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Chapter two Loads on buildings and mechanics and design requires of many factors. These include the loads and load the the strength and of the materials from elements are made, the ways by which the loads and load via the structure to the the and the ground, is therefore important to estimate the loads that a to withstand during its intended useful life, in order to achieve safety in behaviour of under loads depends on the strength of the materials of and the between the and parts of the frame and between the frame, and the ground. Designers in their try to predict this behaviour of the structure and identify the model to be used in the analyses. If they succeed then designs will usually be safe and present, existing knowledge of the loads on of the materials of and analysis of frames is well so that design can usually be to be regard to these aspects. However, future research on of loads on will help to reduce a number of the existing and hence result in safer and more economic design, the loads on buildings and are into based on their frequency of and method of are: 1 dead loads 2 imposed loads 3 wind loads 4 earth and liquid other load effects such as thermal effects; ground movement; creep in concrete; and each type of load, there will be a value and a These will be explained later in this chapter. The design of any Load element of the frame of the structure or of the structure as a whole has to be based on the design load or design load that is likely the most adverse effect on that element or the structure as a whole in terms of tension, bending, moment, shear, loads BS 6399–1: 1996 Loading for Part 1: Code of practice for dead load is the weight of such as floors, walls and finishes, and includes all other permanent to as pipes, conduits, air heating ducts and all items intended to remain in place the life of the It from the unit weights given in BS 648: 1964 Schedule of weights materials or from the actual known weights of the materials used. In the analysis process, although the dead load of parts or mem- bers can be it is usual practice to simplify to reduce the analysis and design time, for example in of beams an uniformly load is usually of the actual the design process, the of the dead load of most load bear- ing parts has to be done in practice by a method of trial and error to determine the required for such parts. most of the common types of elements, for example and columns, there are some simple rules for assessing the required. These rules are explained in the relevant code for example, for concrete and steel see BS 8110: Part 1: 1997 and BS 5950: 2000 loads BS 6399–1: 1996 Loading for Part 1: Code of practice for dead loads are sometimes called live loads or are gravity loads varying in magnitude and location. They are be produced by the intended occupancy or use of the They in- clude impact and snow loads but exclude Such loads are usually caused by human furniture of or their Because of the unknown nature of the location and of imposed load items, are difficult to These values are by both and local building 6399–1: 1996 Loading for Part 1: Code of practice for dead loads gives imposed loads for various occupancy and of such as • domestic and (dwelling houses, flats, hotels, guest houses) • and (schools, colleges and 2 LOADS ON BUILDINGS AND 19 20 PART 1 BEHAVIOUR OF power bridges highway and shopping areas • and storage with this there is still broad variation in the for example within the high school building some space is used and The imposed loads for these various are different and hence different values should be specified for such as highway bridges, it is necessary to consider in terms of both a load and a varying uniformly In addition, the effect of impact forces due to traffic loading must be in total imposed floor code of practice allows for the reduction of imposed loads in the design of certain and should be consulted for full the main are as and girders. Where a single span of a beam or girder less than 46 m2 of floor at one general level, the imposed load may in the design of the beam or girder be reduced by 5 per cent for each 46 m2 supported subject to a maximum reduction of 25 per cent. No however, shall be made for any plant or machinery for provision has been made nor for buildings for storage garages and those office areas that