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Dead loads are typically loads which the program can calculate
by itself based ongeometry (length, width, thickness) and material
properties specified (density). Steelframing member weights are
automatically calculated by the program.
Any dead load which cannot be calculated by the program need to
be input as aSUPERDEAD
. Examples of this could be slab finishes (marble, tile),
masonry loads etc.Live load reduction is typically permitted for
columns supporting multiple floors, asrecommended by ASCE . !he
reduction of li"e load is based on probabilities. Assumethat you
ha"e a #$ story building, with a column supporting a tributary area
of %&'x%&' oneach floor. !he chances that each one of these
floors has a li"e load at the same time isunlikely and the code
allows you to reduce the load based on its nfluence Area .*reater
the influence area, the more reduction you are allowed to take, but
no more than a+& reduction on li"e loads. -inimum o / &.0 .
Some restrictions apply, checkChapter 0 of ASCE (la noi se aplica
pre"ederile 12&&3#&&+)
Dead loads are weights of material, equipment or components that
are relativelyconstant throughout the structure's life. Permanent
loads are a wider category whichincludes dead loads but also
includes forces set up by irreversible changes in a
structure's constraints - for example, loads due to settlement,
the secondary effects of prestress or due to shrinkage and creep in
concrete.
Also, 4ead oads are not limited to walls, floors, roofs,
ceilings, stairways, built5in partitions, finishes, cladding and
other similarly incorporated architectural and structuralitems, and
fixed ser"ices e6uipment, including the weight of cranes. All dead
loads areconsidered permanent loads.
i"e loads, sometimes referred to as probablistic load include
all the forces that are"ariable within the ob7ect's normal
operation cycle. 8sing the staircase example the li"eload would be
considered to be 5
1ressure of feet on the stair treads ("ariable depending on
usage and si9e):ind load (if the staircase happens to be
outside)i"e loads do not include construction or en"ironmental
loads such as;
:ind loadSnow load#?i"e loads (
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!he reason for splitting loads into these categories is not
always apparent, and in terms ofthe actual load on the ob7ect there
is no difference between dead or li"e loading. =or themost part,
the split occurs for use in safety calculations or ease of analysis
on complexmodels.
:hen considering the feasibility of a structure, safety always
takes precedent and because of this, go"erning bodies around the
world ha"e regulations to which structuresha"e to adhere. 8sing the
example of the staircase, if it was intended for use in the 8
itwould ha"e to follow British and European Standards
BS 0$ # 5 ndustrial type flooring and stair treadsBS $% $ 5 Code
of practice for the design of straight stairsDther standards
specific to the application (e.g. BS 202##5%;#&&2 5
1ermanent means ofaccess to machinery. Stairways, stepladders and
guard5rails):ithin these standards a safety factor is usually
determined where the structure should beable to withstand a certain
force abo"e the maximum expected load. Dnce again using the
staircase example, assuming it is an indoor medium5usage
industrial staircase the currentsafety factor would be 2.0 times
the maximum stress imposed by the dead load and 2.+times the
maximum stress imposed by the li"e load. !he reason for the
disparity between"alues, and thus the reason the loads are
initially categorised as dead or li"e is becausewhile it is not
unreasonable to expect a large number of people ascending the
staircase atonce (or the wind speed increasing, snowfall or any
other li"e load increase), it is lesslikely that the structure will
experience much change in its permanent load. !he same can
be said of many structures and so it is con"enient to assess
loading based on itsapplication.
Calculating combined loadst is worth noting that on first
inspection it seems you should find the maximum stress foreach of
dead and li"e, factor them and add them together. !his will gi"e
you a massi"elyo"erestimated stress result. !he combination needs
to be applied with great care andalmost exclusi"ely
programmatically because you may only combine two stress results
atthe same point. Since the maximum stress is "ery rarely at the
same place in a structurefor dead and li"e it may well be the case
that the o"erall increase is a fraction of theaddition of the two
maximum stresses and in a completely different position two either
ofthe two original maximums. !o clarify, take the staircase
analysis. !he maximum stressunder dead load appears at the foot of
a support beam and it is & Fmm5#, at this pointthe stress from
the li"e load is $ Fmm5#. !he maximum stress under li"e load is
+&
Fmm5# and appears at the corner of the second stairtread where
the dead load stress is %& Fmm5#. At a third point the stresses
from both dead and li"e are $& Fmm5#. *i"en thesefigures you
can see that the combined load cases for each point would be;
2. & x 2.0 G $ x 2.+ / 2#. %& x 2.0 G +& x 2.+ /
2%%. $& x 2.0 G $& x 2.+ / 2$&As you can see the
maximum combined stress appears away from both the original
