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NTC-ACERO Diseño y Construcción de Estructuras de Acero
Diseño por Estabilidad
Palacio de Minería – México D.F. 17de septiembre de 2015
CURSO SOBRE ESTRUCTURAS DE ACERO
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Literatura sobre Estabilidad Estructural
1. Galambos and Surovek (2008). Structural stability of steel: concepts and applications for structural engineers. John Wiley and Sons, Inc.
2. Chen, Lui (1987). Structural Stability: Theory Implementation, Prentice Hall. 3. Bazant, Cedolin (2010). Stability of Structures: Elastic, Inelastic, Fracture and Damage Theories. 4. Ziemian (2010). Guide to stability design criteria for metal structures, Wiley. 6th Ed. 5. Timoshenko, Gere (1963). Theory of elastic stability, McGraw Hill. 6. Simitses, Hodges (2005). Fundamentals of Structural Stability, Butterworth-Heinemann. 7. Chen, Lui (1991). Stability Design of Steel Frames, CRC Press.
NTC-ACERO Diseño y Construcción de Estructuras de Acero
2. Diseño por Estabilidad
Contenido: 2.1 Disposiciones generales 2.2 Rigidez lateral 2.3 Factor de longitud efectiva y
efectos de esbeltez de conjunto 2.4 Método de análisis por estabilidad 2.5 Método de análisis directo 2.6 Método de la longitud efectiva 2.7 Determinación aproximada
de los efectos de segundo orden
NTC-ACERO Diseño y Construcción de Estructuras de Acero
2. Diseño por Estabilidad
Los estudios para determinar la estabilidad de la estructura, han de incluir: • Deformaciones axiales, en flexión y en cortante,
de todos los miembros y conexiones, así como cualquier otra deformación que contribuya a los desplazamientos de la estructura
• Efectos de segundo orden, P∆ y Pδ • Imperfecciones geométricas • Reducciones de rigidez debidas a inelasticidad • Incertidumbres en los valores de rigideces y
resistencias
NTC-ACERO Diseño y Construcción de Estructuras de Acero
2. Diseño por Estabilidad
Efecto: Definición
PΔ
Son los que producen las cargas verticales al actuar sobre la estructura desplazada lateralmente (efectos de conjunto en toda la estructura)
Pδ
Son los ocasionados por las cargas originalmente axiales, cuando actúan sobre el miembro deformado entre sus extremos (Efectos individuales en cada columna)
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Métodos de diseño por estabilidad
Concepto Longitud Efectiva Método Directo
Tipo de Análisis Elástico de segundo orden (1)
Elástico de segundo orden (1)
Carga ficticia (2) Ni = 0.003 Wi (o Δo = 0.003L)
Ni = 0.003 Wi (o Δo = 0.003L)
Rigidez efectiva Nominal EI* = EI
EA* = EA
0.8 Nominal: EI* = 0.8 EI
EA* = 0.8 EA Resistencia axial PR con KL (3) PR con L (K=1) Limitaciones I ≤ 0.3 Ninguna (1) Puede realizarse con un método aproximado, iterativo o riguroso. (2) Carga ficticia solo en combinaciones con cargas de gravedad, no se consideran en
cargas que incluyan sismo (3) K = 1 se permite cuando el factor I ≤ 0.08
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Índice de estabilidad de un entrepiso
𝐼= ∑↑▒𝑃↓𝑢 𝑄∆↓𝑂𝐻 /𝐿∑↑▒𝐻
Donde:
ΣPU Fuerza vertical de diseño en el entrepiso en consideración
Q Factor de comportamiento sísmico. En diseño por viento se toma Q=1.0
ΔOH Desplazamiento horizontal relativo de primer orden de los niveles que limitan el entrepiso en consideración, en la dirección de análisis, producido por las fuerzas de diseño
ΣH Suma de todas las fuerzas horizontales de diseño que obran encima del entrepiso en consideración
L Altura del entrepiso
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Métodos de análisis de segundo orden
Métodos simplificados: amplifican análisis 1er orden • Análisis B1-B2 (traslación-impedida y traslación-permitida) • Análisis modificado • Análisis separados (gravedad y lateral) • Amplificación de distorsiones de entrepiso (White 2007)
