# The Stability of a Movable High-Strength Inverted-Triangular Steel Bridge

Authors: Lei Gao; Linyue Bai; Kebin Jiang; Qiang Wang; Xiaohui He
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

The overall stability of a movable high-strength inverted-triangular steel bridge is worth studying because of its new truss structure. In this study, an approach was proposed based on the stiffness equivalence principle in which the inverted-triangle truss structure was modeled as a thin-walled triangular beam. On this basis, the calculation of the critical load of elastic stability of a movable high-strength inverted-triangular steel bridge with variable rigidity at both ends and locally uniformly distributed load was carried out based on the energy theory, which was in good agreement with existing theories. A material performance test at BS700 was carried out to establish the material properties, and then a finite element model of the bridge was established, the results of which were compared with those of the experimental load test, in order to verify the accuracy of the finite element model. Considering material nonlinearity and geometric nonlinearity, nonlinear buckling analysis of the bridge was conducted and the factors influencing the bridge’s ultimate bearing capacity were analyzed.


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## Body

## 1. Introduction

High-strength steel with a yield strength of 460–690 MPa has been commonly used in construction projects in Japan, the United States, and some other countries [1–3]. In addition, the 1100 MPa-level high-strength steel has been used in some military bridges, such as FB48. Over the past few years, an increasing number of high-strength steel buildings have been built in China [4–6], with Q420 and Q460 as the main two types of high-strength steel being used. According to the research, the application of high-strength steel can effectively contribute to self-weight reduction and cost saving [6], which has made it more suitable for the design of movable emergency bridges. This is the reason why high-strength steel with a yield strength greater than 700 MPa (σ0.2≥ 700 MPa) is used in some movable bridges [7]. Movable high-strength inverted-triangular steel bridges developed with BS700 high-strength steel (σ0.2= 700 MPa) are a new type of long-span bridge whose length and deck load can reach 51 m and 60 ton, respectively. This kind of bridge consists of two side bridge segments with variable cross-sections and four identical central bridge segments.Apart from BS700 high-strength steel, a new inverted-triangular truss structure was used in this bridge. Two pieces of inverted-triangular truss structures made up two lanes, which were connected and integrated in between on the top using cross beams. Central bridge segments are shown in Figure1. Normally, a rectangular truss system is used in a truss bridge, yet research findings suggested that a triangular treadway bridge performs better in saving materials and reducing weight than a rectangular truss bridge [8]. Meanwhile, the triangular truss was usually employed on a limited basis for a number of structures such as roof truss girders, transmission towers, and crane booms [9]. In some composite bridges the triangular trusses were also widely used [10]. The mentioned triangular trusses were usually vertically triangular, but an inverted-triangular truss system emerged with the development of long-span spatial structures [11]. Today the inverted-triangular truss is more commonly applied in composite bridges [12, 13]. Inverted-triangular trusses were used in the movable bridge designed in this study. To make it convenient for mechanized bridging, there was no connection to the bottom chord, and the top chord was also weakly connected, which offers a promising area for further studies on the bridge’s overall stability.Figure 1
The members of a typical bridge segment.Considerable research on the stability of triangular or inverted-triangular truss structures has been done. The eigenvalue stability of truss structures was studied through critical point theory [14, 15]. Some laboratory-scale tests and full-scale testing of old steel truss tests were performed to get the load-carrying capacity of truss structures [16, 17]. A full-scale triangular composite truss structure specimen was tested and compared with finite element analysis [13], and the finite element model (FEM) was validated with experimental results. Recently the truss bridge’s linear buckling analysis and nonlinear stability analysis, which can consider material nonlinearity and initial geometric defects, were both done by the finite element method [18–20]. The stability ultimate load influence factors of truss structures were also studied. It was found that the joint rigid connection has more influence on the ultimate bearing capacity of triangular space-truss structure [21]. Height-width ratio has more influence on the integral stability and the proposed height-width ratio is 0.75~0.875 [22]. In comparison with the paper that has already been published, the main purpose of this work is to investigate the stability of a novel high-strength inverted-triangular steel truss bridge that has varied sections in its side bridge segments. In this study, the elastic stability of the movable inverted-triangular steel bridge was analyzed theoretically, and a calculation equation was obtained to predict its elastic load-carrying capacity. The material performance test of BS700 and the field test of load-carrying capacity were conducted. The bridge’s elastic stability load-carrying capacity was calculated using the finite element method and compared with the theoretical result. Both material nonlinearity and geometric nonlinearity were considered in analyzing the nonlinear stability of the bridge by the finite element method. Moreover, factors influencing the stability load-carrying capacity of the movable high-strength inverted-triangular steel bridge were analyzed.


## 2. Analysis of Elastic Stability Load-Carrying Capacity

Analysis of the elastic stability load-carrying capacity lays the foundation for analyzing the nonlinear stability load-carrying capacity. To perform the theoretical analysis, the truss structure is usually modeled as a beam, following the principle of equivalent stiffness [23]. In terms of the truss structure of a movable high-strength inverted-triangular steel bridge, the stiffness parameters are mainly axial rigidity EA, in-plane flexural rigidity EIx, out-plane flexural rigidity EIy, and torsional rigidity GIk.


