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Dissemination of information for training – Brussels, 2-3 April 2009 slide 1 out of 23
EUROCODE 6Background and applications
EUROCODE 6Design of masonry structures
Unreinforced masonry Shear loading
Prof. Dr.-Ing. Wolfram JägerDresden University of Technology
Dr.-Ing. Sebastian OrtleppJaeger Consulting Engineers Ltd., Office for Structural Design
Dipl.-Ing. Carola HauschildJaeger Consulting Engineers Ltd., Office for Structural Design
Dissemination of information for training – Brussels, 2-3 April 2009 slide 2 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Types of actions for verification against shear
In plane Out of plane
• wind action perpendicular to the wall
• pressure of earth.
• transmission of wind loads
• stiffening forces
Dissemination of information for training – Brussels, 2-3 April 2009 slide 3 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Bracing of buildingsThe layout of the walls in the ground plan of the building should be foreseen in such a way that the sufficient bracing of a building should be guaranteed.
The principles of bracing are:• concrete ceiling or ring beam• more than 3 walls• the axes should not intersect in one point
Dissemination of information for training – Brussels, 2-3 April 2009 slide 4 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Bracing of buildings
• favourable
• unfavourable
• instable
M0
high excentricity
M0
M0M0
Dissemination of information for training – Brussels, 2-3 April 2009 slide 5 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification format EN 1996-1-1
filled head joints
unfilled head joints
shell bedded masonry
⎩⎨⎧ ⋅
≤⋅+=vlt
bdvkvk f
fff
065,04,00 σ
⎩⎨⎧ ⋅
≤⋅+⋅=vlt
bdvkvk f
fff
045,04,05,0 0 σ
⎩⎨⎧ ⋅
≤⋅+⋅=vlt
bdvkvk f
ff
tgf
045,04,00 σ
cvdRd ltfV ⋅⋅=
NDP - 1
NDP - 2
Dissemination of information for training – Brussels, 2-3 April 2009 slide 6 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification format
Variables:fvk0: initial shear strength without loadσd: compressive stress, perpendicular with shear loadfb: normalized compressive strength of masonryfvlt: limit of shear strengthg: overall with of mortar stripst: thickness of the wall
Dissemination of information for training – Brussels, 2-3 April 2009 slide 7 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification format EN 1996-3Simplified calculation methods
with: cv = 3 filled head joints cv = 1,5 unfilled head joints eEd: excentricity of load t: thickness of the wallfvd0= fvk0/γMNEd: vertical loadl: length of the wallfvdu: ultimate shear strength (0,065 fb or fvlt/γM)
vduEdM
EdvdoEdvRd ftelN
ftelcV ⋅⋅⎥⎦⎤
⎢⎣⎡ −≤⋅+⋅⋅⎥⎦
⎤⎢⎣⎡ −⋅=
234,0
2 γ
Dissemination of information for training – Brussels, 2-3 April 2009 slide 8 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification modelBackground – theory
Failure due toa) gaping of bed joint b) friction of bed jointc) tensile failure of the unitsd) crushing
σDd
a
σDd
b
σDd
d
σDd
τ
tensile strength
cdba
initial shear strength Shear strength
Dissemination of information for training – Brussels, 2-3 April 2009 slide 9 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification modelBackground – theory
result:friction failure
equilibrium of forces at a masonry unit
theory of principle stresses
tensile failure (anisotrophy)
σ
Δx
τ
τΔy
Δσσ Δσσ
a
bDd
Qb
( ) yx4x
2x22 Δ⋅Δ⋅τ=
Δ⋅
Δ⋅σΔ⋅
( )22
DdDdvb 3,2
42f τ⋅+
σ±
σ=
Ddvk0vk ff σ⋅μ+=bt
Ddbtvk f
1f45,0f σ+=
Dissemination of information for training – Brussels, 2-3 April 2009 slide 10 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Comparison to theory
AAC Brick with vertical holes
Dissemination of information for training – Brussels, 2-3 April 2009 slide 11 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Procedure in plane shear loadingA) Actions• wind action• bracing forces• normal stresses/forces and
