses in the bars are:

- in the central bar:

211380

25,2

3100

cm

daNx

- in the lateral bars:

2222

25,2

50

cm

daNx

Remarks:

1. When the central bar is longer than the lateral ones, the calculus is similar,

but the axial forces and stresses have opposite sign in the assembling stage.

In Fig. 2.18d there are given the final axial forces for the same numerical data.

The axial forces in the deformable bars of such systems depend on the magnitude

and direction of these clearances. More than that, their magnitude could be chosen

so that, in certain bars, the stresses to reach an initially imposed magnitude. These

clearances become a possibility of controlling the stresses in the bars of the system.

As an example, in case of the present problem, when cml 285,00 , in the lateral

bars, in the assembling stage, the axial force would be -1000daN and in the final

stage, 0.


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36

2.2.14 A concentrated force NP 41012 applies on a rigid element AC

(Fig. 2.19) suspended through the deformable bars 11,BBAA , made of steel and

1CC , made of copper.

Knowing the sections of the three bars 21 3cmA ,2

2 5cmA ,2

3 7cmA ,

the length l=2m, the ratio of longitudinal moduli of elasticity 7,0ST

CO

E

E, as well as

and the materials strengths2020

1000,1600cm

daN

cm

daNCOST

, it is required to

check up the strength of this system of bars.

Solution:

In the bars there are only axial forces because they are pin connected bars.

These axial forces may be rendered evident by using the sections method. The

equilibrium of the rigid element AC is expressed by only two equations (the thirdoneprojections along the horizontal direction - leading to 00 )

0230

0320

21

32

aPaNaNM

aNaPaNM

C

A

By simplifying both equations, it is obtained:

PNN 23 32 (a)

PNN 21 23 (b)


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37

Fig. 2.19

It is a system of two equation with three unknowns ),,( 321 NNN , which

shows that the system is statically indeterminate to the first degree. The redundant

equation, which is called compatibility equation, expresses the relation between the

elongations of the deformable bars and it results from the study of the deformed

shape of the system. Fig. 2.19 shows that:

'''

'''

'''

'''

CA

BA

CC

BB

or, by considering the vertical displacements (equal to the deformable bars

elongations) 321 ,, :

a

a

313

12

Finally, the following equation is obtained:

1A 1B 1C

A B C

P

''C''B

'C

'B

'A 32

3N

2N1N

a aa

l

1


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38

032 321

The elongations of the bars may be expressed by using relation (2.9):

0323

3

2

2

1

1 AE

lN

AE

lN

AE

lN

COSTST

By multiplying the equation with the product 1AEST and taking into account that:

7,0,34,23

7,67,1

3

5

1

3

1

2 ST

CO

E

E

A

A

A

A, we obtain:

061,08,12321

NNN (c)

that represents the compatibility equation. Equations (a), (b) and (c) form an

algebraic system of equations with three unknowns whose solutions are the axial

forces in the deformable bars:

PNPNPN 56,0,32,0,12,0 321

These values check the equilibrium equation of projection along the vertical

direction:

056,032,012,00 PPPPY

The strength requirement for each bar is checked up by using relations (2.13)

Bar 1AA :

22

3

11

11 1600480

3

101212,012,0

cm

daN

cm

daN

A

P

A

NSTax

Bar 1BB :

STaxcm

daN

A

P

A

N

2

3

22

22 765

5

101232,032,0

Bar 1CC :


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39

22

3

33

33 1000960

7

101256,056,0

cm

daN

cm

daN

A

P

A

NCOax

These requirements are all satisfied.

2.2.15 A rigid bar AB, supported as presented in Fig. 2.20, is loaded by two

uniformly distributed forces p and g, respectively, in intensity. . Select the

dimensions for the cross-sections of the bars 1AA , 1BB and 1CC , that are made of

steel and have a circular solid section, by using the Limit State Method. Compute

the displacement of end B.

