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INDETERMINATE STRUCTURESMETHOD OF CONSISTENT DEFORMATIONS
(FORCE METHOD)
• If all the support reactions and internal forces (M, Q, and N) can not be determined by using equilibrium equations only, the structure will be referred as STATICALLY INDETERMINATE. A statically indeterminate structure is also referred as a REDUNDANT STRUCTURE, because of redundant reaction components (or redundant members in trusses) which are not necessary for stability considerations.
Statically determinate
doi=0
Statically indeterminate to the first degree
doi=1
Statically indeterminate to the third degree
doi=3
Additional equations to solve statically indeterminate structures come from prescribed conditions of translations and rotations, commonly called conditions of compatibility or consistent displacements. A statically determinate structure which can be obtained by removing enough reaction force possesses no redundant, and will be referred as primary structure. All methods used to analyze indeterminate structures employ equations that relate the forces acting on the structure to the deformations of the structure. If these equations are formed so that the deformations are expressed in terms of the forces, then the forces become independent variables or unknowns in the analysis. Methods of this type are referred to as force methods.
Statically indeterminate structures may be considered as a primary structure with redundant forces. Redundant forces will have values so that the deformations of the primary structure must be equal to the deformations of the actual statically indeterminate structure. Using the principle of superposition we may divide the loading into several cases. At each case only one load will be considered.
ACTUAL STRUCTUREq
R1
R2 R3
q
PRIMARY STRUCTURE WITH REDUNDANT
FORCES
R1
R2 R3
q
33
22
11
0
.
RM
RM
RM
M
M
+
+
+
=q
R1 x1 kN.m
R2 x
1 kN
R3 x
1 kN
Bending moment at an arbitrary section of the actual structure (M) can be written as
..... 3322110 ++++= RMRMRMMMWhere
M0 = Bending moment at the corresponding section of the primary structure due to the external forces
M1 = Moment at the same section of the primary structure due to a unit load in the direction of the first redundant
M2 = Moment at the same section of the primary structure due to a unit load in the direction of the second redundant
M3 = Moment at the same section of the primary structure due to a unit load in the direction of the third redundant
R1 = Value of the first redundant force
R2 = Value of the second redundant force
R3 = Value of the third redundant force
Most of the deflection for beams and frames due to bending moment, for trusses due to axial force. Castigliano’sfirst theorem states that the partial derivative of the strain energy with respect to one of the forces on the structure will give the deflection in the direction of that force.
)(20
2
i
L
iii
dxEIM
PPU
∆=∂∂
=∂∂
∫ δ
To reduce the mathematical work we can change the order of the signs.
i
L
i
L
ii EIdxM
PMdx
EIM
PPU δ=
∂∂
=
∂∂
=∂∂
∫∫0
2
0 2
The bending moment is a function of the redundant forces. If we take the partial derivatives of the strain energy with respect to the redundant forces and equate them to the deflections in the direction of the redundant forces.
..... 3322110 ++++= RMRMRMMM
33
22
11
............. MRMM
RMM
RMM
RM
ii
=∂∂
=∂∂
=∂∂
=∂∂
( ) i
L
i
L
ii EIdxRMRMRMMM
EIdxM
RM
RU δ=++++=
∂∂
=∂∂
∫∫0
33221100
...
A set of linear equation will be obtained to determine the redundant forces.
=++++→∂∂
∫∫∫∫ ntrtSettlemeGivenSuppodx
EIMMRdx
EIMMRdx
EIMMRdx
EIMM
RU LLLL 0
......0
31
03
212
0
111
0
01
1
=++++→∂∂
∫∫∫∫ GSSdx
EIMMRdx
EIMMRdx
EIMMRdx
EIMM
RU LLLL 0
......0
32
03
222
0
121
0
02
2
=++++→∂∂
∫∫∫∫ GSSdx
EIMMRdx
EIMMRdx
EIMMRdx
EIMM
RU LLLL 0
......0
33
03
232
0
131
0
03
3
…………
…………
We can write the equations in the following form
=++++GSS
RRR0
...31321211110 δδδδ
=++++GSS
RRR0
...32322212120 δδδδ
=++++GSS
RRR0
...33323213130 δδδδ
ELASTIC EQUATIONS OF THE STRUCTURE
This set of linear equation is referred as Elastic Equations of the structure. Each equation will give the deflection in the direction of the redundant due to external loading and redundant forces respectively. The right hand side of the equation will be either zero or given support settlement.