compressed area– Bending action due to in plane lateral
forces– Determination of the compressed area– Bending due to deflection of the ceilings
B) Resistance• Determination of shear strength and
shear resistanceC) Verification• Design action ≤ Design resistance
y
x
NEk
z
HEk
NRF
NEF
f
y
x
NEk
z
NRF
NEF
NRK
f
RdEd V V ≤
Dissemination of information for training – Brussels, 2-3 April 2009 slide 12 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Procedure out of plane shear loadingA) Actions• Determination of wind action• Determination of normal stresses/forces
and compressed area– Bending action due to out of plane lateral
forces– Determination of the compressed area
B) Resistance• Determination of shear strength and
shear resistanceC) Verification• Design action ≤ Design resistance
RdEd V V ≤
Dissemination of information for training – Brussels, 2-3 April 2009 slide 13 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Working exampleBuildingBrick with vertical holes
longitudinal section A-A
Dissemination of information for training – Brussels, 2-3 April 2009 slide 14 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Working exampleBuilding
first floor plan
Beams
Dissemination of information for training – Brussels, 2-3 April 2009 slide 15 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Pos. W2 interior wallGeometryt = 0,24ml = 2,24ml1 = 3,60m
Material parametersclay bricks, group 2
fb = 15 N/mm²mortar M 2,5,
fm = 2,5 N/mm²
g = 5,5 kN/m²; q = 1,5 kN/m²Ag= 56,6 kN; Ap= 16,9 kN
zone of beam load
beam
Verification of shear loading acc. EN 1996-1-1
Dissemination of information for training – Brussels, 2-3 April 2009 slide 16 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Vertical loadsbeam
Ag=56,6 kN; Aq=16,9 kNangle of load distribution 60°
line force from dead load of the wall gwall=4,67 kN/m²
line supporting force from dead load of the ceilingg = 5,5 kN/m²; q = 1,5 kN/m²
verification of bending in plane: see 3_am_6_Jäger.pdf on the stickdetermination of input data for shear check from bending in
planelc = l = 2,24 m (linear stress distribution)min N = 22,96 kN
Dissemination of information for training – Brussels, 2-3 April 2009 slide 17 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Load case combinations
LFK 1
1,35g+1,5q
LFK 2
1,35g
LFK 3 LFK 4 LFK 5
LFK 6
1,35g+1,5q 1,35g
1,35g+1,5q 1,35g+1,5q
1,0 g
1,35g+1,5q
1,35g
1,35g 1,35g
LFK LFK 8 LFK 9
1,0g+1,5q 1,35 g+1,5q
1,0g+1,5q
1,0g
1,0g 1,5H,w
1,35 N +1,5Ng q 1,35 N +1,5Ng q 1,35 Ng 1,35 Ng 1,35 Ng
1,0 Ng 1,0 Ng 1,0 Ng 1,35 N +1,5Ng q
1,35 g+1,5q
1,5H,w 1,5H,w 1,5H,w
1,0g
Dissemination of information for training – Brussels, 2-3 April 2009 slide 18 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification to shear loading Design value of the shear resistance (load case 1)
with
(filled head joints)
M
cvkRd
ltfV
γ⋅⋅
=
dvkovk 4,0ff σ⋅+=
²m/MN20,0fvko = unit group 2, mortar M2,5 from table NA
m/MN427,024,224,0
23,0ltNminvorh 2
cd =
⋅=
⋅=σ
²m/MN371,0427,04,020,0fvk =⋅+=
²m/MN975,015065,0f065,0 b =⋅=⋅<
γM =1,7 from NA (see table 1 DIN EN 1996-1-1)
Dissemination of information for training – Brussels, 2-3 April 2009 slide 19 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification to shear loadingDesign value of the shear resistance
(unfilled head joints)
(shell bedded joints)(g/t = 0,5)
(filled head joints)
(unfilled/shell bedded)
kN3,1177,1
24,224,0371,0VRd =⋅⋅
=
RdEd V)kN7,85(kN17,31 kN61,29 V =<=
²m/MN271,0427,04,020,05,0fvk =⋅+⋅=
²m/MN675,015045,0f045,0 b =⋅=⋅<
²m/MN271,0427,04,020,05,0fvk =⋅+⋅=
²m/MN675,015045,0f045,0 b =⋅=⋅<
kN7,857,1
24,224,0271,0VRd =⋅⋅
=
Dissemination of information for training – Brussels, 2-3 April 2009 slide 20 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification to shear loading simplified method Requirements• stores above ground level ≤ 3; • the walls are fixed either through the ceiling or through appropriate