### 2.1. Simplification to Thin-Walled Triangular Beam Based on Equivalent Stiffness

The simplified diagram of the cross section of the movable high-strength inverted-triangular steel bridge is shown in Figure2.Figure 2
Schematic diagram of the cross section of the movable high-strength inverted-triangular steel bridge.There are two different sections on the top chord whose areas areA1 and A2. The area of the bottom chord is A3. The position of a section centroid is determined according to Sx=0. Supposing that point o is the centroid and h1 and h2 are the distances between the centroid and the top and bottom chords, then(1)h1=A3h2A1+A2+A3h2=2A1+A2h2A1+A2+A3.The equivalent area is(2)A=2A1+A2+A3.Neglecting the contribution of the component’s cross section to the overall moment of inertia moment of inertia, the equivalent moment of inertia is(3)Ix=2A1+A2h12+A3h22,Iy=12A1b2.Therefore, the axial rigidity EA, in-plane equivalent flexural rigidity EIx, and out-plane flexural rigidity EIy could be calculated, respectively. In terms of torsional rigidity GIk, since the web member connecting the chord primarily bears the shear force, its contribution to torsion could not be ignored. There are two types of web members used in the movable high-strength inverted-triangular steel bridge: one is connected to the bottom and top chord I and the other is connected to the bottom chord and top chord II (Figure 1). For simplification, the middle chord (top chord II) and web member are evenly distributed to both sides, forming two sections of trusses on the left and right sides (Figure 3).Figure 3
Schematic diagram of the movable high-strength inverted-triangular steel bridge.One section of the movable high-strength inverted-triangular steel bridge is taken as a calculation unit and converted into a thin plate whose thickness iste at each side. The whole section to be calculated is converted into a thin-walled triangular beam, with its torsion strength calculated. If the intersegment length is d and height is bi, according to the equation of strain energy density, the strain energy of each piece of thin plate i is(4)Uei=q2d2G∑i=13bitei.The thickness tei can be determined based on the condition that strain energy Ui and Uei of a truss with a length of d are equal. Suppose the total shear force that results from shear flow q acting upon the truss plane is Vi=qbi. A segment of the bridge is taken as the research object, as shown in Figure 4.Figure 4
One segment of the movable high-strength inverted-triangular steel bridge.The internal force generated by a slanted web member with the action of shear force is(5)D=Visin⁡αi=Vikibi=qbikibi=qki.If the length and area of the web member are l and A, and it bears a constant force P, its strain energy is(6)Δu=P2l2EA.The chord’s axial force varies linearly within the length of l. Assuming the maximum force S=1/2qd, the chord’s strain energy is(7)Δu=12EA∫0lS-2lSx2dx=S2l6EA.The strain energy of each piece of truss unit with a length ofd can be obtained according to (6) and (7):Top chord:(8)Δu0=q2d324EA0where A0=A1/2+A2/4.Slanted web member:(9)Δud=q2k32EAdwhere Ad=Ad1/2 and Ad1 is the area of central slanted web member.Bottom chord:(10)Δuu=q2d324EAuwhere Au=A3/2.Vertical web member:(11)Δuv=q2d32EAvwhere Av is the area of the slanted web member.The equation for theith piece of truss unit could be calculated by summing up the strain energy of the aforementioned four chords and web members:(12)uti=Δuoi+Δuui+Δuvi=q22Ed3121Aoi+1Aui+k3Adi+bi3Avi.The following equation is obtained according to the principle of equal strain energy:(13)q22Ed3121Aoi+1Aui+k3Adi+bi3Avi=q2d2Gbitei.Then, the thickness of equivalent thin-walled beam is(14)tei=EGbidd3/121/Aoi+1/Aui+k3/Adi+bi3/Avi.At this point, the converted thicknesses of the triangle’s three sides, te1, te2, and te3, are calculated. The equivalent torsion moment of inertia is obtained according to the constant equation of free torsion of a thin-walled closed section of arbitrary shape [24].(15)Ik=4A02∮ds/t=4A02S1/te1+S2/te2+S3/te3where A0 is the area of the closed section formed by the medians of the thin-walled triangle; S1, S2, and S3 are the widths of three thin-walled plates, i.e., lengths of the three sides of the triangle’s cross section; and te1, te2, and te3 are the converted thicknesses of three sides. GIk, the torsional rigidity of the thin-walled triangular beam simplified from the movable high-strength inverted-triangular steel bridge, can be obtained based on (14) and (15).The equivalent rigidity can be calculated using MATLAB according to this theoretical derivation [25]. Due to the fact that the rigidities of both end units of the movable high-strength inverted-triangular steel bridge are variable while the central rigidity remains unchanged, two programs are compiled to calculate their respective equivalent rigidities, which can be calculated by entering the bridge’s height and width and the sizes of various chords and bars into the programs. With this calculation, the moments of area of the sections where central rigidity remained unchanged were Ix=0.0096m4, Iy=0.0018m4, and Ik=0.000163m4.


