Loads must be combined dependently in structural design

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Alexina Davis
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1 /(6)/tp Loads must be combined dependently in structural design Introduction A common misconception is that the permanent and the variable load are independent. However, these loads are independent only during one year and almost fully dependent during the normal service time of structures, 50 years. Another misconception is that structural loads are combined dependently if the loads are dependent and independently if the loads are independent. However, loads must always be combined dependently. Extreme distribution In the structural design the load distributions denote the extreme values of the loads during a selected reference time. Consequently, the load combination must also consist of the extreme values of the combination loads. When the loads are combined independently, the load distribution is the sum of extremes but it should be the extreme of the extremes. Linearity, Hook s law Hook s law directs one basic rule in the structural design, linearity: when a load has a certain effect at one load level, the same effect occurs by the same load at any other load level. The independent load combination is in contrast with the Hook s law and the linearity: One realization of the independent load combination is the combination rule 6.0,mod of Eurocode. In this load combination with low permanent load when load increases no effect occurs. The independent combination theory actually contradicts the linearity and the Hook s law and includes a load vanish: Assume the target reliability is 0.98, the loads G and Q have normal distribution, and act in a tension bar, A = 000 mm 2. We see in Table and the load case 5 with the independent combination that ca 0 % load vanishes in the combination; the linearity and the Hook s law are not valid. Table. Five load cases of the permanent G and the variable Q load. The load case 5 shows that the independent combination results in a load vanish of ca 0 %, the linearity and the Hook s law are not valid. load case load [kn] stress [N/mm 2 ] G Q depend. independ

2 /2(6)/tp Equality design equation The basic design equation of the structural design with the permanent load G, the variable load Q and the material property M is G G Q Q M M The equality equation makes definite dependence and correlation between the distributions G, Q as the material property M is constant. G and Q have a full negative correlation when one load is increased the other decreases correspondingly. Due to this correlation the loads G and Q must be combined dependently though the loads were independent. Loads act independently We may deduce the independent and the semi-dependent combination erroneous by an example: When a new load is added to existing loads, the new load has a specific action regardless of the existing loads. This is possible only if the new load is combined dependently i.e. by the accumulation. In the semi-dependent and the independent combination the existing loads affect the combination. This is impossible. There is no link between the existing loads and the new load which would make this effect possible. Numerical example of Eurocode This calculation is a numerical demonstration that the permanent and the variable load are dependent. The permanent load G and the variable load Q are combined here in a load ratio = 0.2, = Q / G + Q. The distributions correspond to Eurocode. The design point is set at unity. Parameters are listed at Table below. The loads are combined in the serviceability state, G = Q =. A similar calculation with equal results can be made in the failure state G =.35, Q =.5. permanent load, G variable load, Q load safety factor, coefficient of variation, V design point value, dp distribution normal gumbel mean, deviation, The loads can be combined independently by using the convolution equation below, =. FGQ( x) FGx r G G fq r Q Q dr In this equation, F denotes the cumulative function and f the density function respectively. This combination results in an equal outcome as the Ferry Borges Castanheta s combination. The dependent combination is obtained from the same equation with = The - factor fits the combination distribution to cross the crossing point (.243, ) of the partial distributions G and Q. This value corresponds to G =.35, V M = 0.

3 /3(6)/tp The Figures below show the distributions: G GQ i 0.6 GQ d Q 0.8G Q GQ i G solid black line, permanent load G solid red line, variable load Q solid brown line, dependent combination load GQ d dotted brown line, independent combination load GQ i dashed black line, 0.8G dashed red line, 0.2Q blue dash-dot line, fractile 0.98 We study the load values in the fractile 0.98, i.e. in a variable load return period 50 years and in a normal service time of a structure. The load values in this fractile are tabulated at Table below: Q GQ d G 0.2Q 0.98 Q G 0.98 GQ i,0.98 GQ d,0.98 G As explained above the permanent load value and the variable load value are simultaneous in each fractile when the time is long. Accordingly the loads G 0.98 and Q 0.98 plus 0.8G 0.98 and 0.2Q 0.98 occur at the same time and the loads are combined dependently by adding up the values as such. The combination distribution has the same fractile as the partial ones. We see at Table, that 0.2. G Q is equal to the dependent combination. Common sense reasoning The permanent load G and the variable load Q can be deduced dependent by common sense reasoning: Assume a material (or a structure) has the survival probability S and the resistance for the permanent load G alone and the variable load Q load alone. Now, if the material is loaded by 0.5G and 0.5Q and the loads are combined independently, the material has the resistance of ca. 4 and if combined dependently, the resistance is. The independent combination is impossible. There is no link between the loads which would affect one load if the other were present. This is shown numerically below in the demonstration. 2 This value applies the load ratio = This value has been calculated from the convolution equation. This equation is approximate when the distributions are normal and gumbel. The quantile sum equation is correct and results in.503, i.e. 0.7 % less value. 4 in Eurocode more precisely.0646, V G,normal = , V Q,gumbel = 0.4

