Steel Structures 6 (2006) 93-105 www.kssc.or.kr
Plastic Analysis of Steel Frames with Unsymmetrical Sections
W. S. King1,*, L. Duan2 and W. F. Chen3
1Department of Construction Engineering, Chaoyang University of Technology, No. 168,
Jifong East Road, Wufong Township, Taichung County, 41349, Taiwan, R.O.C.2California Dept. of Transportation, Sacramento, CA 95816.
3College of Engineering, University of Hawaii at Manoa, Holmes Hall 240, 2540 Dole St., Honolulu, HI 96822, U.S.A.
Abstract
The plastic behavior of steel frames with hot-rolled channel sections is investigated. The stability functions are used in thestiffness matrix of a beam-column element. The effects of both material and geometrical nonlinearity on the structural behaviorare studied. The initial yield surface and the limit surface of a channel section bent in the unsymmetrical weak axis aredeveloped corresponding to the condition that a section is subjected to a compressive or tensile axial load combined with asagging or hogging bending moment. The differences of the limit loads applied on beams or rigid frames with channel sectionsare compared for different conditions. A simple model for semi-rigid connections is proposed. The effect of semi-rigidconnections on the limit loads of steel frames with unsymmetrical channel sections is also studied.
Keywords: Hot-rolled, Unsymmetrical section, Yielding surface, Flexible connection, Frame analysis, Structural engineering.
1. Introduction
Nonlinear analyses of frames with doubly symmetric
sections have been studied extensively. However, nonlinear
analyses of frames with unsymmetrical sections as shown
in Fig. 1 have not been studied systematically and
thoroughly. The nonlinear plastic analysis of a frame with
unsymmetrical sections is much more complex than that
with symmetrical sections. A channel section is one of
these singly symmetric sections. The behavior of frames
with channel sections bent in the unsymmetrical weak
axis is studied in this paper.
The plastic analysis of steel frames using channel
sections bent in the unsymmetrical axis was rarely
published in the past. Chen and Atsuta (1976) developed
several limit surfaces of channel sections bent in strong
axis or weak axis. Horne (1979) discussed the limit
surface of a T-section under the effect of an axial load.
Gjelsvik (1981) studied a limit surface of a channel
section bent in symmetrical strong axis. Rasmussen and
Hancock (1989) investigated the compression tests of
welded channel section columns. However, those specific
sections described above are different from the hot-rolled
channel section. Herein, the channel section of MC3×7.1
is chosen for study. The limit and initial yield surfaces of
this section bent in the unsymmetrical weak axis are first
developed. The elastic-plastic structural analyses of portal
frames with channel sections bent in weak axis are then
carried out.
A complete stiffness matrix is usually composed of the
linear elastic stiffness matrix and the geometric stiffness
matrix for a nonlinear structural analysis. Herein, the
stability functions are used directly for the beam-column
stiffness matrix. The effect of the tensile or compressive
axial force in a member can be represented simply by the
stability functions.
The rigid connections between beams and columns
were usually assumed in frame analyses. However, all
real connections are semi-rigid. Lui (1985) published a
second-order simple plastic hinge method for semi-rigid
frames. Lui and Chen (1988) also derived the stiffness
matrix of a beam-column with semi-rigid connections at
both ends. King etc. (1992) developed a stiffness matrix
for a beam-column with semi-rigid connections at both
ends. In order to represent both the linear elastic and
nonlinear plastic behavior of a connection, a simple
connection model is proposed here. Only four parameters
need to be used in this connection model. The load-
deflection responses of rigid or semi-rigid frames with
channel sections are investigated.
2. Modification of the Previous Stiffness
Matrix
The matrix for the degradation of stiffness due to a
partial plastification in member’s cross sections was
derived previously by King and Chen (1994) where the
decay factor D of a section’s stiffness was calculated
*Corresponding author
Tel: +886-927219646, Fax: +886-4-2374-2325E-mail: [email protected]
94 W. S. King et al.
from the m-φ-p relationship. Those terms in the stiffness
matrix were obtained by simply taking the first three
terms in a Taylor series expansion of the stability functions.
Herein, the stability functions are used directly instead
of taking the first three terms in a Taylor series expansion
of stability functions. These functions can be expressed as
Eqs. (1), (2), (3), and (4) (1987). When the axial load
applied in a member is compressive, Sii, Sij, Sji and Sjj can
be expressed by Eqs. (1) and (2).