### 2.2. Critical Load of Linear-Elastic Stability

It is difficult to find the solution to the torsional rigidity if the movable high-strength inverted-triangle steel bridge is simplified into a thin-walled triangular beam, since the bridge’s rigidity varies at the two ends while the central rigidity stays the same. To this end, it is solved using the energy method [24]. The axial compressive strain energy and shear strain energy are too low to be taken into account. In addition, a study made by Liu [23] suggests that the warping strain energy of a thin-walled triangular beam is negligible. This strain energy contains lateral bending strain energy U1, in-plane bending strain energy U2, and pure torsion strain energy U3.Lateral bending strain energy can be obtained according to the strain energy equation [24].(16)U1=12∫0lEIyu′′2dzIn-plane bending strain energy is(17)U2=12∫0lEIxv′′2dz.Free torsion strain energy is(18)U3=12∫0lGIkϕ′2dzwhere u is the lateral displacement, v is the vertical displacement, and φ is the torsion angle.Therefore, total strain energy is(19)U=U1+U2+U3=12∫0lEIyu′′2+GIkϕ′2+EIxv′′2dz.Chen [24] reported that the flexural bars’ elastic flexural-tensional buckling critical flexural moment was increased to some extent by the effect of large counter-arch due to flexural deformation. Therefore, ignoring the impact of in-plane flexure would result in a slightly smaller critical load and a more conservative result. Equation (19) is thus reduced to(20)U=U1+U2+U3≈12∫0lEIyu′′2+GIkφ′2dz.Without considering residual stress, the potential energy of external force can be calculated with the following equation when any section is subjected to the combined action of axial force and flexural moment [24].(21)W=12∫0l2βyMxφ′2+2Mxu′′φdzThe potential energy of the external force should be (21) combined with the strain energy generated by the changes of the load’s action spot. Supposing the distance between the load’s action spot and shearing center is a (Figure 5), the potential generated is(22)ΔW=-12∫0lqzaφ2dz.Hence, the total potential energy of the external force is(23)W=12∫0l2βyMxφ′2+2Mxu′′φdz-qzaφ2dzwhere βy=∫Ay(x2+y2)dA/2Ix-y0 is the section asymmetry coefficient and βy=-y0 because the section is symmetrical. y0 is the distance between the shearing center and the centroid. According to our calculation, βy=-0.2,a=0.17.Figure 5
Position of the shearing center of the movable high-strength inverted-triangular steel bridge.Total potential energy is(24)Π=12∫0lEIyu′′2+GIkϕ′2dz+2βyMxϕ′2+2Mxu′′ϕdz-qzaϕ2dz.The integral of the total potential energy should be taken in segments as the height of the movable high-strength inverted-triangular steel bridge varies linearly at both ends and the uniformly distributed load acts locally upon the mid-span (Figure 6). Due to the linear variation of height, supposing the flexural rigidity and torsional rigidity are also subject to linear variations, the maximum rigidities are Iy2 and Ik2, respectively, and the minimum rigidities are Iy1 and Ik1, respectively. Assume the width of the local uniformly distributed load, the length of the segment with variable rigidity, the intensity of uniformly distributed load, and the span of the simply supported bridge are B, m, q, and L, respectively.Figure 6
Thin-walled triangular beam under locally uniformly distributed load.The expressions of segmented load and rigidity are shown as follows.(25)Uniformly-distributedload:qz=00≤z≤L-B2qL-B2<z≤L+B20L+B2<z≤L(26)Flexuralmomentofinertia:Iy=Iy1+Iy2-Iy1m0≤z≤mIy2m<z≤L-mIy1-Iy2mz-LIy1-Iy2-mIy1mL-m<z≤L(27)Torsionmomentofinertia:Ik=Ik1+Ik2-Ik1m0≤z≤mIk2m<z≤L-mIk1-Ik2mz-LIk1-Ik2-mIk1mL-m<z≤L(28)Flexuralmoment:Mx=qBz20≤z≤L-B2qBz2-qz-L-B/222L-B2<z≤L+B2qBz2-qBz-L2L+B2<z≤LTherefore, the integral of total potential energy is performed on five intervals along the beam’s lengthwise direction; namely,(29)0,m;m,L-B2;L-B2,L+B2;L+B2,L-m;L-m,L.The Rayleigh–Ritz method is used to solve the critical flexural moment [21]. Suppose the shape function that matches the boundary conditions is(30)u=c1sin⁡πzL,φ=c2sin⁡πzL.Total potential energy is(31)Π=12∫0lπ4EIyL4c12sin2⁡πzL+GIkπ2L2c22+2βyMxπ2L2c22cos2⁡πzL-2Mxπ2L2c1c2sin2⁡πzL-qzac22sin2⁡πzLdz.After the segment integral, total potential energy is(32)Π=f1Iy1+f2Iy2-Iy1+f7Iy2+f11Iy2+f16Iy2+f22Iy1-Iy2-f23LIy1-Iy2-mIy1c12+f3Ik1+Ik2-Ik1f4+qβyf5+Ik2f8+qβyf9+f12Ik2+qβyf13-qaf15+f17Ik2+qβyf18-qβyf19+f24Ik1-Ik2-f25LIk1-Ik2-mIk1+f26qβy-f27qβyc22+q-f6-f10-f14-f20+f21-f28+f29c1c2where f1~f29 are the parameters related to L, B, and m.According to the minimal potential energy principle, we have(33)∂Π∂c1=0,∂Π∂c2=0.Three parameters are introduced:(34)K1=f1Iy1+f2Iy2-Iy1+f7Iy2+f11Iy2+f16Iy2+f22Iy1-Iy2-f23LIy1-Iy2-mIy1.(35)K2=f3Ik1+Ik2-Ik1f4+qβyf5+Ik2f8+qβyf9+f12Ik2+qβyf13-qaf15+f17Ik2+qβyf18-qβyf19+f24Ik1-Ik2-f25LIk1-Ik2-mIk1+f26qβy-f27qβy.(36)K3=q-f6-f10-f14-f20+f21-f28+f29.The buckling condition of the equivalent thin-walled triangular beam is(37)2K1K3K32K2=0.To separate the load q from the equation above, several coefficients, i.e., z1,z2,z3, and z4, are introduced:(38)z1=f3Ik1+Ik2-Ik1f4+Ik2f8+f12Ik2+f17Ik2+f24Ik1-Ik2-f25LIk1-Ik2-mIk1(39)z2=f5+f9+f13+f18-f19+f26-f27(40)z3=f15(41)z4=-f6-f10-f14-f20+f21-f28+f29.The critical load is obtained:(42)qcr=2K1βyz2-2K1az3+2K1βyz2-2K1az32+4K1z1z42z42.To verify the correctness of this expression, a special circumstance, where load is uniformly distributed along the full span and rigidity remained unchanged along the beam’s axial direction, was modeled. In this case, the critical load is as follows according to (42):(43)qcr=12π2π2+3π2EIyL4-6aπ2+3+π2-3π2+3βy+-6aπ2+3+π2-3π2+3βy2+GIkL2π2EIy.According to Chen [24], the elastic critical flexural moment with full-span uniform load distribution is(44)Mcr=3π22π2+3π2EIyL2-6aπ2+3+π2-3π2+3βy+-6aπ2+3+π2-3π2+3βy2+IωIy1+GIkL2π2EIω.Without considering warping the moment of inertia, the result of (43) is consistent with that of (44). Not considering the self-weight, the elastic critical load of the movable high-strength inverted-triangular steel bridge was 2300.6 kN, which was calculated by substituting the data obtained based on the strain energy principle into (43).


