Cambering structural steel I-girders using cold bending

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ARTICLE IN PRESS

Journal of Constructional Steel Research xx (xxxx) xxx–xxxwww.elsevier.com/locate/jcsr

Cambering structural steel I-girders using cold bending

Antoine N. Gergessa,∗, Rajan Senb

a Department of Civil Engineering, University of Balamand, Lebanonb Department of Civil and Environmental Engineering, University of South Florida, Tampa, FL, United States

Received 16 February 2007; accepted 2 October 2007

Abstract

Cambering is often required in structural steel beams to compensate for dead load deflections. In this process, permanent deformations areinduced in the girder after it is fabricated in order to match a required vertical profile under service loads. Traditionally, cambering is achieved byapplication of heat (heat cambering) or force (cold cambering). As in heat curving, heat cambering is time-consuming and costly. Cold camberingis faster and is most commonly used by fabricators. However, in the absence of data relating loads to deformations, the process is based on trialand error and relies on the fabricator’s skills and expertise. Recently, closed form equations and fabrication aids were derived for cold curvingbased on an available proprietary cold bending system. This paper extends this analysis to cambering where a girder is bent about its strong axiscompared to weak axis bending in curving. New equations are derived and guidelines proposed that set limits on the maximum load and permanentresidual strains to ensure they are within acceptable norms and prevent local failure. A comprehensive numerical example is included to illustratethis procedure.c© 2007 Elsevier Ltd. All rights reserved.

Keywords: Cambering; Cold bending; Inelastic; Plastic; Rolled shape; Plate girder; Example; Standardized procedure; Buckling

1. Introduction1

Cambering is usually applied to steel girders used in floor2

construction to compensate for deflections due to permanent3

dead loads (weight of the concrete topping and superimposed4

dead loads) and part of the live load. The deflection curve for5

these uniform loads is approximated as a parabola. Cambering6

consists of developing a reverse parabola by curving the girder7

about its major axis in the opposite direction. It is usually8

defined by the maximum ordinate at mid-span (Fig. 1).9

Steel girders are generally cambered by the application10

of heat (heat cambering) or force (cold cambering) or their11

combination [1,2]. Heat cambering is accomplished by heating12

segments of the web at intervals along the length of the13

girder. Its main drawback is that it is a trial and error14

process since it is impossible to verify the camber until15

the beam has cooled. Cold cambering is accomplished by16

inducing permanent deformations in the girder by applying17

mechanical forces using specialized equipment. The design of18

the equipment relies on the ingenuity of the steel fabricator. The

∗ Corresponding author. Tel.: +961 3 960291; fax: +961 6 930238.E-mail address: [email protected] (A.N. Gergess).

most popular ones consists of a rigid frame in which the girder 19

is mounted horizontally with loads applied to one side of the 20

flange using a single ram (http://www.voortman.net), or double- 21

ram system (http://www.oceanmachinery.com) while the other 22

side of the flange is supported (Fig. 2(a) and (b)). The single 23

ram is recommended for short span girders (up to 5m in length) 24

(Figs. 2(a) and 3(a)) and the two-ram system for spans > 5 m. 25

In the two-ram system, the rams are spaced approximately at 26

one third the distance between supports as it results in a camber 27

curve that is closer to a parabola (Figs. 2(b) and 3(b)). 28

As cambering is more related to fabrication than analysis, 29

there is little published analytical information. The AISC 30

Manual [3] sets limits on camber as a function of the section 31

shape and length as follows: (a) the minimum radius for camber 32

to be induced by cold bending in members up to a nominal 33

depth of 75 cm is between 10–14 times the depth of the 34

member and (b) the practical length of the member to be 35

cambered is in the range of 12–15 m [3]. Cambering cost 36

and the need for compatibility with the fabricator’s equipment 37

are addressed in Downey’s paper [4]. Earlier papers discussed 38

procedures for cambering [1] and loads for which camber 39

must be specified [2]. Recently, Bjorhovde published a state- 40

of-the-art paper on cold bending that also reviewed procedures, 41

0143-974X/$ - see front matter c© 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.jcsr.2007.10.001

Please cite this article in press as: Gergess AN, Sen R. Cambering structural steel I-girders using cold bending. Journal of Constructional Steel Research (2007),doi:10.1016/j.jcsr.2007.10.001

http://www.elsevier.com/locate/jcsr

mailto:[email protected]

http://www.voortman.net

http://www.oceanmachinery.com

http://dx.doi.org/10.1016/j.jcsr.2007.10.001

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ARTICLE IN PRESS2 A.N. Gergess, R. Sen / Journal of Constructional Steel Research xx (xxxx) xxx–xxx

Nomenclature

a: distance from end of girder to ram (two-ramsystem)

a′: distance from support to ram (two-ram system)b: distance from end of girder to supportsb f : flange widthc: half flange widthd: girder depthh: clear distance between flangesE : modulus of elasticityFy : yield stressIx : strong axis moment of inertiak: distance from outer face of the flange to the web

toel: unbraced lengthL: girder’s lengthL ′: load frame spacingM : bending momentN : length of bearing (not less than k) for beam

reactionP: plastic load at which plastic hinging develops —

single ramPfinal: final (increased) plastic load — single ramP ′: plastic load at which plastic hinging develops —

double ramP ′

final: final (increased) plastic load — double ramP ′

mod: inelastic load at which yield occurs withindistance a′ — double ram

R: radius of curvaturet f : flange thicknesstw: web thicknessx2: distance from support to point where yielding is

initiatedy: offset along girder depthα: constant developed for derivation of plastic

deformationβ: constant developed for derivation of plastic

deformationδcamber: camber within supports (distance L ′)∆e: mid-span elastic deformation∆final: mid-span plastic deformation due to increased

loads∆p: mid-span plastic deformation∆res: mid-span residual deformationε: strain along girder depthεmax: maximum strainεres: residual strain (Fig. 5)εy : yield strainη: ratio of maximum load to plastic load (Tables 2

and 3; Fig. 6(a), (b))κ: ratio of maximum deformation to plastic defor-

mation (Tables 2 and 3; Fig. 6(a), (b))

