The document discusses the direct stiffness method for structural analysis. It covers key aspects of applying the direct stiffness method including defining the global stiffness matrix [K], movements matrix [∆], and forces matrix [F]. It also discusses calculating individual bar stiffness matrices and considering transformations when bars are not aligned with the global coordinate system. The document uses a simple frame example to demonstrate applying the direct stiffness method step-by-step.
6. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
1.- DIRECT STIFFNESS METHOD (matrix formulation)
GENERAL METHOD (computers use it)
AUTOMATED PROCESS
WE DO NOT HAVE TO THINK
MANY NUMBERS SO WE HAVE TO BE
VERY CAREFUL ORGANIZING THE
INFORMATION
WE DO NOT NEGLECT AXIAL EFFECT: BARS GET LONGER AND SHORTER
8. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
[K] x [∆] = [F]
GLOBAL
STIFFNESS
MATRIX
MOVEMENTS
MATRIX
FORCES
MATRIX
KEY MATRIX EQUATION
Frame
geometry
all the frame information will be included in it
10. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
MOVEMENTS MATRIX (always 3 numbers per joint in 2D frames)
uA
D A
vA
gA
uC
D C
vC
gC
[D] = uD
D D
vD
gD
uB
D B
vB
gB
In this example which of these elements are ZERO?
11. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
FORCES MATRIX
(made up as the addition of 3 elements or 3 submatrices)
[F] = [-fe] + [P] + [R]
Fixed end forces and
moments (effect of
the load applied
along the bars)
Reaction
forces
Punctual loads
applied at the
joints
12. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
FIXED END FORCES
SUBMATRIX
[-fe] only external load along the
bar in the beam CD
0 0
[-fe]C
-qL/2 -30
-qL^2/12 -30
[-fe]CD
= 0 = 0
[-fe]D
-qL/2 -30
qL^2/12 30
SINGLE FIXED END FORCES AND MOMENTS MATRIX
For every bar we should compute the effect of the loads applied along
the bars in the bars ends: forces and moments
14. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
FIXED END FORCES
SUBMATRIX
[-fe] only external load along the
bar in the beam CD
0 0
[-fe]C
-qL/2 -30
-qL^2/12 -30
[-fe]CD
= 0 = 0
[-fe]D
-qL/2 -30
qL^2/12 30
[-fe] AC = 0
[-fe] DB = 0
SINGLE FIXED END FORCES AND MOMENTS MATRIX
For every bar we should compute the effect of the loads applied along
the bars in the bars ends: forces and moments
15. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
STRUCTURAL ANALYSIS I
2016/2017 CLASS 6
FORCES MATRIX
(the assembled force matrix
has 3 numbers per joint, same
as movements matrix)
[-fe] [R] [P]
0 RxA 0 RxA
A
0 RyA 0 RyA
0 MA 0 MA
0 0 4 4
C
-30 0 0 -30
-30 0 0 -45
[F] = 0 + 0 + 0 = 0
D
-30 0 0 -30
30 0 0 45
0 RxB 0 RxB
B
0 RyB 0 RyB
0 MB 0 MB
16. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
GLOBAL (or master) STIFFNESS MATRIX
[𝐾]𝐴𝐴 [𝐾]𝐴𝐶 [𝐾]𝐴𝐷 [𝐾]𝐴𝐵
[𝐾]𝐶𝐴 [𝐾]𝐶𝐶 [𝐾]𝐶𝐷 [𝐾]𝐶𝐵
[𝐾]𝐷𝐴 [𝐾]𝐷𝐶 [𝐾]𝐷𝐷 [𝐾]𝐷𝐵
[𝐾]𝐵𝐴 [𝐾]𝐵𝐶 [𝐾]𝐵𝐷 [𝐾]𝐵𝐵
𝑥
[𝜕𝐴]
[𝜕𝐶]
[𝜕𝐷]
[𝜕𝐵]
=
[𝐹𝐴]
[𝐹𝐶]
[𝐹𝐷]
[𝐹𝐵]
12 submatrices
Each submatrix 9 elements
[K] x [∆] = [F]
17. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
GLOBAL (or master) STIFFNESS MATRIX
[𝐾]𝐴𝐴 [𝐾]𝐴𝐶 [𝐾]𝐴𝐷 [𝐾]𝐴𝐵
[𝐾]𝐶𝐴 [𝐾]𝐶𝐶 [𝐾]𝐶𝐷 [𝐾]𝐶𝐵
[𝐾]𝐷𝐴 [𝐾]𝐷𝐶 [𝐾]𝐷𝐷 [𝐾]𝐷𝐵
[𝐾]𝐵𝐴 [𝐾]𝐵𝐶 [𝐾]𝐵𝐷 [𝐾]𝐵𝐵
𝑥
[𝜕𝐴]
[𝜕𝐶]
[𝜕𝐷]
[𝜕𝐵]
=
[𝐹𝐴]
[𝐹𝐶]
[𝐹𝐷]
[𝐹𝐵]
[K]AA links movements at A with forces at A
[K]AC links movements at C with forces at A
[K]AD links movements at D with forces at A
[K]AB links movements at B with forces at A
[K]CA links movements at A with forces at C
[K]CC links movements at C with forces at C
[K]CD links movements at D with forces at C
[………………………………………………..