## 2.1. Simplification to Thin-Walled Triangular Beam Based on Equivalent Stiffness

The simplified diagram of the cross section of the movable high-strength inverted-triangular steel bridge is shown in Figure2.Figure 2
Schematic diagram of the cross section of the movable high-strength inverted-triangular steel bridge.There are two different sections on the top chord whose areas areA1 and A2. The area of the bottom chord is A3. The position of a section centroid is determined according to Sx=0. Supposing that point o is the centroid and h1 and h2 are the distances between the centroid and the top and bottom chords, then(1)h1=A3h2A1+A2+A3h2=2A1+A2h2A1+A2+A3.The equivalent area is(2)A=2A1+A2+A3.Neglecting the contribution of the component’s cross section to the overall moment of inertia moment of inertia, the equivalent moment of inertia is(3)Ix=2A1+A2h12+A3h22,Iy=12A1b2.Therefore, the axial rigidity EA, in-plane equivalent flexural rigidity EIx, and out-plane flexural rigidity EIy could be calculated, respectively. In terms of torsional rigidity GIk, since the web member connecting the chord primarily bears the shear force, its contribution to torsion could not be ignored. There are two types of web members used in the movable high-strength inverted-triangular steel bridge: one is connected to the bottom and top chord I and the other is connected to the bottom chord and top chord II (Figure 1). For simplification, the middle chord (top chord II) and web member are evenly distributed to both sides, forming two sections of trusses on the left and right sides (Figure 3).Figure 3
Schematic diagram of the movable high-strength inverted-triangular steel bridge.One section of the movable high-strength inverted-triangular steel bridge is taken as a calculation unit and converted into a thin plate whose thickness iste at each side. The whole section to be calculated is converted into a thin-walled triangular beam, with its torsion strength calculated. If the intersegment length is d and height is bi, according to the equation of strain energy density, the strain energy of each piece of thin plate i is(4)Uei=q2d2G∑i=13bitei.The thickness tei can be determined based on the condition that strain energy Ui and Uei of a truss with a length of d are equal. Suppose the total shear force that results from shear flow q acting upon the truss plane is Vi=qbi. A segment of the bridge is taken as the research object, as shown in Figure 4.Figure 4
One segment of the movable high-strength inverted-triangular steel bridge.The internal force generated by a slanted web member with the action of shear force is(5)D=Visin⁡αi=Vikibi=qbikibi=qki.If the length and area of the web member are l and A, and it bears a constant force P, its strain energy is(6)Δu=P2l2EA.The chord’s axial force varies linearly within the length of l. Assuming the maximum force S=1/2qd, the chord’s strain energy is(7)Δu=12EA∫0lS-2lSx2dx=S2l6EA.The strain energy of each piece of truss unit with a length ofd can be obtained according to (6) and (7):Top chord:(8)Δu0=q2d324EA0where A0=A1/2+A2/4.Slanted web member:(9)Δud=q2k32EAdwhere Ad=Ad1/2 and Ad1 is the area of central slanted web member.Bottom chord:(10)Δuu=q2d324EAuwhere Au=A3/2.Vertical web member:(11)Δuv=q2d32EAvwhere Av is the area of the slanted web member.The equation for theith piece of truss unit could be calculated by summing up the strain energy of the aforementioned four chords and web members:(12)uti=Δuoi+Δuui+Δuvi=q22Ed3121Aoi+1Aui+k3Adi+bi3Avi.The following equation is obtained according to the principle of equal strain energy:(13)q22Ed3121Aoi+1Aui+k3Adi+bi3Avi=q2d2Gbitei.Then, the thickness of equivalent thin-walled beam is(14)tei=EGbidd3/121/Aoi+1/Aui+k3/Adi+bi3/Avi.At this point, the converted thicknesses of the triangle’s three sides, te1, te2, and te3, are calculated. The equivalent torsion moment of inertia is obtained according to the constant equation of free torsion of a thin-walled closed section of arbitrary shape [24].(15)Ik=4A02∮ds/t=4A02S1/te1+S2/te2+S3/te3where A0 is the area of the closed section formed by the medians of the thin-walled triangle; S1, S2, and S3 are the widths of three thin-walled plates, i.e., lengths of the three sides of the triangle’s cross section; and te1, te2, and te3 are the converted thicknesses of three sides. GIk, the torsional rigidity of the thin-walled triangular beam simplified from the movable high-strength inverted-triangular steel bridge, can be obtained based on (14) and (15).The equivalent rigidity can be calculated using MATLAB according to this theoretical derivation [25]. Due to the fact that the rigidities of both end units of the movable high-strength inverted-triangular steel bridge are variable while the central rigidity remains unchanged, two programs are compiled to calculate their respective equivalent rigidities, which can be calculated by entering the bridge’s height and width and the sizes of various chords and bars into the programs. With this calculation, the moments of area of the sections where central rigidity remained unchanged were Ix=0.0096m4, Iy=0.0018m4, and Ik=0.000163m4.


