Plastic Analysis of Steel Fram es with Unsymmetrical Sectionskssc.or.kr/wonmun/KSSC_3_2006_6_2_93(C).pdf · 2018. 1. 8. · Plastic Analysis of Steel Frames with Unsymmetrical Sections

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.


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.


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.


(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.


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)


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


0.508 Compression


0.407 Compression


0.305 Compression


0.203 Compression


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;


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









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.



hogging bending moment and a tensile axial load

between zero and 0.3Py, the initial yield surface is given

in Eq. (26).

for (26)


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.


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.


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


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)


(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)


(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



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

Chen, W. F. and Toma, S. (1994). Advanced Analysis of Steel

Frames. CRC Press, Inc., USA.

Chen, W. F., and Atsuta, T. (1976). Theory of Beam-

Columns. Vol. 2, Space Behavior and Design. McGraw-

Hill Book Co., Inc., New York, N. Y., USA.

Gjelsvik, A. (1981). The Theory of Thin Walled Bars. John

Wiley & Sons, Inc., USA.

Horne, M. R. (1979). Plastic Theory of Structure. 2nd Ed.,

Pergamon Press, Oxford, UK.

King, W. S. and Chen, W. F. (1994). “Practical Second-

Order Inelastic Analysis of Semi-Rigid Frames.” Journal

of Structural Engineering, ASCE, 120(7), pp. 2156-2175.

King, W. S., White, D. W. and Chen, W. F. (1992). “Second-

Order Inelastic Analysis Methods for Steel-Frame

Design.” Journal of Structural Engineering, ASCE,

118(2), pp. 408-429.

Lui, E. M. (1985). Effects of Connection Flexibility and

Panel Zone Deformation on the Behavior of Plane Steel

Frames. Ph.D. Thesis, School of Civil Engineering,

Purdue University, West Lafayette, Indiana, USA.

Lui, E. M. and Chen, W. F. (1987). Structural Stability

Theory and Implementation. Elsevier Science Publishing

Co., Inc., New York, N.Y., USA.

Lui, E. M. and Chen, W. F. (1988). Steel Frame Analysis

with Semi-Rigid Connection. Structural Engineering

Report, CE-STR-88-5, School of Civil Engineering,

Purdue University, West Lafayette, Indiana, USA.

Rasmussen, K. J. R. and Hancock, G. J. (1989).

“Compression Tests of Welded Channel Section

Columns.” Journal of Structural Engineering, ASCE,

115(4), pp. 789-808.

Timoshenko, S. P. and Gere, J. M. (1972). Mechanics of

Materials. Van Nostrand Reinhold Co., New York, N.Y.,

USA.

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


(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


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