(1)
(2)
When the axial load applied in a member is tensile,
those terms can be expressed by Eqs. (3) and (4).
(3)
(4)
where
I = moment inertia of a cross section
L = length of a member
When the axial load applied in a member becomes
zero, then Sii = Sjj = 4, and Sij = Sji = 2. These values of
stability functions can be calculated and substituted into a
stiffness matrix. The tangent elastic modulus Et (1994) is
used to consider material inelasticity. The modulus Et
Sii Sjj
KLSinKL KL( )2CosKL–
2 2CosKL– KLSinKL–---------------------------------------------------------
EI
L-----= =
Sij Sji
KL( )2KLSinKL–
2 2CosKL– KLSinKL–------------------------------------------------------
EI
L-----= =
Sii Sjj
KL( )2CoshKL KLSinhKL–
2 2CoshKL– KLSinhKL+----------------------------------------------------------------
EI
L-----= =
Sij Sji
KLSinhKL KL( )2
–
2 2CoshKL– KLSinhKL+------------------------------------------------------------
EI
L-----= =
KP
EtI------=
Figure 1. Channel sections application.
Plastic Analysis of Steel Frames with Unsymmetrical Sections 95
decreases gradually after the axial load is greater than
0.5Py as shown in Eq. (5).
(5)
E = modulus of elasticity of material
There are many connection models published in the
open literature. In order to simplify the simulation, a new
connection model is proposed below. Four parameters Ki,
Mu, Mcy, and A are needed to describe the behavior of a
semi-rigid connection. When the bending moment M at
connection is less than Mcy, the instantaneous tangent
stiffness Kt of a connection is equal to Ki. Otherwise, the
instantaneous tangent stiffness is expressed in Eq. (6).
When Mcy is not zero, there is a linear elastic response in
the moment-rotation curve of a semi-rigid connection as
shown in Fig. 2. The tangent stiffness Kt of a connection
can be calculated and substituted into the stiffness matrix
directly.
(6)
where
Kt = instantaneous tangent stiffness of a connection
Ki = initial tangent stiffness of a connection
Mu
= ultimate bending moment of a connection
M = bending moment applied at a connection
Mcy
= initial yielded bending moment of a connection
A = shape factor for the decay of tangent stiffness of a
connection (determined by the least square method to
match the experimental M-θ connection curve. (1992))
3. Yield Criteria of a Channel Section
The limit surfaces of a channel section of MC3×7.1 are
studied in what follows. The channel section of MC3×7.1
is symmetric about its strong axis. However, it is not
symmetric about its weak axis. The limit surface of
channel section bent in weak axis is developed in the
following. The behavior of portal frames with channel
sections bent in weak axis is studied later. Residual
stresses are not considered here.
3.1. Limit surface of a channel section
The limit surface of a channel section bent in weak axis
is developed referring to the research work done by Horne
(1981). There are three assumptions in this derivation:
(1) Material is elastic perfect plastic.
(2) In-plane loading is applied at the shear center of a
section.
(3) Effect of shear is neglected.
Derivation of channel section’s limit surface is explained
here. A channel section of MC3×7.1 is shown in Fig.
3(a). The thickness of web is slightly different from that
in the handbook of AISC. In this study, all dimensions of
Et
E
4EP
Py
----- 1P
Py
-----–⎝ ⎠⎛ ⎞
⎩⎪⎨⎪⎧
=
for P 0.5Py<
for P 0.5Py≥
Kt
Ki
Ki 1M Mcy–
Mu Mcy–-------------------⎝ ⎠⎛ ⎞
A
–
⎩⎪⎨⎪⎧
=
for M Mcy≤
for M Mcy>
Figure 2. Connection model.
Figure 3. Neutral axis in the web of a downward channel section MC3×7.1.
96 W. S. King et al.
a channel section MC3×7.1 after simplification are shown
in Fig. 3(a). The area of a cross section is 13.4 cm2
(2.0774 in2). The moment of inertia of a cross section
bent in weak axis is 30.88 cm4 (0.7419 in4). The plastic
modulus Z of a cross section is 17.34 cm3 (1.058 in3). The
distance between edge and centroid is 3.17 cm
(1.249 in). The ratio P/Py is denoted by n where P is the
axial load. Py is the squash load, which is equal to σyA.
