Architectural design and practice Phần 5 doc

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 [...]... further loading of magnitude WA’, WB’ and WC’ respectively The loading in walls A and B will be negative and in wall C will be positive Assume the deflection of walls due to twisting moment is equal to ⌬a, ⌬b and ⌬c as shown in Fig 6.8 As the floor is rigid, (6.9) (6.10) Also (6.11) where K is the deflection constant and (6.12) Substituting the value of ⌬b from (6.9) and ⌬a from (6.11), we get or or (6.13)... be analysed by using a standard computer program or by conventional analysis which may or may not take into consideration the 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... 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 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... threedimensional structure was replaced by an equivalent twodimensional wall and beam system having the same areas and moments of inertia as the actual structure and analysed by the various methods described in this chapter The theoretical and experimental deflections are compared in Fig 6 .5 The strain and thus the stress distribution across the shear wall near ground level was nonlinear, as shown in Fig 6.6... 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 are replaced by columns and beams having the same flexural... walls and floor slabs respectively The span of the beams is taken to be the distance between ©2004 Taylor & Francis Fig 6.3 Idealization of shear walls with opening for theoretical analysis ©2004 Taylor & Francis the centroidal axes of adjacent columns (Fig 6.3(c)) The axial and shear deformations of beams and columns may be neglected or may be included if the structure is analysed by using any standard... the wall This is the opposite effect to that described in relation to settlement and results in an additional precompression on the wall, the value depending on the stiffness of the building against upward thrust As shown in Fig 7.2 the stiffness of a building, however, is highly indeterminate and nonlinear and in practical design this additional precompression may be ignored This will add to the safety... lateral pressure 7.2.4 Boundary conditions In practice, the walls in loadbearing masonry structures will be supported at top and bottom and may, in addition, be supported at the sides by return walls Returns can give extra strength depending on the ratio of the length to the height of the wall attached to the return, the tensile strength of the brick or block and the number of headers tying the wall to... two-dimensional cases The difference between the experimental and theoretical results may be due to the assumptions regarding interactions between the elements, which in a practical structure may not be valid because of the method of construction and the jointing materials To investigate the behaviour of a three-dimensional brickwork structure and the validity of the various analytical methods, a full-scale... disused quarry, and lateral loads Fig 6.4 (a) Test structure ©2004 Taylor & Francis Fig 6.4 (b) Test structure were applied by jacking at each floor level against the quarry face, which had been previously lined with concrete to give an even working face The deflections and strains were recorded at various loads The threedimensional structure was replaced by an equivalent twodimensional wall and beam system . (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 . 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=8.8 35. 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 31.7 85/ 102 .5 and since this is greater than 0. 25 N/mm 2 the code