Department of Structural Mechanics
Faculty of Civil Engineering, VŠB-Technical University of Ostrava
Statics of Building Structures I., ERASMUS
Continuous beam
• Basic properties of a continuous beam
• Solution of a continuous beam by the Force method
• Symmetry of a continuous beam
2 / 40Basic properties of a continuous beam
Supports of the transversally loaded
continuous beam
Continuous beam
2) 1, (0, supports fixed ofnumber
spans ofnumber
n
:acyindetermin statical of Degree
support) (fixed
node endat rotation against b)
movement, rticalagainst ve a)
: restrained isIt
loading. sversal with tranbeamdirect ateindetermin statically a is beam Continuos
s
k
k
v
p
vp 1
3 / 40Solution of a continuous beam by the Force method
First 3 steps of the Force method in
solution of a continuous beam
Continuous beam, derivation of the “Three moments equation”
.conditions naldeformatio writing4)
supports), fixed(at reactionsor nsinteractio
moment with links removed of replacing 3)
hinges), (inserting links internal of removal 2)
acy indetermin statical of degree ofion determinat 1)
:equation moments Three derive tosteps basic method, Force The
s
s
n
n ,
4 / 40Solution of a continuous beam by the Force method
Derivation of the Three moments equation
1,1, :condition nalDeformatio rrrr
5 / 40
Calulation of rotations at the ends of beams
Simply supported beam as a member of a statically indeterminate structure
Deformations of a simply supported beams at the ends
abababbaababbaabbabababa MMMM ,,,,0,,,,,,,0,,,
:rotation clockwise For the
6 / 40
Derivation of 3moments (Clapeyron’s) equation
equation each in moments bendingunknown 3maximally are there
,acy indetermin statical of degree its toequal is beam continuous afor equations ofnumber
rotation), (clockwise #5 and #4 slideson picturesat marking toscorrespond ofsign
:Notes
:conversion andon substitutiAfter
,1rr,ba,
,1rr,ba,
,1rr,
:is 1)r (r,span right For
,1-rr,ab,
,1-rr,ab,
,1-rr,
:is r) 1,-(rspan left For
sn
rr
rrrrrrrM
rrrrrM
rrrM
rrrrM
rrrM
rrrr
rM
abM
rM
baM
ba
rrrM
rrrrM
rrrr
rM
baM
rM
abM
ab
rrrr
0,1,
00,1,0,1,1,1
)1,1,
(1,1
1,1,1,0,1,1,
1,,
,,
1,1,,10,1,1,
1,,
,,
1,1,
7 / 40Solution of a continuous beam by the Force method
Derivation of the Three moments equation
:support For
:is beam theof endright For
:is beam theof endleft For
.( is equations ofnumber
ate,indetermin statically times1)-(p is (b) and (a) pictures ofconnetion by obtained beam continuousA
:composed be illequation w 1Only
1.ancy indetermin statical of degree has (c) below picture at the beam The
0)(
0)(
0)(
)1,0
0)(
.,0,0
0,1,0,1,1,11,1,,11
0,1,0,1,1,1,,11
0,1,20,3,23,233,21,22
11
0,1,20,3,21,22,12
231
rrrrrrrrrrrrrrr
pppppppppppp
p
rs
MMMr
MM
MM
pMM
M
MMMMn
8 / 40Solution of a continuous beam by the Force method
Derivation of the 3moments equation, the beam with a cantilever
s.cantilever theof loading fromdirectly calculated becan moments Those
below) depicted case in the negative are(they
loaded are ends cantilever when zero-non are supportsouter above moments Bending
9 / 40Solution of a continuous beam by the Force method
Derivation of the 3moments equation, fixed end
0,0
0
"
0
0,2,12,10,,1,11,1
0,0,10,1
0,2,12,122,11
,12,1
pppppppppppp
pp
MM
MM
2p1,p
1,0
:(d)) (c), (picture end fixedright for Similarily
is Here
picture).in pole" nulové (ie. span" zero" called-so inserted with (b) picturefor validiscondition same The
:is (a)) (picture end fixedfor condition nalDeformatio
equation. moments Three theof solvingby calculated becan moments The
zero.-non moments bendingbut zero are , ends fixedat Rotations
10 / 40
Variable cross-section within span
0)(2
6,
6
0)(6)(2
6,
633
0)(
1,
1,
0,1,
1,
1,
0,1,
1,
1,
1
,1
,1
1,
1,
,1
,1
1
0,1,
1,
1,
0,1,
1,
1,
0,1,0,1,
1,
1,
1
,1
,1
1,
1,
,1
,1
1
1,
1,
1,
1,
11,
1,
1,
1,
0,1,0,1,1,1,11,,11
rr
rr
rr
rr
rr
rr
rr
rr
r
rr
rr
rr
rr
r
rr
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r
rr
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r
rr
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r
rr
rr
r
rr
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rr
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rr
rr
rrrrrrrrrrrrrrr
l
lZ
l
lZ
I
lM
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lM
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lM
l
IE
l
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EI
lM
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l
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l
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l
MMM
: then Z ZmarkingWhen
:conversionAfter
, ,
:spanevery in section -cross invariablefor is
equation In
1rr,1-rr,
1-rr,1rr,1-rr,1rr,
11 / 40
3moments equation – constant cross-section on all beam
Assumption:
materially and geometrically invariable cross-section
on all beam, ie. E·I = konst.
