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Keywords: Accidental action, impact, explosion, risk analysis, robustness ABSTRACT: Recently the Eurocode on Accidental Actions, officially referred to as EN 1991-1-7, has been completed. The code describes the principles and application rules for the assessment of accidental actions on
buildings and bridges. The leading design principle is that local damage is acceptable, provided that it will not
endanger the structure and that the overall load-bearing capacity is maintained during an appropriate length of
time to allow necessary emergency measures to be taken. As measures to mitigate the risk various strategies
are proposed like prevention of actions, evacuation of persons, physical protection of the structure and suffi-
cient structural redundancy and ductility. The code makes a clear distinction between identified and unidentified accidental actions. For the identified
accidental actions (impact, explosions) a structural analysis is proposed, the level of which may depend on the
envisaged consequences of failure. It may vary from an analysis on the basis of static equivalent forces to a
quantitative risk analysis including nonlinear dynamic structural analysis. Also for unidentified accidental ac-
tions the measures depend on the consequence class. In these cases more general measures are proposed to
ensure a sufficient robustness of the structure. One may think enhanced redundancy, design of special key elements and three-dimensional tying for additional integrity. This paper summarises the code and provides some background information as well as design examples. 1 INTRODUCTION
In September 2004, CEN TC250 Subcommittee ap-
proved the current draft of Eurocode 1, Part 1.7, Acci-
dental Actions" for a formal voting by the EU-
member states. The draft is now in the process of translation into German and French and editorial im-
provements. This paper summarizes this code and provides some background and examples. Accidental actions may be defined as actions with low probability, severe consequences of failure and usually of short duration. Typical examples are fire, explosion, earthquake, impact, floods, avalanches, landslides, and so on. Next to these identified accidental actions, structural members may get damaged for a variety of less identifiable reasons like human errors, improper use, exposure to aggressive agencies, failure of equipment, terrorist attacks and so on. In the Eurocode system, fire and earthquake are dealt with in specific parts. The document EN 1991-
1-7 deals primarily with impact and explosion. In addition, the document gives general guidelines on how to deal with identified and unidentified accidental actions in general. The identified actions may be analysed using classical (advanced) structural analysis. For the unidentified actions more general robustness requirements (e.g. prescribed tying forces) have been introduced.
The objective of design in general is to reduce risks at an economical acceptable price. Risk may be expressed in terms of the probability and the consequences of undesired events. Thus, risk reducing measures consist of probability reducing measures (e.g. lightning rods) and consequence reducing measures (e.g. sprinklers, vent openings and so on). No design, however, will be able or can be expected to counteract all actions that could arise due to an extreme cause. The point is that a structure should not be damaged to an extent disproportionate to the original cause. As a result of this principle, local failure may be accepted. For that reason, redundancy and non-linear effects play a much 3311
ICOSSAR 2005, G. Augusti, G.I. Schuëller, M. Ciampoli (eds)
© 2005 Millpress, Rotterdam, ISBN 90 5966 040 4
larger role in design for accidental actions than in
the case of variable actions.
Design for accidental design situations needs to be primarily included for structures for which a collapse
may cause particularly large consequences in terms of injury to humans, damage to the environment or eco-
nomic losses for the society. A convenient measure
to decide what structures are to be designed for acci-
dental situations is to arrange structures or structural components in categories according to the conse-
quences of an accident. Eurocode 1991 Part 1.7 ar-
ranges structures in the following categories based
on consequences of a failure: class conse-
Example structures 1
2, lower group
3, upper group
low rise buildings buildings up to 4 stories buildings up to 15 stories high rise building,
grand stands etc.
Not only the appropriate measures but also the appropriate method of analysis may depend on the safety category, e.g. in the following manner:
-class 1: no specific consideration of accidental actions;
-class 2: a simplified analysis by static equivalent load models for identified accidental loads
and/or by applying prescriptive design and detailing rules; -class 3: extensive study of accident scenarios and
using dynamic analyses and non-linear analyses if appropriate. It is up to the European member states to decide what is considered as an appropriate strategy in the various cases. 2 UNIDENTIFIED ACCIDENTAL LOADS
The design for unidentified accidental load is pre-
sented in Annex A of EN1991-1-7. Rules of this type were developed from the UK Codes of Practice and regulatory requirements introduced in the early
70s following the partial collapse of a block of flats at Ronan Point in east London caused by a gas ex-
plosion. The rules have changed little over the inter-
vening years. They aim to provide a minimum level of building robustness as a means of safeguarding buildings against a disproportionate extent of col-
lapse following local damage being sustained from
an accidental event (Institution of Structural Engi-
neers).
The rules have proved satisfactory over the past 3
decades. Their efficacy was dramatically demon-
strated during the IRA bomb attacks that occurred in
the City of London in 1992 and 1993. Although the rules were not intended to safeguard buildings against terrorist attack, the damage sustained by those buildings close to the location of the explo-
sions that were designed to meet the regulatory re-
quirement relating to disproportionate collapse was found to be far less compared with other buildings that were subjected to a similar level of abuse. Annex A, in fact, only specifies operational guid-
ance for consequences class 2. A distinction is made between framed structures and load-bearing wall construction. For framed structures in the lower
group of consequences class 2, Annex A recom-
mends horizontal ties in each floor around the pe-
rimeter and internally in two right angle directions between the columns (Figure 1). The capacity of the ties is a function of the self-weight, the imposed load and the geometrical properties.
For the upper class 2, in addition one of the fol-
lowing measures should be taken: • Introduce vertical ties; • Design key elements for an accidental design ac-
= 34 kN/m
• Ensure that upon the notional removal of a sup-
porting column, wall section or beam the damage
does not exceed 15% of the floor in each of 2 ad-
jacent storeys.
perimeter tie L
Figure 1: Example of effective horizontal tying of a
framed office building.