4 /4(6)/tp Numerical demonstration Assume a material has the coefficient of variation V M = In Eurocode G =.35, G =.5, this material has the material safety factor MG =.2309 when it is loaded by the permanent load G and MQ =.078 when loaded by the variable load Q. We find that MG / MQ = Q / G, i.e. a structural component of this material, e.g. a tension bar, has equal resistance in permanent and in variable load. Assume, for instance, the material has characteristic material property x k = and the bar has a cross section area, the bar resists a permanent and a variable load Assume the same bar is a tension member of a truss and it is loaded by 80 % permanent and 20 % variable load, = 0.2. If the loads are combined independently, the material safety factor is.0964 and the bar resists a load i.e. ca 0 % higher load than G or Q alone. The independent combination cuts down a part of the load. Therefore the strength is increased. The reason for the error is that the loads must be combined by accumulation, not by random. When the loads are combined dependently, the bar resists load in all load combinations, = 0..., of G and Q. Numerical demonstration 2 A permanent load G and a variable load Q of Eurocode are combined here. To simplify calculations, we assume that the distribution of the variable load is normal (gumbel in Eurocode) and the design point value of the permanent load is 0.98 (0.5 in Eurocode). The design point is set at unity. The coefficients of variation are V G = and V G = 0.4, i.e. the distribution parameters are G = 0.842, G = 0.077, Q = 0.549, Q = The loads are combined in the serviceability state, i.e., G = Q =. We claim that the design must be fulfilled by 98 % probability. We select the loads to be, i.e. the loads are at the design point and at the validation cut-off point, The loads are combined in a load ratio = 50 %, i.e. the loads are combined 50 % + 50 %. In Eurocode, permanent loads are combined dependently (both loads have the highest value in all fractiles). The combination load distribution is calculated (in this case) by a normal distribution N: N x G NGQ d x 2 Q G Q 2 2 Q G The independent load combination is applied in Eurocode (rules 6.0a,b, 6.0a,mod) to define the safety factors G, Q, M (both loads have a random value, Ferry Borges - Castanheta s model). The combination load distribution is calculated by: N x G NGQ i x 2 Q G Q 2 The figures below show the load distributions, the horizontal axis denotes the load, the vertical axis the fractile, permanent load: circled line; variable load: boxed line; dependent combination: solid line; independent combination: dashed line; target reliability, 0.98: horizontal solid line The combination load value at the validation cut-off point, i.e. in 0.98 fractile, shown by the horizontal solid line, is the key issue. One would

5 /5(6)/tp be a common sense value ( = ). This is true, as the permanent load prevails one during a whole time and the variable load hits the 0.98-fractile, i.e. the unity load, at least once (by a high probability, this topic is discussed later) during the service time, 50 years. The full loads, and their halves, are simultaneous and added up as such. If the loads are combined independently, the combination load is In this model, the load combination lessens the load, <, which is wrong and results in an unsafe design. In this combination, the G and Q loads occur simultaneously in fractile 0.98 by a negligible probability. Thus, the loads must be combined dependently. Summary Loads must always be combined dependently in structural design. In current codes the loads are sometimes combined independently or semi-dependently, which results in an unsafe error in comparison with the target reliability: The combination rules including two permanent load factors e.g. the rules 6.0a,b and 6.0a,mod of Eurocode are induced from the independent combination and these combination rules should be deleted. The total safety factor, t = ( G, Q, M ) is up to 0 % too low, if the permanent and the variable load is combined independently. Two variable loads are currently combined semi-dependently, which results in up to ca 30 % too low combination factors 5 0. The combination factor 0 for snow is currently but should be as the current snow load distribution is equal to its active time distribution. The current imposed load factor is 0.7 but should be. The reason is that the imposed loads on floors are dependent and proportions of the whole live load of the house 6. The combination factor 0 of variable loads with short duration, e.g. live load and wind load is ca. 0.6, i.e. the present values are correct. The unsafe error of the reliability model due to the independent and the semi-dependent load combination is up to ca 5 % in two loads and up to ca 20 % in three loads. The unsafe error of the actual code is less, up to ca 5 %, as the safety factors are normally selected higher than obtained from the independent and semi-dependant calculation. Tampere Tuomo Poutanen Docent, D.Sc. (Civ. Eng.) Tampere University of Technology Department of Civil Engineering P.O. Box 600, FI-330 Tampere mobile: fax: tuomo.poutanen@tut.fi 5 Two variable loads are combined dependently when the distributions are altered in a way the loads are simultaneous with one on the other. 6 The characteristic live loads in current codes are high enough to compensate the 0 error.

Competitive comparison of load combination models

Czech Technical University in Prague, Klokner Institute Competitive comparison of load combination models Milan Holicky and Miroslav Sykora Czech Technical University in Prague, Klokner Institute Introduction