Department of Structural Mechanics
Faculty of Civil Engineering, VŠB-Technical University of Ostrava
Statics of Building Structures I., ERASMUS
Introduction to statically
indeterminate structures
2 / 58
Outline of Lecture
Outline of Lecture
• Statically indeterminate structures, degree of
indeterminacy
• Force method
• Simple statically indeterminate beam
• Simple statically indeterminate beam – axial loading
• Unilaterally fixed beam – transversal loading
• Bilaterally fixed beam – transversal loading
• Simply supported beam as an element of statically
indeterminate structure
• Simple statically indeterminate beam – torsional loading
• Calculation of deformation of the simple statically
indeterminate beam
3 / 58
• free mass point in the plane: nv=2,
coordinates [x, y], 2 different positions
• free mass point in the space:
nv=3, coordinates [x, y, z], 3 different
positions
• free solid bar in the plane:
nv=3, coordinates [x, y, ], 3 different
positions
• solid body in the space: nv=6,
coordinates [x, y, z, ], 6 different
positions
Possibility of free motion of physical objects
+x
+z
m[xm,zm]
x’
z’
Degree of freedom nv : ability to make one right-angled component of
displacement or rotation.
Statics of structures – basic knowledge
4 / 58
External constraints to movement
(a) (b) (c) (d) (e)
(f) (g)
(a) (b) (c) (d) (e)
Displacement constraint – prevents movement of constrained point in the
given direction
Statics of structures – basic knowledge
5 / 58
External constraints to rotation
Complex displacement and rotation constraints
Simple rotation constraints
(a) (b) (c)
(a) (b) (c)
Rotation constraint – prevents rotation of constrained bar in the given
direction
Full fix (clamp) in
space or plane, partial
fix in plane
Statics of structures – basic knowledge
6 / 58
Multiplicity of constraints
External constraints eliminates degrees of freedom of the object.
Name Multiplicity of
constraint
Symbol of constraint
(support), reactions
Pin, swing rod
Roller
Hinge
Sliding clamp
Fixed (clamped)
Examples of simple constraints and their reactions of solid bar in the plane
n–multiple constraint eliminates n degrees of freedom (n=1, 2, 3) of
object
a
Raz
aRaz
aRaz Rax
a Raz Rax May
a Raz May
1
2
2
3
1
Statics of structures – basic knowledge
7 / 58
Ensuring immobility of the bar
To sufficiently support some object it is needed v constraints to eliminate nv
degrees of freedom.
v = nv
Supporting of the object is kinematically determinate and
statically determinate, immobility of building is ensured,
structure can be used as building construction.
v < nvSupporting of the object is kinematically indeterminate and
statically overdeterminate, immobility of building is not
ensured, structure is inadmissible as the building construction.
(insufficient number of constrained degrees of freedom)
v > nvSupporting of the object is kinematically overdeterminate
and statically indeterminate, immobility of building is
ensured, structure can be used as building construction..
Constraints must be properly arranged to ensure immobility of structures –
it must not be Exceptional case of kinematically determinate or
overdeterminate structure.
Statics of structures – basic knowledge
8 / 58
Kinematically and statically determinate structure
Supporting of the structure is kinematically and
statically determinate
b
Rbz
a
Raz
Rax
P1 P2
Raz
RaxMay
P1 P2
a
v = nv
v = 3, nv = 3
Statics of structures – basic knowledge
9 / 58
Statically and kinematically determinate bars
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
nv = 6
nv = 3
nv = 3
nv = 2
Transversal loading
nv = 1
Axial loading
nv = 1
Torsional loading
nv = 3
nv = 3
nv = 3
nv = 3
nv = 3
nv = 6
Statics of structures – basic knowledge
10 / 58
Statically indeterminate structure
Supporting of the structure is kinematically
overdeterminate and statically indeterminate
b
Rbz
a
Raz
Rax
P1 P2
Raz
RaxMay
P1 P2
a
v = 4
nv = 3Rbx
Rbz
Rbx
Mby
b
v = 6
nv = 3
v > nv
Statics of structures – basic knowledge
11 / 58
Kinematically indeterminate structure
b
Rbz
a
Raz
P1 P2
Equilibrium of the object is ensured only for specific
loading
In building practice unusable structure
Supporting of the structure is kinematically
indeterminate and statically overdeterminate
v < nv
Statics of structures – basic knowledge
12 / 58
Exceptional cases of supporting
Constraints must be properly arranged to ensure immobility of structures
– it must not be exceptional case of supporting that is unusable in
construction practice.