Fig. 1. Induced camber in a structural steel girder.

limitations and guidelines on straightening, cambering and 1

curving of wide-flange shapes used in building structures [5]. 2

A detailed study was completed recently for an innovative 3

cold bending system used for horizontally curving steel 4

girders [6–9]. This system used a single-ram mobile loading 5

frame in which bending loads were applied at different 6

locations along the girder’s length. The main goal of this paper 7

is to extend this analysis to provide new solutions relating 8

applied loads to required camber at mid-length for symmetrical, 9

unstiffened steel I-girders (Fig. 4). It also addresses possible 10

adverse effects and specifies measures to prevent localized 11

flange/web buckling and overall lateral torsional buckling. 12

The same procedure for cold curving is adopted excepting 13

that: (1) bending is about the major axis of the girder rather 14

than its weak axis and, (2) the bending equipment and the girder 15

are stationary, e.g. the cambering curve develops based on a 16

one-load application (Figs. 2 and 3). As loads will be higher, 17

large uniaxial compressive stresses are introduced in the web 18

which makes it susceptible to localized failure due to buckling. 19

Therefore, limits on the maximum load and permanent residual 20

strains for both single- (Fig. 2(a)) and double-ram systems 21

(Fig. 2(b)) are specified and preventive measures such as 22

flange and/or web restraint elements introduced. Finally, a 23

comprehensive illustrative numerical example is included for 24

the single- and double-ram systems. 25

2. Problem statement 26

In cold bending, inelastic loads and corresponding residual 27

deformations are calculated based on plastic hinging at mid- 28

span (point of maximum moment). Theoretically, this is the 29

moment at which the section becomes fully plastic with 30

longitudinal flexural stresses across the girder’s section equal 31

to the yield stress Fy . In cold curving, [6], bending load limits 32

were set based on plastic hinging. In cambering, the increased 33

stiffness due to strong axis bending requires cambering loads to 34

be much larger than curving loads and rams of higher capacity 35

(sometimes impractical to find) are required. In order to reduce 36

the loads, the following set-up is required: 37

1. Use a multiple-ram system to reduce the individual ram 38

loads. A two-ram system is analysed in this paper (Fig. 3(b)). 39

2. Allow the cambering curve (approximated as a parabola, 40

Fig. 1) to develop in one load application (Fig. 2(a) and (b)) 41

by maximizing the spacing between the end supports (the 42

larger the span length, the smaller the load). 43

3. Allow increase in the plastification of the girder during 44

cambering, e.g. yield propagation along the length of the 45

member by gradually increasing the applied loads beyond 46


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ARTICLE IN PRESSA.N. Gergess, R. Sen / Journal of Constructional Steel Research xx (xxxx) xxx–xxx 3

(a) Single-ram system.

(b) Two-ram system.

Fig. 2. Layout (plan-view) of the frame of the cambering equipment.

(a) Single-load (ram) system. (b) Multiple-loads (two-ram) system.

Fig. 3. Single/double-ram systems — plan view.

the plastic load limit without damaging the girder or1

adversely affecting the ductile properties of structural steel.2

3. Approach3

The goal of the analysis is to derive closed form solutions4

relating loads to desired camber for cold cambering procedures5

commonly used by fabricators [5]. The outline of the analysis6

is as follows:7

1. Define the process, e.g. girder layout, load configuration and8

parameters.9

2. Develop closed form solutions that relate plastic loads to10

mid-length deformations based on major axis bending and11

plastic hinging at mid-length for the single (Fig. 3(a)) and12

double-ram (Fig. 3(b)) systems, for symmetrical, unstiffened13

structural steel girders (Fig. 4).14

3. Determine the increase in the magnitude of the plastic15

loads for which the desired camber is achieved based on16

the maximum strain in the plastic range of 10–20 times17

Fig. 4. Symmetrical steel girder cross-section.

the yield strain εy , (typical limit for mild steel before the 18

strain-hardening range) [1,3,10] (Fig. 5). The final loads are 19

obtained as a function of the plastic loads using multipliers 20

established from a parametric study conducted using finite 21

elements (analytical solutions are extremely difficult to 22

develop in the postplastic range). 23


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Fig. 5. Idealized stress–strain curve — loading and unloading.

4. Identify possible adverse effects of cold bending such as1

localized flange/web buckling, and overall lateral torsional2

buckling of the shape. Propose appropriate checks [3] and3

fabrication measures to prevent their occurrence.4

4. Cold-cambering system5

Cold cambering equipment commonly used by fabricators6

consists of a rigid steel frame in which the rams (hydraulic7

jacks) are horizontally mounted on one side and a rigid reaction8

beam supported on two points placed on the opposite side9

(Fig. 2). The reaction points (supports) can be adjusted to suit10

the girder’s length.11

4.1. Layout12

The girder is placed horizontally (e.g. self-weight neglected)13

in the steel frame and loads applied until the desired camber is14

achieved. A steel plate (or wide-flange) of sufficient length is15

provided between the hydraulic jacks and the flange plate in16

order to prevent localized flange and web buckling (Fig. 2(a)17

and (b)). The layout of the cambering equipment for a single-18

ram and a double-ram system are shown in Fig. 2(a) and19

(b). The single-ram equipment is similar to the double-ram20

equipment excepting that there is only one hydraulic jack at21

mid-span.22

4.2. Cambering procedure23

The steel girder of length L is placed horizontally with the24

end supports placed at distance b from the ends of the girder25

(Fig. 3). The actual span length between end supports (that is26

the spacing of the load frame bending system) is L ′= L − 2b.27

b is usually set at 0.1L so L ′= 0.8L . Bending loads consist of28

a concentrated load at mid-span for the single ram (Fig. 3(a))29

(defined as P) and two equally spaced concentrated loads for30

the double rams (Fig. 3(b)) (defined as P ′ each). As loads31

induce permanent deformations, camber is checked by using32

a stringline strung between the girder’s ends (shown as a dotted33

line in Fig. 1).34

5. Analytical solution 35

Plastic loads and induced deformations are determined based 36

on inelastic analysis [11,12] using engineering mechanics 37

principles. In cambering, stiffness of the girder is based on 38

strong axis moment of inertia[

Ix ≈ tw d3

12 + t f cd2]

