matrix stiffnes beam, frame
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Transcript of matrix stiffnes beam, frame
1
! Development: The Slope-DeflectionEquations
! Stiffness Matrix! General Procedures! Internal Hinges! Temperature Effects! Force & Displacement Transformation! Skew Roller Support
BEAM ANALYSIS USING THE STIFFNESSMETHOD
2
Slope � Deflection Equations
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
3
� Degrees of Freedom
L
θΑ
A B
M
1 DOF: θΑ
PθΑ
θΒΒΒΒ
A BC 2 DOF: θΑ , θΒΒΒΒ
4
L
A B
1
� Stiffness Definition
kBAkAA
LEIkAA
4=
LEIkBA
2=
5
L
A B1
kBBkAB
LEIkBB
4=
LEIkAB
2=
6
� Fixed-End ForcesFixed-End Forces: Loads
P
L/2 L/2
L
w
L
8PL
8PL
2P
2P
12
2wL12
2wL
2wL
2wL
7
� General Case
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
8
wP
settlement = ∆j
(MFij)∆ (MF
ji)∆
(MFij)Load (MF
ji)Load
+
Mij Mji
θi
θj
+
i jwPMij
Mji
settlement = ∆jθj
θi
ψ
=+ ji LEI
LEI θθ 24
ji LEI
LEI θθ 42
+=
,)()()2()4( LoadijF
ijF
jiij MMLEI
LEIM +++= ∆θθ Loadji
Fji
Fjiji MM
LEI
LEIM )()()4()2( +++= ∆θθ
L
9
Mji Mjk
Pi j kw Cj
Mji Mjk
Cj
j
� Equilibrium Equations
0:0 =+−−=Σ+ jjkjij CMMM
10
+
1
1
i jMij Mji
θi
θj
LEIkii
4=
LEIk ji
2=
LEIkij
2=
LEIk jj
4=
iθ×
jθ×
� Stiffness Coefficients
L
11
[ ]
=
jjji
ijii
kkkk
k
Stiffness Matrix
)()2()4( ijF
jiij MLEI
LEIM ++= θθ
)()4()2( jiF
jiji MLEI
LEIM ++= θθ
+
=
F
ji
Fij
j
iI
ji
ij
MM
LEILEILEILEI
MM
θθ
)/4()/2()/2()/4(
� Matrix Formulation
12
[D] = [K]-1([Q] - [FEM])
Displacementmatrix
Stiffness matrix
Force matrixwP(MF
ij)Load (MFji)Load
+
+
i jwPMij
Mji
θj
θi
ψ ∆j
Mij Mji
θi
θj
(MFij)∆ (MF
ji)∆
Fixed-end momentmatrix
][]][[][ FEMKM += θ
]][[])[]([ θKFEMM =−
][][][][ 1 FEMMK −= −θ
L
13
L
Real beam
Conjugate beam
� Stiffness Coefficients Derivation
MjMi
L/3
θi
LMM ji +
EIM j
EIMi
θι
EILM j
2
EILMi
2
)1(2
0)3
2)(2
()3
)(2
(:0'
−−−=
=+−=Σ+
ji
jii
MM
LEI
LMLEI
LMM
)2(0)2
()2
(:0 −−−=+−=Σ↑+EI
LMEI
LMF jiiy θ ij
ii
LEIM
LEIM
andFrom
θ
θ
)2(
)4(
);2()1(
=
=
LMM ji +
i j
14
� Derivation of Fixed-End Moment
Real beam
8,0
162
22:0
2 PLMEI
PLEI
MLEI
MLFy ==+−−=Σ↑+
P
M
M EIM
Conjugate beamA
EIM
B
L
P
A B
EIM
EIML2
EIM
EIML2
EIPL
16
2
EIPL4 EI
PL16
2
Point load
15
L
P
841616PLPLPLPL
=+−
+−
8PL
8PL
2P
2PP/2
P/2
-PL/8 -PL/8
PL/8
-PL/16-PL/8
-
PL/4+
-PL/16-PL/8-
16
Uniform load
L
w
A B
w
M
M
Real beam Conjugate beam
A
EIM
EIM
B
12,0
242
22:0
23 wLMEI
wLEI
MLEI
MLFy ==+−−=Σ↑+
EIwL8
2
EIwL
24
3
EIwL
24
3
EIM
EIML2
EIM
EIML2
17
Settlements
M
M
L
∆
MjMi = Mj
LMM ji +L
MM ji +
Real beam
2
6LEIM ∆
=
Conjugate beam
EIM
A B
EIM
∆
,0)3
2)(2
()3
)(2
(:0 =+−∆−=Σ+L
EIMLL
EIMLM B
EIM
EIML2
EIMEI
ML2
∆
18
� Typical Problem
0
0
0
0
A C
B
P1P2
L1 L2
wCB
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
P
8PL
8PL
L
w12
2wL
12
2wL
L
19
MBA MBC
A C
B
P1P2
L1 L2
wCB
B
CB
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0
20
A C
B
P1P2
L1 L2
wCB
Substitute θB in MAB, MBA, MBC, MCB
MABMBA
MBC
MCB
0
0
0
0
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
21
A BP1
MAB
MBA
L1
A CB
P1P2
L1 L2
wCB
ByR CyAy ByL
Ay Cy
MABMBA
MBC
MCB
By = ByL + ByR
B CP2
MBCMCB
L2
22
2EI EI
Stiffness Matrix
� Node and Member Identification
� Global and Member Coordinates
� Degrees of Freedom
�Known degrees of freedom D4, D5, D6, D7, D8 and D9� Unknown degrees of freedom D1, D2 and D3
1 21 2 31
2 3
7
8
9
4
5
6
1
2 3
7
8
9
4
5
6
23
i j
E, I, A, L
Beam-Member Stiffness Matrix
d4 = 1
AE/LAE/L AE/L AE/L
64
5
1
2
3
[k] =
1 2 3 4 5 6
3
2
1
4
6
5
d1 = 1
k11 = = k44
0
0
0
0
0
0
0
0
- AE/LAE/L
-AE/L AE/L
k41 k14
24
d2 = 1
[k] =
1 2 3 4 5 6
3
2
1
4
6
5
0
0
0
0
0
0
0
0
- AE/LAE/L
-AE/L AE/L
6EI/L2
6EI/L2
d5 = 1
6EI/L2
6EI/L2
12EI/L312EI/L3 12EI/L3
12EI/L3
6EI/L2
12EI/L3
- 6EI/L2
- 12EI/L3
0
0
6EI/L2
-12EI/L3
0
0
- 6EI/L2
12EI/L3
k22 = = k55
= k52
k62 =
= k32 = k65
= k25
= k35
i j
E, I, A, L6
4
5
1
2
3
25
[k] =
1 2 3 4 5 6
3
2
1
4
6
5
0
0
0
0
0
0
0
0
- AE/LAE/L
-AE/L AE/L
6EI/L2
12EI/L3
- 6EI/L2
- 12EI/L3
0
0
6EI/L2
-12EI/L3
0
0
- 6EI/L2
12EI/L3
4EI/L
6EI/L2 6EI/L2
2EI/L
0
0
2EI/L
-6EI/L2
0
0
-6EI/L2
4EI/L
d6 = 1
d3 = 1 2EI/L 4EI/L
6EI/L26EI/L26EI/L2
4EI/L 2EI/L
6EI/L2
k33 =
k23 = = k53
= k63 = k36
= k56 k26 =
= k66
i j
E, I, A, L6
4
5
1
2
3
26
FFxi
FFyi
MFi
FFxj
FFyj
MFj
+
MFi MF
j
FFxjFF
xi
FFyi FF
yj
1
6EI/L2
6EI/L2
12EI/L312EI/L3
+
x ∆i
12EI/L3
6EI/L2
6EI/L2
12EI/L3
1
+
4EI/L 2EI/L
6EI/L26EI/L2
x θi
6EI/L2
4EI/L 2EI/L
6EI/L2
x δjAE/L
1
AE/L
+
AE/L AE/L
6EI/L2
6EI/L2
1
12EI/L312EI/L3
x ∆j
6EI/L2
6EI/L2
12EI/L3
12EI/L3
1
+
2EI/L
6EI/L26EI/L2
4EI/L
x θj
6EI/L2
2EI/L
6EI/L2
4EI/L
Fxi
Fyi
MiFxj
Fyj
Mj
� Member Equilibrium Equations
Mj
i jFxj
Fxi
Fyi Fyj
=
MiE, I, A, L
x δiAE/L AE/L
1
AE/LAE/L
27
+
−−−−
−−−
−
=
Fj
Fyi
Fxj
Fi
Fyi
Fxi
j
j
j
i
i
i
j
yj
xj
i
yj
xi
MFFMFF
θ∆δθ∆δ
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
MFFMFF
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
22
2323
22
2323
Fjjjjiiij
Fyjjjjiiiyj
Fxijjjiiixj
Fijjjiiixi
Fyijjjiiiyi
Fxijjjiiixi
MθLEI∆LEIδθLEI∆LEIδMFθLEI∆δθLEI∆LEIδFFθ∆δLAEθ∆δLAEFMθLEI∆LEIδθLEI∆LEIδMFθLEI∆LEIδθLEI∆LEIδFFθ∆δLAEθ∆δLAEF
)/4()/6()0()/2()/6()0()/6()0()0()/6()/12()0(
)0()0()/()0()0()/()/2()/6()0()/4()/6()0()/6()/12()0()/6()/12()0(
)0()0()/()0()0()/(
22
223
22
2323
−=−−−=
−=−=−+=
+++−++=
[q] = [k][d] + [qF]
Fixed-end force matrix
End-force matrix
Stiffness matrix
Displacement matrix
28
6x6 Stiffness Matrix
[ ]
−−−−
−−−
−
=×
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
k
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
22
2323
22
2323
66
θi ∆j θj∆i δjδi
Mi
Vj
Mj
Vi
Nj
Ni
4x4 Stiffness Matrix
[ ]
−−−−
−−
=×
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
44
θi ∆j θj∆i
Mi
Vj
Mj
Vi
29
Comment: - When use 4x4 stiffness matrix, specify settlement. - When use 2x2 stiffness matrix, fixed-end forces must be included.