Métodos iterativos: iteraciones de análisis de 1er orden • Carga lateral equivalente (Wood et al. 1976) • Carga de gravedad iterativa (Gaiotti y Smith 1989) • Rigidez negativa (Rutenberg 1981) Métodos rigurosos: análisis 2º orden directo • Funciones de estabilidad
• Ecuaciones pendiente-deflexión [ M = f (θ, Δ) ] • Matriz de rigidez directa [ K2 ]
• Matriz de rigidez geométrica [ K2 ] = [ K1 ] + [ Kg ]
NTC-ACERO Diseño y Construcción de Estructuras de Acero
2.7. Determinación aproximada de los efectos de segundo orden
Los momentos de segundo orden, se obtienen multiplicando los de primer orden por factores de amplificación:
• Momentos de diseño en los extremos de las columnas (PΔ):
𝑀↓𝑢𝑜 = 𝑀↓𝑡𝑖 + 𝐵↓2 𝑀↓𝑡𝑝
• Momentos de diseño en la zona central de la columna (PΔ y Pδ ):
𝑀↑∗ ↓𝑢𝑜 = 𝐵↓1 (𝑀↓𝑡𝑖 + 𝐵↓2 𝑀↓𝑡𝑝 )
𝐵↓1 = 𝐶𝑚/1− 𝑃↓𝑈 ∕𝐹↓𝑅 𝑃↓𝑒1
𝐵↓2 = 1/1−1.2𝐼 = 1/1− 1.2𝑄∆↓𝑂𝐻 𝛴𝑃↓𝑈 /𝐿𝛴𝐻 𝐵↓2 = 1/1− 1.2𝑄𝛴𝑃↓𝑈 /𝛴𝑃↓𝑒2 o bien
Cm = 0.6±0.4Mi
M j
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo – Análisis de 2º orden simplificado
exacting analysis is required. Of course, in those instances, it may be moreappropriate to simply use a direct second-order approach.
8.5.2 Example 8.2: Second-order Amplified Moments
The simple portal frame shown in Figure 8.7 is used in a number of exam-ples throughout the chapter to illustrate the application of the stability provi-sions in the specification. The loads given are the factored loads. We start bydetermining the second-order axial forces and moments in the columns fromthe NT–LT approach using amplification factors B1 and B2. We then com-pare these to the second order forces and moments in the system using adirect second-order analysis.
To begin the NT–LT process, we first analyze the nonsway frame by fix-ing joint C against translation and applying the factored loads as shown inFigure 8.8. After a first-order analysis is run, we record the moments andaxial loads in the columns, as well as the reaction at joint C. The forces andmoments are shown in Table 8.1, and the reaction R ¼ 9:42 kips.
C
D
B
A
w = 2.5 kip/ft.
W21 x 50
H = 4.5k
W10 x 33
30 ft.
18 ft.W10 x 49
Fig. 8.7 Portal frame example.
C
D
B
A
RH = 4.5k
w = 2.5 kip/ft.
Fig. 8.8 NT analysis model.
8.5 SPECIFICATION-BASED APPROACHES FOR STABILITY ASSESSMENT 335
w = 3.72 t/m
9.15 m
5.5
m
V=2 t
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo – Análisis TI-TP
exacting analysis is required. Of course, in those instances, it may be moreappropriate to simply use a direct second-order approach.
8.5.2 Example 8.2: Second-order Amplified Moments
The simple portal frame shown in Figure 8.7 is used in a number of exam-ples throughout the chapter to illustrate the application of the stability provi-sions in the specification. The loads given are the factored loads. We start bydetermining the second-order axial forces and moments in the columns fromthe NT–LT approach using amplification factors B1 and B2. We then com-pare these to the second order forces and moments in the system using adirect second-order analysis.
To begin the NT–LT process, we first analyze the nonsway frame by fix-ing joint C against translation and applying the factored loads as shown inFigure 8.8. After a first-order analysis is run, we record the moments andaxial loads in the columns, as well as the reaction at joint C. The forces andmoments are shown in Table 8.1, and the reaction R ¼ 9:42 kips.
C
D
B
A
w = 2.5 kip/ft.
W21 x 50
H = 4.5k
W10 x 33
30 ft.
18 ft.W10 x 49
Fig. 8.7 Portal frame example.
C
D
B
A
RH = 4.5k
w = 2.5 kip/ft.
Fig. 8.8 NT analysis model.
8.5 SPECIFICATION-BASED APPROACHES FOR STABILITY ASSESSMENT 335
For the LT analysis, an equal and opposite reaction at C is applied to thesway permitted model of the frame, as shown in Figure 8.9.
