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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
1 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
Analysis of statically indeterminate structures
Structure of any type is classified as statically indeterminate when the number of
unknown reactions or internal forces exceeds the number of equations available for its
analysis.
Advantages of statically indeterminate structures
1- For a given loading the maximum stress and deflection of an indeterminate
structure are generally smaller than those of its statically determinate
counterpart.
2- Statically indeterminate structures can redistribute its load to its redundant
supports in cases where faulty design or overloading occurs.
Although from these advantages of selecting statically indeterminate structures, also;
there are some disadvantages such as deformations caused by relative support
displacement, or changes in member length caused by temperature or fabrication errors
will introduce additional stresses in the structure, which must be considered when
designing indeterminate structures.
Methods of analysis
When analyzing any indeterminate structures, it is necessary to satisfy equilibrium,
compatibility, and force-displacement requirements for the structure. Equilibrium is
satisfied when the reactive forces hold the structure at rest, and compatibility is satisfied
when the various segments of the structures fit together without intentional breaks or
overlaps. the force-displacements requirements depends upon the way of material
responds. Generally, there are two different methods to satisfy these requirements: the
force or flexibility method, and the displacement or stiffness method.
In this stage, the following methods will be studied:
1- Consistent method
2- Slope-deflection method
3- Moment distribution method
4- Stiffness matrix method
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
2 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
1-consistent method
1- Determine the degree of indeterminacy of structures.
The above structure is statically indeterminate to third degree.
2- Remove redundant to convert structure from statically indeterminate to stable
statically determinate structure , this structure is named as primary structure.
Primary structure
3- Determine internal moment at each part of primary structure (M)
4- Apply unit load in position of removed redundant, then internal moment at each
part of structure is calculated as shown in figures.
m1
1
m2
1
m3
1
X1 X2
X3
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
3 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
5- Determine redundant by solving the following simultaneous equations
β10 + π 11π₯1 + π 12π₯2 + π 13π₯3 = π ππ π ππ‘π‘πππππ‘ ππ‘ π π’πππππ‘ 1-----1
β20 + π 21π₯1 + π 22π₯2 + π 23π₯3 = π ππ πππ‘ππ‘πππππ ππ‘ π π’πππππ‘ 2---2
β30 + π 31π₯1 + π 32π₯2 + π 33π₯3 = π ππ π ππ‘π‘πππππ‘ ππ‘ π π’πππππ‘ 3---3
Where
β10= β«ππ1 ππ₯
πΈπΌ
β20= β«ππ2 ππ₯
πΈπΌ
β30= β«ππ3 ππ₯
πΈπΌ
π 11 = β«π1. π1 ππ₯
πΈπΌ
π 12 = β«π1. π2 ππ₯
πΈπΌ
π 13 = β«π1. π3 ππ₯
πΈπΌ
π 22 = β«π2. π2 ππ₯
πΈπΌ
π 33 = β«π3. π3ππ₯
πΈπΌ
S12=S21 , S31=S13, S32=S23
Applications of consistent method
1-beams
Example(1):- determine the reaction at the supports
A B L
W
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
4 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
Solution
The beam is statically indeterminate to first degree, the roller is considered as redundant
and is removed
ππππ ππ‘ πππ π‘ππππ π₯ ππππ π =ππ
πΏ
π = βπππ
πΏ.1
2.π
3= β
ππ3
6πΏ
Apply unit load at position of removed support
π1 = βπ
β10 + π 11π₯1 = π
β10= β«ππ1 ππ₯
πΈπΌ= β«
βπ€π₯3
6π
π
π
ββπ₯ππ₯
πΈπΌ=
π€π4
30πΈπΌ
π 11 = β«π1. π1 ππ₯
πΈπΌ= β« βπ₯
π
0
βπ₯ππ₯
πΈπΌ=
π3
3πΈπΌ
π€π4
30πΈπΌ+
π3
3πΈπΌπ₯1 = π
A L
W
X
M
A L
X
m1 1
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
5 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
π₯1 = βπ€π
10=
π€π
10β
After the reaction at roller is known , remaining reactions can be determined by applying
equilibrium equations.