∫=L
jiij dx
EIMM
0
δ Deflection in the direction of the ith redundant due to a unit load in the direction of the jth redundant
Steps in the analysis of a statically indeterminate structure by the force method may be distinguished as follows
- Determine the number of redundant forces (the degree of indeterminacy doi)
- Remove enough redundant forces to obtain primary structure
- Calculate the coefficients of the elastic equations
- Solve the redundant forces
- Use equilibrium equations to determine remaining reaction forces
- Draw the diagrams
Example : Draw the shear force and moment diagrams of the propped cantilever beam
q
L
EI
Primary StructureVertical reaction at roller support taken as a redundant force
qLR
REIL
EIqL
LLLLikLEI
qLLqLLikLEI
R
83
038
3.
33
8244
0
1
1
34
3
11
42
10
11110
=
=+−
===
−=−=−=
=+
δ
δ
δδ
8
2qLq
1
0
M
MqL2
____
2
-qL85 qL
83
qL83
Shear Force Diagram
+
+
-
-
L85
2
1289 qL
1 kN qL85
+L
8
2qL
Bending Moment Diagram
Example : Draw the shear force and moment diagrams of the propped cantilever beamPrimary Structureq
L
EI
Bending moment at fixed end support taken as a redundant force
q
-
1
0
M
M8
2qL+
8
0324
31.1
33
24833
0
2
1
1
3
11
32
10
11110
qLR
REIL
EIqL
LLikLEI
qLqLLikLEI
R
=
=+−
===
−=−==
=+
δ
δ
δδ
q
8
2qL
qL85 qL
83
qL83
Shear Force Diagram
+
+
-
-
L85
2
1289 qL
1
qL85
1
8
2qL
Bending Moment Diagram
Example : Draw the shear force and moment diagrams of the two-span continuous beam
67 kNPrimary Structure
Vertical reaction at roller support C taken as a redundant force
3 m 3 m 3 m
kNR
ikLEI
ikLEI
R
...75.16
273.339
3
25.4523*5.1002366)1(
6
0
1
11
10
11110
−=
===
==+=
=+
δ
αδ
δδ
Negative sign indicates the redundant will be opposite to the unit force
67 kN
100.5 + 0M
3+
1 kN
1M25.125
67 kN
58.625 16.7542.875+
25.125 16.75
+-
75.375
50.25
-Solve the same problem by taking the redundant as vertical reaction at support A and B respectively
-Shear force diagram (kN)
Moment diagram (kN/m)
Example : Draw the shear force and moment diagrams of the two-span continuous beam
kNmREI
EI
R
...61.102
067.3
67.3140
1
11
10
11110
=
=
−=
=+
δ
δ
δδ40 kN/m40 kN 20 kN
1.6 EI EI2.6 2.6 1.5
Take bending moment at internal support as redundant6.40
PrimaryStructurePin
40 kN/m20 kN
40 kN
40 kN40 kN/m
20 kN
204.8
+
-
+
-
144
34156.8
102.6114.21
-Bending Moment
+ +
-
0M52
6
-
112 20301 kN.mShear Force
-
1M30
1
SUPPORT SETTLEMENTIf one of the supports of a simply supported beam settles by a small amount, no major changes occur in the external and internal forces acting on the member. However if one of the supports of a multi-span beam settles a small amount significant changes occur in both the reactions and the bending moments. The problem of support settlement is far more serious for indeterminate structures than it is for determinate ones.
Example : Determine the reactions and draw the bending moment diagram for the beam in the figure, assuming that the center support settles 5 cm. and compare the results thus obtained with those corresponding to zero settlement.
