Deformations of statically determinate bar...
Transcript of Deformations of statically determinate bar...
Department of Structural Mechanics
Faculty of Civil Engineering, VŠB-Technical University of Ostrava
Statics of Building Structures I., ERASMUS
Deformations of statically
determinate bar structures
2 / 62
Outline of Lecture
Outline of Lecture
• Term „deformation“
• Virtual works principle
• Deformation of bar – axial loading
• Deformation of bar – transversal loading
• Deformation of bar – torsional loading
• Deformation of indirect bar
• Deformation of curved bar
• Deformation of plane truss structure
3 / 62
Deformation
Term „deformation“
Deformation:
a) Global deformation of structure
b) Local component of deformation in some point (displacement, rotation)
4 / 62
Deformation
Why to calculate deformations?
1. Usability of structure
2. Solution of statically indeterminate structures
3. Verifying the correctness of the calculation by measurement
Calculation assumptions:
Physical linearity (Hooke's law applies)
Geometric linearity (small deformations theory)
Consequence:
Equilibrium conditions are formulated on the deformed structure –
The First order Theory
Apply the principle of superposition and the principle of
proportionality
Term „deformation“
5 / 62
Deformation
Nonlinear mechanics:
2nd order theory – equilibrium conditions formulated on deformed
structure (small deformations)
Physical nonlinearity (nonlinearly elastic or permanent deformations)
Theory of big deformations
Structures with unilateral links
Cable structures
Term „deformation“
6 / 62
Work of external forces and moments
Virtual works principle
Work of point force and point moment
cos ce PPLWork (external) of a force at point:
Work - scalar, units are Joules (J = N.m), kJ, MJ
.MLe Work of a moment at point:
Notice:
It is assumption that () has
other cause than P (M).
The work is positive when there are same
directions of:
vectors of force and displacement ,
moment and rotation .
7 / 62
Work of continuous force and moment loading
Work of continuous loading
( ) ( )
b
a
e xxwxqL d ( ) ( )
b
a
xe xxxmL d
Assumption – magnitude of loading is constant during movement.
Work of external forces and moments:
Virtual works principle
8 / 62
Virtual work
a) Deformational virtual work
b) Force virtual work
1) Real loading state
2) Virtual loading state
2a) Deformational virtual state
2b) Force virtual state
ce
ce
wPL
wPL
Deformational virtual work described by Lagrange
to study equilibrium of structures
Virtual works principle
real loading statereal deflection curve
virtual deflection curveforce virtual
loading state
9 / 62
Work of internal forces
Coordinate system of the bar
Loaded bar in a space: N, My, Mz, Vz, Vy, T
Virtual works principle
10 / 62
Work of internal forces
Work of internal forces of bar
l
x
l
y
l
z
l
zz
l
yy
l
i TvVwVMMuNL dˆdˆdddd
Positive directions of internal forces
Work of internal forces:
Internal forces restrain deformations, they have opposite direction compared
to picture below, that is reason for negative sign in calculation of Li.
Virtual works principle
11 / 62
Virtual works principle
0 ie LL
Axiom:
Total virtual work on solved structure (i.e. sum of works of
external and internal forces) is equal to zero.
A) Deformational principle of virtual works (principle of virtual displacements)
B) Force principle of virtual works (principle of virtual forces)
Virtual internal forces
Real internal forces, causes deformations
xEA
Nu dd
xGA
Vw
z
z dˆd*
xEI
M
y
y
y dd
xGA
Vv
y
ydˆd
*
xEI
M
z
zz dd
xGI
T
t
x dd
TVVMMN yzzy ,,,,,
Virtual works principle
12 / 62
Deformational loading caused by temperature
Uniform thermal loading and decomposition of linearly changing thermal loading
across cross-section
Force principal of virtual works
l
tztyt
l
ty
yy
z
zz
z
zz
y
yy
e xb
tM
h
tMtNx
GI
TT
GA
VV
GA
VV
EI
MM
EI
MM
EA
NNL
0
210
0
**dd
dxtdu
h
etttt
t
zhdh
0
0 )(
h
dxtd
ttt
ty
hd
1
1
Virtual works principle
13 / 62
Betti's theorem (1872)
Enrico Betti
(1823 - 1892)
The work done by the 1st loading
state through the displacements
produced by the 2nd loading
state is equal to the work done
by the 2nd loading state the
displacements produced by the
2nd loading state.
l
y
yyx
EI
MMPP
0
II,I,
2211 d
l
y
yyx
EI
MMMP
0
I,II,
4433 d
44332211 MPPP
Virtual works principle
1st loading state
2nd loading state
14 / 62
Maxwell’s theorem
Zvláštní případ Bettiho věty, kdy v každém z obou zatěžovacích
stavů působí na konstrukci jediná síla P nebo jediný moment M.
