# Lateral Buckling Analysis of the Steel-Concrete Composite Beams in Negative Moment Region

Authors: Fengqi Guo; Shun Zhou; Lizhong Jiang
Journal: Advances in Materials Science and Engineering
 (2015)

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## Abstract

Distortional buckling is one of the important buckling models of steel-concrete composite beam in negative moment region. Rotation restraining rigidity and lateral restraining rigidity which steel beam web to bottom plate of steel-concrete composite are the key factors to influence the distortional buckling behavior. A comprehensive and intensive study on rotation restraining rigidity and lateral restraining rigidity which steel beam web to bottom plate of I-shaped steel-concrete composite beam in negative moment region is conducted in this paper. Energy variation principle is adopted to deduce the analytical expressions to calculate the rotation restraining rigidity and lateral restraining rigidity. Combined with the buckling theory of axial compression thin-walled bars in elastic medium, the buckling moment is obtained. Theoretical analysis shows that the rotation restraining rigidity and lateral restraining rigidity of steel beam web appear to have a linear relationship with the external loads and could also be negative. Compared with other methods, the results calculated by the proposed expressions agree well with the numerical results by ANSYS. The proposed expressions are more concise and suitable than the existing formulas for the engineering application.


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## Body

## 1. Introduction

The steel-concrete composite beam is composed of profiled steel or welded I-shaped beam and concrete slab through shear connectors, which can resist external loads together. Because this composite structure combines the tensile resistance of steel with the compressive resistance of concrete, it has the advantages of higher bearing capacity, better plasticity and ductility, constructing conveniently, and lower cost, which makes it widely used in long-span bridges and high-rise buildings. In practical engineering, it is unnecessary to verify the lateral buckling of composite beams in positive bending moment region because of the enough bending rigidity and torsional rigidity of concrete slab. However, with bigger variable loads and unfavorable loads, the bottom flange of steel beam in the negative moment region would yield lateral buckling associated with web transverse deformation. The distortion buckling is then likely to occur.In recent years, several scholars used the energy variational methods to analyze global stability of composite beam. Some authors computed the critical buckling load [1–3] and others compiled the corresponding specifications [4]. These specifications only consider global bending instability of the steel beam web but fail to take into account the distortional buckling. Besides, the critical load formulas by these methods are a little tedious for engineering calculations. Based on the elastic foundation compressive bar method, Williams and Jemah [5], Svensson [6], Goltermann and Svensson [7], and Ronagh [8] successively study the stability of composite beam under constant axial force, increasing the contribution of torsional rigidity of the concrete slabs and participating area of the steel girder web plate. In order to consider the effect of bending moment gradient compressive bar variable, the axial force is introduced. Diansheng and Xiaomin [9] presented a model for analyzing the local buckling property of cold-formed thin-wall steel-concrete composite beam. The elastic buckling stresses at steel beam web and flange wall are calculated by the energy method. Jiang [10, 11] researched the local stability in the negative moment region for steel beam web of continuous composite beams in attempt to establish the simplified calculation model of local stability under various loads and propose the critical local buckling stress formula under a variety of stress states. Ye and Chen [12] improved Svensson compressive bar model appropriately. Considering the steel beam web effective participation part, two variable axial forces distortional buckling stability critical load calculation formulas were deduced based on the improved model. Making use of the finite element, the accuracy of the above method was analyzed by calculating the composite beam constraint distortional buckling load. Zhou et al. [13, 14] used energy variation principle to deduce the calculation method of rotation restraining rigidity and lateral restraining rigidity.In this paper, a comprehensive and intensive study on rotation restraining rigidity and lateral restraining rigidity which steel beam web to bottom plate of steel-concrete composite beam in negative moment region is conducted. The energy method is used to deduce the theoretical formula for rotation restraining rigidity and lateral restraining rigidity which the steel beam web provides for bottom plate. Energy variation principle is adopted to derive the steel beam critical stress of positive symmetry bending buckling and dissymmetry bending-torsion buckling in order to calculate the buckling moment. In the end of the paper, the proposed formulas are discussed and analyzed.