2x2 Stiffness Matrix
θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
30
21 312
1
2
3
4
5
6
General Procedures: Application of the Stiffness Method for Beam Analysis
wP P2y
M2M1
Local
1´
2´
3´
4´
1
1´
2´
3´
4´
2
Global
31
Member
1
2
3
4
1
3
4
5
6
2
1
2
3
4
5
6
3121 2
Global
wP P2y
M2M1
Local
1´
2´
3´
4´
1
1´
2´
3´
4´
2
32
(q´F4 )2
(q´F3 )2
(q´F1)2
(q´F2)2
w
2
P
(q´F4)1
(q´F3)1
(q´F2)1
(q´F1 )1
1[FEF]
1
2
3
4
5
6
3121 2
Global
Local
1´
2´
3´
4´
1
1´
2´
3´
4´
2
wP P2y
M2M1
33
[q] = [T]T[q´]
1
2
3
4
5
6
3121 2
Global
Local
1´
2´
3´
4´
1
[k] = [T]T[q´] [T]
=
'4
'3
'2
'1
4
3
2
1
1000010000100001
qqqq
qqqq
1´ 4´2´ 3´12341
34
[q] = [T]T[q´]
1
2
3
4
5
6
3121 2
Global
Local
1´
2´
3´
4´
2
[k] = [T]T[q´] [T]
=
'4
'3
'2
'1
6
5
4
3
1000010000100001
qqqq
qqqq
1´ 4´2´ 3´34562
35
Stiffness Matrix:
21 312
1
2
3
4
5
6
[k]1 =
1
2
3
4
1 2 3 4
[k]2 =
3
4
5
6
3 4 5 6
Member 1
[K] =
1
23
4
5
6
1 2 3 4 5 6
Member 2
Node 2
1 2
36
+
=
F
F
F
F
F
F
QQQQQQ
DDDDDD
QQQQQQ
6
5
1
4
3
2
6
5
1
4
3
2
6
5
1
4
3
2 234156
5 612 3 4
000
M1z
-P2y
M3z
KBA
KAB
KBB
KAA
21 312
1
2
3
4
5
6
ReactionQu
Joint LoadQk
Du=Dunknown
Dk = DknownQF
B
QFA
[ ] [ ][ ] [ ]FAuAAk QDKQ +=
[ ] [ ] [ ] [ ])(1 FAkAAu QQKD −+= −
[ ] [ ][ ] [ ]FqdkqForceMember +=:
37
+
6
5
4
3
2
1
DDDDDD
Global:
21 312
1
2
3
4
5
6
1 2
2
3
4
51
6
(q´F6 )2
(q´F5 )2
(qF4)2
(qF3)2
w
2
P
(qF4)1
(qF3)1
(qF2)1
(qF1 )1
1[FEF]
=−=
=
24
23
12
MQPQ
MQ
y
4
3
2
DDD
+F
F
F
QQQ
4
3
2
2 3 4
Node 2
=
234
Member 1
123456
1 2 3 4 5 6
Member 2
Node 2=
6
5
4
3
2
1
QQQQQQ
+=+=
F
F
F4
F4
F
F3
F3
F
F
F
qqQqqQ
6
5
214
213
2
1
)()()()(
M1
-P2y
M2
0
00
38
1 2 312
1
2
3
4
5
6
4
3
2
DDD
2 3 4
Node 2
=
234
−
=−=
=
F
F
F
QQQ
MQPQ
MQ
4
3
2
24
2y3
12
39
Member 1:
21 312
1
2
3
4
5
6
P
qF4
qF3
qF2
qF1
1
[FEF]
4
3
2
1
qqqq
====
44
33
22
1 0
DdDdDd
d
+
F
F
F
F
qqqq
4
3
2
11234
1 2 3 4
= 1k
40
21 312
1
2
3
4
5
6
Member 2:
qF6
qF5
qF4
qF3
w
2
[FEM]
6
5
4
3
qqqq
==
==
00
6
5
44
33
dd
DdDd
+
F
F
F
F
qqqq
6
5
4
31234
1 2 3 4
= 1k
41
AC
B
9 m 3 m
10 kN1 kN/m
1.5 m
Example 1
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.
42
21Global
123
1Members
23
2
12
AC
B
9 m 3 m
10 kN1 kN/m
1.5 m
wL2/12 = 6.75wL2/12 = 6.75 PL/8 = 3.75PL/8 = 3.75
10 kN
1.5 m1.5 m
2
1 kN/m
9 m 1[FEF]
43
4/3
2/3
2/3
21
123
9 m 3 m
[k]1= EI4/9
2/9
3
4/9
2/92
3
2[k]2 = EI
4/3
4/3
2/3
2/3
2 12
1
[K] = EI
2 12
1
(4/9)+(4/3)
21
123
Stiffness Matrix: θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
44
A CB
10 kN1 kN/m
123
Equilibrium equations: MCB = 0MBA + MBC = 0
Global Equilibrium: [Q] = [K][D] + [QF]
=2.423/EI
0.779/EI
θC
θB
9 m 1.5 m1.5 m6.756.75
+-3.75
-6.75 + 3.75 = -3MBA + MBC = 0
MCB = 01
2
θC
θB
4/3
2/3
2/3 = EI
2 12
1
(4/9)+(4/3)
3.753.756.75
45
= -6.40
6.92
Member 1 : [q]1 = [k]1[d]1 + [qF]1
MBA
MAB
2
3
= 0.779/EIθB
θA
0
+ -6.75
6.75
Substitute θB and θC in the member matrix,
1
23
wL2/12 = 6.75wL2/12 = 6.75
1 kN/m
9 m 1[FEF]
4.56 kN 4.44 kN
6.92 kN�m 6.40 kN�m
wL2/12 = 6.75wL2/12 = 6.75
= EI4/9
2/9
3
4/9
2/92
1 kN/m
19 m
46
Member 2 : [q]2 = [k]2[d]2 + [qF]2
Substitute θB and θC in the member matrix,
MCB
MBC
1
2 = EI
4/3
4/3
2/3
2/3
2 1
= 0
6.40
θC
θB
= 2.423/EI
= 0.779/EI
2
12
[FEF]PL/8 = 3.75PL/8 = 3.75
10 kN
1.5 m1.5 m
2
7.13 kN 2.87 kN
6.40 kN�m
10 kN
2
PL/8 = 3.75PL/8 = 3.75
+-3.75
3.75
47
A CB
10 kN1 kN/m
9 m 1.5 m1.5 m
6.92 kN�m
11.57 kN 2.87 kN4.56 kN
4.56 kN 4.44 kN 7.13 kN 2.87 kN
1 kN/m6.92 6.40
6.40
10 kN
M (kN�m) x (m)
-6.92 -6.40
3.48 4.32
V (kN) x (m)
4.56
-4.44 -2.874.56 m
7.13
θB = +0.779/EIθC = +2.423/EI
48
20 kN
A CBEI2EI
4 m 4 m
9kN/m 40 kN�m
Example 2
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.