James Clerk
Maxwell
(1831 - 1879)
Displacement done by the first force in the place and direction of
second force is equal to displacement done by second force in the
place and direction of the first force.
IIIIII PP PPP III III
A special case of Betti's theorem. In each loading state acts only
one force P or moment M.
Virtual works principle
1st state
2nd state
15 / 62
Unit force method
Unit force method
.1eL
l
ty
yy
z
zz
z
zz
y
yyx
GI
TT
GA
VV
GA
VV
EI
MM
EI
MM
EA
NN
0
**d
l
tztyt xb
tM
h
tMtN
0
210 d
Force loading
Thermal loading
Virtual works principle
16 / 62
Deformation of bar – axial loading
Deformation of bar – axial loading
Deformation of bar exposed to axial loading
l
e xA
NN
Eu
0
d1
Nt
l
te AtxNtu 0
0
0 d
Force loading
Thermal loading
EA
AxNN
EAu N
l
e
0
d1
Constant cross-section
Variable cross-section
Simpson’s rule( ) ( )
324d 42310
0
dfffffxxf
l
17 / 62
Example 2.1
xN
R
R
ax
ax
.4,813
kN13
085,2.4,8
Deformation of bar – axial loading
Problem definition and solution of example 2.1
A = 64 mm2,
E = 2,1.108 kPa, t = 1,2.10-5K-1
Calculate horizontal displacement uc
for force and thermal loading state
Force loading state:
m000685,010.4,6.10.1,2
2,9
d
58
0
l
Nc
EA
Ax
EA
NNu
18 / 62
Example 2.1
Deformation of bar – axial loading
Problem definition and solution of example 2.1
mm48,0m00048,02).20.(10.2,1
dd
5
0
0
0
0
0
c
Nt
l
t
l
tc
u
AtxNtxtNu
Displacement caused by thermal
loading:
19 / 62
Example 2.2
Deformation of bar – axial loading
Problem definition and solution of example 2.2
Concrete
r = 2400 kg.m-3
E = 2.107 kPa
Calculate vertical
displacement wb of
top of column for
loading by self-weight
20 / 62
Example 2.2
Deformation of bar – axial loading
Problem definition and solution of example 2.2
E
Zz
z
zz
E
zA
N
Ez
EA
NNw
i
ni
i i
ii
b
1
2
4
0
4
0
.2,08,0
.4,2.2,191
d1
d
2.4,2.2,19
2)..8,42,192,19()(
zz
zzzN
zz
zA .2,08,0)4
.8,08,0.(1)(
33kNm24Nm240002400.10
z
zAzn
.8,42,19
24)..2,08,0()(
21 / 62
Example 2.2
Solution using: 1) Simpson’s rule
1
4
0
kNm895,1543
)7240.2)571,56600,21.(40(d
z
A
N
Deformation of bar – axial loading
i z A N N/A
m m2 kN kNm-2
0 0 0,8 0 0,.0000
1 1 1 -21,6 -21,.6000
2 2 1,2 -48 -40,0000
3 3 1,4 -79,2 -56,5714
4 4 1,6 -115,2 -72,0000
3)2)(4(d)( 4231
4
0
0
dfffffxxf 1
4
4d
mm007745,0m10.745,710.2
895,154 6
7
bw
22 / 62
Example 2.2
Solution using:
m10.749,710.2
9756,154
4,010
4
10
6
7
E
Zw
n
lzz
n
b
i
Deformation of bar – axial loading
2) Rectangular method
(numerical integration)
i zi Ni /Ai
m kNm-2
1 0,2 1,874286
2 0,6 5,384348
3 1,0 8,64
4 1,4 11,69778
5 1,8 14,59862
6 2,2 17,3729
7 2,6 20,04364
8 3,0 22,62857
9 3,4 25,14162
10 3,8 27,59385
S Ni /Ai 154,9756
23 / 62
Axially loaded bar – column
Structure with axially loaded bar
Graded cross-section of column on tall building, Chicago, USA
24 / 62
Deformation of bar – transversal loading
Deformation of direct beam – transversal loading
Types of transversally loaded direct beams
ll
xA
VV
Gx
I
MM
E0
*
0
d1
d1
l
t xM
t0
1 dh
Force loading
Thermal loading
ll
xVVGA
xMMEI
0
*
0
d1
d1
Constant
cross-section
25 / 62
Vereshchagin’s rule
TM
l
MAxMM .d0
Aid for computation of the integral
