conférences cecobois
Effective Use of Timber in Architectural Applications
David Bowick, P.Eng. February 20, 2014
Taught the answer is usually Repetition, but is that true?
Efficiency – The great Myth
Repetition
Piece Count
Tonnage
More complex can be cheaper
What would Gary Williams Say?
Complexity Curve
Complexity
Slab Joists Trusses Opt. Joists Reciprocal
Quantity
Cost
Unit Cost
“But Dave, we want an example of efficient, complex construction!”
Reciprocal Framing - “A structure made up of mutually supporting beams in a closed circuit”
- Reciprocal Frame Architecture Olga Popovic Larsen, 2008
“He lifted himself by the seat of his pants and I'll never forget the grim look on his face
as he hoisted himself and took leave of this place “ - The Lorax Dr Suess, 1971
Reciprocal Frames
How it Works
How do you span 16’ joists 20’?
Example
Image from minktoast.net
Example
Fort York Armoury - Toronto Built in 1935
Example
NCFS Longhouse – Toronto, 2009 Levitt Goodman Architects Blackwell Engineers
Case Study Indian River Pavilion – PEI David Sisam Blackwell Engineers Construction Documents
Options 1 – Valley Framing Piece Count – 15 Total Volume – 5.8 m3 Largest Piece – 731 kg
Options 2 – Reciprocal Beam Piece Count – 37 Total Volume – 4.8 m3 Largest Piece – 268 kg
Options 3 – Reciprocal Joist Piece Count – 47 Total Volume – 5.6 m3 Largest Piece – 60 kg
Summary
Piece Count
Total Volume (m3)
Largest Piece (kg)
Valley Beam
Reciprocal Beam
Reciprocal Joist
15
5.8
731
37
4.8
268
47
5.6
60
Prestress is a method of strengthening or stiffening a structural element by introducing internal stresses. It can be applied during the manufacturing process or during the installation process.
What is Pre-Stress?
Tempered Glass Prestressed Concrete Tensile Structures
Prestressing can be used to couple a strong flexible element (steel tendon) with a weak stiff element (concrete) to create a strong stiff element (post tensioned concrete)
Why Pre-Stress?
Strength of Material
Stiffn
ess
of Sy
stem
*
* Strength of system held constant.
• Simple Span Roof
• L = 18m
• Structure at 2.0 m c/c
• d = 0.8 kPa
• l = 1.2 kPa
• w = 1.2 kPa (gross uplift)
Case Study
Sim
ple
Bea
m
Pin
ned
Ch
ord
Tru
ss
“But Dave… What about uplift?”
Pin
ned
Ch
ord
Tru
ss -
Up
lift
Co
nti
nu
ou
s C
ho
rd T
russ
We have a valid load path and it's strong, but to satisfy L/300 gross uplift deflection we need a section almost as big as the simple beam section. The cable utilization is negligible and the trussing is just “bling”.
The Dilemma
Co
nti
nu
ou
s C
ho
rd T
russ
w/
Pre
stre
ss
Simple Beam – 265x646
Pinned Chord Truss – 265x380 But it's irrelevant because it doesn't work for uplift. We could ballast for uplift, but increased load results in increased size. Cable oversized to achieve stiffness.
Continuous Chord Truss – 265x570 Governed by uplift stiffness.
Prestressed Truss – 265x342 40% Savings in glulam relative to non-prestressed truss. Remains rigid under load reversal.
Summary
An Aside About Uplift While we reasonably consider net uplift when designing for strength, we need to consider gross uplift when designing for stiffness.
An Aside About Deflection Criteria The “L-over” criteria for deflection is completely irrelevant for cantilevers.
A Possible Criteria
“So Dave… How do we design using prestress?”
Designing with Prestress Most non-linear versions of analysis programs such as ETABS and SAP can accommodate prestress directly.
L = LoT
where = coefficient of thermal expansion
L = TLo/AE
Therefore
LoT = TL
o/AE, and
T = T/AE or T = TAE/ *
Designing with Prestress Most simple analysis programs have the ability to accommodate stresses due to temperature changes. We can use this to apply prestress:
This is valid for a member restrained against perfectly rigid supports. Against elastic supports the force will be reduced by the movement of the supports and the temperature change will have to be adjusted.
“Dave… How do they erect this?”
Applying Prestress
Manufacture elements intentionally short by a specific amount. Force fit. This method is imprecise and probably appropriate for nominal prestress only.
DRIFTING
Elements can be pulled directly using a calibrated tension jack such as those used in prestressed concrete. This is probably most appropriate where prestress forces are large.
CONTROLLED STRETCH
The structure can be deformed intentionally using simple tools such as bottle jacks and come-alongs. The prestressing tendon can be attached under zero stress, then the jack or come-along relieved, transferring the load to the tendon. The measured impact of the prestressed element can be predicted fairly accurately so the deformation of the structure itself becomes the gauge.
SPRUNG STRUCTURE
“Dave, I'm still not comfortable. This is fine in principal, but I
have no feel for it?”
Think about Bow Saws Direct tensioning using lever
Think about Long Bows Tension by Springing Structure
Hydraulic Jacks
A Real Job
1. Install perimeter beams
and temporary jack at
mid-span
2. Install top chord, attaching
at one end and letting
bear on temporary jack
3. Pull down on other
end of beam to achieve
force noted
4. Attach cables and
connect to other beam,
then remove jack.
To Brace or Not to Brace
“Dave, I find tensile systems are complex. How do we apply these
methods to more complex systems?”
Load Case 1
100 kN Load Case 2
50 kN
Analyze the frame with cables as though the cables
were capable of resisting compression. Step 1:
Determine and apply prestrain to each cable that will
(hopefully) result in prestress to overcome the
compression.
