loads estimetion 1

8/13/2019 Loads Estimetion 1

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Evaluation of the loads and load combinations

According Limit state design method (LSDM- in the EU)or - load and resistance factor design method (LRFDM-

in the USA)

Limit state designrequires the structure to satisfy two principal criteria:the ultimate limit state (ULS)and the serviceability limit state (SLS).

A limit state is a set of performance criteria (e.g. vibration levels,deflection, strength), stability (buckling, twisting, collapse) that must bemet when the structure is subject to loads.

Any design process involves a number of assumptions. The loads towhich a structure will be subjected must be estimated, sizes ofmembers to check must be chosen and design criteria must be

selected. All engineering design criteria have a common goal: that ofensuring a safe and functional structure.

Ultimate Limit State

To satisfy the ultimate limit state, the structure must not collapse whensubjected to the peak design loadfor which it was designed. A structureis deemed to satisfy the ultimate limit state criteria if all factoredbending, shear and tensile or compressive stresses are below thefactored resistance calculated for the section under consideration. Thelimit state criteria can also be set in terms of stress rather than load.

Thus the structural element being analysed (e.g. a beamor a columnorother load bearing element, such as walls) is shown to be safe whenthe factored loads are less than their factored resistance.

Serviceability Limit State

To satisfy the serviceability limit state criteria, a structure must remainfunctional for its intended use subject to routine (read: everyday)loading, and as such the structure must not cause occupant discomfortunder routine conditions. A structure is deemed to satisfy the

serviceability limit state when the constituent elements do not deflect by


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more than certain limits laid down in the building codes, the floors fallwithin predetermined vibration criteria, in addition to other possiblerequirements as required by the applicable building code. Examples offurther serviceability limit requirements may include crack widths inconcrete, which typically must be kept below specified dimensions. Astructure where the serviceability requirements are not met, e.g. the

beams deflect by more than the SLS limit, will not necessarily failstructurally. The purpose of SLS requirements is to ensure that peoplein the structure are not unnerved by large deflections of the floor,vibration caused by walking, sickened by excessive swaying of thebuilding during high winds, or by a bridgeswaying from side to side andto keep beam deflections low enough to ensure that brittle finishes onthe ceiling above do not crack, affecting the appearance and longevityof the structure. Many of these limits depend on the finish materials(sheetrock, acoustical tile) selected by the architect, as such, the limitsin the building codes on deflections are generally descriptive and leavethe choice to the engineer of record (this may not be as true outside theU.S.)

Loading Effects Combination / Groupings for UltimateLimit State1st Combination / Grouping

ikikjk QQG ,,01,, 5.15.135.1 ++ With:

1.35 Gk, j Design Dead Load

f=1.35 Partial Safety Coefficient for Dead Load for unfavorableexception

1.5 Qk,1 Design Dominant Variable Load

f=1.5 Partial Safety Coefficient for Variable Load for unfavorableexception

1.5 0,i Qk,i Design Variable Load

0,i=0,7 Concomitance (Reduction) Factor for combination value ofa variable action


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Symbols: The notation used is based on ISO 3898:1987.

F k characteristic value of a load (force) F d= (f ) (F k) design value of a load

f =1.35 (1.40 in USA, Canada, Japan, Australia etc) for Gk and

=1.50 (1.60 in USA, Canada, Japan, Australia etc) for Qkpartial safety coefficients for loadsDead (Permanent) Loads Gk or Wk or SWk - concentrated permanent load, or

weight, or self-weight load, in N or daN or kN g k or wk - weight per unit area in N .m -2, or weight per unit length

in N .m -1 = () (g) - unit weight in N.m-3 or kN.m-3 = m.V-1- unit mass or density of material in kg.m-3 g=9.81(~10) m.s-2 gravitational acceleration Densities (Unit Masses) of building materials and stored materials:

Definitions. Tables. Unit Weights of building materials

Permanent (Dead) Load

by the old RomanianStandard 10101-1

Load type by EC1Part 2.1 asGor Wor SW

[ ]

[ ]

[ ][ ]

[ ]

8.0;9.0

3.1;2.1;1.1

.