constructions, such as ring beams; • load depth of the ceiling and the roof on the wall is at least 2 / 3 of wall
thickness, but not less than 85 mm; • floor height ≤ 3.0 m; • the smallest dimensions in the building floor plan is at least 1 / 3 of
building height; • the characteristic values of the variable loads on the ceiling and the roof
≤ 5.0 kN / m²; • span of the ceiling ≤ 6.0 m;
EN 1996-3
Dissemination of information for training – Brussels, 2-3 April 2009 slide 21 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification to shear loading - simplified method Design value of the shear resistance
cv = 3 (filled head joints)
vduEdM
EdvdoEdvRd fte
2l3
N4,0fte
2lcV ⋅⋅⎥⎦
⎤⎢⎣⎡ −≤
γ⋅+⋅⋅⎥⎦
⎤⎢⎣⎡ −⋅=
7,1371,024,0376,0
224,23
7,1229,04,0
7,12,024,0376,0
224,23VRd
⋅⋅⎥⎦⎤
⎢⎣⎡ −≤
⋅+⋅⋅⎥⎦⎤
⎢⎣⎡ −⋅=
kN117kN180VRd ≤=
RdEd VkN171 kN61,29 V =<=
m376,0585,229031,77
NMe
Ed
EdEd ===
Dissemination of information for training – Brussels, 2-3 April 2009 slide 22 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Verification to shear loading - simplified methodDesign value of the shear resistancecv = 1,5 (unfilled head joints)
7,1371,024,0376,0
224,25,1
7,1229,04,0
7,12,024,0376,0
224,25,1VRd
⋅⋅⎥⎦⎤
⎢⎣⎡ −≤
⋅+⋅⋅⎥⎦⎤
⎢⎣⎡ −⋅=
kN5,58kN90VRd ≤=
RdEd VkN8,55 kN61,29 V =<=
Dissemination of information for training – Brussels, 2-3 April 2009 slide 23 out of 23
EUROCODE 6Background and applications Unreinforced masonry – shear loading
Applikation of fvlt
tensile failure of the unit
fbt: tensile strength of the unitfbt = 0,025 · fb hollow blocks;fbt = 0,033 · fb vertically perforated bricks and units with grip hole;fbt = 0,040 · fb full blocks without grip hole;
simplified(filled head joints)(unfilled head joints)
bt
Ddbtvklt f
ff σ+= 145,0
bvklt f065,0f =
bvklt f045,0f =
Dissemination of information for training – Brussels, 2-3 April 2009 slide 24 out of 23
EUROCODE 6Background and applications Hint to ESECMaSE-Project
Reason: 1,5⋅H/1.0⋅N = 1,5⋅eprevious
Aim: description of the shearfailure closer to reality to usethe reserves
Done:a) Matrial tests/test methodsb) Static/cyclic shear testsc) Pseudodynamic testsd) Shaking tabel testse) Theoretical analysesf) Engineering modelg) Recommandations for the practice
Test on shaking table in Athens (CS)
Collapse analyses
ESECMaSEEnhanced Safety and Efficient Construction of Masonry Structures in Europe
Dissemination of information for training – Brussels, 2-3 April 2009 slide 25 out of 23
EUROCODE 6Background and applications
for AAC:Tensile
failure of the unit
Friction
Gaping
Bending
Proposal from WP 4
⎟⎟⎠
⎞⎜⎜⎝
⎛⋅⋅
−⋅⋅
=flt
NNVv
bending
2
21λ
⎟⎟⎠
⎞⎜⎜⎝
⎛+
⋅=
hhNlV
b
bgaping
112
NV friction ⋅= μ
Hint to ESECMaSE-Project
( ) ( ) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛++⋅⋅⋅⋅=
− 1111
,
2*2*,
calbt
dcalbtunit f
FFtlfc
V σ
( ) ( )⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛++⋅⋅⋅⋅⋅=
− 114
121
,
2*2*
,,calbt
dcalbtAACunit f
FFtlfc
V σ
btfF 85.02.1* +=
Eng.-Model: From stress level to force level
ESECMaSEEnhanced Safety and Efficient Construction of Masonry Structures in Europe
Dissemination of information for training – Brussels, 2-3 April 2009 slide 26 out of 23
EUROCODE 6Background and applications Anauncement of 8th IMC Dresden 2010
Dissemination of information for training – Brussels, 2-3 April 2009 slide 27 out of 23
EUROCODE 6Background and applications Anauncement of 8th IMC Dresden 2010
Dissemination of information for training – Brussels, 2-3 April 2009 slide 28 out of 23
EUROCODE 6Background and applications Anauncement of 8th IMC Dresden 2010
Dissemination of information for training – Brussels, 2-3 April 2009 slide 29 out of 23
EUROCODE 6Background and applications Anauncement of 8th IMC Dresden 2010