49
1 2 3
0.5625
3
-0.375
-0.375
1
2
3
4
5
6
[K] = EI
3 4
3
4
Use 4x4 stiffness matrix,
[k]1
0.375EI
-0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
[k]2
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
2EI EI
1 2
1
2
5
6
θi ∆j θj
Mi
∆i
Vj
Mj
Vi
[ ]
−−−−
−−
=
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
50
20 kN9kN/m 40 kN�m
4 m 4 m
12 kN�m18 kN
12 kN�m
18 kN 18 kN
D4
D3
Global:
D4
D3=
9.697/EI
-61.09/EI
Q4 = 40
Q3 = -203
4 + -12
18 3
4
20 kN
40 kN�m
12 kN�m
1 2
1
2
3
4
5
6
1 2 3
2EI EI
[Q] = [K][D] + [QF]
0.5625
3
-0.375
-0.375= EI
3 4
3
4
51
Member 1:
1
9 kN/m
A B12 kN�m
[qF]1
12 kN�m
18 kN 18 kN
1
9 kN/m
A B
[q]1
53.21 kN�m
48.18 kN67.51 kN�m
12.18 kN
1
1
2
3
4
1 2
2EI 12 kN�m 12 kN�m
18 kN 18 kN
q2
q1
q4L
q3L
67.51
48.18
53.21
-12.18
12
18
-12
18
12(2EI)/43
-0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
[q]1 = [k]1[d]1 + [qF]1
d3 = -61.09/EI
d4 = 9.697/EI
d2 = 0
d1 = 0
52
2B C
[q]2
Member 2:
[q]2 = [k]2[d]2 + [qF]2
2
3
4
5
6
2 3
EI
q4R
q3R
q6
q5
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
0
0
0
0
d3 =-61.09/EI
d4 = 9.697/EI
d6 = 0
d5 = 0
18.06 kN�m
7.82 kN
13.21 kN�m
7.82 kN
-13.21
-7.818
-18.06
7.818
53
V (kN) x (m)
-7.818
48.18
-7.818-
M (kN�m) x (m)
13.21
-18.08-
-67.51
-
1
9 kN/m
A B
[q]1
53.21 kN�m
48.18 kN67.51 kN�m
12.18 kN2
B C
[q]2
18.06 kN�m
7.82 kN
13.21 kN�m
7.82 kN
12.18+
53.21
+
20 kN
4 m 4 m
9kN/m 40 kN�m
7.818 kN
18.06 kN�m67.51 kN�m
48.18 kN
D3 = ∆B = -61.09/EI
D4 = θB = +9.697/EI
54
20 kN
A CBEI2EI
4 m 2 m
9kN/m 40 kN�m10 kN
2 m
Example 3
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.
55
Global1 2
1
2
3
4
5
6
1 2 3
20 kN
A CB EI2EI
9kN/m 40 kN�m10 kN
4 m 2 m 2 m
10 kN5 kN�m
5 kN
5 kN�m
5 kN
212 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
θi ∆j θj
Mi
∆i
Vj
Mj
Vi
[ ]
−−−−
−−
=×
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
44
56
1 2
1
2
3
4
5
6
1 2 3 4 m 2 m 2 m
1
2
5
1 2
0.375
0.50.375 0.5 1
[K] = EI
3 4 6
3
46
0.5625
3
-0.375
-0.375
12(2EI)/43
[k]1 =-0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI/L
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
[k]2 =0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
57
MBA+MBC = 40
VBL+VBR = -20
MCB = 0θB
∆B
θC
-12 + 5 = -7-5
θB
∆B
θC
= -7.667/EI
-116.593/EI
52.556/EI
Global: [Q] = [K][D] + [QF]
= EI
0.5625
3
-0.375
-0.375
0.375
0.50.375 0.5 1
3 4 6
3
46
20 kN
A CB
9kN/m 40 kN�m10 kN
21
1
2
3
4
5
6
1 2 3 10 kN5 kN�m
5 kN
5 kN�m
5 kN
212 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
20 kN
40 kN�m
5 kN�m5 kN�m
5 kN
12 kN�m
18 kN
+18 + 5 = 23
58
0.375EI
= -0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI/L
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
=91.78
55.97
60.11
-19.97
91.78 kN�m
55.97 kN
Member 1: [q]1 = [k]1[d]1 + [qF]1
MAB
VA
MBA
VBL+
12
18
-12
18=-116.593/EI
=-7.667/EI
0
0
θB
∆B
1
1
2
3
4
1 2
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1
[FEF]
19.97 kN
60.11 kN�m4 m
9kN/m
1
12 kN�m
18 kN
12 kN�m
18 kN18 kN
59
VC
MBC
VBR
MCB
+ 5
5
5 -5
=-20.11
-0.0278
0
10.03=52.556/EI
=-116.593/EI
=-7.667/EIθB
∆B
θC
0
10 kN5 kN�m
5 kN
5 kN�m
5 kN
2
[FEM]
10.03 kN
2 m
10 kN
2 m2
= 0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
0.0278 kN
20.11 kN�m
Member 2: [q]1 = [k]1[d]1 + [qF]1
223
5
6
3
45 kN�m
5 kN
5 kN�m
5 kN
60
20 kN
4 m 2 m
9kN/m 40 kN�m10 kN
2 m
10.03 kN
91.78 kN�m
55.97 kN
V (kN) x (m)
55.97
-0.0278
19.97+
-10.03 -10.03-
91.78 kN�m
55.97 kN 19.97 kN60.11 kN�m4 m
9kN/m
110.03 kN0.0278 kN
20.11 kN�m2 m
10 kN2 m
2
91.78 kN�m
60.11 kN�m
20.11 kN�m
M (kN�m) x (m)
-91.78
-
+20.11
60.11
θB = -7.667/EI
θC = 52.556/EI
∆B = -116.593/EI
61
40 kN
8 m 4 m 4 m
2EI EI
6 kN/m
AB
C∆B = -10 mm
Example 4
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support B settlement 10 mm.