T1 = T/EA
= (43,000N)/[(11.7x10-6)(135,000MPa)(283mm2)]
=96°C
T2 = T/EA
= (75,000N)/[(11.7x10-6)(135,000MPa)(283mm2)]
=167°C
Step 2: Or Misstep # 1
The prestress is not in internal
equilibrium. When we apply T
and analyze, the structure deforms
slightly to achieve equilibrium and
our forces change.
Make a note of the relationship between the
prestress forces.
P1/P2 = 40kN / 56kN
= 0.71
Analyze the system. Are the Forces what we expected?
When you apply prestress in a system, the forces will redistribute to achieve
equilibrium regardless of how you applied the prestrain.
Apply T to cable 1 only, and leave cable 2 alone.
Make a note of the relationship between the
prestress forces.
P1/P2 = 16.5 kN / 23.2 kN
= 0.71
Try This:
Apply T to cable 2 only, and leave cable 1 alone.
Again, make a note of the relationship between
the prestress forces.
P1/P2 = 19.1 kN / 26.8 kN
= 0.71
Now This:
Preq/P = 43.4/16.5
=2.63
Preq/P = 75/23.2
What Next: Determine the ratio of required prestress to the prestress resulting from an arbitrary application of prestrain for each cable.
Multiply your arbitrary T by the highest of these ratios. 3.23 in this case.
=3.23
But if we actually apply it this way, our frame drifts. The geometry
changes when the forces redistribute to generate equilibrium.
=23mm
This level of prestress is in equilibrium and adequate to
overcome the worst case cable compression. Apply T
Why isn’t the result exact?
T1=53300/[(11.7x10-6)(135000)(284)]
= 119 °C
T2=75000/[(11.7x10-6)(135000)(284)]
= 167 °C
Apply T to each element that will result in prestress that is distributed in proper proportion and is adequate to overcome the compression in the worst case.
Step 3:
The movement of the other elements under prestress all result in small changes
in force. In this case the top beam sags slightly relieving the prestress.
Load Case 1 + Prestress
=22.7 mm
Now, we are done!
Final Design Envelope Load Case 2 + Prestress
Load Case 1
100 kN
Load Case 2
50 kN
“Why go to all this trouble? Why wouldn’t I just use tension only
bracing, and let the compression side go slack? It looks like the final
design force is the same”
Load Case 1
100 kN
=37.4 mm
=>22.7 mm
With a conventional tension only system the “compression”
cable doesn’t contribute to stiffness, so if the system is
stiffness controlled, members have to be made larger to
compensate.
“Dave, you’ve convinced me. The technical arguments are compelling.
But how do I make it beautiful?”
Swage Eye and Fork
Cable Hardware
Open Spelter Socket
Cable Hardware
Take-up assembly, Open and Closed
Cable Hardware
Strut Hardware
Universal Pin Connectors
Architectural Swage Fitting
Rod Hardware
Forged Clevis
Rod Hardware
“Dave, what does this look like in real world projects?”
Milton Leisure Centre
Circa 1991
Circa 1999
Cawthra Community Centre
Completed 2004
Sir Sandford Fleming College
Rock community Church
Circa 2006
Lakefield College School
Circa 2006
Completed 2012
Local Church of the Saints
St. Catharines Aquatic Centre
Completed 2012
“Dave, those are all trusses. What about slabs and beams? And is it only useful
for pretty stuff?”
Stress Laminated Timber Bridge Decks
US Department of Transportation http://www.fhwa.dot.gov/publications/publicroads/97winter/p97wi32.cfm
Proposal for Two Way Prestressed Composite Wood/Concrete Bridge - Masters Thesis by Andrew Lehan, U of T
Prestressed Timber/FRP Bowstring Arch Bridge - Robert Widmann,
URS Meier, EMPA Swiss Federal Laboratories for Material Science and Technology, Structural Engineering Research Laboratory
“Dave, every client wants to work with CLT. How does this help?”
Using prestress we can join CLT panels together to span in their weak direction creating genuine two-way action. In this example a 7 ply assembly spans 6 m supporting 1.2 kPa Dead and 1.2 kPa snow.
In this example, we increased the bending capacity of a 215x380 beam by 22% using
prestress. Coincidentally this is almost exactly the increase achieved by increasing the beam
size by one ply. This is likely a useful technique only in applications where depth is
constrained. It could be the basis for full strength field splices of glulam beams, however.
“But wait, there’s MORE!”
Belleville (Spring) Washers
Some Things About Springs
Beam Reaction, R = 72 kN
Objective: Insert a stack of springs such that following shrinkage, the tension rod will retain nominal tension under load (to remain rigid). Say Tresidual = 15% R.
Maximum deformation of each spring to be 75% h0.
R = 72 kN Fresidual = R + Tresidual
= 72 kN + 15% * 72 kN = 82.8 kN
Consider nesting 3- DSC 112x57x6 Spring Washers
Now F75% = 43.8 kN Each Fresidual = 1/3 * 82.8 = 27.6 kN 25% = 0.63 mm @ F = 15.9 kN 50% = 1.25 mm @ F = 30.2 kN 75% = 1.88 mm @ F = 43.8 kN
By linear interpolation,
residual = (27.6-15.9)/(30.2-15.9)*1.25 + (30.2-27.6)/(30.2-15.9)*0.63 = 1.14mm
/stack = 75% - residual = 1.88 - 1.14 = 0.74 mm #stacks = residual / residual = 6.35/0.74 = 8.6
Say 9 Stacks
“Good judgment comes from experience. Experience comes from poor judgment”
-unknown
Thank You!
Bla
ckw
ell
This concludes The American Institute of Architects
Continuing Education Systems Course
Canadian Wood Council
Wood WORKS! Alberta
www.cwc.ca
www.wood-works.org