4. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
1.- DIRECT STIFFNESS METHOD (matrix formulation)
GENERAL METHOD (computers use it)
AUTOMATIC 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
6. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
[K] x [∆] = [F]GLOBAL
STIFFNESS
MATRIX
MOVEMENTS
MATRIX
FORCES
MATRIX
KEY MATRIX EQUATION
Frame
geometry
all the frame information will be included in it
8. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
MOVEMENTS MATRIX (always 3 numbers per joint in 2D frames)
uA
D AvA
gA
uC
D CvC
gC
[D] = uD
D DvD
gD
uB
D BvB
gB
In this example which of these elements are ZERO?
9. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
10. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
12. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
13. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
A0 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
B0 RyB 0 RyB
0 MB 0 MB
14. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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]
[𝐹 𝐷]
[𝐹𝐵]
12 submatrices
Each submatrix 9 elements
[K] x [∆] = [F]
15. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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]
[𝐹 𝐷]
[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]
16. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
MATRIXbar
geometry
18. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
19. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
32. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
34. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
35. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
ONCE WE KNOW THE REACTION FORCES WE CAN
CALCULATE THE INTERNAL FORCES
A B
C D
6 m
3m
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
36. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
STRUCTURAL ANALYSIS I
2017/2018 CLASS 7
A B
C D
-13,94 kN
-29,25 kN -30,75 kN
-19,92 kN15,92 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
37. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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?
39. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
[kAA]barra DA [kAA]barra DD
[k]AD
41. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
42. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
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
44. cperez.eps@ceu.es
molina.eps@ceu.es
STRUCTURAL ANALYSIS I
DEGREE IN ARCHITECTURE
Year 3 Term 1
18/19 CLASS 9
STRUCTURAL ANALYSIS I
2017/2018 CLASS 7
A B
C
D
-15,38 kN
-29,79 kN -31,65 kN
-12,42 kN11,38kN
-29,79 kN
30,21 kN
A B
C
A B
C
D
ESFUERZOSAXILESESFUERZOSCORTANTES
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