## 2. Basic Assumptions

The schematic diagram of steel-concrete composite beam is shown in Figure1. The lateral buckling model of steel I-shaped beam in composite beam is different from the free simply supported steel beam (unconstrained steel beam). The top flange of steel beam constituted by concrete slab has big stiffness, so the lateral deformation and torsional deformation are restricted to a certain degree. The bottom flange of steel beam is under compression. Although it can yield lateral displacement and torsion angle, the bottom flange constrained by the web is not perfectly free. Therefore, the lateral buckling of composite beam can be regarded as the distortional buckling in company with lateral bending deformation of steel beam web.Figure 1
Cross section dimensions of steel-concrete composite beams and axes.The right handed coordinate systemxyz is fixed to the centroid of steel beam bottom flange. As shown in Figure 1, the monosymmetrical composite beam bears a bending moment Mx in the yz plane which shows big stiffness. In order to analyze the rotation restraining rigidity kφ of steel beam bottom flange to web and the buckling moment, some assumptions are made as follows:(1)
The materials are isotropic and perfectly elastic body.(2)
The element is constant section beam and there were no initial imperfections.(3)
The cross-sectional shape of steel beam bottom flange does not change during distortional buckling yielding.(4)
The lateral deformation and torsional deformation of steel beam top flange could not happen because of enough stiffness of concrete slab.(5)
Due to the negative moment, most of concrete in composite beam has been cracked when buckling yields. Therefore, the bending capacity of concrete is ignored, which means that only the bending capacity of the steel reinforcements in concrete slab is considered.(6)
The vertical restraining rigidity which webs to bottom flangeky=∞.Based on above assumptions, the problem to be analyzed can be simplified as a thin-walled constraint distortion problem which is restricted by spring restraint and vertical rigid constraint in horizontal and distortion direction. The simplified model is plotted in Figure2.Figure 2
Simplified calculation model of steel-concrete composite beams.


## 3. Web Constraint Factorkϕ and kx

### 3.1. Rotation Constraint Rigiditykϕ

Figure3 presents a half-wave length of web section under consideration. The width and thickness of web section are hw and tw, respectively. λ refers to the half-wave length of web caused by distortion buckling in longitudinal direction (called as the half-wave length hereafter). Two transversal opposite sides are simply supported. The side connected to top flange is fixed and the other side connected to bottom flange is simply supported. The two simply supported sides bear the longitudinal linear distributed stress σ in Z direction (compressive stress is positive and tension stress is negative). The side connected to bottom flange bears the equivalent spring constraint moment mz which bottom flange exerted on web. The coordinate of gravity centre of steel beam total cross section is represented by -yc, and the moment of inertia is I. According to the assumptions mentioned above, when the negative moment Mx acts on the reinforcement in concrete slab, the axial compressive stress at bottom edge of web is σ1=Mxyc/I and the axial compressive stress at top edge of web is σ2=σ1yc-hw/yc. Therefore, the axial compressive stress at other points of web is σ=σ1yc+y/yc.Figure 3
Rectangular plate subjected to compression and moments.AssumingD=Etw3/121-μ2, μ is Poisson’s ratio of steel; E is the elastic modulus of steel; u denotes the deformation function of web. The boundary conditions of u can be expressed as(1)uz=0,λ=0,uy=0,-hw=0,∂u∂yy=-hw=0,-D∂2u∂z2+μ∂2u∂y2z=0,λ=0.With (1), the displacement functions are written as(2)u=cyhw+2yhw2+yhw3sinπzλ.The strain energy of half-wave length web in the case of small deformations is [16–18](3)
U
1
=
D
2
∫
0
λ
∫
-
h
w
0
∂
2
u
∂
y
2
2
+
∂
2
u
∂
z
2
2
+
2
μ
∂
2
u
∂
y
2
∂
2
u
∂
z
2
+
2
1
-
μ
∂
2
u
∂
y
∂
z
2
d
y
d
z
.Substituting (2) into (3) leads to the fact that(4)U1=λD22c2hw3+2c215hwπλ2+c2hw210πλ4.The elastic potential energy caused by half-wave length longitudinal side spring constraint is(5)U2=kφ2∫0λ∂u∂yy=02dz.Substituting (2) into (5),(6)U2=λkφc24hw2.The external force work of half-wave length web can be computed by [16–18](7)W=tw2∫0λ∫-hw0σ∂u∂z2dydz=tw2∫0λ∫-hw0σ1yc+yyc∂u∂z2dydz.Substituting (5) into (8),(8)W=λc2σ1twhw420πλ2-λc2σ1twhw21120ycπλ2.The total potential energy of half-wave length web is(9)Π=U1+U2-W.Substitution of (4), (6), and (8) into (9) results in(10)Π=λD22c2hw3+2c215hwπλ2+c2hw210πλ4+λkφc24hw2-λc2σ1twhw420πλ2+λc2σ1twhw21120ycπλ2.Based on principle of resident potential energy, we obtained the following:(11)D22hw3+215hwπλ2+hw210πλ4+kφ4hw2-σ1twhw420πλ2+σ1twhw21120ycπλ2=0.By solving (11), one can obtain(12)kφ=twhw3105-twhw4280ycπλ2σ1-D4hw+4hw15πλ2+hw3105πλ4.