2.1 Design example As a design example consider a framed structure, 5
storeys with story height h = 3.6 m, consequences class 2, upper group (as originally prepared by the Dutch Studiecel Stufip (2004)). Let the span be L = 7.2 m and the span distance s = 6 m. The characteris-
tic values for the self weight and floor loads are g
respectively and the combination fac-
tor Ψ = 1.0. In that case, according to Annex A5, the required internal tie force may be calculated from:
= 0.8 {g
+Ψ q
} sL (1)
= 0.8 {4+4} (6 x 7.2) = 276 kN For Steel quality FeB 500 this force corresponds to a steel area A = 550 mm
or 2 ø 18 mm. The perimeter
tie is simply half the value. Note that in continuous
beams this amount of reinforcement usually is al-
ready present as upper reinforcement anyway. For the vertical tying force we find in a similar way T
= (4 + 4) (6 x 7.2) = 350 kN/column This correspondents to A = 700 mm
or 3 ø 18 mm. 3 IDENTIFIED ACCIDENTAL LOADS
The general principles and combination rules, to be applied in design situations for identified accidental
actions, are defined in EN 1990 Basis of Design.
Partial load factors to be applied in accidental design
situation are defined to be 1.0 for all loads (perma-
nent, variable and accidental).
Combinations for accidental design situations ei-
ther involve an explicit accidental action A (e.g. fire or impact) or refer to a situation after an accidental
event (A = 0). After an accidental event the structure will normally not have the required strength in per-
sistent and transient design situations and will have
to be strengthened for a possible continued applica-
tion. In temporary phases there may be reasons for a relaxation of the requirements e.g. by allowing wind or wave loads for shorter return periods to be applied in the analysis after an accidental event.
Chapter 4 of EC1-Part 1.7 deals with impact from vehicles, ships, trains, fork lift trucks and helicopters.
In this paper we will only discuss the impact from ve-
3.1 Impact from vehicles
The mechanics of a collision may be rather complex.
The initial kinetic energy of the colliding object can be transferred into many other forms of kinetic energy and into elastic-plastic deformation or fracture of the structural elements in both the building structure and the colliding object. Small differences in the impact location and impact angle may cause substantial chan-
ges in the effects of the impact. This, however, is ne-
glected in Eurocode EN 1991-1-7 and the analysis is confined to the elementary case, where the colliding
object hits a structural element under a right angle.
Even then, impact is still an interaction phenome-
non between the object and the structure. To find the forces at the interface, one should consider object and structure as one integrated system. Approximations, of course, are possible, for instance by assuming that the
structure is rigid and immovable and the colliding ob-
ject can be modelled as a quasi elastic, single degree of freedom system (see Figure 2). In that case the ma-
ximum resulting interaction force equals (CIB, 1993): F = v
√(km) (2)
is the object velocity at impact, k the equiva-
lent stiffness of the object and m its mass. In practice the colliding object will not behave elastically. In most cases the colliding object will respond by a mix of ela-
stic deformations, yielding and buckling. The load de-
formation characteristic may, however, still have the nature of a monotonic increasing function (see Figure
3). As a result one may still use equation (2) to obtain useful approximations. In the background documenta-
tion, the model (2) has been successfully compared
with experiments by Popp (1961) and Chiapetta and
Pang (1981). Also later (unpublished) tests initiated by the British High Way Agency gave results in the same order.
The formula (2) gives the maximum force value on the outer surface of the structure. Inside the struc-
ture these forces may give rise to dynamic effects.
Amplification factors between 1.0 and 1.4 are consid-
ered as a more or less realistic. Figure 2 Spring and rod models for the colliding objects
Figure 3: General load displacement diagram of colliding object The design values in EC1, Part 1.7, Table 4.3, have for political reasons been chosen in accordance with
Eurocode 1, Part 3. In the following table some corre-
sponding input values for the parameters m
of equation (2) are presented.
Proceedings ICOSSAR 2005, Safety and Reliability of Engineering Systems and Structures
Table 1: Calculation of design value of EC1, 1.7.
) a H ( a −
25 1 0 5 25
) . . ( . − 1
1000=940 kNm type of road mass velocity equivalent
collision force based on (2) m
motorway 20000 40 300 1000
urban area 20000 30 300 500
- only cars - also trucks
a H −
25 . 1 00 . 5 −
1000 = 750 kN Similar for the direction perpendicular to the driving direction:
25 1 0 5 25 1
) . . ( . −
500=470 kNm Q
= 00 . 5
500 = 375 kN The input design values for masses and velocity are relatively low values. As a consequence also the F
value in Eurocode 1, Part 1.7 is low. However, if
combined with a conservative linear classic static structural model, the overall design could still very well be over designed.
Other loads are not relevant in this case. The self
weight of the bridge deck and the traffic loads on the bridge only lead to a normal force in the column.
Normally this will increase the load bearing capacity
of the column. So we may confine ourselves to the accidental load only. Using a simplified model, the bending moment capacity can conservatively be es-
timated from:
3.2 Design example for vehicle impact Consider by way of example the reinforced concrete bridge pier of Figure 4. The cross sectional dimen-
sions are b = 0.50 m and h = 1.00 m. The column
height h = 5 m and is assumed to be hinged to both the bridge deck and to the foundation structure. The reinforcement ratio is 0.01 for all four groups of bars as indicated in Figure 4, right hand side. Let the steel
yield stress be equal to 300 MPa and the concrete strength 50 MPa. The column will be checked for
impact by a truck under motorway conditions.
