Note that some designers include the above modification factors in the basic equation (5.11) where they appear as a multiplication factor on the right-hand side, e.g. for narrow walls, equation (5.11) could be rewritten 5.9 EXAMPLES 5.9.1 Example 1: Internal masonry wall (Fig. 5.15) (a) Using BS 5628 Loading (per metre run of wall) Safety factors For material strength, ␥ m =3.5 For loading, ␥ f (DL)=1.4 ␥ f (LL)=1.6 Fig. 5.15 Plan and section details for example 1. ©2004 Taylor & Francis Design vertical loading (Fig. 5.16) Loading from above (W 1 )=1.4×105+1.6×19=177.4 kN/m Load from left (W 2 ) dead load only=1.4×4.1=5.7kN/m imposed load=5.7+1.6×2.2=9.2 kN/m Load from right (W 3 ) dead load only=1.4×4.1=5.7kN/m imposed load=5.7+1.6×2.2=9.2kN/m Wall self-weight=1.4×17=23.8kN/m Slenderness ratio Effective height=0.75×2650=1988 mm Effective thickness=actual thickness=102.5 mm Slenderness ratio=1988/102.5=19.4 Eccentricity See section 5.5.1 • With full DL+IL on each slab there will be no eccentricity since W 2 =W 3 . Fig. 5.16 Loading arrangement for eccentricity calculation. ©2004 Taylor & Francis • With only one slab loaded with superimposed load, W 2 =9.2 and W 3 =5.7. Taking moments about centre line From equation (5.2) So that, since e t is greater than e x , e m =e t= 0.145t, which is greater than 0.05t, with the result that: Design vertical load resistance Assume t in mm and f k in N/mm 2 : Determination of f k We have design vertical load=design vertical load resistance Modification factors for f k • Horizontal cross-sectional area of wall=0.1025×4.25=0.44 m 2 . Since A>0.2 m 2 , no modification factor for area. • Narrow masonry wall. Since wall is one brick thick, modification factor=1.15. Required value of f k ©2004 Taylor & Francis Selection of brick/mortar combination Use Fig. 4.1 to select a suitable brick/mortar combination. Any of the following would provide the required value of f k . (b) Using ENV 1996–1–1 The dimensions, loadings and safety factors used here are the same as those given above in section (a). The reinforced concrete floor slabs are assumed to be of the same thickness as the walls (102.5 mm) and the modular ratio E slab /E wall taken as 2. Loading As for section (a). Safety factors For material strength, ␥ m =3.0 For loading, ␥ f (DL)=1.35 ␥ f (LL)=1.5 Design vertical loading (Fig. 5.16) Loading from above (W 1 )=1.35×105+1.5×19=170.25kN/m Load from left (W 2 ) dead load only=1.35×4.1=5.535kN/m imposed load=5.535+1.5×2.2=8.835kN/m Load from right (W 3 ) dead load only=5.535 kN/m imposed load=8.835 kN/m Wall self-weight=1.35×17=22.95kN/m ©2004 Taylor & Francis Eccentricity Because of the symmetry equation (5.8) can be rewritten: or Taking E c /E w =2 , I c /I w =1, h=2650 mm and the clear span L 3 =2797.5 mm Taking e hi =0 and e a= h ef /450=1.988/450=0.004m equation (5.4) be- comes The design vertical stress at the junction is 207.57/102.5 and since this is greater than 0.25 N/mm 2 the code allows the eccentricity to be reduced by (1-k/4) where k is given by equation (5.9). For this example and the factor so that the eccentricity can be reduced to 0.0049 and Slenderness ratio As for section (a). Design vertical load resistance In this section the value of Φ i =0.90 must replace the value of ß=0.78 used in section (a) and ␥ m =3.0, resulting in a value of 30.87 f k for the design vertical load resistance. ©2004 Taylor & Francis Determination of f k As for section (a) Modification factors for f k There are no modification factors since the cross-sectional area of the wall is greater than 0.1m 2 and the Eurocode does not include a modification factor for narrow walls. Required value of f k f k =6.83N/mm2 (compared with 8.35 in section (a)) Note that in ENV 1996–1–1 an additional assumption is required for the calculation in that the modular ratio is used. This ratio is not used in BS 5628. It can be shown that for this symmetrical case the value assumed for the ratio does not have a great influence on the final value obtained for f k . In fact for the present example taking E slab /E wall =1 would result in f k =7.0N/mm 2 whilst taking E slab /E wall =4 would result in f k =6.7N/mm 2 . Selection of brick/mortar combination This selection can be achieved using the formula given in section 4.4.3.