### 3.2. Lateral Constraint Rigiditykx

The half-wave length of web section is shown in Figure4. Two transversal opposite sides are simply supported. The side connected to top flange is fixed and the other side connected to bottom flange can move laterally. The two simply supported sides bear the longitudinal linear distributed stress σ in Z direction (similarly, compressive stress is taken as positive and tension stress is negative). The side connected to bottom flange bears the equivalent spring constraint distributed force fz which bottom flange exerted on web.Figure 4
Rectangular plate subjected to compression and lateral stress.Based on above analysis, the boundary conditions ofu can be expressed as(13)uz=0,λ=0,uy=-hw=0,∂u∂yy=0,-hw=0,-D∂2u∂z2+μ∂2u∂y2z=0,λ=0.According to above boundary conditions, the displacement functions can be written as(14)u=c-3cyhw2-2cyhw3sinπzλ.Substituting (14) into (3), the strain energy of half-wave length web in the case of small deformations is then obtained as follows:(15)U1=λD26c2hw3+13c2hw70πλ4+6c25hwπλ2.The elastic potential energy caused by half-wave length longitudinal side spring constraint is(16)U2=kx2∫0λuy=02dz.Substituting (14) into (16) leads to the fact that(17)U2=λkxc24.Substituting (14) into (7), the external force work of half-wave length web can be obtained as follows:(18)W=13λc2σ1twhw140πλ2-3λc2σ1twhw2140ycπλ2.Substituting (15), (17), and (18) into (9), the total potential energy of half-wave length web is(19)Π=λD26c2hw3+13c2hw70πλ4+6c25hwπλ2+λkxc24-13λc2σ1twhw140πλ2+3λc2σ1twhw2140ycπλ2.Based on principle of resident potential energy, one can have(20)D26hw3+65hwπλ2+13hw70πλ4+kx4-13σ1twhw140πλ2+3σ1twhw2140ycπλ2=0.By solving (20), we obtained the following:(21)kx=13twhw35-3twhw235ycπλ2σ1-D12hw3+125hwπλ2+13hw35πλ4.


### 3.3. Discussion aboutkx and kϕ

(1)
Equations (12) and (21) indicated that both kϕ and kx show a linear relationship with the longitudinal compressive stress σ1. Generally, hw/yc is less than 2, so the coefficient before σ1 is positive for most situations. The bigger σ1 is the higher kϕ and kx are. At the same time, it is of interest to note that both kϕ and kx which steel beam bottom flange to web are determined by the compressive stress σ1 but not by composite beam section properties.
(2)
Since the polynomials on right-hand side of (12) and (21) have negative terms, kϕ and and kx could be negative. This is not consistent with regular positive definite rigidity and rigidity matrix. If the rotation constraint rigidity and lateral constraint rigidity are negative, the rotation and lateral displacement of steel beam bottom flange will be restricted by web. Namely, the steel beam web will restrict bottom flange to buckle, but the steel beam bottom flange will induce the web to buckle. According to [16], the lateral constraint rigidity k=Etw3/4hw3 is obtained by using strip method in the elastic constraint compression member buckling model. However, the restraint action of two adjacent strips is not considered in this method. Therefore, the lateral constraint rigidity which has nothing to do with external forces is always positive. But this does not agree with the actual situation. Furthermore, the neglected rotation constraint rigidity will lead to certain errors when calculating buckling load of composite beam.
(3)
The ratios of the first term, second term, and third term on the right side of (21) and (12) are:(22)13σtwhw/35π/λ2σtwhw3/105π/λ2=39hw2,3σ1twhw2/35ycπ/λ2σ1twhw4/280ycπ/λ2=24hw2,D12/hw3+12/5hwπ/λ2+13hw/35π/λ4D4/hw+4hw/15π/λ2+hw3/105π/λ4≈3hw2.From (22), hw2kx/kφ is not an infinitesimal value, so the lateral constraint rigidity of bottom flange to web cannot be disregarded. Namely, in the calculation, the equation kx=0 is not available. Therefore the lateral constraint rigidity of bottom flange to web cannot be approximated by zero. This is different from [16] in which the lateral constraint rigidity which the cold-formed thin-walled lipped channel steel webs to the top and bottom flange is taken as zero.


