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Al Nageim - Chapter 2
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Understanding structural mechanics and structural design requires knowledge of many inter-linked factors. These include the loads and load actions on the structure, the strength and properties of the materials from which structural elements are made, the ways by which the loads and load actions are transferred via the structure to the foundations, the interaction between the foundations and the supporting ground, structural stability, durability and environmental conditions. It is therefore important to estimate accurately the loads that a structure has to withstand during its intended useful life, in order to achieve safety and economy in design. The behaviour of structures under loads depends on the strength properties of the materials of construction and the interaction between the components and parts of the structural frame and between the structural frame, its foundations and the supporting ground. Designers in their structural analyses try to predict this behaviour of the structure and identify the model to be used in the structural analyses. If they succeed then designs will usually be safe and economic. At present, existing knowledge of the loads on structures, properties of the materials of construction and analysis of structural frames is well advanced so that structural design can usually be considered to be economic with regard to these aspects. However, future research on understanding the actions of loads on structures will help to reduce a number of the existing uncertainties and hence result in safer and more economic designs.
In design, the loads on buildings and structures are classified into different types based on their frequency of occurrence and method of assessment. These are: 1 2 3 4 5 dead loads imposed loads wind loads earth and liquid pressures other load effects such as thermal effects; ground movement; shrinkage and creep in concrete; and vibration.
For each type of load, there will be a characteristic value and a design value. These will be explained later in this chapter. The design of any
Part 1: Code of practice for dead and imposed loads. flats. shear. Dead load is the weight of structural components. for most of the common types of structural elements. Part 1: Code of practice for dead and imposed loads. beams and columns. However. it is usual practice to simplify complicated load distributions to reduce the analysis and design time.CHAPTER 2
particular 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 combination that is likely to produce the most adverse effect on that element or the structure as a whole in terms of compression. location and distribution of imposed load items. such as floors. Part 1: Code of practice for dead and imposed loads gives imposed loads for various occupancy and functional requirements of buildings. heating ducts and all items intended to remain in place throughout the life of the structure. torsion and overturning. for example in the design of beams an approximate uniformly distributed load is usually used instead of the actual stepped-type loading. deflection. there are some simple rules for assessing the approximate dimensions required. for example slabs. the assessment of the dead load of most load bearing structural parts has to be done in practice by a method of trial and error to determine the approximate dimensions required for such parts. hotels. colleges and universities)
. They are gravity loads varying in magnitude and location. and includes all other permanent attachments to structures such as pipes. Such loads are usually caused by human occupancy. These rules are explained in the relevant code of practice. although the dead load of structural parts or members can be calculated accurately. These values are prescribed by both government and local building codes. tension. for example. or their combinations. electrical conduits. moment. such as • domestic and residential (dwelling houses. furniture and storage of materials. bending. for reinforced concrete and steel structures see BS 8110: Part 1: 1997 and BS 5950: 2000 respectively. They include distributed. BS 6399–1: 1996 Loading for buildings. In the design process.
BS 6399–1: 1996 Loading for buildings. walls and finishes. Imposed loads are sometimes called live loads or superimposed loads. guest houses) • institutional and exhibitions (schools. They are assumed to be produced by the intended occupancy or use of the structure. It is calculated from the unit weights given in BS 648: 1964 Schedule of weights of building materials or from the actual known weights of the materials used. impact and snow loads but exclude wind loads. In the analysis process. realistic values are difficult to determine. concentrated.
BS 6399–1: 1996 Loading for buildings. air conditioning. Because of the unknown nature of the magnitude.
warehouses. In structures such as highway bridges. 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. walls. In addition.
Reduction in total imposed floor loads
The code of practice allows for the reduction of imposed loads in the design of certain structural components and should be consulted for full details. their supports and foundations. factories.