Numerical data:

mdaNg

mdaNp

n

n

/105

/102

3

9,02,231,12,1 mmlmafgfp

23

2

1

2 21002,18,0cm

daNR

A

A

A

A

Solution:

The same procedure as at the previous problem, (2.2.14) is followed. The

axial forces in the three deformable bars and the two reactive forces (vertical and

horizontal) at support D represent the five unknowns of the problem and only three

equilibrium equations can be written. The system is statically indeterminate to the

second degree.


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40

Fig. 2.20

By removing two constraints, as example the bars 2 and 3, the system

becomes a statically determinate one (Fig. 2.21), loaded by forces p, g, 2N and

3N .

Fig. 2.21

A

BCD

1

2 3np

a a a5,1

1A

1B1C

A

D C B

aaa5,1

'A

'B'C

1N

2N3N

DV

DH

l2,1

l

ng


41/54

41

The condition of static equilibrium is expressed by the following equations:

000 AD MMX

Obviously, the first equation leads to 0DH , because all forces are vertical

ones. The third equation contains the reactive forceD

V , that is not the object of our

interest., so that we might renounce to it. Finally, the condition of static equilibrium

is reduced to a single equation:

075,15,12

5,20 321 aagaapaNaNaNMcc

D

or 062,25,05,2 321 agapNNNcc

But 25303)1051,162,2102,15,0(62,25,0 23 agap cc

And the preceding equation becomes:

25305,2 321 NNN (a)

The two redundant equations, which express the displacements

compatibility, are obtained by noticing the geometrical dependence of

displacements on the deformed shape of the system. From Fig. 2.21 it results:

a

a

BB

CC

5,2'

'

anda

a

AA

CC

'

'

But, ',' CCAA and 'BB are the elongations of bars 1,2 and 3, which are denoted by

321 ,, . By using these notations, from the previous relations we obtain:

23 5,2 and 21

or

2

2

3

3 5,2A

lN

A

lNcc

and2

2

1

1 2,1

A

lN

A

lNcc


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By multiplying both relations with 2A , it results:

ccNN 23 5,22,1 and

ccNN 2196,0 (b) (c)

Solving the equations (a), (b), (c), the axial forces are obtained:

daNN

daNN

daNN

c

c

c

727

349

363

3

2

1

The selection of bars cross-sections:

cmdcmA

A

cmdcmAA

cmdcmmRNA

c

nec

85,05775,08,0

462,0

8,0

76,0462,0385,02,12,1

7,0385,04385,021009,0

727

3

221

2

2

32

323

3

The areas of bars 1 and 2 are obtained from the initially imposed ratios, that

is why their strength requirements must be checked up.

RcmdaNAm

N

RcmdaNAm

N

ef

c

x

ef

c

x

2

2

22

2

1

11

/839462,09,0

349

/6985775,09,0

363

The selection has been correctly performed as these conditions are satisfied.

When these requirements are not satisfied, the design must be started from thestrength requirement of the bar with underestimated area.

The displacement of end B, being equal to the elongation of bar 1BB , is:

cmAE

lNc

B 198,0385,0101,2

2207276

3

3


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2.2.16 a) Select the necessary cross-sectional areas 1A and 2A and the

corresponding cross-sectional dimensions for the two portions of the member

shown in Fig. 2.22.

b) Determine the global length change of the member and the length change

for each component portion.

Fig. 2.22

Numerical data:

,11 ml ,2,12 ml ,101 KNP ,152 KNP mKNp /4 ,2/160 mmNo ,

25 /101,2 mmNE , 75.0e

i

D

D

Solution:

a) The selection of the necessary cross-sectional areas is performed by

expressing the strength requirement for each portion of the member at limit.

1

1

2

2

1P2P

1l 2l

p

11 22

iD

eDD


44/54

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For this reason the axial force diagram must be plotted, to find the most

loaded sections.