.

.

.

3

3

2

1

=

=

=

===

=

=

=

=

=

n

orn

mNing

VgVmaF

mkginV

m

mNindg

ormNinAg

NinVG

nGG

n

n

n

nc

- concentrated design dead load

- (nominal) characteristic load

- load per unit length

- load per unit area or load(ing)per squared meter

-material unit mass (density)

- force

- material unit weight

- (overloading) partial safetyfactor for dead load

[ ]

[ ]

[ ][ ]

[ ]

9.0

40.135.1

.

.

.

.

3-

3-

2-

1-

=

==

=

===

=

=

=

=

=

f

ff

k

k

k

kfd

mNing

VgVmaF

mkginV

m

mNindg

ormNinAg

NinVG

GG


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Dead Load Acting on Building Element

for Outer Garret for Inner GarretRoof Roof with Thermal InsulationLast Floor with Thermal Insulation Last FloorFloors including Staircase Floors including Staircase

First Floor above Basement First Floor above BasementOuter Walls; Inner Walls Outer Walls; Inner WallsBasement Walls Basement Walls

No layerthickness

unitweight

characteristicvalue

ultimatevalue

d =g gk=d gd= F gk=1.35gk

m N.m-3 N.m-2 - N.m-2

Roof Load Estimation (and example)

Gable roof dead load


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Snow Load on roof

Sd= FSk=1.50 Sk (N.m-2)


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SNOW LOAD (according STAS CR1-1-3/2005)

sk= i ce ct s0,ksk

= facturated intensity of the load given by the snow

i= shape coefficient of the surface exposed the snowce = exposure coefficient ce =1,0 for the normal exposed buildings with flat

roofs.Ct= thermal factor c7=1,0S0.k= weight of the snow layer: (kN/m

2) for recovering period of 50 years .


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values of s0,kin (kN/m2)


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Steps for the snow computation:

Loads Computation by SR EC1 CR -1.1.3:2005, Design Code.

Snow Load Evaluation on Constructions, Romnia, published in


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Monitorul Oficial al Romniei, 1st part, year 174(XVIII), no.148encore, February, the 16th, 2006.

Example:

Two slopes pitched roof with iron sheet covering having theslope of p%=tg180 and the length of half roof in horizontal projection

L 0=6000 mm, with timber support (king post truss) as structure, willbe designed to support a snow variable load (Sk) as uncrowded(neaglomerat) load for a partial exposure of placement in Iai(Suceava).

Dwelling functional solution has 2 floors level (ground floor andthe1st floor) and the outer garret. In the assumptions of:

snow density () of 235...350...400 kg.m-3 function ofbreaking state (starea de afnare) and unfavorable snowfalling (depunerii nefavorabile de zpad),

gravitational constant (g) of 10 [m.s-2],

snow depth at the earth level (t0) of 600 mm, friction coefficient between snow and the iron sheet covering(cfr) of 0.05,

shape coefficient () of 2 slopes pitched roof is 0,8 for,

exposure coefficient of placement (Ce) is 1 for a partialexposure,

thermal coefficient (Ct) is 1 for thermal insulation applied onlast floor plate,

characteristic value of snow load on earth level (s0,k) is 2,0(2,5) kN.m-2for Iai (Suceava),

coefficient function of unsteady (depunerea neregulat) snowfalling (k) of 2,5,

partial safety coefficient for Dead Load is F=1,35 in EU and1,40 in the USA,

partial safety coefficient for Variable Load is F= 1,50 in EUand 1,60 in the USA,

concomitance coefficient for variable loads is 0=0,7 in EUand in the USA,

possible variable loads Qk are: snow load Sk, wind load Wk,and live load Lk.