## 2.2. Critical Load of Linear-Elastic Stability

It is difficult to find the solution to the torsional rigidity if the movable high-strength inverted-triangle steel bridge is simplified into a thin-walled triangular beam, since the bridge’s rigidity varies at the two ends while the central rigidity stays the same. To this end, it is solved using the energy method [24]. The axial compressive strain energy and shear strain energy are too low to be taken into account. In addition, a study made by Liu [23] suggests that the warping strain energy of a thin-walled triangular beam is negligible. This strain energy contains lateral bending strain energy U1, in-plane bending strain energy U2, and pure torsion strain energy U3.Lateral bending strain energy can be obtained according to the strain energy equation [24].(16)U1=12∫0lEIyu′′2dzIn-plane bending strain energy is(17)U2=12∫0lEIxv′′2dz.Free torsion strain energy is(18)U3=12∫0lGIkϕ′2dzwhere u is the lateral displacement, v is the vertical displacement, and φ is the torsion angle.Therefore, total strain energy is(19)U=U1+U2+U3=12∫0lEIyu′′2+GIkϕ′2+EIxv′′2dz.Chen [24] reported that the flexural bars’ elastic flexural-tensional buckling critical flexural moment was increased to some extent by the effect of large counter-arch due to flexural deformation. Therefore, ignoring the impact of in-plane flexure would result in a slightly smaller critical load and a more conservative result. Equation (19) is thus reduced to(20)U=U1+U2+U3≈12∫0lEIyu′′2+GIkφ′2dz.Without considering residual stress, the potential energy of external force can be calculated with the following equation when any section is subjected to the combined action of axial force and flexural moment [24].(21)W=12∫0l2βyMxφ′2+2Mxu′′φdzThe potential energy of the external force should be (21) combined with the strain energy generated by the changes of the load’s action spot. Supposing the distance between the load’s action spot and shearing center is a (Figure 5), the potential generated is(22)ΔW=-12∫0lqzaφ2dz.Hence, the total potential energy of the external force is(23)W=12∫0l2βyMxφ′2+2Mxu′′φdz-qzaφ2dzwhere βy=∫Ay(x2+y2)dA/2Ix-y0 is the section asymmetry coefficient and βy=-y0 because the section is symmetrical. y0 is the distance between the shearing center and the centroid. According to our calculation, βy=-0.2,a=0.17.Figure 5
Position of the shearing center of the movable high-strength inverted-triangular steel bridge.Total potential energy is(24)Π=12∫0lEIyu′′2+GIkϕ′2dz+2βyMxϕ′2+2Mxu′′ϕdz-qzaϕ2dz.The integral of the total potential energy should be taken in segments as the height of the movable high-strength inverted-triangular steel bridge varies linearly at both ends and the uniformly distributed load acts locally upon the mid-span (Figure 6). Due to the linear variation of height, supposing the flexural rigidity and torsional rigidity are also subject to linear variations, the maximum rigidities are Iy2 and Ik2, respectively, and the minimum rigidities are Iy1 and Ik1, respectively. Assume the width of the local uniformly distributed load, the length of the segment with variable rigidity, the intensity of uniformly distributed load, and the span of the simply supported bridge are B, m, q, and L, respectively.Figure 6
Thin-walled triangular beam under locally uniformly distributed load.The expressions of segmented load and rigidity are shown as follows.(25)Uniformly-distributedload:qz=00≤z≤L-B2qL-B2<z≤L+B20L+B2<z≤L(26)Flexuralmomentofinertia:Iy=Iy1+Iy2-Iy1m0≤z≤mIy2m<z≤L-mIy1-Iy2mz-LIy1-Iy2-mIy1mL-m<z≤L(27)Torsionmomentofinertia:Ik=Ik1+Ik2-Ik1m0≤z≤mIk2m<z≤L-mIk1-Ik2mz-LIk1-Ik2-mIk1mL-m<z≤L(28)Flexuralmoment:Mx=qBz20≤z≤L-B2qBz2-qz-L-B/222L-B2<z≤L+B2qBz2-qBz-L2L+B2<z≤LTherefore, the integral of total potential energy is performed on five intervals along the beam’s lengthwise direction; namely,(29)0,m;m,L-B2;L-B2,L+B2;L+B2,L-m;L-m,L.The Rayleigh–Ritz method is used to solve the critical flexural moment [21]. Suppose the shape function that matches the boundary conditions is(30)u=c1sin⁡πzL,φ=c2sin⁡πzL.Total potential energy is(31)Π=12∫0lπ4EIyL4c12sin2⁡πzL+GIkπ2L2c22+2βyMxπ2L2c22cos2⁡πzL-2Mxπ2L2c1c2sin2⁡πzL-qzac22sin2⁡πzLdz.After the segment integral, total potential energy is(32)Π=f1Iy1+f2Iy2-Iy1+f7Iy2+f11Iy2+f16Iy2+f22Iy1-Iy2-f23LIy1-Iy2-mIy1c12+f3Ik1+Ik2-Ik1f4+qβyf5+Ik2f8+qβyf9+f12Ik2+qβyf13-qaf15+f17Ik2+qβyf18-qβyf19+f24Ik1-Ik2-f25LIk1-Ik2-mIk1+f26qβy-f27qβyc22+q-f6-f10-f14-f20+f21-f28+f29c1c2where f1~f29 are the parameters related to L, B, and m.According to the minimal potential energy principle, we have(33)∂Π∂c1=0,∂Π∂c2=0.Three parameters are introduced:(34)K1=f1Iy1+f2Iy2-Iy1+f7Iy2+f11Iy2+f16Iy2+f22Iy1-Iy2-f23LIy1-Iy2-mIy1.(35)K2=f3Ik1+Ik2-Ik1f4+qβyf5+Ik2f8+qβyf9+f12Ik2+qβyf13-qaf15+f17Ik2+qβyf18-qβyf19+f24Ik1-Ik2-f25LIk1-Ik2-mIk1+f26qβy-f27qβy.(36)K3=q-f6-f10-f14-f20+f21-f28+f29.The buckling condition of the equivalent thin-walled triangular beam is(37)2K1K3K32K2=0.To separate the load q from the equation above, several coefficients, i.e., z1,z2,z3, and z4, are introduced:(38)z1=f3Ik1+Ik2-Ik1f4+Ik2f8+f12Ik2+f17Ik2+f24Ik1-Ik2-f25LIk1-Ik2-mIk1(39)z2=f5+f9+f13+f18-f19+f26-f27(40)z3=f15(41)z4=-f6-f10-f14-f20+f21-f28+f29.The critical load is obtained:(42)qcr=2K1βyz2-2K1az3+2K1βyz2-2K1az32+4K1z1z42z42.To verify the correctness of this expression, a special circumstance, where load is uniformly distributed along the full span and rigidity remained unchanged along the beam’s axial direction, was modeled. In this case, the critical load is as follows according to (42):(43)qcr=12π2π2+3π2EIyL4-6aπ2+3+π2-3π2+3βy+-6aπ2+3+π2-3π2+3βy2+GIkL2π2EIy.According to Chen [24], the elastic critical flexural moment with full-span uniform load distribution is(44)Mcr=3π22π2+3π2EIyL2-6aπ2+3+π2-3π2+3βy+-6aπ2+3+π2-3π2+3βy2+IωIy1+GIkL2π2EIω.Without considering warping the moment of inertia, the result of (43) is consistent with that of (44). Not considering the self-weight, the elastic critical load of the movable high-strength inverted-triangular steel bridge was 2300.6 kN, which was calculated by substituting the data obtained based on the strain energy principle into (43).


## 3. Finite Element Analysis of Elastic Stability Load-Carrying Capacity

To further verify the accuracy of the equation for the elastic stability ultimate load, a model was built using the finite element method to perform elastic buckling analysis.


### 3.1. Establishment and Verification of the Finite Element Model

The beam element 188 of the ANSYS software, which can import a user’s section data, was adopted in the finite element model. In order to analyze many modes used for studying the influence factors on stability ultimate load capacity, the model uses many parameters which can be easily changed in an APDL document (Analysis Parameter Design Language) [15]. The whole finite element model contains a basic model, eigenvalue buckling analysis, and the results of picking-up. The material model made use of the multilinear hardening model.In order to get the material model data, uniaxial tension tests were carried out as shown in Figure7(a). No obvious plastic flow or necking was observed in the tests. The failure planes were nearly placed along the 45° cross section of the specimens as shown in Figure 7(b). Shear failure is dominant in the material.Figure 7
Material mechanical property test. (a) Test procedure. (b) The members’ failure.
(a) (b)As shown in Figure7, the complete stress-strain curve of the BS700 steel is gently strain-hardening. According to the test data, the elastic modulus of the material is 210.9 GPa. The nominal yield stress σ0.2 is 734 MPa and the nominal yield strain ε0.2 is 0.00426. For the material model we chose the multilinear isotropic hardening (MISO) model. The MISO model needs points, as shown in Table 1.Table 1
MISO data used for ANSYS models.
Strain 0.001 0.00178 0.00203 0.003 0.0037 0.00385 0.00426 0.00451 0.008 Stress [MPa] 209 370 419 592 689 704 734 754 790