b Rbxa
Raz
Rax
P1 P2
P1 P2
c
Rcz
a
Raz Rbz
b
Statics of structures – basic knowledge
13 / 58
Exceptional cases of supporting
Exceptional cases of supporting
(a)
(b)
(c)
(d)
(e)
(c) beam is not secured against rotation
(d) 3 constraints against movement intersect at one point
(e) 3 constraints against movement lying in one direct line
Statics of structures – basic knowledge
14 / 58
Equilibrium conditions of loaded bar
Number of equilibrium conditions depends on the type of solved task. It is
equal to degree of freedom of unsupported bar nv.
Supported bar has to be motionless and in equilibrium.
Number of reactions is equal to number of constrained degrees of freedom v.
v = nv Number of unknown reactions is equal to number of
equilibrium conditions, the bar is statically determinate
v < nv
v > nv
Number of unknown reactions is smaller than number of
equilibrium conditions, the bar is kinematically indeterminate
and cannot be used as building construction.
Number of unknown reactions is bigger than number of
equilibrium conditions, the bar is statically indeterminate, it can
be building construction. There must be supplemented
deformation conditions to statical equilibrium conditions.
If the determinant of system of equations is equal to zero, then it is exceptional case
of supporting of structure.
Statics of structures – basic knowledge
15 / 58
Equilibrium conditions of loaded bar system
For every separate bar can be written 3 equilibrium conditions
Number of reactions is equal to number of constrained degrees of freedom
of a structure.
Number of internal and external constraints: v = ve + vi
Statics of structures – basic knowledge
16 / 58
Kinematic and static determinacy of truss
Truss as a system of mass-points and
external and internal links
evps.2Condition of kinematical determinacy:
Practical approach – computational model is made from mass-points (at
joints) and internal links (bars), that constrain mutual displacement of both
connected joints.
Statics of structures – basic knowledge
bar (internal link) joint (mass-point)
external constraint
(support)
17 / 58
Kinematic and static determinacy
Raz
Raxa
Rbz
b
F3F2F1
N1 N5 N9
N3 N7 N11
N2 N6 N10
N4 N8
c d
e f g
s=7 number of joints (2 equilibrium conditions in each of them)
p=11 number of bars (1 unknown axial force in each of them)
a1=1
a2=1number of constraints with 1 and 2 unknown reactions
14.2.2 21 aaps
Statics of structures – basic knowledge
18 / 58
Kinematic and static determinacy
Raz
Rax
a
Rbz
b
N1
N3
N2
c d
s=4
p=5
a1=1
a2=1
8.2 s 8.2 21 aap
N4
N5
Statically and kinematically determinate truss
Kinematically indeterminate truss
F2F1
s.221 .2 aap
Statics of structures – basic knowledge
19 / 58
Kinematic and static determinacy
s=4
p=6
a1=0
a2=2
8.2 s 10.2 21 aap 2x statically indeterminate truss
Raz
Rax
a
Rbz
b
N1
N3
N2
c d
N4
N5
N6
This is not joint
(bars are
overlapping
here)
F2F1
Rbx
Statics of structures – basic knowledge
20 / 58
Calculation of the degree of static indeterminacy
Plane frame structures and beams
1. Open bar systems:
ns = v – 3 – pk = a1 + 2.a2 + 3.a3 – 3 – pk
v number of external constraints (reactions)
ai number of external constraints with 1, 2 or 3 reactions
pk number of internal hinges recalculated to simple
hinges
2. Closed bar systems:
ns = 3.u + v – 3 – pk
u number of closed areas
21 / 58
Force method
Force method
Force method: • is used for solution of statically indeterminate structures, ns ≥ 1,
• is a basic method for solution of statically indeterminace bar
structures
• is a direct method
Force method utilizes• equilibrium conditions
• deformational conditions
• principal of superposition and principal of proportionality
22 / 58
Force method
Routine of solution of structures by the Force method:
1. Determination of degree of statical indeterminacy ns
2. Creating of Basic statically determinate structure by
removing of ns constraints (external or internal)
3. Removed constraints are replaced by statically
indeterminate forces or moments (statically indeterminate
reactions or interactions)
4. Formulating of the system of ns deformational conditions
in the form of linear equations
5. Solving of the system of linear equations. The solution are
values of statically indeterminate reactions or interactions.