Then the Three moments equation has form:
0)(2 1,1,1,1,1,11,1,1,1 rrrrrrrrrrrrrrrrrrr lZlZlMllMlM
12 / 40Solution of a continuous beam by the Force method
Loading members
Formulas for loading members of
the Three moments equation
beam supportedsimply aon determined is
a Rotation
:loading thermaland forcefor members Loading
a,0b,b,0a,
0,,
,
,
,
0,,
,
,
,
6
6
ab
ba
ba
ab
ba
ba
ba
ba
l
IEZ
l
IEZ
13 / 40
Internal forces of the continuous beam
ba
bbaa
xbaaxx
ba
abxabxx
ba
ababababab
ba
abbabababa
ba
l
xMxlMMxVMMM
l
MMVVVV
l
MMVVVV
l
MMVVVV
MM
,
,
0,,0,
,
0,,0,
,
0,,,0,,,
,
0,,,0,,,
)(
: writtenbecan moments bendingFor
:forcesshear for can write We
. a momentsby b) beam, supportedsimply a as a)
:loaded isIt beam. continuous theofpart a is b-a beam The
14 / 40
Reactions of a continuous beam
1,1,
1
rrV
rrVrR
rR : writtenbecan r support aat reaction verticalaFor
sideright at 1rr,span theb)
sideleft at r ,-rspan thea)
:separatesr support The
15 / 40
Loading members for loading by displacement of supports
1p
1p
1
1
1,0rr,1,0rr,
: is end fixed at the 1)("support right theof rotationa clockwiseFor
: is end fixed at the "support left theofrotation clockwiseFor
:is )( nt displaceme alFor vertic
:span) 1ithin constant w(but spansin section -crossdifferent with beam continuousfor equation moments Three The
El
wwE
I
lM
I
lM
p
El
wwE
I
lM
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lM
l
ww
l
wwE
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lM
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l
ww
w
EI
lM
I
l
I
lM
I
lM
pp
pp
pp
pp
p
pp
pp
p
rr
rr
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r
rr
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r
rrrr
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r
6)(62
662
0)(6)(2
0)(6)(2
1.
1
1,
1,
1
1,
1,
1,2
12
2,1
2,1
2
2,1
2,1
1
1,
1
1,
1
1,
1,
1,
1,
,1
,1
,1
,1
1
1,
1
1,
1
0,1,0,1,
1,
1,
1,
1,
,1
,1
,1
,1
1
"
"1
16 / 40Solution of a continuous beam by the Force method
Problem definition and solution
of the example 4.1 (part one)
Example 4.1
loading Thermal
(a), picture toaccording loading Force
:definition Problem
mhmhh
mImII
6,03,0
1040,106
3,24,32,1
44
3,2
44
4,32,1
17 / 40Solution of a continuous beam by the Force method
Solution of example 4.1 (part two)
Example 4.1, displacement of supports
kPaEm,m , l,,hmI
m,lm,,m, l,h,hmII
)mm(,w), mm(w
,,
-
,
,,,,
-
,,
7
3232
44
32
43214321
44
4321
32
104,246601040
138230106
532
,
:definition Problem
18 / 40
Example 4.1, displacement of supports, solution
.10129032,1101,3
05,3,10234375,010
4,6
25,3
,10234375,0104,6
25,3,10714286,010
6,2
02
,10714286,0108,2
02
,,,,,
0)(6)(2
0)(6)(2
062
33
0,4,3
33
0,2,3
33
0,3,2
33
0,1,2
33
0,2,1
4,3
340,4,3
3,2
230,2,3
3,2
230,3,2
2,1
120,1,2
2,1
120,2,1
0,4,30,4,3
4,3
4,3
3,2
3,2
3
3,2
3,2
2
0,1,20,3,2
3,2
3,2
3
3,2
3,2
2,1
2,1
2
2,1
2,1
1
0,2,1
2,1
2,1
2
2,1
2,1
1
:onsubstitutiafter
:isIt
:equations ofn Compositio
l
ww
l
ww
l
ww
l
ww
l
ww
EI
l
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lM
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lM
EI
lM
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lM
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lM
19 / 40
Example 4.1, displacement of supports, solution continuation
kNVRkNVVR
kNVVRkNVR
kNl
MMVVkN
l
MMVV
kNl
MMVV
kNmMkNmMkNmM
MMM
MMM
MMM
MM
MMM
MM
309,4,877,4309,4568,0
,576,8568,0144,9144,9
309,41,3
0358,13568,0
4,6
722,9358,13
144,98,2
722,9881,15
,358,13,722,9,881,15
6,1958983,1353316000
2,6910716003,125337,4666
041,2202
010)234375,0129032,1(104,26)106
1,3
1040
4,6(2
1040
4,6
010)714286,0234375,0(104,261040
4,6)
1040
4,6
106
8,2(2
106
8,2
010714286,0104,26106
8,2
106
8,22
3,444,32,33
3,21,222,11
4,3
433,44,3
3,2
322,33,2
2,1
211,212
321
321
321
321
37
44342
37
4344241
37
4241
:
:
:úpravě Po
:
reactions and forcesShear
is equations 3 of system theofSolution
equations of system intoon Substituti
20 / 40Solution of a continuous beam by the Force method
Problem definition and solution
of the example 4.2
Example 4.2
21 / 40Employment of the symmetry of a continuous beam
Symmetry of geometry and supports of continuous beam
Symmetry of a continuous beam
)cantileveror fixed ended,roller or ended(pin
same are beam continuous a of endsboth - supports ofsymmetry b)
sections-cross identical andlenght identical have spans lsymmetrica -geometry ofsymmetry a)
:assumes beam continuous a ofSymmetry
spans ofnumber even b)
spans ofnumber odd a)
: withbeams lsymmetricafor symmetry of axis ofposition different is There
22 / 40Employment of the symmetry of a continuous beam
Symmetrical, antisymmetrical and
general loading
Loading of a symmetrical continuous beam
spans ofnumber even loading, ricalantisymmet AS
spans ofnumber even loading, lsymmetrica SS
spans ofnumber even loading, ricalantisymmet AL
spans ofnumber odd loading, lsymmetrica SL
:
ryantisymmetor symmetry
ofcharacter without loading c) ad
forces ofdirection oposite of pictures mirrored b) ad
forces ofdirection same of pictures mirrored a) ad
general c)
A- ricalantisymmet b)
S - lsymmetrica a)
:
onsAbbraviati
:by formed
is sidesboth on Loading
becan beam continuous lsymmetrica a of loading The
23 / 40Employment of the symmetry of a continuous beam
Utilization of a symmetry for symmetrical and antisymmetrical
loading of a continuous beam.
Symmetrical and antisymmetrical loading
2
1 , ´,:AS ,
2 ́ , ́ :AL
2
1 , ´, :SS ,
2 ́ , ́ :SL
3223322
3223322
sAS
s
sAL
s
sSS
s
sSL
s
nnMMM
nnMMMM
nnMMM
nnMMMM
24 / 40Employment of the symmetry of a continuous beam
Problem definition and solution
of the example 4.3 (part one)
Example 4.3
beam. lsymmetrica
same theof states loading ricalantisymmet
and lsymmetrica theof solutions of
ionsuperpositby given issolution Final
moments..) bending (shear, forces internal of
evaluation including ,separately solved are
beam continuous a of states loadingBoth
loading ricalantisymmet b)
loading symetrical a)
:into decomposed becan beam
continuous lsymmetrica a of Loading
25 / 40Employment of the symmetry of a continuous beam
Solution of example 4.3 (part two)
Example 4.3
Resulting diagrams of internal forces - by superposition of SL+AL
26 / 40
Donau-wald bridge, Winzer, Germany
Examples of real continuous beam structures
27 / 40
Donau-wald bridge, Winzer, Germany
Examples of real continuous beam structures
28 / 40
Bogenberg bridge, Bogen, Germany
Examples of real continuous beam structures
29 / 40
Kingstone Bridge, Glasgow, Scotland
Examples of real continuous beam structures
30 / 40
Kingstone Bridge, Glasgow, Scotland
Examples of real continuous beam structures
31 / 40
Kingstone Bridge, Glasgow, Scotland
Examples of real continuous beam structures
32 / 40
Bridge in Nusle quarter, Prague
Examples of real continuous beam structures
33 / 40
Bridge in Nusle quarter, Prague
Examples of real continuous beam structures
34 / 40
Construction of highway D47, Ostrava
Examples of real continuous beam structures
35 / 40
Construction of highway D47, Ostrava
Examples of real continuous beam structures
36 / 40
Construction of highway D47, Ostrava
Examples of real continuous beam structures
37 / 40
Energetic Research Centre, VŠB-TU Ostrava
Examples of real continuous beam structures
38 / 40
Energetic Research Centre, VŠB-TU Ostrava
Examples of real continuous beam structures
39 / 40
Energetic Research Centre, VŠB-TU Ostrava
Examples of real continuous beam structures
40 / 40
Energetic Research Centre, VŠB-TU Ostrava
Examples of real continuous beam structures