(Fig. 4). 39

The driving points of the analytical solution are listed as 40

follows: 41

1. Inelastic load P is determined by equating the externally 42

applied moment at mid-span to the moment that develops 43

due to internal stresses for a symmetrical steel plate 44

girder of web depth d, thickness tw, flange width 2c 45

and thickness t f (Fig. 4). The external moment is equal 46

to P L ′

4 (Fig. 3(a)) for a single ram and P ′ [a − b] = 47

P ′a′, where a′= [a − b] (Fig. 3(b)) for double rams. 48

Based on a fully plastic moment at mid-span (stress 49

equal to the yield stress Fy across the girder’s depth), 50

the moment that develops due to internal stresses is: 51

2Fy

(t f × 2c ×

(d+t f

2

)+ tw ×

( d2

)×

( d4

)). Note that self- 52

weight moment is neglected since the girder is mounted 53

horizontally. 54

2. Neglecting the flange thickness t f with respect to the girder 55

depth d leads to inelastic loads P and P ′ determined from 56

Eqs. (1a) and (1b) as follows: 57

Single Ram: P =8Fyd

L ′

(ct f +

twd

8

)(1a) 58

Double Ram: P ′=

2Fyd

[a − b]

(ct f +

twd

8

). (1b) 59

3. Inelastic loads P and P ′ (Fig. 1) introduce plastic 60

deformations at mid-span noted as ∆p. Deformation ∆p is 61

determined from double integration of curvature(

1R

), where 62(

1R

)is equal to strain variation

(εy

)in the inelastic range 63

and flexural moment to stiffness ratio(

ME Ix

)in the elastic 64

range [11,12]. 65

4. ∆p is expressed by Eq. (2) for single ram and Eq. (4) for 66

double rams in terms of constants α, β, C1, C2, C ′

1, C ′

2, C ′

3 67

and C ′

4 developed for ease of calculations: 68

Single-ram 69

∆p = C1L ′

2+ C2 (2) 70

where 71

C1 =P

48E Ix

[3L ′2

− 12x22

]+

2Fy

Eb(α − βx2)

0.572

C2 =Px2

48E Ix

[3L ′2

− 4x22

]− C1x2 −

4Fy

3Eβ2 (α − βx2)1.5

73

α =

(3d2

4+

6t f cd

tw

), β =

12tw L ′

(t f cd +

twd2

8

), 74

x2 =4Fy Ix

Pd. 75


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(Fig. 3(a) and (b))

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Double-ram1

In the double-ram system, the moment (P ′a′) that devel-2

ops due to plastic load P ′ (Eq. (1b)) is constant between3

the points of load application (Fig. 3(b)). For P ′, a state4

of constant yield stress Fy develops along the girder depth,5

within distance (L ′− 2a′). This leads to formation of a se-6

ries of plastic hinges e.g. instability compared to one plas-7

tic hinge that develops at mid-length only for single ram8

(Fig. 3(a)). Analytically, closed form solutions due to P ′9

from Eq. (1b) are difficult to derive. Consequently, they are10

developed for inelastic loads smaller than P ′ (this leads to11

elasto-plastic stress variation across the girder depth). New12

load P ′

mod < P ′ is introduced and selected so that yielding is13

initiated within distance a′ (Fig. 3(b)), determined from Eq.14

(3) as follows:15

P ′

mod =2Fy Ix

x2d(3)16

where 0 < x2 ≤ a′.17

The corresponding plastic deformation is now determined18

from Eq. (4) as:19

∆p =Fy L ′2

8E√

α − βa′+ C ′

3L ′

2+ C ′

4 (4)20

where21

C ′

1 =P ′

mod

2E Ix

[L ′a′

− a′2− x2

2

]+

2Fy

Eβ(α − βx2)

0.522

C ′

2 =P ′

modx2

6E Ix

[3L ′a′

− 3a′2− x2

2

]− C ′

1x2

−4Fy

3Eβ2 (α − βx2)1.5

23

C ′

3 = C ′

1 −Fya′

E√

α − βa′−

2Fy

Eβ

(α − βa′

)0.524

C ′

4 = C ′

1a′+ C2 − C3a′

−Fya′2

2E√

α − βa′

+4Fy

3Eβ2

(α − βa′

)1.525

α =

(3d2

4+

6t f cd

tw

), β =

3P ′

mod

tw Fy, x2 =

2Fy Ix

P ′

modd.26

5. The corresponding residual deformation ∆res is obtained by27

subtracting the elastic deflection during unloading ∆e [3]28

from the plastic deflection ∆p. ∆res should be compared29

against the camber that develops within the load frame30

supporting ends (Fig. 1) and is calculated from Eq. (5)31

as:32

∆res = ∆p − ∆e (5)33

where34

∆e =P L ′3

48E Ixfor single ram35

∆e =P ′

moda′

24E Ix

[3L ′2

− 4a′2]