One the analyses are complete, we can calculate the B1 and B2 factors.B1 is given by
B1 ¼Cm
1" aPr
Pe1
# 1:0
where
Cm ¼ 0:6" 0:4M1
M2
! "¼ 0:6" 0:4 $ 0
303:8
! "¼ 0:6
and
Pe1 ¼p2EI
ðK1LÞ2¼ p2 $ ð29; 000Þð272Þ½ð1:0Þð18Þð12Þ(2
¼ 1;670 kips ðusing K1 ¼ 1:0Þ
For column AB
Pr ¼ Pnt þ Plt ¼ "40:5þ 5:7 ¼ 34:8 kips
B1 ¼0:6
1" ð1:0Þ34:8
1670
¼ 0:61!B1 ¼ 1:0
C
D
B
A
R = 9.42k
Fig. 8.9 LT analysis model.
TABLE 8.1 NT–LT Analysis Results
NT Analysis LT Analysis
PNT MNT PLT MLT
(kip) (in-kip) (kip) (in-kip)
Col AB "40:5 "1; 063 5.7 2035
Col CD "34:5 0 "5:7 0
336 SPECIFICATION-BASED APPLICATIONS OF STABILITY IN STEEL DESIGN
For the LT analysis, an equal and opposite reaction at C is applied to thesway permitted model of the frame, as shown in Figure 8.9.
One the analyses are complete, we can calculate the B1 and B2 factors.B1 is given by
B1 ¼Cm
1" aPr
Pe1
# 1:0
where
Cm ¼ 0:6" 0:4M1
M2
! "¼ 0:6" 0:4 $ 0
303:8
! "¼ 0:6
and
Pe1 ¼p2EI
ðK1LÞ2¼ p2 $ ð29; 000Þð272Þ½ð1:0Þð18Þð12Þ(2
¼ 1;670 kips ðusing K1 ¼ 1:0Þ
For column AB
Pr ¼ Pnt þ Plt ¼ "40:5þ 5:7 ¼ 34:8 kips
B1 ¼0:6
1" ð1:0Þ34:8
1670
¼ 0:61!B1 ¼ 1:0
C
D
B
A
R = 9.42k
Fig. 8.9 LT analysis model.
TABLE 8.1 NT–LT Analysis Results
NT Analysis LT Analysis
PNT MNT PLT MLT
(kip) (in-kip) (kip) (in-kip)
Col AB "40:5 "1; 063 5.7 2035
Col CD "34:5 0 "5:7 0
336 SPECIFICATION-BASED APPLICATIONS OF STABILITY IN STEEL DESIGN
A value of B1 < 1 indicates that no amplification of the MNT moment isrequired.
Next, we determine the second-order amplification of the sway momentsby calculating
B2 ¼1
1" aP
PNTPPe2
# 1:0; where a ¼ 1:0 ðfactored loading usedÞ
XPNT ¼ ð40:5þ 35:5Þ ¼ 75 kips
XPe2 ¼
X p2EI
ðK2LÞ2¼ RM
PHL
DH
where Rm ¼ 0:85 for moment framesUsing the second form of the
PPe2 equation based on the LT analysis
results, where DH ¼ 5:87 in:
XPe2 ¼
PHL
DH¼ 0:85ð9:42Þð18
0Þð1200=ft:Þ
5:8700¼ 295 kips
B2 ¼1
1" 75
295
¼ 1:34
The final forces are given by
Mr ¼ B1Mnt þ B2Mlt
Pr ¼ Pnt þ B2Plt
and are tabulated in Table 8.2, where it can be seen that, in this case, theycompare conservatively to the direct second-order elastic analysis results.