= 0.8 ω h
= 0.8 0.01 1.00
0.50 300 000 = 1200 kNm
= 0.8 ω h b
= 0.8 0.01 1.00 0.50
300 000 = 600 kNm
As no partial factor on the resistance need to be used in the case of accidental loading, the bending mo-
ment capacities can be considered as sufficient. The shear capacity of the column, based on the concrete tensile part only is approximately equal to (say f
= 1200 kN/m2):
= 0.3 bh f
= 0.3 1.00 0.50 1200 = 360 kN. This is almost sufficient for the loading in y-
direction, but not for the x-direction. An additional
shear force reinforcement is necessary. 3.3 Explosion loads Gas explosions account for by far the majority of ac-
cidental explosions in buildings. Gas is widely used and, excluding vehicular impact, the incidents of oc-
currence of gas explosions in buildings is an order of magnitude higher than other accidental loads caus-
ing medium or severe damage (Mainstone (1978), Leyendecker and Ellingwood (1977)). Figure 4 Bridge pier under impact loading
According to the code, the forces F
should be taken as 1000 kN and 500 kN respectively and act at a height of a = 1.25 m. The design value of the bending moments and shear forces resulting from the static force in longitudinal direction can be cal-
In this context an explosion is defined as a rapid chemical reaction of dust or gas in air. It results in
high temperatures and high overpressure. Explosion pressures propagate as pressure waves. The follow-
ing are necessary for an explosion to occur (NFPA (1988)):
- fuel, in the proper concentration; - an oxidant, in sufficient quantity to support the combustion;
- an ignition source strong enough to initiate com-
The pressure generated by an internal explosion de-
pends primarily on the type of gas or dust, the con-
centration of gas or dust in the air and the uniformity of gas or dust air mixture, the size and shape of the enclosure in which the explosion occurs, and the amount of venting of pressure release that may be available. In completely closed rooms with infinitely strong walls, gas explosions may lead to pressures up to 1500 kN/m
, dust explosions up to 1000 kN/m
, depending on type of gas or dust. In practice, pressures generated are much lower due to imperfect mixing and the venting which occurs due to failure of doors, windows and other openings. According to Annex D of EN1991, Part 1.7, ele-
ments of a structure should be designed to withstand the effects of an internal natural gas explosion, using a nominal equivalent static pressure given by (Dra-
gosavic (1972, 1973), Leyendecker and Ellingwood (1977)):
= 3 + 0.5 p
+0,04/(A
whichever is the greater, where p
is the uniformly distributed static pressure in kN/m
at which venting components will fail, A
is the area of venting com-
ponents and V is the volume of room. The explosive pressure acts effectively simultaneously on all of the bounding surfaces of the room. The expressions are valid for rooms up to a volume of 1000 m
and vent-
ing areas over volume ratios of 0,05 m
The venting pressure p
may be determined from tests or may be calculated from uniformly loaded plate bending formulae. Guidance can be obtained in Harris at all (1977). An important issue is further raised in the note ot clause 5.3.3. It states that the peak pressures in the main text may be considered as having a load dura-
tion of 0.2 s. The point is that in reality the peak will generally be larger, but the duration is shorter. So combining the loads from the above equations with a duration of 0.2 s seems to be a reasonable approxi-
mation. The conditions following an explosion, and the reaction of the various elements of the structure to these conditions, are obviously extremely complex. Deciding on a design pressure is only part of the dif-
ficulty for a designer. There is then the difficulty of how to assess the probable response of the structure to loading which is short term, dynamic and omni-
3.4 Design example for explosion loading Consider by way of example a living compartment in a multistory flat building. Let the floor dimen-
sions of the compartment be 8 x14 m and let the height be 3 m. The two small walls (the facades) are made of glass and other light materials and can be considered as venting area. These walls have no load bearing function in the structure. The two long walls are concrete walls; these walls are responsible for carrying down the vertical loads as well as the lateral stability of the structure. This means that the volume V and the area of venting components A
for this ca-
se are given by: A
= 2 x 8 x 3 = 48 m
V = 3 x 8 x 14 = 336m
So the parameter A
/ V can be calculated as: A
/ V = 48 / 336 = 0.144 m
As V is less then 1000 m
/ V is well within the limits of 0.05 m
and 0.15 m
lowed to use the loads given in the code. The col-
lapse pressure of the venting panels p
is estimated as 3 kN/m
. Note that these panels normally can re-
sist the design wind load of 1.5 kN/m
. The equivalent static pressure for the internal natural gas explosion is given by: p
Ed = 3 + p
= 3 + 3 = 6 kN/m
/2+0,04/(A
= 3+1.5+0.04/0.144
= 3+1.5+2.0 = 6.5 kN/m
This means that we have to deal with the latter. The load arrangement for the explosion pressures is presented in Figure 5. According to Eurocode EN 1990, Basis of Design, these pressures have to be combined with the self weight of the structure and the quasi-permanent values of the variables loads. Let us consider the design consequences for the va-
rious structural elements. 3315
where ∆t = 0.2 s is the load duration, g = 10 m/s
the acceleration of the gravity field and u
design value for the midspan deflection at collapse.
This value, of course, depends on the ductility prop-
erties of the floor slab and in particular of the con-
nections to the rest of the structure. It is beyond the scope of this paper to discuss the details of that as-
sessment, but assume that u
= 0.20 m is consid-
ered as being a defendable design value. In that case the resistance against explosion loading can be as-
sessed as: H=3m p
) 2 . 0 ( 10
20 . 0 * 2
]*7.7=12.5kN/m
So the bottom floor system is okay in this case.
Figure 5: Load arrangement for the explosion load
3.4.2 Upper floor 3.4.1 Bottom floor
Let the self weight be 3 kN/m
and the characteristic live load 2 kN/m
. This means that the design load
for the explosion is given by: Let us next consider the upper floor. Note by the way that the upper floor for one explosion could be the bottom floor for the next one. The design load for the explosion in that case is given by (upward value positive!): p
= 3.00 + 6.50 + 0.5*2.00 = 10.50 kN/m
= - 3.00 + 6.50 + 0 = 3.50 kN/m
The design for normal conditions is given by: So the load is only half the load on the bottom floor, but will nevertheless give rise to larger problems.