Table 2. The imposed loads for these various buildings are different and hence different values should be specified for design. The imposed floor loads contributing to the total loads for the design of such structural elements may be reduced in accordance with Table 2.1 Reduction in total distributed imposed floor loads
Number of floors. carried by member under consideration
Reduction in total distributed imposed load on all floors carried by the member under construction (%) 0 10 20 30 40 50
1 2 3 4 5 to 10 Over 10
. for example within the high school building some space is used in classrooms and laboratories. including the roof. however.20
PART 1 BEHAVIOUR OF STRUCTURES
industrial (warehouses.
Even with this classification there is still broad variation in the imposed loads. garages and those office areas that are used for storage and filing purposes. Columns. No reduction. This reduction is allowed because of the reduced probability that the full imposed loads will occur at all the floors simultaneously. power stations) bridges (pedestrian. the effect of impact forces due to traffic loading must be accounted for. Briefly the main reductions are as follows: Beams and girders. highway and railway) shopping areas warehousing and storage areas. shall be made for any plant or machinery for which specific provision has been made nor for buildings for storage purposes.1. it is necessary to consider traffic loads in terms of both a concentrated load and a varying uniformly distributed load. Where a single span of a beam or girder supports not less than 46 m2 of floor at one general level. piers.
Part 2: Code of practice for wind loads. Where permanent partitions are shown in the construction plans their actual weights shall be included in the dead load. cranes and other plant supported by or connected to the structure. To provide for demountable partitions it is normal practice to consider an equivalent uniformly distributed load of not less than one-third of the per metre run of the finished partitions and treat it as an imposed load in design. it has to change speed and direction to keep flowing around the building and over it. In structural analysis and design it is necessary to consider the design wind loads due to these pressures in combination with other applied imposed and dead loads. the dynamic effects of the wind. have to be considered in design. Wind loads on structures are dynamic loads due to changes in wind speed.0 kN/m2. such as induced oscillations.1. such as a building or a structure. Wind loads depend on the wind environment and on the aerodynamic and aeroelastic behaviour of the building. Imposed loads from demountable partitions. Dead loads from permanent partitions. However. Allowance is made for these dynamic effects. and a static analysis is then carried out for these increased loads and the computed load effects used in the design. such as metal chimneys. since the wind direction varies with time the wind loads on structures have to be considered as of possible application from all directions.CHAPTER 2
Dynamic loads are those that produce dynamic effects from machinery. runways. sides and roof of the building. To allow for such effects in practical design. for some light tall structures. it is common practice in most cases to increase the dead-weight value of machinery or plant by an adequate amount to cater for the additional dynamic effect.
BS 6399–2: 1997 Loading for buildings. in the design of the relevant structural parts. For convenience in design it is usual practice to consider the wind loads as static loads.4 of BS 6399–1: 1996. In view of the complexity of the assessment of wind loads on structures it is not possible to give the subject full treatment here and the reader is advised to consult one of the references at the end of the book. For floors of offices. including impact. Furthermore. When the wind flow meets an obstruction. this additional uniformly distributed partition load should be not less than 1.
Load from partitions
Clause 5. The appropriate dynamic increase for all affected members is ascertained as accurately as possible and must comply with the relevant code of practice. Owing to the change in direction when wind flow encounters stable structures. the induced wind pressure can vary in direction such that the resultant wind loads are horizontal and vertical.
. In this process of change in direction it exerts pressures of varying magnitudes on the face.
and are caused by the effect of gravity. Figure 2. see the relevant code of practice or. Where there is doubt as to the permanency of dead loads. see BS 6399–1: 1996 and BS 648: 1964. geographical location of structure or building. then the change in wind speed is reduced and hence the dynamic wind pressure will also be reduced (see the relevant code of practice).3 of BS 6399: 1997 Loading for buildings – Part 2: Code of practice for wind loads. Pa (N/m2)) e 2 the air density r = 1.22
Fig. For example from section 2.226 kg/m3 Ve effective wind speed from section 2. in the UK. size.3 of BS 6399: 1997 Loading for buildings.613 V2 e (1)
The wind speed to be used in equation (1) is not the maximum recorded value. Dead loads are calculated from the actual known weights of the materials used (see Table 2. and its direction depends on the wind environment.2). Dead loads are the unit weight multiplied by the volume.