62
[k]1 =4
8
8
4
1 2
1
2
EI8
1 2
1 2 3
2EI EI
1
2
42
2 3
2
3[K] = EI
8
12
2
4
4
2
2 3
2
3[k]2 = EI
8
Use 2x2 stiffness matrix:θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
63
75 kN�m
6(2EI)∆/L2 = 75 kN�m10 mm
1 37.5 kN�m
6(EI)∆/L2 = 37.5 kN�m10 mm
2[FEM]∆∆∆∆
6(EI)∆/L2 = 37.5 kN�m
D2
D3
=-61.27/EI rad
185.64/EI rad
40 kN
2EI EI
6 kN/m
A
B C
∆B = -10 mm 1 2
1 2 3
[FEM]load
32 kN�mwL2/12 = 32 kN�m
6 kN/m
1 40 kN�mPL/8 = 40 kN�m 2
40 kN8 m 4 m 4 m
32 kN�m 40 kN�mPL/8 = 40 kN�m
75 kN�m 37.5 kN�m
+-32 + 40 + 75 -37.5 = 45.5Q2 = 0
Q3 = 0 -40 - 37.5 = -77.5
D2
D3
2
42
2 3
2
3
12
8EI
=
Global: [Q] = [K][D] + [QF]
64
6(2EI)∆/L2 = 75 kN�m
[FEM]load
32 kN�mwL2/12 = 32 kN�m
6 kN/m
1
75 kN�m
6(2EI)∆/L2 = 75 kN�m10 mm
1[FEM]∆∆∆∆
6 kN/m
18 m
1
2
2EI
1
q1
q2
d1 = 0
d2 = -61.27/EI=
76.37 kN�m
-18.27 kN�m
Member 1: [q]1 = [k]1[d]1 + [qF]1
32 kN�mwL2/12 = 32 kN�m
75 kN�m
+32 + 75 = 107
-32 + 75 = 43
31.26 kN
76.37 kN�m
16.74 kN
18.27 kN�m
4
8
8
4
1 2
1
28EI
=
65
Member 2: [q]2 = [k]2[d]2 + [qF]2
2
2 3
40 kN�mPL/8 = 40 kN�m 2
40 kN
37.5 kN�m
6(EI)D/L2 = 37.5 kN�m10 mm
2
[FEM]load
[FEM]∆∆∆∆
40 kN�m
37.5 kN�m
6(EI)D/L2 = 37.5 kN�m
PL/8 = 40 kN�m
=18.27 kN�m
0 kN�m
q2
q3
+40 - 37.5 = 2.5
-40 - 37.5 = -77.5
d2 = -61.27/EI
d3 = 185.64/EI
17.72 kN
18.27 kN�m
22.28 kN
40 kN
2
2
4
4
2
2 3
2
38EI
=
66
Alternate method: Use 4x4 stiffness matrix
2EI EI
1 2
2
5
6
3
4
1
8EI
= 1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
[k]1
8EI
=0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
[k]2
2-0.75
-0.1875-0.75
0.1875-0.75
0.752
-0.754
-0.18750.75
1.5 4
1.54
0.375
-0.3751.5
1.58
-1.5
-0.375-1.5
00
00
00
00
-0.7512
0.5625-0.758
EI=[K]
123456
1 4 5 62 3
67
Global: [Q] = [K][D] + [QF]
40 kN
8 m 4 m 4 m
2EI EI
6 kN/m
A
B C
∆B = -10 mm 1 2
2
5
6
3
4
1
40 kN40 kN�m
20 kN
40 kN�m
20 kN
232 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
1[FEF]
53
2
1
D4
D6=
-61.27/EI = -1.532x10-3 rad
185.64/EI = 4.641x10-3 rad
+(200x200/8)(-0.75)(-0.01) = 37.5
(200x200/8)(0.75)(-0.01) = -37.5
D4
D6
Q4 = 0
Q6= 0
2
4
12
2+
8
-40
4 6
4
68EI
=
-0.75-0.75 D5 = -0.01)
8200200( ×
+ 46
5
D4
D6
Q4 = 0
Q6= 0
2
4
12
2
4 6
4
68EI
= +8
-40
68
+32
24
-32
24
q1
q2
q4
q3
1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
= (200x200)8
q1
q2
q4
q3=
76.37 kN�m
31.26 kN
-18.27 kN�m
16.74 kN
Member 1: [q]1 = [k]1[d]1 + [qF]1
1
2
3
4
1∆B = -10 mm 32 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
1[FEF]load
d2 = 0
d1 = 0
d4 = -1.532x10-3
d3 = -0.01
6 kN/m
18 m
31.26 kN
76.37 kN�m
16.74 kN
18.27 kN�m
69
Member 2:
+40
20
-40
20
q3
q4
q6
q5
d4 = -1.532x10-3
d3 = -0.01
d6 = 4.641x10-3
d5 = 0
0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
= (200x200)8
q3
q4
q6
q5=
18.27 kN�m
22.28 kN
0 kN�m
17.72 kN17.72 kN
18.27 kN�m
22.28 kN
40 kN
2
2
5
6
3
4
-10 mm = ∆B
40 kN40 kN�m
20 kN
40 kN�m
20 kN
2[FEF]
70
40 kN
8 m 4 m 4 m
2EI EI
6 kN/m
A
B C
∆B = -10 mm31.26 kN
76.37 kN�m
16.74 + 22.28 kN 17.72 kN
V (kN)
x (m)+ +--
31.26
-16.74
22.28
-17.725.21 m
M (kN�m) x (m)
-76.37
5.06
-18.27
70.85
+
--
D6 = θC = 4.641x10-3 rad
d4 = θB = -1.532x10-3 rad
Deflected Curve
∆B = -10 mm
71
Example 5
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support C settlement 10 mm.
4 kN
8 m 4 m 4 m
2EI EI
0.6 kN/m
AB
C∆C = -10 mm
20 kN�m
72
2EI EI
1 2-10 mm
2
5
6
3
4
1
2
51
Q3
Q6
Q4
0.75 2
0.7524
1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
[k]1
8EI
=
8EI
=0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
[k]2
D3
D6
D4
-0.1875
-0.75-0.75 D5 = -0.01
� Global: [Q] = [K][D] + [QF]
�Member stiffness matrix [k]4x4
0.5625-0.75
-0.7512
346
3 4 6
8EI
= )8
200200( ×+
346
5QF
3
QF6
QF4+
73
4 kN4 kN�m
2 kN
4 kN�m
2 kN
23.2 kN�m
2.4 kN
3.2 kN�m
2.4 kN
0.6 kN/m
1[FEF]
4 kN
8 m 4 m 4 m2EI EI
0.6 kN/m
A
BC
10 mm
20 kN�m2EI EI
1 2
6
3
42
51
346
0.5625
0.75-0.75
-0.75122
0.752
3 4 6
48EI
=
D3
D6
D4 +9.375
37.537.5
2.4+2 = 4.4
-4.0-3.2+4 = 0.8+
D3
D6
D4
-377.30/EI = -9.433x10-3 m
+74.50/EI = +1.863x10-3 rad-61.53/EI = -1.538x10-3 rad=
Global: [Q] = [K][D] + [QF]
Q3 = 0
Q6 = 20Q4 = 0
74
+3.2
2.4
-3.2
2.4
q1
q2
q4
q3
q1
q2
q4
q3=
43.19 kN�m
8.55 kN
6.03 kN�m
-3.75 kN
Member 1: [q]1 = [k]1[d]1 + [qF]1
d2 = 0
d1 = 0
d4 = -1.538x10-3
d3 = -9.433x10-3
0.6 kN/m
18 m
8.55 kN
43.19 kN�m
3.75 kN
6.03 kN�m
1
2
3
4
13.2 kN�m
2.4 kN
3.2 kN�m
2.4 kN
0.6 kN/m
1[FEF]load 8 m
1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
8200200×
=
75
Member 2:
+4
2
-4
2
q3
q4
q6
q5
d4 = -1.538x10-3
d3 = -9.433x10-3
d6 = 1.863x10-3
d5 = -0.01
q3
q4
q6
q5=
-6.0 kN�m
3.75 kN
20.0 kN�m
0.25 kN0.25 kN
6.0 kN�m
3.75 kN
4 kN
2
10 mm2
5
6
3
4 4 kN4 kN�m
2 kN
4 kN�m
2 kN
2[FEF]
8 m
0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
8200200×
=
20 kN�m
76
4 kN
8 m 4 m 4 m2EI EI
0.6 kN/m
A
BC
10 mm
20 kN�m
Deflected Curve
0.6 kN/m
18 m
8.55 kN
43.19 kN�m
3.75 kN
6.03 kN�m
V (kN)x (m)
+
8.553.75
-0.25
0.25 kN
6.0 kN�m
3.75 kN
4 kN
2
20 kN�m
M (kN�m) x (m)
-43.19
621+
-
20
D3 = -9.433 mm
D6 = +1.863x10-3 radD4 = -1.538x10-3 rad
D5 = -10 mm
77
30 kN
A CB
EI2EI
4 m 2 m
9kN/mHinge
2 m
Example 6
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.