for double ram.36

6. Maximum limits 37

In cold curving, the maximum loads were capped by the 38

plastic moment capacity of the steel girder at mid-length [6]. 39

This provided enough bending capacity for curving since: (1) 40

the girder was bent about its weak axis, (2) bending load 41

was distributed due to multiple load applications at different 42

intervals along the girder’s length and (3) the curving operation 43

is sometimes achieved in more than one pass [6]. 44

In cambering, the loading concept is similar, but strong axis 45

bending requires much larger inelastic loads than in curving, 46

which might be impractical for the steel fabricator in the 47

steel shop because of the large capacity of the jack required. 48

Therefore, bending loads should be kept to a minimum in order 49

to minimize possible adverse effects such as web buckling 50

under large axial strains and subsequent overstraining and 51

fracture of the steel in the web-flange area. As mentioned 52

earlier, this is usually accomplished by: (1) optimizing the 53

span length and developing the camber in a single bending 54

operation and (2) using a two-ram system. Still, the residual 55

deformation calculated by subtracting the elastic deformation 56

from the plastic deformation, is usually much smaller than 57

the required camber and this necessitates multiple applications 58

of the bending loads to achieve the desired camber that is 59

time consuming for fabricators. One way of overcoming this 60

problem is by taking advantage of the stress–strain properties 61

(Fig. 5) of mild ductile steel by increasing the bending loads 62

beyond the plastic limit where maximum strains are still within 63

the flat plateau of the stress–strain curve [1,10], without altering 64

the properties of steel. However, this necessitates taking extra 65

measures against possible local failure by conducting checks 66

for the shape of the girders for web compression buckling (Eq. 67

(6a)), web sidesway buckling (Eq. (6b)), web crippling (Eq. 68

(6c)), web local yielding (Eq. (6d)) and flange local bending 69

(Eq. (6e)) [3,5]. Therefore, the upper bound for the magnitude 70

of the bending loads (noted as Ψ ) is determined based on 71

service loads as the smallest value from Eqs. (6a)–(6e) [3] 72

(note that the coefficients of the equations differ from AISC 73

for metrification and the numbers in the denominators indicate 74

the service load safety factor): 75

Ψ =2.4t3

w

√E Fy

1.67h(6a) 76

Ψ =331,000t3

wt f

1.76h2

[1 + 0.4

(h/twl/b f

)3]

(6b) 77

Ψ =0.08t2

w

2

[1 + 3

(N

d

) (twt f

)1.5] √

E Fy t f

tw(6c) 78

Ψ =0.1 (5k + N ) Fy tw

1.5(6d) 79

Ψ =0.625Fy t2

f

1.67(6e) 80

where Ψ is in (kN), E (MPa), Fy (MPa), tw (cm), t f (cm), h 81

(cm), d (cm), b f (cm), l (cm), N (cm), k (cm). For definition of 82

terms, refer to Nomenclature. 83


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Table 1W-shapes analysed in finite element model