Py = 36 ksi×2.0774 in2 = 74.79 kips = 332.82 kN.
3.1.1. Downward channel section
When the web of a channel section is on the top, it
means that the tip of flange is downward as shown in Fig.
3. This kind of section is called the downward channel
section.
a. Sagging bending moment
When a sagging bending moment is applied in the
downward channel section, the bending stresses in the
web are compressive and the bending stresses in the tips
of flanges are tensile.
(1) Neutral axis in the web
When the downward channel section is subjected to a
sagging bending moment and the neutral axis is in the
web, the distance between the neutral axis and the
centroid of a section is a1. A stress diagram is shown in
Fig. 3(b). The average axial compressive stress is nσy,
where n = P/Py. The axial load P = Anσy is given in Eq.
(7).
2.0774nσy =
3(0.689 − a1)σy− 3(a1− 0.377)σy− 2×0.351×1.626σy(7)
Solved for a1, we get . Since
for the neutral axis in the web,
. If the sagging plastic moment Mpc under the
axial load is considered to act about the centroidal axis
XX, then, we have
(8)
The fully plastic moment Mp under a zero axial load is
1.058σy. If Mpc is divided by Mp, we get
(9)
After simplification, Eq. (9) becomes Eq. (10).
(10)
When n = −1, a1 is 0.689 in (about 1.75 cm), and Mpc/
Mp is zero. It means that the whole section is subjected to
a tensile axial force Py.
(2). Neutral axis in the flange
When the neutral axis remains in the flange as shown
in Fig. 4(a), the stress diagram is shown in Fig. 4(b). The
axial load in a section is given in Eq. (11).
2.0774nσy = 3×0.312σy + 0.702×(0.377 − a2)σy −
0.702a2σy − 0.702×1.249σy (11)
Solved for a2, we get a2 = 0.23 − 1.479n. Since −1.249
≤ a2≤0.377 for the neutral axis to be in the flange, −0.1
≤ n ≤ 1. Taking the equilibrium of moment about the
centroid axis XX, we obtain the plastic moment Mpc.
After Mpc is divided by Mp, We get
(12)
After simplification, we get
(13)
When n = 1, Mpc/Mp = 0. It means that the whole
section is subjected to a compressive load Py.
b. Hogging bending moment
When a hogging bending moment is applied in a
downward channel section, the bending stresses in the
Y
a1
0.342 0.346n–=
0.377 a1
0.689≤ ≤
1– n 0.1–≤ ≤
Mpc
3 0.6892a1
2–( )
2--------------------------------σ
y
3 a1
20.377
2–( )
2--------------------------------σ
y– 0.497σ
y+=
Mpc
Mp
---------0.688
2a1
2–
0.352------------------------ 0.982 1 n+( ) 1.03 0.346n–( )= =
Mpc
Mp
--------- 1 0.67n 0.33n2
–+=
Mpc
Mp
---------1.248
2a2
2–
1.507------------------------ 0.981 1 n–( ) 1.018 1.479n+( )= =
Mpc
Mp
--------- 1 0.45n 1.45n2
–+=
Figure 4. Neutral axis in the flange of a downward channel section MC3×7.1.
Plastic Analysis of Steel Frames with Unsymmetrical Sections 97
web are tensile and the bending stresses in the tips of
flanges are compressive. The limit surface can be written
in Eq. (14) for 0.1 ≤ n ≤ 1.
(14)
The limit surface can be written in Eq. (15) for −1 ≤
n ≤ 0.1.
(15)
Complete results are represented graphically in Fig. 5.
The graph covers the full possible range for axial tension
as well as axial thrust, and for both hogging and sagging
bending moments.
3.1.2. Upward channel section
When the web of a channel section is on the bottom, it
means that the tips of flanges are upward as shown in Fig.
6. This kind of section is known as the upward channel
section. Following the same procedure, we can get the
limit surfaces.
a. Sagging bending moment
When a sagging bending moment and an axial load are
applied in the upward channel section, the bending
stresses in the tips are compressive, but the bending
stresses in the web may be tensile. The limit surface is
written in Eq. (16) for 0.1 ≤ n ≤ 1. When n = −1, Mpc/Mp
= 0. It means that the whole section is subjected to a
compressive load Py.
