stiffness 9
TRANSCRIPT
Lecture No. : 9 ال تاسعةالمحاضرة
F = K Dl l l
T K =g K l T
T
m m
Drive the member local stiffness matrix
Obtain the member global stiffness matrix
Drive the member transformation matrixT
Solution Steps of assembly method :Remember
Make assembly F = K D
Kuu
Kru
Kur
Krr
Fu
Fr
Du
Dr=
Make partition
Kgm K
gm K
gm K
gm
Remember
Kuu
Kru
Kur
Krr
Fu
Fr
Du
Dr=
Extract the stiffness equation
KuuFu Du= Kur Dr+
KuuDu =-1{ }Fu Kur Dr-
Obtain the deformation
Remember
Find internal forces in members
Calculate the reactions
KruFr Du= Krr Dr+
gF = K l lm m mT D
T
Remember
d1
d2
d3
d1
d2
Normal Force doesn’t taken
Drive the member local stiffness matrix
k11F1
F2
=k21
F3 k31
k12
k22
k32
k13
k23
k33
F4k41 k42 k43
k14
k24
k34
k44
d1
d2
d3
d4
d1
d2 d4
d3
First column inLocal Stiffness matrix
d1 =1
6 EI
L2
6 EIL2
12 EIL3
12 EIL3
d3
d4
d2
d2
12 EIL3
F2 =
F3 = -
F4 =
F1 =12 EI
L3
6 EIL3
6 EIL3
6 EI
L2
6 EIL2
12 EIL3
12 EIL3
=k31
12 EIL3
6 EIL2
-12EIL3
6 EIL2
k11
k21
k41
Second column inLocal Stiffness matrix
d2 =1
4 EIL
2 EIL
6 EIL2
6 EIL2
d3
d4
d1
d2
F3 =F1 =
F4 =F2 =
6 EIL2
6 EIL2
-
2 EIL
4 EIL
4 EIL
2 EIL
6 EIL2
6 EIL2
Second column inLocal Stiffness matrix
=
k12
k22
k32
k42
6 EI
L2
4 EIL
6 EIL2-
2 EIL
Third column inLocal Stiffness matrix
d3 =1
6 EIL2
6 EIL2
12 EIL3
12 EIL3
d3
d4
d1
d2
12 EIL3
6 EIL2
6 EIL2
12 EIL3F1 = F3 =
F4 =F2 = - -
-
6 EI
L2
6 EIL2
12 EIL3
12 EIL3
Third column in Local Stiffness matrix
=
k13
k23
k33
k33
12 EIL3
6 EIL2
12 EIL3
6 EIL2
-
-
-
Fourth column inLocal Stiffness matrix
d4 =1
4 EIL
2 EIL
6 EIL2
6 EIL2
d3
d4
d1
d2
F1 = F3=
F2 = F4=
6 EIL2
2 EIL
4 EIL
6 EIL2
-
4 EIL
2 EIL
6 EIL2
6 EIL2
Fourth column inLocal Stiffness matrix
=
k14
k24
k34
k44
6 EI
L2
2 EIL
6 EIL2-
4 EIL
K l
12 EIL3
6 EIL2
6 EIL2
4 EIL
-12 EIL3
=-12 EI
L3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EIL2
2 EIL
6 EIL2
2 EIL
-6 EIL2
4 EIL
K l
12 EIL3
6 EI
L2
6 EI
L2
4 EIL
-12 EIL3
=-12 EI
L3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EI
L2
2 EIL
6 EI
L2
2 EIL
-6 EIL2
4 EIL
K l = EIL3
K l
12 EIL3
6 EIL2
6 EIL2
4 EIL
-12 EIL3
= -12 EIL3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EIL2
2 EIL
6 EIL2
2 EIL
-6 EIL2
4 EIL
12
6 L
-12
6 L
6 L
-6 L
4 L2
2 L2
-12
6 L
12
6 L
-6 L
2 L2
4 L2
6 L
d1
d2
Normal Force doesn’t taken
d1
If Shear is omitted
Drive the member local stiffness matrix
k11F1
F2
=k21
k12
k22
d1
d2
d1 d2
K l
12 EIL3
6 EIL2
6 EIL2
4 EIL
-12 EIL3
=-12 EI
L3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EIL2
2 EIL
6 EIL2