W-shape

1000 ×

883920 ×

381760 ×

220610 ×

92410 ×

60310 × 74

b f2t f

2.6 3.5 4.4 6.0 6.9 6.3htw

19.1 33.8 41.5 50.1 46.6 26.2

When the required strength of the web exceeds the above1

limits, restraining elements such as stiffeners or lateral bracing2

should be provided at the load point and plastic hinge locations.3

Those are left to the discretion of the fabricator.4

Current practice in the steel shop is based on trial and error5

for determining the magnitude of the bending load, e.g. the6

loads are applied, released and reapplied until the required7

camber is within fabrication tolerances. Theoretically, if the8

order of magnitude of the bending load beyond the plastic loads9

from Eqs. (3) and (1a) is predicted, it reduces the extent of the10

trial and error operation.11

Structurally, once a plastic hinge forms at mid-length,12

additional deformation is induced due to yield propagation13

along the length of the member after section plastification has14

occurred, due to a finite increase in the applied loads. Such15

effects beyond the plastic load limit are determined based on16

a parametric study performed using MSC/NASTRAN finite17

element model [13]. It is noted in the finite element analysis that18

beyond plastic hinging, deformations become sensitive to loads,19

e.g. small increases in the plastic loads induce disproportionate20

increases in deformation.21

7. Finite element model22

The purpose of the finite element model was to assess the23

effect of increase in loads beyond the plastic limit on induced24

deformations. Consequently, the analysis was performed for25

full-sized W-shape girders. Various girder dimensions with26

different flange width to thickness ratio(

b ft f

)and web depth27

to thickness ratio(

dtw

)were used in the model based on Grade28

250 (Fy = 250 MPa) and Grade 345 (Fy = 345 MPa) steel (for29

girder details, refer to Table 1).30

In the model, four-noded isoparametric plate elements 31

(2.5 cm × 2.5 cm elements for the flanges and 2.5 cm × 5 cm 32

elements for the web) built-in MSC/NASTRAN finite element 33

software was used [13]. At the supporting ends of the load 34

frame, the displacement along the web depth was restrained. 35

Lateral displacement and rotations were restrained at a single 36

node (point of application of the load) for stability purposes. 37

At the supporting end locations and at the point of application 38

of the concentrated load, stiff beam elements were provided 39

along the flange width in order to limit stress concentration and 40

deformations. 41

In the models, the W shapes used are listed in Table 1. One of 42

the W shapes used was a W410 × 60 girder. The frame spacing 43

L ′= (L−2b) (Fig. 3) was set at 8 m and the girder length (L) at 44

10 m. Single- and double-ram loads were applied to the girder. 45

An illustrative numerical example for this girder is provided 46

later on. 47

Results are summarized in Table 2 and Fig. 6(a) (single ram) 48

and Table 3 and Fig. 6(b) (double ram) in the form of multipliers 49

that show the increase in loads and corresponding increase in 50

deformations and strains. The ratio of the increased load beyond 51

the plastic limit to the plastic load is noted as η =Pfinal

P =P ′

finalP ′

mod. 52

The ratio of the induced deformation due to increased load to 53

the plastic deformation is noted as κ =∆final∆p

. In addition, max- 54

imum strains εmax are provided function of the yield strain εy . 55

Multipliers in Tables 2 and 3 are averaged values. It should 56

be noted that the difference in values for different size girders 57

is not critical. This is attributed to the fact that additional 58

deformations are induced after cross-section plastification is 59

initiated, e.g. a finite change in the applied load causes a 60

significant change in deformation. It may be seen from Table 2 61

and Fig. 6(a) that a 5% increase in the applied plastic load (P 62

for single ram) doubles the plastic deformation while a 10% 63

increase quadruples it. The increases are somewhat smaller 64

for two-ram systems (Table 3 and Fig. 6(b)). Here a 5% 65

increase the plastic loads (P ′

mod for double rams) increases the 66

plastic deformation by 25% while a 10% increase induces a 67

deformation equal to 2.25 times the plastic deformation. This 68

is a characteristic of ductile steel that enables it to continue to 69

deform at yield (deformation is of the order of 10 to 20 times 70

Table 2Bending loads multipliers (unstiffened girders) — single-ram system

Parametersη =

PfinalP η = 1.01 η = 1.025 η = 1.05 η = 1.075 η = 1.10

κ =∆final∆p

κ = 1.05 κ = 1.40 κ = 2.00 κ = 2.87 κ = 4.00εmaxεy

εmaxεy

= 2.5 εmaxεy

= 5.0 εmaxεy

= 10.0 εmaxεy

= 15.0 εmaxεy

= 20.0

Table 3Bending loads multipliers (unstiffened girders) — double-ram system

Parameters

η =P ′

finalP ′

modη = 1.01 η = 1.025 η = 1.05 η = 1.075 η = 1.10 η = 1.125

κ =∆final∆p

κ = 1.025 κ = 1.075 κ = 1.25 κ = 1.50 κ = 2.25 κ = 5.00εmaxεy

εmaxεy

= 1.1 εmaxεy

= 1.2 εmaxεy

= 1.5 εmaxεy

= 2.0 εmaxεy

= 3.0 εmaxεy

= 7.5


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(a) Single-ram system. (b) Double-ram system.

Fig. 6. Bending loads/deformations multipliers (unstiffened girders).

Fig. 7. Deformed shapes due to applied loads.

the yield deformation, Fig. 5) without adversely affecting its1

material properties.2

The larger increase in plastic deformations for the single-ram3

is due to the fact that its plastic load (Eq. (1a)) was determined4

based on plastic hinging at mid-span while for double ram, it5

was determined based on a modified inelastic load (Eq. (3))6

smaller than the plastic load (Eq. (1b)). The increase in the7

corresponding strains was also more noticeable for single ram,8

since the deformed shape develops as two straight lines with9

respect to the plastic hinge at mid-length, while in the double-10

ram system it develops as a smooth parabola between points11

of load applications (Fig. 7) (note that strains are determined12

function of curvature 1/R).13

8. Procedure for cambering14

The procedure for cambering based on the analysis is15

outlined as follows for a steel girder Grade Fy , length L , web16

depth d and thickness tw, flange width 2c and thickness t f , and17

strong axis moment of inertia Ix (Fig. 4).18

Step 1. Setup the load frame criteria19

The load frame spacing L ′ is usually set at 0.8L20

(Fig. 3) e.g. b ≈ 0.1L , that ensures a smooth21

parabolic curve develops (Fig. 1).The single-ram22

system is recommended for load frame spacing L ′ <23

5 m. For larger spacing, the double-ram system is24

recommended. In the double-ram system, the rams are 25

set at 1/3L ′ from the supporting ends ⇒ a′≈ 0.33L ′. 26

Step 2. Determine the magnitude of the plastic bending loads 27

From Eqs. (3) and (1a), determine the magnitude 28

of the plastic loads P (for single ram) and P ′

mod for 29

double ram: 30

Single ram: P =8Fyd

L ′

(ct f +

twd

8

)31

Double ram: P ′

mod =2Fy Ix

x2d32

where 0 < x2 ≤ a′ (usually x2 is set equal to a′). 33

Step 3. Determine the magnitude of the corresponding plastic 34

deformation 35

From Eqs. (2) and (4), determine the magnitude of 36

the plastic deformations ∆p due to P (for single ram) 37

and P ′

mod for double ram at mid-length: 38

Single ram: 39

∆p = C1L ′

2+ C2 40

C1 =P

48E Ix

[3L ′2

− 12x22

]+

2Fy

Eb(α − βx2)

0.541

C2 =Px2

48E Ix

[3L ′2

− 4x22

]− C1x2

−4Fy

3Eβ2 (α − βx2)1.5

42


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α =

(3d2

4+

6t f cd

tw

),1

β =12

tw L ′

(t f cd +

twd2

8

), x2 =

4Fy Ix

Pd.2

Double-ram:3

∆p =Fy L ′2

8E√

α − βa′+ C ′

3L ′

2+ C ′

44

C ′

1 =P ′

mod

2E Ix

[L ′a′

− a′2− x2

2

]+

2Fy

Eβ(α − βx2)