### 3.2. Finite Element Model

In order to verify the basic finite element model validity, the load capacity test of the bridge was performed. A 51 m-long bridge, where the bearing length at both ends was 2 m and clearance height under the span was 0.6 m, was constructed. Two sections of planks were placed at the middle of bridge span to simulate two decks. Steel plates weighing 25.42 t→48.97 t→58.42 t→63.15 t→67.87 t→71.17 t→74.56 t→80.84 t (1 t≈10 kN) were loaded to and piled up on the planks. There were three deformation points atL/6, 5L/6, andL/2 (Lis 51 m). Measuring points and the loading diagram of the test and field test images are shown in Figures 8 and 9, respectively.Figure 8
Diagram of test points and loading.Figure 9
The load test scene.The test results compared with finite element analysis (FEA) results are shown in Table2.Table 2
Comparison of the displacement results of the test points and FEM calculation.
p  [kN] u 1   [ m m ] u s 1   [ m m ] u s 1 - u 1 u 1 % u 2   [ m m ] u s 2   [ m m ] u s 2 - u 2 u 2 % u s 3   [ m m ] u s 3   [ m m ] u s 3 - u 3 u 3 % 254.2 38 40 5.3 97 105 8.2 40 42 5.0 489.7 73 78 6.8 202 211 4.5 80 83 3.6 584.2 89 93 4.5 247 262 5.7 95 100 5.0 631.5 95 100 5.3 272 282 3.5 105 108 2.8 678.7 102 107 4.9 292 304 3.9 115 116 0.9 711.7 107 112 4.7 312 319 2.2 120 121 0.8 745.6 112 117 4.5 327 334 2.1 130 127 -2.4 808.4 122 127 4.1 342 362 5.5 135 138 2.2 Note: p is load, ui is test deformation, usi is FEA calculation deformation, (usi-ui)/ui is ratio of FEA results to test results, and i=1, 2, 3 represents the test points D1, D2, and D3.According to Table2 it can be declared that the finite element model can simulate the bridge accurately; the errors were below 5%, excepting a few points.


### 3.3. Eigenvalue Buckling Analysis

The typical equation for eigenvalue buckling analysis was [26](45)K0+λKσD=0where K0 is the matrix of elastic rigidity; Kσ is the matrix of stress rigidity; D is the characteristic displacement vector; and λ is the eigenvalue.Without considering self-weight, the eigenvalueλ obtained from eigenvalue buckling analysis multiplied by the load applied is the eigenvalue buckling load. In fact, the load kept changing, so that(46)K0+λKσ=0.When the self-weight was not considered, the first-order eigenvalue buckling mode was obtained, as shown in Figure 10. The first-order eigenvalue buckling load was 2405.4 kN, which was 4.6% larger than the 2300.6 kN load that was derived by the thin-walled beam equivalent theory in this study. This suggests a highly accurate value of the eigenvalue buckling load of the movable high-strength inverted-triangular steel bridge was derived in this study.Figure 10
First-order eigenvalue buckling mode of the movable high-strength inverted-triangular steel bridge.


## 3.1. Establishment and Verification of the Finite Element Model

The beam element 188 of the ANSYS software, which can import a user’s section data, was adopted in the finite element model. In order to analyze many modes used for studying the influence factors on stability ultimate load capacity, the model uses many parameters which can be easily changed in an APDL document (Analysis Parameter Design Language) [15]. The whole finite element model contains a basic model, eigenvalue buckling analysis, and the results of picking-up. The material model made use of the multilinear hardening model.In order to get the material model data, uniaxial tension tests were carried out as shown in Figure7(a). No obvious plastic flow or necking was observed in the tests. The failure planes were nearly placed along the 45° cross section of the specimens as shown in Figure 7(b). Shear failure is dominant in the material.Figure 7
Material mechanical property test. (a) Test procedure. (b) The members’ failure.
(a) (b)As shown in Figure7, the complete stress-strain curve of the BS700 steel is gently strain-hardening. According to the test data, the elastic modulus of the material is 210.9 GPa. The nominal yield stress σ0.2 is 734 MPa and the nominal yield strain ε0.2 is 0.00426. For the material model we chose the multilinear isotropic hardening (MISO) model. The MISO model needs points, as shown in Table 1.Table 1
MISO data used for ANSYS models.
Strain 0.001 0.00178 0.00203 0.003 0.0037 0.00385 0.00426 0.00451 0.008 Stress [MPa] 209 370 419 592 689 704 734 754 790


## 3.2. Finite Element Model

In order to verify the basic finite element model validity, the load capacity test of the bridge was performed. A 51 m-long bridge, where the bearing length at both ends was 2 m and clearance height under the span was 0.6 m, was constructed. Two sections of planks were placed at the middle of bridge span to simulate two decks. Steel plates weighing 25.42 t→48.97 t→58.42 t→63.15 t→67.87 t→71.17 t→74.56 t→80.84 t (1 t≈10 kN) were loaded to and piled up on the planks. There were three deformation points atL/6, 5L/6, andL/2 (Lis 51 m). Measuring points and the loading diagram of the test and field test images are shown in Figures 8 and 9, respectively.Figure 8
Diagram of test points and loading.Figure 9
The load test scene.The test results compared with finite element analysis (FEA) results are shown in Table2.Table 2
Comparison of the displacement results of the test points and FEM calculation.
p  [kN] u 1   [ m m ] u s 1   [ m m ] u s 1 - u 1 u 1 % u 2   [ m m ] u s 2   [ m m ] u s 2 - u 2 u 2 % u s 3   [ m m ] u s 3   [ m m ] u s 3 - u 3 u 3 % 254.2 38 40 5.3 97 105 8.2 40 42 5.0 489.7 73 78 6.8 202 211 4.5 80 83 3.6 584.2 89 93 4.5 247 262 5.7 95 100 5.0 631.5 95 100 5.3 272 282 3.5 105 108 2.8 678.7 102 107 4.9 292 304 3.9 115 116 0.9 711.7 107 112 4.7 312 319 2.2 120 121 0.8 745.6 112 117 4.5 327 334 2.1 130 127 -2.4 808.4 122 127 4.1 342 362 5.5 135 138 2.2 Note: p is load, ui is test deformation, usi is FEA calculation deformation, (usi-ui)/ui is ratio of FEA results to test results, and i=1, 2, 3 represents the test points D1, D2, and D3.According to Table2 it can be declared that the finite element model can simulate the bridge accurately; the errors were below 5%, excepting a few points.