15 kN/m
6 m 6 m
The degree of indeterminacy is one. Redundant must be chosen as a vertical reaction at the internal support.
kNR
REIEI
kNR
REIEI
ikLEI
ikLEI
GSSR
m
...5.112
0364050
...70.52
05.0364050
363
4050)1(3
1
1
1
1
11
10
11110
=
=+−
=
−=+−
==
−=+=
=+
δ
αβδ
δδ
Right hand side is taken zero for zero settlement
Negative sign indicates the settlement is
opposite to the unit load
Primary Structure
270
+0M
1 kN
-1M3
0.05 m15 kN/m 15 kN/m
63.6533.75 33.7563.65 112.5
52.7063.65 33.75
63.65 33.75
56.25
- -
+ +
26.35
26.35
Shear Force
Diagram
56.25134.94 134.94
37.97111.9
+
37.97BendingMomentDiagram
67.5
+ +
-
Comparison of the two moment diagrams indicates that the maximum bending moment at B increases from -67.5 to 111.9 kNm as a result of the support settlement. The support settlement can give rise to significant increases in stress in an indeterminate structure.
C
6 m 3 m
A B
5 kN
Example : Determine the support reactions and draw the shear force and bending moment diagrams of the given structure.
CableE=200 GPaA=20 mm2
L= 10 m
BeamI=100.10-6
The beam is statically indeterminate to the first degree. Let the force in the cable (T) be the redundant. Point B moves down by an amount equal to the elongation of the cable.
kNTEATT
EIEI
EIik
EIL
EIikk
EIL
cabletheofElongationT
...164.5
1072630
723
630)2(6
...
11
2110
1110
=
−=+−
==
−=+=
=+
δ
δ
δδ 5 kN5.164 kNPrimary Structure
5 kN
451 kN
-
1M+6
0M
-
5
Shear Force Diagram
0.164
Bending Moment Diagram
14.016 -
15
Example : Determine the horizontal reaction component and draw the shear force and moment diagrams of the given portal frame
A
4mDCB
2 m
4 m
50 kN
F
66.67
16.67 33.33
50 +
-
1M
0M
kNREIEI
kiLikEIL
EIEIikL
R
m
...412.4
67.9021..
32
02.40021)2(
6
0
1
11
10
11110
=
=+=
−=++=
=+
δ
βαδ
δδ
16.67
33.33
4.41
2
4.41
2
+
-+
ShearForce
diagram
16.6733.33
4.412
4.412
Primary StructureAnd External loading
17.64817.648
49.03
Moment diagram
+
2EI
EI EI
4 41
Primary StructureAnd Unit loading
Example : Determine the bending moment at B and draw the shear force and moment diagrams of the frame
A
4mDCB
2 m
4 m
50 kN
F
2EI
EI EI
-
1M
66.67 50 +
0M16.67 33.33
Primary StructureAnd External loading
1
1 kNm
0.25
10 11 1
10
11
1
0100
5.667
17.648... .
R
EI
EIR kNm
δ δ
δ
δ
+ =
=
=
= 16.67
33.33
4.41
2
4.41
2
+
-+
ShearForce
diagram
16.6733.33
4.412
4.412
Each term in elastic equation becomes relative rotation
17.64817.648
49.03
Moment diagram
+
0.25
PS And Unit loading
Example : It is required to evaluate the redundant forces and to draw the bending moment diagram of the continuous beam under the given loading. The stiffness is constant throughout the beam.
The beam is statically indeterminate to the second degree. Let the bending moment at support B be the first redundant and the bending moment at support A be the second redundant.
34 kN/m
A B C
Primary Structure6.90 7.50
34 kN/m
kNmRkNmRRR
RREIEI
EIEI
EIEI
EIEI
RRRR
73.70........46.141030.215.1
015.18.466.597
30.21*1*390.6
15.11*1*690.6
80.41*1*3
50.790.60
66.597)1(*06.23935.7
00
21
21
21
22
12
11
20
10
22212120
21211110
−===+
=++−
==
==
=+
=
=
−=−=
=++=++
δ
δ
δ
δ
δ
δδδδδδ
+
239.06
1 kN/m
0M
1 kN/m
-1 1M
-12M
34 kN/m
A B C
70.73
30.75 146.36 108.64
108.64
-
+
3.19 mShear Force Diagram
30.75
+
146.36
-
+
+
141.46
173.57
70.73Bending Moment Diagram
Example: It is required to evaluate the redundant forces and to draw the bending moment diagram of the portal frame.
