Steel Towers: Lateral forces
STEEL TOWER
arc422 Building Technology VI _ Structures III _ Spring 2010
This laboratory project challenges students to empirically investigate the means and methods of resisting lateral
forces. Students iteratively design, fabricate and test these means and methods in the context of a structural
tower. Towers were selected as the structures most vulnerable to lateral forces generated by seismic and wind
conditions. The tower models are eight feet in height and are dimensionally constrained at the bottom and top.
Materials are restricted to 1/8” diameter steel rod and 16 gauge sheet steel; the method of assembly is restricted to welding. Projects were categorized into four groups, each with a specific cross-sectional geometric
basis [equilateral triangle, circle, square and rectangle]. The towers are evaluated on a floor mounted linear inertial
shaker apparatus.
This project is executed by teams of four students and consists of two iterations. Each stage consists of a drawing
component and a construction component for collective testing and analysis. This laboratory project is preceded
by a precedent study through which students research and diagram three high-rise structures.
Objectives: following the completion of this project students should be able to:
- Understand lateral forces and employ strategies for rendering lateral stability
- Understand variations in the magnitudes of stress relative to the given conditions of base and apex
- Understand the inherent opportunities and challenges of the diverse geometric schemes
- Establish a design strategy that considers the variable conditions, locations and magnitudes of stress, inclusive
of quantity and geometric distribution of material
- Develop diagramming methods effective in the conception, development and evaluation of your tower’s
performance
- Address the challenges inherent to steel fabrication and the importance of jigging
Course Context:
The Steel Tower exercise is administered concurrently with lectures on lateral forces and with computational
exercises related to the sizing of steel beams, columns and connections.
Project Conditions
| teaching
| structures
| steel towers
219
Equilateral Triangle Scheme - Example Project - Iteration 2
96”
Top Level
96”
Top Level
P.
P.
180°
P.
P.
180°
P.
P.
P.
Top Level Plan:
Scale: 1-1/2”=1’-0”
P.
Force one
P.
Top Level Plan:
Scale: 1-1/2”=1’-0”
Force two
P.
•Equilateral triangle scheme.
Force one perpendicular to the
face of triangle. Force two oblique
to the face of the triangle.
P.
P.
Level Seven Plan:
88”
Level Twelve
P.
Scale: 1-1/2”=1’-0”
48”
Level 7
96”
Top Level
80”
Level Eleven
P.
Diaz.Alcantar.Frederickson.Owen.Lehtinen.
180°
P.
96”
Top Level
72”
Level Ten
88”
Level Twelve
P.
P.
Scale: 1-1/2”=1’-0”
0°
P.
P.
Top Level Plan:
P.
P.
80”
Level Eleven
P.
9”
Scale: 1-1/2”=1’-0”
48”
Level 7
64”
Level Nine
180°
P.
56”
Level Eight
P.
Top Level Plan:
Scale: 1-1/2”=1’-0”
P.
Decrease of lateral bracing
Level Seven Plan:
Force one
Decrease of lateral bracing
P.
•Tapering form responds
to forces.
•Interconnected triangular
floorplates respond to torsion.
•The top is rotated 180º Forc
from the bottom.
•Equilateral triangle sc
Force one perpendicula
face of triangle. Force t
to the face of the triang
96”
Top Level
P.
P.
•Adding torsion to introduce a bias
and to equilibrate the
forces acting on the triangle.
Concept:
96”
Top Level
90°
P.
Forc
Concept:
90°
Force one
P.
Force two
72”
•Equilateral triangle scheme.
Level Ten
Concept:
48”
Level Seven
90°
Base Level Plan:
P.
P.
P.
Concept:
Scale: 1-1/2”=1’-0”
90°
96”
Top Level
P.
0°
P.
Level Seven Plan:
•Equilateral triangle scheme.
Force40”one perpendicular to the
Level
Six
face of
triangle.
Force two48”oblique
Level 7
to the face of the triangle.
P.
Scale: 1-1/2”=1’-0”
P.
P.
P.
P.
•Adding torsion to introduce a bias
and to equilibrate the
64”
forces acting on the triangle. Level Nine
88”
Level Twelve
80”
Level Eleven
32”
Level Five
0°
P.
•Floorplate datum every 8 inches.
•Each plate is rotated counter-clockwise
15º.
56”
Level Eight
64”
Level Nine
P.