6. Calculation of remaining reactions and internal forces
Force method
23 / 58
Simple statically indeterminate beam
Simple statically indeterminate beam
Assumptions:
a) direct beam of constant or variable cross-section
b) axis of the beam is identical with x-axis
c) beam is supported in 2 points,
d) every of external constraints is parallel to one of
coordinate axis,
e) every of external rotational constraints works in the
plane that is perpendicular to some of the coordinate
axis
f) a beam prut can be spatially loaded
24 / 58
Simple statically indeterminate beam
Simple statically indeterminate beam
Spatial problem of simple direct beam
The degree of statical indeterminacy ns = v – nv is number
of redundant constraints, i.e. number of constraints that
has to be removed to make the beam statically determinate
25 / 58
Simple statically indeterminate beam
Simple statically indeterminate beam
Every problem of a simple statically indeterminate beam
in a space (3D) can be divided into 4 simplifier
problems:
1. Axial problem…nv=1
2. Transversal problem in xz plane…nv=2
3. Transversal problem in xy plane…nv=2
4. Torsional problem…nv=1
26 / 58
Simple statically indeterminate beam – axial loading
ll
bx
bx
vs
dxAE
dxAE
NN
RX
X
RX
nvn
0
1
0
11
11
101
10111
1
11
0
112
Silové zatížení
lll
dxA
N
Edx
AE
NNdx
AE
NN
0
0
0
10
0
010
1
Simple statically indeterminate beam – axial loading
Force method in axial problem
Silové
zatížení
Silové zatížení
Oteplení
ltdxNt t
l
t 0
0
1010
27 / 58
Simple statically indeterminate beam – axial loading
Displacement of supports
Displacement in x-axis direction
Deformational condition: 110111 dX
In this case (X1 parallel to x-axis):
ūa and Ra1 have oposite direction
ūb and X1 have same direction
b
aaaa
ud
uuuRR
1
110 )1()()(
,au bu
110
11
1
111
XRRR
Ruu
X
uuX
axaxax
bxab
ba
Force method in axial problem
28 / 58
Example 3.1
kNEA5
10681,9
)(9,2)9,1(18,4
)(9,1
)(9,13
7,5
10681,9
7,5
10681,9
2
122,1
2
11)2,48,4(
10681,9
1
10681,9
3
)(1
)(8,4
0
110
11
101
5510
3
0
05
0
010
5
0
111
1
0
10111
kNXRRR
kNR
kNXR
dxNdxAE
NN
AE
ldx
AE
NN
R
kNR
X
axaxax
bx
bx
l
l
a
a
Force loading
Simple statically indeterminate beam – axial loading
Problem definition and solution
of the example 3.1
Deformational condition
29 / 58
Example 3.1
Simple statically indeterminate beam – axial loading
Problem definition and solution
of the example 3.1
Thermal loading
The beam is loaded by change of temperature t0=15oC. The temperature is constant
along length of beam.
)(258,174 )(258,174
)(258,174)258,174(10
)(258,1743
10681,9104,5
10681,9
3
104,5315102,1
0
15 ,1021 ,10681,9
110
54
11
10
511
45
00
0
10
10111
0
0
55
kNRkNR
kNXRRR
kNR
EA
l
ltdxtN
X
C Δt,αkNEA
bxax
axaxax
bx
tt
l
-
t
Deformational condition:
30 / 58
Example 3.1, supports shifting
Ūa= 5 mm = 0,005 m (to the right)
Ūb= 8 mm = 0,008 m (to the right)
Displacement Ūa and reaction Ra1 are in opposite direction
Displacement Ūb and unit force X1=1 are in same direction
Deformational condition:
)(1,968 )(1,968
)(1,968
0
)(1,96810681,93
005,0008,0
008,0
005,0)1()()(
10681,9
3
5
11
1
1
110
511
kNRkNR
kNRR
RR
kNuu
XR
ud
uuuRR
EA
l
bxax
bxax
bxax
abbx
b
aaaax
110111 dX
Problem definition and solution
of the example 3.1
31 / 58
Transversally loaded statically indeterminate beams
110111
10111
:supports of shiftingby loading -condition nalDeformatio
0
:loading thermaland loading force -condition nalDeformatio
123 :acyindetermin statical of Degree
dX
X
nvn vs
Simple statically indeterminate beam – transversal loading
One-end fixed statically indeterminate beam
32 / 58
Bilaterally fixed beam – transversal loading
Force method - bilaterally fixed beam with
transversal loading
224vs nvn
baab MXMX
XX
XX
21
20222112
10212111
0
0
Deformational conditions for
loading by lifting of supports:
220222112
110212111
dXX
dXX
Deformational conditions for force
loading and thermal loading:
33 / 58
Example 3.8
The beam of steel cross-section I200 is loaded by:
forces according to pict. (a)
linear change of temperature t1=15o K – pict. (e)
supports shifting – pict. (h)
Deformational conditions for force and thermal loading:
0
0
20222112
10212111
XX
XX
Deformational conditions for loading by supports shifting:
220222112
110212111
dXX
dXX
Problem definition and solution
of the example 3.8
34 / 42
Example 3.8 – solution, force loading
Force loading
Basic statically determinate structure: simply supported beam
Statically indeterminate values: X1 = Mab X2 =- Mba
„0“ loading state:
Reactions Raz0, Rbz0:
Bending moment Mo:
Can be also written:
lx xxx
qMxxFM
lx x
qMxRM
MMMM
xxqxxF
xqxRxM
xx
qxRxM
kNR
kNR
F
F
qFF
qazRa
qFqRa
F
Faz
az
bz
az
for 2
)( )(
0for 2
where
M
4,8x1,6for
2
)()(
2)(
6,10for 2
)(
6,278,4/)8,06,166,1142,32,310(
288,4/)6,12,3102,31446,16(
2
220
2
11000
2001000
2
2
2
100
2
100
0
0
35 / 42
Example 3.8 – solution, force loading
beam supportedsimply of angles naldeformatio basic , , ,
8,0
6
8,0
6
8,4
6
section -crossconstant for EI
1,6
3
6,1
3
8,4
3
:tscoefficien naldeformatio ofn Calculatio
1 state loading 2nd
1 state loading1st
ba ab
0
12
21
0
21
12
0
22
22
0
11
11
2
2
1
1
EIEI
ldx
IE
MM
EIEIEI
ldx
IE
MM
EI
ldx
IE
MM
EIEIEI
ldx
IE
MM
xlM
X
xM
X
l
ba
l
ab
baab
l
ba
l
ab
36 / 42
Example 3.8 – solution, force loading
Calculation of deformational coefficients 10, 20
14044,58
30222,60
: isit case In this
2.2 Table of use with 3)
rule, sin'Vereshchag of use with 2)
y,analytical 1) :realized becan n Integratio
)(1
)(1
20
10
2000
0
0
0
20
20
1000
0
0
0
10
10
21
21
EI
EI
dxMMMMMEI
dxEI
MM
dxMMMMMEI
dxEI
MM
qFq
l
R
l
qFq
l
R
l
a
a
37 / 42
Example 3.8 – solution, force loading
Solution of linear equations is:
kNR
XRXRRR
kNR
XRXRRR
kNmX
kNmX
b
bbbb
a
aaaa
037,278,4
3244,23
8,4
0266,266,27
563,288,4
3244,23
8,4
0266,2628
:Reactions
32444,23
02667,26
22110
22110
2
1
Problem definition and solution
of the example 3.8
38 / 42
Example 3.8 – thermal loading
0)8,4
1
8,4
1(0
alsoor ,0 and . because ,0Reaction
(g). pict. see -moment bending of Diagram
4494 supplied For
0446,44,2
10216
8,06,1
10216
102162,0
15102,1
0)(
:one into reduced becan equationslinear twoofSolution
thereforel,symmetrica is problem The
15tsection -cross theacross re temperatuof changeLinear
210
2
55
1211
10
5
8,4
0
5
0
1
110
101211
21
1
XXRXRRR
VkonstMRR
kNmEIEI
kNmEI
EIEI
X
dxl
xldx
h
Mt
X
XXX
C
azazazaz
bzaz
l
t
o
Problem definition and solution
of the example 3.8
39 / 42
Example 3.8 – loading by supports shifting
(j) and (i) pictures see -Solution
144,68,4
)491,18(
8,4
)001,11(0
18,491kNm-MX -11,001kNM :is equations ofSolution
10625,41025,6004,04494
6,1
4494
8,0
1025,64494
8,0
4494
6,1
:onsubstitutiAfter
1025,68,4
006,0003,0)
8,48,4()()(
1025,68,4
006,0003,0)
8,48,4()()(
direction) same have and ( 004,0 ,0 :is system edeterminat statically basic selectedFor
:conditions nalDeformatio
22110
ab2ab1
44
20221
4
10121
4
2220
4
1110
b121
220222121
110212111
kNXRXRRR
X
dXX
dXX
wwwRwRR
wwwRwRR
Xdd
dXX
dXX
abababab
ba
bbaa
ba
bbaa
Problem definition and solution of the example 3.8
40 / 58
Table 3.2
Unilaterally fixed beam – transversal loading
Fixed-end moment reactions of
the beam with constant cross-
section
41 / 58
Simply supported beam as an element of statically indeterminate
structure
Simply supported beam as an element of statically indeterminate structure
Decomposition of simply supported beam into separate loading states
42 / 58
Simply supported beam as an element of statically indeterminate
structure – diagrams of internal forces
Simply supported beam as an element of statically indeterminate structure
Diagrams of bending moments in loading state “0” and in “moment’s” loading state
Moment’s loading state:0. loading state (q): 0. loading state (P):
44 / 58
Variable cross-section - example 3.5
kNmnxq
nx
baRM
mq
baR
nx
kNl
abMlq
abR
kNl
abMlq
baR
abMl
ql
baR
kNmlqab
M
71,162
2
545,114545.163,21
2
2
,,max
545,114
25,26,
:max
M ofn Calculatio
37,485
85,66
2
514,
2, similarly
63,215
85,66
2
514,
2,
0,
2
2, :reacions ofn Calculatio
85,6625141910,02
,
1910,0 obtained becan ion interpolatlinear by ,1855,0 ,1921,0 tableFrom
125,00,6
0,3 4,0
5
2
:3.3 Tableby Solution
15,012,0
3
3
Unilaterally fixed beam – transversal loading
45 / 58
Table 3.3
Coefficients
for moment
reaction
(uniformly
distributed
loading)
Unilaterally fixed beam – transversal loading
46 / 58
Table 3.4
Unilaterally fixed beam – transversal loading
Coefficients
for moment
reaction
(nodal force
loading)
47 / 58
Table 3.6
Unilaterally fixed beam – transversal loading
Coefficients
for moment
reaction
(uniformly
distributed
loading)
48 / 58
Table 3.7
Unilaterally fixed beam – transversal loading
Coefficients
for moment
reaction
(nodal force
loading)
49 / 58
Force method and torsional loading
Simple statically indeterminate beam – torsional loading
110111
10111
vs
dδXδ
:shifting supportsby loadingfor b)
0δXδ
:loading forcefor a)
condition nalDeformatio
112nνn
:acyindetermin of Degree
Simple statically indeterminate beam – torsional loading
50 / 58
Torsional problem – example 3.11
2
4433
9013
1075,936,024,0196,0
: tab.)see ( is stiffness Torsional
kNmIG
mhbI
t
t
h/b 1,0 1,1 1,2 1,3 1,4 1,5
0,1406 0,1540 0,1661 0,1771 0,1869 0,1958
h/b 1,6 1,7 1,8 1,9 2 3
0,2037 0,2109 0,2174 0,2233 0,2287 0,2633
Problem definition:
Reinforced-concrete beam (G=9,24.106kPa) of constant rectangular cross-
section, width b=0,24m, height h=0,36m. It is loaded by:
a) torsional loading according pict. (a)
b) supports shifting rad. 002,0 rad, 001,0 ba
Simple statically indeterminate beam – torsional loading
Problem definition and solution
of the example 3.11
51 / 58
Torsional problem – example 3.11, force loading
(d) pict. see - T moments torsionalof diagram Resulting
25,675,7114
75,74
31
312
5,13
2
5,2)914(
1
4
moments) loading tocompareddirection oposite has (X 1XMoment
0
:condition nalDeformatio
110
1
11
10
1
4
0
0
0
10
10
0
11
11
11
10111
kNmXTTT
XTkNmIG
IGX
IGIGdxT
IGdx
IG
TT
IGIG
ldx
IG
TT
X
aaa
b
t
t
ttt
l
t
l
ttt
Simple statically indeterminate beam – torsional loading
Problem definition and solution
of the example 3.11
52 / 58
Torsional problem – example 3.11, supports shifting
kNm 76,60
kNm 76,610438,4
003,0
10438,4
)001,0(002,0
rad 002,0
rad 001,0))001,0(1()(
rad/kNm10438,49013
44
44
11
1011
1
110
4
11
110111
baba
b
aa
t
TTTT
dTX
d
T
GI
dX
Supports shifting:
Deformational condition:
rad. 002,0 rad, 001,0 ba
Simple statically indeterminate beam – torsional loading
Problem definition and solution
of the example 3.11
53 / 58
Calculation of deformation of simple statically indeterminate beam
Force loading state
Reduction theorem:When calculation deformation of statically indeterminate structure of degree of indeterminacy ns than the unite force loading state can be made on:
original statically indeterminate beam
statically indeterminate beam with lower degree of indeterminacy nsj < ns, (removed less than ns links),
statically determinate beam (removed ns links).
Reduction theorem can be used to calculate the deformation of any statically indeterminate structure exposed to force loading.
Calculation of deformation of the simple statically indeterminate beam
54 / 58Calculation of deformation of the simple statically indeterminate beam
Calculation of deformation – force loading state
Examples of virtual load case options to calculate the displacement of
statically indeterminate beam.
55 / 58
Example 3.12 and its solution
Calculation of deformation – force loading state, example 3.12
2.2 Table using c)
rule sin'Vereshchag using b)
lyanalytical a)
:solved becan above formula The
1296
51
(c). pictureon as choosed state forceunit Virtual
3.2, tableusing calculated weremoments Bending
s..point in beam theofnt displaceme Calculate
3
0IE
PldxMM
IEdx
IE
MMw
ll
s
Calculation of deformation of the simple statically indeterminate beam
56 / 58
Calculation of deformation – force loading state, example 3.12
EI
PlPl
IEw
PlPl
IEw
llPlllPl
IEw
MAMAIE
w
dxMMIE
dxIE
MMw
s
s
s
TMTMs
ll
s
1296
5
12183
)149(1
)12183
7
818(
1
)23
2
2
1
254
7
4218(
1
)(1
1
33
33
2211
0
Calculation using Vereshchagin’s rule
Calculation of deformation of the simple statically indeterminate beam
Problem definition and solution
of the example 3.12
57 / 58
Calculation of deformation – thermal loading
tpr w ww
Resulting deformation is given by superposition of:
a) elastic deformation wpr
b) thermal deformation wt on basic statically determinate
structure
Calculation of deformation of the simple statically indeterminate beam
58 / 58
Example 3.13 and its solution
Calculation of deformation – thermal loading, example 3.13
mw
mdxh
Mtw
mw
EIw
ww
kNmEI
c
l
t
t
pr
pr
tpr
333
35
0
1
3
c
2
10931,010326,310395,2
10326,32
232,15
2,0
18102,1
10395,2
4494/)2
2,2135,2232,1562.4
2
8,2232,1(
1
w
4494 :Solution
The beam is loaded by change of temperature.
Calculate displacement in point c.
Calculation of deformation of the simple statically indeterminate beam