How to Estimate Designs Loads

8/11/2019 How to Estimate Designs Loads

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Loads

Dead loads

Imposed loads

oor

roofDetermining load per m and m

Wind


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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 ty

Also note values are forces per unit volume not mass per unit volume.

Unit Weights of baisc construction materials kN/m

Aluminium 24

Brick 22

Concrete 24

Steel 70

Timber 6


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Precast concrete beam length

1. Calculate weight of beam per unit leng

2. Calculate total weight of beam in it is 1

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

Weight per unit length = 0.09 x 24 = 2.16

Total weight = 2.16 x 10.5 = 22.68kN


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Often we are dealing with sheet materials or we know a layer thickness of oor or roof build up

Figures here are per unit area

Again when using these type of charts some care is needed to ensure you have the correct g

that it corresponds with your design.

Unit weight of basic sheet materials kN/m

Asphalt (19mm) 0.45

Aluminium roof sheeting 0.04

Glass (single glazing) 0.1

Plasterboard and skim 0.15

Rafters battens roong felt 0.14Sand/cement screed (25mm) 0.6

Slates 0.6

Timber oorboards 0.15

Plaster on wall face 0.3


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Calculate the dead load in kN/mof

oor build up:

Timber oor boards

40mm sand/cement screed

125mm reinforced concrete slab

timber oor boards = 0.15screed = 0.6 x 40/25

concrete slab = 0.125 x 24

dead load /m


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if we are dealing with a wall acting on a beam we are interested in load per linear unit of the b


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In this example calculate the load per metre on th

The build up is a double glazed window on a cavi

102.5mm brick outer face and 100mm plastered l

work which is 12kN/m.

brickwork = 1.2 x 0.1025 x 22 =

blockwork = 1.2 x 0.1 x 12 =

plaster = 1.2 x 0.3 =

double glazing = 2 x 1.3 x 0.1 =

load on beam =


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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/mArt galleries 4.0

Banking halls 3.0

bars 5.0

Car parks 2.5

Classrooms 3.0

Churches 3.0

Computer labs 3.5Dance halls 5.0

Factory workshop 5.0

Foundaries 20.0

Hotel bedrooms 2.0

Ofces (general) 2.5

Ofces (ling) 5.0

Private houses 1.5Shops 4.0

Theatres (xed seats) 4.0


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If a bar should be designed with live load of 5.0 kN/mand if an average person is 80kg how m

are expected to be standing in one square metre of oor?

Force exerted by one person = 80 x 9.81 = 785 NNumber of people per m = 5000 / 785 = 6.4 people/m

equivalently if your house is designed with 1.5 kN/mand the total area was 22mhow many p

you invite to a party?

Force exerted by one person = 785NNumber of people per m = 1500 / 785 = 1.9 people/ m

Total number of people at party = 1.9 x 22 = 42 and a bit.

certain types of dancing can cause dynamic effects that increase the effect of load.


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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


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Estimating ground snow loads in Canada. Info from Canadian Cryospheric Information Netwo

Find worst case depth and multiply by density (kg/m) and 9.81

Tables in National Building Code provide further details


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In UK snow load varies from 0.3kN/mon south coast to 3.0kN/min Scotland

Calculating a snow load in Canada. National Building Code Part 4 4.1.7.

S = Ss (Cbx Cwx Csx Ca) + Sr

Snow load per m

ground snow load in kPa (kN/m)

roof snow load factor = 0.8???

wind exposure factor

slope factor

accumulation factor

associated rain load

National Building Code of Canada appendix c for tables of clim


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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 30

is 0 if a > 70

if roof is a slippery surface (where snow and ice slide off)

slope factor is 1.0 if a 15

is 0 if a > 60


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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


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Theres more:

in reality full and partial loading has to be considered

In addition to the load calculation above roofs of slope less than 15and arched or curved roo

designed with accumulation factor 1.0 on one portion while half that load is applied to the rem


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Calculate snow load

structure

What is the snow loa

length of truss?

What is the total snow

roof truss?