100 kN The structure is statically indeterminate to the second degree. Let the horizontal reaction at left-end support and the bending moment at right-end support are be the redundant forces.1.8 EI
EI6 m8 mEI
12 m
0M
800100 kNPrimary Structure
100
1M 1
2M
166.6766.67
86 -1
1/121 kN1/12
1/6 1/6
10 11 1 12 2
20 21 1 22 2
1 2
00
35.14 ........ 293
R RR R
R kN R kNm
δ δ δδ δ δ
+ + =+ + =
= − = −
EIEIEI
EIEIEI
EIEIEIEI
EIEIEI
EIEIEI
22.101*1*811*1312
8.11
44.561*82811*)8*26(
6121
56.5718*8381)8*28*6*26*2(
612
8.116*6
361
78.49771*800281800
312
8.11
22.366228*800381800*)8*26(
612
8.11
22
12
2211
20
10
=+=
=++=
=++++=
=+=
=++=
δ
δ
δ
δ
δ
86.64)0(*)293()1(*)14.35(100
4.36)121(*)293()
61(*)14.35(67.66
4.36)121(*)293()
61(*)14.35(67.66
293)1(*)293()0(*)14.35(088.255)1(*)293()8(*)14.35(800
84.2100*)293()6(*)14.35(00
22110
=−+−+=
=−+−+=
−=−
−+−−+−=
=−−+−+=−=−−+−−+−=
=−+−−+==
++=
X
Y
Y
D
C
B
A
D
D
A
MMMM
MRMRMM
293
+
MOMENT DIAGRAM IS DRAWN ON
COMPRESSION SIDE
225.88210.84
35.14
64.86
36.4 kN
36.4 kNBending Moment Diagram
20 kN/mExample : Draw the shear force and moment diagrams of the frame.
A
2.4 EI
D
CB
8 m EI
14 m
The frame is statically indeterminate to the third degree. Let chose the redundant forces as the bending moments at A and D, and the horizontal reaction at support D.
20 kN/m 1
A D EI
+
4900M
2.4 EI CB
8 m
1M
+
12M
3M1
+
97...........97.......35.36
78.95278.95244.15244
94.9972.033.55972.094.933.5533.5533.5567.714
000
321
3
2
1
33323213130
32322212120
31321211110
==−=
−=
=+++=+++=+++
RRR
RRR
RRRRRRRRR
δδδδδδδδδδδδ+ 88
1
1
8.193)0(*97)1(*97)8(*)35.36(08.193)1(*97)0(*97)8(*)35.36(0
97)1(*97)0(*97)0(*)35.36(03322110
−=++−+=−=++−+=
=++−+=+++=
C
B
A
MMM
MRMRMRMM MOMENT DIAGRAM IS DRAWN ON
TENSION SIDE
D
CB+
193.80193.80
296.20
A
97 97
Bending Moment Diagram
STRUCTURES WITH INTERNAL REDUNDANTS (STATICALLY INDETERMINATE TRUSSES)
AB C
D
EF
P Q
NA
B CD
EF
P Q
oN
2R
AB C
D
E F
2N
11
AB C
D
EF
1N
1R
Internal force at an arbitrary member of a statically indeterminate truss can be written as
0 1 1 2 2 3 3. ....N N N R N R N R= + + + +Where
N0 = Force at the same member due to the external forces on the primary structure
Ni = Force at the same member due to a couple of unit load in the direction of the ith redundant member on the primary structure
Ri = Force at the ith redundant member
Strain energy for an axially loaded member 2
2N LUA E
=
2
2i ii i i
U N L N LNR R AE R AE
δ δ∂ ∂ ∂= → = =
∂ ∂ ∂∑ ∑According to the Castigliano’s theorem
1 2 31 2 3
.... ..... ....ii
N N N NN N N NR R R R∂ ∂ ∂ ∂
= = = =∂ ∂ ∂ ∂
( )0 1 1 2 2 3 3 ...i ii i
U N L LN N N N R N R N RR R EA EA
δ∂ ∂= = + + + + =
∂ ∂∑ ∑
1 0 1 31 1 1 21 2 3
1
0. . . ...N N L N N LN N L N N LU R R R
GivenElongationR EA EA EA EA∂
→ + + + + = ∂
∑ ∑ ∑ ∑
2 0 2 32 1 2 21 2 3
2
0. . . ...N N L N N LN N L N N LU R R R
GER EA EA EA EA∂
→ + + + + = ∂
∑ ∑ ∑ ∑
…………
…………
3 0 3 1 3 2 3 31 2 3
3
0. . . ...N N L N N L N N L N N LU R R R
GER EA EA EA EA∂
→ + + + + = ∂
∑ ∑ ∑ ∑
10 11 1 12 2 13 3
0...R R R
GivenElongationδ δ δ δ
+ + + + =
20 21 1 22 2 23 3
0...R R R
GivenElongationδ δ δ δ
+ + + + =
30 31 1 32 2 33 3
0...R R R
GivenElongationδ δ δ δ
+ + + + =
This set of linear equation is referred to as ELASTIC EQUATIONS of the truss
Example: Calculate the member forces of the truss if the roller support settles down by 1 mm. Chose the force in member AD and the vertical reaction at support E as the redundant forces. E=200 GPa, A=5000 mm2.