(b) Using the previously calculated value of f k and an appropriate value for f m , the compressive strength of the mortar, the formula can be used to find f b, the normalized unit compressive strength. This value can then be corrected using δ , from Table 4.6, to allow for the height/width ratio of the unit used. 5.9.2 Example 2: External cavity wall (Fig. 5.17) (a) Using BS 5628 Loads on inner leaf ©2004 Taylor & Francis From equation (5.2) So that, since e t is greater than e x , e m =e m =0.088t which is greater than 0.05t, with the result that: Design vertical load resistance Assume t in mm and f k in N/mm 2 . design vertical load resistance Determination of f k We have design vertical load=design vertical load resistance Fig. 5.18 Loading arrangement for eccentricity calculations. ©2004 Taylor & Francis Modification factors for f k • Horizontal cross-sectional area=4.25×0.1025=0.44m 2 . This is greater than 0.2m 2 . Therefore no modification factor for area. • Narrow masonry wall. Wall is one brick thick; modification factor=1.15. Required value of f k Selection of brick/mortar combination Use Fig. 4.1 to select a suitable brick/mortar combination-nominal in this case. (b) Using ENV 1996–1–1 The dimensions, loadings and safety factors used here are the same as those given above in section (a). The reinforced concrete floor slabs are assumed to be of the same thickness as the walls (102.5 mm) and the modular ratio E slab /E wall is taken as 2. Loading As for section (a). Safety factors Design vertical loading (Fig. 5.18) Load from above=1.35×21.1+1.5×2.2=31.785kN/m Self-weight of wall=1.35×17=22.95kN/m Total vertical design load W 1 =54.735kN/m Load from slab W 2 =1.35×4.1+1.5×2.2=8.835 kN/m Eccentricity Equation (5.8) can be rewritten: ©2004 Taylor & Francis or Taking and the clear span L 2 = 2797.5mm As shown in section (a) Taking e hi =0 and e a =h ef /450=1.988/450=0.004m equation (5.4) be- comes The design vertical stress at the junction is 31.785/102.5 and since this is greater than 0.25 N/mm 2 the code allows the eccentricity to be reduced by (1-k/4) where k is given by equation 5.9. For this example and the factor so that the eccentricity can be reduced to and Slenderness ratio Effective height=0.75×2650=1988 mm Effective thickness=(102.53+102.53) 1/3 =129 mm Slenderness ratio=1988/129=15.4 ©2004 Taylor & Francis Design vertical load resistance In this section the value of Φ i =0.58 must replace the value of ß=0.91 used in section (a) resulting in a value of 19.82 f k for the design vertical load resistance. Determination of f k As for section (a) Modification factors for f k There are no modification factors since the cross-sectional area of the wall is greater than 0.1m 2 and the Eurocode does not include a modification factor for narrow walls. Required value of f k f k =3.20N/mm 2 (compared with 2.16 in section (a)) Note that in ENV 1996–1–1 an additional assumption is required for the calculation in that the modular ratio is used. This ratio is not used in BS 5628. It can be shown that for the present example taking E slab /E wall =1 would result in f k =4.7N/mm 2 whilst taking E slab /E wall =4 would result in f k =2.44N/mm 2 . To obtain the same result from BS 5628 and ENV 1996–1– 1 would require a modular ratio of 6 approximately. Selection of brick/mortar combination This selection can be achieved using the formula given in section 4.4.3(b) Using the previously calculated value of f k and an appropriate value for f m , the compressive strength of the mortar, the formula can be used to find f b the normalized unit compressive strength. This value can then be corrected using δ , from Table 4.6, to allow for the height/width ratio of the unit used. ©2004 Taylor & Francis [...]