## 3.3. Eigenvalue Buckling Analysis

The typical equation for eigenvalue buckling analysis was [26](45)K0+λKσD=0where K0 is the matrix of elastic rigidity; Kσ is the matrix of stress rigidity; D is the characteristic displacement vector; and λ is the eigenvalue.Without considering self-weight, the eigenvalueλ obtained from eigenvalue buckling analysis multiplied by the load applied is the eigenvalue buckling load. In fact, the load kept changing, so that(46)K0+λKσ=0.When the self-weight was not considered, the first-order eigenvalue buckling mode was obtained, as shown in Figure 10. The first-order eigenvalue buckling load was 2405.4 kN, which was 4.6% larger than the 2300.6 kN load that was derived by the thin-walled beam equivalent theory in this study. This suggests a highly accurate value of the eigenvalue buckling load of the movable high-strength inverted-triangular steel bridge was derived in this study.Figure 10
First-order eigenvalue buckling mode of the movable high-strength inverted-triangular steel bridge.


## 4. Nonlinear Ultimate Bearing Capacity of the Bridge

### 4.1. Theories on Nonlinear Analysis

Values of bridge’s elastic ultimate bearing capacity obtained by theoretical derivation and eigenvalue buckling analysis were usually so large that they could only be used for qualitative analysis of the structure’s load-carrying capacity instead of genuinely reflecting it. Therefore, nonlinear buckling analysis was required. Considering factors such as the structure’s geometric defect and the nonlinearity of materials, a nonlinear buckling analysis was able to more truthfully present the structure’s load-carrying capacity. In consideration of the material-geometry dual nonlinearity, the basic buckling equation could be converted into(47)K0+Kσ+KεU=Pwhere K0 is the small-displacement elastic stiffness matrix, Kσ is the initial stress stiffness matrix, Kε is the large displacement elastic stiffness matrix, also known as the initial strain matrix, and U and P are nodal displacement and load vector, respectively. Solutions were found by means of the iteration method by gradually applying load increments. Structure stiffness changed as load rose. The structure reached its maximum load-carrying capacity when detK0+Kσ+Kε=0. Ultimate bearing capacity was tracked using the arc-length method.


### 4.2. Ultimate Bearing Capacity of the Bridge

The geometric deflection was introduced to the analysis model by the consistent-deflection-mode method, which simulates deflection distribution by the lowest buckling mode shape. The maximal displacement of buckling analysis was multiplied by a coefficient to amplify to the 1/1000 span of bridge.Based on the above finite element model and considering material and geometric dual nonlinearity factors, the stability ultimate load of the bridge can be obtained by using the arc-length method [16]. Firstly, an elastic buckling analysis was performed when the load was 600 kN, and then the elastic buckling load, which was the top limit load of bridge, was found. Secondly, setting the elastic buckling load as the initial load, the accumulated loads were then introduced gradually until the solution was emanative. The load-displacement curve can be traced through time post-process (post 26). The highest point of the load-displacement curve reveals the stability ultimate load capacity of a movable high-strength inverted-triangular steel truss bridge. The load-displacement curves are shown in Figure 11.Figure 11
Load-displacement at middle span. (a) Load-vertical. (b) Load-lateral.
(a) (b)From Figure11 we can obtain the stability ultimate load capacity (Pu) of the bridge, which is 1637.6 kN. The load-displacement curve is approximately linear, and the vertical displacement is bigger than the lateral displacement before the load arrives at the ultimate load capacity. Therefore, the bridge’s deformation is mainly in the moment plane; it is a kind of bending buckling when the load is small. When the load arrives at the stability ultimate load capacity of the bridge, the lateral deformation becomes bigger, and the bridge exhibits true bending-torsion buckling.


## 4.1. Theories on Nonlinear Analysis

Values of bridge’s elastic ultimate bearing capacity obtained by theoretical derivation and eigenvalue buckling analysis were usually so large that they could only be used for qualitative analysis of the structure’s load-carrying capacity instead of genuinely reflecting it. Therefore, nonlinear buckling analysis was required. Considering factors such as the structure’s geometric defect and the nonlinearity of materials, a nonlinear buckling analysis was able to more truthfully present the structure’s load-carrying capacity. In consideration of the material-geometry dual nonlinearity, the basic buckling equation could be converted into(47)K0+Kσ+KεU=Pwhere K0 is the small-displacement elastic stiffness matrix, Kσ is the initial stress stiffness matrix, Kε is the large displacement elastic stiffness matrix, also known as the initial strain matrix, and U and P are nodal displacement and load vector, respectively. Solutions were found by means of the iteration method by gradually applying load increments. Structure stiffness changed as load rose. The structure reached its maximum load-carrying capacity when detK0+Kσ+Kε=0. Ultimate bearing capacity was tracked using the arc-length method.


## 4.2. Ultimate Bearing Capacity of the Bridge

The geometric deflection was introduced to the analysis model by the consistent-deflection-mode method, which simulates deflection distribution by the lowest buckling mode shape. The maximal displacement of buckling analysis was multiplied by a coefficient to amplify to the 1/1000 span of bridge.Based on the above finite element model and considering material and geometric dual nonlinearity factors, the stability ultimate load of the bridge can be obtained by using the arc-length method [16]. Firstly, an elastic buckling analysis was performed when the load was 600 kN, and then the elastic buckling load, which was the top limit load of bridge, was found. Secondly, setting the elastic buckling load as the initial load, the accumulated loads were then introduced gradually until the solution was emanative. The load-displacement curve can be traced through time post-process (post 26). The highest point of the load-displacement curve reveals the stability ultimate load capacity of a movable high-strength inverted-triangular steel truss bridge. The load-displacement curves are shown in Figure 11.Figure 11
Load-displacement at middle span. (a) Load-vertical. (b) Load-lateral.
(a) (b)From Figure11 we can obtain the stability ultimate load capacity (Pu) of the bridge, which is 1637.6 kN. The load-displacement curve is approximately linear, and the vertical displacement is bigger than the lateral displacement before the load arrives at the ultimate load capacity. Therefore, the bridge’s deformation is mainly in the moment plane; it is a kind of bending buckling when the load is small. When the load arrives at the stability ultimate load capacity of the bridge, the lateral deformation becomes bigger, and the bridge exhibits true bending-torsion buckling.