•Floorplate datum every 8 inches.
•Each plate is rotated counter-clockwise
15º.
•Tapering form responds
to forces.
•Interconnected triangular
floorplates respond to torsion.
•The top is rotated 180º
from the bottom.
72”
Level Ten
9”
96”
Top Level
•Tapering form responds
to forces.
•Interconnected triangular
floorplates respond to torsion.
•The top is rotated 180º
from the bottom.
•Adding torsion to introduce a bias
and to equilibrate the
forces acting on the triangle.
Force one perpendicular to the
face of triangle. Force two oblique
to the face of the triangle.
Force two
9”
Level Seven Plan:
88”
Level Twelve
P.
Base Level Plan:
24”
Level Four
P.
48”
Level Seven
Scale: 1-1/2”=1’-0”
P.
Scale: 1-1/2”=1’-0”
P.
Base Level Plan:
Scale: 1-1/2”=1’-0”
P.
48”
Level 7
40”
Level Six
P.
P.
Equilateral Triangle Diagram
24”
Level Four
P.
P.
P.
16”
Level Three
8”
Level Two
72”
Level Ten
Scale: 1-1/2”=1’-0”
Equilateral Triangle Diagram
0°
180°
8” TYP.
P.
P.
Elevation:
Scale: 2-1/2”=1’-0”
9”
Elevation:
Scale: 1-1/2”=1’-0”
32”
Level Five
P.
P.
Elevation:
P.
P.
P.
P.
P.
P.
[ m. ] Moment
[ v. ] Shear
Elevation: primary vertical members
Scale: 1-1/2”=1’-0”
P.
P.
Elevation: diagonal cross bracing
Scale: 1-1/2”=1’-0”
56”
Level Eight
P.
Elevation: diagonal cross bracing
Scale: 1-1/2”=1’-0”
P.
P.
[ v. ] Shear
Elevation: primary vertical members
Scale: 1-1/2”=1’-0”
Scale: 1-1/2”=1’-0”
24”
Level Four
P.
P.
Top Level Force Diagram:
P.
P.
P.
Elevation:
Scale: 2-1/2”=1’-0”
0”
Base Level
0”
Base Level
180°
180°
180°
8” TYP.
0”
Base Level
P.
64”
Level Nine
P.
40”
Level Six
8”
Level Two
P.
Scale: 1-1/2”=1’-0”
0”
Base Level
P.
48”
Level Seven
32”
Level Five
16”
Level Three
80”
Level Eleven
ARC 422: Building Technology VI
Lateral Forces [Equilateral Triangle Scheme]
P.
P.
Diaz.Alcantar.Frederickson.Owen.Lehtinen.
56”
Level Eight
Scale: 3”=1’-0”
P.
Base Level Plan:
P.
48”
Level Seven
Scale: 1-1/2”=1’-0”
16”
Level Three
P.
40”
Level Six
Equilateral Triangle Diagram
P.
20 gauge steel shear planes
to resolve highest lateral forces on
the bottom three floors
8”
Level Two
32”
Level Five
Scale: 1-1/2”=1’-0”
P.
P.
P.
8” TYP.
0”
Base Level
24”
Level Four
Corners rounded to allow
room for diagonal bracing
0”
Base Level
P.
90°
90°
16”
90°
Level Three
90°
Elevation:
Shear Plane Detail:
P.
Equilateral Triangle Diagram
P.
Elevation:
Scale: 6”=1’-0”
Scale: 2-1/2”=1’-0”
P.
Elevation: diagonal cross bracing
Scale: 1-1/2”=1’-0”
Scale: 1-1/2”=1’-0”
P.
8”
Level Two
P.
Scale: 1-1/2”=1’-0”
8” TYP.
0”
Base Level
Level 7 Force Diagram:
0”
Base Level
Scale: 3”=1’-0”
P.
P.
P.
Elevation:
Elevation:
P.
Elevation: diagonal cross bracing
Scale: 1-1/2”=1’-0”
Scale: 2-1/2”=1’-0”
Scale: 1-1/2”=1’-0”
P.
P.
Elevation: primary vertical members
Scale: 1-1/2”=1’-0”
P.
P.
[ v. ] Shear
P.
P.
P.
[ m. ] Moment
P.
P.
P.