The point is that the explosion load is in the opposite direction of the normal dead and live load. This me-
ans that the resistance against the explosion may
simply be close to zero. What is needed is top rein-
forcement in the field and bottom reinforcement a-
bove the supports. The required resistance can be found by solving p
0.85 * 1.35 * 3.00 + 1.5*2.00 = 6.4 kN/m
We should keep in mind that for accidental actions
there is no need to use a partial factor on the resis-
tance side. So for comparison we could increase the
design load for normal conditions by a factor of about 1.2. The result could be conceived as the resis-
tance of the structure against accidental loads, if it designed for normal loads only: ϕ
= [1 + Rd
= 1.2*6.4 = 7.7 kN/m
Using again p
, ∆t = 0.2 s, g=10 m/s
we arrive at p
= 1.5 kN/m
. This would require about 25 percent of the reinforcement for normal
conditions on the opposite side.
So a floor designed for normal conditions only should be about 30 percent too light. It is now time
to remember the clause 5.3.3 mentioned earlier. If we take into account the short duration of the load we may increase the load bearing capacity by a fac-
tor ϕ
given by (see EC1-Part1.7, Background Do-
cument (2005)): ϕ
= 1 + Rd
An important additional point to consider is the
reaction force at the support. Note that the floor could be lifted from its supports, especially in the upper two stories of the building where the normal
forces in the walls are small. In this respect edge
walls are even more vulnerable. The uplifting may
change the static system for one thing and lead to different load effects, but it may also lead to free-
standing walls. We will come back to that in the next section. If the floor to wall connection can resist the lift force, one should make sure that also the wall it-
self is designed for it. 3.4.3 Walls
Finally we have to consider the walls. Assume the wall to be clamped in on both sides. The bending moment in the wall is then given by: m = 1/16 p H
= (1/16) 6.5 3
= 4 kNm/m If there is no normal force acting in the wall this would require a central reinforcement of about 0.1 percent. The corresponding bending capacity can be estimated as: m
= 0.4 ω d
= 0.4 0.001 0.2
300.000 = 5 kNm/m Normally, of course, normal forces are present. As detailed calculations are out of the scope of this do-
cument, the following scheme looks realistic. If the explosion is in a top floor apartment and there is an adequate connection between roof slab and top wall, we will have a tensile force in the wall, requiring some additional reinforcement. In our example the tensile force would be (p
) B/2 = (6.5-2x3) * 4 = 2 kN/m for a centre column and (p
) B/2 = (6.5-3) * 4 = 14 kN/m for an edge column. If the ex-
plosion is on the one but top story, we usually have no resulting axial force and the above mentioned reinforcement will do. Going further down, there will probably be a resulting axial compression force and the reinforcement could be diminished or even left out completely. 4 CLOSURE The Eurocode system has been enlarged by an im-
portant document on accidental actions. It will be in-
teresting to see how the document will be used in the various member states and how the various national annexes will look like. On almost no other subject the differences between the design rules in the vari-
ous European member states were so large. This do-
cument should for sure be seen as a first step only. Both from a theoretical and a practical point of view improvements need to be made. REFERENCES
CIB .W81 (1993), "Accidental Actions", Publication 167, CIB, Rotterdam Chiapella, R.L., Costello, J.F. (1981) "Automobile Impact Forces on Concrete Walls", Transactions of the 6th International Conference on Structural Me-
chanics in Reactor Technology, Vol. J(b), Paris. Dragosavic, M. (1972), "Research on Gas Explo-
sions in Buildings in the Netherlands", Notes Reference No. 354/72, Symposium on Buildings and the Hazards of Explosion, Building Research Station, Garston, England. Dragosavic, M.,(1973) "Structural Measures Against Natural-Gas Explosion in High-Rise Blocks of Flats", Heron, Vol. 19, No. 4 (1973). Harris, R.J., Marshall, M.R., Moppett, D.J., (1977) "The Response of Glass Windows to Explosion Pressures", British Gas Corporation. I.Chem.E. Symposium Serious No. 49 Institution of Structural Engineers "Safety in tall buildings and other buildings with large occu-
pancy". London, UK. Leyendecker, E.V. and Ellingwood, B.R., (1977), "Design Methods for Reducing the Risk of Pro-
gressive Collapse in Buildings"; US National Bu-
reau of Standards, Washington Mainstone, R.J., Nicholson, H.G., Alexander, S.J.,(1978), "Structural Damage in Buildings caused by Gaseous Explosions and Other Acci-
dental Loadings, 1971 - 1977", Building Research Establishment, England. NFPA (1988), "Venting of Deflagrations", National Fire Protection Association Popp, dr.-ing. Camillo (1961) "Der Querstoss beim Aufprall" (in German)", Forschungshefte aus dem Gebiete des Stahlbaues, Deutschen Stahlbau Ver-
band, Köln am Rhein. Project Team ENV 1991-2-7, (2005) (under prepara-
tion), "Background document to ENV 1991-2-7 Accidental actions due to impact and explosions" Stufib Studiecel (2004), "Incasseringsvermogen van bouwconstructies" (Robustness of building struc-
tures, in Dutch), Delft. 3317
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‚Ä˜_‚¶‚Ä‚¢‚é•B •¡Œã‚Í•AŠe•‘‚ª‚»‚ê‚¼‚ê‚Ì Nat ional Annex ‚Å•A‚±‚±‚ÅŒˆ‚ß‚ç‚ê‚½‚±‚Æ‚ð‚Ç‚Ì‚æ‚¤‚ÉŽæ‚èˆµ
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Figure 1: Example of effective horizontal tying of a framed office building. -class 3: extensive study of accident scenarios and using dynamic analyses and non-linear analyses if appropriate. upper group (as originally prepared by the Dutch Studiecel Stufip (2004)). In that case. in the following manner: -class 1: no specific consideration of accidental actions. e. -class 2: a simplified analysis by static equivalent load models for identified accidental loads and/or by applying prescriptive design and detailing rules. according to Annex A5.6 m.1 Design example As a design example consider a framed structure.