80 kN
A
B D
E
C
80 kN
D
0
00
80
-113.1
37
CA 80 4 m
0N
EB4 m 4 m
-1.414
11
01.4
14 2N
CA
B
D
-2 A
0
0-0.707
-0.7071
1
1
1N
C-0.707
B
D
1 kN
46.6410.8317.32-1545.46-1092.85Sum/EA
400001004DE
11.320000-1.414005.66CE
0020-226.240-0.707804CD
4-2.832001-0.70704BD
11.3285.66-905.46-640.371.4141-113.1375.66BC
165.662-640-226.24-2-0.707804AC
005.66000105.66AD
LN22LN1N2LN1
2LN2N0LN1N0N2N1N0LMember
1 2
1 2
1
2
1092.85 17.32 10.83 0
1545.96 10.83 46.64 0.001
65.263.46
R REA EA EA
R R mEA EA EA
R kNR kN
−+ + =
−+ + = −
== −
10 11 1 12 2 0R Rδ δ δ+ + =
20 21 1 22 2R R GivenSupportSettlementδ δ δ+ + =
Solving redundant forces, the bar forces can be obtained
-3.46
4.89
33.86
-49.60
-52.77
40.78
65.26
N
+ 65.26
+R1
0
0
-0.707
-0.707
1
-0.707
1
N1
10DE
-1.4140CE
080CD
10BD
1.414-113.137BC
-280AC
=
0
- 3.46
0AD
N2+R2N0Member
Example: Compute the forces in diagonals BF and CE. All members are 2000 mm2. in area. E=200 GPa, What would be the effect on these forces of a rise in temperature of 20 degrees of Celsius in member EF relative to other members. Take the coefficient of thermal expansion as 12.10-6 1/oC
Let Force in member BF be the redundant force.FE
D
240 kN
240 kN
A CB D
E -80
80160
80
-113.1370N
F
160
113.13
7
-113.137
5 m
A
B C
5 m 5 m5 m
-0.7
07
A CB D
E-0.707
0-0.707
-0.7
071 1N
F
0
0 0
1
1
24.14/EA-1365.6/EASUM
7.07-8001-113.147.07CE
2.5282.8-0.707-805EF
000-113.147.07DF
2.5-282.8-0.707805CF
000805CD
7.070107.07BF
2.50-0.70705BE
2.5-565.6-0.7071605BC
000113.147.07AE
0001605AB
LN1N1LN1N0N1N0LengthMem
10 11 1
1
01365.6 56.5724.1456.57113.14 56.57 *1 56.57
R
R kN
BF kNCE kN
δ δ+ =
= =
== − + = −
Let force in member EF be the redundant force. Elongation of member EF due to the temperature rise of 20 degrees6 3
6 6
. . 5*20*12.10 1.2*10200.10 *2000.10 400000
L L T mEA kN
α − −
−
∆ = ∆ = =
= =
1
A B D
E 1
0
1-1.414
1N
F
0
0
0
1
-1.41
4
B 1 C
48.28/EA0SUM
14.1360-1.41407.07CE
50105EF
00007.07DF
50105CF
00005CD
14.1360-1.41407.07BF
50105BE
50105BC
00007.07AE
00005AB
LN1N1LN1N0N1N0LengthMem
10 11 1
31
1
48.280 1.2*10
9.9421.414*( 9.942) 14.06
. .56.57 14.06 42.5156.57 14.06 70.63
R GivenElongation
R mEA
R kNBF CE kNExternal load Temperature ChangeCE kNBF kN
δ δ
−
+ =
+ = −
= −= = − − =
+= − + = −= + + = +
THREE MOMENTS EQUATION (CLAPEYRON’S EQUATION)
A general equation based on the force method can be developed for continuous beams. The equation relates the moments at the three consecutive support points to the loading on the intermediate support. This theorem was represented by Clapeyron in 1857 for the analysis of continuous beams.