0.55

C ′

2 =P ′

modx2

6E Ix

[3L ′a′

− 3a′2− x2

2

]− C ′

1x2

−4Fy

3Eβ2 (α − βx2)1.5

6

C ′

3 = C ′

1 −Fya′

E√

α − βa′−

2Fy

Eβ

(α − βa′

)0.57

C ′

4 = C ′

1a′+ C2 − C3a′

−Fya′2

2E√

α − βa′

+4Fy

3Eβ2

(α − βa′

)1.58

α =

(3d2

4+

6t f cd

tw

), β =

3P ′

mod

tw Fy,9

x2 =2Fy Ix

P ′

modd.10

Step 4. Determine the magnitude of the corresponding11

residual deformation12

From Eq. (5), the magnitude of the residual13

deformation ∆res is calculated as:14

∆res = ∆p − ∆e15

where ∆p is calculated from Step 3 and ∆e is16

determined as follows:17

Single Ram: ∆e =P L ′3

48E Ix18

Double Ram: ∆e =P ′

moda′

24E Ix

[3L ′2

− 4a′2].19

Step 5. Compare ∆res from Step 4 to the required camber20

∆res from Step 4 should be compared to the camber21

value within span L ′ (Fig. 8) δcamber, calculated from22

Eq. (7) function of the desired camber as:23

δcamber = camber

[1 −

4b

L+ 4

(b

L

)2]

. (7)24

If ∆res is equal to δcamber, then the bending loads25

required from cambering are those calculated in26

Step 2.27

If δcamber (Fig. 8) is larger than ∆res from Step 4,28

then go to Step 6.29

Step 6. Determine the increase in the plastic loads30

The multiplier for increase in the bending load is31

defined as η.32

The multiplier for increase in the plastic deforma-33

tion is defined as κ .34

Fig. 8. Permanent deformation within load frame supporting arms (L ′).

η and κ should be determined from Table 2 or 35

Fig. 6(a) for single ram and Table 3 or Fig. 6(b) for 36

double ram so that: 37

δcamber = [κ × ∆p − η × ∆e] 38

where ∆p is determined from Step 3 and ∆e from 39

Step 4. 40

The modified cambering loads are determined as: 41

Single Ram: Pfinal = η8Fyd

L ′

(ct f +

twd

8

)(8a) 42

Double Ram: P ′

final = κ2Fy Ix

x2d. (8b) 43

Step 7. Check web buckling and flange local bending 44

Check that loads Pfinal (Eq. (8a)) and P ′

final (Eq. 45

(8b)) are smaller than the limiting buckling loads Ψ 46

determined using Eqs. (6a)–(6e). Otherwise, restraints 47

should be provided at the plastic hinge location (size 48

and type to be kept to the discretion of the steel 49

fabricator). 50

9. Numerical example 51

A grade 250, 10 m long W410 × 60 is to be cambered to 52

4 cm. Determine the magnitude of the applied loads for single- 53

and double-ram bending systems. 54

9.1. Solution 55

Properties: Fy = 250 MPa; E = 200,000 MPa 56

d = 40 cm; tw = 0.775 cm; c = 8.9 cm; t f = 1.28 cm; 57

Ix = 21,560 cm4; L = 10 m 58

Step 1. Setup the load frame criteria 59

Set L ′ at 0.8L = 0.8×10 m = 8 m ⇒ b = 0.1L = 60

1 m (Fig. 3). Since L ′= 8 m > 5 m, the double-ram 61

system is recommended. For illustration, both single 62

and double ram systems are used in this example. 63

In the double-ram system, the rams are set at a′= 64

1/3L ′= 1/3 × 8 m = 2.66 m, say 2.65 m from 65

the supporting ends (3.65 m from the girder’s ends) 66

⇒ a = 3.65 m. 67


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Step 2. Determine the magnitude of the plastic bending loads1

Single Ram, Eq. (1a):2

P =8Fyd

L ′

(ct f +

twd

8

)3

=8 × 250 × 103

× 0.48

4

×

(0.089 × 0.0128 +

0.00775 × 0.48

)5

= 152.7 kN.6

Double Ram, Eq. (1b):7

P ′=

2Fyd

[a − b]

(ct f +

twd

8

)8

=2 × 250 × 103

× 0.4

[2.65]9

×

(0.089 × 0.0128 +

0.00775 × 0.48

)10

= 115.2 kN.11

From Eq. (3): P ′

mod =2Fy Ix

x2d where 0 < x2 < a′. Let12

x2 = a′= 2.65 m.13

P ′

mod =2 × 250,000 × 21,560 × 10−8

2.65 × 0.4= 101.7 kN.14

Step 3. Determine the magnitude of the corresponding plastic15

deformation16

Single Ram:17

∆p = C1L ′

2+ C218

α =

(3d2

4+

6t f cd

tw

)=

(3 × 0.42

4

+6 × 0.089 × 0.0128 × 0.4

0.00775

)= 0.47319

β =12

tw L ′

(t f cd +

twd2

8

)=

120.00775 × 8

×

(0.089 × 0.0128 × 0.4 +

0.00775 × 0.42

8

)= 0.11820

x2 =4Fy Ix

Pd=

4 × 250,000 × 21,560 × 10−8

152.7 × 0.4= 3.53 m21

C1 =P

48E Ix

[3L ′2

− 12x22

]+

2Fy

Eβ(α − βx2)

0.5

= 0.00811

=152.7

48 × 200,000,000 × 21,560 × 10−8

×

[3 × 82

− 12 × 3.532]

+2 × 250,000

200,000,000 × 0.118× (0.473 − 0.118 × 3.53)0.5

22

C2 =Px2

48E Ix

[3L ′2

− 4x22

]− C1x2

−4Fy

3Eβ2 (α − βx2)1.5

= 0.00682

=152.7 × 3.53

48 × 200,000,000 × 21,560 × 10−8

×

[3 × 82

− 4 × 3.532]