## 3.1. Rotation Constraint Rigiditykϕ

Figure3 presents a half-wave length of web section under consideration. The width and thickness of web section are hw and tw, respectively. λ refers to the half-wave length of web caused by distortion buckling in longitudinal direction (called as the half-wave length hereafter). Two transversal opposite sides are simply supported. The side connected to top flange is fixed and the other side connected to bottom flange is simply supported. The two simply supported sides bear the longitudinal linear distributed stress σ in Z direction (compressive stress is positive and tension stress is negative). The side connected to bottom flange bears the equivalent spring constraint moment mz which bottom flange exerted on web. The coordinate of gravity centre of steel beam total cross section is represented by -yc, and the moment of inertia is I. According to the assumptions mentioned above, when the negative moment Mx acts on the reinforcement in concrete slab, the axial compressive stress at bottom edge of web is σ1=Mxyc/I and the axial compressive stress at top edge of web is σ2=σ1yc-hw/yc. Therefore, the axial compressive stress at other points of web is σ=σ1yc+y/yc.Figure 3
Rectangular plate subjected to compression and moments.AssumingD=Etw3/121-μ2, μ is Poisson’s ratio of steel; E is the elastic modulus of steel; u denotes the deformation function of web. The boundary conditions of u can be expressed as(1)uz=0,λ=0,uy=0,-hw=0,∂u∂yy=-hw=0,-D∂2u∂z2+μ∂2u∂y2z=0,λ=0.With (1), the displacement functions are written as(2)u=cyhw+2yhw2+yhw3sinπzλ.The strain energy of half-wave length web in the case of small deformations is [16–18](3)
U
1
=
D
2
∫
0
λ
∫
-
h
w
0
∂
2
u
∂
y
2
2
+
∂
2
u
∂
z
2
2
+
2
μ
∂
2
u
∂
y
2
∂
2
u
∂
z
2
+
2
1
-
μ
∂
2
u
∂
y
∂
z
2
d
y
d
z
.Substituting (2) into (3) leads to the fact that(4)U1=λD22c2hw3+2c215hwπλ2+c2hw210πλ4.The elastic potential energy caused by half-wave length longitudinal side spring constraint is(5)U2=kφ2∫0λ∂u∂yy=02dz.Substituting (2) into (5),(6)U2=λkφc24hw2.The external force work of half-wave length web can be computed by [16–18](7)W=tw2∫0λ∫-hw0σ∂u∂z2dydz=tw2∫0λ∫-hw0σ1yc+yyc∂u∂z2dydz.Substituting (5) into (8),(8)W=λc2σ1twhw420πλ2-λc2σ1twhw21120ycπλ2.The total potential energy of half-wave length web is(9)Π=U1+U2-W.Substitution of (4), (6), and (8) into (9) results in(10)Π=λD22c2hw3+2c215hwπλ2+c2hw210πλ4+λkφc24hw2-λc2σ1twhw420πλ2+λc2σ1twhw21120ycπλ2.Based on principle of resident potential energy, we obtained the following:(11)D22hw3+215hwπλ2+hw210πλ4+kφ4hw2-σ1twhw420πλ2+σ1twhw21120ycπλ2=0.By solving (11), one can obtain(12)kφ=twhw3105-twhw4280ycπλ2σ1-D4hw+4hw15πλ2+hw3105πλ4.


## 3.2. Lateral Constraint Rigiditykx

The half-wave length of web section is shown in Figure4. Two transversal opposite sides are simply supported. The side connected to top flange is fixed and the other side connected to bottom flange can move laterally. The two simply supported sides bear the longitudinal linear distributed stress σ in Z direction (similarly, compressive stress is taken as positive and tension stress is negative). The side connected to bottom flange bears the equivalent spring constraint distributed force fz which bottom flange exerted on web.Figure 4
Rectangular plate subjected to compression and lateral stress.Based on above analysis, the boundary conditions ofu can be expressed as(13)uz=0,λ=0,uy=-hw=0,∂u∂yy=0,-hw=0,-D∂2u∂z2+μ∂2u∂y2z=0,λ=0.According to above boundary conditions, the displacement functions can be written as(14)u=c-3cyhw2-2cyhw3sinπzλ.Substituting (14) into (3), the strain energy of half-wave length web in the case of small deformations is then obtained as follows:(15)U1=λD26c2hw3+13c2hw70πλ4+6c25hwπλ2.The elastic potential energy caused by half-wave length longitudinal side spring constraint is(16)U2=kx2∫0λuy=02dz.Substituting (14) into (16) leads to the fact that(17)U2=λkxc24.Substituting (14) into (7), the external force work of half-wave length web can be obtained as follows:(18)W=13λc2σ1twhw140πλ2-3λc2σ1twhw2140ycπλ2.Substituting (15), (17), and (18) into (9), the total potential energy of half-wave length web is(19)Π=λD26c2hw3+13c2hw70πλ4+6c25hwπλ2+λkxc24-13λc2σ1twhw140πλ2+3λc2σ1twhw2140ycπλ2.Based on principle of resident potential energy, one can have(20)D26hw3+65hwπλ2+13hw70πλ4+kx4-13σ1twhw140πλ2+3σ1twhw2140ycπλ2=0.By solving (20), we obtained the following:(21)kx=13twhw35-3twhw235ycπλ2σ1-D12hw3+125hwπλ2+13hw35πλ4.