0°
0°
0°
ARC 422: Building Technology VI
Lateral Forces [Equilateral Triangle Scheme]
0°
Base Force Diagram:
Scale: 3”=1’-0”
P.
220
steel towers |
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teaching |
P.
LATERAL FORCES SQUARE
Our original concept was based upon a strong core with a
supporting outer frame that added lateral depth. For the second
iteration we wanted to eliminate unnecessary weight while reinforcing
critical sections within the tower. The core’s hierarchy is now that a
cluster of 3 vertical rods rise up and at the 5th plate a rod stops so
that only 2 core rods rise up to the top of the tower. Lateral bracing is
more concentrated at the base of the tower and at the critical area in
the middle of the tower. Cross bracing is also more concentrated at
the base of the tower and spans greater lengths in the upper portions
of the tower where reinforcement is not as greatly needed. Unneeded
material in the plates was eliminated and replaced with rod that
distributes the loads from inner plate to outer plate.
TOP (12) PLATE
4.5” x 4.5”
OUTER VERTICAL
MEMBERS 8’
4.5”
.25”
.25”
1”
.5”
.5”
1”
PLATE 11
5_5/8” x 5_5/8”
3 7/8”
PLATE 10
DIAGONAL BRACING
MEMBERS 2’
5/16”
4.5”
5/16”
1”
6.75” x 6.75”
PLATE 9
7_7/8” x 7_7/8”
CROSS SECTION OF TOP (12) PLATE 6” =1’
BIRD’S EYE VIEW OF STRUCTURE 3” =1’
TOP (12) PLATE
DIAGONAL BRACING
MEMBERS 2’
8”
4.5” x 4.5”
PLATE 8
8”
MODULE I: STEEL STRUCTURES
9” x 9”
PLATE 11
1’ 1.5”
5_5/8” x 5_5/8”
5/16”
5/16”
1.75”
1.75”
CORE MEMBERS
8.5’
3 3/16”
.75”
OUTER
PLATE 10
3/8”
1”
3/8”
3 3/16”
PLATE 7
10_1/8” x 10_1/8”
OUTER SIDE BRACING
MEMBERS AT
CRITICAL SECTIONS 8”
3 1/16”
8”
6.75” x 6.75”
OUTER
PLATE 9
PLATE 6
11.25 “ x 11.25”
8”
7_7/8” x 7_7/8”
1’ 1.5”
DIAGONAL BRACING
MEMBERS 1’5”
8”
9” x 9”
WORM’S EYE VIEW OF STRUCTURE 3” =1’
5 7/8”
OUTER
PLATE 8
OUTER
PLATE 7
DIAGONAL BRACING
MEMBERS 1’5”
8”
8”
10_1/8” x 10_1/8”
OUTER SIDE BRACING
MEMBERS AT
CRITICAL SECTIONS 2’
3 1/16”
8’
OUTER
PLATE 6
OUTER
PLATE 5
CROSS SECTION OF PLATE 4
6” =1’
DIAGONAL BRACING
MEMBERS 10”
8”
1’_3/8” x 1’_3/8”
OUTER
PLATE 4
DIAGONAL BRACING
MEMBERS 9.75”
1’ 1.5” x 1’ 1.5”
8”
1’ _3/8” x 1’_3/8”
PLATE 4
1’ 1.5” x 1’ 1.5”
.75”
8”
11.25 “ x 11.25”
PLATE 5
GROUP 2: CLAYTON CALKINS JESSICA ENGLUND BRITTANY PETERS
Square Scheme - Example Project - Iteration 2
CONCEPT AND CONSTRUCTION:
9”
.75”
1.25”
2”
1”
2”
1.25”
DIAGONAL SECTION
VERTICALLY THRU PLATES
2”= 1’
4 3/16”
8”
OUTER
PLATE 2
1’ 1.5” x 1’ 1.5”
PLATE 2
1’ x 1’
.75”
5/16”
OUTER
PLATE 3
1’ 1.5” x 1’ 1.5”
PLATE 3
PLATE 1
10.5” x 10.5”
BASE PLATE
CROSS SECTION OF BASE PLATE 6” =1’
ELEVATION 2” = 1’
SHEAR DIAGRAM
MOMENT DIAGRAM
The form of our structure was designed to
relieve and distribute some of the moment
stresses near the bottom of the tower.
