Chapter Four---------------------------------------------------------------------------------------------------------------Steel Beams ============================================================================================= ------------------------------------------------------------------------------------------------------------------------------------------ Assist. Prof. Dr.Thaar S. Al-Gasham , Wasit University, Eng. College 64 4-1 types of beams Beams are members supporting transverse loads. They are probably thought of as being used in horizontal position and subjected to gravity or vertical loads, but there are frequent exceptions- roof rafter, for example. Below the well-known types of beams; Joist: - they are closely spaced beams supporting the floors and roofs of buildings. Lintels: - they are beams over the openings in masonry walls, such as windows and doors. Spandrel: - they support the exterior walls of building and perhaps a part of floor or hallway loads. Stringer: - they are beams in the floor bridge running parallel to the roadway. Floor beams: - they are perpendicular to the roadway of the bridge and are used to transfer the loading from the stringers to the supporting girder or trusses. Girders:- they are the largest beams in the bridge floor, by them the loadings are transferred from the floor beams to the columns or supporting soil.
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4-1 types of beams
Beams are members supporting transverse loads. They are probably
thought of as being used in horizontal position and subjected to gravity or
vertical loads, but there are frequent exceptions- roof rafter, for example.
Below the well-known types of beams;
Joist: - they are closely spaced beams supporting the floors and roofs
of buildings.
Lintels: - they are beams over the openings in masonry walls, such as
windows and doors.
Spandrel: - they support the exterior walls of building and perhaps a
part of floor or hallway loads.
Stringer: - they are beams in the floor bridge running parallel to the
roadway.
Floor beams: - they are perpendicular to the roadway of the bridge and
are used to transfer the loading from the stringers to the supporting
girder or trusses.
Girders:- they are the largest beams in the bridge floor, by them the
loadings are transferred from the floor beams to the columns or
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4-2 sections used as beams
The most economical shapes to be used as beams are W-shapes.
Sometimes channels are used as beams when light loadings are applied or
small flanges are required due to available of narrow clearness. The W-
shapes have more steel concentrated in their flange than do S-shapes and
thus have larger moment of inertia and resisting moment for the same
weights. They are relatively wide and have appreciable lateral stiffness.
4-3 bending stress and the plastic moment
Initially, when the moment is applied to a beam, the stress will vary linearly
from the neutral axis to the extreme fiber of the beam as shown in fig (b), if
the applied moment is increased, there will to be a linear variation in the
stress until the yielding stress is reached in the outermost beam fiber as
shown in part c, this moment is known as the yielding moment.
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If the moment is increased beyond the yielding moment, the outermost fiber
will have same yield stress, and the duty of providing the necessary
additional resisting moment will fall on the fiber nearer to the neutral axis.
This procedure will continue, with more and more fibers stressed to the
yielding stress as shown in part d and e. until finally plastic distribution is
approached, as shown in part f. note that the strain variation form the neutral
axis to the outer fiber remain linearly for all of these cases.
The applied moment is called plastic moment, when all beam fibers reached
to the yielding. The ratio of plastic to yielding moment is referred as shape
factor, which is 1.5 for rectangular sections and ranging from 1.1 to 1.2 for
standard rolled- beam section.
4-4 PLASTIC HINGES
This section is devoted to a description of the development of a plastic hinge
as in the simple beam shown in Fig. The load shown is applied to the beam
and increased in magnitude until the yield moment is reached and the
outermost fiber is stressed to the yield stress. The magnitude of the load is
further increased, with the result that the outer fibers begin to yield. The
yielding spreads out to the other fibers, away from the section of maximum
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moment, as indicated in the figure. The distance in which this yielding occurs
away from the section in question is dependent on the loading conditions
and the member cross section. For a concentrated load applied at the center
line of a simple beam with a rectangular cross section, yielding in the extreme
fibers at the time the plastic hinge is formed will extend for one-third of the
span. For a W shape in similar circumstances, yielding will extend for
approximately one-eighth of the span. During this same period, the interior
fibers at the section of maximum moment yield gradually, until nearly all of
them have yielded and a plastic hinge is formed, as shown in Fig.
Although the effect of a plastic hinge may extend for some distance along
the beam, for analysis purposes it is assumed to be concentrated at one
section. For the calculation of deflections and for the design of bracing, the
length over which yielding extends is quite important.
For plastic hinges to form, the sections must be compact. The compact
sections are defined as being those which have sufficiently stocky profiles
such that they are capable of developing fully plastic stress distributions
before they buckle locally.
The student must realize that for plastic hinges to develop the members must
not only be compact but also must be braced in such a fashion that lateral
buckling is prevented.
Finally the effects of shear, torsion, and axial loads must be considered. They
may be sufficiently large as to cause the members to fail before plastic
hinges can form. In the study of plastic behavior, strain hardening is not
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When steel frames are loaded to failure, the points where rotation is
concentrated (plastic hinges) become quite visible to the observer before
collapse occurs.
4-5 elastic design
Until recent years, almost all steel beams were designed on the basis of the
elastic theory. The maximum load that a structure could support was
assumed to equal the load that first caused a stress somewhere in the
structure to equal the yield stress of the material. The members were
designed so that computed bending stresses for service loads did not exceed
the yield stress divided by a safety factor. Engineering structures have been
designed for many decades by this method, with satisfactory results. The
design profession, however, has long been aware that ductile members do
not fail until a great deal of yielding occurs after the yield stress is first
reached. This means that such members have greater margins of safety
against collapse than the elastic theory would seem to indicate.