## 3.3. Discussion aboutkx and kϕ

(1)
Equations (12) and (21) indicated that both kϕ and kx show a linear relationship with the longitudinal compressive stress σ1. Generally, hw/yc is less than 2, so the coefficient before σ1 is positive for most situations. The bigger σ1 is the higher kϕ and kx are. At the same time, it is of interest to note that both kϕ and kx which steel beam bottom flange to web are determined by the compressive stress σ1 but not by composite beam section properties.
(2)
Since the polynomials on right-hand side of (12) and (21) have negative terms, kϕ and and kx could be negative. This is not consistent with regular positive definite rigidity and rigidity matrix. If the rotation constraint rigidity and lateral constraint rigidity are negative, the rotation and lateral displacement of steel beam bottom flange will be restricted by web. Namely, the steel beam web will restrict bottom flange to buckle, but the steel beam bottom flange will induce the web to buckle. According to [16], the lateral constraint rigidity k=Etw3/4hw3 is obtained by using strip method in the elastic constraint compression member buckling model. However, the restraint action of two adjacent strips is not considered in this method. Therefore, the lateral constraint rigidity which has nothing to do with external forces is always positive. But this does not agree with the actual situation. Furthermore, the neglected rotation constraint rigidity will lead to certain errors when calculating buckling load of composite beam.
(3)
The ratios of the first term, second term, and third term on the right side of (21) and (12) are:(22)13σtwhw/35π/λ2σtwhw3/105π/λ2=39hw2,3σ1twhw2/35ycπ/λ2σ1twhw4/280ycπ/λ2=24hw2,D12/hw3+12/5hwπ/λ2+13hw/35π/λ4D4/hw+4hw/15π/λ2+hw3/105π/λ4≈3hw2.From (22), hw2kx/kφ is not an infinitesimal value, so the lateral constraint rigidity of bottom flange to web cannot be disregarded. Namely, in the calculation, the equation kx=0 is not available. Therefore the lateral constraint rigidity of bottom flange to web cannot be approximated by zero. This is different from [16] in which the lateral constraint rigidity which the cold-formed thin-walled lipped channel steel webs to the top and bottom flange is taken as zero.


## 4. Theoretical Derivation of Critical Moment

### 4.1. Derivation of Critical Moment

The lateral constraint rigidity and rotation constraint rigidity which steel beam web to bottom flange can be simplified, respectively, as(23)kx=α1βσ+Dα2,(24)kφ=α3βσ+Dα4,(25)β=πλ2,(26a)α1=twhw0.086hwyc-0.37,(26b)α2=12hw3+2.4βhw+0.37hwβ2,(26c)α3=twhw30.0036hwyc-0.0095,(26d)α4=4hw+0.27hwβ+0.0095hw3β2,(27)σ=MxycI=PAf.Therein,D=Etw3/121-μ2; μ is Poisson’s ratio of steel; σ represents the compressive stress of bottom flange; P refers to the compressive force of bottom flange; -yc is the gravity centre coordinate of steel beam total cross section; I is the inertia moment of steel beam; λ=l/n; n is the buckling half-wave number; Af is the area of bottom flange.As shown in Figure2, the thin-walled member is doubly symmetric along x-axis and y-axis and the centre of origin O coincides with the flexural center. The displacements of origin O in x direction and y direction are denoted as u and v, respectively. Because the rigidity in y direction is infinity, v is equal to zero. The equivalent distributed forces caused by elastic medium as a result of displacements of thin-walled member can be written as(28)rx=kxu,ry=kyv,where kx and ky are the lateral and vertical constraint rigidity which web to bottom flange, respectively.The torsional angle which the member rotates around the bending center is assumed to beφ. The equivalent distributed moment of torsional thin-walled member induced by equivalent spring is(29)m=kφφ.Neutral balance differential equation of thin-walled member can be expressed as [13, 14](30)EIyuIV+Pu′′+yaφ′′+kxu-yd-yaφ=0,EIxvIV+Pv′′-xaφ′′+kyv+xd-xaφ=0,EIwφIV+r02P-GJφ′′-Pxav′′-yau′′-kxu-yd-yaφyd-ya+kyv+xd-xaφxd-xa+kφφ=0.Therein,Iy=tfbf3/12;  Ix=bftf3/12;  J=bftf3/3;  r02=xa2+ya2+Ix+Iy/As,As being the area of steel beam; xa is the horizontal coordinate of the bottom flange section bending centre, xa=0; ya is the vertical coordinate of the bottom flange section bending centre, ya=0; xd is the horizontal coordinate of the bottom flange section rotation axis, xd=0; yd is the vertical coordinate of the bottom flange section rotation axis, yd=0; Iw is the fan-shaped inertia moment of bottom flange section, Iw=0; E is the tensile elastic modulus of steel; G is the shear elastic modulus of steel.With substitution ofya=0, yd=0, xa=0, v=0, Iw=0, kx=0, xa=0, ya=0, and yd=0 into (30), one can obtain(31)EIyIuIV+ycAfMxu′′+α1βycMx+IDα2u=0,(32)kyv+xd-xaφ=0,(33)r02ycAfMx-GJIφ′′+α3βycMx+IDα4φ=0.When the steel-concrete composite beam in negative moment region bears lateral bending buckling, its neutral balance equation is shown in (31) and the corresponding boundary conditions are(34)uz=0,l=0,u′′z=0,l=0.Assuming u=Asin⁡βz, which satisfies the boundary conditions, according to Galerkin’s method,(35)β2-MxycAfEIyIβ+α1βMxyc/I+Dα2EIy=0.By solving (35),(36)Mcr1=EIyβ+Dα2/βAf-α1ycI.Due to dMcr1/dβ=0, we obtained the following:(37)βcr1=3.46EIyhw3/D+0.37hw4,ncr1=lβcr1π.If lβcr1/π is an integer, substitution of βcr1 into (36) leads to the lateral bending critical moment. If lβcr1/π is not an integer, the two values of βcr1 which makes lβcr1/π be two integers most near to lβcr1/π are then substituted into (36), and the smaller value is chosen to be the lateral bending critical moment.When the steel beam bottom flange of steel-concrete composite beam in negative moment region suffers rotational buckling, the steel-concrete composite beam will yield lateral bending and torsional buckling. The neutral balance equation is seen in (33) and the boundary conditions are(38)φz=0,l=0,φ′′z=0,l=0.Assume ξ2=α3βMxyc/I+Dα4/r02MxycAf/I-GJ. Solving (33), we obtained the following:(39)φ=Asin⁡ξz+Bcos⁡ξz.Substituting (39) into (38) and with φ≠0, one can have(40)sin⁡ξl=0,(41)ξ=nπl=β,(42)α3βMxyc/I+Dα4r02MxycAf/I-GJ=β.Solving (42),(43)Mcr2=Dα4/β+GJr02Af-α3ycI.Because of the fact that dMcr2/dβ=0, then(44)βcr2=20.5hw2,ncr2=lβcr2π.Whenlβcr2/π is an integer, the substitution of βcr2 into (43) gets the lateral bending critical moment. When lβcr2/π is not an integer, a similar calculation mentioned before is taken to obtain the lateral bending critical moment.After getting the lateral bending buckling critical momentMcr1 and the bending and torsional buckling critical moment Mcr2 the smaller one of these two is taken as the buckling load of composite beam in negative moment region. The analytical results indicate that the composite beam in negative moment region yields as a result of lateral bending and torsional buckling acting together. Therefore, in [11, 12] the method in which only one case is considered is questionable. The presented work is an improvement to them.To sum up, the calculation formula of buckling moment can be expressed as(45)Mcr=min⁡EIyβcr1+Dα2/βcr1Af-α1ycI,Dα4/βcr2+GJr02Af-α3ycI,βcr1=3.46EIyhw3/D+0.37hw4,βcr2=20.5hw2,α1=twhw0.086hwyc-0.37,α2=12hw3+2.4βcr1hw+0.37hwβcr12,α3=twhw30.0036hwyc-0.0095,α4=4hw+0.27hwβcr2+0.0095hw3βcr22.Whenlβcr1/π is not an integer, substituting βcr1 that corresponds to two integers of the left and right side of  lβcr1/π into (36), the smaller resulting value of βcr1 is the desired value. And when lβcr2/π is not an integer, substituting βcr2 that corresponds to two integers of the left and right side of lβcr2/π into (43), the smaller resulting value of βcr2 is the desired value.