TABLE 8.2 Second-order Moment and Axial Load Values
B1-B2 Analysis Second-order Analysis
Pr Mr Pr Mr
(kip) (in-kip) (kip) (in-kip)
Col AB "33:4 1664 "33 1534
Col CD "42:1 0 "42 0
8.5 SPECIFICATION-BASED APPROACHES FOR STABILITY ASSESSMENT 337
TI TP
Análisis TI Análisis TP
w = 3.72 t/m V=2 t
14.97 cm
R=4.27 t
k=R/∆ =28.66t/m
-147 t-m 281.35 t-m
PTI (t)
MTI (t-m)
PTP (t)
MTP (t-m)
-18.4 t -15.6 t
+2.6 t -2.6 t
K ≤ 1.0
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo – Factor B1 B1 para Columna A-B:
𝐵↓1 = 𝐶𝑚/1− 𝑃↓𝑈 ∕𝐹↓𝑅 𝑃↓𝑒1 = 0.6/1− 34∕(0.9∗756.9) =0.63 → 𝐵1=1.0
𝐶𝑚=0.6±0.4𝑀𝑖/𝑀𝑗 =0.6−0.4(0/−147 )=0.6
𝑃↓𝑒1 = 𝜋2𝐸𝐼/(𝐾𝐿𝑐)2 = 𝜋2(2,039,000)(11,321.5)/(1∙549)2(1,000) =756.9 t
𝑃↓𝑢 =𝑤𝐿𝑣=3.72(9.15)=34 t
K ≤ 1.0
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo – Factor B2 y momento de 2o orden
𝐵↓2 = 1/1−1.2𝐼 = 1/1−1.2(0.22) =1.35
B2 para Columna A-B: 𝐼= ∑↑▒𝑃↓𝑢 𝑄∆↓𝑂𝐻 /𝐿∑↑▒𝐻 = 34(1)/550 14.97/4.27 =0.2167
𝑀↓𝑢𝑜 = 𝑀↓𝑡𝑖 + 𝐵↓2 𝑀↓𝑡𝑝 =−147+1.35(281.35)=𝟐𝟑𝟑 𝐭∙𝐦
𝑀↑∗ ↓𝑢𝑜 = 𝐵↓1 (𝑀↓𝑡𝑖 + 𝐵↓2 𝑀↓𝑡𝑝 )=1(233)=233 t∙m
Momento de segundo orden en Col. AB
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo – Momento en columna AB
Análisis de segundo orden Simplificado TI-TP vs. Riguroso
𝑀↓𝑢 =134.4 t∙m
𝑀↓𝑢𝑜 /𝑀↓𝑢 = 233/134.4 =1.73
𝑀↓𝑢𝑜 =233 t∙m ( Error = 9.9% )
Análisis de primer orden
𝑀↓𝑢𝑜 =212 t∙m
𝑀↓𝑢𝑜 /𝑀↓𝑢 = 212/134.4 =1.58
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Métodos de diseño por estabilidad
Concepto Longitud Efectiva Método Directo
Tipo de Análisis Elástico de segundo orden (1)
Elástico de segundo orden (1)
Carga ficticia (2) Ni = 0.003 Wi (o Δo = 0.003L)
Ni = 0.003 Wi (o Δo = 0.003L)
Rigidez efectiva Nominal EI* = EI
EA* = EA
0.8 Nominal: EI* = 0.8 EI
EA* = 0.8 EA Resistencia axial PR con KL (3) PR con L (K=1) Limitaciones I ≤ 0.3 Ninguna (1) Puede realizarse con un método aproximado, iterativo o riguroso. (2) Carga ficticia solo en combinaciones con cargas de gravedad, no se consideran en
cargas que incluyan sismo (3) K = 1 se permite cuando el factor I ≤ 0.08
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Tolerancias Código de Practicas Generales
118
(a) AISC Code of Standard Practice (AISC, 2005d) (b) ACI-117 (2006)
Figure 5.1. Column tolerances as specified in the AISC and the ACI Specifications
Thus, nominal compression capacity of a CFT column (Pn) can be calculated as:
0.658 / 2.25
0.877 / 2.25
o
e
PP
o o en
e o e
P if P PP
P if P P
§ ·° d¨ ¸¨ ¸ ® © ¹°
!¯
(5.1)
(I2-2, I2-3, AISC-10)
The cross-section compressive strength (Po) and the Euler load (Pe) are computed by:
� �2 ' for compact sections
0.7 ' for non-compact sectionss y r yr c c
os cr c c sr s c
A F A F C A fP
A F f A A E E� �
� �
(5.2)
(I2-9, AISC-10)
2
2( )eff
e
EIP
KLS
(5.3)
(I2-5, AISC-10)
Inclusión directa de las imperfecciones (desplome)
3 m
6 m
3 m
9 m
m
18 mm
Δo = 1.5 ( L / 500 ) = 0.003 L IMCA (2014)
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Imperfecciones iniciales *
Inclusión directa de las imperfecciones (desplome)
3 m
6 m
3 m
9 m
m
18 mm
3 m
6 m
3 m
0.12 t
0.09 t
Ni = 0.003 Wi Δo = 0.003 L
Fuerzas ficticias
15 t 15 t
20 t 20 t
* Solo en combinaciones sin carga lateral
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Métodos para determinar K
§ Análisis de carga crítica ( Pcr = PeK = Pe / K2 ) § Ecuaciones teóricas simplificadas
(ojo: hay hipótesis que no siempre se cumplen) § Sin corrección § Con corrección (por hipótesis que no se cumplen)
§ Nomogramas (Julian-Lawrence 1959) § Ecuaciones semi-empíricas aproximadas § Metodologías para el pandeo de entrepisos:
§ Método de rigidez del entrepiso (LeMessurier 1976) § Método de pandeo del entrepiso (Yura 1971) § Método de LeMessurier
NTC-ACERO Diseño y Construcción de Estructuras de Acero
important to note that neither of these values for K2 should be used incalculation Pe2 to determine B2.
