Author: Abdy Kermani Its a comprehensive textbook which provides in depth guidance on design methods of structural timber to BS 5268 Part 2 and proposed Eurocode 5.Deskripsi lengkap
timber
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Design of Timber StructuresDeskripsi lengkap
this example will guiding student to calculate timber structure.Deskripsi lengkap
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Structural Design of Timber to BS 5268 Timber engineering Design of timber beams Design of timber joists Design of timber columnsFull description
Descripción: Design of Timber Structures
PRACTICE PROBLEMS IN TIMBER DESIGN
Descripción: h
STRUCFull description
Design of Bending Members in Timber
An Example of Timber Beams
What can go wrong ? TIMBER BEAMS:
• Lateral torsional buckling • Shear failure • No Notc tch h fa fail ilur ure e • Bearing failure • Excessive deflections
max
y
M
=
y
=
I S M = σ max S
for rectangular sections
=
bh
Design Equation: r
= ϕ
b
Where F is the characteristic bendin stren th For timber it is Fb = f b (KDKHKSbKT)
6
2
Bending failure in compression • • Benign failure mode
Logging bridge near Pemberton, BC Glulam I-beam
• • Brittle •
om na on o ens on an s ear, although tension fracture is the n a ng mo e
Mr = φ Fb S KZb KL where
φ
and
Fb = f b (KD KH KSb KT )
= 0.9
Lateral torsional buckling
Glulam beams in a Gerber system
-
20f-E and 24f-E grades
-
better
20f-EX and 24f-EX grades
Lateral torsional buckling of Elastic buckling: cr = π e Torsional stiffness
y Lateral bending stiffness y y
Le
∆x
x
y x
o e: e warp ng s ness or rectangular shapes is small compared to the torsional and bending stiffness
y y
θ
x
Lateral torsional buckling of -
Capacity of a timber beam subject
Mr = φ Fb S KZb material failure
u
Mr = φ Fb S KZb KL
r combination of material failure and lateral torsional buckling
elastic lateral torsional buckling
Le
Lateral torsional buckling factor KL
L 1.0
KL = 1 L
=
–
B
K
0.67
prac ca
m
0.5 KL = (0.65 E KSE KT) / (CB2 Fb KX)
CK = ( 0.97 E KSE KT / Fb )0.5 0
0
10
20
30
40
50
C
Slenderness ratio CB = ( Le d / b2 )0.5
Deep glulam beam
d/b
Lateral support
d b
of lateral ors ona
at spacing:
purlins or tie rods
KL = 1.0 when lateral support is prov e as shown
< 610 mm
< .
compression edge held by
decking or joists < 610 mm
< 7.5 <9
op e ge < 8d
plus bridging
both edges
A y
d
max
N.A.
= V(0.5A)(d/4) (bd3/12)b =1.5 V/A
b
τ
=
τ
VAy Ib
=
VQ Ib
s
σv(max)
Vr = φ Fv 2/3 A KZv where
φ
= 0.9
and
F
=f
K K K
σv(avg)
K
σv(max)
= . σv(avg) = 1.5 V / A
UNBC Prince Geor e, BC
• • Shrinkage cracks often occur at the ends of beams in the zone of maximum shear stress This part of the load transferred in
• transfer of loads in the end zones reduces the total shear force to be carr e .
compression
45o critical section
volume < 2.0 m3: Vr = φ Fv 2/3 A KN where
φ
and
Fv = f v (KD KH KSv KT )
= 0.9
KN = notch factor (see next section)
For larger beams this is usually quite approach is used (see clause 6.5.7.3)
n
d dn
e
KN = ( 1 – dn /d )2
For e > d : KN = ( 1 – dn /d ) For e < d : KN = 1 – dne/[d(d – dn)]
• For notches on the tension side of supports (sawn lumber)
Notch effect in sawn um er
• In new code: Reaction calculation Fr =
NEW !!
Ft A KN
= 0.9
Area A
Ft = f t (KD KH KSt KT) where f t = specified reaction force strength = 0.5 MPa for sawn lumber K = 1.0 for dr and 0.7 for wet service conditions A = gross cross-section area KN = notch factor
Notch factor KN
d1
e
d
dn
Mechanics theory
⎛ N
⎝ α = −
⎛
. n
⎛ 1 ⎞ . − ⎝ ⎝ α ⎠ an η = e
2
⎛ 1 ⎞ ⎞ ⎞ − 3 ⎝ α ⎠ ⎠ ⎠
−0.5
•
“
”
• Often governs • y u grain but also tension of the fibres
compression perpendicular to grain tension of fibres along the edges
