large span timber structures - lth...large span timber structures roberto crocetti division of...

Large span timber structures

Roberto CrocettiDivision of Structural Engineering

Lund University

Table of contents

- Material efficiency

- Shape efficiency

- Common shapes

- Trusses

- Arches

- Bracing

Buckling due to only self-weight

Trees can reach considerable heights

How tall can we build a column before it buckles? (due to its ownweight)

Critical buckling length due to self-weight

3

2

25,1

rELcr

E: column’s E-modulr: radius of column’s cross section: column specific weight(r*g)

Specific strength and stiffness and material efficiency

(1) In case of compression, the values are usable only for members restrained against buckling

(2) Applies only for members in compression

(3) Applies only for member in tension

It is not a coincidence that among the largest spans, for roof

structures, are made by timber

Geodetisk kupol

Arena i Northern Michigan

University, Michigan, USA.

Diameter: 163 m

Pilhöjd: 49 m

Byggår: 1995

The superior dome

Two geodesic domes (coal power supply)Brindisi, Italy Diameter: 143 m (largest in Europe)Rise: 44 mBuilt in: 2014

The domes in Brindisi

Efficient shapes

Efficient shapes

Arch action Cable action

Efficient shapes

W

d

C

T

The efficiency of the beam is not too high because:- The parts of the beam close to neutral axis are almost unstressed- The lever arm is small (depth of the beam is approx. 1/20 of span)

Structural efficiency – Planar structures

In

cre

asin

g s

tructu

ral e

fficie

ncy

Beam

Truss

Arch

Suspension system

Structural efficiency – Spatial structures

3D-system

2D-system

If structural efficiency is the goal, one should avoid

bending and choose forms where members work

only in tension or in compression

Show an example of bending stress vs axial stress

Suppose that a force “F” acts in between the two supports

L/2

F

Support 1Support 2

L/2

Suppose that the force can be taken by means of the 2 structures below

L

f

F

b

h1

L

F b

h21 2

Hypothesis:- Bending strength = compression strength = f- Disregard bending in structure (1) (immovable

supports) and assume that buckling is not an issue- For the beam case, assume L/(h2)=20

Determine :1. The ratio of (Volume 1)/ (Volume 2) as a function of the slope a2. What is the value of that ratio when f/L=0,15?

Efficient shapes: observe this

Structural Engineering - Lund University 23

The shape of a hanging cable subjected to a set o load is similar to the shape of

the bending moment of a corresponding beam subjected to the same set of loads.

Efficient shapes: observe this


Cable shape vs bending moment diagram



The shape of a hanging cable subjected to a set o load is

similar to the shape of the bending moment of a corresponding

beam subjected to the same set of loads.

OK, so what?


If we give the structure the same shape as the hanging cable (or the same shape as the

bending moment of the corresponding beam), then we will have only tension in the

structure! (Or only compression if we turn the structure upside down)

Hanging rope subjected to “uniformly” distributed load

Thrust exerts a “pull” in the hands

This shape gives no bending moments!

Upside down rope (arch) subjected to “uniformly” distributed load

Thrust exerts a “push” in the hands

This shape gives no bending momentseither!

Structural efficiency – Spatial structures

Pedestrian bridge, Essing, Germany

Grandview Heights Aquatic, BC, Canada

Attachment to the

concrete supports

Parabola-shaped trusses

Beams with ”constant stress”


Parabola-shaped trusses

A few example from Nature…

(On Growth and Form, by Sir D’arcy Thompson, 1917)


Trusses

Typical shapes –truss with parallel chords

Typical shapes –pitched trusses

Typical shapes –curved trusses

Forces in the members

Ratio of:

lq

membertheinForce

Preliminary design

Preliminary design

In preliminary design, consider a reduced area of the cross section of the members:

Ared 0,7 A

Preliminary design

Choose:• Relatively wide cross sections (to allocate connection)• Relatively shallow cross section of lower and upper chord (to reduce bending

stiffness and thus to reduce magnitude of bending moments)

Preliminary design

Avoid eccentricity at the nodes


Clamps or hinges?

Hinges or clamps?


Model 1: Diagonals hinged to chords

55

Model 1: Diagonals hinged to chords

In this case the bending moment in the diagonals is obviously zero

56

Model 2: Diagonals clamped to chords

In this case the bending moment in the diagonals is ≠ 0

N M

57

Model 2: Diagonals clamped to chords

Note

1. The ratio above is not influenced by the width of the diagonal members

2. In timber structures the diagonals will never be completely clamped, thus bending

moment (and thus bending stresses) will be lower in practice

0,0

5,0

10,0

15,0

20,0

25,0

30,0

0 100 200 300 400 500 600 700

Depth of the diagonal h [mm]

100 MN

M

h

F


How do we design the truss in practice?

- Model the truss with continuous upper and lower chords and hinged web members

- Determine the Axial forces and the bending moments

- The nodes can be designed by considering pure axial force increase by approx. 15-20% in order to take into

consideration the presence of moment and shear

Buckling of trusses

In-plane

Out-of-plane (a)

Out-of-plane (b)


Buckling


How to increase stiffness and lateral stability


Nodes

Olympic hall in Hamar, 1994 L=71m

Courtesy Moelven Limtre A/SFd7000 kN!

Lintel truss beam

Courtesy Moelven Töreboda AB

The Perkolo bridge

47.5 m

The Perkolo bridge –the failure

The Perkolo bridge – the critical joint

47.5 m

Truss node and force polygon

The design force assumed by the designer for the

dimensioning of the node in the lower chord

The actual design force in the lower chord

Ductile failure mode

Glulam GL30c

Dowel: fy=900 MPa

Report:

Kollapsen av Perkolo bru – hva gikk galt?