### 4.2. Practical Example Analysis

Nine cases of I-shape steel-concrete composite beam are shown in Table1. The length of specimen in negative moment region is equal to 4000 mm. Finite element method is adopted to calculate these nine examples by using ANSYS, in which the concrete part of composite beam is substituted by lateral restraint and the steel I-beam is simulated by SHELL 43. The method proposed by British Steel Construction Institution [4] and the buckling model method [15] are also used to calculate the buckling loads. The obtained results are shown in Table 2.Table 1
Geometric dimension of examples.
Number of example
h
w/mm
t
w/mm
t
f/mm
b
f/mm
1
188.6
7
11.4
100
2
188.6
7
10.4
100
3
188.6
7
9.4
100
4
188.6
8
9.4
100
5
188.6
9
9.4
100
6
150
8
9.4
100
7
200
8
9.4
100
8
200
8
9.4
80
9
200
8
9.4
120Table 2
Calculation results.
Number of example
ANSYS/MPa
Literature method [4]/MPa
Literature method [15]/MPa
The proposed method/MPa
1
5.81 × 102
2.21 × 102
8.38 × 102
5.83 × 102
2
5.46 × 102
2.05 × 102
8.32 × 102
5.51 × 102
3
5.13 × 102
1.87 × 102
8.27 × 102
5.18 × 102
4
6.13 × 102
2.13 × 102
12.2 × 102
6.28 × 102
5
7.03 × 102
2.46 × 102
17.3 × 102
7.69 × 102
6
6.32 × 102
2.56 × 102
19.2 × 102
6.29 × 102
7
6.08 × 102
2.06 × 102
10.9 × 102
6.09 × 102
8
4.80 × 102
1.38 × 102
10.8 × 102
4.89 × 102
9
7.89 × 102
2.92 × 102
10.4 × 102
8.24 × 102As shown in Table2, the results obtained by buckling model method [15] tend to be excessively unsafe. The results by using the method proposed by British Steel Construction Institution [4] are too conservative. However, the results of the proposed method are in good agreement with those of ANSYS, which means that the proposed method is a reasonable and effective method.