## 5. Stability Ultimate Load Influence Factors Analysis

There are many factors which can affect the stability ultimate load capacity of a movable high-strength inverted-triangular steel truss bridge, such as bridge span, bridge height, track width, and member’s section. In order to study the influence of these factors on the stability ultimate capacity of the bridge, many FEA (finite element analysis) models with different parameters were analyzed.


### 5.1. Influence of Height and Track Width

The height and track width are two important parameters for design. When the analysis was performed, one parameter was changed each time. Through FEA analysis the stability ultimate load capacityPu was obtained. Based on the actual bridge’s ultimate load capacity Pu (which is 1637.6 kN) the change percentage can be determined Pb-Pu/Pu×100. The results are shown in Figure 12.Figure 12
Change percentage of the bridge’s ultimate load capacity when track width varies.From Figure12 a conclusion can be made that the height has more influence on the stability ultimate load capacity than the track width. The reason for this is that the bridge’s height has more influence on section bending stiffness than track width does. Meanwhile, the track width changes but the whole bridge’s width does not change; the lateral torsion stiffness changes a little. Thus, track width has less influence on the stability ultimate load capacity of the bridge.


### 5.2. Influence of Chord

The top and bottom chord are the important members in the bridge. How a chord influences the stability ultimate load was also studied in this paper. The top chord has two different box sections, which were labeled chord I and chord II. The top chord area changed at the same time of the width and height changes of the section. The multiplied area of the top chord can be found through adjusting the width and height of section gradually. The bottom chord is a kind of complex section. The multiplied area of the bottom chord can be found through adjusting the main dimensions of the section gradually. The results are shown in Figure13.Figure 13
Change percentage of the bridge’s ultimate load capacity when chord area varies.From Figure13 it can be concluded that the top chord has more influence on the stability load capacity than the bottom chord. The top chord I has more influence than chord II. The bottom chord has a little influence on the stability load capacity that changes below 5%. Even top chord I has more influence than height when both two parameters change by the same percentage, as shown by comparing Figure 12 with Figure 13.


### 5.3. Influence of Web Member

The web member mainly bears shear force in the bridge structure but also provides some bending stiffness. There are two different web members shown in Figure1, the sections of which are box shapes. Thus, the influence of the web member on the stability ultimate load was studied using the same method as the chord. The results are shown in Figure 14.Figure 14
Change percentage of the bridge’s ultimate load capacity when web member area varies.According to Figure14, the web member has a little influence on the stability ultimate load capacity of the bridge. The influence of the upright web member is more than the slanted web member.


## 5.1. Influence of Height and Track Width

The height and track width are two important parameters for design. When the analysis was performed, one parameter was changed each time. Through FEA analysis the stability ultimate load capacityPu was obtained. Based on the actual bridge’s ultimate load capacity Pu (which is 1637.6 kN) the change percentage can be determined Pb-Pu/Pu×100. The results are shown in Figure 12.Figure 12
Change percentage of the bridge’s ultimate load capacity when track width varies.From Figure12 a conclusion can be made that the height has more influence on the stability ultimate load capacity than the track width. The reason for this is that the bridge’s height has more influence on section bending stiffness than track width does. Meanwhile, the track width changes but the whole bridge’s width does not change; the lateral torsion stiffness changes a little. Thus, track width has less influence on the stability ultimate load capacity of the bridge.


## 5.2. Influence of Chord

The top and bottom chord are the important members in the bridge. How a chord influences the stability ultimate load was also studied in this paper. The top chord has two different box sections, which were labeled chord I and chord II. The top chord area changed at the same time of the width and height changes of the section. The multiplied area of the top chord can be found through adjusting the width and height of section gradually. The bottom chord is a kind of complex section. The multiplied area of the bottom chord can be found through adjusting the main dimensions of the section gradually. The results are shown in Figure13.Figure 13
Change percentage of the bridge’s ultimate load capacity when chord area varies.From Figure13 it can be concluded that the top chord has more influence on the stability load capacity than the bottom chord. The top chord I has more influence than chord II. The bottom chord has a little influence on the stability load capacity that changes below 5%. Even top chord I has more influence than height when both two parameters change by the same percentage, as shown by comparing Figure 12 with Figure 13.


## 5.3. Influence of Web Member

The web member mainly bears shear force in the bridge structure but also provides some bending stiffness. There are two different web members shown in Figure1, the sections of which are box shapes. Thus, the influence of the web member on the stability ultimate load was studied using the same method as the chord. The results are shown in Figure 14.Figure 14
Change percentage of the bridge’s ultimate load capacity when web member area varies.According to Figure14, the web member has a little influence on the stability ultimate load capacity of the bridge. The influence of the upright web member is more than the slanted web member.


## 6. Conclusions

This paper investigated the stability of a movable high-strength inverted-triangular steel bridge by the theoretical, experimental, and finite element methods. From this study the following conclusions can be drawn:( 1 ) The movable high-strength inverted-triangular steel bridge can be modeled as equivalent to a thin-walled triangular beam according to the stiffness equivalence principle. The equivalent rigidities of the end units which are variable and the central units rigidity which remains unchanged were calculated.( 2 ) The equation which can calculate the elastic stability critical load of the inverted-triangular steel bridge with variable rigidities at both ends was put forward. This equation was well agreed with existing theoretical calculation. The calculated elastic stability load of the bridge was 2300.6 kN by substituting equivalent rigidity into the equation.( 3 ) The material performance test of BS700 (σ0.2= 700 MPa) high-strength steel was carried out, and the material properties used for finite element analysis were obtained. The load capacity test of the movable high-strength inverted-triangular steel bridge whose length is 51m was performed.( 4 ) A finite element model for the elastic buckling analysis of the movable high-strength inverted-triangular steel bridge was established. The finite element simulated results well agreed with field test results. The elastic buckling critical load of the bridge is 2405.4 kN, which further verifies the accuracy of the equation of the bridge’s elastic stability critical load proposed in this study.( 5 ) The nonlinear buckling analysis model was put forward with consideration of material and geometric nonlinearity. The material model made use of the multilinear hardening model. The geometric deflection was introduced to the analysis model by the consistent-deflection-mode method. According to calculations, the bridge’s ultimate bearing capacity is 1637.6 kN which is lower than the elastic stability critical load.( 6 ) Analysis of factors influencing the bridge’s ultimate bearing capacity suggests that top chord and bridge height are two major factors, of which top chord 1 exerts the dominant influence. Track width, bottom chord, and web member have less influence on the bridge’s ultimate bearing capacity.



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*Source: 1568629-2018-09-19.xml*