Internal Hinges
78
[K] = EI 1
2
1 2
0.0
0.0
1.0
2.0
A C
B
2EI
4 m 2 m 2 m
EI1 2
43
1
2
[k]1 =EI
2EI
2EI
EI
3 1
3
1[k]2 =
0.5EI
1EI
1EI
0.5EI
2 4
2
4
3 4
Use 2x2 stiffness matrix,θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
79
0.0
0.0
D1
D2
15 kN�mB C
15 kN�m30 kN
2
12 kN�m
A B
12 kN�m 9kN/m
1[FEF]
= EI 1
2
1 2
0.0
0.0
1.0
2.0
D1
D2=
0.0006 rad
-0.0015 rad
-12
15+
Global matrix:
30 kN9kN/m
Hinge
A C
B1 2
43
12 kN�m 15 kN�m1
2
80
Member 1:
q3
q1
12
-12+=
EI
2EI
2EI
EI
3 1
3
1
18
0.0=
d3
d1
= 0.0
= 0.0006
12 kN�m
A B
12 kN�m 9kN/m
1[FEF]A
B1
31
18 kN�m22.5 kN 13.5 kN
A B
9 kN/m
1
81
B C30 kN
CB2
4
2
15 kN�mB C
15 kN�m 30 kN
2
q2
q4
= -0.0015
= 0.0
d2
d4
15
-15+=
0.5EI
1EI
1EI
0.5EI
2 4
2
4
0.0
-22.5=
Member 2:
22.5 kN�m
20.63 kN9.37 kN
82
30 kN
A CB
EI2EI
4 m 2 m
9kN/mHinge
2 m
-20.63
V (kN) x (m)
22.59.37
-13.5
+
2.5 m
M (kN�m) x (m)
-18
10.13
-22.5
18.75
+-
+--
θBL= 0.0006 rad θBR= -0.0015 rad
13.5 kN22.5 kN
22.5 kN�mB C
30 kN
20.63 kN9.37 kN
A B18 kN�m
9 kN/m
9.37 kN13.5 kN
22.87 kN
83
20 kN
A CBEI2EI
4 m 4 m
9kN/m
Hinge
Example 7
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.
84
A CB1 2
2EI EI 6
7
4
5
4
5
6
7
0.375
0.375
1.0
-0.75
-0.75 2.0
0.0
0[K] = EI 1
2
3
1 2 3
0.5625
[k]1 =0.75EI
EI-0.75EI
2EI/L
0.375EI0.75EI
0.75EI-0.375EI
2EI0.75EI
EI-0.75EI
-0.75EI- 0.375EI
0.375EI-0.75EI
4 5 1 24
51
2
0.375EI
0.375EI-0.1875EI
0.5EI -0.375EI
-0.375EI
[k]2 =
0.1875EI
-0.375EIEI0.375EI - 0.1875EI
0.1875EI
0.375EI0.5EI
-0.375EI EI
1 3 6 71
36
7
1
2
3
85
Q1 = -20
Q3 = 0.0
Q2 = 0.0
D1
D3
D2
18
0.0
-12+
D1
D3
D2 =-0.02382 m
0.008933 rad
-0.008333 rad
[FEM]
12 kN�mA B
12 kN�m
18 kN18 kN1
9kN/m
20 kN
A CB
9kN/m
A CB1 2
6
7
4
5
2EI EI
Global:
= EI 1
2
3
1
0.5625
0.375
-0.75
2
-0.75
2.0
0.0
3
0.375
0
1.0
12 kN�m
18 kN
20 kN
1
2
3
86
A B1
2EI1
24
5
[FEF]
12 kN�mA B
12 kN�m
18 kN18 kN1
9kN/m
+12
18
-12
18
Member 1:
q4
q5
q1
q2
=
0.75EI
EI
-0.75EI
2EI/L
0.375EI
0.75EI
0.75EI
-0.375EI
2EI
0.75EI
EI
-0.75EI
-0.75EI
- 0.375EI
0.375EI
-0.75EI
4 5 1 2
4
5
1
2
= 0.0
= 0.0
= -0.02382
= -0.00833
d5
d4
d2
d1
= 107.32
44.83
0.0
-8.83
A B
107.32 kN�m
8.83 kN44.83 kN
9kN/m
1
87
Member 2:
CB2
EI1
3
6
7
44.66 kN�mB C
11.16 kN11.16 kN
2
+0
0
0
0
= -0.02382
= 0.008933
= 0.0
= 0.0
d3
d1
d7
d6
q1
q3
q7
q6
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
=
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
1 3 6 71
3
6
7
= 0.0
-11.16
-44.66
11.16
88
20 kN
A CB4 m 4 m
9kN/m
V (kN) x (m)
44.83
8.83
-11.16 -11.16
+
-
M (kN�m) x (m)
-107.32
-44.66-
-
A B
107.32 kN�m
8.83 kN44.83 kN
9kN/m
144.66 kN�m
B C
11.16 kN11.16 kN
2
θBL= -0.008333 rad θBR= 0.008933 rad
89
30 kN
A C
BEI2EI
4 m 2 m
9kN/mHinge
40 kN�m
2 m
Example 8
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4.
90
[K] = EI 1
2
1 2
0.0
0.0
1.0
2.0
A C
B
2EI
4 m 2 m 2 m
EI1 2
43
[k]1 =EI
2EI
2EI
EI
3 1
3
1[k]2 =
0.5EI
1EI
1EI
0.5EI
2 4
2
4
3 4
1
2
Use 2x2 stiffness matrix: θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
91
15 kN�mB C
15 kN�m30 kN
2
12 kN�m
A B
12 kN�m 9kN/m
1[FEM]
Global matrix:
A C
B1 2
43
1
2
12 kN�m 15 kN�m
30 kN
A C
BEI2EI
9kN/mHinge
40 kN�m
Q1 = 40
Q2 = 0.0
D1
D2
-12
15+
D1
D2=
0.0026 rad
-0.0015 rad
= EI 1
2
1 2
0.0
0.0
1.0
2.0
40 kN�m
92
12 kN�m
A B
12 kN�m 9kN/m
1[FEF]A
B1
31
40 kN�mA B38 kN�m
9 kN/m
1.5 kN37.5 kN
Member 1:
38
40=
q3
q1
= 0.0
= 0.0026
d3
d1
12
-12+=
EI
2EI
2EI
EI
3 1
3
1
93
CB2
4
2
15 kN�mB C
15 kN�m 30 kN
2
Member 2:
q2
q4
= -0.0015
= 0.0
d2
d4
15
-15+=
0.5EI
1EI
1EI
0.5EI
2 4
2
4
0.0
-22.5=
22.5 kN�mB C
30 kN
20.63 kN9.37 kN
94
9.37 kN
1.5 kN
30 kN
A C
B
4 m 2 m
9kN/mHinge
40 kN�m
2 m
-20.63
V (kN) x (m)
37.5
9.371.5+
M (kN�m) x (m)
-38
40
-22.5
+-
18.75+
--
θBL= 0.0026 radθBR=- 0.0015 rad
7.87 kN40 kN�mA B
38 kN�m
9 kN/m
1.5 kN37.5 kN
22.5 kN�mB C
30 kN
20.63 kN9.37 kN
95
20 kN
A CBEI2EI
4 m 4 m
9kN/m
Hinge40 kN�m
Example 9
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4
96
A CB1 2
6
7
4
51
2
3
20 kN
A CBEI2EI
4 m 4 m
9kN/m
Hinge40 kN�m
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEM]θi ∆j θj
Mi
∆i
Vj
Mj
Vi
[ ]
−−−−
−−
=×
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
44
97
1 22A CB
6
7
4
5
[K] = EI 1
2
3
1 2 3
6
7
4
5
0.5625
0.0
0
0.375
0.375
1.0
-0.75
-0.75
2.0
[k]1 = 0.75EI
EI-0.75EI
2EI
0.375EI0.75EI
0.75EI-0.375EI
2EI0.75EI
EI-0.75EI
-0.75EI- 0.375EI
0.375EI-0.75EI
4 5 1 24
51
2
1
0.375EI
0.375EI-0.1875EI
0.5EI -0.375EI
-0.375EI
[k]2 =
0.1875EI-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI0.5EI
-0.375EI EI
1 3 6 71
36
7
2EI EI1
2
3
98
Q1 = -20
Q3 = 0.0
Q2 = 40
D1
D3
D2
18
0.0
-12+
D1
D3
D2 =-0.01316 m
0.0049333 rad
-0.002333 rad
A CB1 2
6
7
4
5
2EI EI
20 kN
EI
9kN/m
40 kN�m
20 kN
40 kN�m
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
= EI 1
2
3
1 2 3
0.5625
0.0
0
0.375
0.375
1.0
-0.75
-0.75
2.0
12 kN�m
18 kNGlobal:
1
2
3
99
1
2EI
[q]1
1
12
1
2EI4
5
+12
18
-12
18
q4
q5
q1
q2
=
0.75EI
EI
-0.75EI
2EI
0.375EI
0.75EI
0.75EI
-0.375EI
2EI
0.75EI
EI
-0.75EI
-0.75EI
- 0.375EI
0.375EI
-0.75EI
4 5 1 2
4
5
1
2
= 87.37
49.85
40
-13.85
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
Member 1: [q]1 = [k]1[d]1 + [qF]1
13.85 kN
40 kN�m87.37 kN�m
49.85 kN
= 0.0
= 0.0
= -0.01316
= -0.002333
d5
d4
d2
d1
12 kN�m
18 kN
12 kN�m
18 kN1
100
22 CB
1
3
EI 6
7
Member 2: [q]2 = [k]2[d]2 + [qF]2
+0
0
0
0
q1
q3
q7
q6
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
=
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
1 3 6 71
3
6
7
= 0.0
-6.18
-24.69
6.18
24.69 kN�m
6.18 kN6.18 kN
= -0.01316
= 0.004933
= 0.0
= 0.0
d3
d1
d7
d6
22
[q]2
101
20 kN
A CB
EI2EI
4 m 4 m
9kN/m 40 kN�m
24.69 kN�mB C
6.18 kN6.18 kN
A B
87.37 kN�m 13.85 kN49.85 kN
40 kN�m
M (kN�m) x (m)
-
-87.37
+40
--24.69
θBL= -0.002333 rad θBR = 0.004933 rad
49.85
-6.18
V (kN) x (m)-6.18
-13.85+
102
� Fixed-End Forces (FEF)- Axial- Bending
� Curvature
Temperature Effects
103
Tm> TR
� Thermal Fixed-End Forces (FEF)
Tl
Tt
2bt
mTTT +
=
Tt
Tb > Tt
βF
AFF
BF
BFBMF
AM
Rmaxial TTT −=∆ )(
tbbending TTT −=∆ )(
)()(dTyTy
∆=∆
y β
)(tandT
dTT tb ∆
=−
=β
A
A B
σaxialσaxial
A B
σy
y
TR Tm
Tt
Tb
Tb - Tt
Room temp = TR
ct
cbd
TR
NA
104
- Axial
A B
σaxialσaxial
FAF F
BF
dAFA
axialF
A ∫= σ
dAE axial∫= ε
dATE axial∫ ∆= )(α
axialTEA )(∆= α
AETTF RmAF
axial )()( −= α
dATE axial ∫∆= )(α
Tt
Tb > Tt
Tm> TR
Rmaxial TTT −=∆ )(
TR Tm
105
- Bending
A B
σy
y
dAyMA
yFA ∫= σ
dATyE y∫ ∆= )(α
dAdTyyE∫
∆= )(α
∫∆
= dAydTE 2)(α
EId
TTF ulA
Fbending )()( −
= α
FBMF
AM
dAyE∫= ε
tbbending TTT −=∆ )(
)()(dTyTy
∆=∆
y β Tt
Tb
106
O´
ds´
dx
Afterdeformation
dθ
dθ
Before deformation
y dx
ds = dx y
ρ)()( θρ ddx =
)( θdy )(1dxdθ
ρ=
� Elastic Curve: Bending
107
dxdTyyd )()( ∆
= αθ
dxdTd )()( ∆
= αθ
Tb
Tt
yy
dT∆
� Bending Temperature
dx
Tt
Tb
Tl > Tu
y
Tl > Tu
Tu
O
dθ
2bt
mTTT +
= dθ
MM
Tb
cb
ct
Tt
∆T = Tb - Tt
dT∆
=β
EIM
dT
dxd
=∆
== )(1)( αρ
θ
108
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
Example 10
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 32.5oC, α = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4.
109
12 3
1 2
Mean temperature = (40+25)/2 = 32.5Room temp = 32.5oC
FEM+7.5 oC-7.5 oC
a(∆T /d)EI19.78 kN�m = 19.78 kN�m
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
mkNEIdTF F
bending •=××−
×=∆
= − 78.19)502002)(182.0
2540)(1012()2)(( 6α
110
[M1] = 3EIθ1 + (1.978x10-3)EI0
Element 1:
M1
M2
q1
q2
[q] = [k][d] + [qF]
+ 1.978
-1.978(10-3) EI= 1
2
2
1
2 1
2
1EI
Element 2:
M3
M1
q3
q1 + 0
0= 0.5
1
1
0.5
1 3
1
3EI
θ1 = -0.659x10-3 rad
12 3
A C
BEI2EI
4 m 4 m
1 2
19.78 kN�m19.78 kN�m182 mm
111
Element 2:
M3
M1
q3
q1= + 0
0
0.5
1
1
0.5
1 3
1
3EI
= -0.659x10-3
= 0= 6.59 kN�m
-26.37 kN�m
= -0.659x10-3
= 0= -3.30 kN�m
-6.59 kN�m
Element 1:
M1
M2 = q1
q2 + 19.78
-19.78
1
2
2
1
2 1
2
1EI
4.95 kN4.95 kN
26.37 kN�m 6.59 kN�m
A B
6.59 kN�m 3.3 kN�m
B C
2.47 kN2.47 kN
12 3
A C
BEI2EI
4 m 4 m
1 2
19.78 kN�m19.78 kN�m182 mm
112
A CB
4 m 4 m
+7.5 oC
-7.5 oC 3.3 kN�m
26.37 kN�m
4.95 kN2.47 kN
2.47 kN
M (kN�m) x (m)
26.37
6.59
-3.30
+-
V (kN) x (m)
-4.95-2.47
- -
θB = -0.659x10-3 radDeflected curve x (m)
113
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
Example 11
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 28 oC, a = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4,A = 20(10-3) m2
114
A
CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC1
2 3
7
8
9
1 24
5
6
Element 1:
mkNm
mkNmLAE /)10(2
)4()/10200)(1020(2)2( 6
2623
=××
=−
mkNm
mmkNLEI
•=×××
=−
)10(20)4(
)1050)(/10200(24)2(4 34626
mkNm
mmkNLEI
•=×××
=−
)10(10)4(
)1050)(/10200(22)2(2 34626
kNm
mmkNLEI )10(5.7
)4()1050)(/10200(26)2(6 3
2
4626
2 =×××
=−
mkNm
mmkNL
EI /)10(75.3)4(
)1050)(/10200(212)2(12 33
4626
3 =×××
=−
115
Fixed-end forces due to temperatures
Mean temperature(Tm) = (40+25)/2 = 32.5 oC ,TR = 28 oC
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
mkNEIdTF F
bending •=××−
×=∆
= − 78.19)502002)(182.0
2540)(1012()2)(( 6α
19.78 kN�m19.78 kN�m
432 kN432 kNA B
+12 oC
-3 oC
kNmkNmAETF Faxial 432)/10200)(310202)(285.32)(1012()( 2626 =×−××−×=∆= −α