PLATE PLANS 1.5” = 1’
These force diagrams assume a loading
condition portrayed in the deflection
diagram. For the vertical members, the
same pattern continues the full height.
| structures
ARC 422
SPRING 2010
Scale: 2” = 1’
5/16”
9” x 9”
| teaching
9” x 9”
4 3/16”
8”
BASE PLATE
GROUP 2: CLAYTON CALKINS JESSICA ENGLUND BRITTANY PETERS
PLATE 1
10.5” x 10.5”
ARC 422
SPRING 2010
MODULE I: STEEL STRUCTURES LATERAL FORCES SQUARE
9”
8”
1’ x 1’
| steel towers
221
Rectangle Scheme - Example Project - Iteration 2
Rectangle extruded into
an inverse of the base
rectangle. Both the top
and bottom measuring
1’ by 6”.
The same concept of
inversion was used
but repeated into
four modules.
Modules spaced farther
apart as the tower grows
in height.
A Taper was introduced
to reduce the dead load
of the steel. The top plate
measures 1/2 of the base.
The floor plates were
also introduced every
8”, also helping with
overall stiffness.
Spacing between modules:
(16”, 24”, 24”, 32”)
In order to respond to
the bias given from the
tapering modules; tension
lines were added along the
outside using triangular
geometries to attatch to.
This aspect of the project
has beendeveloped further by
extending them further laterally.
Moment Elev 1 (stage D)
A
B
6”
C
Moment Elev 1 (stage E)
D
E
Bending moment weakness
and response (tension lines)
3”
Pretensioning
Elev. 2 1 1/2” = 1’0”
Elev. 1 1 1/2” = 1’0”
February 23, 2010
Design Development
1 Rod spot welded as point of reference for placement of tension
2 Tension rods welded in place for pre-tensioning
3 Turnbuckle replaces orginal rod and as turnbuckle jack expands
outward, it forces rods into tension
1
2
3
4
TEAM 5 - Reid Olson, Romiar Karamooz, Sean Bollinger, Emily Akaba, Dani Alvarez
4 Triangular compression pieces are added to hold tension rods in place
and turnbuckle jack is removed
7
Axonometric 3” = 1’0”
6”
13
13
February 23, 2010
1’-0”
442 LAB iteration 1
8’-0”
Process
8” typ.
12
12
11
6
11
10
5
9
10
8
4
7
6
3
8
5
4
2
1
2
1
Plan sections at floor plates (@8”) 2-1/2” = 1’0”
222
steel towers |
Elevation Force Diagrams/Deformation 2-1/2” = 1’0”
structures |
P
Leveled jigs and welded using them to maintain plum verticality
Cardboard jigs placed on tube stock core
442 LAB iteration 1
3
TEAM (4,5 or 6) - Reid Olson, Romiar Karamooz, Sean Bollinger, Emily Akaba, Dani Alvarez
9
teaching |
LATERAL FORCES - LABORATORY PROJECT
Module I: Steel Structure - ARC 422 Building Technology VI - Spring 2010
Circle Scheme - Example Project - Iteration 2
LATERAL
FORCES - LABORATORY PROJECT
ELEVATION/
PLAN/ SECTION
GEOMETRIC RELATIONSHIP
Module I: Steel Structure
- ARC 422 Building Technology VI 5”
Spring 2010
GEOMETRIC RELATIONSHIP
ELEVATION/ PLAN/ SECTION
5”
Tier 3
8’-0”
Single 1/8” Steel Rod
Tier 3
8’-0”
Single 1/8” Steel Rod
5”-4”
Tier 2
5”-4”
Double 1/8” Steel Rod
Tier 2
Double 1/8” Steel Rod
ISOMETRIC
ISOMETRIC
2’-8”
2’-8”
STRATEGY
The overall strategy of the tower utilizes the diagrid
structure due to the light weight and redundant
functionality of the members.
STRATEGY
The overall
strategy
of the tower
utilizes
the diagrid
The tower
is built
of 24 individual
rods,
12 traveling
structure due
to the12light
weight and redundant
in a clockwise
rotation,
counterclockwise,
each
functionality
of the
members.
rod making
one full
revolution
around the tower.