8.6.6 Example 8.4 Effective Length Calculation with LeanerColumn
Reconsidering the frame in Figure 8.7, we determine the K-factor for col-umn AB. For this frame, we consider three possible variations from the as-sumptions built into the alignment chart:
1. The behavior is purely elastic.
2. All joints are rigid.
3. For columns in sidesway inhibited frames, rotations at opposite endsof the restraining beams are equal in magnitude and opposite in di-rection, producing single curvature bending.
From Example 8.3, we know that Pr ¼ 33k, so we can check whether aninelastic stiffness reduction factor is applicable
aPr
fc
! "
ðFyAgÞ¼
1:0ð33Þ0:9
! "
50ð14:4Þ¼ 0:051 $ 0:39
therefore, t ¼ 1:0, and no inelastic stiffness reduction is applied.The pinned joint at C requires the length of the beam to be modified be-
fore using the alignment chart approach:
L0g ¼ Lg 2%MF
MN
! "¼ 30 2% 0
77:76
! "¼ 600
Next we calculate the sway K-factor:
Gtop ¼
P taI
L
! "
cP I
L
! "
g
¼
1:0ð272Þ18
! "
984
60
! " ¼ 0:92
Gbottom ¼ 10:0ðpinned-endÞ
From alignment charts with Gtop ¼ 0:92 and Gbottom ¼ 10:0, we obtain
Kxffi 1:9
352 SPECIFICATION-BASED APPLICATIONS OF STABILITY IN STEEL DESIGN
important to note that neither of these values for K2 should be used incalculation Pe2 to determine B2.
8.6.6 Example 8.4 Effective Length Calculation with LeanerColumn
Reconsidering the frame in Figure 8.7, we determine the K-factor for col-umn AB. For this frame, we consider three possible variations from the as-sumptions built into the alignment chart:
1. The behavior is purely elastic.
2. All joints are rigid.
3. For columns in sidesway inhibited frames, rotations at opposite endsof the restraining beams are equal in magnitude and opposite in di-rection, producing single curvature bending.
From Example 8.3, we know that Pr ¼ 33k, so we can check whether aninelastic stiffness reduction factor is applicable
aPr
fc
! "
ðFyAgÞ¼
1:0ð33Þ0:9
! "
50ð14:4Þ¼ 0:051 $ 0:39
therefore, t ¼ 1:0, and no inelastic stiffness reduction is applied.The pinned joint at C requires the length of the beam to be modified be-
fore using the alignment chart approach:
L0g ¼ Lg 2%MF
MN
! "¼ 30 2% 0
77:76
! "¼ 600
Next we calculate the sway K-factor:
Gtop ¼
P taI
L
! "
cP I
L
! "
g
¼
1:0ð272Þ18
! "
984
60
! " ¼ 0:92
Gbottom ¼ 10:0ðpinned-endÞ
From alignment charts with Gtop ¼ 0:92 and Gbottom ¼ 10:0, we obtain
Kxffi 1:9
352 SPECIFICATION-BASED APPLICATIONS OF STABILITY IN STEEL DESIGN
Ejemplo: Factor K de Columna AB
The K-value can be calculated more accurately using the transcendentalequation that defines the sway-permitted alignment chart
p
K
tanp
K
! "!
p
K
! "2ð10Þð0:95Þ ! 36
6ð10:95Þ ¼ 0
which gives a value of Kx ¼ 1:89.This initial value of Kx must be adjusted to account for the leaning col-
umn, column CD. This can be done using either the story-stiffness or thestory-buckling approach in the commentary. Using story stiffness, where K2
is given by
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
The ratio DHPHL
was previously calculated to find B2 in Example 8.2.