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4.6 The Plastic Modulus
The yield moment My equals the yield stress times the elastic modulus. The
elastic modulus equals (I/c) or (bd2/6) for a rectangular section, and the yield
moment equals (Fy b d2 /6). This same value can be obtained by considering
the resisting internal couple shown in Fig.
The elastic section modulus can again be seen to equal (bd2/6) for a
rectangular beam.
The resisting moment at full plasticity can be determined in a similar manner.
The result is the so-called plastic moment, Mp It is also the nominal moment
of the section, Mn. This plastic, or nominal, moment equals T or C times the
lever arm between them. For the rectangular beam of Fig.
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The plastic moment is said to equal the yield stress times the plastic section
modulus. From the foregoing expression for a rectangular section, the plastic
section modulus Z can be seen to equal bd2/4. The shape factor, which
equals Mp/My or Z/S, is 1.50 for a rectangular section. A study of the plastic
section modulus determined here shows that it equals the statically moment
of the tension and compression areas about the plastic neutral axis. Unless
the section is symmetrical, the neutral axis for the plastic condition will not
be in the same location as for the elastic condition. The total internal
compression must equal the total internal tension. As all fibers are
considered to have the same stress (Fy) in the plastic condition, the areas
above and below the plastic neutral axis must be equal. This situation does
not hold for unsymmetrical sections in the elastic condition.
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𝑠 =𝐼
𝑐
c for compression= 1.875, and for tension=6.5-1.875=4.625'' (control)
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4-7 The Virtual-Work Method
Although a plastic hinge may have formed in a statically indeterminate
structure, the load can still be increased without causing failure if the
geometry of the structure permits. The plastic hinge will act like a real hinge
insofar as increased loading is concerned. As the load is increased, there is
a redistribution of moment, because the plastic hinge can resist no more
moment. As more plastic hinges are formed in the structure, there will
eventually be a sufficient number of them to cause collapse. Actually, some
additional load can be carried after this time, before collapse occurs, as the
stresses go into the strain hardening range, but the deflections that would
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4-8 analysis and design of steel beams
The steel beams are classified into three categories; compact, non-compact
and slender sections.
1- Compact sections; they are sections having sufficiently stocky profile
so that it is capable of developing fully plastic stress distribution before
their web or flange buckle locally.
2- Non-compact sections: they are sections for which the yield stress can
be reached in some, but in all, of its compression elements.
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4-8-1 Flexural strength of compact sections
These section can be failed in three zones depending on the
unbraced length as shown in figure below
The term Lb is used throughout this chapter to describe the length
between points which are either braced against lateral displacement of the
compression flange or braced against twist of the cross section.
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Example :-
Solution
1- Specify the type of beam
2- Specify the failure zone
The beam is continuous lateral support Lb=0, thus it will fail in zone 1
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𝑀𝑛 = 𝑀𝑝 = 𝐹𝑦𝑍𝑥 =50(54.0)
12= 225 𝑓𝑡 − 𝑘
LRFD 𝑀𝑢 = 0.9 𝑀𝑛 = 0.9𝑥225 = 202.5 𝑓𝑡 − 𝑘
ASD 𝑀𝑎 =𝑀𝑛
1.67=
225
1.67= 134.7 𝑓𝑡 − 𝑘
Note;- you can use table (3-2) instead of table (1-1) for W-shapes with
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Cb is called bending coefficient and can be calculated as following
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Example:-
Referring to Table (3-2), the section W14x90 have footnote ''f'', which means
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Another way of checking as follows
The applied moment is 821.4 ft-k is smaller than LRFD flexural strength of
W24x84 that is 840 ft-k as referred in Table (3-2)
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2. Evaluating the applied moment
𝑀𝑢 =𝑤𝑢𝑙2
8=
1.88(20)2
8= 94𝑘𝑖𝑝 − 𝑓𝑡
3. Referring to table (3-2) to select the lightest section with flexural
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𝑀𝑢 =𝑤𝑢𝑙2
8=
1.906 (20)2
8= 95.3 𝑘𝑖𝑝 − 𝑓𝑡 < 97.5 𝑜. 𝑘
5- Checking for shear
𝑉𝑢 =𝑤𝑢𝑙
2=
1.906 (20)
2= 19.06 𝑘𝑖𝑝 < 73.2 𝑜. 𝑘
---- use section W10x22 according to LRFD method.
ASD Solution
1. Evaluating the wa as follows;
𝑤𝑎 = 𝐿. 𝐿 + 𝐷. 𝐿 = (8)(100) + (8)(150)5
12= 1300
𝐼𝑏
𝑓𝑡= 1.3
𝑘𝑖𝑝
𝑓𝑡
2. Evaluating the applied moment
𝑀𝑎 =𝑤𝑎𝑙2
8=
1.3(20)2
8= 65𝑘𝑖𝑝 − 𝑓𝑡
3. Referring to table (3-2) to select the lightest section with flexural
strength equal to or greater than Mu
Both above sections do have ASD moment greater than the applied
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4-15 design of steel beams using manual charts
In part 3 of manual there are figures named as Table (3-1). As shown in
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