Deformation of direct beam – transversal loading
26 / 62
Vereshchagin’s rule
Parabolic parts of moment diagrams for use within Vereshchagin’s rule
Deformation of direct beam – transversal loading
27 / 62
Example 2.3
Problem definition and solution of example 2.3
Reinforced concrete cantilever
E = 2,2.107 kPa
Calculate vertical
deflection = wa. Use
Vereshchagin’s rule.
Neglect the work of
shear forces.
Deformation of direct beam – transversal loading
28 / 62
Example 2.3
Problem definition and solution of example 2.33
3
1
0
33
3
2
2
1
22
3
1
1
0
11
3
321
2
0
23
37
3
kNm667,21)667,1.(13
d
kNm15)5,1.(10
d
kNm5,2)75,0.(333,3
d
m005407,010.24416,7
667,21155,2
)(1
d
kNm1024416,7
12
28,0.18,0.10.2,2
12
MAxMMS
MAxMMS
MAxMMS
SSSEI
xEI
MMw
bhEEI
M
M
M
a
Deformation of direct beam – transversal loading
29 / 62
Example 2.4
Real and virtual shear forces in example 2.4
Reinforced concrete cantilever
G = 9,24.106 kPa
The same problem as in Example 2.3 but the
work of shear forces is taken into account
00
´
5
5
´
2,2
1,1
56*
2*
21*
2
0
*
´
7,1100.407,5
093,0100.
mm093,0m10.276,910.8808,3
2610
kNm26)1.(1.26
kNm10)1.(2
1.20
kN10.8808,3042,0.10.24,9
m042,02,1
28,0.18,0
)(1
w
w
w
VAS
VAS
GA
bhA
SSGA
dxGA
VVw
c
c
V
V
c
Deformation of direct beam – transversal loading
30 / 62
Table 2.2
Equations for calculation of
integrall
xMM0
d
Deformation of direct beam – transversal loading
31 / 62
Example 2.5
Problem definition and solution of example 2.5
Calculate vertical
displacement wc
and rotation a
Wood
E = 107 kPa
Deformation of direct beam – transversal loading
32 / 62
Table 2.3
Local deformations of a cantilever beam and simply supported beam
Deformation of direct beam – transversal loading
33 / 62
Example 2.6
Problem definition and solution of example 2.6
M
t
l
t Ah
txM
h
t 1
0
1 d
Steel t = 1,2.10-5 K-1
Thermal loading – the change of temperature is linear along height of the cross-section
Calculate deflections wc a ws.
h = 0,24 m
Deformation of direct beam – transversal loading
34 / 62
Example 2.6
Problem definition and solution of example 2.6
cM
tttc A
h
txM
h
tx
h
Mtw 1
9
0
9
0
11 dd
92
2.9
cMA
mm2,7m0072,024,0
)9.(16.10.2,15
cw
sM
ts A
h
tw 1
125,6
2
75,1.7
sMA mm9,4m0049,0
24,0
125,6.16.10.2,15
sw
Deformation of direct beam – transversal loading
35 / 62
Example 2.6
Problem definition and solution of example 2.7
Shape, loading – see Example 2.3
Variable cross-section
Reinforced concrete cantilever
E = 2,2.107 kPa
Deformation of direct beam – transversal loading
36 / 62
Research energetic center, VŠB-TU Ostrava
Example of the structure with variable cross-section
Cantilever beam:
• Steel welded and rolled I- profile
• Trapeze metal plate
• Concrete floor
37 / 62
Research energetic center, VŠB-TU Ostrava
Example of the structure with variable cross-section
Cantilever beam:
• Steel welded and rolled I- profile
• Trapeze metal plate
• Concrete floor
38 / 62
Deformation of bar – torsional loading
t
T
l
t
l
t
cGI
AxTT
GIx
GI
TT
00
d1
d
Torsional rotation
Force virtual state
Deformation of direct beam – torsional loading
39 / 62
Example 2.8
Determine torsional rotation of
the right end b. Use unit force
method.