Tier 1
Theoftower
built ofdirection
24 individual
rods, 12
traveling
The 12
each is
rotation
are further
paired
a clockwise
12 counterclockwise,
each
into in
6 goups
of tworotation,
each. Each
of these two paired
making
onetogether
full revolution
the tower.
rodsrod
become
closer
as thearound
height also
decreases in radius as it rises. These two strategies
Thethe
12decreasing
of each rotation
direction
further
reflect
moment
of the are
tower
as it paired
into 6ingoups
of two each. Each of these two paired
increases
height.
rods become closer together as the height also
decreases
radius
as ithelp
rises.
two strategies
Rings
encirclinginthe
tower
toThese
resist outward
reflect
theasdecreasing
moment
of thethe
tower
forces,
acting
tension rings
to restrain
out-as it
wardincreases
bucklinginofheight.
the tower when an axial load is
8”
Triple 1/8” Steel Rod
Tier 1
8”
Triple 1/8” Steel Rod
9”
9”
422_LAB 1/2_TEAM 11
422_LAB 1/2_TEAM 11
| teaching
| structures
imposed.
Rings encircling the tower help to resist outward
forces, acting as tension rings to restrain the outward buckling of the tower when an axial load is
imposed.
BACHELIER, GARCIAVONBORSTEL, HANSEN, LUTHALA, TAYLOR
BACHELIER, GARCIAVONBORSTEL, HANSEN, LUTHALA, TAYLOR
| steel towers
223
Diverse Schemes - Diverse Example Projects
PLAN SECTIONS,
PLAN, SECTIONS, ELEVATIONS
CRITICAL DIMENSIONS
SCALE: 3” = 1’
PLAN SECTIONS,
SCALE: 3” = 1’
DIMENSIONS OF EACH EQUILATERAL
TRIANGLE DATUM
SCALE: 3” = 1’
ELEVATION,
ELEVATION,
FROM ANGLED SIDE OF BASE
SCALE: 3” = 1’
Concept Statement:
Our project is based on the idea of a torsion tower which uses a rotating triangular form. As
thirteen levels of triangular plates are stacked on top of one another, each subsequent triangle is
reduced by 3/16" and is rotated by 15°. This torsion throughout the structure implements a
form-bias that helps to resolve lateral forces.
Another resolution to lateral forces comes from cross bracing, which continues from the
bottom of the tower to the top. This is different from our previous iteration, in which cross bracing
stopped in the center of the tower. Additionally, our triangular plates are now strengthened by
rods along the bottom of their perimeter which act as tension rings.
Hierarchy is implemented by separate vertical members that move through the three points of
the triangle and rotate with the tower. The rods are in bundles that reduce from three to two to one
as they move upward. This is in observance of the idea that the bottom of the tower receives the most
forces and thus needs the most material.
The other resolution to vertical hierarchy and lateral stability is the shear plates added to
the edges of the first seven levels. In our last iteration, the tower failed in its seventh floor so we
chose to strengthen this level and those below it. Our hope is that with a strong base, the tower
can withstand the forces which would otherwise cause it to buckle.
There are two moves happening as the tower rises vertically: rotation and tapering. The
following concept sketches depict those ideas separately and combined.
Concept A: Triangles are
consistently the same size
and are rotating
Concept B: Triangles are
tapering in size and are oriented
in the same direction
FROM FLAT SIDE OF BASE
SCALE: 3” = 1’
8”
8”
Combination: Triangles
are both rotating and
tapering
8”
concept sketches
P
P
P
P
plan
P
3D force diagrams
2.5” = 1’-0”
P
deformation (plan) diagram
P
1.5” = 1’-0”
1.5” = 1’-0”
3D force diagram
1’-0” = 1’-0”
8”
8”
8”
8”
8”
8”
8”
triangulation
1’-0” = 1’-0”
Iteration 2 - concept, 3D force diagrams, deformation
8”
8”
m_bassey s_elliott p_wojtczak e_shallcross d_terrasi
sp_2010 arc_422
8’
80”
Bundle detail 3” = 1’0”
The bundle connects the four
towers by the four inner corners.
56”
8ʼ
32”
80”
64”
48”
9”
32”
4.5”
8ʼ
9”
9”
8ʼ
Concept Diagram - Four individual towers that are
bundled and held together by exterior bracing.
9”
ARC 422_lab1_TEAM 1
224
J_Hollis, A_Jalalian, E_Lindberg, N_Rodriguez
steel towers |
ARC 422_lab2_TEAM 1
structures |
3”
J_Hollis, A_Jalalian, E_Lindberg, N_Rodriguez
teaching |