Loads on structures — summary
• Dead loads or permanent actions according to the Eurocodes They are the self-weight of structures or buildings. such loads should be considered as imposed loads.1 shows a typical graph of speed versus time during a gale.2. The wind pressure.1 Wind speed versus time
Wind speed v (m/s)
80 60 40 20 0 0 5 10 15 20 Time (s)
Average speed 46 m/s @ 25 s gust
The effective wind loads on structures are dependent on the wind speed. Part 2: Code of practice for wind loads. It should be calculated from the relevant section of the code of practice. is given by qs: dynamic pressure qs = 1rV2 (in pascals. Therefore: qs = 0. as occurs when the wind meets a building and has to change direction. shape and height. 2. and so act downwards. which is caused by changes of wind speed from Ve in m/s (metres/second) to zero. The wind normally blows in gusts of varying speed. For more information.2.
. If the shape of the structure is streamlined.
see the appropriate code of practice or Table 1 of BS 6399–1: 1996. stone aggregate 55 kg/m2 Aerated per 25 mm thick 15 kg/m2 Board Blackboard per 25 mm thick 12.5 kg/m2 0. Also called superimposed loads or live loads. • Wind loads Due to dynamic wind movements.0 kg/m2 Gypsum panels and partitions Building panels 75 mm thick 44 kg/m2 Lead Sheet. act downwards. Such as vehicles. trains.
.2 Weights of building materialsa
Material Plaster Two coats gypsum. 19 mm thick 44 kg/m2 Bitumen roofing felts Mineral surfaced bitumen 3. people. per mm thick Tarmacadam 25 mm thick Terrazzo 25 mm thick Tiling. Caused by gravity. cranes. etc. per 25 mm thick 2. 19 mm thick 42 kg/m2 Damp-proofing. Moveable imposed loads. roof Clay Timber Softwood Hardwood Water Woodwool Slabs. Moving imposed loads.5 kg/m2 Brickwork Clay.
• Imposed loads or variable actions according to Eurocodes They are gravity loads which vary in magnitude and location and are appropriate to the types of activity or occupancy for which a floor area will be used in service. etc. British Standards for Students of Structural Design. solid per 25 mm thick medium 55 kg/m2 density Concrete. 13 mm thick Screeding Cement : sand (1 : 3). Considered in structural design and analysis as static loads. these depend on the wind environment and on the aerodynamic and aeroelastic behaviour of the structure or building.160 Flagstones Concrete.7 kg/m2 2400 kg/m3 30 kg/m2 30 kg/m2 24–78 kg/m2 7850 kg/m3 10 kg/m2 60 kg/m2 54 kg/m2 70 kg/m2 590 kg/m2 1250 kg/m3 1000 kg/m2 15 kg/m2
Ashphalt Roofing 2 layers. 13 mm thick Plastic sheeting (corrugated) Plywood per mm thick Reinforced concrete Rendering Cement : sand (1 : 3). 19 mm thick 41 kg/m2 Road and footpaths. Their dynamic effects should be considered in addition to their static effects. 25 mm thick
Weight 22 kg/m2 4. stored material.CHAPTER 2
Table 2.0–5. 2. PP 7312:2002 (British Standards Institute)) aSee also BS 648: 1964 Schedule of weight of building materials.5 kg/m2 Blockwork Solid per 25 mm thick. 13 mm thick Slate tiles (depending upon thickness and source) Steel Solid (mild) Corrugated roofing sheets. 50 mm thick 120 kg/m2 Glass fibre Slab. Such as furniture.5 mm thick 30 kg/m2 Linoleum 3 mm thick 6 kg/m2
(Source: Adapted from Various extracts. solid per 25 mm thick 59 kg/m2 Cast stone 2250 kg/m3 Concrete Natural aggregates 2400 kg/m3 Lightweight aggregates (structural) 1760 + 240 kg/m3 .
in the UK. Fk. imposed and wind loads and BS 2573 for crane loads. imposed loads and wind loads. design load = Fk * gf where gf = the partial factor of safety for loads. and the probability of particular load combinations occurring. See the relevant national code of practice or BS 6399: 1997 – Part 2: Code of practice for wind loads. However.
A structure is usually exposed to the action of several types of loads. for stability or the behaviour of continuous members. directional and topographic effects. minor inaccuracies in calculation. BS 5950: 2000 and BS 8110: 1997 give recommendations for practical partial factors of safety for loads. • Others Soil pressure. such as. The partial factor of safety also takes into account the importance of the sense of the limit state under consideration. hydraulic pressure. is a statistically determined load value above which not more than x per cent of the measured values fall. such as dead loads. For the ultimate limit state. thermal effects. Using the principles of probability and standard deviation. At the present state of knowledge. They should be considered separately and in such realistic combinations as to take account of the most critical effects on the structural elements and on the structure as a whole. BS 6399: Parts 1–3: 1996 and 1997 for dead. shrinkage and creep in concrete. unusual increases in loads and construction inaccuracies. the loads should be multiplied by the appropriate factor of safety given in the relevant table of the code of practice. and when x 5 per cent. ground movement.
Characteristic load.64S standard deviation for load (2)
The plus sign is ‘commonly’ used since in most cases the characteristic load is the maximum load on a critical structural member. Depend on: 1 shape of structure/building 2 height of structure/building above its base 3 location of structure/building. the characteristic load is that obtained from the relevant national codes of practice. and vibration are determined by special methods found in specialist literature. readers are referred to the relevant code of practice. The factored loads
. which is introduced to take into account the effects of errors in design assumptions.
Design loads and partial factors of safety
The design load is calculated by multiplying the characteristic load Fk by the appropriate partial safety. characteristic loads can be defined as: charateristic load S mean load ± 1.24
Variable in intensity and direction. i.e.
4Gk 1. 3 Dead.2Gk + 1.9Gk acting on some parts of the structure to give the most severe condition.2Qk + 1. gf (dead loads) 1.0Gk
1. Obviously.6Qk (vertical load) 2 Dead and wind loads (a) design dead load = 1.2 (case 3.0Gk (b) design wind load = 1. 2.0 and 1 .4En where Qk = imposed load. Table 2. Different load combinations are recommended by the codes of practice.6Qk
1.4Wk 1.0Gk 1.2Qk = vertical load. Partial factors for loads gf.0Gk
1. For example.4Gk or 1.0 and for case 2.6Qk (c) design earth and water load = 1. 2. Some examples on load combinations are as follows: 1 Dead and imposed load (a) design dead load = 1. 2.4Wk where Gk = dead load (vertical load). for case 1.0Gk or 0.2 Load combinations
1. see BS 5950: Part 1: 2000.6Qk
1.4Gk
Maximum hogging
Sagging (case 1) (case 2) (case 3)
. and Wk = wind load.4 for load resisting uplift or overturning.4Gk + 1. For example.0Gk 1.CHAPTER 2
should be applied in the most unfavourable realistic combination to the part of the structure or the effect under consideration. and 1.4Wk 1.4Gk + 1.4Gk + 1. 2 In the design of a continuous beam. the worst load combination should be associated with the design dead load of 1.0Gk 1.
Fig.2Gk + 1. in the design of a simply supported beam the following load combination is commonly used: design load = 1.2Wk = wind load. gf (dead loads) = 1.0Gk (b) design imposed load = 1.0Gk 1.0Gk 1. Gk = dead load and En = design earth and water load.0Gk 1. more load combinations are possible in this case).2Wk where 1.0Gk 1.
1 The criterion for any load combination is that it is likely to produce the worst effect on a structure or structural element for design and/or analysis purposes. 3 In Fig.0Gk 1. imposed and wind loads design loads = 1.4Gk 1.2. only possible design load combinations should be considered.4Gk or 1. see Fig.0Gk
see clause Cl 2.3.4) Cross-sectional area = 0. Calculate the following: (a) the weight of each beam per unit length (the uniformly distributed loads per unit length) (b) the total weight of each beam (c) the design dead load for each beam.064 kN
Fig. (b) steel
4 Other realistic combinations that give the most critical effects on the individual structural elements or the structure as a whole are shown in the relevant code of practice. EC2 and Cl 2.2 m
1 Reinforced concrete beam (see Fig.4 m
0.08 m2 * 24 kN/m3 = 1.1 beams:
(a) reinforced concrete. 2.4 * 5. 2.3.3 shows a 3 m long reinforced concrete beam and a 914 mm deep 419 mm wide universal steel beam that is 6 m long.6 (8Q) are 1. unit weight of concrete = 24 kN/m3 Therefore the unit weight per unit length = 0.3 Example 2.92 kN/m
w = 8.08 m2 From Table 2.
Fig. EC3.4.4 = 0.4Gk = 1.064 kN
3m Dead load per metre (UDL) 3m Total design dead load
.76 = 8.92 kN/m Total weight of beam = 1.4 Example 2.76 kN Design dead load of the beam = 1.1
Figure 2. If the design loads are wrongly assessed at the beginning then all the subsequent structural design and/or analysis calculations will also be wrong. for example see Table 2 of BS 5950: Part 1: 2000.2 * 0.35 (8g) and 1. 5 The comparative values in Eurocodes for 1.92 kN/m * 3 m = 5.5 (8q). It is important that the design loads are assessed accurately.
Example 2.2.4 (8G) and 1.1. 2.1 loads on
Calculate the design loads on the reinforced concrete beam A (including self-weight).1 loads
w = 3.602 kN
Fig. p.7) Design loads = 1 .4 * 23. Unit weight of concrete = 24 kN/m3.5 kN/m2.287 kN Design dead load of the beam = 1.5 kN/m3 = 3.6 m
3.8 m2 Dead load: asphalt = 42 * 10 = 420 N/m2 = 0.5 kN/m3 The weight per unit length = (49 400>106) m2 * 78.4Gk = 1. 260) From Table 2. unit weight of steel (mild steel) = 78.21.2.88 kN/m * 6 m = 23. 2.42 kN/m2
.88 kN/m w = 32.6Qk Area carried by the beam = 3. 2.6 m
roof details of single-storey extension to an existing house
Existing house Beam A
Cavity brick wall 25 mm timber board 175 × 50 timber joists @ 400 mm C to C L L 400 mm 400 mm section X .5 Example 2.X 19 mm asphalt.6 shows plan and roof details of a flat roof single-storey extension to an existing house. two layers Plaster board Skim plaster
8m Garage
(See Fig. softwood 50 175 mm timber joists spaced at 400 mm centre to centre plaster board and skim (plaster finish) 42 kg/m2 590 kg/m3 590 kg/m3 15 kg/m2
Fig.2: plan and
Figure 2. which is 300 mm wide and 600 mm deep.602 kN
on steel beam
6m Dead load per metre (UDL) 6m Total design dead load
Example 2. Roof construction: asphalt (two layers) 19 mm thick 25 mm timber boards.287 = 32. Access is to be provided to the roof.6 Example 2. 2.CHAPTER 2
2 Steel beam (see Fig. since 1 kN is equivalent to a mass of 100 kg) Total weight of beam = 3.4Gk + 1. therefore use an imposed load of 1.88 kN/m (i.6 * 8 = 28. mass per metre of the beam = 388 kg/m.5) Cross-sectional area = 49 400 mm2 (from Table 11. 2.
56 kN = (1. 2.655 kN
Area carried by beam A
Fig.300 * 0.15 kN/m2
Total dead loads (excluding self-weight of the beam) = 0.9 kN/m2 Qk = 1.5 kN/m2
Fig.6 m
Example 2.384 + 69.025 147.6 * 1.2 kN Imposed load = 1.5 * 28.151 + 48.
Gk = 0. steel decking.129 kN/m2 15 * 10 = 150 N/m2 0.5 N/m2 0.6 * 12) = 29.8 Part of a roof in a steel
Beam A 29.5 * 4 * 2 = 12 kN Design loads = (1. The flat roof consists of felt.7 Example 2.12 = 151.28 kN
.847 kN/m2 = 24 * 0. 2.050 * 0.06 N/m2 0.600 * 8 = 34. Calculate the design loads acting on one steel beam (joist).4 * 34.4 * (0.56) + (1.4 (number of joists in 1 m width) 129.2 solution
151.147 kN/m2 590 * 10 * 0.655 kN
Design loads on beam A
3. insulation boards and a suspended ceiling below the rolled steel joists.28 kN = design load
Dead load = 0.9 * 4 * 2 = 7.175 * 1>0.8) = 34.2) + (1.8 shows part of a roof plan for a small steel building.3
Construction of a roof beam using a rolled steel joist Figure 2.8)) + (1.847 * 28.28
= = plaster board and skim = = Beam self-weight Design loads
590 * 10 * 0.4 * 7.
Two-way span
When the above rules are not satisfied then the loads on the roof or the floor can be distributed as shown in Fig.10 shows the relevant details at side walls and columns for a fully braced (in both directions) four-storey steel office building comprising a steel frame with reinforced concrete slabs on profiled metal decking.9 Example 2.14). 2.3:
Slab spans in this direction
A Beam B
(a) and (b) one-way action of the slab or floor.
. see Fig.9(b)). 2.
One-way span
1 Where the ratio L1>L2 2 (see Fig. lower length 2–5–8 (for column reference numbers. The load on the roof ABCD can be assumed to be carried equally between beams AB and DC. (c) two-way action of the slab
A B Area on beam BC
Area carried by beam DC
Example 2.CHAPTER 2
General notes on the calculation of loading from roofs or slabs onto supporting beams
A roof or slab can be designed and detailed to span one way so that load is physically transmitted only to supporting beams and not directly to the beams running at right angles to them.9(a)). 2. Calculate the design loads acting on roof beam C2–D2. and normally when the slab is made from concrete which is cast in situ. two-way action of the slab or roof must be taken into account. 2 When the roof or the slab is constructed of precast concrete units with the ratio L1>L2 2 (see Fig. External cladding is brick/breeze block and double-glazing. floor beam C2–D2. and inner column. The loads on the roof ABCD may be assumed to be carried equally between beams AB and DC.4
Figure 2. If this is not the case. 2.9(c) to take account of two-way action.
(side column)
(ceiling at side wall)
(internal column)
. side column and internal column
Lift. 2. side walls.5 m
r.5 mm facing brickwork (parapet) 1000 mm 0.9 m
Inner leaf Fixing straps at 600 mm centres
Doubleglazed window Inner leaf 180 mm r. (d) building details at parapet. services 3m Splice 3 at 4 m 17 m 5m 1m 6 at 8 m
Galvanized steel proprietar y edge strip with built-in stainless steel channel for fixing M12 stainless steel head bolts Stainless steel angle support fabricated to suit brickwork coursing Soft joint with pistol block over 102. (c) end elevation.10 Fully braced steel
Centre column 4 3 2 Braced bay 1 A B C 8m D E F G Braced bay 4m
building: (a) plan. slab on profiled metal decking
170 mm thick slab Shims for horizontal adjustment Steel beam Ceiling
1.c. slab on profiled metal decking Topping 1.30
Fig. (b) side elevation.c.
p.5 kN/m = 6.5 0.32 0.11) Dead loads Imposed loads Weight per unit length
= 6 kN/m2 = 1. dead loads external column.3 kN/m
Dead loads on the floor (kN/m2) tile screed 0.2 7. 2. dead load External steel beam at roof level cavity wall dead load concrete slab steel ceiling
1.5 kN/m2 On floors = 3.68 kN/m glazing (double) 0.1 0.2 0. 180 mm thick steel partitions and ceiling (0.0 4.32 kN/m2 steel 0.30 1.5 kN/m2
Roof beam C2–D2 (see Fig. 20).0 kN/m2
4.1.6 kN/m ceiling 0.58 0.58) services
External side wall at floor level cavity wall 5.1 kN/m2 0.0 kN/m2 concrete slab 4.8 kN/m2 = 1.0 kN/m2 4.5 kN/m2 Reduce the imposed loads in accordance with number of stories as recommended in the practical code of practice (see Table 2.5 kN/m 0.6 concrete slab. 170 mm thick steel ceiling services Total dead load Parapet wall (cavity wall) Column and casing internal column.5 kN/m2 = (6 * (4 * 1 m (long))) + (1.5 * (4 * 1 m (long))) = 24 + 6 = 30 kN/m
.0 kN/m2
external wall 1 m high
5.2 6.CHAPTER 2
Imposed loads (BS 6339: Part 1) On roof = 1.
Dead loads on the flat roof (kN/m2) topping materials (asphalt and screed) concrete slab.8 kN/m2 4.
5 4.52 kN 1.6 kN)
Floor beam C2–D2 (see Fig.6 = 43.4 kN 135.6 * 1.2 + 22.5 kN/m2 Weight per unit length = (7 * (4 * 1 m (long))) + (3.0 kN/m
Area carried by beam C2–D2 at both roof level and floor level
Fig.92 kN
.8 = 345.5 * * * * 0. 2.5 * 2 * 8 = 38.4 * 7 * (4 * 1 m)) + (1.2 kN/m 2 m (wide strip) = 1.32
Design loads per unit length Total design loads
= (1.4 * 7 * (4 * 8)) + (1.6 + 179.6 * 3.1 * 8 = 135 .12) Dead loads = 7.5 kN/m 2 m (wide strip) = 8.6 * 3.1 0.4 kN/m 1 m (one unit length) = 0.4 = 61.5 * (4 * 1 m (long))) = 28 + 14 = 42 kN/m = (1.8 + 76.6 + 9.4 * 12.8 kN
61.8 kN) C2 D2 4m
Design loads per unit length
Total design loads
Fig. 2.12 Design loads on beam
Area on beam C2–D2
C2–D2 at floor level
C2 8m
C1 8m
External steel beam C1–D1 at roof level (see Fig.5 m (high) = 2.13) Loads from: cavity wall steel concrete slab ceiling Total dead load Total design dead load Imposed design load Total design loads = = = = = = = = 4.4 = 173.1 kN/m 1.6 * 1.2 = 492. 2.5 * (4 * 1 m)) = 39.6 kN/m (total = 492.4 * 6 * (4 * 1 m)) + (1.8 0.6 kN/m = (1.11 Design loads on beam
4 3 2 1 A Braced bay B C 8m D E 2m 2m
C2–D2 at roof level
43.5 * (4 * 1 m)) = 33.4 * 6 * (4 * 8)) + (1.5 * (4 * 8)) = 313.5 * (4 * 8)) = 268.6 * 1.0 kN/m2 Imposed loads = 3.2 kN/m (total design load = 345. 2.2 kN/m = (1.52 + 38.
4 * 3.8 (20% reduction.8 kN 2 floors = 1.8 kN
.4 * 7 * (4 * 4) + zero imposed loads = 156. calculate the loads on the beams as follows: Load on beam 5–4 = 1.5) = 25.4 * (3 * 4 * 1. see the code) = 286. lower length 2–5–8 (see Fig.92 kN C2 D2 21.52 kN Note: For the calculation of the moment acting on the column at joint 5.14 Column reference
Area carried by the column at the roof and floor levels
13 10 7 4 1 Outside column
15 12 9 6 3 Centre column 4m 4m
Loads from beam 4–5 and beam 5–6 Design dead loads Design imposed loads = 1.CHAPTER 2
Fig.4 * (2 * 7) * (4 * 8) = 627.5 * (4 * 4) [from beam 5–64 = 179.74 kN/m
external roof beam C1–D1
Centre column.13 Design loads on
173.5 * (4 * 4) [from beam 4–5] + 1.2 kN Design imposed loads for: Roof = 1.6 kN = 1.5 * (8 * 4) = 76.4 * 7 * (4 * 4) [from beam 4–5] + 1.6 * 2 * 3.5 * (8 * 4) * 0.14) Load on column above joint 5 Design dead loads for: Roof = 1.2 kN 3 columns (4 m high) and casing = 1. 2.6 * 3.2 kN
Total design loads on the column below joint 5 = 1777.6 * 1. 2. 2.8 kN 2 floors = 1.4 * 7 * (4 * 4) [from beam 5–6] = 313.4 * 6 * (4 * 8) = 268.72 kN
steel beam A and steel beam B.34
Load on beam 5–6 (usually the longest beam of 5–4 or 5–6) = (1. Roof construction: asphalt (two layers) 19 mm thick 25 mm timber boards 50 175 mm timber joists spaced at 400 mm centre to centre plaster board and skim (plaster finish) 3 42 590 kg/m3 590 kg/m3 15 kg/m2 kg/m2
concrete weights 24 kN/m3 and the coping weighs 0.’ Discuss the above statements.15.6 m Cavity brick wall
Garage 3m Beam B 19 mm asphalt.52 kN
246.4 kN The design loads and bending moments are shown in Fig. Calculate and sketch the total design loads on one timber joist.4 * 7 * (4 * 4)) + (1.5 kN/m2 in addition to its own weight.15 Design loads and
246.4 kN 4 246. 2. all relevant loads should be considered separately and in such realistic combinations as to comprise the most critical effects on the elements and the structure as a whole.Q2 Exercise 2: plan and roof details of
a single-storey extension to an existing house
A reinforced concrete bridge between two buildings spans 8 m.5 kN/m2.6 m 3. The floor of the bridge has to carry a uniformly distributed imposed load of 1. two layers Plaster board Skim plaster
25 mm timber board 175 50 timber joist @400 mm C to C L L 400 mm 400 mm Section ×–×
Fig. 2. The brickwork weighs 20 kN/m3.4 kN 6
Design loads at central column 2–5–6
For moment at 5 use the above loads
1 ‘In the design and/or assessment of an individual structural element or a structure as a whole.
Fig. therefore use imposed load of 1.5 kN/m. Access is to be provided to the roof.6 * 3.4 kN
1284. Figure 2. The magnitude and frequency of fluctuating loads should also be considered. For beam A shown in
. 2.72 kN
156.5 * (4 * 4)) = 246. 2. Use annotated sketches to support your answers.8 kN 6 4 5 5 1777.Q3. The cross-section of the bridge is shown in Fig.Q2 shows plan and roof details of a flat roof single-storey extension to an existing house.
floor beam B1–B2 and outer column.10 (Example 2. 2. 2.CHAPTER 2
a 8m Beam A Building (A) Beam A Building (B)
Fig.Q3 Reinforced concrete bridge
.4) shows the relevant details at side walls and columns for a fully braced (in both directions) four-storey steel office building that is made of a steel frame with a reinforced concrete slab on profiled metal decking. floor beam A1–A2.c. 2. 4 Figure 2.Q3.3 m
150 mm 254 mm headroom available
203 × 133 × UB25 Window frame 3m Section a–a
Fig. The beam is simply supported at both ends. calculate and sketch the total uniformly distributed design loads in kN/m and the total design loads in kN. External cladding is brick/breezeblock and double-glazing. lower length 1–4–7 (see Fig. slab
225 mm 1. Calculate and sketch the design loads acting on roof beam A1–A2.14).
Plan Coping
225 mm brickwork r.
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