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(a) Compute the snow weight in the defavorable situation (=g) inkN.m-3.(b) Draw the loading patterns.(c) Compute the snow characteristic value on roof horizontalprojection (sk=CeCts0,k) in kN.m-2. Use Snow Load Eq. and inputdata.(d) Compute the snow characteristic value on roof horizontalprojection (sk) in kN.m-2. Use the Dead Load Eq. for a snow layerhaving the known dimensions at the earth level (s0,k=t0g) andinput data. Compare the results.(e) Compute the characteristic value of snow pushing force (foreide mpingere) (Fk=Sk,L,X - cfrSk,L,Y) in kN.m-1 on the eavesgutters (asupra opritorilor de zpad dispui n vecintatea

jgheabului).(f) Compute snow load hanged down at roof eaves and distributedon roof length run (atrnatde marginea acoperiului i distribuitpe lungimea acoperiului) (Se=kSk2/) in kN.m-1.(g) Show the Loads Grouping with snow dominant variable load forthe check-up of roof structure at ultimate limit state of strength(pentru verificarea structurii acoperiului la starea limit ultimde rezisten) (FGk+FQk,1+F0Qk).

Example:(a) =g=(235)(10)=2350kg.m-3m.s-2=2350 N.m-3=2.35 kN.m-3(b) snow loading patterns:

L0

Se

Fk

Sk

S0,k


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A:(c) sk=0.8x1x1x2.5

=2.00 kN.m-2(d) so,k=tg=t =0.60x235x10x10-3 =1.41 kN/m

2

=0.60x350x10x10-3 =2.10 kN/m2=0.60x400x10x10-3 =2.40 kN/m2

sk=0.8x1x1x2.10=1.68 kN/m2

Comentary: sk=1.68 kN/m2suits to the placement in Iasi.

(e) skL0=sk,LL where L0/L=cos where L=L0x1/cossk,L=sk cos =2.00x0.95 =1.90 kN/m2Sk,L=sk,LL =1.90x6x1/0.95 =12 kN/m-1Sk,L,Y=Sk,Lcos=12x0.95 =11.4 KN/mSk,L,X=Sk,Lsin=12x0.30 =3.60 KN/mFk=Sk,L,X - cfrSk,L,Y =3.60 - 0.05x11.4 =3.03 KN/m

(f) 2=0.8+0.8/30 =0.8+0.8x18/30 =1.28

sk=2CeCts0,k =1.28x1x1x2.5 =3.2 kN/m2Se=2.5x3,22/2.35 =10.9 KN/m

(g) 1.35Gk+1.50Sk+1.05LkComentary: W=0.


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II - Design of a Timber Rafter by LSDM-ULS of Strength (according EC5 )

(i) Rafter cross section:

(ii) Loading alternatives :

Dead Load: (in N/m) for outer garret

tiles covering: gkg/1issue10m/s2

noof issues/m2dr

battens : bbhbtimberdrnoof battens/ Lrafter : br . hr . timber

gk

Dead Load: (in N.m-1) for inner garret

shingle covering: g kg/m210m/s2dr

boarding : hbtimberdr


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thermal insulation: d insulationdr

vapor barrier: g kg/m210m/s2dr

rafter: brhr timberceiling if any:

gk

(iii) Patterns:

(iv) Static Analysis of beam ABsubjected to an uniformly load wand a single concentrated load P acting at mid-span (at point C)


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(v) Material Properties: Structural Timber Strength Classes: conifer

& deciduous species classes: C14 C16 C18 C22 C24 C27 C30 C35

C40 D30 D35 D40 D50 D60 D70

Stiffness properties (kN/mm2) Mean modulus of elasticity parallel tograin E0, mean 10


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(vi) Sizing i.e. estimation of a suitable section using timber ofstrength class C22 subjected to a bi-axial bending

dm

ymx

reqz

y

zmdmymx

z

dm

dm

y

y

z

xdm

f

MkMW

W

WkfMkM

W

fW

M

W

M

,

,

,,

,,

)(1

==

=

In which,W (Z by EC5) : section modulus (in mm3); Wz=bh2/ 6 andWy=hb2/6 ; M z (y) : design bending moments (in N. m)

m,,d: design normal bending stress (in N.mm-2)fm,d : design bending strength (in N.mm-2),km=0.7(1.0) : bending factor for rectangular (other) section.