Bell, K. 2016

Cross section of the lower chord

Arches

Preliminary design

Support conditions – tied arch

2 M30, 8.8

Support conditions – tied arch

Support conditions: directely on fundation

Statical systems

Three-hinged

Two-hinged

Zero-hinged

In general:- Concrete arches are usually “zero-hinged”

- Steel arches are usually “two or three-hinged”- Timber arches are usually “three-hinged” (when they are massive)

Three-hinged arch

Bridge over railway, Haninge

Three-hinged arch, Span: 35 m

Erection: 25 march 2017

Two-hinged trussed arch

(span: 71m) for road traffic

Tynset bridge

pre- assembling of the

arches in the factory

Fixed (zero-hinged)

Skubbergsenga bridge, Norway

Total length 40 m

Arch span 32 m

Bridge width 4 m

Building year 1997

Geometry considerations

Common rise to span ratio:

f/L=0,15 (up to 0,20)

circular arches

2

2

41 x

lfzEquation of parabola,

origin in the middle

2

3

81

l

fls

l

f4arctan

180

a

Forces in the cross section

Effect of symmetric and non-symmetric loading

Effect of non- uniformly distributed loads

How can we reduce the effect of bending moment in three-hinged arches?

Bandy hall in Nässjö (75m)

Bandy hall in Nässjö: Erection

Buckling of arches

• In plane buckling

• Out-of-plane buckling

Buckling modes

Different types of buckling analysis

The buckling load can be determined by 2nd- order analysis, by giving the arch initial

imperfection and increasing the load stepwise until instability is reached

By using simplified formulas to determine the buckling load. In this case the arch is

considered as a compressed strut with an appropriate buckling length

Buckling factor (according to Timoshenko)

&

for = 32, the buckling length factor becomes: b 1,17

Buckling Load at ¼ of the span

Inclined suspenders increase both in plane buckling

(and they also reduce bending moments in the arch)

After: Kolbein Bell

Bending in the arch with different hanger configuration

Buckling of the arch with different hanger configuration

Non-linear analysis

Out-of-plane buckling

Students at LTH getting acquainted with 1. lateral buckling and 2. bracing of arches


Buckling analysis is conducted as for a compressed bar with relevant buckling lengths

Bridge in Hägernäs, total length: 42m, arch span: 34m


Bracing of the bottom part of the arch

Extra compression strut

Bracing

Improper bracing during construction

Conversion of an unstable structure into a stable structure Unstable configuration Stable configurations

or(a)

(b)

(c)

(d)

(e)

(d’)

Bracing system’s main functions

• Transmission of horizontal loads

• Reduction of lateral deformations

• Enhancing buckling strength

Complete bracing

Bracing system for heavy timber structures

Bracing elements

• Longitudinal wall bracing (A)

• Transversal roof bracing (B)

• End wall bracing (C)

• Longitudinal roof bracing (D)

Stable or unstable?

Stable or unstable?

Equilibrium?

Stable or unstable?

Prerequisites for stability

• Wall bracings must be able to resist horizontal forces along three different directions in the plane

• The three directions shall not converge in the same point

• At least two of the three directions shall not be parallel one to another

Brace stiffness

Brace stiffness

Hp: Neglect the axial

deformation of beam and

column

Bracing of high walls

System (b) is in general more efficient than system (a), due to ab < aa

40o<a<55 o is a good compromise between economy and efficiency

System (b) is however more expensive than system (a)

Strength and stiffness requirements for bracing systems

Perfectly straight column


Equilibrium about “C”:

Equilibrium about “C”:

This tends to

unstabilise- Munst

This tends to

stabilise- Mst


Critical stiffness “CE”

The maximum axial load that the column can resist cannot be larger than the buckling load of the column

Column with several braces

a

PC E

E 4,3

(for 3 brace points)

Column with several braces

(for more than 4 braces)

a

PC E

E 4

Beam bracing

Beam bracing

H

MN d

d

3

2

Lateral forces in beams -modelR

qNd

Rq

Nd

Rq

LH

R

NqdRqdN d

rrd aa

Nd

R

Nd

qr

dadl=R·da

What is “R”

Assumed initial deformation: parabolic shape

DT

x

yL

R

2

4

D

L

xy T

Assumed shape

2

21

dx

yd

R

2

2

2

2

2

2 84

1

LL

x

dx

d

dx

yd

R

TT

D

D

Lateral forces

Nd

Rq

Nd

Rq

R

Nq d

r 2

81

LR

TD

T

dr

LNq

D

8

2

&

Estimation of lateral loads for glulam structures

• Initial out-of-straightness: D0=L/500

• Additional deformation D (e.g. due to wind load) shall not exceed L/500.

• This means that the final deformation shall be (maximum value)

250)( 0

LT DDD

Lateral forces

Nd

Rq

Nd

Rq

T

dr

LNq

D

8

2

H

MN d

d

3

2

HL

Mq d

r

20

250

LT D

Brace forces for a series of bent members

EC5 approach

crit

f

dh k

LHk

Mnq

1

3,

Md= design moment in the beam

H = depth of beam

L= span of the beam

n= number of laterally braced beams

kf,3= modification factor (kf,3=30-80)

kcrit= reduction factor for lateral buckling when the beam is unbraced

Erection of timber structures

Timber struts as bracing system

Erection: pair of fully braced trusses

Bracing by diaphragm action

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