... proportion to their flexural rigidities This is the most commonly used method for the design of masonry structures The deflection of the wall is given by (6 .5) (6.6) where w=total uniformly distributed wind load/unit height, h=height of building, x=distance of section under consideration from the top, and I1, I2=second moments of areas (Fig 6.3(b)) 6.3.3 Equivalent frame In this method, the walls and slabs... height of the building, so that the intensity of the equivalent uniformly distributed load varies accordingly In the United Kingdom, wind loads on buildings are calculated from the provisions of the Code of Practice CP 3, Chapter V, Part 2, 1970 Whilst masonry is strong in compression, it is very weak in tension; thus engineering design for wind loading may be needed, not only for multi-storey structures, ... axial and shear deformation of the beams and columns 6.3 .5 Continuum In this method, the discrete system of connecting slabs or beams is replaced by an equivalent shear medium (Fig 6.3(e)) which is assumed continuous over the full height of the walls, and a point of contra-flexure is taken at the centre of the medium Axial deformation of the medium and shear deformations of the walls are neglected Basically,... and (6.12) Substituting the value of ⌬b from (6.9) and ⌬a from (6.11), we get or or (6.13) Similarly, (6.14) Now the sum of the moments of all the forces about the shear centre of all the walls must be equal to the twisting moment Hence or ©2004 Taylor & Francis 7 Lateral load analysis of masonry panels 7.1 GENERAL In any typical loadbearing masonry structure two types of wall panel resist lateral pressure,... Francis Fig 6.2 A system of shear walls resisting wind force (6.3) (6.4) where W 1 , W 2 =lateral forces acting on individual walls, ⌬ 1 , ⌬2 =deflections of walls, A=area of walls, h=height, E=modulus of elasticity, G=modulus of rigidity, I1, I2=second moments of areas and ␭=shear deformation coefficient (1.2 for rectangular section, 1.0 for flanged section) The proportion of the lateral load carried... Application of the equations of equilibrium of the forces acting at these nodal points leads to a number of simultaneous equations which can be solved with the aid of a computer The method provides a very powerful analytical tool, and suitable computer programs are readily available which can deal with any type of complex structure However, this may prove to be a costly exercise in practical design situations... situations 6.3.7 Selection of analytical method Although these methods are used in practice for analysis and design of rows of plane walls connected by slabs or beams, the analysis of a ©2004 Taylor & Francis complex three-dimensional multi-storey structure presents an even more difficult problem Furthermore, it has been observed experimentally that the results of these methods of analysis are not necessarily... single-storey structures Figure 6.1 shows how a typical masonry building resists lateral forces It can be seen that two problems in wind loading design need to be considered: (1) overall stability of the building and (2) the strength of individual wall panels In this chapter overall stability of the building will be considered 6.2 OVERALL STABILITY To provide stability or to stop ‘card-house’ type of collapse,... precompression Stiffness of the building against upward thrust Boundary conditions 7.2.1 Flexural tensile strength The flexural tensile strength of masonry normal to the bed joint is very low, and therefore it may be ignored in the lateral load design of panels with precompression without great loss of accuracy 7.2.2 Initial precompression As will be shown in section 7.3, the lateral strength of a wall depends... basic methods of analysis for the estimation of wind stresses and deflection in such shear walls, namely: (i) cantilever approach, (ii) equivalent frame, (iii) wide column frame, (iv) continuum, and (v) finite element Figures 6.3(b) to (f) show the idealization of shear walls with openings for each of these methods 6.3.2 Cantilever approach The structure is assumed to consist of a series of vertical . (DL)=1. 35 ␥ f (LL)=1 .5 Design vertical loading (Fig. 5. 16) Loading from above (W 1 )=1. 35 1 05+ 1 .5 19=170.25kN/m Load from left (W 2 ) dead load only=1. 35 4.1 =5. 535kN/m imposed load =5. 5 35+ 1 .5 2.2=8.835kN/m . above=1. 35 21.1+1 .5 2.2=31.785kN/m Self-weight of wall=1. 35 17=22.95kN/m Total vertical design load W 1 =54 .735kN/m Load from slab W 2 =1. 35 4.1+1 .5 2.2=8.8 35 kN/m Eccentricity Equation (5. 8) can. load =5. 5 35+ 1 .5 2.2=8.835kN/m Load from right (W 3 ) dead load only =5. 5 35 kN/m imposed load=8.8 35 kN/m Wall self-weight=1. 35 17=22.95kN/m ©2004 Taylor & Francis Eccentricity Because of the symmetry