Now consider m span continuous beam, the degree of indeterminacy will be m-1 . If the moments at the internal supports are chosen as the redundant forces the primary structure will be m simple beams.
Pi j k
jLiL
Relative rotation at support j will be the function of external forces at the two adjacent span ( Li,, Lj )and the support moments at i, j and k. Loading over the other spans and the other support moments will not contribute to the relative rotation at support j. Summation of the relative rotations at j due to the external load and the redundant moments Mi , Mjand Mk must be equal to the relative rotation of the actual continuous beam at j which is zero.
i j kP The relative rotation due to the external forces can be
written by using Moment Area Theorem.
0 0 0
, .. , ..6 3 3 6
j jL R i k i ij j j
i j i i j j
j ji iji jj jk
i i j j
A xt t A xL L L EI L EI
L LL LEI EI EI EI
δ θ θ
δ δ δ
= + = + = +
= = + =
jLiL
jA+iA+
ix jxRelative rotation is in clockwise direction (positive)
itkt
i kj
0L
jθ 0R
jθ
0jδ6 3 6
j j k j j ji i i i i
i i j j i i j j
M L M L A xM L L A xEI E I I EI L EI L EI
+ + + = − +
Rearranging the last equation
62 j k j j ji i i i ij
i i j j i i j j
L M L A xM L L A xMEI EI EI EI E L I L I
+ + + = − +
1
*iM
jiδ
1 jjδ
*jM
If EI is constant throughout the beam. The equation simplifies to
( )2 6 j ji ii i j i j k j
i j
A xA xM L M L L M LL L
+ + + = − +
Application of the three moments equation to a continuous beam results in a set of simultaneous equations with the moments over the supports as the unknowns.
1jkδ
*kM
In applying the three moments equation to a particular beam, we locate the interior supports successively and write as many equations as the unknown redundant support moments. A simultaneous solution of the equations for the unknown moments yields the required results.
321
1 0M = 3 0M =2 ?M =
1
One redundant moment (M2)
321
1M 3M2 ?M =
M1 and M3 can be calculated
1
1 0M = 2 ?M = 3 ?M = 4 0M =
21
Two redundant moments and two equations
1 0M = 2 ?M = 3 ?M = 4 ?M = 5 0M =Unknowns M2 , M3 , M4 . Three equations can be written
31 2
Example: It is required to draw shear and bending moment diagrams of the two span continuous beam. I is constant.
24 kN/m
12 3
20 kN/m31
265 5 6 6 208.33* 2.5 432*32
5 6MM M
EI EI EI EI E I I + + + = − +
5 m 6 m 1 3
2
2
0; .... 022 1921
87.32
M MM
M kNm
= == −
= −2
62.58qL
=
2
1088qL
=
++
87.32
12 62.5*5 208.333
A = = 22108*6 4323
A = = 50 7250 72
32.54
67.46
86.55
57.45
−
+ +
−
1.63m
2.39m
87.32 17.465
= 87.32 14.556
=
68.7626.47
−
++
87.32
Example: It is required to determine the support moments and reactions for a continuous beam fixed at one end and having an overhang at the other as shown.
20kN 20kN 8. /kN m
1.5EI EIA B C
',... 'L I = ∞
'A
10kN The beam is statically indeterminate to the second degree and requires two equations. For the purpose of writing three moments equations an imaginary span to the left of fixed support A having an arbitrary length L’ and moment of inertia I’=∞ may be considered.13*2.20 6.60m m= 5
+ +44 25
193.6A = 83.33A =
' ' 6.60' 6.60 6 193.6*3.321.5 1.5 1.5 *6.6
56.6 6.60 5 6 193.6*3.3 83.33*2.521.5 1.5 1.5 *6.6 5
A BA
CAB
M L MLMEI EI E I
MM MEI EI EI EI E I I
+ + + = − ∞ ∞
+ + + = − +
8.80 4.40 387.24.40 18.8 587.19
10 ,..... 32.14,.... 23.71
A B
A B
C A B
M MM M
M kNm M M
+ = −+ = −
= − = − = −
23.7120
81032.14 10
20 20
1.28
20 30
2.74
20
18.72 17.26
+
−
+
− 1.2810
21.28 22.74
23.71
8.64
32.14
−
++
14.67 16.96
10
Three moments equation can be modified to take into account the support settlements. For example consider that supports i, j, and k settle downward by amount ∆i , ∆j , ∆k respectively.