are used for storage piers, walls, their supports and The loads to the total loads for the design of such elements may be reduced in with Table reduction is allowed because of the reduced that the loads will occur at all the floors of floors, including the Reduction in total carried by member under imposed load on all carried by the member under (%) 1 0 2 10 3 20 4 30 5 to 10 40 Over 10 50 Table 2.1 Reduction in total imposed floor 2 LOADS ON BUILDINGS AND loads are those that produce dynamic effects from cranes and other plant supported by or connected to the is made for these dynamic effects, including impact, in the design of the relevant allow for such effects in practical design, it is common practice in to increase the value of machinery or plant by an to cater for the dynamic effect, and a static analysis is out for these increased loads and the computed load effects used in the design. The dynamic increase for all affected members is as as possible and must comply with the relevant code of from 5.1.4 of BS 6399–1: loads from permanent Where permanent shown in the plans their actual weights shall be included in the dead load. For floors of offices, this uniformly load should be not less than 1.0 loads from To provide for it is normal practice to consider an uniformly load of not less than one-third of the per metre run of the and treat it as an imposed load in loads on 6399–2: 1997 Loading for Part 2: Code of practice for wind loads depend on the wind and on the behaviour of the building. Wind loads on are dy- namic loads due to changes in wind speed. When the wind flow meets an such as a building or a it has to change speed to keep flowing around the building and over it. In this process of change in direction it exerts pressures of varying on the and roof of the building. In analysis and design it is consider the design wind loads due to these pressures in applied imposed and dead loads. For in design it is to consider the wind loads as static loads. However, for some such as metal chimneys, the dynamic effects of the wind, such as induced have to be in to the change in direction when wind flow stable the induced wind pressure can vary in direction such that the loads are and vertical. since the wind with time the wind loads on have to be as of from all view of the of the of wind loads on it is not possible to give the subject full treatment here and the reader is advised to consult one of the at the end of the PART 1 BEHAVIOUR OF effective wind loads on are dependent on the wind location of structure or building, size, shape and wind normally blows in gusts of varying speed, and its direction de- pends on the wind Figure 2.1 shows a typical graph of time during a wind pressure, which is caused by changes of wind speed from Ve in m/s to zero, as occurs when the wind meets a building and has to change is given by qs: Ve � effective wind speed from section 2.2.3 of BS 6399: 1997 Loading – Part 2: Code of practice for wind wind speed to be used in equation (1) is not the maximum It should be from the relevant section of the code of For example from section 2.2.3 of BS 6399: 1997 Loading for 2: Code of practice for wind the shape of the structure is then the change in wind speed is reduced and hence the dynamic wind pressure will also be reduced (see code of on — Dead loads or permanent actions according to the are the of or and are caused by the effect of gravity, and so act Dead loads are from known weights of the materials used (see Table 2.2). Where there is doubt as to the of dead loads, such loads should be imposed loads. Dead loads are the unit weight by the more see the relevant code of practice or, in the UK, see BS 6399–1: 1996 and BS 648: 1964. qs = 0.613 V2e the air density r = 1.226 pressure qs = 12rV 2 e (in pascals, Pa sp ee d v speed 46 m/s @ 25 s gust 0 5 10 15 20 25 30 35 40 Fig. 2.1 Wind speed versus 2 LOADS ON BUILDINGS AND 23 • Imposed loads or variable actions according to are gravity loads which vary in magnitude and location and are to the types of activity or occupancy for which a floor area will be used in service; see the code of practice or Table 1 of BS imposed loads. Such as stored material, people, by gravity, act in design and as static loads. Also called loads or live imposed loads. Such as vehicles, cranes, trains, etc. Their should be in addition to their static Wind loads Due to