Vierendeel girder and frame
Vierendeel Bridge Grammene Belgium
Vierendeel structures Copyright Prof Schierle 2011 2
Arthur Vierendeel (1852–1940) born in
Leuven, Belgium was a university
professor and civil engineer.
The Vierendeel structure he developed
was named after him.
His work, Cours de stabilité des
constructions (1889) was an important
reference during more than half a
century. His first bridge was built 1902
in Avelgen, crossing the Scheldt river
Vierendeel structures Copyright Prof Schierle 2011 4
1 Base girder
2 Global shear
3 Global moment
4 Bending
5 Chord forces
6 Pin joints
7 Strong web
8 Strong chord
9 Shear
10 Chord shear
1 1-bay girder
2 Gravity load
3 Lateral load
4 Articulated
Inflection points
5 3-bay girder
6 Gravity load
7 Lateral load
8 Articulated
Inflection points
One-way girders
1 Plain girder
2 Prismatic girder
3 Prismatic girder
Space frames
4 2-way
5 3-way
6 3-D
Vierendeel girder and frame
Vierendeel structures Copyright Prof Schierle 2011 6
Salk Institute, La Jolla
Architect: Louis Kahn
Engineer: Komendant and Dubin
Perspective section and photo, courtesy Salk Institute
Viernedeel girders of 65’ span, provide adaptable
interstitial space for evolving research needs
Yale University Library
Architect/Engineer: SOM
1
Vierendeel facade
2
Vierendeel elements
3
Cross section
•
The library features five-story Vierndeel frames
•
Four concrete corner columns support the
frames
•
Length direction span: 131 feet
•
Width direction span: 80 feet
•
Façades are assembled from prefab steel
crosses welded together at inflection points
•
The tapered crosses visualize inflection points
Vierendeel structures Copyright Prof Schierle 2011 8
Commerzbank, Frankfurt
Architect: Norman Foster
Engineer: Ove Arup
Floors between sky gardens are
supported by eight-story high
Vierendeel frames which also
resist lateral load
Commerzbank, Frankfurt
Architect: Norman Foster
Engineer: Ove Arup
Vierendeel elevation / plan
Vierendeel / floor girder
joint detail
Vierendeel steel girder
Assume:
10” tubing, allowable bending stress F
b
= 0.6x46 ksi
F
b
= 27.6 ksi
Girder depth d = 6’, span 10 e = 10x10’
L = 100’
DL=
18 psf
LL =
12 psf
=
30 psf
Uniform load
w = 30 psf x 20’ / 1000
w = 0.6 klf
Joint load
P = 0.6 x 10’
P= 6 k
Max shear
V = 9 P/2 = 9 x 6/2
V = 27 k
CHORD BARS
Shear (2 chords)
V
c
= V/2 = 27/2
V
c
= 13.5 k
Chord bending (k’)
M
c
= V
c
e/2 = 13.5x5
M
c
= 67.5 k’
Chord bending (k”) M
c
= 67.5 k’ x12”
M
c
= 810 k”
Moment of Inertia
I = M
c
c/F
b
= 810 k” x 5”/27.6 ksi
I = 147 in
4
2nd bay chord shear V
c
= (V–P)/2 = (27-6)/2
V
c
= 10.5 k
2nd chord bending M
c
= V
c
e/2 = 10.5 x 5
M
c
= 52.5 k’
2nd chord bending
M
c
= 52.5 k’ x 12”
M
c
= 630 k”
WEB BAR (2nd web resists bending of 2 chords)
Web bar bending
M
w
= M
c
end bay + M
c
2nd bay
Load
Shear
Vierendeel structures Copyright Prof Schierle 2011 14
Chord bars
Moment of Inertia required I= 147 in
4
Use ST10x10x5/16
I= 183>147
Web bars
Moment of Inertia required I= 261 in
4
Sport Center, University of California Davis
Architect: Perkins & Will
Engineer: Leon Riesemberg
Given the residential neighborhood, a major objective was to
minimize the building height by several means:
• The main level is 10’ below grade
