Survey results for fire loads and live loads in office buildings
Loads on Industrial Buildings
Transcript of Loads on Industrial Buildings
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Objectivesj By the end of this lecture, you should be able to:
1. Understanding the different design philosophies2. Identify the different types of loads acting on
industrial buildings3. Understand the procedure used to calculate the
loads acting on different structural elements
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Types of Structural Steel MembersTypes of Structural Steel Members Structural steel Structural steel members may be found in several imagesimages:
Rods Plates Plates Hot rolled sections (Angles, ( g ,pipes, I sections, channels…etc)C ld f d Cold formed sections (Z, C, ,…etc), )
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Specifications and building codesp g Specifications and building
codes are the result of bi d j d t f combined judgment of
researchers and practicing engineers.
Th i bj ti f th The main objectives for the codes is to provide simple design procedure to get safe and economic structureand economic structure.
Two design philosophies are supported by the ECP:L d d R i F 1. Load and Resistance Factor Design (LRFD)
2. Allowable stress design (ASD)(ASD)
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Load and Resistance Factor DesignLoad and Resistance Factor Design (LRFD)( )
QR :where
..iin QR
factorreduction strength theis : where
factoroverloadtheiscapacitymember or strength nominal theis nR
loadservicerelevant theis
factoroverload theis
i
i
Q
iQ
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Load combinations for LRFD 1.4*D
*D 6L L 1.2*D+1.6L+0.5Lr 1.2D+1.6Lr+(0.5L or 0.8W) 1.2D+1.3W+0.5L+0.5Lr 1.2D±1.0EQ+0.5L 0.9D±(1.3W or 1.0EQ)Where, D: Dead load, L: Live load, Lr: roof live load, W: Where, D: Dead load, L: Live load, Lr: roof live load, W: wind load, EQ: earthquake load
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Allowable Stress DesignAllowable Stress Design (ASD)( ) This method of design focus on ensuring that the stresses due to straining actions of service load should stresses due to straining actions of service load should not exceed the allowable stresses specified by the code i ei.e.
n QR
.i
i
n Q
i
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Load combinations for ASD DD L D+L
D+L+W D+L+EQ 0.9D±(W or EQ)Where, D: Dead load, L: Live load, Lr: roof live load, W: wind load, EQ: earthquake loadq
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Loadsf l d d lTypes of loads acting on industrial
buildingsbuildings Dead LoadLi L d Live Load
Wind load Crane load Special loads
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Dead loads The dead loads acting on industrial buildings includes:
O i h f f i1. Own weight of roof covering2. Own weight of different structural steel elements3. Weight of R.C. slab (case of Mezzanine)4. Collateral loads (from Mechanical, elect.,
HVAC,…etc) Equipments acting on the top roof
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Dead Loads:•Dead Loads:Own weight of roof coveringg g The own weight of roof covering depends on the type of roof cladding i e :of roof cladding i.e.:‐
1. For steel corrugated sheets wt 5‐10kg/m2
F d i h l k /2. For sandwich panels, wt 10‐20kg/m2
3. This weight is acting on inclined projection
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For example, when calculating the dead load acting on the purlin, Wdl (roof cladding)= (5‐‐>10kg/m2) x adl
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C i D d L dContinue, Dead LoadsOwn weight of steel structuresg The own weight of steel structures composing the industrial
building represents the most important factor in competition b t t t l i i dditi t th t f i between structural engineers in addition to the cost of easier fabrication.
Usually the weight of steel structures depends on the span of the b ildi i h i h l d i f i li i building, its height, loads acting, roof inclination, cranes loading,….etc.
However, for open free areas, with (no cranes) spans up to 30m d h i h f 6 8 i h l d h and eave heights from 6m‐8m with steel corrugated sheets as a
cladding, the weight of steel structures (including purlins, wind bracing, columns, rafters, struts,…etc 25‐40 kg/m2
F i l l i h R C h i h f l For mezzanine level with R.C. the weight of steel str. 70‐120 kg/m2
This load is acting on horizontal projectiong p j
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For example, when we calculate the D.L acting on a single joint on the main supporting steel truss then Wsteel structure=(2540 kg/m2) x b x spac. Bet. trusses
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Cont Dead LoadsCont. Dead LoadsD.L Weight of deck for Mezzanine levelg In case of R.C. deck in Mezzanine level then Mezzanine level, then WR.C. deck = ts x R.C + Wfl Wfloor cover (100150kg/m2)+ Wpartitionspartitions
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C ti D d L dContinue, Dead LoadsCollateral loads
Collateral loads is usually referred to the extra dead load due to Hvac ducts, fire fighting system, fans,
k l bl k t th t rock wool blanket that are extra applied to the roof of the industrial roof of the industrial building.