From Table, b (in mm) x h (in mm) would be suitable Wz=xxx. 10 3mm3, Iz =xx.x .106mm4,

A =xx .103mm2, since h>150 mm & kh=1 as assumed.

(vii) Checking If members are not fail in bending, the followingcondition should be satisfied:

1,,

,,

,,

,,

ydm

ydmm

zdm

zdm

fk

f


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Load Estimation (for walls and floor)

Walls Load Estimation (example)

1. DEAD LOADS for each building element (we recommend to organizethe estimation on tables for each building element).

a)WARM FLOOR PLATENo. Layer Type d g G

k Gd

(m) (daN/m3) (daN/m2) (daN/m2)

0. 1 2 3 4=2x3 5 6=4x51. Parquet (50x30x3.5): (Parchet L.U.

stejar, 22 cm gros.,)0.022 900 1.35

2. Wood board 0.02 600 1.35

3. Sand 0.035 160 1.35

4. Reiforced Concrete Floor 0.12 2500 1.35

5. Plaster 0.02 1900 1.35

Sgk= Sg

d=

b) COLD FLOOR0 1 2 3 4 5 6

1. Mosaic Mortar ( or clay plates) 0.02 2100 1.35

2. Cement Mortar 0.03 2100 1.35

3. Reinf. Concrete Floor 0.12 2500 1.35

4. Plaster 0.02 1900 1.35

Sgk= Sg

d=

c) EXTERNAL WALLS TYPE - I0 1 2 3 4 5 6

1. Internal Cement / lime plaster 0.02 1800 1.352. Brick-work masonry 0.25 1700 1.35

3. Polystyrene 0.10 20 1.35

4. External plaster 0.015 1800 1.35

Sgk= Sg

d=

d) BEARING INTERNAL WALLS OF 20 cm0 1 2 3 4 5 61. M 100 Plaster 0.015 2100 1.35

2. Brick-work masonry 0.25 1700 1.35

3. M 100 Plaster 0.015 2100 1.35

Sgk= Sg

d=


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e) NON BEARING WALLS OF CERAMIC STRIPS (300 X 300 X75) mmNo. 1 2 3 4 5 6

1. M 100 Plaster 0.015 1800 1,35 35.12. Ceramic Strips Masonry

300 x 300 x 750.075 1300 1,35 1.7

3. M 100 Plaster 0.015 1800 1,35 35.1 Sgk= Sgd=

f) STAIRS ROOM FLOOR0 1 2 3 4 5 6

1. M100 Plaster 0.015 1800 1,35

2. Reinforced ConcretePlatform

0.150 2500 1,35

3. Mosaic Mortar 0.020 2100 1,35

4. Cement Mortar 0.030 2100 1,35

Sgk= Sgd=

Partitions Load on the Floors

The internal dividing walls will be assimilated with uniformdistributed loads over the floor surface of the room where there areconstructed. The uniform distributed load will be considered with thefollowing intensity :

The wall weight on the 1m lengthof the wall measured in ( N/m)

The intensity of the equivalentuniform distributed loadsmeasured in ( N/m2)

Gk1500 gk,e= 5001500< Gk3000 gk,e= 10003000< Gk5000 gk,e= 1500

5000< Gh gk,e=(Vnet ) / (L1L2)

ge,d=Fgk,e=(1.50)gk,e (N/m2)

LIVE LOADS on the floor : qd

=F0qk

=(1.50)(0.7)qk


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a) At the current floorqf=1500 N/m2 and pc=1500 x 1.5x0.7 =. N/m2

b) Stairspf=3000 N/m2 and pc=3000 x 1.5x0.7 =. N/m2

c) Balconypn=5000 N/m2 and pc=5000 x 1.5x0.7 =. N/m2

loads estimetion 1

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