internal ties perimeter tie L
The design for unidentified accidental load is presented in Annex A of EN1991-1-7. in addition one of the following measures should be taken: • Introduce vertical ties. Although the rules were not intended to safeguard buildings against terrorist attack. grand stands etc. Eurocode 1991 Part 1. 2 UNIDENTIFIED ACCIDENTAL LOADS
strated during the IRA bomb attacks that occurred in the City of London in 1992 and 1993. in fact. ISBN 90 5966 040 4
.0. Rotterdam. the damage sustained by those buildings close to the location of the explosions that were designed to meet the regulatory requirement relating to disproportionate collapse was found to be far less compared with other buildings that were subjected to a similar level of abuse. For the upper class 2. A convenient measure to decide what structures are to be designed for accidental situations is to arrange structures or structural components in categories according to the consequences of an accident. Annex A recommends horizontal ties in each floor around the perimeter and internally in two right angle directions between the columns (Figure 1). The capacity of the ties is a function of the self-weight.
2. Let the span be L = 7.larger role in design for accidental actions than in the case of variable actions. only specifies operational guidance for consequences class 2. damage to the environment or economic losses for the society. For framed structures in the lower group of consequences class 2. wall section or beam the damage does not exceed 15% of the floor in each of 2 adjacent storeys. consequences class 2. upper group 4
low rise buildings buildings up to 4 stories buildings up to 15 stories high rise building.g.
Not only the appropriate measures but also the appropriate method of analysis may depend on the safety category. the required internal tie force may be calculated from: Ti = 0. • Design key elements for an accidental design action Ad = 34 kN/m2. the imposed load and the geometrical properties. 5 storeys with story height h = 3.2 m and the span distance s = 6 m. lower group 3. Annex A. The rules have proved satisfactory over the past 3 decades. Design for accidental design situations needs to be primarily included for structures for which a collapse may cause particularly large consequences in terms of injury to humans.8 {gk +Ψ qk } sL (1)
© 2005 Millpress. The rules have changed little over the intervening years. The characteristic values for the self weight and floor loads are gk = qk = 4 kN/m2 respectively and the combination factor Ψ = 1.7 arranges structures in the following categories based on consequences of a failure:
class consequences Limited Medium Medium Large Example structures
1 2. A distinction is made between framed structures and load-bearing wall construction. Rules of this type were developed from the UK Codes of Practice and regulatory requirements introduced in the early 70s following the partial collapse of a block of flats at Ronan Point in east London caused by a gas explosion. • Ensure that upon the notional removal of a supporting column. They aim to provide a minimum level of building robustness as a means of safeguarding buildings against a disproportionate extent of collapse following local damage being sustained from an accidental event (Institution of Structural Engineers). It is up to the European member states to decide what is considered as an appropriate strategy in the various cases.
7 deals with impact from vehicles. For the vertical tying force we find in a similar way Tv = (4 + 4) (6 x 7.0 and 1. In the following table some corresponding input values for the parameters m1. trains. The formula (2) gives the maximum force value on the outer surface of the structure. is neglected in Eurocode EN 1991-1-7 and the analysis is confined to the elementary case. This.2) = 276 kN For Steel quality FeB 500 this force corresponds to a steel area A = 550 mm2 or 2 ø 18 mm. single degree of freedom system (see Figure 2). are possible.
Proceedings ICOSSAR 2005. for instance by assuming that the structure is rigid and immovable and the colliding object can be modelled as a quasi elastic. Safety and Reliability of Engineering Systems and Structures
. Partial load factors to be applied in accidental design situation are defined to be 1. impact is still an interaction phenomenon between the object and the structure. In most cases the colliding object will respond by a mix of elastic deformations. Part 1. however. Part 3. The load deformation characteristic may. are defined in EN 1990 Basis of Design. yielding and buckling. Amplification factors between 1. 3 IDENTIFIED ACCIDENTAL LOADS
course. ko and ν of equation (2) are presented. where the colliding object hits a structural element under a right angle. one should consider object and structure as one integrated system. by allowing wind or wave loads for shorter return periods to be applied in the analysis after an accidental event. In temporary phases there may be reasons for a relaxation of the requirements e. fire or impact) or refer to a situation after an accidental event (A = 0).g. ships. In that case the maximum resulting interaction force equals (CIB. variable and accidental). to be applied in design situations for identified accidental actions. Chapter 4 of EC1-Part 1. Table 4. Note that in continuous beams this amount of reinforcement usually is already present as upper reinforcement anyway. Inside the structure these forces may give rise to dynamic effects. In practice the colliding object will not behave elastically.2) = 350 kN/column This correspondents to A = 700 mm2 or 3 ø 18 mm. Also later (unpublished) tests initiated by the British High Way Agency gave results in the same order. 3. still have the nature of a monotonic increasing function (see Figure 3). As a result one may still use equation (2) to obtain useful approximations. the model (2) has been successfully compared with experiments by Popp (1961) and Chiapetta and Pang (1981).3.7. In this paper we will only discuss the impact from vehicles. In the background documentation. have for political reasons been chosen in accordance with Eurocode 1. however. The initial kinetic energy of the colliding object can be transferred into many other forms of kinetic energy and into elastic-plastic deformation or fracture of the structural elements in both the building structure and the colliding object. The perimeter tie is simply half the value. Even then. 1993): F = vr √(km) (2)
The general principles and combination rules.g. Small differences in the impact location and impact angle may cause substantial changes in the effects of the impact. After an accidental event the structure will normally not have the required strength in persistent and transient design situations and will have to be strengthened for a possible continued application.1 Impact from vehicles The mechanics of a collision may be rather complex.