2 EIL
-6 EIL2
4 EIL
K l4 EI
L=
2 EIL
2 EIL
4 EIL
K l = 2EIL 1
2
2
1
Construct the stiffness matrix for the shown beam where EI is
constant for all members
Example 1:
8 10
A CB
First element : (A-B )Start Joint : A End Joint : B
K l = EIL3
12
6 L
-12
6 L
6 L
-6 L
4 L2
2 L2
-12
6 L
12
6 L
-6 L
2 L2
4 L2
6 L
K l K g=
K l
12 EIL3
6 EI
L2
6 EI
L2
4 EIL
-12 EIL3
=-12 EI
L3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EI
L2
2 EIL
6 EI
L2
2 EIL
-6 EIL2
4 EIL
K l = EIL3
12
6 L
-12
6 L
6 L
-6 L
4 L2
2 L2
-12
6 L
12
6 L
-6 L
2 L2
4 L2
6 L
K l = EI
.023
.0937
.-023
.0937
.0937
.5
.0937
.25
.-023
.0937
.023
.-0937
.0937
.25
.-0937
.5
A
B
�ِA
B
Second element : ( B-c)Start Joint : B End Joint : c
K l
12 EIL3
6 EI
L2
6 EI
L2
4 EIL
-12 EIL3
=-12 EI
L3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EI
L2
2 EIL
6 EI
L2
2 EIL
-6 EIL2
4 EIL
K l = EI
.012
.06
.-012
.06
.06
.4
.-06
.2
.-012
.-06
.012
.-0937
.06
.20
.-06
.4
B C
C
B
Assembly :
K =g1 EI
.023
.0937
.-023
.0937
.0937
.5
.0937
.25
.-023
.0937
.023
.-0937
.0937
.25
.-0937
.5
A B
B
A
EI .06
.-012
.06
.06
.4
.2
.-012
.-06
.012
.-06
.06
.2
.-06
.012
.-06
K =g2
.4
B C
B
C
Ks = EI
.023
.0937
.-023
.093700
.0937
.5
.0937
.2500
.-023
.0937
.035
.-0337.-012.06
.0937
.25
.-0337
.9
.-06
.2
0
0
.-012
.-06
.012.-06
0
0
.06
.2
.-06
.4
A B C
B
A
C
Partition
KuuK =Kru
Kur
Krr
u ru
r
Kuu = EI .90
Example 2:Construct the stiffness matrix for the shown beam where EI is constant for all members
4254
ECB DA
= E I
12/L^3
6/L^2
-12/L^3
6/L^2
6/L^2
2/L
6/L^2-
4/L
12/L^3-
6/L^2-
12/L^3
6/L^2-
6/L^2
-6/L^2
4/L
2/L
K gK l =
K l
First element : (A-B )Start Joint : A
End Joint : BAngle : 0s = sin = 0c = cos = 1
Is conastant EA
LAB = 400 cm
Kl= EI
.1875
.375
.-1875
.375
.375
1
.-375
.5
.-1875
.-375
.1875
.-375
.375
.5
.-375
1
A B
B
A
Second element : B-C( Start Joint : B
End Joint : cAngle : 0s = sin = 0c = cos = 1
Is conastant EA
LBC = 500 cm
EI .24
.-096
.24
.24
.8
.4
.-096
.-24
.096
.-24
.24
.4
.-24
.096
.-24
.8
B C
B
C
K =g1
End Joint : D
Third element : (C-D )
Start Joint : C
Angle : 0s = sin = 0c = cos = 1
LCD = 200 cm
Is conastant EA
= EI
1.5
1.5
-1.5
1.5
1.5
2
-1.5
.2
-1.5
-1.5
1.5
-1.5
1.5
1
-1.5
.2
C D
D
CK l
Fourth element : (D-E )
Start Joint :D
End Joint : EAngle : 0s = sin = 0c = cos = 1
Is conastant EA
LD-E = 400 cm
Kl= EI
.1875
.375
.-1875
.375
.375
1
.-375
.5
.-1875
.-375
.1875
.-375
.375
.5
.-375
1
D "ُE
D
E
K =s
.1875 .375.375 1
.-1875 .-375
.375 .5
0 0
0 0
0 0
0 0
.-096 .-24.24 .4
0 00 0
0
0
0 00
00
0 0
0 0
0
0
0
0
0
0 00
0.-1875 .375.-375 .5
.2835 .-135.-135 1.8
0000
.-096
.-24
.24.4
1.596 1.261.26 2.8
-1.5 1.51.5 1
0 00 0
00
-1.5 1.51.5 1
1.687 -1.125-1.125 3.-1875.375
.-375.5
0
00
0
0
0
.-1875 .375.-375 .5
.1875.-375 1
.-375
A B C D E
AB
CDE
KUU1.8 EI0.4 EI
0
0.4 EI2.8 EI
EI
0EI
3 EI=
1.80.4
0
0.42.8
1
01
3EI=
Example 3:Calculate the deformation of the shown beam where EI = 105 kN.m2 for all members
4254DCB
EA 30 kNm50 kNm60 kNm
KUU
1.8 EI0.4 EI
0
0.4 EI2.8 EI
EI
0EI
3 EI=
1.80.4
0
0.42.8
1
01
3EI
The stiffness equation
F = K D
Stiffness matrix From Exampl (2)
F1
F2
=-60
-50
F3 30
= d1
d2
-60
-50
1.8
.4
.4
2.8
F = K D
30 d3
0
1
0 1 3
D =
B
C
D
D = K-1 F- 60- 50
30=
1.80.4
0
0.42.8
1
01
3
B
C
D
1EI
-1
- 60- 50
30=
7.4-1.2
.4
-1.25.4
-1.8
.4-1.8
4.88
B
C
D
1105
112.84
=
- 0.290- 0.196
0.165
X10-3
rad
Example 4:Draw B.M.D for the shown beam where EI = 105 kN.m2 for all members
4
A
B
E
5 2 4C D
60 kNm 50 kNm 30 kNm
From the previous example
=
B
C
D
- 0.290
- 0.196
0.165
X10-3
rad
4
AB
E
5 2 4C D
=
B
C
D
- 0.290- 0.196
0.165
X10-3
rad
EI = 105 kN.m2
For member AB
LAB
MAB = == - 14.5 kN.m
2 EILAB
MBA= (2 B + A ) = 5x104(2X-0.029)x10-3
5x104(0-0.29)x10-3(2 A + B )2 EI
= - 29 kN.m
4
AB
E
5 2 4C D
=C
D
- 0.290- 0.196
0.165
X10-3
rad
B
EI = 105 kN.m2
For member BC
LBC
MBC = (2 B + C ) =
2 EI
LBC
MCB= (2 C + B ) =
4x104(2x-0.29-0.196)x10-3
= - 31 kN.m
4x104(2x-0.196-0.029)x10-3
= - 27.3 kN.m
MAB = - 14.5 kN.mMBA= - 29 kN.m
MBC = - 31 kN.mMCB= - 27.3 kN.m
MCD = - 22.7 kN.mMDC= 13.5 kN.m
MDE = 16.5 kN.mMED= 8.3 kN.m
A B B C C D D E
14.5 29 31 27.3 22.7 13.5 16.5 8.3
A B B C C D D E
14.5 29 31 27.3 22.7 13.5 16.5 8.3
14.531
27.3
22.713.5
16.5
B.M.D
14.5 13.5
16.527.3
22.731
29
A
B
E
C D
60 kNm 50 kNm 30 kNm
Example 5:Draw B.M.D for the shown beam where EI is constant for all members
2
A B C D
1 2 2 1 2
100 kN 200 kN 150 kN
Solution Steps of assembly method :
Drive the member local stiffness matrixLocal
k11F1
F2
=k21
F3 k31
k12
k22
k32
k13
k23
k33
F4k41 k42 k43
k14
k24
k34
k44
d1
d2
d3
d4
d3
d1
d2 d4
First column inLocal Stiffness matrix
6 EI
L2
6 EI
L212 EI
L3
F1 = 12 EIL3
F2 = 6 EI
L2
6 EI
L2
F3 = 12 EIL3
-
F4 =
first column in Local Stiffness matrix
=
k31
k41
12 EIL3
6 EIL2
-12 EIL3
6 EIL2
k11
k21
Second column inLocal Stiffness matrix
d2 =1
4 EIL
2 EIL
6 EIL2
6 EIL2
F3 =F1 =
F4 =F2 =
6 EIL2
6 EIL2
-
2 EIL
4 EIL
Second column inLocal Stiffness matrix
=
k12
k22
k32
k42
6 EI
L2
4 EIL
6 EIL2
- 2 EIL
Third column in Local Stiffness matrix
d3 =1
12 EIL3
12 EIL3
6 EIL2
6 EIL2
Third column in Local Stiffness matrix
=
k13
k23
k33
k33
-12 EIL3
-6 EIL2
12 EIL3
-6 EIL2
12 EIL3
-6 EIL2
-6 EIL2
F4 =
F3 = F1 = -12 EIL3
F2 =
Fourth column inLocal Stiffness matrix
6 EIL2
6 EIL2
4 EIL
2 EIL
F1 = F3=
F2 = F4 =
6 EIL2
2 EIL
-6 EIL2
4 EIL
K l
12 EIL3
6 EIL2
6 EIL2
4 EIL
-12 EIL3
=
-12 EIL3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EIL2
2 EIL
6 EIL2
2 EIL
-6 EIL2
4 EIL
First element : (A-B )
Start Joint : A
End Joint : BAngle : 0s = sin = 0c = cos = 1EA Is conastant
LAB = 300 cm
K l K g=
K l = E I
12/L^3
6/L^2
-12/L^3
6/L^2
6/L^2
2/L
6/L^2-
4/L
12/L^3-
6/L^2-
12/L^3
6/L^2-
6/L^2
-6/L^2
4/L
2/L
K l = EI
0.44
0.67
-0.44
0.67
0.67
1.33
-0.67
0.67
-0.44
-0.67
0.44
-0.67
0.67
0.67
-0.67
1.33
A B
"ِ"ِA
B
Second element : ( B-c)Start Joint : B End Joint : c
= E I
12/L^3
6/L^2
-12/L^3
6/L^2
6/L^2
2/L
6/L^2-
4/L
12/L^3-
6/L^2-
12/L^3
6/L^2-
6/L^2
-6/L^2
4/L
2/LK l
K l = EI
0.1875
0.375
-0.1875
0.375
0.375
1
-0.375
0.5
-0.1875
-0.375
0.1875
-0.375
0.375
0.5
-0.375
1
B C
C
B
Second element : ( C-D )Start Joint : C End Joint : D
= E I
12/L^3
6/L^2
-12/L^3
6/L^2
6/L^2
2/L
6/L^2-
4/L
12/L^3-
6/L^2-
12/L^3
6/L^2-
6/L^2
-6/L^2
4/L
2/LK l
K l = EI
0.44
0.67
-0.44
0.67
0.67
1.33
-0.67
0.67
-0.44
-0.67
0.44
-0.67
0.67
0.67
-0.67
1.33
C D
�ِC
D
K
g = EI
0.44
0.67
-0.44
0.67
0.67
1.33
-0.67
0.67
-0.44
-0.67
0.44
-0.67
0.67
0.67
-0.67
1.33
A B
�Aِ
B
K g= EI
0.1875
0.375
-0.1875
0.375
0.375
1
-0.375
0.5
-0.1875
-0.375
0.1875
-0.375
0.375
0.5
-0.375
1
B C
C
B
K g= EI
0.44
0.67
-0.44
0.67
0.67
1.33
-0.67
0.67
-0.44
-0.67
0.44
-0.67
0.67
0.67
-0.67
1.33
C D
�ِC
D
Ks = EI
0.44
.0937
0.67
-0.44
0.67
0
0
0.67
1.33
-0.67
0.67
0
0
0.67
0.67
-.0.295
2.33
-0.375
0.5
0
0
-0.1875
.-375
.2525
0.295
0
0
0.375
0.5
0.295
2.33
0
0
.-44
.-67
A B C
B
A
C
K l K g=
00 0 .-44 .-67 .-67
0 0 0 .67 0.67 1.33
D
D
-0.44
-0.67
.0.6275
-0.295
-0.1875
0.375
0
0
0
0
.67
0.67
0
0
0
0
.-67
0.44
Partition
KuuK =Kru
Kur
Krr
u ru
r
Kuu = EI 2.33
2.33
0.5
0.5
Force vectorTransformation from member forces to Joint forces
L
P
8LP
8LP
90
P a b2
L2
L
P
a b P b a2
L2
2
A B C D
1 2 2 1 2
100 kN 200 kN 150 kN
100 kN 150 kN
200 kN
100 kNm100 kNm
44.4 kNm
22.2 kNm
66.7 kNm
33.3 kNm
Fixed End Reaction
(FER)
2
A B C D
1 2 2 1 2
100 kN 200 kN 150 kN
100 kN 150 kN
200 kN
100 kNm100 kNm
44.4 kNm
22.2 kNm
66.7 kNm
33.3 kNm
Fixed End Action (FEA)
2
A B C D
1 2 2 1 2
100 kN 200 kN 150 kN
100 kNm
44.4 kNm 66.7 kNm
100 kNm
33.3 kNm55.6 kNm
2
A B C D
1 2 2 1 2
100 kN 200 kN 150 kN
33.3 kNm55.6 kNm
F1
F2
=-55.6
33.3
F = K Dk11F1
F2
=k21
k12
k22
d1
d2
-55.6
33.3=
7/3
0.5 7/3
0.5EI
B
C
B
C=
1EI
7/3
0.5 7/3
0.5-1
-55.6
33.3
B
C=
1EI
7/3
0.5 7/3
0.5-1
-55.6
33.3
B
C=
1EI
-28.18
20.31
Internal forces in beam elements2 EI
LMAB= ( 2 + )
MBA=2 EI
L( + 2 )
MBA=2 EI
L( + 2 ) M(FER) BA +
MAB= ( 2 + ) M(FER) AB +2 EI
L
2
A B CD
1 2 2 1 2
100 kN 200 kN 150 kN
100 kN 150 kN
200 kN
100 kNm100 kNm
44.4 kNm
22.2 kNm
66.7 kNm
33.3 kNm
Fixed End Reaction
(FER)
B
C=
1EI
-28.18
20.31
2 EILMAB= ( 2 + ) M(FER) AB +
MBA=2 EI
L( + 2 ) M(FER) BA +
100 kN
44.4 kNm22.2 kNm
= 22.2 + 2/3 (-28.18) = 3.4
= -44.4 + 2/3 (2x-28.18) = - 82
A B
B
C=
1EI
-28.18
20.31
2 EILMBC= ( 2 + ) CM(FER) BC +
MCB=2 EI
L( + 2 ) CM(FER) CB +
= 100 + 2/4 (2x-28.18+20.31) = 82
= -100 + 2/4 (-28.18+2x20.31) = - 93.8
200 kN
100 kNm100 kNm
B C
B
C=
1EI
-28.18
20.31
2 EILMCD= ( 2 + )C DM(FER) CD +
MDC=2 EI
L( + 2 )C DM(FER) DC +
= 66.7 + 2/3 (2x20.31) = 93.8
= -33.3 + 2/3 (20.31) = - 19.8
150 kN
33.3 kNm66.7 kNm
C D
MAB= 3.4MBA= -82
MBC= 82MCB= -93.8
MCD= 93.8MDC= -19.8
3.4
82
19.8
93.8
B.M.D
3.4
82
19.8
93.8
B.M.D
2
A B C D
1 2 2 1 2
100 kN 200 kN 150 kN
55.887.9
69.1
66.7
10.9
200
112.1
100
30.9
Example 6:Draw B.M.D for the shown beam where EI is shown in figure
3
A B C D
5 5 2
100 kN 200 kN 100 kN
3 2
250 kN
III2 I2 I2 I
E
Solution Steps of assembly method :
Drive the member local stiffness matrixLocal
k11F1
F2
=k21
F3 k31
k12
k22
k32
k13
k23
k33
F4k41 k42 k43
k14
k24
k34
k44
d1
d2
d3
d4
d3
d1
d2 d4
First column inLocal Stiffness matrix
6 EI
L2
6 EI
L212 EI
L3
F1 = 12 EIL3
F2 = 6 EI
L2
6 EI
L2
F3 = 12 EIL3
-
F4 =
first column in Local Stiffness matrix
=
k31
k41
12 EIL3
6 EIL2
-12 EIL3
6 EIL2
k11
k21
Second column inLocal Stiffness matrix
d2 =1
4 EIL
2 EIL
6 EIL2
6 EIL2
F3 =F1 =
F4 =F2 =
6 EIL2
6 EIL2
-
2 EIL
4 EIL
Second column inLocal Stiffness matrix
=
k12
k22
k32
k42
6 EI
L2
4 EIL
6 EIL2
- 2 EIL
Third column in Local Stiffness matrix
d3 =1
12 EIL3
12 EIL3
6 EIL2
6 EIL2
Third column in Local Stiffness matrix
=
k13
k23
k33
k33
-12 EIL3
-6 EIL2
12 EIL3
-6 EIL2
12 EIL3
-6 EIL2
-6 EIL2
F4 =
F3 = F1 = -12 EIL3
F2 =
Fourth column inLocal Stiffness matrix
6 EIL2
6 EIL2
4 EIL
2 EIL
F1 = F3=
F2 = F4 =
6 EIL2
2 EIL
-6 EIL2
4 EIL
K l
12 EIL3
6 EIL2
6 EIL2
4 EIL
-12 EIL3
=
-12 EIL3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EIL2
2 EIL
6 EIL2
2 EIL
-6 EIL2
4 EIL
First element : (B-C )
Start Joint : B
End Joint : CAngle : 0s = sin = 0c = cos = 1EA Is conastant
LAB = 1000 cm
K l K g=
K l =E I
12/L^3
6/L^2
-12/L^3
6/L^2
6/L^2
2/L
6/L^2-
4/L
12/L^3-
6/L^2-
12/L^3
6/L^2-
6/L^2
-6/L^2
4/L
2/L
K l = EI
0.024
0.12
-0.024
0.12
0.12
0.8
-0.12
0.4
-0.024
-0.12
0.024
-0.12
0.12
0.4
-0.12
0.8
B C
�ِB
C
First element : (C-D )
Start Joint : C
End Joint : DAngle : 0s = sin = 0c = cos = 1EA Is conastant
LAB = 500 cm
K l K g=
K l = E I
12/L^3
6/L^2
-12/L^3
6/L^2
6/L^2
2/L
6/L^2-
4/L
12/L^3-
6/L^2-
12/L^3
6/L^2-
6/L^2
-6/L^2
4/L
2/L
K l = EI
0.096
0.24
-0.096
0.24
0.24
0.8
-0.24
0.4
-0.096
-0.24
0.096
-0.24
0.24
0.4
-024
0.8
C D
C
D
Assembly :
K g= EI
0.024
0.12
-0.024
0.12
0.12
0.8
-0.12
0.4
-0.024
-0.12
0.024
-0.12
0.12
0.4
-0.12
0.8
B C
�ِB
C
K g= EI
0.096
0.24
-0.096
0.24
0.24
0.8
-0.24
0.4
-0.096
-0.24
0.096
-0.24
0.24
0.4
-024
0.8
C D
C
D
Ks = EI
0.024
.0937
0.12
-0.024
0.12
0
0
0.12
0.8
-0.12
0.4
0
0
-0.024
-0.12
0.12
0.12
-0.096
0.024
0.12
0.4
0.12
1.6
-0.024
0.4
0
0
-0.096
-0.24
0.096
-0.24
0
0
0.24
0.4
-0.24
0.8
B C D
C
B
D
K l K g=
Partition
KuuK =Kru
Kur
Krr
u ru
r
Kuu = EI 0.8
0.4
0
0.4
1.60.4 0.8
0.4
0
Force vectorTransformation from member forces to Joint forces
L
P
8LP
8LP
128
P a b2
L2
L
P
a b P b a2
L2
100 kN
100 kN200 kN
250 kNm250 kNm
300 kNm
200 kNm
Fixed End Reaction (FER)
3
A B C D
5 5 2
100 kN 200 kN 100 kN
3 2
250 kN
E
250 kN180 kNm 120 kNm
100 kN
100 kN200 kN
250 kNm250 kNm
300 kNm
200 kNm
Fixed End Action (FEA)
3
A B C D
5 5 2
100 kN 200 kN 100 kN
3 2
250 kN
E
250 kN180 kNm 120 kNm
250 kNm250 kNm
300 kNm
200 kNm
3
A B C D
5 5 2
100 kN 200 kN 100 kN
3 2
250 kN
E
180 kNm 120 kNm
50 kNm 70 kNm 80 kNm
3
A B C D
5 5 2
100 kN 200 kN 100 kN
3 2
250 kN
E
50 kNm 70 kNm 80 kNm
F1
F2 =
50
70
F3 - 80
F = K DThe stiffness equation
=
50
70- 80
0.80.4
0
0.41.6
0.4
00.4
0.8EI
B
C
D
=
50
70- 80
0.80.4
0
0.41.6
0.4
00.4
0.8
B
C
D
1EI
-1
=1EI
27.08
70.83- 135.42
=
B
C
D
1EI
27.08
70.83- 135.42
Internal forces in beam elements
2 EILMAB= ( 2 + )
MBA=2 EI
L( + 2 )
M(FER) AB +
M(FER) BA +
100 kN
100 kN200 kN
250 kNm250 kNm
300 kNm
200 kNm
Fixed End Reaction (FER)
3
A B C D
5 5 2
100 kN 200 kN 100 kN
3 2
250 kN
E
250 kN180 kNm 120 kNm
2 E(2I)LMBC= ( 2 + ) CM(FER) BC +
MCB=2 E(2I)
L( + 2 ) CM(FER) CB +
= 250 + 4/10 (2x27.08+70.83) = 300
= -250 + 4/10 (27.08+2x70.83) = - 182.5
200 kN
250 kNm250 kNm
B C
=
B
C
D
1EI
27.08
70.83- 135.42
2 E(2I)LMBC= ( 2 + ) CM(FER) BC +
MCB=2 E(2I)
L( + 2 ) CM(FER) CB +
= 250 + 4/10 (2x27.08+70.83) = 300
= -250 + 4/10 (27.08+2x70.83) = - 182.5
200 kN
250 kNm250 kNm
B C
=
B
C
D
1EI
27.08
70.83- 135.42
