Unreinforced masonry Shear loading - Eurocodes

EUROCODE 6 

Background and applications 

Dissemination of information for training – Brussels, 2-3 April 2009 slide 1 out of 23 

EUROCODE 6 

Design of masonry structures 

Unreinforced masonry 

Shear loading 

Prof. Dr.-Ing. Wolfram Jäger 

Dresden University of Technology 

Dr.-Ing. Sebastian Ortlepp 

Jaeger Consulting Engineers Ltd., Office for Structural Design 

Dipl.-Ing. Carola Hauschild 

Jaeger Consulting Engineers Ltd., Office for Structural Design

EUROCODE 6 


Unreinforced masonry – shear loading 


Types of actions for verification against shear 

In plane 

• transmission of wind 

loads 

• stiffening forces 

Out of plane 

• wind action perpendicular to 

the wall 

• pressure of earth.

EUROCODE 6 




Bracing of buildings 

The 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

EUROCODE 6 




Bracing of buildings 

• favourable 

• unfavourable 

• instable 

M 0 

M 0 

M 0 

M 0 

high 

excentricity

EUROCODE 6 




Verification format EN 1996-1-1 

V 

Rd 

= 

filled head joints 

f 

vd 

⋅t 

⋅l 

f 

vk 

c 

= 

f 

vk 

0 + 0,4 ⋅σ 

d 

NDP - 2 

⎧0,065⋅ 

≤ ⎨ 

⎩ f vlt 

f 

b 

unfilled head joints 

f 

vk 

= 

0,5 

⋅ f 0 + 0,4 ⋅σ 

vk 

d 

≤ 

⎧0,045⋅ 

⎨ 

⎩ f vlt 

f 

b 

shell bedded masonry 

NDP - 1 

f 

vk 

= 

g 

t 

⋅ 

f 

vk 

0 + 0,4 ⋅σ 

d 

≤ 

⎧0,045 

⋅ 

⎨ 

⎩ f vlt 

f 

b

EUROCODE 6 




Verification format 

Variables: 

f vk0 : initial shear strength without load 

σ d : compressive stress, perpendicular with shear load 

f b : normalized compressive strength of masonry 

f vlt : limit of shear strength 

g: overall with of mortar strips 

t: thickness of the wall

EUROCODE 6 




Verification format EN 1996-3 

Simplified calculation methods 

V 

Rd 

= 

c 

v 

⎡ l ⎤ 

N 

⋅ ⎢ − eEd 

⎥ ⋅ t ⋅ f vdo + 0,4 ⋅ 

⎣ 2 ⎦ 

γ 

Ed 

M 

≤ 

⎡ l 

3⎢ 

⎣ 2 

− 

e 

Ed 

⎤ 

⎥ 

⎦ 

⋅ 

t 

⋅ 

f 

vdu 

with: 

c v = 3 filled head joints 

c v = 1,5 unfilled head joints 

e Ed : excentricity of load 

t: thickness of the wall 

f vd0 =f vk0 /γ M 

N Ed : vertical load 

l: length of the wall 

f vdu : ultimate shear strength (0,065 f b or f vlt 

/γ M 

)

EUROCODE 6 




Verification model 

Background – theory 

σσ σ Dd Dd Dd 

τ 

a 

b 

d 

c 

a 

b 

d 

initial shear strength Shear strength 

tensile strength 

σ Dd 

Failure due to 

a) gaping of bed joint 

b) friction of bed joint 

c) tensile failure of the units 

d) crushing

EUROCODE 6 




Verification model 

Background – theory 

σa 

σ 

σ 

Dd 

Δy 

equilibrium of forces at a 

masonry unit 

Δσ 

b Δσ 2 ( 2 ⋅ Δσ) ⋅ ⋅ = τ ⋅ Δx 

⋅ Δy 

2 4 

τ 

Q b 

τ 

Δx 

Δx 

theory of principle stresses 

Δx 

result: 

friction failure 

f 

= 

vk f vk0 

+ μ ⋅ 

σ 

Dd 

f 

2 

σDd 

σDd 

vb = ± + 3 

2 

( 2, ⋅ τ) 2 

tensile failure (anisotrophy) 