dynamic wind these depend on the wind on the and behaviour of the structure or Weight Material 2 layers, 19 mm thick 42 kg/m2 Two coats gypsum, 13 mm thick 22 19 mm thick 41 kg/m2 Plastic sheeting 4.5 and 19 mm thick 44 kg/m2 roofing felts per mm thick 0.7 surfaced bitumen 3.5 kg/m2 concrete 2400 per 25 mm thick, stone aggregate 55 kg/m2 Cement : sand (1 : 3), 13 mm thick 30 per 25 mm thick 15 kg/m2 Cement : sand (1 : 3), 13 mm thick 30 per 25 mm thick 12.5 kg/m2 Slate upon thickness and source) 24–78 solid per 25 mm thick medium 55 kg/m2 Solid (mild) 7850 solid per 25 mm thick 59 kg/m2 roofing sheets, per mm thick 10 stone 2250 kg/m3 25 mm thick 60 2400 kg/m3 25 mm thick 54 Tiling, 50 mm thick 120 kg/m2 Clay 70 fibre per 25 mm thick 2.0–5.0 kg/m2 Softwood 590 panels and Hardwood 1250 panels 75 mm thick 44 kg/m2 Water 1000 2.5 mm thick 30 kg/m2 Slabs, 25 mm thick 15 mm thick 6 + 240 - 160 2.2 Weights of building Adapted from Various extracts, British Standards for Students of Design, PP 7312:2002 (British also BS 648: 1964 Schedule of weight of building PART 1 BEHAVIOUR OF in intensity and Depend on: 1 shape of height of above its base 3 location of and the relevant national code of practice or BS 6399: 1997 – Part 2: Code of practice for wind loads. • pressure, hydraulic pressure, thermal effects, ground and creep in concrete, and vibration are by found in load, Fk, is a load value above which not more than x per cent of the measured values fall. Using the and standard and when x � 5 per cent, can be defined plus sign is used since in most cases the is the maximum load on a critical for stability or the behaviour of members, referred to the relevant code of the present state of the load is that the relevant national codes of practice, such as, in the UK, BS 1–3: 1996 and 1997 for dead, imposed and wind loads and BS 2573 for crane loads. S � standard deviation for load load � mean load ± load The design load is by the load Fk by partial safety, the partial factor of safety for loads, which is to take into account the effects of errors in design minor unusual increases in loads and factor of safety also takes into account the of the sense of the limit state under and the of BS 5950: 2000 and BS 8110: 1997 give for practical partial factors of safety for = design load = Fk * loads and of safety A structure is usually exposed to the action of several types of loads, such as dead loads, imposed loads and wind loads. They should be and in such realistic as to take account of the most effects on the elements and on the structure as a whole. For limit state, the loads should be by the factor of safety given in the relevant table of the code of practice. The factored 2 LOADS ON BUILDINGS AND be applied in the most realistic to the part of the structure or the effect under Different load are by the codes of practice. For example, see BS 1: 2000, Table 2, Partial factors for loads . Some examples on are as Dead and imposed load (a) design dead (b) design imposed (c) design earth and water where imposed load, dead load and design earth and water example, in the design of a simply supported beam the following is commonly used: 2 Dead and wind design dead (b) design wind where load (vertical load), and load. 3 Dead, imposed and wind where load, and The criterion for any load is that it is likely to produce the worst effect on a structure or element for design purposes. only possible design load be In the design of a beam, the worst load should be with the design dead load of 1.0Gk or 0.9Gk acting on some parts of the structure to give the most severe see Fig. 2.2 (case 3, more load are possible in this case). 3 In Fig. 2.2, for case 1, (dead loads) � 1.0 and for case 2, (dead loads) for load resisting uplift or 1.0 and 1 = wind1.2Gk + 1.2Qk = = 1.2Gk + 1.2Qk + 1.2Wk Wk = windGk = = = 1.4Gk or load = 1.4Gk + 1.6Qk (vertical load) En = Gk = Qk = load = = = 1.4Gk or 1) (case + 1.6G Qk k 1.4 + 1.6G Qk 3) 1.0Gk 2.2 Load PART 1 BEHAVIOUR OF Other realistic that give the most critical effects on the elements or the structure as a whole are shown in the relevant code of practice, for example see Table 2 of BS 5950: Part 1: 2000. 