(16)
The limit surface is given in Eq. (17) for −1 ≤ n ≤ 0.1.
When n = −1, Mpc/Mp is zero. It means that the whole
section is subjected to a tensile axial force Py.
Mpc
Mp
--------- 1– 0.67n 0.33n2
+ +=
Mpc
Mp
--------- 1– 0.45n 1.45n2
+ +=
Mpc
Mp
--------- 1 0.67– n 0.33n2
–=
Figure 5. Limit surface of a downward channel section MC3×7.1.
98 W. S. King et al.
(17)
b. Hogging bending moment
When a hogging bending moment and an axial load are
applied in the upward channel section, the bending
stresses in tips may be tensile, the bending stresses in web
are compressive. The limit surface can be written in Eq.
(18) for −0.1 ≤ n ≤ 1.
(18)
The limit surface can be written in Eq. (19) for −1 ≤
n ≤ −0.1.
(19)
These limit surfaces of the upward channel section are
plotted in Fig. 6.
3.2. Initial yield surfaces
The initial yield of a section is defined as the edge of
a section is equal to the yielding stress σy when a member
is subjected to bending moments and an axial load. There
are two methods to derive the initial yield surface of a
channel section bent in the weak axis.
Method 1:
The initial yield surfaces can be derived by using the
following formula (1972).
(20)
Mpc
Mp
--------- 1 0.45– n 1.45n2
–=
Mpc
Mp
--------- 1– 0.45n– 1.45n2
+=
Mpc
Mp
--------- 1.01– 0.67n– 0.34n2
+=
σy
Mycyi
I------------
P
A---±=
Figure 6. Limit surface of an upward channel section MC3×7.1.
Plastic Analysis of Steel Frames with Unsymmetrical Sections 99
where
Myc = initial yielded bending moment
I = moment inertia of a channel section bent in weak
axis
yi = distance between centroid of a cross section and
edges of cross section (i = 1, 2)
P = axial load (tension is positive; compression is
negative)
A = area of a channel section
The initial yielded bending moments under different
axial loads can be calculated by using Eq. (20). The initial
yield surface of a channel section can be obtained from
the corresponding values of axial loads and initial yielded
bending moments. When a channel section is subjected to
a bending moment and an axial load, the position of
initial yield in the section may change due to the
magnitude of an axial load.
When the downward channel section is subjected to a
sagging bending moment and a compressive axial load,
the position of initial yield is at the tips of flanges if the
axial load P is between zero and 0.3Py. The position of
initial yield is at the top edge of web if the axial load P
is between 0.3Py and Py. These calculated values of the
initial yielded bending moment Myc are expressed in
Table 1.
When a downward channel section is subjected to a
sagging bending moment and a compressive axial load
between zero and 0.3Py, the initial yield surface is given
in Eq. (21).
for (21)
When a downward channel section is subjected to a
sagging bending moment and a compressive axial load
between 0.3Py and Py, the initial yield surface is given as
Eq. (22).
for (22)
Equations (21) and (22) can be simulated by the curve
fitting method and represented as a single Eq. (23). The
Eq. (23) in the first quadrant is shown in Fig. 7 by the
dashed line. This equation is convenient for computer
programming.