f 

vk 

= 

0,45 f 

bt 

4 

1+ 

σ 

f 

Dd 

bt

EUROCODE 6 




Comparison to theory 

AAC 

Brick with 

vertical holes

EUROCODE 6 




Procedure in plane shear loading 

A) 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 resistance 

C) Verification 

• Design action ≤ Design resistance 

VEd ≤ V Rd 

H E 

k 

y 

z 

y 

z 

f 

k 

N E 

F 

N E 

k 

N E 

K 

N R 

F 

N E 

F 

N R 

N R 

F 

x 

f 

x

EUROCODE 6 




Procedure out of plane shear loading 

A) 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 resistance 

C) Verification 

• Design action ≤ Design resistance 

VEd ≤ V Rd

EUROCODE 6 




Working example 

Building 

Brick with 

vertical holes 

longitudinal section A-A

EUROCODE 6 




Working example 

Building 

Beams 

first floor plan

EUROCODE 6 




Pos. W2 interior wall 

Geometry 

t = 0,24m 

l = 2,24m 

l 1 = 3,60m 

Material parameters 

clay bricks, group 2 

f b =15N/mm² 

mortar M 2,5, 

f m = 2,5 N/mm² 

Verification of shear loading 

acc. EN 1996-1-1 beam 

zone of 

beam load 

g = 5,5 kN/m²; q = 1,5 kN/m² 

A g = 56,6 kN; A p = 16,9 kN

EUROCODE 6 




Vertical loads 

beam 

A g =56,6 kN; A q =16,9 kN 

angle of load distribution 60° 

line force from dead load of the wall 

g wall =4,67 kN/m² 

line supporting force from dead load of the ceiling 

g = 5,5 kN/m²; q = 1,5 kN/m² 

verification of bending in plane: see 3_am_6_Jäger.pdf on the stick 

determination of input data for shear check from bending in 

plane 

l c = l = 2,24 m (linear stress distribution) 

min N = 22,96 kN

EUROCODE 6 




Load case combinations 

1,35 N 

g+1,5Nq 1,35 N 

g+1,5Nq 1,35 N g 

1,35 N g 

1,35 N g 

1,35g+1,5q 

1,35g 

1,35g+1,5q 

1,35g+1,5q 

1,35g 

1,5H,w 1,5H,w 1,5H,w 

1,35g 1,35g+1,5q 1,35g 1,35g 

1,35g+1,5q 

LFK 1 

LFK 2 

LFK 3 LFK 4 LFK 5 

1,0 N g 1,0 N g 1,0 N g 

1,35 N g+1,5Nq 

1,0 g 

1,0g+1,5q 

1,0g 

1,35 g+1,5q 

1,5H,w 

1,0g 

1,0g 

1,0g+1,5q 

1,35 g+1,5q 

LFK 6 

LFK LFK 8 LFK 9

EUROCODE 6 




Verification to shear loading 

Design value of the shear resistance (load case 1) 

V 

Rd 

with 

f 

= 

f 

vk 

⋅ t ⋅ l 

γ 

M 

c 

vk = fvko 

+ 0, 4 

⋅ 

σ 

f vko = 0,20MN/ m² 

d 

γ M =1,7 from NA 

(see table 1 DIN EN 1996-1-1) 

unit group 2, mortar M2,5 from table NA 

vorhσ 

minN 

t ⋅ l 

0,23 

0,24 ⋅ 2,24 

d = = = 

c 

0,427MN/ m 

2 

f vk = 0,20 + 0,4 ⋅ 0,427 = 0,371MN/ m² 

< 0,065 

⋅ fb = 0,065 ⋅15 

= 0,975 MN / m² 

(filled head joints)

EUROCODE 6 




Verification to shear loading 

Design value of the shear resistance 

f vk = 0,5 ⋅ 0,20 + 0,4 ⋅ 0,427 = 

0,271MN / m² 

(unfilled head joints) 

< 

0,045 

⋅ fb = 0,045 ⋅15 

= 

0,675 MN / m² 

f vk = 0,5 ⋅ 0,20 + 0,4 ⋅ 0,427 = 

< 0,045 

⋅ fb = 0,045 ⋅15 

= 

0,271MN / m² 

0,371⋅ 

0,24 ⋅ 2,24 

V Rd = 

= 117,3 kN 

1,7 

0,271⋅ 

0,24 ⋅ 2,24 

V Rd = 

= 85,7 kN 

1,7 

V = 29,61kN < 117,3kN (85,7kN) = 

0,675 MN / m² 

Ed V Rd 

(shell bedded joints) 

(g/t = 0,5) 

(filled head joints) 

(unfilled/shell bedded)

EUROCODE 6 




Verification to shear loading simplified method 

Requirements 

EN 1996-3 

• 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;

EUROCODE 6 




Verification to shear loading - simplified method 


V 

Rd 

= 

c 

v 

⎡ 

⋅ ⎢ 

⎣ 

l 

2 

Ed 

⋅ t ⋅ f 

vdo 

c v = 3 (filled head joints) 