## 4.1. Derivation of Critical Moment

The lateral constraint rigidity and rotation constraint rigidity which steel beam web to bottom flange can be simplified, respectively, as(23)kx=α1βσ+Dα2,(24)kφ=α3βσ+Dα4,(25)β=πλ2,(26a)α1=twhw0.086hwyc-0.37,(26b)α2=12hw3+2.4βhw+0.37hwβ2,(26c)α3=twhw30.0036hwyc-0.0095,(26d)α4=4hw+0.27hwβ+0.0095hw3β2,(27)σ=MxycI=PAf.Therein,D=Etw3/121-μ2; μ is Poisson’s ratio of steel; σ represents the compressive stress of bottom flange; P refers to the compressive force of bottom flange; -yc is the gravity centre coordinate of steel beam total cross section; I is the inertia moment of steel beam; λ=l/n; n is the buckling half-wave number; Af is the area of bottom flange.As shown in Figure2, the thin-walled member is doubly symmetric along x-axis and y-axis and the centre of origin O coincides with the flexural center. The displacements of origin O in x direction and y direction are denoted as u and v, respectively. Because the rigidity in y direction is infinity, v is equal to zero. The equivalent distributed forces caused by elastic medium as a result of displacements of thin-walled member can be written as(28)rx=kxu,ry=kyv,where kx and ky are the lateral and vertical constraint rigidity which web to bottom flange, respectively.The torsional angle which the member rotates around the bending center is assumed to beφ. The equivalent distributed moment of torsional thin-walled member induced by equivalent spring is(29)m=kφφ.Neutral balance differential equation of thin-walled member can be expressed as [13, 14](30)EIyuIV+Pu′′+yaφ′′+kxu-yd-yaφ=0,EIxvIV+Pv′′-xaφ′′+kyv+xd-xaφ=0,EIwφIV+r02P-GJφ′′-Pxav′′-yau′′-kxu-yd-yaφyd-ya+kyv+xd-xaφxd-xa+kφφ=0.Therein,Iy=tfbf3/12;  Ix=bftf3/12;  J=bftf3/3;  r02=xa2+ya2+Ix+Iy/As,As being the area of steel beam; xa is the horizontal coordinate of the bottom flange section bending centre, xa=0; ya is the vertical coordinate of the bottom flange section bending centre, ya=0; xd is the horizontal coordinate of the bottom flange section rotation axis, xd=0; yd is the vertical coordinate of the bottom flange section rotation axis, yd=0; Iw is the fan-shaped inertia moment of bottom flange section, Iw=0; E is the tensile elastic modulus of steel; G is the shear elastic modulus of steel.With substitution ofya=0, yd=0, xa=0, v=0, Iw=0, kx=0, xa=0, ya=0, and yd=0 into (30), one can obtain(31)EIyIuIV+ycAfMxu′′+α1βycMx+IDα2u=0,(32)kyv+xd-xaφ=0,(33)r02ycAfMx-GJIφ′′+α3βycMx+IDα4φ=0.When the steel-concrete composite beam in negative moment region bears lateral bending buckling, its neutral balance equation is shown in (31) and the corresponding boundary conditions are(34)uz=0,l=0,u′′z=0,l=0.Assuming u=Asin⁡βz, which satisfies the boundary conditions, according to Galerkin’s method,(35)β2-MxycAfEIyIβ+α1βMxyc/I+Dα2EIy=0.By solving (35),(36)Mcr1=EIyβ+Dα2/βAf-α1ycI.Due to dMcr1/dβ=0, we obtained the following:(37)βcr1=3.46EIyhw3/D+0.37hw4,ncr1=lβcr1π.If lβcr1/π is an integer, substitution of βcr1 into (36) leads to the lateral bending critical moment. If lβcr1/π is not an integer, the two values of βcr1 which makes lβcr1/π be two integers most near to lβcr1/π are then substituted into (36), and the smaller value is chosen to be the lateral bending critical moment.When the steel beam bottom flange of steel-concrete composite beam in negative moment region suffers rotational buckling, the steel-concrete composite beam will yield lateral bending and torsional buckling. The neutral balance equation is seen in (33) and the boundary conditions are(38)φz=0,l=0,φ′′z=0,l=0.Assume ξ2=α3βMxyc/I+Dα4/r02MxycAf/I-GJ. Solving (33), we obtained the following:(39)φ=Asin⁡ξz+Bcos⁡ξz.Substituting (39) into (38) and with φ≠0, one can have(40)sin⁡ξl=0,(41)ξ=nπl=β,(42)α3βMxyc/I+Dα4r02MxycAf/I-GJ=β.Solving (42),(43)Mcr2=Dα4/β+GJr02Af-α3ycI.Because of the fact that dMcr2/dβ=0, then(44)βcr2=20.5hw2,ncr2=lβcr2π.Whenlβcr2/π is an integer, the substitution of βcr2 into (43) gets the lateral bending critical moment. When lβcr2/π is not an integer, a similar calculation mentioned before is taken to obtain the lateral bending critical moment.After getting the lateral bending buckling critical momentMcr1 and the bending and torsional buckling critical moment Mcr2 the smaller one of these two is taken as the buckling load of composite beam in negative moment region. The analytical results indicate that the composite beam in negative moment region yields as a result of lateral bending and torsional buckling acting together. Therefore, in [11, 12] the method in which only one case is considered is questionable. The presented work is an improvement to them.To sum up, the calculation formula of buckling moment can be expressed as(45)Mcr=min⁡EIyβcr1+Dα2/βcr1Af-α1ycI,Dα4/βcr2+GJr02Af-α3ycI,βcr1=3.46EIyhw3/D+0.37hw4,βcr2=20.5hw2,α1=twhw0.086hwyc-0.37,α2=12hw3+2.4βcr1hw+0.37hwβcr12,α3=twhw30.0036hwyc-0.0095,α4=4hw+0.27hwβcr2+0.0095hw3βcr22.Whenlβcr1/π is not an integer, substituting βcr1 that corresponds to two integers of the left and right side of  lβcr1/π into (36), the smaller resulting value of βcr1 is the desired value. And when lβcr2/π is not an integer, substituting βcr2 that corresponds to two integers of the left and right side of lβcr2/π into (43), the smaller resulting value of βcr2 is the desired value.