(23)
Myc
Mp
-------- 0.503P
Py
-----⎝ ⎠⎛ ⎞
0.561+= 0 P 0.3Py
≤ ≤
Myc
Mp
-------- 1.017–P
Py
-----⎝ ⎠⎛ ⎞
1.017+= 0.3PyP P
y≤ ≤
Myc
Mp
-------- 0.56 0.6P
Py
-----⎝ ⎠⎛ ⎞
2.5P
Py
-----⎝ ⎠⎛ ⎞
2
15P
Py
-----⎝ ⎠⎛ ⎞
3
–+ + +=
Table 1. Initial yielded bending moment for a downward channel section subjected to a sagging bending moment and acompressive axial load
Axial load ratio P/Py Initial yielded bending moment Myc; #: controlled #Myc/Mp Type of initial yield
0.0σy×S1 = 38.76 in-kips=438.10 cm-kNσy×S2 = 21.38 in-kips=241.66 cm-kN #
0.561 Tension
0.10.9σy×S1 = 34.88 in-kips= 394.25cm-kN1.1σy×S2 = 23.51 in-kips = 265.73cm-kN #
0.617 Tension
0.20.8σy ×S1 = 31.00 in-kips= 350.39cm-kN1.2σy ×S2 = 25.65 in-kips = 289.92cm-kN #
0.673 Tension
0.30.7 σy ×S1 = 27.13 in-kips = 306.65cm-kN #1.3 σy ×S2 = 27.79 in-kips= 314.11cm-kN
0.712 Compression
0.40.6 σy ×S1 = 23.25 in-kips= 262.79cm-kN #1.4 σy ×S2 = 29.93 in-kips= 338.30cm-kN
0.610 Compression
0.50.5 σy ×S1 = 19.38 in-kips = 219.05cm-kN #1.5 σy ×S2 = 32.07 in-kips= 362.49cm-kN
0.508 Compression
0.60.4 σy ×S1 = 15.5 in-kips= 175.20cm-kN #1.6 σy ×S2 = 34.2 in-kips= 386.56cm-kN
0.407 Compression
0.70.3 σy ×S1 = 11.62 in-kips= 131.34cm-kN #1.7 σy ×S2 = 36.34 in-kips= 410.75cm-kN
0.305 Compression
0.80.2 σy ×S1 = 7.75 in-kips = 87.60cm-kN #1.8 σy ×S2 = 64.8 in-kips= 732.43cm-kN
0.203 Compression
0.90.1 σy ×S1 = 3.87 in-kips = 43.74cm-kN #1.9 σy ×S2 = 68.40 in-kips= 773.13cm-kN
0.101 Compression
1.00 σy ×S1 = 0 in-kips = 0 cm-kN #2 σy ×S2 = 42.7 in-kips= 482.64cm-kN
0.000 Compression
S1 = I/y1 = 0.7419/0.689 = 1.0767 in3 = 17.64 cm3,S2 = I/y2 = 0.7419/1.249 = 0.5939 in3 = 9.73 cm3,σy = 36 ksi = 248 Mpa; Mp=38.08 in-kips = 430.42 cm-kN;
100 W. S. King et al.
Following the same procedure, the initial yield surface
for a downward channel section that is subjected to a
sagging bending moment and a tensile axial load is given
in Eq. (24), where f is the shape factor. These calculated
values of initial yielded bending moment Myc are
expressed in Table 2.
(24)
Equation (24) also can be written as the following
equation for the downward channel section MC3×7.1.
for (25)
20.04P
Py
-----⎝ ⎠⎛ ⎞
4
8.7P
Py
-----⎝ ⎠⎛ ⎞
5
–M
yc
Mp
-------- 1P
Py
-----–⎝ ⎠⎛ ⎞1
f---=
Myc
Mp
-------- 0.561–P
Py
-----⎝ ⎠⎛ ⎞
0.561+= 0 P Py
≤ ≤
Table 2. Initial yielded bending moment for a downward channel section subjected to a sagging bending moment and atensile axial load
Axial load ratio P/Py Initial yielded bending moment Myc; #: controlled #Myc/Mp Type of initial yield
0.0σy×S1 = 38.76 in-kips = 438.1 cm-kNσy×S2 = 21.38 in-kips = 241.66 cm-kN #
0.561 Tension
0.1 0.9σy ×S2 = 19.21 in-kips = 217.13 cm-kN# 0.504 Tension
0.2 0.8σy ×S2 = 17.1 in-kips = 193.28 cm-kN# 0.449 Tension
0.3 0.7σy ×S2 = 14.94 in-kips = 168.87 cm-kN# 0.392 Tension
0.4 0.6σy ×S2 = 12.8 in-kips = 144.68 cm-kN# 0.336 Tension
0.5 0.5σy ×S2 = 10.67 in-kips = 120.6 cm-kN# 0.280 Tension
0.6 0.4σy ×S2 = 8.54 in-kips = 96.53 cm-kN# 0.224 Tension
0.7 0.3σy ×S2 = 6.40 in-kips = 72.34 cm-kN# 0.168 Tension
0.8 0.2σy ×S2 = 4.26 in-kips = 48.15 cm-kN# 0.112 Tension
0.9 0.1σy ×S2 = 2.13 in-kips = 20.08 cm-kN# 0.056 Tension
1.0 0σy ×S2 = 0 in-kips = 0 cm-kN# 0.000 Tension
Figure 7. Initial yield surfaces of a downward channel section MC3×7.1.
