concentric tension or compression. problems
TRANSCRIPT
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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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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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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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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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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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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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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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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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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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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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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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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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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
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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
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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
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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
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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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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
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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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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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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
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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
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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