5 The values in Eurocodes for 1.4 (8G) and 1.6 (8Q) are 1.35 (8g) and 1.5 (8q), see clause Cl 2.3.1, EC2 and Cl 2.4.3, EC3. It is important that the design loads are assessed If the are wrongly assessed at the beginning then all the and/or analysis will also be 2.1 Figure 2.3 shows a 3 m long concrete beam and a 914 mm deep � 419 mm wide universal steel beam that is 6 m the the weight of each beam per unit length (the uniformly per unit the total weight of each beam (c) the design dead load for each 1 concrete beam (see Fig. From Table 2.2, unit weight of the unit weight per unit Total weight of Design dead load of the beam = 1.4Gk = 1.4 * 5.76 = 8.064 kN beam = 1.92 kN/m * 3 m = 5.76 kN = 1.92 = 0.08 m2 * 24 = 24 = 0.2 * 0.4 = 0.08 m2 0.2 m 3m 2.3 Example 2.1 beams: (a) concrete; (b) steel w = 1.92 kN/m 3 m Dead load per = 8.064 kN 3 m Total design dead 2.4 Example 2.1 loads concrete 2 LOADS ON BUILDINGS AND 27 2 Steel beam (see Fig. (from Table 11.21, p. Table 2.2, unit weight of steel The weight per unit (i.e. mass per metre of the beam = 388 kg/m, since 1 kN is to a mass of 100 weight of Design dead load of the beam = 1.4Gk = 1.4 * 23.287 = 32.602 kN beam = 3.88 kN/m * 6 m = 23.287 = (49 400>106) m2 * 78.5 kN/m3 = 3.88 kN/m (mild steel) = 78.5 = 49 400 mm2 w = 3.88 kN/m 6 m Dead load per = 32.602 kN 6 m Total design dead 2.5 Example 2.1 loads on steel 2.2 Figure 2.6 shows plan and roof details of a flat roof extension to an existing house. Calculate the design loads on the concrete beam A which is 300 mm wide and 600 mm deep. Access is to be provided to the roof, therefore use an imposed load of 1.5 kN/m2. of concrete . Roof (two layers) 19 mm thick 42 kg/m2 25 mm timber boards, softwood 590 kg/m3 50 � 175 mm timber joists spaced at 400 mm centre to centre 590 board and skim (plaster finish) 15 kg/m2 = 24 m 3.6 m Room wall 25 mm timber board 8 m 175 joists @ 400 mm 400 X - X 19 mm CL Fig. 2.6 Example 2.2: plan and roof details of to an existing (See Fig. 0.42 kN/m2 = 42 * 10 = 420 carried by the beam = 3.6 * 8 = 28.8 loads = 1 .4Gk + PART 1 BEHAVIOUR OF of joists in 1 m board and dead loads of the loads = 34.151 + 48.384 + 69.12 = 151.655 kN + (1.4 * 34.56) + (1.6 * 1.5 * 28.8) = (1.4 * (0.847 * 28.8)) = 24 * 0.300 * 0.600 * 8 = 34.56 kN = 0.847 kN/m2 = 0.15 kN/m2 = 15 * 10 = 150 N/m2 = 0.129 kN/m2 = 129.06 N/m2 = 590 * 10 * 0.050 * 0.175 * 1>0.4 = 0.147 kN/m2 = 147.5 N/m2 = 590 * 10 * 2.3 of a roof beam using a rolled steel 2.8 shows part of a roof plan for a small steel building. The flat of felt, steel decking, boards and a suspended the rolled steel joists. Calculate the design loads acting on one m 8 m Area carried by beam A Design loads on beam A Fig. 2.7 Example 2.2 = 0.9 kN/m2k Qk = 1.5 A 29.28 kN = design A Fig. 2.8 Part of a roof in a Dead load Imposed load Design loads = (1.4 * 7.2) + (1.6 * 12) = 29.28 kN = 1.5 * 4 * 2 = 12 kN = 0.9 * 4 * 2 = 7.2 2 LOADS ON BUILDINGS AND notes on the of loading from roofs or slabs onto beams A roof or slab can be designed and detailed to span one way so that load only to beams and not directly to the beams running at right angles to them. If this is not the case, two-way action of the slab or roof must be taken into span 1 Where the ratio (see Fig. 2.9(a)), and normally when the slab is made from concrete which is cast in situ. The load on the roof ABCD can be assumed to be carried equally between beams AB and DC. 2 When the roof or the slab is of precast concrete units with the ratio (see Fig. 2.9(b)). The loads on the roof ABCD may be assumed to be carried equally between beams AB and the above rules are not satisfied then the loads on the roof or the floor can be as shown in Fig. 2.9(c) to take account of two-way � 2 L1>L2 � B sp an s on beam BC Area beam 2.9 Example 2.3: (a) and (b) one-way action of the slab or floor; (c) two-way action of the 2.4 Figure 2.10 shows the relevant details at side walls and columns for a (in both steel office building a steel frame with concrete slabs on profiled metal decking. cladding is block and Calculate the acting on roof beam C2–D2, floor beam C2–D2, and inner length 2–5–8 (for column reference numbers, see Fig. PART 1 BEHAVIOUR OF m Braced column 8 m Braced at 8 m 1 m 5 m 3 at 4 m 3 m (b) mm 0.9 m 0.5 steel strip with built-in channel for stainless steel steel to joint with mm straps at 600 mm slab on mm thick for mm r.c. slab at side m 1.0 B C D E F G Fig. 2.10 Fully braced (a) plan; (b) (c) end (d) building details at walls, side column 2 LOADS ON BUILDINGS AND loads (BS 6339: Part 1) On roof On floors Reduce the imposed loads in with number