Fig. 2.23

Along domain AB, the axial force has a linear variation according to

relations (1.1).

KNPPN

KNPPlpN

bleft

A

251510

21151014

21

211

Along domain BC, the axial force diagram is a constant one:

KNPNCB 101

The strength requirement expressed for the domain AB:

0

1

maxmax

A

NAB

AB

X

For 23

0

max10max 25,156

160

1025mm

NA

AB

nec

AB

X

mmDmm

AD

DA e

nec

enec

e

nec 2232,2175,01

25,1564

1

4)1(

4 2212

2

1

The strength requirement expressed for the domain BC:

1EA2EA

KNP 152

KNP 101 p

1l 2l

21

2510

( ) N KN


45/54

45

0

2

A

NBC

BC

x

For 24

0

20 5,62160

10mm

NA

BC

nec

BC

x

mmDmmA

DD

A necnecnec 992,85,6244

4

22

2

b) Along a domain i of constant cross-sectional area, the length change is:

ii

N

i

i AEl

WhereN

i is the area of the axial force diagram afferent to the considered

portion and iiAE is the corresponding axial rigidity.

In our case:

222

2

5

33

2

22

222

2

1

5

33

1

1

1

6,63

4

9

4

9,06,63101,2

102,11010

31,16675,014

221

4

66,031,166101,22

101025212

mmD

A

mmEA

lNl

mmD

A

mmEA

lNN

l

BC

e

BleftA

The global length change of the bar is: mmlll 56,121 .

2.2.17. The stepped bar shown in (Fig. 2.24) is acted by a concentrated force P and

a temperature change 0t .


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Fig. 2.24

a) Plot the axial force diagram;b) Check the strength requirement for the considered bar.

Numerical data:

,20KNP 1010 Ct , 25 /101,2 mmNE , 20 /160 mmN , ,2002

1 mmA

22 100mmA ,

16102,12 Co

t

Solution:

a) The bar is a hyperstatic one, to the first degree, because:112)( eund si

u- number of unknowns;

e- number of equilibrium equations.

In our case there are two unknowns, the reactive forcesA

R andB

R and only one

equilibrium equation (of projection along the longitudinal axis of the member).

P

ml 5,11 ml 12

1EA2EA

0t


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Fig. 2.25

1) The equilibrium equations:0, BAxi RPRoF

An additional equation is needed and this equation is named compatibility

equation.

2) The compatibility equationA statically determined bar is adopted by removing a number of constraints

that equals the degree of indeterminacy and substituting them with the

corresponding reactive forces, which are called static redundants.

We shall remove the fixed connection from end B.

The statically determinate bar (Fig. 2.25.b) is called primary structure.

This bar must have the same deformations as the original, double fixed one:

0l - the compatibility equation

ARBR)a

)b

)c

A BP

0t

C

1l 2l

0t

P

)( BRX

18,15

82,4

( ) N KN


48/54

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By applying the superposition principle: 0 XtP llll

Fig. 2.26

The length change produced by load P (Fig. 2.26.a) is:

mmEA

Pll

P 714,0200101,2

105,110205

33

1

1

The length change (contraction) produced by the temperature change is:

mmllll t 3125,0105,210105,12)( 3621

The length change (contraction) produced by the unknown force X is:

XXAl

Al

EX

EAXl

EAXll X 5

33

5

2

2

1

1

2

2

1

1 1033,810010

200105,1

101,2

KNNXXl 82,4482001033,83125,0714,0 5

The axial force diagram is pictured in Fig. 2.25.c.

b) the strength requirement for portion 1:

P

)( BRX

P

)a

)b

X

N

N


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49

2

0

2

1

11 /160/9,75200

15180mmNmmN

A

Nx

The strength requirement for portion 2:

2

0

2

2

22 /160/2,48100

4820mmNmm

A

Nx

2.2.18. Determine the carrying capacity of the following system consisting in a rigid

bar supported by two rods, made of the same material.