What is the load per m

supporting wall? Ass

from trusses are even


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Calculate snow load for Halifax S

ground snow Halifax = 1.7

snow load factor = 0.8

wind exposure factor = 1.0

slope factor = (70 - 40) / 40 = 0.75

accumulation factor = 1.0

rain load Halifax = 0.5

S = 1.7 x (0.8 x 1 x 0.75 x 1) + 0.5

S = 1.52kN/m

Snow load per m

ground snow load in kPa (kN/m)

roof snow load factor = 0.8???


slope factor

accumulation factor

associated rain load

S = Ss (Cbx Cwx Csx Ca) + S


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Trusses are at 0.6m c

So snow load per me

truss is:

0.6 x 1.52 = 0.9kN/m

Note load is vertical s

sion is measured horz

For 7m truss load is

7 x 0.9 = 6.4 kN


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Load per m on wall = 1.52 x 3.5 = 5.32 kN/m


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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 t

equately 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


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Faster moving air creates lower pressure (bernoulli effect) as in plane wings.

The same principle causes forces to act on building surfaces.


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Structure for resisting wind loads


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For structural design it is often necessary to consider several load cases due to the wind blow

different directions.

Designing a building in Halifax calculating wind loads. National Building Code of Canada Part

p = q x Cex Cgx 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


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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 Cex Cgx Cp

internal pressure acting statically and

normal to surface

reference velocity pressure

exposure factor

gust effect factor

internal pressure coefcient


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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 exposure factor > 0 and 6 and 12 and 20 and 30 and 44 and


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gust factor

1.0 or 2.0 for internal pressures to be found somewhere in the 500 pages of appendix A

well 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 well use 1.0 for now.


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Forces due to wind on simple building

external pressure p = q x Cex Cgx 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/macting normal to

vertical surfaces windward

and 0.936 + 0.468 = 1.4kN/mleeward


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Forces due to wind on simple building

external pressure p = q x Cex Cgx 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/macting 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)


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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


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A couple of other load types (to know about)

Uniform and point loads

Safety factors

Calculating load on beams

Load paths

H d t ti l d f il d li id


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Hydrostatic pressure loads from soils and liquids

Increases linearly with depth.

A li ti f f t f t t l d


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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

The result is the design load.Safety factors

load combination dead imposed wind

dead and imposed 1.4 or 1.0 1.6 -

dead and wind 1.4 or 1.0 - 1.4

dead, imposed and wind 1.2 1.2 1.2

For exampleimposed roof


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For example

to obtain the maximum compressive design in the suppor

two load combinations should be checked and the larger

1.4 x dead + 1.6 imposed

or

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 using

1.0 x dead + 1.4 x wind

A B

wind

p

imposed oor

dead load


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Point load (kN)

Uniformly distributed load (kN/m)

Calculating loads on beams


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Calculating loads on beams

Example

Building type - ofce

Floor construction = 4.11kN/m

Perimeter wall construction = 4.77kN/mself 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

design load kN/m

oor 1.4 x 4.11 5.75

wall 1.4 x 4.77 6.68

beams 1.4 x 0.6 0.84

imposed 1.6 x 2.5 4.00

total design floor load = 5.75 + 4.00 = 9.75kN/m

B B1 (8 )


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Beam B1 (8m span)

supports a total width of 6m

load from oor = 9.75 x 6 = 58.50kN/m

self weight of beam = 0.84 kN/m

design UDL = 59.34 kN/m

symmetry indicates reactions will be equal

reaction = (59.34 x 8) / 2 = 237.4 kN

beam B2 is the same as B1

Beam B3 suports a 3m width of oor plus the p


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Beam B3 suports a 3m width of oor plus the p

wall

Load from oor = 3 x 9.75 = 29.25 kN/m

load from wall = 6.68kN/m

self weight of beam = 0.86kN/m

total design UDL = 36.79kN/m

symmetry indicates reactions will be equal

reaction = (36.9 x 8) /2 = 147.6kN

B6


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Beam B4 has same UDL as B1 and B2 but span is 6m

B6


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Beam B6

Supports perimeter wall and a point load from th

Load from wall = 6.68kN/m

self weight of beam = 0.86kN/m

design UDL = 7.54kN/m

design point load = 237.4kN

Reaction from symmetry

= ((7.54 x 12) + 237.4) / 2 = 163.9kN

B6

Now work out reactions for B5


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Load paths


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p


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Working through load path for a simple sign.


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How to Estimate Designs Loads

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