π π¦π΄ =π€π
2β
π€π
10=
2
5π€π β
ππ΄ =π€π
10π β
π€π
2β
1
3= β
π€π2
15=
π€π2
15 πππ’ππ‘πππππππ π€ππ π
H.W: repeat previous example and choose end moment at A as redundant
Example (2): analysis the following beam
The beam is statically indeterminate to first degree due to symmetry
π =ππΏ
2π β
ππ2
2=
π
2[πΏπ β π2]
A L B
w
A L B
w M
X
X
A L B
m 1 1
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
6 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
m=1
β10= β«ππ1 ππ₯
πΈπΌ= β«
π€
2πΈπΌ[ππ₯ β π₯2]
π
0
ππ₯ =π€π3
12πΈπΌ
π 11 = β«π1. π1 ππ₯
πΈπΌ= β«
ππ₯
πΈπΌ
π
0
=π
πΈπΌ
β10 + π 11π₯1 = π
π€π3
12πΈπΌ+
π
πΈπΌπ₯1 = π
π₯1 = βπ€π2
12
Example (3):- determine the internal moments acting in the beam at support B and c .
the wall at A moves upward 30mm. take E=200GPa,I=90(106)mm4.
The structure is statically indeterminate to first degree.
MAB=0
MBC=-20X
MCD=-40X
10 m 10 m 5 m
40 kN A B C D
M M M
60
10 m 10 m 5 m
A B C D 40 kN
X X X
20
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
7 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
mAB=-X
mBC=-(10+X)+2X=-10+X
mCD=0
β10= β«ππ1 ππ₯
πΈπΌ=
1
πΈπΌ[β« β20π₯(β10 + π₯)ππ₯
10
0
] =10000
3πΈπΌππ2. π3
π 11 = β«π12 ππ₯
πΈπΌ=
1
πΈπΌ[β« π₯2ππ₯ +
10
0
β« (β10 + π₯)2ππ₯10
0
] =2000
3πΈπΌππ2. π3
β10 + π 11π₯1 = π ππ‘π‘ππππππ‘
10000
3 β 200 β 90+
2000
3 β 200 β 90π₯1 = β30 β 10
β3
π₯1 = β5.81 ππ
RA=5.81 KN upward
RB=-20+2*-5.81=-31.62=31.62 downward
RC=60-1*5.81=65.81 upward
MB=5.81*10=58.1 kN.m
MC=40*5=200 kN.m
10 m 10 m 5 m
A B C D
1
X X X
2 1
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
8 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
2-frame
Example(4):- Analysis the frame shown. There is rotational slip of 0.003 rad counter
clockwise at support B. take EI=104 kN.m2
The structure above is statically indeterminate to 2nd degree, there is part of structure is
statically determinate
4kN/m
3m
4m
4m 2m 1m 1m
2kN
4kN/m
3m
4m
4m
1kN
2kN.m
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
9 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
Primary structure
β10 + π 11π₯1 + π 12π₯2 = π
β20 + π 21π₯1 + π 22π₯2 = 0.003
β10= β«ππ1 ππ₯
πΈπΌ=
1
πΈπΌ[β« (7.6π β 1.6π2)(0.8π)ππ₯
5
0
] =53.33
πΈπΌ
4kN/m 3m
4m
4m
1kN
2kN.m
11.5
9.5
0
X
M
M=0
X
M
M=9.5(0.8x)-
4x(0.4x)=7.6x-
1.6x2
3m
4m
4m 1.75
1.75
1
X
M
M=X
X
M
M=1.75(0.8X)-
1(0.6X)=0.8X
1
3m
4m
4m 0.25
0.25
1 X
M
M=-1
X
M
M=-0.25(0.8X)=-
0.2X
0
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
10 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
β20= β«ππ1 ππ₯
πΈπΌ=
1
πΈπΌ[β« (7.6π β 1.6π2)(β0.2π)ππ₯
5
0
] =β13.33
πΈπΌ
π 11 = β«π12 ππ₯
πΈπΌ=
1
πΈπΌ[β« (0.8π)2ππ₯ +
5
0
β« (π₯)2ππ₯4
0
] =48
πΈπΌ
π 12 = β«π1. π2 ππ₯
πΈπΌ=
1
πΈπΌ(β« (0.8π)(β0.2π)ππ₯ + β« βπππ₯) =
4
0
5
0
β14.67
πΈπΌ
π 22 = β«π22 ππ₯
πΈπΌ=
1
πΈπΌ[β« (β0.2π)2ππ₯ +
5
0
β« (β1)2ππ₯4
0
] =5.67
πΈπΌ
53.33
πΈπΌ+
48
πΈπΌπ1 β
14.67
πΈπΌπ2 = 0
48π1 β 14.67π2 = β53.33 β β β β β 1
β13.33