Plastic Analysis of Steel Frames with Unsymmetrical Sections 101
When a downward channel section is subjected to a
hogging bending moment and a tensile axial load
between zero and 0.3Py, the initial yield surface is given
in Eq. (26).
for (26)
When a downward channel section is subjected to a
hogging bending moment and a tensile axial load
between 0.3Py and Py, the initial yield surface is given in
Eq. (27).
for (27)
Following the same procedure, the initial yield surface
for a downward channel section that is subjected to a
hogging bending moment and a compressive axial load is
given in Eq. (28).
(28)
Equation (28) also can be written as the following
equation for the downward channel section MC3×7.1.
for (29)
The values of the initial yielded sagging bending
moment Myc can be calculated under different axial
compressive loads. Then, the initial yield surface can be
drawn by curve fitting method. These initial yield
surfaces from Eqs (21) to (29) are drawn in Fig. 7. If we
put the initial yield surfaces and the limit surfaces
together, we obtain Fig. 8. These initial yield surfaces of
the upward channel section can be derived following the
same process. We will not repeat here again.
3.3. Application of yield surfaces
A member that is subjected to sagging bending
moments and a compressive or tensile axial load is shown
in Fig. 9(a). A member that is subjected to a hogging
bending moment and a compressive or tensile axial load
is shown in Fig. 9(b). A member whose both ends are
subjected to clockwise bending moments and a compressive
or tensile axial load is shown in Fig. 9(c). A member
whose both ends are subjected to counterclockwise
bending moments and a compressive or tensile axial load
is shown in Fig. 9(d). The initial yield surface and limit
surface in one of the four quadrants in Fig. 8 will be
chosen corresponding to the case in Fig. 9.
4. Numerical Analysis of Beams and Rigid Frames
A simply supported beam using the channel section
MC3×7.1 as shown in Fig. 10 is analyzed. The channel
section is downward or upward. The axial load P from
zero to Py is applied first in the beam. Then, a vertically
concentrated load H is applied at the mid-span of the
Myc
Mp
-------- 0.503P
Py
-----⎝ ⎠⎛ ⎞
0.561+= 0 P 0.3Py
≤ ≤
Myc
Mp
-------- 1.017–P
Py
-----⎝ ⎠⎛ ⎞
1.017+= 0.3PyP P
y≤ ≤
Myc
Mp
-------- 1P
Py
-----–⎝ ⎠⎛ ⎞1
f---=
Myc
Mp
-------- 0.561–P
Py
-----⎝ ⎠⎛ ⎞
0.561+= 0 P Py
≤ ≤
Figure 8. Limit surface and initial yield surface of a downward channel section MC3×7.1.
102 W. S. King et al.
beam from zero to the limit load of the beam. The results
of analyses are shown in Table 3. Except when the axial
load ratio is zero or one, the limit loads H of the beam
using a downward channel section are greater than that of
the beam using an upward channel section. The
difference of limit load H between the beam using an
upward channel section and the beam using a downward
channel section is increased, when the axial load ratio P/
Py is increased.
Several portal frames whose members are made of
channel sections are studied in the following. Two columns
of the portal frame are simply supported. A vertical gravity
load P = 0.25Py is applied to each column first. The lateral
load H on the top of the left column is applied from zero
to the limit load until the portal frame as shown in Fig. 11.
There are five portal frames as shown in Table 4. These
directions of channel sections in columns and beams of
those five portal frames are different. Some tips of section
in beams are downward, but some are upward. Some tips
of sections in columns are leftward, but some are
rightward. Hence, we can distinguish the differences of
responses between these frames.
The load-deflection curves of the five portal frames are
Table 3. The limit loads H of the simple beam
(1)Axial load ratio P/Py
(2)H of the upward channel
section, kN (Kips)
(3)H of the downward channel
section, kN (Kips)
(4)Difference between
(2) & (3), %
0.0 22.606 (5.080) 22.606 (5.080) 00
0.1 19.602 (4.405) 22.005 (4.945) 12
0.2 17.199 (3.865) 21.138 (4.750) 23
0.3 14.707 (3.305) 19.914 (4.475) 35
0.4 12.304 (2.765) 17.778 (3.995) 44
0.5 09.990 (2.245) 15.246 (3.426) 53
0.6 07.663 (1.722) 12.264 (2.756) 60
0.7 05.251 (1.180) 08.775 (1.972) 67
0.8 02.808 (0.631) 04.851 (1.090) 73
0.9 00.472 (0.106) 00.828 (0.186) 75
1.0 0.00 0.00 00
Figure 9. Loading cases of members subjected to bendingmoments and axial loads.