− 0.00811 × 3.53

−4 × 250,000

3 × 200,000,000 × 0.1182

× (0.473 − 0.118 × 3.53)1.523

∆p = C1L ′

2+ C2 = 0.00811 ×

82

+ 0.00682

= 0.0393 m = 3.93 cm. 24

Double Ram: ∆p =Fy L ′2

8E√

α−βa′+ C ′

3L ′

2 + C ′

4 25

Set x2 = a′= 2.65 m 26

α =

(3d2

4+

6t f cd

tw

)= 0.473; 27

β =3P ′

mod

tw Fy=

3 × 101.70.00775 × 250,000

= 0.157 28

C ′

1 =P ′

mod

2E Ix

[L ′a′

− a′2− x2

2

]+

2Fy

Eβ(α − βx2)

0.5= 0.0122

=101.7

2 × 200 × 106× 21,560 × 10−8

×

[8 × 2.65 − 2.652

− 2.652]

+2 × 250

200,000 × 0.157(0.473 − 0.157 × 2.65)0.5

29

C ′

2 =P ′

modx2

6E Ix

[3L ′a′

− 3a′2− x2

2

]− C ′

1x2 −4Fy

3Eβ2 (α − βx2)1.5

= 0.00384

=101.7 × 2.65

6 × 200 × 106× 21,560 × 10−8

×

[3 × 8 × 2.65 − 3 × 2.652

− 2.652]

− 0.0122 × 2.65 −4 × 250

3 × 200,000 × 0.1572

× (0.473 − 0.157 × 2.65)1.530

C ′

3 = C ′

1 −Fya′

E√

α − βa′−

2Fy

Eβ

(α − βa′

)0.5

= −0.00562

= 0.0122 −250 × 2.65

200,000√

0.473 − 0.157 × 2.65

−2 × 250

200,000 × 0.157(0.473 − 0.157 × 2.65)0.5

31

C ′

4 = C ′

1a′+ C ′

2 − C3a′−

Fya′2

E√

α − βa′

+4Fy

3Eβ2

(α − βa′

)1.5= 0.0333 32


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= 0.0122 × 2.65 + 0.00384 + 0.00562 × 2.65

−250 × 2.652

2 × 200,000√

0.473 − 0.157 × 2.65

+4 × 250

3 × 200,000 × 0.1572

× (0.473 − 0.157 × 2.65)1.51

∆p =250 × 82

8 × 200,000√

0.473 − 0.157 × 2.65

− 0.00562 ×82

+ 0.0333 = 0.0532 m

= 5.32 cm.2

Step 4. Determine the magnitude of the corresponding3

residual deformation4

Single Ram:5

∆e =P L ′3

48E Ix=

152.7 × 83× 100

48 × 200,000,000 × 21,560 × 10−86

= 3.78 cm.7

The corresponding residual deformation ∆res, is8

obtained from Eq. (4) as:9

∆res = ∆p − ∆e = 3.93 cm − 3.78 cm = 0.15 cm.10

Double Ram:11

∆e =P ′

moda′

24E Ix

[3L ′2

− 4a′2]

=107.8 × 2.5

24 × 200,000,000 × 21, 560 × 10−8

×

[3 × 82

− 4 × 2.52]

= 0.0427 m12

∆res = ∆p − ∆e = 5.32 cm − 4.27 cm = 1.05 cm.13

Step 5. Compare ∆res from Step 4 to the required camber14

Eq. (7): δcamber = camber

[1 −

4b

L+ 4

(b

L

)2]

15

= 4

[1 −

4 × 110

+ 4(

110

)2]

16

= 2.56 cm17

δcamber = 2.56 cm, 17 times the residual deformation18

of 0.15 cm determined based on plastic load for single19

ram and 2.5 times the residual deformation of 1.05 cm20

determined for double ram. Consequently, the bending21

load should be increased beyond the plastic limit. Go22

to Step 6.23

Step 6. Determine the increase in the plastic loads24

Single Ram:25

∆res = 2.56 cm ⇒ κ × ∆p − η × ∆e26

= 2.56 cm ⇒ κ × 3.93 − η × 3.78 = 2.56 cm.27

From Table 2, if the load is increased by 2.5% (e.g.28

Pfinal = 156.5 kN), the deformation is increased by29

40% ⇒ 1.4 × 3.93 − 1.025 × 3.78 = 1.63 cm <30

2.56 cm ⇒ load shall be increased further.31

From Table 2, if the load is increased by 5% (e.g. 32

Pfinal = 160.3 kN), the deformation is increased by 33

100% ⇒ 2 × 3.93 − 1.05 × 3.78 = 3.89 cm > 34

2.56 cm ⇒ load increase should be between 2.5 and 35

5%. By interpolation, the increase is noted as 3.5% e.g. 36

Pfinal = 1.035 × 152.7 kN = 158 kN. 37

Double Ram: 38

∆res = 2.56 cm ⇒ κ × ∆p − η × ∆e 39

= 2.56 cm ⇒ κ × 5.32 − η × 4.27 = 2.56 cm. 40

From Table 3, if the load is increased by 5% (e.g. 41


25% ⇒ 1.25 × 5.32 − 1.01 × 4.27 = 2.34 cm < 43

2.56 cm ⇒ load shall be increased further. 44

From Table 3, if the load is increased by 7.5% (e.g. 45


50% ⇒ 1.5 × 5.32 − 1.025 × 4.27 = 3.67 cm > 47

2.56 cm ⇒ load increase should be between 5% and 48

7.5%. By interpolation, the increase is noted as 5.5% 49

e.g. Pfinal = 1.055 × 101.7 kN = 107.3 kN. 50

Step 7. Check web buckling and flange local bending 51

Fy = 250 MPa; E = 200,000 MPa 52

d = 40 cm; tw = 0.775 cm; b f = 17.8 cm; t f = 53

1.28 cm; h = 36.1 cm 54

l = 8 m (distance between end supports since no 55

lateral bracing is provided) 56

k = 3 cm; set N = k = 3 cm 57

Eq. (6a):Ψ =2.4t3

w

√E Fy

1.67h

=2.4 × (0.775)3 √

200,000 × 2501.67 × 36.1

= 131 kN. 58

Eq. (6b):Ψ =331,000t3

wt f

1.76h2

[1 + 0.4

(h/twl/b f

)3]

