Numerical examples

11.1. Verification of masonry wall subjected to vertical loading

Verify the load bearing capacity of the masonry wall given in Figure 48. The wall is loaded by the NEd uniformly distributed vertical load. The wall is made of Category I.

and Group 2. clay units, and prescribed general purpose mortar of class M5. Class of the masonry wall is 3.

NEd

t

h/2

C C

NEd

h

Side view Cross section

A A

B

h B

e0

Figure 48: Masonry wall subjected to vertical compression

Parameters for the calculation:

Type and dimensions of masonry unit: Porotherm 38; ℓ_u=250 mm, w_u=380 mm, h_u= 238 mm Mean compressive strength of the applied masonry unit: f = 11 N/mm²

Volumetric weight of masonry: ρ = 8.21 kN/m³ (given by the manufacturer) NEd = 300 kN/m

e0 = 15 mm h = 3.00 m t = 0.38 m

ρ2 = 0.75 (reduction factor for the wall, see Chapter 9.1.2) φ∞ = 1,00 (final creep coefficient, see Chapter 8.4.2)

γM = 2.2 (units of Category I., prescribed mortar, Class 3. masonry, see Table 9) Calculation:

A) Calculation of the design compressive strength of masonry

Coefficient depending on the dimensions of the unit: δ = 1.138 (see Table 11) Normalised mean compressive strength: fb δ f⋅ 12.52 N

mm²

⋅

=

=

K = 0.45 (Group 2. clay units, general purpose mortar, see Table 12)

Compressive strength of mortar: fm 5 N

mm²

=

Characteristic compressive strength of masonry: fk K fm⋅ ^0.3⋅fb^0.7 4.28 N mm²

⋅

=

=

Design compressive strength of masonry: fd f_k

γM 1.94 N

mm²

⋅

=

=

B) Calculation of the capacity reduction factor at the top of the wall (section A-A)

First order eccentricity: e0 = 15mm

Effective height of the wall: hef = ρ2 h⋅ =2.25m Initial eccentricity at the top of wall: einit hef

450 = 5 mm⋅

=

Eccentricity at the top of wall: ei ⁼ max e0 einit

(

+ , 0.05t

)

^{= 20 mm⋅}

Capacity reduction factor at the top of the wall: Φi.A 1 2 ei⋅

− t = 0.895

=

C) Calculation of the capacity reduction factor at the bottom of the wall (section C-C)

First order eccentricity: e0 = 15mm

Initial eccentricity at the top of wall: einit = 0mm

Remark: We asssume that the bottom of the wall is in the desired position, so no initial eccentricity is considered here.

Eccentricity at the top of wall: ei ⁼ max e0 einit

(

+ , 0.05t

)

^{= 19 mm⋅}

Capacity reduction factor at the top of the wall: Φi.C 1 2 ei⋅

− t =0.900

=

D) Calculation of the capacity reduction factor in the middle of the wall (section B-B)

First order eccentricity: e0 = 15mm

Initial eccentricity in the middle of the wall: einit 0.5 hef

⋅450 =2.5 mm⋅

=

Remark: We asssume that the initial eccentricity varies linearly between the bottom and the top of the wall, thus between 0 and h_ef/450.

The eccentricity due to loads: em = e0 einit+ = 17.5 mm⋅ Effective width of the wall (single-leaf wall): tef = t = 380 mm⋅

The eccentricity due to creep: ek 0.002⋅ϕ∞ hef

⋅ tef⋅ t em⋅ =1.0 mm⋅

=

Eccentricity at the middle height of the wall: emk ⁼ max em ek

(

+ , 0.05t

)

^{= 19 mm⋅}

The short term secant modulus of elasticity: E 1000 fk⋅ 4277 N mm²

⋅

=

= (clay units)

The long term modulus of elasticity: Elongterm E

1+ϕ∞ 2139 N mm²

⋅

=

=

Slenderness of the wall: λ hef

tef

fk Elongterm

⋅ = 0.265

=

u λ 0.063−

0.73 1.17 emk

⋅ t

−

0.301

=

=

A1 1 2 emk

⋅ t

− = 0.9

=

Reduction factor in the middle of the wall: Φm A1 e u²

−

⋅ 2 =0.860

=

D) Verification of the appropriate cross sections of the wall At the top of the wall (section A-A):

NRd.A Φi.A t⋅ fd⋅ 661.05kN

= m

= > NEd 300kN

= m satisfactory

At the middle height of the wall (section B-B):