πΈπΌβ
14.67
πΈπΌπ1 +
5.67
πΈπΌπ2 = 0.003
β14.67π1 + 5.67π2 = 43.33 β β β β β 2
Solving Eqs 1&2,gives
X1=5.85kN , X2=22.78 kN.m
Final result
4kN/m
3m
4m
4m 2m 1m 1m
2kN
1
6.96
14.04
5.85
5.85
22.78
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
11 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
Example(5):- using method of consistent deformation, analysis the frame shown
EI=104 kN.m2, settlement at support A=0.002m βand 0.003mβ , rotational slip at
A=0.003 rad counter clock wise , settlement at B=0.004m β
The structure is statically indeterminate to 1st degree, the deformations of support
A must be included by using virtual work
3m
4m
4m
A
B
3m
4m
4m
A
B
1
1
4
X
m=0.8X
X m=4
βequ
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
12 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
1π₯βππ= 4π₯0.003 β 1π₯0.002 + 0π₯0.003 = 0.01
βπ‘ππ‘ππ= 0.01 + 0.004 = 0.014π
β10 + π 11π₯1 = 0.014
β10= π ππ ππππ πππππππ
π 11 = β«π12 ππ₯
πΈπΌ=
1
πΈπΌ(β« 16 ππ₯ + β« (0.8π₯)2ππ₯) =
90.67
πΈπΌ
5
0
4
0
90.67
πΈπΌπ₯1 = 0.014
π₯1 = 1.54 β ππ
3-Arch
Example(6):- Analysis the semi-circular arch shown in figure by the method of consistent
deformation. Radius=R, EI constant
3m
4m
4m
A
B
1.54
1.54
6.16
P
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
13 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
The arch is statically indeterminate to 1st degree
β10 + π 11π₯1 = 0
β10= β«ππ1 ππ₯
πΈπΌ= 2 β«
π
2π (1 β πππ π)(βπ π πππ)π ππ
πΈπΌ
0.5π
0
=βππ 3
πΈπΌ[βπππ π β
π ππ2π
2]0
0.5π
=βππ 3
2πΈπΌ
π 11 = β«π12 ππ₯
πΈπΌ
= 2 β«(βπ π πππ)2π ππ
πΈπΌ=
2π 3
πΈπΌ
0.5π
0
[β«(1 β πππ 2π)
2ππ]
0.5π
0
=2π 3
πΈπΌ[π
2β 0.25π ππ2π]0.5π =
ππ 3
2πΈπΌ
βππ 3
2πΈπΌ+
ππ 3
2πΈπΌπ₯1 = 0
π₯1 =π
π
0.5P
P
0.5P
M
ΞΈ
π =π
2π (1 β πππ π)
m
ΞΈ
π = βπ π πππ
1 1
P
0.5P 0.5P
P/Ο P/Ο
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
14 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
4-Trusses
Same procedure for beam and frame can be followed for truss, the following equations
are used in the analysis of trusses. The truss may externally indeterminate or internally
indeterminate or both cases
β10 + π 11π₯1 + π 12π₯2 + π 13π₯3 = π
β20 + π 21π₯1 + π 22π₯2 + π 23π₯3 = π
β30 + π 31π₯1 + π 32π₯2 + π 33π₯3 = π
Where
β10= βπ1ππΏ
πΈπ΄+ β π1 β βππΏ + β π1 πππππ
β20= βπ2ππΏ
πΈπ΄+ β π2 β βππΏ + β π2 πππππ
β30= βπ3ππΏ
πΈπ΄+ β π3 β βππΏ + β π3 πππππ
π 11 = βπ1
2πΏ
πΈπ΄
π 12 = βπ1π2πΏ
πΈπ΄
π 13 = βπ1π3πΏ
πΈπ΄
π 22 = βπ2
2πΏ
πΈπ΄
π 33 = βπ3
2πΏ
πΈπ΄
π 33 = βπ2π3πΏ
πΈπ΄
S12=S21 , S31=S13, S32=S23
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
15 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
Example(7):- Analysis the truss shown. EA constant
The truss is statically indeterminate to 2nd degree; it is internally and externally
indeterminate
b+r 2j 6+4 2(4) 10 8
4m
3m
3m
A
B
D C
E
10kN
4m 10
3m
A
B
D C 13.33
10
13.33
10
13.33
10
0
-16.67
N
4m
3m
A
B
D C 1.6
0.6
0.8
0
-1.6
-0.6
0
1