116
FEM
+12 oC
-3 oC
19.78 kN�m
A B
19.78 kN�m
432 kN432 kN
Element 1:
[q] = [k][d] + [qF]
0
0
0
d1
d2
d3
q4
q6
q5
q1
q2q3
=
4
5
6
1
2
3
4 1
- 2x106
0.00
0.00
2x106
0.00
0.00
2
0.00
-3750
-7500
0.00
3750
-7500
3
0.00
7500
10x103
0.00
-7500
20x103
2x106
0.00
0.00
-2x106
0.00
0.00
5
0.00
3750
7500
0.00
-3750
7500
6
0.00
7500
20x103
0.00
-7500
10x103
432
-19.78
0.00
-432
0.00
19.78
+
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm1
2 3
7
8
9
1 24
5
6
117
Element 2:
[q] = [k][d] + [qF]
d1
d3
d2
0
0
0
0
0
0
0
0
0
+
q1
q3
q2
q7
q8q9
- 1x106
0.00
0.00
1x106
0.00
0.00
0.00
-1875
-3750
0.00
1875
-3750
0.00
3750
5x103
0.00
-3750
10x103
1x106
0.00
0.00
-1x106
0.00
0.00
0.00
1875
3750
0.00
-1875
3750
0.00
3750
10x103
0.00
-3750
5x103
7 8 91 2 3
=
1
2
3
7
8
9
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm1
2 3
7
8
9
1 24
5
6
118
D1
D3
D2 =0.000144 m
-719.3x10-6 rad
-0.0004795 m
Q1 = 0.0
Q3 = 0.0
Q2 = 0.0
D1
D3
D2
-432
19.78
0.0+= 1
2
3
1
3x106
0.0
0.0
2
0.0
5625
-3750
3
0.0
-3750
30x103
Global:
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm1
2 3
7
8
9
1 24
5
6
119
= 144x10-6
= -479.5x10-6
= -719.3x10-6
Element 1:
0
0
0
d1
d2
d3
q4
q6
q5
q1
q2q3
=
4
5
6
1
2
3
4 1
- 2x106
0.00
0.00
2x106
0.00
0.00
2
0.00
-3750
-7500
0.00
3750
-7500
3
0.00
7500
10x103
0.00
-7500
20x103
2x106
0.00
0.00
-2x106
0.00
0.00
5
0.00
3750
7500
0.00
-3750
7500
6
0.00
7500
20x103
0.00
-7500
10x103
432
-19.78
0.00
-432
0.00
19.78
+
=
q4
q6
q5
q1
q2q3
144.0 kN
-23.38 kN�m
-3.60 kN
-144.0 kN
3.60 kN
9.00 kN�m
A B 1
2 3
4
5
6
A B144 kN
3.60 kN
23.38 kN�m
144 kN
3.60 kN
9 kN�m
120
= 144x10-6
= -479.5x10-6
= -719.3x10-6
144 kN
-9 kN�m
-3.6 kN
-144 kN
3.6 kN
-5.39 kN�m
=
q1
q3
q2
q7
q8q9
B C 7
8 9
1
2
3
144 kNB C
3.60 kN
9 kN�m
144 kN
3.60 kN
5.39 kN�m
Element 2:
d1
d3
d2
0
0
0
0
0
0
0
0
0
+
q1
q3
q2
q7
q8q9
=
- 1x106
0.00
0.00
1x106
0.00
0.00
0.00
-1875
-3750
0.00
1875
-3750
0.00
3750
5x103
0.00
-3750
10x103
1x106
0.00
0.00
-1x106
0.00
0.00
0.00
1875
37500
0.00
-1875
3750
0.00
3750
10x103
0.00
-3750
5x103
7 8 91 2 3
1
2
3
7
8
9
121
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
Isolate axial part from the system
RCRA
RA RA = RC+
Compatibility equation: dC/A = 0
RA = 144 kN
RC = -144 kN
0)4)(285.32(1012)4(2
)4( 6 =−×++ −
AER
AER AA
122
A B
144 kN
3.60 kN
23.38 kN�m
144 kN
3.60 kN
9 kN�m
144 kN
B C3.60 kN
9 kN�m
144 kN
3.60 kN
5.39 kN�m
V (kN) x (m)
-3.6
-
M (kN�m) x (m)
23.388.98
-5.39
+-
D3 = -719.3x10-6 radDeflected curve x (m)
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
D2 = -0.0004795 mm
123
Skew Roller Support
- Force Transformation- Displacement Transformation- Stiffness Matrix
124
m
i
j
m
i
j
x´y´
x
y
� Displacement and Force Transformation Matrices
12
3
45
6
θyθx
4´
5´
6´
1´
2´
3´
125
λx λy
i
jθy
θx
y´
x´
4´5´
6´
1´
2´
3´
θy
θx
m
i
j
x
y
12
3
45
6
Force Transformation
Lxx ij
x
−=λ
Lyy ij
y
−=λ
yx qqq θθ coscos 5'4'4 −=
xy qqq θθ coscos 5'4'5 −=
6'6 qq =
6
5
4
qqq
−=
10000
xy
yx
λλλλ
6'
5'
4'
qqq
−
−
=
6'
5'
4'
3'
2'
1'
6
5
4
3
2
1
1000000000000000010000000000
qqqqqq
λλλλ
λλλλ
qqqqqq
xy
yx
xy
yx
[ ] [ ] [ ]'qTq T=
126x
y
θyθx
x
y
θyθx
d4 d 4 cos θ x
d 4 cos θ yθy
θx
d5
θx
θy
d 5 cos θ x
d 5 cos θ y
x´
Displacement Transformation
j4´
5´
6´j4
5
6λx λy
yx θdθdd coscos' 544 +=
xy θdθdd' coscos 545 +−=
66' dd =
6'
5'
4'
ddd
−=
10000
xy
yx
λλλλ
6
5
4
ddd
−
−
=
6
5
4
3
2
1
6'
5'
4'
3'
2'
1'
1000000000000000010000000000
dddddd
λλλλ
λλλλ
dddddd
xy
yx
xy
yx
[ ] [ ][ ]dTd ='
127
[ ] [ ] [ ]'qTq T=
[ ] [ ][ ] [ ])'( FT qd'k'T +=
[ ] [ ][ ] [ ] [ ]FTT qTd'k'T '+=
[ ] [ ] [ ][ ][ ] [ ] [ ] [ ][ ] [ ]FFTT qdkqTdTk'Tq +=+= '
[ ] [ ] [ ][ ]TkTkTherefore T ', =
[ ] [ ] [ ]FTF qTq '=
[ ] [ ] [ ]'qTq T=
[ ] [ ][ ]dTd ='
[ ] [ ] [ ][ ]TkTk T '=
128
i j
θjθi 4 *
5 * 6*
1*
2*3 *
λix = cos θi
λiy = sin θi
λjx = cos θj
λjy = sin θj
[q*] = [T]T[q´]
1
Stiffness matrix
i j
4´
5´ 6´
1´
2´3´
*6
5*
*4
*3
2*
*1
qqqqqq
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
λλλλ
λλλλ
1 4 5 62 31*2*3*4*5*6*
[ T ]T
6'
5'
4'
3'
2'
1'
qqqqqq
129
[ ]
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
λλλλ
λλλλ
T
1 4 5 62 3123456
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
k
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
'
22
2323
22
2323
1
34
6
2
5
1 2 3 4 5 6
130
[ k ] = [ T ]T[ k´ ][T] =
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
131
40 kN
4 m4 m 22.02 o
Example 12
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.