Figure 2 Spring and rod models for the colliding objects
Figure 3: General load displacement diagram of colliding object
The design values in EC1.0 for all loads (permanent. Approximations. Combinations for accidental design situations either involve an explicit accidental action A (e.4 are considered as a more or less realistic. of
where vr is the object velocity at impact.Leading to: Ti = 0. k the equivalent stiffness of the object and m its mass.8 {4+4} (6 x 7. fork lift trucks and helicopters. To find the forces at the interface.
01 for all four groups of bars as indicated in Figure 4. if combined with a conservative linear classic static structural model. ISBN 90 5966 040 4
.3 1.8 ω h2 b fy = = 0.00 0.50 300 000 = 1200 kNm MRdy = 0. Gas is widely used and. So we may confine ourselves to the accidental load only.01 1.0 − 1. 3.8 0. the forces Fdx and Fdy should be taken as 1000 kN and 500 kN respectively and act at a height of a = 1.
1. The follow-
Figure 4 Bridge pier under impact loading
According to the code. right hand side.8 ω h b2 fy = = 0. The reinforcement ratio is 0. the bending moment capacity can conservatively be estimated from: MRdx = 0.00 − 1. The cross sectional dimensions are b = 0.25 m.00 H
Similar for the direction perpendicular to the driving direction:
m [kg] 20000
v [km/h] 40
equivalent collision force stiffness based on (2) k F [kN/m] [kN] 300 1000
Mdx=
1. Rotterdam. Part 1.3 bh fctk = 0.2 Design example for vehicle impact Consider by way of example the reinforced concrete bridge pier of Figure 4.00 H
Other loads are not relevant in this case.7 is low.3 Explosion loads Gas explosions account for by far the majority of accidental explosions in buildings.25 H −a 500 = 375 kN Fdy = Qxy = 5. The self weight of the bridge deck and the traffic loads on the bridge only lead to a normal force in the column.50 1200 = 360 kN. 0 5. Explosion pressures propagate as pressure waves. Leyendecker and Ellingwood (1977)). 3.8 0.25( 5.
H Fdy a
This is almost sufficient for the loading in ydirection. the bending moment capacities can be considered as sufficient. Let the steel yield stress be equal to 300 MPa and the concrete strength 50 MPa. As a consequence also the Fd value in Eurocode 1.00 m.25( 5.00 0. 0 5. the incidents of occurrence of gas explosions in buildings is an order of magnitude higher than other accidental loads causing medium or severe damage (Mainstone (1978). based on the concrete tensile part only is approximately equal to (say fctk = 1200 kN/m2): QRd = 0. In this context an explosion is defined as a rapid chemical reaction of dust or gas in air. but not for the x-direction.also trucks 20000
The input design values for masses and velocity are relatively low values. An additional shear force reinforcement is necessary.002 0. The shear capacity of the column.502 300 000 = 600 kNm
As no partial factor on the resistance need to be used in the case of accidental loading.25 H −a 1000 = 750 kN Fdx = Qdx = 5.01 1.50 m and h = 1.7. excluding vehicular impact. 1. It results in high temperatures and high overpressure. The column will be checked for impact by a truck under motorway conditions. the overall design could still very well be over designed. The design value of the bending moments and shear forces resulting from the static force in longitudinal direction can be calculated as follows:
© 2005 Millpress.25 ) a( H − a ) 500=470 kNm Fdy = H 5. However.25 ) a( H − a ) 1000=940 kNm Fd = H 5. The column height h = 5 m and is assumed to be hinged to both the bridge deck and to the foundation structure. Using a simplified model.only cars 1500 .Table 1: Calculation of design value of EC1. Normally this will increase the load bearing capacity of the column.00 − 1.0 − 1.
windows and other openings. Let the floor dimensions of the compartment be 8 x14 m and let the height be 3 m. Safety and Reliability of Engineering Systems and Structures
. The load arrangement for the explosion pressures is presented in Figure 5.5 kN/m2 This means that we have to deal with the latter. elements of a structure should be designed to withstand the effects of an internal natural gas explosion.an ignition source strong enough to initiate combustion. In practice. where pv is the uniformly distributed static pressure in kN/m2 at which venting components will fail. This means that the volume V and the area of venting components Av for this case are given by: Av = 2 x 8 x 3 = 48 m2 V = 3 x 8 x 14 = 336m3 So the parameter Av / V can be calculated as: Av / V = 48 / 336 = 0.4 Design example for explosion loading Consider by way of example a living compartment in a multistory flat building. using a nominal equivalent static pressure given by (Dragosavic (1972. 3.2 s seems to be a reasonable approximation. these pressures have to be combined with the self weight of the structure and the quasi-permanent values of the variables loads.7.5+2. depending on type of gas or dust. Av is the area of venting components and V is the volume of room. The collapse pressure of the venting panels pv is estimated as 3 kN/m2. In completely closed rooms with infinitely strong walls. The two long walls are concrete walls.144 m-1 As V is less then 1000 m3 and Av / V is well within the limits of 0. in the proper concentration. There is then the difficulty of how to assess the probable response of the structure to loading which is short term. The equivalent static pressure for the internal natural gas explosion is given by: pEd = 3 + pv = 3 + 3 = 6 kN/m2 or pEd = 3 + pv/2+0. . dynamic and omnidirectional.
whichever is the greater.3. The pressure generated by an internal explosion depends primarily on the type of gas or dust. Leyendecker and Ellingwood (1977)): or pd= 3 + pv pd = 3 + 0. Part 1.2 s. the concentration of gas or dust in the air and the uniformity of gas or dust air mixture. in sufficient quantity to support the combustion. the size and shape of the enclosure in which the explosion occurs.05 m-1 < Av / V < 0.ing are necessary for an explosion to occur (NFPA (1988)): . According to Annex D of EN1991. The conditions following an explosion. but the duration is shorter. Guidance can be obtained in Harris at all (1977).5 pv+0.fuel. It states that the peak pressures in the main text may be considered as having a load duration of 0.0 = 6. these walls are responsible for carrying down the vertical loads as well as the lateral stability of the structure.