• Landscaped berms reduce the visual façade height
• Along the edge the roof is attached to bottom chords
to articulates the façade and reduce bulk
Assume
Bar cross sections 16”x16” tubing, 3/16” to 5/8” thick
Frame depth d = 14’ (max. allowed for transport)
Module size:
21 x 21 x 14 ft
Width/length:
252 x 315 ft
Structural tubing F
b
= 0.6 Fy = 0.6x46 ksi
F
b
= 27.6 ksi
DL = 22 psf
LL = 12 psf (60% of 20 psf for tributary area > 600 ft
2
)
= 34 psf
Note: two-way frame carries load inverse to deflection ratio:
r = L1
4
/(L1
4
+L2
4
) = 315
4
/(315
4
+252
4
) r = 0.71
Vierendeel structures Copyright Prof Schierle 2011 16
Design end chords
Joint load
P = w x 21’ = 0.5klf x 21’
P = 10.5 k
Max. shear
V = 11 P /2 = 11 x 10.5 / 2
V = 58 k
Chord shear (2 chords)
Vc = V/2 = 58 k / 2
Vc = 29 k
Chord bending
Mc = Vc e/2 = 29x 21’x12”/2
Mc= 3654 k”
Moment of Inertia required
I = Mc c /F
b
= 3654 x 8”/27.6 ksi I = 1059 in
4
Check mid-span compression
Global moment
M = w L
2
/8 = 0.5 x 252
2
/8
M = 3969 k’
Compression (d’=14’–16”=12.67’)
C = M/d’= 3969 k’/ 12.67
C = 313 k
Modules:
21x21x14’
Chord bars
Moment of Inertia required
I= 1059 in
4
Use ST16x16x1/2
I= 1200
Check mid-span chord stress
Compression
C = 313 k
Allowable compression
P
all
= 728 k
313 <<728
Note:
Vierendeel structures Copyright Prof Schierle 2011 18
Commerzbank, Frankfurt
Design edge girder
Assume:
Tributary area
60’x20’
End bay width
e = 20’
Loads: 70 psf DL+ 30 psf LL
∑=100 psf
Allowable stress F
b
=0.6 x36
F
b
= 21.6 ksi
Girder shear
V = 60’x20’x 100 psf/1000
V = 120 k
Bending moment
M = V e/2 = 120x20/2
M = 1200 k’
Required section modulus
S = M/F
b
= 1200 k’ x 12”/ 21.6 ksi
S = 667 in
3
Use W40x192
S = 706 in
3
Note: check also lateral load
Variable bay widths equalize bending stress
Load at corners increases stability
Vierendeel steel girder
Assume:
10” tubing, allowable bending stress F
b
= 0.6x46 ksi
F
b
= 27.6 ksi
Girder depth d = 6’, span 10 e = 10x10’
L = 100’
DL=
18 psf
LL =
12 psf
=
30 psf
Uniform load
w = 30 psf x 20’ / 1000
w = 0.6 klf
Joint load
P = 0.6 x 10’
P= 6 k
Max shear
V = 9 P/2 = 9 x 6/2
V = 27 k
CHORD BARS
Shear (2 chords)
V
c
= V/2 = 27/2
V
c
= 13.5 k
Chord bending
M
c
= V
c
e/2 = 13.5 x (10’x12”)/ 2 M
c
= 810 k”
Moment of Inertia
I = M
c
c/F
b
= 810 k” x 5”/27.6 ksi
I = 147 in
4
2nd bay chord shear V
c
= (V–P)/2 = (27-6)/2
V
c
= 10.5 k
2nd chord bending
M
c
= V
c
e/2 = 10.5 x 120”/2
M
c
= 630 k”
WEB BAR (2nd web resists bending of 2 chords)
Web bar bending
M
w
= M
c
end bay + M
c
2nd bay
M
w
= 810 + 630
M
w
=1,440 k”
Vierendeel structures Copyright Prof Schierle 2011 20
Commerzbank, Frankfurt
Design edge girder
Assume:
Tributary area
60’x20’
End bay width
e = 20’
Loads: 70 psf DL+ 30 psf LL
∑=100 psf
Allowable stress F
b
=0.6 x36
F
b
= 21.6 ksi
Girder shear
V = 60’x20’x 100 psf/1000
V = 120 k
Bending moment
M = V e/2 = 120x20/2
M = 1200 k’
Required section modulus
S = M/F
b
= 1200 k’ x 12”/ 21.6 ksi
S = 667 in
3
Use W40x192
S = 706 in
3
Note: check also lateral load
Variable bay widths equalize bending stress
Load at corners increases stability
Vierendeel structures Copyright Prof Schierle 2011 22
Vierendeel structures Copyright Prof Schierle 2011 28