[K] x [∆] = [F]
18. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
SINGLE BAR STIFFNESS MATRIX
How many single bar stifnees matrices do you have to consider in this frame?
How many numbers does it contain?...
[k] x [∂] = [f]
BAR
STIFFNESS
MATRIX
BAR
MOVEMENTS
MATRIX
BAR
FORCES
MATRIX
bar
geometry
20. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
every single stiffness matrix have 36 element
organized in 4 submatrices
𝐸𝐴
𝐿
0 0
−𝐸𝐴
𝐿
0 0
0
12𝐸𝐼
𝐿3
6𝐸𝐼
𝐿2 0
−12𝐸𝐼
𝐿3
6𝐸𝐼
𝐿2
0
6𝐸𝐼
𝐿2
4𝐸𝐼
𝐿
0
−6𝐸𝐼
𝐿2
2𝐸𝐼
𝐿
−𝐸𝐴
𝐿
0 0
𝐸𝐴
𝐿
0 0
0
−12𝐸𝐼
𝐿3
−6𝐸𝐼
𝐿2 0
12𝐸𝐼
𝐿3
−6𝐸𝐼
𝐿2
0
6𝐸𝐼
𝐿2
2𝐸𝐼
𝐿
0
−6𝐸𝐼
𝐿2
4𝐸𝐼
𝐿
𝑥
𝑢1
𝑣1
𝜃1
𝑢2
𝑣2
𝜃2
=
𝑁1
𝑉1
𝑀1
𝑁2
𝑉2
𝑀2
1 2
21. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
When the bar is horizontal and we consider the first joint in the
left and the second in the right, the bar local coordinate system is
the same as the global coordinate system
𝐸𝐴
𝐿
0 0
−𝐸𝐴
𝐿
0 0
0
12𝐸𝐼
𝐿3
6𝐸𝐼
𝐿2 0
−12𝐸𝐼
𝐿3
6𝐸𝐼
𝐿2
0
6𝐸𝐼
𝐿2
4𝐸𝐼
𝐿
0
−6𝐸𝐼
𝐿2
2𝐸𝐼
𝐿
−𝐸𝐴
𝐿
0 0
𝐸𝐴
𝐿
0 0
0
−12𝐸𝐼
𝐿3
−6𝐸𝐼
𝐿2 0
12𝐸𝐼
𝐿3
−6𝐸𝐼
𝐿2
0
6𝐸𝐼
𝐿2
2𝐸𝐼
𝐿
0
−6𝐸𝐼
𝐿2
4𝐸𝐼
𝐿
𝑥
𝑢1
𝑣1
𝜃1
𝑢2
𝑣2
𝜃2
=
𝑁1
𝑉1
𝑀1
𝑁2
𝑉2
𝑀2
1 2
23. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
TRANSFORMATION MATRIX
𝑇 =
𝑐𝑜𝑠 𝛼 𝑠𝑒𝑛 𝛼 0 0 0 0
−𝑠𝑒𝑛 𝛼 cos 𝛼 0 0 0 0
0 0 1 0 0 0
0 0 0 cos 𝛼 𝑠𝑒𝑛 𝛼 0
0 0 0 −𝑠𝑒𝑛 𝛼 cos 𝛼 0
0 0 0 0 0 1
THE ANGLE IS THE KEY!
24. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
𝑘11 𝑘12 𝑘13 −𝑘11 −𝑘12 𝑘13
𝑘12 𝑘22 𝑘23 −𝑘12 −𝑘22 𝑘23
𝑘13 𝑘23 𝑘33 −𝑘13 −𝑘23 𝑘36
−𝑘11 −𝑘12 −𝑘13 𝑘11 𝑘12 −𝑘13
−𝑘12 −𝑘22 −𝑘23 𝑘12 𝑘22 −𝑘23
𝑘13 𝑘23 𝑘36 −𝑘13 −𝑘23 𝑘33
𝑥
𝑢1
𝑣1
𝜃1
𝑢2
𝑣2
𝜃2
=
𝑓𝑥1
𝑓𝑦1
𝑀1
𝑓𝑥2
𝑓𝑦2
𝑀2
[k]coordenadas globales x [∂]globales = [f]globales
k11 = EA cos2 α / L + 12 EI sen2 α /L3
k12 = (EA/L - 12 EI /L3 ) x sen α x cos α
k13 = - 6EI sen α /L2
k22 = EA sen2 α / L + 12 EI cos2 α /L3
k23 = 6 EI cos α /L2
k33 = 4EI /L
k36 = 2EI /L
SINGLE BAR MATRIX
IN GLOBAL COORDINATES
THE ANGLE IS THE KEY!
25. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
[kAA]barra AC [kAC]barra AC
6000 0 -9000 -6000 0 -9000
0 600000 0 0 -600000 0
-9000 0 18000 9000 0 9000
-6000 0 9000 6000 0 9000
0 -600000 0 0 600000 0
-9000 0 9000 9000 0 18000
[kCA]barra AC [kCC]barra AC
[k]AC
26. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
[kCC]barra CD [kCD]barra CD
300000 0 0 -300000 0 0
0 750 2250 0 -750 2250
0 2250 9000 0 -2250 4500
-300000 0 0 300000 0 0
0 -750 -2250 0 750 -2250
0 2250 4500 0 -2250 9000
[kDC]barra CD [kDD]barra CD
[k]CD
27. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
[kDD]barra DB [kDB]barra DB
6000 0 9000 -6000 0 9000
0 600000 0 0 -600000 0
9000 0 18000 -9000 0 9000
-6000 0 -9000 6000 0 -9000
0 -600000 0 0 600000 0
9000 0 9000 -9000 0 18000
[kBD]barra DB [kBB]barra DB
[k]BD
28. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
[𝐾]𝐴𝐴 [𝐾]𝐴𝐶 [𝐾]𝐴𝐷 [𝐾]𝐴𝐵
[𝐾]𝐶𝐴 [𝐾]𝐶𝐶 [𝐾]𝐶𝐷 [𝐾]𝐶𝐵
[𝐾]𝐷𝐴 [𝐾]𝐷𝐶 [𝐾]𝐷𝐷 [𝐾]𝐷𝐵
[𝐾]𝐵𝐴 [𝐾]𝐵𝐶 [𝐾]𝐵𝐷 [𝐾]𝐵𝐵
GLOBAL STIFFNESS MATRIX [K] // ASSEMBLE MATRIX
WHICH SUBMATRICES ARE ZERO?
WHICH SUBMATRICES ARE THE ADDITION OF TWO?
34. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
STRUCTURAL ANALYSIS I
2017/2018 CLASS 7
SOLVED EQUATION: MOVEMENTS VALUES
uC 0,000607 m
vC -0,000049 m
qC -0,001510 rad
uD = 0,000561 m
vD -0,000051 m
qD 0,001176 rad
36. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
GLOBAL EQUATION: ONCE WE KNOW THE MOVEMENTS
WE CAN CALCULATE THE REACTION FORCES
RxA 9,94 kN
RyA 29,25 kN
MA -8,12 kNm
RxB = -13,94 kN
RyB 30,75 kN
MB 15,626 kNm
37. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
ONCE WE KNOW THE REACTION FORCES WE CAN
CALCULATE THE INTERNAL FORCES
A B
C D
6 m
3
m
30 x 30 cm 30 x 30 cm
30 x 30 cm
4 kN
29,25 kN 30,75 kN
9,94 kN 13,94 kN
8,12 kNm 15,626 kNm
38. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
STRUCTURAL ANALYSIS I
2017/2018 CLASS 7
A B
C D
-13,94 kN
-29,25 kN -30,75 kN
-19,92 kN
15,92 kN
-29,25 kN
30,75 kN
A B
C
A B
C D
ESFUERZOSAXILES
ESFUERZOSCORTANTES
DEFORMADA
A B
C D
-21,7 kNm
8,12 kNm
-26,194 kNm
21,053 kNm
15,63 kNm
MOMENTOSFLECTORES
-12,5 mm
v = -0,049 mm
u = 0,607 mm
v = -0,051 mm
u = 0,560 mm
21,078 kNm
39. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
A B
C D
UPN80
30 x 30 cm 30 x 30 cm
30 x 30 cm
E= 20 kN/mm 2
E= 210 kN/mm 2
10kN/m
4 kN
WHAT’S THE EFFECT OF THE DIAGONAL BRACING?
WHICH MATRICES WOULD VARY?
41. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
THE ONLY BAR SINGLE MATRIX WE SHOULD INCLUDE IS AD
[kAA]barra AD [kAD]barra AD
27550 13771 -13 -27550 -13771 -13
13771 6894 27 -13771 -6894 27
-13 27 133 13 -27 66
-27550 -13771 13 27550 13771 13
-13771 -6894 -27 13771 6894 -27
-13 27 66 13 -27 133
[kDA]barra DA [kDD]barra DD
[k]AD
43. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
306000 0 9000 -300000 0 0 uC 4
0 600750 2250 0 -750 2250 vC -30
9000 2250 27000 0 -2250 4500 qC -30
-300000 0 0 333550 13771 9013 x uD = 0
0 -750 -2250 13771 607644 -2277 vD -30
0 2250 4500 9013 -2277 27133 qD 30
[K]reducida, en coordenadas globales [ Δ]reducida [F]reducida
[K]reduced, in global coordinates [ Δ]reduced [F]reduced
44. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
uC 0,000185 m
vC -0,000050 m
qC -0,001388 rad
uD = 0,000133 m
vD -0,000053 m
qD 0,001291 rad
RxA 8,42 kN
RyA 28,35 kN
MA -10,74 kNm
RxB = -12,42 kN
RyB 31,65 kN
MB 12,82 kNm
45. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
A
B
C D
UPN80
30 x 30 cm 30 x 30 cm
30 x 30 cm
E= 20 kN/mm 2
E= 210 kN/mm 2
10kN/m
4 kN
28,34 kN 31,65 kN
8,42 kN 12,42 kN
10,74 kNm 12,82 kNm
46. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
DIRECT STIFFNESS METHOD
STRUCTURAL ANALYSIS I
2017/2018 CLASS 7
A B
C
D
-15,38 kN
-29,79 kN -31,65 kN
-12,42 kN
11,38kN
-29,79 kN
30,21 kN
A B
C
A B
C
D
ESFUERZOSAXILES
ESFUERZOSCORTANTES
DEFORMADA
A B
C
-23,32 kNm
10,83 kNm
-24,61 kNm
21,02 kNm
12,82 kNm
MOMENTOSFLECTORES
-12,5 mm
v = -0,049 mm
u = 0,18 mm
v = -0,052 mm
u = 0,130 mm
+3,29 kN
-0,039 kN
21,05 kNm
0,17 kNm
0,08 kNm
ANALYZE THE EFFECT OF THE BRACING
ANALYZE THE VERTICAL DISPLACEMENTS AT C AND D IN BOTH FRAMES