It should be assessed by ythe EM engineer, but can be assumed 10‐20 kg/m2
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Live Loads For roofs, it depends on the proof inclination and roof accessibilityaccessibility.
For floors in Mezzanine levels, ,the live loads is to be assessed accord To the accord. To the Egyptian codes of practice for loads.
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Wind LoadWind Load Steel structures are usually
light weight structures and the effect of wind should be the effect of wind should be taken into consideration.
Wind acting on the building may cause pressure on one may cause pressure on one side and suction on the other side.
The effect of the wind load The effect of the wind load is the sum of the effect of pressure and suction.
Wind load is calculated by Wind load is calculated by the following eqn. W=Cexkxq
Where Where Ce= coefficient of exposureq= wind pressure kg/m2
K = factor depends on height of bldg from the natural ground
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Crane LoadsCrane Loads The cranes
usually affect the t t i th structure in the 3D as follows:
1. Vertical component component representing dead and live loads + impactp
2. Horizontal component in the plan of f d t frame due to lateral shock
3. Horizontal component in component in longitudinal direction due to braking force g
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Location and values of MaxLocation and values of Max. straining actions for crane Loadsg
Max. Reaction is Rmax.= P(2‐a/L) Max. bending moments1. If a<0.586L, thenM ill b t t l d Mmax will be at nearest load
to the support at a distance x=0.5(L‐a/2)
d M P(Land Mmax=P(L‐a/2)^2/(2L)
2. If a>0.586L, then Mmax5 , maxwill be with one load at center of span and Mmax=PL/4Mmax /4
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Continue crane loads1. The vertical component acting on each wheel is
{the dead weight (wt of crane bridge) + the live ={the dead weight (wt of crane bridge) + the live loads ((weight of trolley + wt of cargo) positioned at the wr0est case location)} * the impact factor The the wr0est case location)} the impact factor. The impact factor for electrical cranes is 25%
2 The horizontal lateral shock is taken 10% of the live 2. The horizontal lateral shock is taken 10% of the live loads without impact
3 The braking force is taken 1/7 of the total vertical 3. The braking force is taken 1/7 of the total vertical component without impact.
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Special loadsp These special loads includes extra requirements according to project technical specifications or due to according to project technical specifications or due to site conditions. For example, the client may ask for a structural system to withstand explosions or extra live structural system to withstand explosions or extra live load more than that required in load code.
Other kinds of special loads includes effect of Other kinds of special loads includes effect of temperature, settlement of supports, loads during construction, terrorist attacks, aircraft attacks…….etc.construction, terrorist attacks, aircraft attacks…….etc.
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ExampleExample Given:Corrugated steel sheets for roof covering weigh=7kg/m2
Spacing between frames S=6mSt l i ht k / 2Steel own weight =30kg/m2
Collateral loads (HVAC, lightings,..etc.)=30kg/m2
Wind pressure=70kg/m2p 7 gWeight of crane bridge=5tWeight of trolley= 2tMaximum crane capacity=6t
h dMin. approach crane distance=1.8mRoof is in‐accessible Required: Calculate the loads acting on one Calculate the loads acting on one
intermediate frame
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Solution1. Dead loads acting on each
frame: WDL= S*(Wroof cov./cosoo + Wsteel struct.+Wcollateral loads)
WDL=6m*(0.007/cos5 71+0 03+0 03)=0 402 t/m’5.71+0.03+0.03)=0.402 t/m
2. Live loads acting on each frame, tan =0.1, then LL k / LL=53.3kg/m2, WLL=S*LL=6*53.3=0.32t/m’
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3. For wind loads we have two columns and two rafter partitions with different partitions with different exterior wind coefficients depending on wind directionsdirections.
Assuming wind blow from left to right, then
Wwind=S*Ce*q*kK=1 as column height <10m
k /q=70 kg/m2W1=6m*0.8*0.070*1=0.336t/m’W2=6m*‐0.5*0.070*1=‐0.21t/m’W2 6m 0.5 0.070 1 0.21t/mFor W3, as tan =0.1, then from
chart, Ce=‐0.8W3=6m*‐0.8*0.070*1=‐0.336t/m’W4=6m*‐0.5*0.070*1=‐0.21t/m’
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4.For crane loading, as shown the max. loading on crane girder will be when crane girder will be when the trolley carries full capacity and approach to the nearest distance from the nearest distance from one columnIn this case, Rmax =2P=
b id crane bridge weight/2+(trolley weight +LL)*(28‐1.8)/28= 5/2+(2+6)*(28‐1.8)/28=9.986 tons, Rmin=3.014t3 4Then P=9.986/2=4.99t‐RmaxLL+I=1.25*9.986=12.48t‐Lateral Shock <‐>=0.1*9.986=0.9986t‐Braking Braking force=1/7*9.986=1.43t
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