DHPHL¼ 1
Pe2¼ 1
295¼ 0:0034
The termP
Pr is the summation of the total axial load on all columns in thestory.
XPr ¼ 75 kips
RL ¼P
Pr leaning columnsPPr all column
¼ 42 kips
75 kips¼ 0:56
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi75
ð0:85þ 0:15ð0:56ÞÞ33
p2ð29;000Þð272Þð18& 12Þ2
!&0:0034
'vuut ¼ 3:7
8.6 EFFECTIVE LENGTH FACTORS, K-FACTORS 353
The K-value can be calculated more accurately using the transcendentalequation that defines the sway-permitted alignment chart
p
K
tanp
K
! "!
p
K
! "2ð10Þð0:95Þ ! 36
6ð10:95Þ ¼ 0
which gives a value of Kx ¼ 1:89.This initial value of Kx must be adjusted to account for the leaning col-
umn, column CD. This can be done using either the story-stiffness or thestory-buckling approach in the commentary. Using story stiffness, where K2
is given by
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
The ratio DHPHL
was previously calculated to find B2 in Example 8.2.
DHPHL¼ 1
Pe2¼ 1
295¼ 0:0034
The termP
Pr is the summation of the total axial load on all columns in thestory.
XPr ¼ 75 kips
RL ¼P
Pr leaning columnsPPr all column
¼ 42 kips
75 kips¼ 0:56
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi75
ð0:85þ 0:15ð0:56ÞÞ33
p2ð29;000Þð272Þð18& 12Þ2
!&0:0034
'vuut ¼ 3:7
8.6 EFFECTIVE LENGTH FACTORS, K-FACTORS 353The K-factor calculated by the story-buckling approach is given by
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2EI
L2
" #
Pr
vuuut PPr
P p2EI
ðKn2LÞ2
!
0
BBBB@
1
CCCCA¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2ð29000Þð272Þð2162Þ
!
33
75
p229000ð272Þð1:95ð216ÞÞ2
!
vuuuuuuut
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1:95Þ2ð75Þ
33
s
¼ 2:94
Based on an elastic buckling analysis of the frame, the elastic critical buck-ling load, Pcr ¼ 165 kips, and the actual K-factor for column, is given by
K ¼ffiffiffiffiffiffiPe
Pcr
r¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2EI
L2
" #
Pcr
vuuut¼
ffiffiffiffiffiffiffiffiffiffiffi1669
165:3
r¼ 3:18
The actual value lies between the calculated story-stiffness and story-buckling values.
8.7 DESIGN ASSESSMENT BY TWO APPROACHES
In this section, we compare the direct-analysis approach to the critical loador K-factor approach using the portal frame introduced in Example 8.2. Inthis example, we cannot use the first-order approach because the B2 factor(calculated in Example 8.2) is greater than 1.1. While D2=D1 and B2 are notthe exact same value (drift amplification versus moment amplification), B2 isallowed as an approximation of the drift ratio.
8.7.1 Critical Load Approach
In order to perform the strength check, we use the results of Example 8.2with the B1 and B2 amplifiers and the K-factor values determined in Exam-ple 8.4. For this example, we assume the frame is braced in the out-of-planedirection. We check the sidesway stability of the frame by checking the lat-eral resisting member, Column AB. As with the previous example, we focuson the factored lateral load case. The steps are provided in section 8.5.3:
Step 1: Verify that the second- to first-order drift ratio for all stories is lessthan 1.5.
354 SPECIFICATION-BASED APPLICATIONS OF STABILITY IN STEEL DESIGN
exacting analysis is required. Of course, in those instances, it may be moreappropriate to simply use a direct second-order approach.
8.5.2 Example 8.2: Second-order Amplified Moments
The simple portal frame shown in Figure 8.7 is used in a number of exam-ples throughout the chapter to illustrate the application of the stability provi-sions in the specification. The loads given are the factored loads. We start bydetermining the second-order axial forces and moments in the columns fromthe NT–LT approach using amplification factors B1 and B2. We then com-pare these to the second order forces and moments in the system using adirect second-order analysis.