i∆
j∆ k∆
i j k
j i
iL∆ −∆
j k
jL∆ −∆
Rjθ
Settlement of support j decrease the relative rotation at j
( ) j i j kss sL sRj j j
i jL Lθ θ θ
∆ − ∆ ∆ −∆= − + = − +
If we substitute this expression of relative rotation into the equation of consistent deformation.
06 3 6
j j k j j j j i j ki i i i i
i i j j i i j j i j
M L M L A xM L L A xE I E I I E I L E I L E I L L
∆ − ∆ ∆ − ∆+ + + + + − + =
If I is constant throughout the beam, this expression simplifies to
( )2 6 6j j j i j ki ii i j i j k j
i j i j
A xA xM L M L L M L E IL L L L
∆ − ∆ ∆ − ∆+ + + = − + + +
The support B of the previous example settles down by 20 mm under the loading. E=200 GPa=200 106 kN/m2 I=80.106 mm4
'
'
' 6.60' 6.60 6 193.6*3.3 0.022 6 01.5 1.5 1.5 *6.6 6.60
56.6 6.60 5 6 193.6*3.3 83.33*2.5 0.02 0.022 61.5 1.5 1.5 *6.6 5 6.60 5
A BA
CAB
M L MLMEI EI E I
MM MEI EI EI EI E I I
− + + + = − + + ∞ ∞
+ + + = − + + +
8.80 4.40 678.114.40 18.8 87.71
10 ,..... 89.91,.... 25.70
A B
A B
C A B
M MM M
M kNm M M
+ = −+ =
= − = − = −
SYMMETRICAL STRUCTURES WITH SYMMETRICAL LOADING
If the structure and loading are both symmetrical, symmetrical joints rotate by the same amount but in opposite direction and the structure will have a skew-symmetrical shear force and a symmetrical bending moment diagrams. Rotation at the symmetry axis is always zero. If symmetry axis passes from a common member and there is no concentrated load at the symmetry axis the shear force at the same section be zero. Otherwise the shear at the symmetry axis is equal to the half of the concentrated load. By choosing the internal forces at the symmetry axis as redundant forces one half of the structure may be analyzed.
00
Qθ==
/ 20
Q Pθ==
Symmetry axis passes from a common member
P
00θ
∆ ==
00θ
∆ ==
Symmetry axis passes from a support or column
Fixed end support
Fixed end support
Example: Draw the bending moment diagram of the given frame. EI is constant.
60 KN/m
A D
CB
4 m
8 m
1R
2R
A
B B
Q=0A
A D
CB+
1 2
1 2
1
2
8 8 25608 21.33 3840223.9896.02
R RR R
R kNmR kN
− =− + = −
== −
MOMENT DIAGRAM IS DRAWN ON
TENSION SIDE
−
−
0M
480
1M
+
1
+
256.02 256.02
1
−
4
223.98
2M Bending Moment Diagram
128.06 128.06
SYMMETRICAL STRUCTURES WITH SKEW-SYMMETRICAL LOADING
If a symmetrical structure is subjected to a skew- symmetrical loading, symmetrical joints rotate by the same amount in the same direction and the structure will have a symmetrical shear force and a skew-symmetrical bending moment diagrams. If symmetry axis passes from a common member horizontal force and bending moment at the symmetry axis is always zero and a roller support may be considered at the center section to analyze half of the structure. If symmetry axis passes from a column, half of the moment of inertia of that vertical member must be considered to analyze half of the structure.
Symmetry axis passes from a common member
Symmetry axis passes from a support or column
000
MN
==
∆ =
I
Roller support
2I
Example: Draw the bending moment diagram of the given frame. EI is constant.