Proceedings ICOSSAR 2005. According to Eurocode EN 1990. The venting pressure pv may be determined from tests or may be calculated from uniformly loaded plate bending formulae. dust explosions up to 1000 kN/m2. Let us consider the design consequences for the various structural elements. The expressions are valid for rooms up to a volume of 1000 m3 and venting areas over volume ratios of 0. An important issue is further raised in the note ot clause 5.05 m-1 and 0. . gas explosions may lead to pressures up to 1500 kN/m2.04/0. 1973).04/(Av/V)2 = 3+1. So combining the loads from the above equations with a duration of 0. 15 m-1.15 m-1 it is allowed to use the loads given in the code. and the reaction of the various elements of the structure to these conditions.3.an oxidant. The point is that in reality the peak will generally be larger. The two small walls (the facades) are made of glass and other light materials and can be considered as venting area. The explosive pressure acts effectively simultaneously on all of the bounding surfaces of the room. Note that these panels normally can resist the design wind load of 1. Basis of Design.5+0. are obviously extremely complex.1442 = 3+1.5 kN/m2. These walls have no load bearing function in the structure.04/(Av/V)2 (3) (4)
Deciding on a design pressure is only part of the difficulty for a designer. pressures generated are much lower due to imperfect mixing and the venting which occurs due to failure of doors. and the amount of venting of pressure release that may be available.
50 g ( ∆t ) 2
ϕd = 1 +
p SW p Rd
2u max g ( ∆t ) 2
Using again pSW = 3 kN/m2.85 * 1. g = 10 m/s2 is the acceleration of the gravity field and umax is the design value for the midspan deflection at collapse. So for comparison we could increase the design load for normal conditions by a factor of about 1.50 kN/m2 So the load is only half the load on the bottom floor.2. It is now time to remember the clause 5.20 ]*7.7.35 * 3. depends on the ductility properties of the floor slab and in particular of the connections to the rest of the structure. Note by the way that the upper floor for one explosion could be the bottom floor for the next one. This means that the design load for the explosion is given by: pda = pSW + pE + ψ1LL pLL = = 3. If we take into account the short duration of the load we may increase the load bearing capacity by a factor ϕd given by (see EC1-Part1. but it may also lead to free-
© 2005 Millpress. Rotterdam.3.5kN/m2 10 (0.50 + 0 = 3. An important additional point to consider is the reaction force at the support. What is needed is top reinforcement in the field and bottom reinforcement above the supports.2*6. The required resistance can be found by solving pRd from:
3.3. This value. especially in the upper two stories of the building where the normal forces in the walls are small.4 = 7.2 s is the load duration.2) 2
So the bottom floor system is okay in this case. The result could be conceived as the resistance of the structure against accidental loads.5 kN/m2.7
2 * 0. In that case the resistance against explosion loading can be assessed as:
pREd=ϕdpRd=[1+
3 7 .2 Let us next consider the upper floor.7 kN/m2 So a floor designed for normal conditions only should be about 30 percent too light. This means that the resistance against the explosion may simply be close to zero.50 kN/m2 The design for normal conditions is given by: pd = γG ξ pSW + γQ pLL = 0.50 + 0.5*2. In this respect edge walls are even more vulnerable.2 s.1 Bottom floor Let the self weight be 3 kN/m2 and the characteristic live load 2 kN/m2. The uplifting may change the static system for one thing and lead to different load effects. but will nevertheless give rise to larger problems. if it designed for normal loads only:
ϕd pRd = [1 +
pRd = 1.H=3m
where ∆t = 0.4. of course. but assume that umax = 0.5*2. The point is that the explosion load is in the opposite direction of the normal dead and live load.4 kN/m2 We should keep in mind that for accidental actions there is no need to use a partial factor on the resistance side. ISBN 90 5966 040 4
.00 = 6.7=12. g=10 m/s2 we arrive at pRd = 1.4. This would require about 25 percent of the reinforcement for normal conditions on the opposite side.00 = 10.3 mentioned earlier. Background Document (2005)):
2u max ] pRd = 3. The design load for the explosion in that case is given by (upward value positive!): pda = pSW + pE + γQ ψ pLL = . ∆t = 0. Upper floor 3.00 + 1.20 m is considered as being a defendable design value. It is beyond the scope of this paper to discuss the details of that assessment.00 + 6.00 + 6. Note that the floor could be lifted from its supports.
Nicholson. "Background document to ENV 1991-2-7 Accidental actions due to impact and explosions" Stufib Studiecel (2004). 1971 . Notes Reference No. National Fire Protection Association Popp. 354/72. The corresponding bending capacity can be estimated as: mp= 0. Dragosavic..L. Garston. S. "Design Methods for Reducing the Risk of Progressive Collapse in Buildings". Both from a theoretical and a practical point of view improvements need to be made. If the explosion is on the one but top story.G. It will be interesting to see how the document will be used in the various member states and how the various national annexes will look like. "Incasseringsvermogen van bouwconstructies" (Robustness of building structures. requiring some additional reinforcement. we usually have no resulting axial force and the above mentioned reinforcement will do.. Moppett. B.(1973) "Structural Measures Against Natural-Gas Explosion in High-Rise Blocks of Flats". (1972). We will come back to that in the next section.. In our example the tensile force would be (pE – 2 pSW) B/2 = (6.. 49 Institution of Structural Engineers "Safety in tall buildings and other buildings with large occupancy".4 0. Paris. Building Research Establishment. Project Team ENV 1991-2-7.22 300. R.5-2x3) * 4 = 2 kN/m for a centre column and (pE – pSW) B/2 = (6.4 ω d2 fy = = 0. 4 (1973). Assume the wall to be clamped in on both sides.J. "Accidental Actions". England. Transactions of the 6th International Conference on Structural Mechanics in Reactor Technology.