Figure 10. A simply supported beam using channel section.
Figure 11. Loading conditions and dimensions of a portalframe.
Plastic Analysis of Steel Frames with Unsymmetrical Sections 103
shown in Fig. 12. The No. 3 portal frame has the largest
lateral limit load capacity. However, the No. 4 portal
frame has the least lateral limit load capacity. It is because
both top ends of columns of portal frame No. 3 are bent
in the first quadrant of Fig. 5. Both top ends of columns
of portal frame No. 4 are bent in the first quadrant of Fig.
6. The limit surface in the first quadrant of Fig. 5 is larger
than that of Fig. 6. Hence, the lateral limit load of portal
frame No. 3 is 19% larger than that of portal frame No. 4.
Portal frames No. 2 and No. 5 have the same lateral
limit load. It is because the columns in these two frames
have the same direction. The directions of channel
sections of beams in these two frames are different. One
is upward, but the other is downward. However, the
beams in these two frames are bent equivalently. The
lateral limit load of Portal frame No. 1 is slightly smaller
than that of portal frame No. 2 due to some numerical
deviation. The lateral limit load of frame No. 1, 2, and 5
is 11% greater than that of frame No. 4.
In the next example, the columns of the above portal
frames are fixed supported instead of simply supported as
shown in Fig. 13. Only the boundary conditions of
support change. All the other conditions of the above
portal frames keep the same. The load-deflection curves
of these five portal frames are shown in Fig. 13. The No.
4 portal frame has the largest lateral limit load capacity.
However, the No. 3 portal frame has the least lateral limit
load capacity. It is reversed as compared to the result of
those portal frames in the first example. This is because
the fixed ends of two columns are subjected to
counterclockwise bending moments and yielded first.
Both fixed ends of columns of portal frame No. 4 are bent
in the first quadrant of Fig. 5. Both fixed ends columns of
portal frame No. 3 are bent in the first quadrant of Fig. 6.
The lateral limit load of portal frame No. 4 is 8% larger
than that of the portal frame No. 3.
Two types of portal frames are studied. The first type is
the frame in Fig. 11 whose sections are the case 3 in
Table 4. The other type is the frame in Fig. 11 whose
sections are the case 4 in Table 4. The vertical loads P are
applied first to the columns. The axial loads P vary from
zero to 0.5Py. The lateral load H is applied on the top of
the left column. The lateral load H starts from zero to the
limit load of the portal frame. The results of analyses are
Case 1, 2, 3, 4, 5 see Table 4Figure 13. Load-deflection curves of a fixed-base portalframe with channel sections.
Case 1, 2, 3, 4, 5 see Table 4Figure 12. Load-deflection curves of a pinned-base portalframe with channel sections.
Table 4. Direction of channel section in portal frames
CaseRight column
(Top view)Beam
(Side view)Left column(Top view)
1
2
3
4
5
104 W. S. King et al.
shown in Table 5. The limit loads H of the portal frames
whose sections are the case 3 are greater than that of the
portal frames whose sections are the case 4. The
difference of limit loads H between these two types of
frames is increased, when the axial load ratio P/Py is
increased.
Two types of two-story frames are studied. The first
type is the frame in Fig. 14 whose sections are the case
3 in Table 4. The other type is the frame in Fig. 14 whose
sections are the case 4 in Table 4. The vertical loads P are
applied first on the columns. The axial loads P vary from
zero to 0.4Py. The lateral load H is applied on the top of
the left column. The lateral load H starts from zero to the
limit load of the portal frame. The results of analyses are
shown in Table 6. The limit loads H of the two-story
frames whose sections are the case 3 are greater than that
of the frames whose sections are the case 4. The
difference of limit loads H between these two types of
frames is increased, when the axial load ratio P/Py is
increased.