1 0.6
0.8
n1
n2
4m
3m
A
B
D C 0
0
0
-0.6
-0.8
-0.6
-0.8
1 1
1
0.6
0.8
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
16 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
β10= βπ1ππΏ
πΈπ΄=
β186.66
πΈπ΄
β20= βπ2ππΏ
πΈπ΄=
β162.01
πΈπ΄
π 11 = βπ1
2πΏ
πΈπ΄=
21.32
πΈπ΄
π 12 = βπ1π2πΏ
πΈπ΄=
11.2
πΈπ΄
π 22 = βπ2
2πΏ
πΈπ΄=
17.28
πΈπ΄
21.32π₯1 + 11.2π₯2 = 186.66 β β β β1
11.2π₯1 + 17.28π₯2 = 162.01 β β β β2
π₯1 = 5.8 , π₯2 = 5.6
Force in each member=N+n1X1+n2X2
member L(m) N n1 n2 N n1 L N n2 L n12 L n22 L n1 n2 L DC 4 13.33 -1.6 -0.8 -85.312 -42.66 10.24 2.56 5.12
AB 4 0 0 -0.8 0 0 0 2.56 0 AD 3 10 -0.6 -0.6 -18 -18 1.08 1.08 1.08
BC 3 10 0 -0.6 0 -18 0 1.08 0
AC 5 -16.67 1 1 -83.35 -83.35 5 5 5 EC 5 0 1 0 0 0 5 0 0
DB 5 0 0 1 0 0 0 5 0 β -
186.66 -162.01 21.32 17.28 11.2
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
17 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
member F
DC -0.43 AB -4.48
AD 3.16
BC 6.64 AC -5.27
EC 5.8 DB 5.6
Example(8):- repeat example 7, if member BC subjected to rise of temperature 40o take
Ξ±=12(10-6) and member DC has error of 1cm too short.
β10= βπ1ππΏ
πΈπ΄+ β π1 β βππΏ + β π1 πππππ
β10=β186.66
πΈπ΄+ 0π₯40π₯3π₯12πΈ β 6 β 1.6π₯(β0.01) =
β186.66
πΈπ΄+ 0.016
β20= βπ2ππΏ
πΈπ΄+ β π2 β βππΏ + β π2 πππππ
β20=β162.01
πΈπ΄β 0.6π₯40π₯3π₯12πΈ β 6 β 0.8π₯(β0.01) =
β162.01
πΈπ΄+ 0.00713
Take EA=105
21.32π₯1 + 11.2π₯2 = 186.66 β 0.016π₯10πΈ5 β β β β1
11.2π₯1 + 17.28π₯2 = 162.01 β 0.00713π₯10πΈ5 β β β β2
X1=-75.11 X2=16.801
Then find the force in each bar by same procedure of Example 7
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
18 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
5-composite structure
Example(8):- Analysis the composite structure shown in figure. The beam has moment
of inertia of 240x106mm4,the member AB&BC each have a cross-sectional area 0f 1250
mm2, and BD has across-sectional area 0f 2500 mm2. Take E=200 GPa.
The structure is statically indeterminate to 1st degree
A
B
D C
0.9m
0.9m
1.2m 150kN
A
B
D C
0.9m
0.9m
1.2m 150kN
200kN 350kN
M=-150x
x x
M=-200x 0 0
A
B
D C
0.9m 1.2m
0.00033
m=0.53x
x x
m=0.7067x
1
0.707
0.707
0.707
0.53
0.884
1.24
0.53
0.707
0.707 0.707
1.24
0.00033
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
19 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT
β10= β«ππ1 ππ₯
πΈπΌ+ β
π1ππΏ
πΈπ΄
=1
200π₯240[β« (β150π)(0.53π)ππ + (β200π)(0.7067π)ππ]
1.2
0
= β1.67π₯10β3
π 11 = β«π12 ππ₯
πΈπΌ
+ βπ1
2πΏ
πΈπ΄
=1
200π₯240[β« (0.53π)2ππ + (0.7067π)2ππ] +
0.8842π₯1.5
200π₯1250+
1.242π₯0.9
200π₯2500
1.2
0
+12π₯1.27
200π₯1250= 1.84π₯10β5
β10 + π 11π₯1 = 0
π₯1 = 90.76 π
FAB=90.76T
FBD=1.24x90.76=112.5 C
FBC=0.884x90.76=80.2 T
A
B
D C
0.9m
-112.5
1.2m 150kN
90.76 80.2
200.03 350.03
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THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM
20 WASSIT UNIVERSITY β ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT