Lateral Load Manual SAP2000v19 for analysisFull description
Computation of LoadsFull description
Manufacturing
Lightning CalculationFull description
Deskripsi lengkap
Drug CalculationsFull description
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Loads Dead loads Imposed loads oor roof Determining load per m and m� Wind
Structures transmit loads from one place to another Where do loads come from Dead loads - permanent and stationary Structure itself Plant and equipment Some rough gures (note that values are subject to variation depending on specifc material type) Also note values are forces per unit volume not mass per unit unit volume. Unit Weights of baisc construction materials Aluminium Brick Concrete Steel Timber
kN/m� 24 22 24 70 6
Precast concrete beam length 1. Calculate weight of beam per unit length. 2. Calculate total weight of beam in it is 10.5m. First cross sectional area of beam = (0.6 x 0.25) - (0.4 x 0.15) = 0.09m� From table unit weight of concrete = 24kN/m� Weight per unit length = 0.09 x 24 = 2.16kN/m Total weight = 2.16 x 10.5 = 22.68kN
Often we are dealing with sheet materials or we know a layer t hickness of oor or roof build up. Figures here are per unit area Again when using these type of charts some care is needed needed to ensure you have the correct gure, or that it corresponds with your design. Unit weight of basic sheet materials Asphalt (19mm) Aluminium roof sheeting Glass (single glazing) Plasterboard and skim Rafters battens roong felt Sand/cement screed (25mm) Slates Timber oorboards Plaster on wall face
kN/m� 0.45 0.04 0.1 0.15 0.14 0.6 0.6 0.15 0.3
Calculate the dead load in kN/m� of the following oor build up: Timber oor boards 40mm sand/cement screed 125mm reinforced concrete slab timber oo timber oorr boa boards rds screed concrete slab dead load /m�
if we are dealing with a wall acting on a beam we are interested in load per linear unit of the beam
In this example calculate the load per metre on the beam. The build up is a double glazed window on a cavity wall of 102.5mm brick outer face and 100mm plastered lightweight blockwork which is 12kN/m�.
brickwork blockwork plaster double glazing
= 1.2 x 0.1025 x 22 = 1.2 x 0.1 x 12 = 1.2 x 0.3 = 2 x 1.3 x 0.1 load on beam
= 2.71 = 1.44 = 0.36 = 0.26 = 4.77kN/m
Imposed loads - or live loads, movable loads that act on the structure when it is in use. People, furniture, cars, computers and machinery are all imposed loads. Normally we consider imposed loads as oor and roof loads Typical oor loads kN/m� Art galleries 4.0 Banking halls 3.0 bars 5.0 Car parks 2.5 Classrooms 3.0 Churches 3.0 Computer labs 3.5 Dance halls 5.0 Factory workshop 5.0 Foundaries 20.0 Hotel bedrooms 2.0 Ofces (general) 2.5 Ofces (ling) 5.0 Private houses 1.5 Shops 40
If a bar should be designed with live load of 5.0 kN/m� and if an average person is 80kg how many people are expected to be standing in one square metre of oor?
Force exerted by one person Number of people per m�
= 80 x 9.81 = 5000 / 785
= 785 N = 6.4 people/m�
equivalently if your house is designed with 1.5 kN/m� and the total area was 22m� how many people could you invite to a party? Force exerted by one person Number of people per m� Total number of people at party
= 785N = 1500 / 785 = 1.9 x 22
= 1.9 people/ m� = 42 and a bit.
certain types of dancing can cause dynamic effects that increase the effect of load.
Calculating imposed roof loads. What you need to know: 1. Is access to the roof provided? (a load of adjacent oor area is required) 2. Predominant load is snow. • which is dependant on • geographical location • height above sea level • shape of roof • wind that redistributes snow into drifts
Estimating ground snow loads in Canada. Info from Canadian Cryospheric Information Network Find worst case depth and multiply by density (kg/m�) and 9.81 Tables in National Building Code provide further details
In UK snow load varies from 0.3kN/m� on south coast to 3.0kN/m� in Scotland Calculating a snow load in Canada. National Building Code Part 4 4.1.7.
wind exposure factor is 1.0 but can be reduced to 0.75 or in exposed areas north of treeline to 0.5 if building is an exposed location and exposed on all sides no obstructions around building no obstructions on roof such as parapet snow cannot drift onto roof from adjacent surfaces slope factor based on roof angle a and surface type. is 1.0 if a <= 30⁰ is (70⁰ - a)/40⁰ when a> 30⁰ is 0 if a > 70⁰ if roof is a slippery surface (where snow and ice slide off) slope factor
is 1.0 if a <= 15⁰ is (60⁰ - a)/40⁰ when a > 15⁰ is 0 if a > 60⁰
accumulation factor is 1.0 except when for large at roofs when 1.2 x [1-(30/l)�] but not less than 1.0 for roofs with wind factor = 1.0 1.6 x [1-(120/l)�] but not less than 1.0 for roofs with wind factor = 0.75 or 0.5 w = smaller plan dimension L = larger plan dimension and l is 2 x w - ( w�/L) in metres can be assigned other values when: roof shapes are arched, curved or domes snow loads in valleys snow drifts from another roof projections on adjacent roofs snow sliding or drainage from adjacent roofs
Theres more: in reality full and partial loading has to be considered In addition to the load calculation above roofs of slope less than 15⁰ and arched or curved roofsmust be designed with accumulation factor 1.0 on one portion while half that load is applied to t he remainder .
Calculate snow load on this roof structure What is the snow load per metre length of truss? What is the total snow load on one roof truss? What is the load per metre on the supporting wall? Assume that loads from trusses are evenly distributed
Calculate snow load for Halifax S
ground snow Halifax = 1.7
S = Ss (Cb x Cw x Cs x Ca) + Sr Snow load per m � ground snow load in kPa (kN/m�)
S = 1.7 x (0.8 x 1 x 0.75 x 1) + 0.5 S = 1.52kN/m�
associated rain load
Trusses are at 0.6m centres So snow load per metre length of truss is: 0.6 x 1.52 = 0.9kN/m Note load is vertical so 1m dimension is measured horzontally For 7m truss load is 7 x 0.9 = 6.4 kN
Load per m on wall = 1.52 x 3.5 = 5.32 kN/m
Wind loads act normal (or perpendicular) to building surfaces winds can cause pressure or suction. For this reason building structures must resist horizontal forces as well as vertical forces. In addition some light weight structures can be subject to uplift forces from the wind so need to be adequately held down. Wind loads like snow loads vary depending on: geographic location degree of exposure building height and size building shape wind direction in relations to structure positive or negative pressures in the building
Faster moving air creates lower pressure (bernoulli effect) as in plane wings. The same principle causes forces to act on building surfaces.
Structure for resisting wind loads
These principles should be well understood by now if not Look at: Francis Ching. Building Construction Illustrated Edward Allen. Architect’s Studio Companion
For structural design it is often necessary to consider several load cases due to the wind blowing from different directions. Designing a building in Halifax calculating wind loads. National Building Code of Canada Part 4 4.1.8. p = q x Ce x Cg x Cp external pressure acting statically and normal to surface reference velocity pressure exposure factor gust effect factor external pressure coefcient
National Building Code of Canada appendix c for tables of climatic information
net pressure on a surface is the difference between internal and external similar to external pressure internal pressure is calculated according to the NBC
p = q x Ce x Cg x Cp internal pressure acting statically and normal to surface reference velocity pressure exposure factor gust effect factor internal pressure coefcient
reference velocty pressure three are shown in table 1 in 10, 1in 30, 1in 100 these are probabilities of pressure occuring so 1 in 10 is used for cladding and stuctural design for vibration and deection 1 in 30 for structural strength post - disaster buldings use the 1 in 100 pressure values. exposure factor exposure increase with height height m > 0 and <=6 > 6 and <=12 > 12 and <=20 > 20 and <=30 > 30 and <=44 > 44 and <=64 and so on more heights given in nbc
exposure factor 0.9 1.0 1.1 1.2 1.3 1.4
gust factor 1.0 or 2.0 for internal pressures to be found somewhere in the 500 pages of appendix A!! we’ll use 1.0 for now. 2.0 for the building as a whole and main structural members 2.5 for small elements external and internal pressure coefcients again appendix A we’ll use 1.0 for now.
Forces due to wind on simple building external pressure p = q x Ce x Cg x Cp 1 in 30 year Pressure Halifax = 0.52 kPa ( kN/m�) Walls below 6m so exposure factor is 0.9 Gust factor = 2.0 external and 1.0 internal Pressure coefcient = 1.0 external p = 0.52 x 0.9 x 2.0 x 1.0 = 0.936 kN/m� internal p = 0.52 x 0.9 x 1.0 x 1.0 = 0.468 kN/m� so 0.936 - 0.468 = 0.468 kN/m� acting normal to vertical surfaces windward and 0.936 + 0.468 = 1.4kN/m� leeward
Forces due to wind on simple building external pressure p = q x Ce x Cg x Cp 1 in 30 year Pressure Halifax = 0.52 kPa ( kN/m�) Roof above 6m so exposure factor is 1.0 Gust factor = 2.0 external and 1.0 internal Pressure coefcient = 1.0 external p = 0.52 x 1.0 x 2.0 x 1.0 = 1.04 kN/m� internal p = 0.52 x 1.0 x 1.0 x 1.0 = 0.52 kN/m� 1.04 kN/m� acting normal to vertical surfaces at roof level normal to roof 1.04 x Sin(40) = 0.67 windward = 0.67 - 0.52 = 0.15 kN/m� leeward = 0.67 + 0.52 = 1.19(suction)
Next up: A couple of other load types (to know about) Uniform and point loads Safety factors Calculating load on beams Load paths Pin Jointed structures
A couple of other load types (to know about) Uniform and point loads Safety factors Calculating load on beams Load paths
Hydrostatic pressure loads from soils and liquids Increases linearly with depth.
Application of safety factors to loads Loads discussed are realistic estimates of loads or characteristic loads when checking ultimate strength characteristic loads are increased by multiplying by a safety factor. The result is the design load. Safety factors load combination dead and imposed dead and wind dead, imposed and wind
dead 1.4 or 1.0 1.4 or 1.0 1.2
imposed 1.6 1.2
wind 1.4 1.2
imposed roof
For example to obtain the maximum compressive design in the support at B two load combinations should be checked and the larger value used
wind imposed oor
1.4 x dead + 1.6 imposed or
dead load
1.2 x dead + 1.2 x imposed + 1.2 wind to obtain maximum tensile design load in the support at A we need to minimise the effect of the dead and imposed loads by using 1.0 x dead + 1.4 x wind
Point load (kN)
Uniformly distributed load (kN/m)
Calculating loads on beams Example Building type - ofce Floor construction = 4.11kN/m � Perimeter wall construction = 4.77kN/m � self weight of beams = 0.6 kN/m safety factors are 1.4 for dead load and 1.6 for live load to find design load