5. Numerical Analyses of Semi-Rigid Frames
The initial tangent stiffness Kt of connection is reduced
from 800,000,000 kN-cm/rad to 80,000 kN-cm/rad. The
plastic moment Mp of the channel section is about 430
kN-cm. The ultimate bending moment Mu of connection
is reduced from 430 kN-cm to 282 kN-cm. The initial
yielded bending moment of connection Mcy is assumed to
be one-third of the corresponding Mu from 143 kN-cm to
94 kN-cm. The Ki, Mu and Mcy are listed in Table 7.
We can observe the effect of semi-rigid connection on
the elastic-plastic behavior of the portal frame as shown
in Fig. 8. The parameter A in Eq. (6) is assumed one. The
directions of channel sections of beam and columns are
arranged as the case 3 or 4 in Table 4. The lateral limit
load of a portal frame decreases when Kt, Mu and Mcy
reduce. The limit loads of portal frames with different
connections are shown in Table 7. The difference of the
limit load H between Frame A and B becomes small as
the connection is close to a pinned connection. It is
because that the connection will fail first before the
section fail. The Frame No. 1 and 2 have the same limit
load either in Frame A or B, because these connections
are close to the rigid connection.
Figure 14. Two-story frame using channel section.
Table 5. The limit loads H of the portal frame in Fig. 11
(1)Axial load ratio, P/Py
(2)H of the portal frame Aa,
kN (Kips)
(3)H of the portal frame Bb,
kN (Kips)
(4)Difference between
(2) & (3), %
0.0 11.289 (2.537)0 12.068 (2.712)0 07
0.1 7.739 (1.739) 8.264 (1.857) 07
0.2 5.313 (1.194) 6.150 (1.382) 16
0.3 3.382 (0.760) 4.410 (0.991) 30
0.4 1.825 (0.410) 2.692 (0.605) 48
0.5 0.672 (0.151) 1.170 (0.263) 74
aThe channel sections in frame A are the case 4 in Table 4.bThe channel sections in frame B are the case 3 in Table 4.
Table 6. The limit loads H of the two-story frame in Fig. 14
(1)Axial load ratio, P/Py
(2)Limit load H of the frame Aa,
kN (Kips)
(3)Limit load H of the frame Bb,
kN (Kips)
(4)Difference between
(2) & (3), %
0.1 7.129 (1.602) 7.565 (1.700) 06
0.2 4.775 (1.073) 5.300 (1.191) 11
0.3 3.057 (0.687) 3.658 (0.822) 20
0.4 1.669 (0.375) 2.692 (0.605) 61
aThe channel sections in frame A are the case 4 in Table 4.bThe channel sections in frame B are the case 3 in Table 4.
Plastic Analysis of Steel Frames with Unsymmetrical Sections 105
6. Summary and Conclusions
The elastic stability functions are used successfully
instead of Taylor series expansion in the stiffness matrix.
The initial yield surfaces and limit surfaces of an
unsymmetrical section are developed completely.
The initial and limit surfaces in four quadrants have
been applied to the lateral limit load analyses of steel
frames reasonably well. The same sagging or hogging
bending moment applied in the weak axis of a channel
section in a beam or frame may result in different ultimate
loads, depending on the directions of channel sections.
The difference of limit loads between the beam using
an upward channel section and the beam using a
downward channel section is increased, when the axial
load ratio P/Py is increased. The difference of limit loads
between different types of frames is also increased, when
the axial compressive load in columns is increased.
The effect of semi-rigid connections on the lateral limit
load of steel frames is obvious when the initial tangent
stiffness Ki and the ultimate moment Mu of connections
degrade. The lateral limit load of a portal frame decreases
when Kt, Mu and Mcy reduce. The difference of the limit
load between different types of semi-rigid frames
becomes smaller as the connection is approaching to a
pinned condition.
References
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Table 7. Limit loads of portal frames with channel sections and semi-rigid connections
(1)Frame No.
(2)Kt (cm-kN/rad)
(3)Mu (cm-kN)
(4)Mcy (cm-kN)
(5)H (kN) of frame Aa
(6)H (kN) of frame Bb
(7)Difference between
(5) & (6), %
1 800,000,000 430 143 4.29 5.28 23
2 80,000,000 387 129 4.29 5.28 23
3 8,000,000 349 116 4.28 5.19 21
4 800,000 314 105 4.15 4.66 12
5 80,000 282 094 3.00 3.12 04
aThe channel sections in frame A are the case 4 in Table 4.bThe channel sections in frame B are the case 3 in Table 4.