The Eurocode system has been enlarged by an important document on accidental actions. Symposium on Buildings and the Hazards of Explosion. Publication 167.F.R.. 19. M.1977". Deutschen Stahlbau Verband. M. I.. British Gas Corporation. D. J(b). Harris. and Ellingwood. Rotterdam Chiapella.1 percent.J.R. Costello. Vol. Going further down. Leyendecker. 4 CLOSURE
REFERENCES CIB . (1981) "Automobile Impact Forces on Concrete Walls". normal forces are present. in Dutch). Dragosavic.3 Finally we have to consider the walls. As detailed calculations are out of the scope of this document.000 = 5 kNm/m Normally. Alexander.. (2005) (under preparation). If the explosion is in a top floor apartment and there is an adequate connection between roof slab and top wall. The bending moment in the wall is then given by: m = 1/16 p H2 = (1/16) 6. "Venting of Deflagrations". If the floor to wall connection can resist the lift force. Vol.W81 (1993). dr.
Proceedings ICOSSAR 2005. Safety and Reliability of Engineering Systems and Structures
. of course. Marshall. "Structural Damage in Buildings caused by Gaseous Explosions and Other Accidental Loadings. we will have a tensile force in the wall. R. R. Köln am Rhein. On almost no other subject the differences between the design rules in the various European member states were so large. M. Camillo (1961) "Der Querstoss beim Aufprall" (in German)". Heron..standing walls. the following scheme looks realistic.(1978). Building Research Station.5-3) * 4 = 14 kN/m for an edge column. Delft.5 32 = 4 kNm/m If there is no normal force acting in the wall this would require a central reinforcement of about 0. England.. "Research on Gas Explosions in Buildings in the Netherlands". H.001 0. J. Washington Mainstone.V. London. Symposium Serious No. No. Walls 3. E.J. (1977) "The Response of Glass Windows to Explosion Pressures". This document should for sure be seen as a first step only. there will probably be a resulting axial compression force and the reinforcement could be diminished or even left out completely. one should make sure that also the wall itself is designed for it. US National Bureau of Standards. CIB. Forschungshefte aus dem Gebiete des Stahlbaues.E. (1977). UK. NFPA (1988).-ing.4.Chem.J.
7 は特に衝突と爆発が主な対象である。さらにこのコードでは、衝撃 作用を、特定されたもの(identified)と、特定されないもの(unidentified)に分類してい る。前者に対しては、伝統的な信頼性解析の枠組が、後者に対しては、構造物の強靭 性(robustness)などの要求が考慮される。 対策としては、リスクの低減（例、避雷針の設置）と被害の低減（例、スプリンクラ ーの設置）があるが、すべての偶発作用に対処する方法はない。肝要な点は、作用に 対して不釣合いに大きな被害を招かないようにするということである。従って構造物 の冗長性や非線形性が大きな役割を演じる。 このコードでは、結果の重大性に基づいて、構造物に範疇を設けている。 クラス１ 特に考慮しない。 クラス２ クラス３ クラス４ 中間的影響 中間的影響 重大な影響 ４階までの建物 15 回までの建物 超高層建物、競技施設 静的平衡に基づい 偶発作用について 同上 た検討、慣用的詳細設計の利用など。 も十分に検討する。動的解析、非線形解析の利用も行う。 限定的影響 低層建物 偶発作用の影響は
特定されない(unidentified)偶発作用 この部分は付録 A に記述されており、1970 年代から開発されてきたイギリスのコードを参 考にしている。 これはロンドン東部の Ronan Point における爆発事件に源を持つコードで、 その結果は、 1992 年 1993 年の IRA の爆弾事件に対して効果を示した。 基本的な考え方は、
.7.3311-3317 Eurocode 1.7. pp. Part 1. Accidental Actions By Ton Vrouwenvelder ICOSSAR 2005 in Rome. Part 1. Accidental Actions が最近最終決定し、現在仏訳よ及び独訳が行われ ている状態である。この偶発作用に関する設計コードについて、その概要を紹介する。 偶発作用は、低い生起確率（質問に対し、構造物の供用期間中に 10-4 以下の生起確率 を目安とするという答えだった。 、その影響が大きく、また持続時間が短い作用とい ） うのが、一般的な定義である。 火災、爆発、地震、衝突、洪水、雪崩、地すべりなどが、含まれる。さらに原因が明 確ではない部材の劣化、ヒューマンエラー、不適切な使用、強い作用への曝露、機器 の故障、テロ攻撃なども考えられる。 この中で、Part1.文献紹介 Eurocode 1.
0 である。 このとき、 永続おより変動荷重にかけ る部分係数も 1.0 である。 Part 1.7 では、自動車、船舶、列車、フォークリフト、ヘリコプターによる衝撃荷重につい て論じている。
今後は、各国がそれぞれの National Annex で、ここで決められたことをどのように取り扱 ってゆくかが興味深い。
.構造物の一体性を増すことにより、爆発に対して不釣合いな建物の倒壊を防止するもので ある。 特定された(identified)偶発作用 特定された偶発作用に乗じる荷重係数は 1.
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