Steel - G = 8,1.107 kPa
o
2
00
277
4544
4
1
4
2
20,2rad0384,08466,60
336,2
kNm336,26,0.72,0.2
11).52,172,2.(
2
1
d1
d
kNm8466,6010.5119,7.10.1,8
mm10.5119,7)2430.(2
)(2
b
T
t
T
l
t
l
t
t
pt
A
GI
AxTT
GIx
GI
TT
GI
rrII
Deformation of direct beam – torsional loading
Problem definition and solution of example 2.7
40 / 62
Deformation of polyline beam – plane loading
Deformation of polyline beam – plane loading
m
j
l
j
j
l
j
j
l
j
j
jjj
xA
VV
Gx
I
MM
Ex
A
NN
E1 0
*
00
d1
d1
d1
3 local components of deformation: u, v a
m
j
l
j
j
j
xI
MM
E 1 0
d1
At most of statically determinate cases the work of shear and normal forces
is neglected
22
ccc uw At the point cc
c
w
utan
Thermal
loading
m
j
l
j
j
j
l
jjt
jj
xh
MtxNt
1 0
,1
0
,0 dd
Constant
cross-section
m
j
l
j
j
j
xMMIE 1 0
d11
41 / 62
Example 2.9
Problem definition and solution of example 2.9
Calculate
ud , wd , , d
Steel
I1 = 16.10-5 m4
I2 = 3,8.10-5 m4
I3 = 9,2.10-5 m4
E = 2,1.108 kPa
Deformation of polyline beam – plane loading
42 / 62
Steel frame structure of industrial hall
Examples of structures of polyline beams
Span 20,5 m
43 / 62
Industrial hall, Vítkovice
• Floor plan dimensions 130 x 320 m
• Cranes capacity 80 a 200 t
• Undermined region
Examples of structures of polyline beams
44 / 62
Multipurpose hall, Frýdek - Místek
Examples of structures of polyline beams
45 / 62
Multipurpose hall, Frýdek - Místek
Rámová ocelová konstrukce
Examples of structures of polyline beams
46 / 62
Platform of football stadium, Ostrava Bazaly
• Undermined region
Examples of structures of polyline beams
47 / 62
Platform of football stadium, Ostrava Bazaly
Detail of moment hinge
Examples of structures of polyline beams
48 / 62
Deformation of curved beam – plane loading
Deformation of curved beam – plane loading
Shape and supporting of a plane curved beam
Span l, deflection f, relative deflection Fl
fΦ
49 / 62
Deformation of curved beam – plane loading
Deflection and relative deflection on different curved beams.
l
fΦ
Deformation of curved beam – plane loading
Span l, deflection f, relative deflection F
50 / 62
Deformation of curved beam – plane loading
Force loading
L jLL
sGA
VVs
EI
MMs
EA
NNddd
*
Thermal loading
L
t
L
t sh
MtsNt dd 10
cos
dd
xs Solution After modification:
Force loading b
a
b
a
b
a
x
x
x
x
x
x
xA
VV
Gx
I
MM
Ex
A
NN
Ed
cos
1d
cos
1d
cos
1*
Thermal loading
b
a
b
a
x
x
t
x
x
t xh
Mtx
Nt d
cosd
cos10
Use of Unit force method
Deformation of curved beam – plane loading
51 / 62
Deformation of curved beam – plane loading
Calculation of deformation
Numerical integration
Simpson’s rule
Rectangular method
3))...(2)...(4(d)( 2421310
0
dffffffffxxf nnn
l
n
i
i
i
iin
i
i
i
ii
n
i
i
ii
iin
i
i
ii
ii
x
x
x
x
sI
MM
Es
A
NN
E
xI
MM
Ex
A
NN
Ex
I
MM
Ex
A
NN
E
b
a
b
a
11
11
11
cos
1
cos
1d
cos
1d
cos
1
Deformation of curved beam – plane loading
52 / 62
Example 2.10
Problem definition and solution of example 2.10
( ) 2.xkxz
22
b
b
a
a
x
z
x
zk
Parabolic central line
xkxkx
z..2.