To begin the NT–LT process, we first analyze the nonsway frame by fix-ing joint C against translation and applying the factored loads as shown inFigure 8.8. After a first-order analysis is run, we record the moments andaxial loads in the columns, as well as the reaction at joint C. The forces andmoments are shown in Table 8.1, and the reaction R ¼ 9:42 kips.
C
D
B
A
w = 2.5 kip/ft.
W21 x 50
H = 4.5k
W10 x 33
30 ft.
18 ft.W10 x 49
Fig. 8.7 Portal frame example.
C
D
B
A
RH = 4.5k
w = 2.5 kip/ft.
Fig. 8.8 NT analysis model.
8.5 SPECIFICATION-BASED APPROACHES FOR STABILITY ASSESSMENT 335
Gb = 10
The K-value can be calculated more accurately using the transcendentalequation that defines the sway-permitted alignment chart
p
K
tanp
K
! "!
p
K
! "2ð10Þð0:95Þ ! 36
6ð10:95Þ ¼ 0
which gives a value of Kx ¼ 1:89.This initial value of Kx must be adjusted to account for the leaning col-
umn, column CD. This can be done using either the story-stiffness or thestory-buckling approach in the commentary. Using story stiffness, where K2
is given by
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
The ratio DHPHL
was previously calculated to find B2 in Example 8.2.
DHPHL¼ 1
Pe2¼ 1
295¼ 0:0034
The termP
Pr is the summation of the total axial load on all columns in thestory.
XPr ¼ 75 kips
RL ¼P
Pr leaning columnsPPr all column
¼ 42 kips
75 kips¼ 0:56
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi75
ð0:85þ 0:15ð0:56ÞÞ33
p2ð29;000Þð272Þð18& 12Þ2
!&0:0034
'vuut ¼ 3:7
8.6 EFFECTIVE LENGTH FACTORS, K-FACTORS 353
The K-value can be calculated more accurately using the transcendentalequation that defines the sway-permitted alignment chart
p
K
tanp
K
! "!
p
K
! "2ð10Þð0:95Þ ! 36
6ð10:95Þ ¼ 0
which gives a value of Kx ¼ 1:89.This initial value of Kx must be adjusted to account for the leaning col-
umn, column CD. This can be done using either the story-stiffness or thestory-buckling approach in the commentary. Using story stiffness, where K2
is given by
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
The ratio DHPHL
was previously calculated to find B2 in Example 8.2.
DHPHL¼ 1
Pe2¼ 1
295¼ 0:0034
The termP
Pr is the summation of the total axial load on all columns in thestory.
XPr ¼ 75 kips
RL ¼P
Pr leaning columnsPPr all column
¼ 42 kips
75 kips¼ 0:56
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPPr
ð0:85þ 0:15RLÞPr
p2EI
L2
$ %DHP
HL
$ %s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi75
ð0:85þ 0:15ð0:56ÞÞ33
p2ð29;000Þð272Þð18& 12Þ2
!&0:0034
'vuut ¼ 3:7
8.6 EFFECTIVE LENGTH FACTORS, K-FACTORS 353
The K-factor calculated by the story-buckling approach is given by
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2EI
L2
" #
Pr
vuuut PPr
P p2EI
ðKn2LÞ2
!
0
BBBB@
1
CCCCA¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2ð29000Þð272Þð2162Þ
!
33
75
p229000ð272Þð1:95ð216ÞÞ2
!
vuuuuuuut
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1:95Þ2ð75Þ
33
s
¼ 2:94
Based on an elastic buckling analysis of the frame, the elastic critical buck-ling load, Pcr ¼ 165 kips, and the actual K-factor for column, is given by
K ¼ffiffiffiffiffiffiPe
Pcr
r¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2EI
L2
" #
Pcr
vuuut¼
ffiffiffiffiffiffiffiffiffiffiffi1669
165:3
r¼ 3:18
The actual value lies between the calculated story-stiffness and story-buckling values.
8.7 DESIGN ASSESSMENT BY TWO APPROACHES
In this section, we compare the direct-analysis approach to the critical loador K-factor approach using the portal frame introduced in Example 8.2. Inthis example, we cannot use the first-order approach because the B2 factor(calculated in Example 8.2) is greater than 1.1. While D2=D1 and B2 are notthe exact same value (drift amplification versus moment amplification), B2 isallowed as an approximation of the drift ratio.