A D
CB
4 m
10 m
10 KN
1RA
B−
−
1M
55
0M−
20
5
1
A D
12.94 12.94
+ C
B
7.067.06
Bending Moment Diagram
MOMENT DIAGRAM IS DRAWN ON
TENSION SIDE
10
11
1
200 /141.67 /1.412
EIEI
R kN
δδ
=== −A D
D
C
5
CB
4 m
A
B
4 m
10 m
5 5
5
Neglecting axialdeformationNo bending
Moment
ANY LOAD SYSTEM CAN BE CONSIDERED ONE SYMMETRICAL AND THE OTHER SKEW-SYMMETRICAL LOADINGS
A
P
A
2P
A
2P
2P
2P
Skew-Symmetrical
loading
B C
D
D
CB
D
CB
Symmetrical loading
1RA
B
1R
2RB
A
A D
CBP
A D
CB2P
A D
CB
2P
2P
2P
Symmetrical loading
Skew-Symmetrical
loading
q
2q
2q
2q
2q
Example: Draw the bending moment diagram of the given frame
70
24 42
3
6
70 70
0M
280
70
1M
1
1
1
97.8
49.07
121.4
85.6
1
2M
31 2
1 2
1
2
12.71 46.065 2305.846.065 254.13 11540.249.0736.52
R RR R
R kNmR kN
− =− + = −
== −
9
FIXED-END MOMENTS & FIXED-END FORCES
If a member is fixed at its ends, the moments at the ends of the member are referred as the fixed-ends moments.
q
LA B
The member is assumed to be subjected transverse loading and the effect of the axial deformation will be neglected therefore change in length will be zero and R3 =0. Fixed end moments will be determined by the force method.
q1R 2R
q
Primary structureRedundant ForcesExternal loading
3R
2qL
+
q
− −
2
1 2qL 2
1 2qL
2qL
2
1 2q L 2
1 2qL
2
2 4qL
2qL
2qL
2
1 2
2
1 2
2
1 2
03 8 3 6
03 8 6 3
12
L qL L LR REI EI EIL qL L LR REI EI EI
qLR R
+ + =
+ + =
= = −
+
2
8qL
0M
+11M
+ 12M
Example: It is required to determine the fixed end moments by using three moments equation.
A Bba
P
A B
B′A′I = ∞ I = ∞
PabL
+
( )1 ( )2L
'
2 2
2 2
6 222 3 2 3
6 222 3 2 3
.....
A BA
A BB
A B
M M LL L Pab a a b bM bEI EI EIL L
M L M LL L Pab a a b bM aEI EI EIL L
Pab Pa bM ML L
′ + + + = − + + ∞ ∞ ′ ′′′′ + + + = − + + ∞ ∞
= − → =−
2
2
PabL
2
2
Pa bL
P
+
2
2
Pa bL
2
2
PabL − −
2
2 21Pab ab a bL L L L
+ − −
RELATIONS BETWEEN THE MEMBER END FORCES AND MEMBER END DISPLACEMENTS
Example: It is required to determine a) fixed end moment at the left end support, b) rotation at end B for the propped cantilever beam as shown.
Let bending moment at A be the redundant force.
AB
BM10 11 1
1
1
0
06 3
2
B
B
RL LM REI EI
MR
δ δ+ =
+ =
= −
L
EI is constant
+ BM
0M
0
3*3 2 2 2
4
LB B
B
BB
M MM M L LdxEI EI EI
M LEI
θ
θ
′= = −
=
∫+1
1M
−+ 3
2BM+
BM
= +2BM−
2BM
+ 1M ′ 4 2,...B A
EI EIM ML L
= =For unit rotation at B
Example: It is required to determine the fixed end moments due to the settlement of right end support (∆) by using three moments equation.
A
B
A′
B′
( )1( )2
I = ∞
I = ∞
∆L∆
L∆
Draw the tangent from the left side of the center support
A
BL∆
2
6A
EIML
= − ∆
2
6B
EIML
= + ∆
03 6
06 3
A B
A B
M L M LEI EI LM L M LEI EI L
∆+ + =
∆+ − =
2
2
6
6
A
B
EIMLEIML
= − ∆
= + ∆
A
B I = ∞L∆
A
B
2
3B
EIML
= − ∆2
03
3
B
B
M LEI L
EIML
∆+ =
= − ∆B′