d
dtg
2
2tg1
1cos
2tg1
tgsin
Calculate ub
EI = 6,72.104 kNm2
Deformation of curved beam – plane loading
53 / 62
Example 2.10
Problem definition and
solution of example 2.10
i x
[m]
tg
ψ
cos ψ M [kNm] [m] M/ cos ψ
[kNm2]
0 -
5,00
-
0,8
0,7808
7
0,0000 0,000 0,000
1 -
3,75
-
0,6
0,8574
9
28,4375 0,875 29,018
2 -
,2,5
0
-
0,4
0,9284
8
47,5000 1,500 76,739
3 -
1,25
-
0,2
0,9805
8
57,1875 1,875 109,350
4 0,00 0,0 1,0000
0
57,5000 2,000 115,000
5 1,25 0,2 0,9805
8
43,1250 1,875 82,461
6 2,50 0,4 0,9284
8
28,7500 1,500 46,447
7 3,75 0,6 0,8574
9
14,3750 1,875 14,668
8 5,00 0,8 0,7808
7
0,0000 0,000 0,000
Deformation of curved beam – plane loading
54 / 62
Curved beam
Gateway Arch, span of the steel arch from the year 1966 is 192,5 m, Saint Louis, Missouri.
Examples of planary loaded curved structures
55 / 62
Gateway Arch, rozpětí a vzepětí ocelového oblouku z roku 1966 192,5 m, Saint Louis, Missouri.
Curved beam
Examples of planary loaded curved structures
56 / 62
Curved beam
Rovinně zakřivený vazník, Výzkumné energetické centrum VŠB-TU Ostrava
Examples of planary loaded curved structures
57 / 62
Deformation of plane truss structure
Deformation of plane truss
p
j j
jjjp
j
l
j
j
jjp
j
l
j
j
jj
A
lNN
Ex
A
NN
Ex
EA
NN jj
11 01 0
.1d
1d
Thermal loading
Virtual work of Normal forces only
p
j
jjt
p
j
l
jjt
p
j
l
jjt ltNxtNxtN
jj
1
,0
1 0
,0
1 0
,0 dd
58 / 62
Example 2.11
Problem definition and solution of example 2.11
Calculate wc
A1 = 24.10-4 m4
A2 = 12.10-4 m4
A3 = 18.10-4 m4
A4 = 18.10-4 m4
A5 = 12.10-4 m4
A6 = 12.10-4 m4
A7 = 18.10-4 m4
l2 = l3 = l6 = 2,236 m
Calculation in table
Deformation of plane truss
59 / 62
Example 2.11 Calculation in table
mm62,5m1062,5101,2
10192,11801 3
8
37
1
j j
jjj
cA
lNN
Ew
j Aj [m2] lj [m] Nj [kN] Ñj [1] (NjÑjlj/Aj).10-3 [kN/m]
1 0,0024 2,000 -90,000 -1,000 75,000
2 0,0012 2,236 134,164 2,236 559,017
3 0,0018 2,236 -67,082 0,000 0,000
4 0,0018 2,000 -60,000 -2,000 133,333
5 0,0012 1,000 0,000 0,000 0,000
6 0,0012 2,236 67,082 2,236 279,508
7 0,0018 2,000 -60,000 -2,000 133,333
1180,192
Deformation of plane truss
60 / 62
Railway bridge, Polanka’s conjunction
from 1964
Examples of plane truss structures
61 / 62
Railway bridge, Polanka’s conjunction
Examples of plane truss structures
62 / 62
Road bridge, Ostrava - Hrabová
Ukázky kloubových příhradových konstrukcí
Truss bridge over Ostravice river