MBC= 300MCB= -182.5
MCD= 182.5MDC= -200
300200
B.M.D
A
B C D
E
182.5
B.M.D
3
A B C D
5 5 2
100 kN 200 kN 100 kN
3 2
250 kN
E
300200
A
B C D
E
182.5241.25
500
258.75
189.5
300
110.5
Example 7:Draw B.M.D for the shown beam where EI is constant for all members
A B
5 5
240 kN
C
5 5
120 kN
K l
12 EIL3
6 EIL2
6 EIL2
4 EIL
-12 EIL3
=
-12 EIL3
-6 EIL2
-6 EIL2
12 EIL3
-6 EIL2
6 EIL2
2 EIL
6 EIL2
2 EIL
-6 EIL2
4 EIL
4 EIL
First element : (A-B )
EA conastant
LAB = 10 m
12 EIL3
=0.012EI
6 EIL2
4 EIL
2 EIL
=0.06 EI
=0.4 EI
=0.2 EI
K l =
0.012 EI
0.012 EI 0.012 EI
0.012 EI
0.06 EI
0.06 EI
0.06 EI0.06 EI
0.06 EI
0.06 EI
0.06 EI
0.06 EI
0.4 EI
0.4 EI
0.2 EI
0.2 EI
A
A
B
B
Second element : ( B-c)
EI conastant LAB = 10 m
12 EIL3
=0.012EI
6 EIL2
4 EIL
=0.06 EI
=0.4 EI
=0.2 EI2 EI
L
K l =
0.012 EI
0.012 EI 0.012 EI
0.012 EI
0.06 EI
0.06 EI
0.06 EI0.06 EI
0.06 EI
0.06 EI
0.06 EI
0.06 EI
0.4 EI
0.4 EI
0.2 EI
0.2 EI
A
A
B
B
K gK l =
Assembly :
=
K gg1
0.012 EI 0.06 EI 0.012 EI 0.06 EI
0.06 EI 0.4 EI 0.06 EI 0.2 EI
0.012 EI 0.06 EI 0.06 EI0.012 EI
0.06 EI 0.2 EI 0.4 EI0.06 EI
2=
K gg
0.012 EI
0.012 EI 0.012 EI
0.012 EI
0.06 EI
0.06 EI
0.06 EI0.06 EI
0.06 EI
0.06 EI
0.06 EI
0.06 EI
0.4 EI
0.4 EI
0.2 EI
0.2 EI
A B
A
B
=
0.012
-0.012
-0.012
0.06
0.06
-0.06 0.06
0.06
-0.06
0.06
-0.06
0.4
0.4
0.2
0.2
A
A
B
B
K s
0.0
0.8
0.024
0.0
0.0
0.0
0.0 0.0
0.0
0
0.0
0.0
0.012
-0.012
-0.012
0.06
-0.06
0.2
0.2
-0.06
-0.06
EI
c
c
Partition
KuuK =Kru
Kur
Krr
u ru
r
Force vectorTransformation from member forces to Joint forces
L
P
8LP
8LP-
240 kN300 kNm300 kNm
Fixed End Reaction
(FER)
A B
5 5
240 kNC
5 5
120 kN
120 kN150 kNm150 kNm
300 kNm300 kNm
A B
5 5
240 kNC
5 5
120 kN
150 kNm150 kNm
120120 6060
F1
F2
=
120
-300
F3
F6
F5
F4
180
150
60
150
120
-300
180
150
60
150
0.012
-0.012
-0.012
0.06
0.06
-0.06 0.06
0.06
-0.06
0.06
-0.06
0.4
0.4
0.2
0.2
0.0
0.8
0.024
0.0
0.0
0.0
0.0 0.0
0.0
0
0.0
0.0
0.012
-0.012
-0.012
0.06
-0.06
0.2
0.2
-0.06
-0.06
d1
d2
d3
d4
d5
d6
= EI
120
-300
180
150
60
150
-0.012 0.06
-0.06
0.06
-0.06
0.06
-0.06
0.4
0.4
0.2
0.2
0.0
0.8
0.024
0.0
0.0
0.0
0.0 0.0
0.0
0.0
0.0
0.0
0.012
-0.012
-0.012
0.06
-0.06
0.2
0.2
-0.06
-0.06
d1
d2
d3
d4
d5
d6
=EI
=B
C
150
150
0.8
0.2 0.4
0.2EI
B
C=
1EI
-1150
150
0.8
0.2 0.4
0.2
B
C=
1EI
107.14
321.43
B
C=
1EI
-1150
150
0.8
0.2 0.4
0.2
Internal forces in beam elements
2 EILMAB= ( 2 + )
MBA=2 EI
L( + 2 )
M(FER) AB +
M(FER) BA +
240 kN300 kNm300 kNm
Fixed End Reaction
(FER)
A B
5 5
240 kNC
5 5
120 kN
120 kN150 kNm150 kNm
2 EILMAB= ( 2 + )
MBA=2 EI