NRd.B Φm t⋅ fd⋅ 635.58kN

= m

= > NEd ρ t⋅ h

⋅2

+ 304.68kN

= m satisfactory

At the bottom of the wall (section C-C):

NRd.C Φi.C t⋅ fd⋅ 664.94kN

= m

= > NEd ρ t⋅ h+ ⋅ 309.36kN

= m satisfactory

11.2. Verification of shear resistance

Verify the shear resistance at the bottom (section K-K) of the masonry wall given in Figure 49. The wall is loaded by the NEd vertical and HEd horizontal edge loads. The wall is made of Category I. and Group 2. clay units, and prescribed general purpose mortar of class M10. Class of the masonry wall is 2. Self weight of the wall can be neglected during the calculation.

N_Ed

H_Ed

t

h

K K

N_Ed

h

Side view Cross section

K K

L

Figure 49: Masonry wall subjected to compression and shear

Parameters for the calculation:

Type and dimensions of masonry unit: Porotherm 38; ℓu=250 mm, wu=380 mm, hu= 238 mm Mean compressive strength of the applied masonry unit: f = 11 N/mm²

Normalized mean compressive strength: fb = 12.52 N/mm² (see previous example) NEd = 200 kN/m

HEd = 100 kN h = 3.00 m L = 6.00 m t = 0.38 m

γM = 2.0 (units of Category I., prescribed mortar, Class 2. masonry, see Table 9)

Calculation:

A) Calculation of internal forces in section K-K

Axial force: NEd = NEd L⋅ = 1200 kN⋅ Bending moment: MEd = HEd h⋅ = 300 kNm⋅ Shear force: VEd = HEd 100 kN= ⋅

B) Verification of the eccentricity of the axial force in section K-K Cross sectional area: A = t L⋅ = 2.28m²

Moment of inertia: I t L⋅ ³

12 =6.84m⁴

=

Kern distance: d I

L 2⋅A

1000 mm⋅

=

= > eN MEd

NEd = 250 mm⋅

=

The eccentricity of the axial force is smaller than the kern distance (i.e. the force remains within the core of the cross section) so the complete K-K section is under compression.

The core of a rectangular cross section is illustrated in Figure 50.

C) Calculation of the characteristic shear strength Average vertical stress in section K-K: σd NEd

A 0.53 N

mm²

⋅

=

=

Initial shear strength of the masonry: fvk0 0.3 N mm²

= (see Table 13)

Characteristic shear strength: fvk min fvk0 0.4 σd

(

+ ⋅ , 0.065 fb⋅

)

^{0.51 N}

mm²

⋅

=

= D) Verification of the shear resistance in section K-K

VRd fvk

γM⋅A=582 kN⋅

= > VEd 100 kN= ⋅ satisfactory

h h/3

y

z b/3

b

Figure 50: The core of a homogeneous rectangular cross section

12. References

[1] C. Hartsuijker, J. W. Welleman: “Engineering Mechanics: Volume 2: Stresses, Strains, Displacements”, Springer Science & Business Media, 2007.

[2] EN 1995-1-1:2010. Eurocode 5: Design of timber structures. Part 1-1: General.

Common rules and rules for buildings.

[3] EN 1996-1:2005+A1:2013. Eurocode 6: Design of masonry structures. Part 1-1: General rules for reinforced and unreinforced masonry structures.

[4] EN 338:2010. Structural timber. Strength classes.

[5] I. Bódi, L. Erdődi, N. Friedman: “Verification of traditional timber connections” (in Hungarian), Electronic lecture note of the Department of Structural Engineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2011.

[6] I. Bódi: “Structural behavior and application of reinforced masonry”, (in Hungar-ian), Electronic lecture note of the Department of Structural Engineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2011.

[7] K. Koris, I. Bódi: “Basis of the design of timber structures according to MSZ EN 1995-1-1 (EC5) ”, (in Hungarian), Electronic lecture note of the Department of Structural Engineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2012.

[8] L. Varga: “Design of masonry structures according to MSZ EN 1996-1-1 (Euro-code 6)”, (in Hungarian), Electronic lecture note of the Department of Structural En-gineering, Budapest University of Technology and Economics (www.hsz.bme.hu), 2010.

[9] Structural fixings – On-site guide for building code compliance, MiTek New Zea-land Ltd, 2009.

[10] “Porotherm application and design guide”, (in Hungarian), Wienerberger Ltd.

(www.wienerberger.hu), 2014.

In document Basis of the design of timber and masonry structures according to Eurocode (Page 56-61)