Fig. 2.27

Numerical data: ma 5,0 ; 21 40mmA ; 12 5,1 AA ;25 /101,1 mmNE ;

20 /100 mmN ;

030

Solution:

The system is a statically indeterminate one to the first degree.

1N

2N

1

A

A

VR

A

HR

a

a

a5,1

p

2


50/54

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There are four unknowns:A

H

A

V RRNN ,,, 21 and three equilibrium

equations:

134)( si nd

1) The equilibrium equations:

0coscos0 21, NNRapFA

Hxi 1

0sinsin0 21, NNRFA

Vyi 2

05,1cos5,2cos30 21 aNaNapaMA 3

As we are interested only in 1N and 2N , the last equation, 3 is the single

useful one.

2. The compatibility equation

In Fig. 2.28, the deformed shape of the system is pictured.

Fig. 2.28

a

a

5,2

5,1

1

2

1l

2l

1

2


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51

But coscos

22

11

land

l

Finally 6,01

2

ll ; or 12 6,0 ll - the compatibility equation.

According to the constitutive law of the material:

sin

5,1,

sin

5,2

6,0

;

21

1

11

2

22

2

222

1

111

al

al

A

lN

A

lN

EA

lNl

EA

lNl

1

1

212

1

1

2

2

5,1

5,26,0

5,1

NA

ANN

A

aN

A

aN

By substituting the 2N in equation 3:

pN

pppa

N

aNaNpa

96,546

64,36430cos75,4

5003

cos75,4

3

05,1cos5,15,2cos3

2

01

11

2

The strength requirement expressed for the rods:

0

1

121

A

Nxx

For: 011021

ANxx

but: pN 64,3641

So that:mm

NApcap 11

64,364

10040

64,364

01


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52

2.2.19. Check up the strength requirements for the bars of the system shown in

Fig.2.29.

Fig.2.29

Numerical data:

mh 2 ; 025 ; 060 ;20

1600cm

daNsteel ; 20 80

cm

daNwood ;

2

6101,2cm

daNEsteel ; 2

510cm

daNEwood .

Solution:

The checking up of strength requirements presumes to determine firstly the

axial forces in the rods.

1

1

2 2

1NA

h

800P

11

28I

22 102U

cm10

2N


53/54

53

At node A, two equilibrium equations of projection along two orthogonal

axes are expressed.

0cossin

0sincos

0

0

21

1

,

,

NNP

NP

F

F

yi

xi

daNNPN

daNP

N

1649825cos9464860sin80000cossin

9464825sin

60cos80000

sin

cos

0

12

0

0

1

The sign of axial forces show that the rod 1 is indeed subjected to

compression (as shown in Fig. 2.29) but the rod 2 is subjected to tension (the

opposite direction then shown in Fig. 2.29).

The strength requirement for the rod 1, which has a homogeneous section is:

steelxcm

daN

A

N02

1

11 07,15491,61

94648

The road 2 has a non-homogeneous section, made of two materials that

work together.

1). The equilibrium equation

2NNN WS

SN - the axial force in steel,

WN - the axial force in wood

2). The compatibility equation:

ws ll

According to the constitutive laws for the involved materials:

ww

w

W

ss

s

s

AE

hNl

AE

hNl


54/54

So:

wwSswwss

ws

ww

w

ss

s

AEAE

N

AEAE

NN

AE

N

AE

N

2

The axial forces in the two materials result:

wwss

ss

sAEAE

AENN

2 ;

wwss

ww

wAEAE

AENN

2

The strength requirements are:

os

wwss

s

s

sxs

cm

daN

AEAE

EN

A

N

256

62 4,519

100105,132101,2

101,216498

wwss

w

w

w

xw cmdaNAEAE

EN

A

N0

2

56

52 /73,24

100105,132101,2

1016498

Concentric Tension or Compression. Problems

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Transcript of Concentric Tension or Compression. Problems