of stories as in the practical code of practice (see Table 2.1, p. 20). = 3.5 kN/m2 = 1.5 loads on Dead loads the flat roof (kN/m2) on the floor materials 1.0 tile screed and slab, 170 mm thick 4.1 concrete mm thick 0.2 steel 0.5 (0.58) 0.2 services dead load 6.0 kN/m2 7.0 wall external wall) 4.8 kN/m2 1 m high 5.0 and column, dead column, dead steel beam at roof wall dead side wall at floor wall 5.0 slab 4.32 0.68 (double) 0.6 0.5 kN/m2 = 0.5 kN/m2 = 0.5 kN/m = 4.1 kN/m2 = 4.8 kN/m2 = 6.3 kN/m = 1.5 Roof beam C2–D2 (see Fig. per unit length = 24 + 6 = 30 kN/m + (1.5 * (4 * 1 m (6 * (4 * 1 m 1.5 kN/m2 = 6 PART 1 BEHAVIOUR OF loads per unit design loads = 268.8 + 76.8 = 345.6 kN + (1.6 * 1.5 * (4 * 8)) = (1.4 * 6 * (4 * 8)) = 33.6 + 9.6 = 43.2 kN/m + (1.6 * 1.5 * (4 * 1 m)) = (1.4 * 6 * (4 * 1 bay 8 m 4 1 A B C D carried by at both and floor level 2 m 2 m 43.2 kN/m (total )= 345.6 kN Fig. 2.11 Design loads on at roof beam C2–D2 (see Fig. loads Imposed loads Weight per unit loads per unit design loads = 313.6 + 179.2 = 492.8 kN + (1.6 * 3.5 * (4 * 8)) = (1.4 * 7 * (4 * 8)) = 39.2 + 22.4 = 61.6 kN/m + (1.6 * 3.5 * (4 * 1 m)) = (1.4 * 7 * (4 * 1 m)) = 28 + 14 = 42 kN/m + (3.5 * (4 * 1 m (7 * (4 * 1 m 3.5 kN/m2 = 7.0 kN/m2 8 m 8 m 4 m 61.6 = 492.8 kN) Area on beam 2.12 Design loads on at floor steel beam C1–D1 at roof level (see Fig. dead design dead design design loads = 135.52 + 38.4 = 173.92 kN = 1.6 * 1.5 * 2 * 8 = 38.4 kN = 1.4 * 12.1 * 8 = 135 .52 kN = 12.1 kN/m = 0.5 * 2 m (wide strip) = 1.0 kN/m = 4.1 * 2 m (wide strip) = 8.2 kN/m = 0.5 * 1 m (one unit length) = 0.5 kN/m = 4.8 * 0.5 m (high) = 2.4 2 LOADS ON BUILDINGS AND column, lower length 2–5–8 (see Fig. on column above joint 5 Design dead loads floors 3 columns (4 m high) and imposed loads see the code) = 286.72 kN = 1.6 * 2 * 3.5 * (8 * 4) * 0.8 = 1.6 * 1.5 * (8 * 4) = 76.8 kN = 1.4 * (3 * 4 * 1.5) = 25.2 kN = 1.4 * (2 * 7) * (4 * 8) = 627.2 kN = 1.4 * 6 * (4 * 8) = 268.8 kN 8 m 8 m 2 m 21.74 2.13 Design loads roof beam C1–D1 4 m 4 m 4 m Area carried by at the roof and floor 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Fig. 2.14 Column from beam 4–5 and beam dead imposed design loads on below joint 5 Note: For the of the moment acting on the column at joint 5, the loads on the beams as on beam 5–4 = 156.8 kN = 1.4 * 7 * (4 * 4) + zero imposed loads = 1777.52 kN = 179.2 kN 5–64 + 1.4 * 3.5 * (4 * 4) [from beam = 1.6 * 3.5 * (4 * 4) [from beam 4–5] = 313.6 kN + 1.4 * 7 * (4 * 4) [from beam 5–6] = 1.4 * 7 * (4 * 4) [from beam PART 1 BEHAVIOUR OF on beam 5–6 longest beam of 5–4 or 5–6) The design loads and bending moments are shown in Fig. 2.15. = 246.4 kN + (1.6 * 3.5 * (4 * 4)) = (1.4 * 7 * (4 * loads at central column 2–5–6 For moment at 5 use the above kN246.4 kN 2 4 6 5 4 kN 246.4 kN Fig. 2.15 Design loads ‘In the design and/or of an element or a structure as a whole, all should be and in such as to comprise the most critical effects on the elements and the structure as a magnitude and frequency of also be Discuss the above Use annotated sketches to support Figure 2.Q2 shows plan and roof details of a flat extension to an existing house. sketch the total design loads on one timber beam A and steel beam is to be provided to the roof, therefore load of 1.5 (two layers) 19 mm thick 42 kg/m2 25 mm timber boards 590 kg/m3 50 � 175 mm timber at 400 mm centre to centre 590 board and skim (plaster finish) 15 kg/m2 3 A concrete bridge between two 8 m. The of the bridge is shown in Fig. 2.Q3. The brickwork weighs 20 kN/m3, weights 24 kN/m3 and the coping weighs 0.5 kN/m. The floor of the bridge has to carry a imposed load of 1.5 kN/m2 to its own weight. For beam A shown m 3.6 m Room wall 3 m Beam mm timber mm 400 ×–× 19 mm × CL CL Fig. 2.Q2 Exercise 2: plan and roof details of a extension to an existing 2 LOADS ON BUILDINGS AND 35 Fig. 2.Q3, calculate and sketch the total uniformly design loads in kN/m and the total in kN. The beam is simply supported at Figure 2.10 (Example 2.4) shows the relevant details at side walls and columns for a fully braced (in both steel office building that is made of a steel frame with a concrete slab on decking. External cladding is Calculate and sketch the acting on roof beam A1–A2, floor beam beam B1–B2 and outer column, lower (see Fig. A 8 A Plan mm brickwork 225 mm 150 mm r.c. slab 1.3 m 203 × ×133 a–a Fig. 2.Q3 concrete bridge