=331,000 × (0.775)3

× 1.28

1.76 × (36.1)2

×

[1 + 0.4

(36.1/0.775800/17.8

)3]

= 125 kN. 59

Eq. (6c):Ψ =0.08t2

w

2

[1 + 3

(N

d

) (twt f

)1.5]

×

√E Fy t f

tw=

0.08 × 0.7752

2

×

[1 + 3

(3

40

) (0.7751.28

)1.5]

×

√200,000 × 250 × 1.28

0.775= 242 kN 60

Eq. (6d):Ψ =0.1 (5k + N ) Fy tw

1.5

=0.1 (5 × 3 + 3) × 250 × 0.775

1.5= 232.5 kN. 61


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Eq. (6e):Ψ =0.625Fy t2

f

1.67=

0.625 × 250 × 1.282

1.67= 152 kN.1

The smallest value of 125 kN from Eq. (6b) is selected. Note2

that for the double ram system, Pfinal = 107.3 kN < 125 kN;3

therefore, restraint elements are not required. However, for the4

single ram system, Pfinal = 158 kN > 125 kN ⇒ bracing at the5

point of load application is necessary (note that it was stated6

earlier that for L ′= 8 m > 5 m, the single-ram system should7

not be recommended and was only used for illustration).8

9.2. Discussion9

It may be seen from the numerical example that for a 10 m10

long steel girder, bending loads of 158 kN (single ram) and11

107.3 kN (double rams, loads spaced at 2.7 m) are required12

to induce a camber of 4 cm at mid-span. Those loads are13

slightly greater than the plastic loads of 152.7 kN (single-14

ram) and 101.7 kN (double ram). The corresponding maximum15

strains may be obtained from Table 2 (single-ram) and Table 316

(double ram) as 7εy and 1.6εy respectively. These are much17

smaller than the maximum strain in the plastic range of 10–20εy18

(Fig. 5) [10]. As noted earlier, the smaller strain in the double-19

ram system is attributed to: (1) the modified bending load ap-20

plied is less than the plastic load that causes plastic hinging21

and, (2) the smoother parabolic shape that develops when the22

bending loads are spaced apart compared to a single mid-span23

bending load in the single-ram system (Fig. 7). Also, it was24

shown that the bending loads for the double-ram system are25

smaller than the limiting buckling loads; therefore no restraint26

or bracing is required. For the single ram system, the bending27

load of higher magnitude exceeds the limiting buckling loads;28

consequently, the use of restraint elements at the point of load29

application such as braces or stiffeners is necessary to avoid30

localized failures to the section. It is clearly shown that the ef-31

fects of web compression buckling and web sidesway buckling32

are more critical than web crippling and web local yielding.33

10. Conclusions34

A cold bending analysis previously developed for curving35

structural steel girders is extended to provide an estimate of

the loads required to induce a specified camber. Equations 36

relating loads to deformations were developed for both single 37

and double-ram systems in the elasto-plastic and plastic range. 38

Possible adverse effects due to concentrated load applications 39

during cambering such as flange local bending, web local 40

buckling and overall lateral torsional buckling were discussed 41

and prevention measures identified. Numerical analyses were 42

conducted to determine the increase in loads, deformations and 43

strains in the post-plastic range. A comprehensive numerical 44

example illustrating the application of the proposed equations 45

is presented. 46

The proposed analysis provides better correlation with the 47

actual cambering operation thereby reducing trial and error and 48

making the operation more efficient. 49

References 50

[1] Ricker DT. Cambering steel beams. American Institute of Steel 51

Construction (AISC), Engineering Journal 1989;26(4):136–42. 52

[2] Kloiber LA. Cambering steel beams. In: Structures congress 1989, steel 53

structures proceedings. p. 101–10. 54

[3] AISC. Steel construction manual, load and resistance factor design. 55

13th ed. USA: AISC Inc.; 2005. p. 1–118, 1–119, 2–28, 2–30, 56

3–213. 57

[4] Downey EW. Specifying camber. Modern Steel Construction 2006;46(7): 58

53–6. 59

[5] Bjorhovde R. Cold bending of wide-flange shapes for construction. 60

American Institute of Steel Construction (AISC), Engineering Journal 61

2006;43(4):271–86. 62

[6] Gergess A, Sen R. Cold bending symmetric un-stiffened I-girder. Journal 63

of Constructional Steel Research 2005;61:473–92. 64

[7] Gergess A, Sen R. Fabrication of unsymmetrical curved plate girders 65

by cold bending. Journal of Constructional Steel Research 2005; 66

61:1353–72. 67

[8] Gergess A, Sen R. Refined analysis for cold bending steel plate girders. 68

Journal of Constructional Steel Research 2006;63:667–76. 69

[9] Gergess A, Sen R. Full-scale cold bending demonstration test. Journal of 70

Constructional Steel Research 2007;63:1295–304. 71

[10] Salmon CG, Johnson JE. Steel structures: Design and behavior. 4th ed. 72

New York: Harper-Collins; 1996. p. 51. 73

[11] Popov EV. Mechanics of materials. 2nd ed. Englewood Cliffs (NJ): 74

Prentice-Hall Inc; 1976. p. 135–42, 377–93. 75

[12] Byars FE, Snyder DR. Engineering mechanics of deformable bodies. 3rd 76

ed. Scranton (PA): International Textbook Co; 1975. 77

[13] MSC/NASTRAN for Windows, The MacNeal-Schwendler Corporation 78

(2000). In: Finite element modeling and postprocessing system. Los 79

Angeles (CA). 80


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Cambering structural steel I-girders using cold bending

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