8.7.1 Critical Load Approach
In order to perform the strength check, we use the results of Example 8.2with the B1 and B2 amplifiers and the K-factor values determined in Exam-ple 8.4. For this example, we assume the frame is braced in the out-of-planedirection. We check the sidesway stability of the frame by checking the lat-eral resisting member, Column AB. As with the previous example, we focuson the factored lateral load case. The steps are provided in section 8.5.3:
Step 1: Verify that the second- to first-order drift ratio for all stories is lessthan 1.5.
354 SPECIFICATION-BASED APPLICATIONS OF STABILITY IN STEEL DESIGN
The K-factor calculated by the story-buckling approach is given by
K2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2EI
L2
" #
Pr
vuuut PPr
P p2EI
ðKn2LÞ2
!
0
BBBB@
1
CCCCA¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2ð29000Þð272Þð2162Þ
!
33
75
p229000ð272Þð1:95ð216ÞÞ2
!
vuuuuuuut
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1:95Þ2ð75Þ
33
s
¼ 2:94
Based on an elastic buckling analysis of the frame, the elastic critical buck-ling load, Pcr ¼ 165 kips, and the actual K-factor for column, is given by
K ¼ffiffiffiffiffiffiPe
Pcr
r¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2EI
L2
" #
Pcr
vuuut¼
ffiffiffiffiffiffiffiffiffiffiffi1669
165:3
r¼ 3:18
The actual value lies between the calculated story-stiffness and story-buckling values.
8.7 DESIGN ASSESSMENT BY TWO APPROACHES
In this section, we compare the direct-analysis approach to the critical loador K-factor approach using the portal frame introduced in Example 8.2. Inthis example, we cannot use the first-order approach because the B2 factor(calculated in Example 8.2) is greater than 1.1. While D2=D1 and B2 are notthe exact same value (drift amplification versus moment amplification), B2 isallowed as an approximation of the drift ratio.
8.7.1 Critical Load Approach
In order to perform the strength check, we use the results of Example 8.2with the B1 and B2 amplifiers and the K-factor values determined in Exam-ple 8.4. For this example, we assume the frame is braced in the out-of-planedirection. We check the sidesway stability of the frame by checking the lat-eral resisting member, Column AB. As with the previous example, we focuson the factored lateral load case. The steps are provided in section 8.5.3:
Step 1: Verify that the second- to first-order drift ratio for all stories is lessthan 1.5.
354 SPECIFICATION-BASED APPLICATIONS OF STABILITY IN STEEL DESIGN
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo: Factor K de columna CD
ASCE (1997), de Buen (2000)
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo: Factor K de columnas (7)
(8)
(10)
(11)
ASCE (1997), de Buen (2000)
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo: Factor K de columnas (7)
(8)
(10)
(11)
ASCE (1997), de Buen (2000)
NTC-ACERO Diseño y Construcción de Estructuras de Acero
En varios casos, el cálculo de K no es simple ni preciso
knee kneeridge
lean-to frame
D. White (2012)
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Método directo Este método puede utilizarse para todas las estructuras. Las acciones de diseño de los componentes de la estructura se determinan con un análisis que incluya imperfecciones (directas o fuerzas ficticias) y ajustes de rigideces (80%) en el análisis (de segundo orden).
Imperfecciones directas
3 m
6 m
3 m
9 m
m
18 mm
3 m
6 m
3 m
0.12 t
0.09 t Ni = 0.003 Wi Δo = 0.003 L
Fuerzas ficticias
15 t 15 t
20 t 20 t
EI * = 0.8EIEA* = 0.8EAK =1
¡ No más nomogramas !
NTC-ACERO Diseño y Construcción de Estructuras de Acero
Ejemplo – Comparación de métodos
Concepto NTC (elástico) Riguroso
(inelástico 2o orden) Long. E. Directo
Del análisis de 2o orden Pu 15 t 14.7 t 14.8 t
Mu 212.1 t-m 244.7 t-m 240 t-m
Cargas ficticias Ni 0.003Wi 0.003Wi 0.003Wi
Rigidez a flexión EI* EI 0.8 EI EI
Del análisis de carga crítica K 3.18 NA (K=1) NA (K=1)
Resistencia a compresión Rc 59 t 173.7 kip 173.7 kip
Resistencia a flexión MR 376.6 t-m 376.6 t-m 376.6 t-m
Revisión P-M para W10x49 UC 0.75 0.69 0.68