132
40 kN40 kN�m
20 kN
40 kN�m
20 kN
[ q´F ][FEF]
1Global
i j
1
i j
Local
40 kN
4 m4 m 22.02o
20sin22.02=7.520cos22.02=18.54
1´
2´
3´
4´
5´
6´
4
5
6
x*
x´1 *
2* 3 *
λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
22.02 o
133
x*
x´1
4
5
6
1 *
2* 3 *
� Transformation matrix
Member 1: [ q ] = [ T ]T[q´]
1
i j1´
2´
3´
4´
5´
6´
Global Local
λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
*3
2*
*1
6
5
4
qqqqqq
6'
5'
4'
3'
2'
1'
qqqqqq
−=
10000009271.03749.000003749.09271.0000000100000010000001
1´ 4´ 5´ 6´2´ 3´4561*
2*
3*
[ ]
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
TT
λλλλ
λλλλ
134
[k´]1 = 103
1´2´ 3´4´5´6´
1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
1
i j1´
2´
3´
4´
5´
6´
Local
� Local stiffness matrix
[ ]
−−−−
−−−
−
=×
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
k
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
22
2323
22
2323
66
θi ∆j θj∆i δjδi
Mi
Vj
Mj
Vi
Nj
Ni
135
Stiffness matrix [k´]:
[k´]1 = 103
1´2´ 3´4´5´6´
1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
Stiffness matrix [k*]: [ k* ] = [ T ]T[ k´ ][T]
[k*]1 = 103
4561*
2*
3*
4 1* 2* 3*5 6-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.75051.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.75010.00
x*
x´1
4
5
6
1 *
2* 3 * 1
i j1´
2´
3´
4´
5´
6´
Global Local
4
5
6
2*
136
Global Equilibrium:
Q1 = 0.0
Q3 = 0.0
D1*
D3*=
36.37x10-6 m
0.002 rad
[Q] = [K][D] + [QF]
= 1031*
3*
1*
129
1.406
3*
1.406
20.0
D1*
D3*
-7.5
-40+
x*
x´14
5
6
1 *
2* 3 *
4 m4 m 22.02 o
40 kN
x*
x´4
5
6
2*
40 kN40 kN�m
20 kN
40 kN�m
20 kN
[ q´F ]20sin22.02=7.5
20cos22.02=18.54
137
Member Force : [q] = [k*][D] + [qF]
0
40.020
-7.5418.54-40.0
+
q4
q6
q5
q1*
q2*q3*
= 103
4561*
2*
3*
4 1* 2* 3*5 6-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.75051.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.75010.00
0
00
36.4x10-6
02x10-3
-5.06
6027.5
013.48
0
=
40 kN
13.48 kN
60 kN�m
27.5 kN
5.06 kN
4 m4 m 22.02 o
40 kN
x*
x´1
4
5
6
1 *
2* 3 *
40 kN40 kN�m
7.518.54
40 kN�m
20 kN
4
5
6
2*
138
4 m4 m 22.02 o
40 kN
5.05 kN27.51 kN
60.05 kN�m
13.47 kN
40 kN
+
Bending moment diagram (kN�m)
Deflected shape
50.04
-
-60.05
0.002 rad
139
40 kN
22.02o
8 m 4 m 4 m2EI, 2AE EI, AE
6 kN/m
Example 13
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.
140
40 kN
22.02 o8 m 4 m 4 m
2EI, 2AE EI, AE
6 kN/m
1 21
2
3
4
5
6
7 *
8 * 9 *
Global
21´
2´3´
4´
5´6´
11´
2´
3´
4´
5´
6´
Local
mkNm
mkNmL
AE
•×=
×=
3
262
10150)8(
)/10200)(006.0(
mkNm
mmkNLEI
•×=
×=
3
426
1020)8(
)0002.0)(/10200(44
mkNLEI
•×= 310102
kNm
mmkNLEI
3
2
426
2
1075.3)8(
)0002.0)(/10200(66
×=
×=
mkNm
mmkNLEI
/109375.0)8(
)0002.0)(/10200(1212
3
3
426
2
×=
×=
141
[k]1 = [k´]1 = 2x103
456123
4 1 2 35 6-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
1 21
2
3
4
5
6
7 *
8 * 9 *
Global Local : Member 1
14
5
6
1
2
3
[ q ]
11´
2´
3´
4´
5´
6´
[ q´]
[q] = [q´] Thus, [k] = [k´]
142
Member 2: Use transformation matrix, [q*] = [T]T[q´]
q1´
q3´
q2´
q4´
q5´
q6´
q1
q3
q2
q7*
q8*q9*
0.92710.3749
-0.37490.9271
000
000
000
0 0
001
10
01
=
1237*
8*
9*
1´ 4´ 5´ 6´2´ 3´
0 0
001
000
000
000
1 21
2
3
4
5
6
7 *
8 * 9 *
λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
21
2
3
x*
x´7*
8 * 9* [ q* ]
21´
2´3´
4´
5´6´
[ q´ ]
143
Stiffness matrix [k´]:
[k´]2 = 103
1´2´ 3´4´5´6´
1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
Stiffness matrix [k*]: [k*] = [T]T[k´][T]
[k*]2 = 103
1237*
8*
9*
1 7* 8* 9*2 3-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.47751.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.47710.00
21
23
x*
x´7 *
8 * 9 * [ q* ]
21´
2´3´
4´
5´6´
[ q´ ]
144
1 21
2
3
4
5
6
7 *
8 * 9 *
1
2
3
4
5
6
8 *
[k]1= 2x103
456123
4 1 2 35 6-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
[k*]2 = 103
1237*
8*
9*
1 7* 8* 9*2 3-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.47751.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.47710.00
145
Global:
=-509.84x10-6 rad
18.15x10-6 m
0.00225 rad
58.73x10-6 mD3
D1
D9*
D7*
00
00 +
-32 + 40 = 80
-40-7.5
D3
D1
D9*
D7*
32 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
1
40 kN6 kN/m
1 21
2
3
4
5
6
7*
8 *
9*
1
2
3
4
5
6
8 *
= 1030
0-139
450
101.406
600
1.406
1.406-139
129
010
1.40620
1 3 7* 9*
137*
9*
40 kN40 kN�m
7.518.54
40 kN�m
20 kN
40 kN�m
7.5
40 kN�m32 kN�m
146
Member 1: [ q ] = [k ][d] + [qF]
0
3224
024-32
+
0
00
d1=18.15x10-6
0d3=-509.84 x10-6
q4
q6
q5
q1
q2q3
= 2x103
456123
4 1 2 35 6-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
-5.45 kN
21.80 kN�m20.18 kN
5.45 kN27.82 kN-52.39 kN�m
=
q4
q6
q5
q1
q2q3
6 kN/m
5.45 kN
20.18 kN
21.80 kN�m
5.45 kN
27.82 kN
52.39 kN�m
32 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
114
5
6
1
2
3
147
q1
q3
q2
q7*
q8*q9*
21
3
x*
x´7*
9 *
= 103
1237*
8*
9*
1 7* 8* 9*2 3-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.75051.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.75010.00
0
4020
-7.518.54-40
+
18.15x10-6
-509.84x10-6
0
58.73x10-6
00.00225
q1
q3
q2
q7*
q8*q9*
-5.45 kN
52.39 kN�m26.55 kN
0 kN14.51 kN0 kN�m
= 5.45 kN
26.55 kN
52.39 kN�m
14.51 kN
40 kN
20sin22.02=7.520cos22.02=18.54
40 kN40 kN�m
20 kN
40 kN�m
20 kN
[ q´F ]
2
2 8 *
Member 2: [ q ] = [k ][d] + [qF]
148
5.45 kN26.55 kN
52.39 kN�m
14.51 kN
40 kN6 kN/m
5.45 kN
27.82 kN
52.39 kN�m
5.45 kN
20.18 kN
21.80 kN�m
40 kN
22.02o
8 m 4 m 4 m
6 kN/m
5.45 kN
20.18 kN
21.80 kN�m
14.51 kN54.37 kN
3.36 m
M (kN�m) x (m)
53.81
-21.8-
V (kN)x (m)
26.55
+20.18
+- -
-27.82
14.51cos 22.02o
=13.45 kN
-52.39-
12.14 +
-13.45