L( + 2 )
M(FER) AB +
M(FER) BA +
= 300 + 2/10 (107.14) = 321.4
= -300 + 2/10 (2x107.14) = - 257.1
B
C=
1EI
107.14
321.43
240 kN
300 kNm300 kNm
A B
2 EILMBC= ( 2 + ) C
MCB=2 EI
L( + 2 ) C
M(FER) BC +
M(FER) CB +
= 150 + 2/10 (2x107.14+321.43) = 257.1
= -150 + 2/10 (107.14+2x321.43) = 0
120 kN
150 kNm150 kNm
B C B
C=
1EI
107.14
321.43
MAB= 321.4MBA= -257.1
MBC= 257.1MCB= 0
321.4257.1
B.M.D
93.8
128.55
300
171.45
A B
5 5
240 kNC
5 5
120 kN
321.4257.1
B.M.D
289.25
600
310.75
Beams with settlement
6 EIL2
6 EIL2 12 EI
L3
12 EIL3
Fixed End Reaction
(FER)
Beams with settlement
6 EIL2
6 EIL2
12 EIL3
12 EIL3
Fixed End Reaction
(FER)
Beams with settlement
3 EIL2 3 EI
L3
3 EIL3
Fixed End Reaction
(FER)
Beams with settlement
3 EIL2
3 EIL3
3 EIL3
Fixed End Reaction
(FER)
169
Example 8:Draw B.M.D for the shown beam due to the shown loads and vertical downward settlement at support B (2000/EI) and at support C (1000/EI) where EI is constant for all members
A B
5 5
240 kN
C
5 5
120 kN
=0.8
0.2 0.4
0.2K EI
From example 7 :
240 kN300 kNm300 kNm
Fixed End Reaction
(FER)
A B
5 5
240 kNC
5 5
120 kN
120 kN150 kNm150 kNm
ForLoads
Fixed End Reaction (FER)
A B
5 5
C
5 5
Forsettlement
1000EI
2000EI
6 EIx1000102 EI
6 EIx1000102 EI
6 EIx2000102 EI
6 EIx2000102 EI
120
120 6060
300 kNm300 kNm
A B
5 5
240 kNC
5 5
120 kN
150 kNm150 kNmFor
Loads
Forsettlement 120 120 60 60
Fixed End Reaction
(FER)
210 kNmTotal 90 kNm
420 180 90 210
A B
5 5
240 kNC
5 5
120 kN
Fixed End Reaction
(FER)210 kNm90 kNm
Fixed End Action (FEA)
210 kNm90 kNm
A B
5 5
240 kNC
5 5
120 kN
Fixed End Action (FEA)
210 kNm90 kNm
F1
F2
=90
210
F = K Dk11F1
F2
=k21
k12
k22
d1
d2
=B
C
B
C=
1EI
-1
90
210
90
210
0.8
0.2 0.4
0.2EI
0.8
0.2 0.4
0.2
B
C=
1EI
-21.43
535.71
B
C=
1EI
-190
210
0.8
0.2 0.4
0.2
Internal forces in beam elements
2 EILMAB= ( 2 + )
MBA=2 EI
L( + 2 )
M(FER) AB +
M(FER) BA +
300 kNm300 kNm
A B
5 5
240 kNC
5 5
120 kN
150 kNm150 kNmFor
Loads
Forsettlement 120 120 60 60
Fixed End Reaction
(FER)
210 kNmTotal 90 kNm
420 180 90 210
2 EILMAB= ( 2 + )
MBA=2 EI
L( + 2 )
M(FER) AB +
M(FER) BA +
= 420 + 2/10 (-21.43) = 415.7
= -180 + 2/10 (2x-21.43) = - 188.6
180 kNm420 kNm
A B B
C=
1EI
-21.43
535.71
2 EILMBC= ( 2 + ) C
MCB=2 EI
L( + 2 ) C
M(FER) BC +
M(FER) CB +
= 90 + 2/10 (2x-21.43+535.7) = 188.6
= -210 + 2/10 (-21.43+2x535.71) = 0
210 kNm90 kNm
B C B
C=
1EI
-21.43
535.71
MAB= 321.4MBA= -257.1
MBC= 257.1MCB= 0
415.7188.6
B.M.D
93.8
94.3
300
205.7
A B
5 5
240 kNC
5 5
120 kN
415.7188.6
B.M.D
302.15
600
297.85