Cold Bending Steel Beams: a State-of The-Art Engineering Solution that
Meets Industry Challenges
By: Antoine N. Gergess, PhD, PE, F.ASCE
Professor, University of Balamand, Lebanon
www.balamand.edu.lb
World Congress and Exhibition on Steel Structure, Dubai
November 15 - 17, 2015
Horizontal Curves
Heat Curving
Continuous heat
V heat (R > 300m)
Drawbacks
 Procedure is complex
 Numerical computations have to account for material and
geometric non-linearity
 Temperature distribution along the flange width are not
exactly uniform
 Analysis has to account for the girder behavior during heating
and during cooling (natural cooling).
 During heating, heated zone elongates to form a larger outer
radius
 During cooling, the heated portion shortens to form a reverse
curvature
Material Properties
1.0
Yield Stress
Modulus of Elasticity
0.8
Ratio
E/E
0.6
Fy/Fy
0.4
0.2
0
135
270
405
Temperature, C
540
675
Cut-Curving
Flange
Flange
Flange
Scrap
Pressure Applied
During Fit-up
Web
Flange
Flange
Fitting Jig
Draw-Backs
 Too costly because of excessive waste
 Too much scrap for sharp curvatures
 Used for mild curvatures (R  300m)
 Fit-up operation too complicated
3-ROLLERS BENDING
 Mainly used in buildings
 Fit-up difficult for larger
size bridge girders
 Perfect curve
OTHER IDEAS?
118
COLD BENDING
Analysis
 = a’/L;  = x2/a’
Pflange  3.75t f Fy (limit state of flange local buckling)
2
P
Fy t f b f
3χL
2
L p  0.255b f
E
(lateral bracing limit)
Fy
Stress-strain curve
Idealization
0.125
 res (m)
0.1
0.075
0.05
0.025
0
0.0
1.0
2.0
a' = 0.333L
3.0
4.0
5.0
6.0
Distance x (m)
a' = 0.25L
Circular
7.0
8.0
9.0
10.0
a' = 0.1L
a
Residual Stresses
 Flexural stiffness is reduced by the presence of cold
bending residual stresses.
ROLLED BEAMS
WELDED SHAPES
Influence of Residual Stress on Average Stress-Strain Curve
Tampa Steel Concept
Apply smaller loads at uniform intervals to achieve
desired curvature
Idealization of Curve
L
1
2
3
3
4
5
max = 4
6
7
5
2
6
x
δ  δmax  R  R  x
2
2
L2
δmax 
8R
S
S
Load P
Li
OFFSETS ii, ij
Load: 2
Load: 3
1
2
3
4
6
5
22
23
33
24
34
44
35
45
Load: 4
Load: 5
25
Load: 6
55
max
26
36
46
56
66
7
Test Girder
Loading Details
Test Girder after Curving
Idealized Stress-Strain Curve
Fy
  2.27y
  1.89y
max  10 y
STRESS
R = 255 m
r  0.39y
r  0.77y
R = 115 m
y
residual  8.5 y
STRAIN
1.5 y
FORMULATION
PARAMETERS:
 Load Frame Spacing S
 Bending Loads Ptf, Pbf
 Deflection  within span S
S
 Segment Length Li
 Number of Segments n
 Offsets
Ptf
Li
LOAD FRAME SPACING (S)
Based on lateral bracing limit:
S  1.76r t
S = 14.4c for Grade 250 steel
S = 12.2c for Grade 345 steel
For unsymmetrical sections use ctf
Flange
Width 2c
Load P
Comp. side
Flange
S = Lp
E
Fy
BENDING LOADS (Ptf, Pbf )
From simple beam plastic load analysis:
P
4Fy t f c
Top Flange: Ptf based on ttf, ctf
Bot. Flange: Pbf based on tbf, cbf
2
S
Constant
P
Mp
Fytfc
PS

4
Fytfc
S
c
c
Mp = Fytfc2
DEFLECTION 
Δ
13Fy S
ctf  cbf
2
Ptf  Pbf
216Ec
Load P
tf  bf
set  = cte = tf
bf
   bf ???
Plastic
Hinge
tf
S
Set P = Pbf,
P  Pbf
Load in cycles
(on-going research)
m=tf/bf
SEGMENT LENGTH Li
tf
4RΔ
Li 
S
S
[m]
Constant
[m-1]
2
2Li/S
2Li
l

8R




[m+1]
2Li
Radius R
2Li
S
NUMBER OF SEGMENTS
Length L
nLi
a
1
2
3
4
a
n
5
n+1
Line of symmetry
Round-down to
the nearest
even integer
L-S
n
Li
adjust
(L - S)
Li 
n
Challenges ?
 Cracking, Fracture
 Flange Upsets
 Dimples
 Web Crippling
Challenges ?
- Effects on Steel mainly fracture characteristics
- Does it lead to a permanent reduction
in the ductility and fracture resistance of steel?
- Effects if steel is loaded beyond plastic limits
- ANSWERS ? Specific need for full-scale
testing.
How could it be assessed - 1?
• Perform visual
inspection using
NDT techniques
• Instrument to
measure loads,
offsets, strains and
web movement
How could it be assessed - 2?
Perform in-depth
material testing of
the bent zone, e.g.




Charpy V-notch test
Fracture sensitivity
Tensile strength
Brinell hardness
How could it be assessed – 3?
Check for compliance with AASHTO Requirements
ASTM A709
ASTM A709
Grade 345
HPS 485W
Plate Thickness
up to 5 cm (2 in.)
up to 10 cm (4 in.)
Yield Strength
345 MPa (50 ksi)
485 MPa (70 ksi)
min. 404 MPa
620 – 758 MPa
(min. 58 ksi)
(90 – 110 ksi)
21%
19%
Toughness: CVN
35 m-N @ -12C
47 m-N @ -23C
Fracture Critical
(25 ft-lbf @ 10F)
(35 ft-lbf @ -10F)
Property
Tensile Strength
Min. Elongation
5 cm (2 in.)
Zone 3
 6.35 cm (to 2.5 in.)
 6.35 cm (to 4in.)
HPS 485W
Results of FHWA Tensile Tests on HPS 485W Specimens.
Wright W, Candra H, Albrecht P. “Fracture Toughness of A709 Grade HPD
-485W Steel”. Draft FHWA Report, Washington, D.C, 2005.
How could it be assessed – 4?
3D Finite
Element
Modeling
Specifications
 Develop criteria and specifications for
consideration by AASHTO
 Limits on strain
 Limits on applied load
 Limits on minimum radius
 Guidelines for localized damage repair
 Guidelines for inspection
 Fabrication aids
Acknowledgment
Dr. Rajan Sen, PhD, PE, F.ASCE, Samuel and Julia Flom Professor,
University of South Florida, Tampa
Tampa Steel Erecting Company, Tampa, Florida
Thank You!
www.balamand.edu.lb