## 4.2. Practical Example Analysis

Nine cases of I-shape steel-concrete composite beam are shown in Table1. The length of specimen in negative moment region is equal to 4000 mm. Finite element method is adopted to calculate these nine examples by using ANSYS, in which the concrete part of composite beam is substituted by lateral restraint and the steel I-beam is simulated by SHELL 43. The method proposed by British Steel Construction Institution [4] and the buckling model method [15] are also used to calculate the buckling loads. The obtained results are shown in Table 2.Table 1
Geometric dimension of examples.
Number of example
h
w/mm
t
w/mm
t
f/mm
b
f/mm
1
188.6
7
11.4
100
2
188.6
7
10.4
100
3
188.6
7
9.4
100
4
188.6
8
9.4
100
5
188.6
9
9.4
100
6
150
8
9.4
100
7
200
8
9.4
100
8
200
8
9.4
80
9
200
8
9.4
120Table 2
Calculation results.
Number of example
ANSYS/MPa
Literature method [4]/MPa
Literature method [15]/MPa
The proposed method/MPa
1
5.81 × 102
2.21 × 102
8.38 × 102
5.83 × 102
2
5.46 × 102
2.05 × 102
8.32 × 102
5.51 × 102
3
5.13 × 102
1.87 × 102
8.27 × 102
5.18 × 102
4
6.13 × 102
2.13 × 102
12.2 × 102
6.28 × 102
5
7.03 × 102
2.46 × 102
17.3 × 102
7.69 × 102
6
6.32 × 102
2.56 × 102
19.2 × 102
6.29 × 102
7
6.08 × 102
2.06 × 102
10.9 × 102
6.09 × 102
8
4.80 × 102
1.38 × 102
10.8 × 102
4.89 × 102
9
7.89 × 102
2.92 × 102
10.4 × 102
8.24 × 102As shown in Table2, the results obtained by buckling model method [15] tend to be excessively unsafe. The results by using the method proposed by British Steel Construction Institution [4] are too conservative. However, the results of the proposed method are in good agreement with those of ANSYS, which means that the proposed method is a reasonable and effective method.


## 5. Conclusions

Based on energy method, a comprehensive and intensive study on rotation restraining rigiditykϕ and lateral restraining rigidity kx which steel beam web to bottom plate of steel-concrete composite beam in negative moment region is performed in this paper. The expressions of lateral bending buckling stress, lateral bending, and torsional buckling stress and buckling moments of steel-concrete composite beam in negative moment region are deduced. Some conclusions are drawn as follows:(1)
Both the rotation constraint rigiditykϕ and the lateral constraint rigidity kx show a linear relationship with longitudinal compressive stress σ1 at bottom flange.(2)
The rotation constraint rigiditykϕ and the lateral constraint rigidity kx could be negative. When the rotation constraint rigidity and the lateral constraint rigidity are negative that means the bottom flange of steel beam can be restrained by rotation or lateral constraint when steel beam was in negative moment region buckling.(3)
Sincehw2kx/kφ is not infinitesimal, the lateral constraint rigidity of bottom flange to web cannot be neglected. In other words, in the calculation, the equation kx=0 cannot be used. Therefore it is proved theoretically that the lateral constraint rigidity of bottom flange to web cannot be approximated to be zero. This point is different with the steel structure.(4)
The results, obtained by literatures [4, 15] show some deviations with the ANSYS results. But the results by the presented expressions agree well with the ANSYS results. The reason is that the proposed expressions consider the lateral bending buckling and lateral bending-torsion buckling simultaneously. The proposed method is more reasonable and clearer in physical concepts in comparison with those methods which consider only one buckling mode. Besides, the proposed expressions are more concise and suitable for the engineering application.



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*Source: 763634-2015-10-11.xml*