V Rd 

V Rd 

⎡2,24 

≤ 3⎢ 

⎣ 2 

− 

⎡2,24 

= 3 ⋅ ⎢ 

⎣ 2 

e 

⎤ 

⎥ 

⎦ 

⎤ 

− 0,376⎥ 

⎦ 

⎤ 

− 0,376⎥ 

⎦ 

= 180kN ≤ 117kN 

⋅ 0,24 ⋅ 

⋅ 0,24 ⋅ 

V = 29,61kN < 117kN = 

Ed V Rd 

N 

+ 0,4 ⋅ 

γ 

0,2 

1,7 

0,371 

1,7 

Ed 

M 

⎡ 

≤ 3⎢ 

⎣ 

+ 0,4 ⋅ 

e 

l 

2 

− 

0,229 

1,7 

M 

= 

N 

e 

Ed 

⎤ 

⎥ 

⎦ 

⋅ t ⋅ f 

vdu 

77,031 

229,585 

Ed 

Ed 

= = 

Ed 

0,376m

EUROCODE 6 




Verification to shear loading - simplified method 


c v = 1,5 (unfilled head joints) 

V Rd 

V Rd 

⎡2,24 

= 1,5 ⋅ ⎢ 

⎣ 2 

⎡2,24 

≤ 1,5 ⎢ 

⎣ 2 

= 

90kN 

≤ 

⎤ 

− 0,376⎥ 

⎦ 

⎤ 

− 0,376⎥ 

⎦ 

58,5kN 

⋅ 0,24 ⋅ 

⋅ 0,24 ⋅ 

V = 29,61kN < 58,5kN = 

Ed V Rd 

0,2 

1,7 

+ 

0,371 

1,7 

0,4 

⋅ 

0,229 

1,7

EUROCODE 6 




Applikation of f vlt 

tensile failure of the unit 

f 

vklt 

σ 

= 0,45 fbt 

1+ 

f 

Dd 

bt 

f bt : tensile strength of the unit 

f bt = 0,025 · f b hollow blocks; 

f bt = 0,033 · f b vertically perforated bricks and units with grip hole; 

f bt = 0,040 · f b full blocks without grip hole; 

simplified 

f vklt = 0,065 f b 

f vklt = 0,045 f b 

(filled head joints) 

(unfilled head joints)

EUROCODE 6 


Hint to ESECMaSE-Project 

ESECMaSE 

Enhanced Safety and Efficient Construction of Masonry Structures in Europe 


Reason: 

1,5⋅H/1.0⋅N = 1,5⋅e previous 

Aim: 

description of the shear 

failure closer to reality to use 

the reserves 

Done: 

a) Matrial tests/test methods 

b) Static/cyclic shear tests 

c) Pseudodynamic tests 

d) Shaking tabel tests 

e) Theoretical analyses 

f) Engineering model 

g) Recommandations for the practice 

Collapse analyses 

Test on shaking 

table in Athens (CS)

EUROCODE 6 


Hint to ESECMaSE-Project 

ESECMaSE 

Enhanced Safety and Efficient Construction of Masonry Structures in Europe 


Eng.-Model: From stress level to force level 

Bending 

Proposal from WP 4 

V 

bending 

2 

1 ⎛ N ⎞ 

= ⋅ 

⎜ N − 

⎟ 

2⋅λv 

⎝ t ⋅l 

⋅ f ⎠ 

Gaping 

V 

gaping 

= 

l 

b 

⋅ N 

2 

⎛ 1 

⎜ 

⎝ hb 

+ 

1 ⎞ 

⎟ 

h ⎠ 

Friction 

V friction 

= μ ⋅ 

N 

Tensile 

failure of 

the unit 

V 

unit 

unit, 

AAC 

1 

− 

= ⋅ fbt, 

cal ⋅ l ⋅ t ⋅ 

c 

* 

F = 1.2 + 0. 85 f bt 

V 

⎛ 

* 2 

( ) ( ) 

⎜ 

⎜ + 

* 2 

⎜ σ 

d 

F 1 F 1 

⎟ −1 

⎟ ⎟ ⎝ fbt, 

cal ⎠ ⎠ 

⎝ 

for AAC: 

1 

− 

= ⋅ fbt, 

cal ⋅ l ⋅ t ⋅ 2 ⋅ 

c 

⎛ + 

⎛ 

* 

* 2 

( ) 

⎜ + 

( F ) 

F 1 

⎜ 

⎝ 

4 

2 

⎛ 

⎜ 

1 + 

⎝ 

⎞ 

σ 

d 

f 

bt, 

cal 

⎞ 

⎞ 

⎟ 

⎠ 

⎞ 

−1 

⎟ 

⎟ 

⎠

EUROCODE 6 


Anauncement of 8 th IMC Dresden 2010 

Dissemination of information for training – Brussels, 2-3 April 2009 slide 26 out of 23

EUROCODE 6 




EUROCODE 6 




EUROCODE 6

Unreinforced masonry Shear loading - Eurocodes

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