Chapter27

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TRUSSES
1.0 INTRODUCTION
Trusses are triangular frame works in which the members are subjected to essentially axial
forces due to externally applied load. They may be plane trusses [Fig. 1(a)], wherein the
external load and the members lie in the same plane or space trusses [Fig. 1(b)], in which
members are oriented in three dimensions in space and loads may also act in any direction.
Trusses are frequently used to span long lengths in the place of solid web girders and such
trusses are also referred to as lattice girders.
Steel members subjected to axial forces are generally more efficient than members in
flexure since the cross section is nearly uniformly stressed. Trusses, consisting of
essentially axially loaded members, thus are very efficient in resisting external loads. They
are extensively used, especially to span large gaps. Since truss systems consume relatively
less material and more labour to fabricate, compared to other systems, they are particularly
suited in the Indian context.
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(b) Space Truss
(a) Plane Truss
(c) Bracings in
Multi-storey Building
(d) Bracings in Single
Storey Buildings
Transverse Vertical
(wall)
Bracings
Rafter (Plan)
Bracings Longitudinal Vertical
(Wall) Bracings
(e) Truss Bridge
Fig. 1 Types of Trusses

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Trusses are used in roofs of single storey industrial buildings, long span floors and roofs of
multistory buildings, to resist gravity loads [Figs. 1(a) and 1(b)]. Trusses are also used in
multi-storey buildings and walls and horizontal planes of industrial buildings to resist
lateral loads and give lateral stability [Figs. 1(c) and 1(d)]. Trusses are used in long span
bridges to carry gravity loads and lateral loads [Fig. 1(e)].
Trusses often serve the action of the girder in transferring the gravity load over larger
span, and are referred to also as lattice girders. Such lattice girders are usually deeper
and much lighter than regular girders and hence are economical, particularly when
repetitive fabrication is taken advantage of. These are used as flooring support systems in
multi-storey buildings, within which depth all the ducts can be easily accommodated
without increasing the depth of the ceiling.
Steel trusses can also be efficiently used along with concrete slabs in buildings and bridges
by mobilising composite action between structural steel and concrete. In this chapter,
initially, the details of structural steel trusses are discussed. Subsequently, the behaviour
and design of structural steel - concrete composite trusses are discussed.
2.0 LOADS
The loads on trusses would depend upon the application for which the trusses are used.
The loads may be static, as in the case of buildings, or dynamic, as in the case of bridges.
These are briefly reviewed below.
2.1 Industrial Buildings
The roof trusses in industrial buildings are subjected to the following loads:
2.1.1 Dead load
Dead load on the roof trusses in single storey industrial buildings consists of dead load of
claddings and dead load of purlins, self weight of the trusses in addition to the weight of
bracings etc. Further, additional special dead loads such as truss supported hoist dead
loads, special ducting and ventilator weight etc. could contribute to roof truss dead loads.
As the clear span length (column free span length) increases, the self weight of the
moment resisting gable frame increases drastically. In such cases roof trusses are more
economical.
2.1.2 Live load
The live load on roof trusses consist of the gravitational load due to erection and servicing
as well as dust load etc. and the intensity is taken as per IS:875-1987 (Reaffirmed 1992).
Additional special live loads such as snow loads in very cold climates, crane live loads in
trusses supporting monorails may have to be considered.

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2.1.3 Wind load
Wind load on the roof trusses, unless the roof slope is too high, would be usually uplift
force perpendicular to the roof, due to suction effect of the wind blowing over the roof.
Hence the wind load on roof truss usually acts opposite to the gravity load, and its
magnitude can be larger than gravity loads, causing reversal of forces in truss members.
The horizontal and vertical bracings employed in single and multi-storey buildings are also
trusses [Fig. 1(d)], used primarily to resist wind and other lateral loads. These bracings
minimize the differential deflection between the different frames due to crane surge in
industrial buildings. They also provide lateral support to columns in small and tall
buildings, thus increasing the buckling strength.
2.1.4 Earthquake load
Since earthquake load on a building depends on the mass of the building, earthquake loads
usually do not govern the design of light industrial steel buildings. Wind loads usually
govern. Howev er, in the case of industrial buildings with a large mass located at the roof,
the earthquake load may govern the design. These loads are calculated as per IS:1893-
1985.
2.2 Multi-Storey Buildings
The lateral load due to wind or earthquake may be resisted by vertical bracings acting as
trusses. These bracings, properly designed, make these buildings very stiff in resisting
lateral loads. Hence they are economical in the buildings of intermediate height ranges. In
the case of earthquake loading, stiff buildings may attract larger inertia force and hence
use of bracings may not be desirable.
2.3 Bridge Trusses
Trusses are used in bridges to transfer the gravity load of moving vehicles to supporting
piers. Depending upon the site conditions and the span length of the bridge, the truss may
be either through type or deck type. In the through type, the carriage way is supported at
the bottom chord of trusses. In the deck type bridge, the carriage way is supported at the
top chord of trusses. Usually, the structural framing supporting the carriage way is
designed such that the loads from the carriage way are transferred to the nodal points of
the vertical bridge trusses. More details of the trusses bridges are discussed in the chapter
on bridges.
3.0 ANALYSIS OF TRUSSES
Generally truss members are assumed to be joined together so as to transfer only the axial
forces and not moments and shears from one member to the adjacent members (they are

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regarded as being pinned joints). The loads are assumed to be acting only at the nodes of
the trusses. The trusses may be provided over a single span, simply supported over the
two end supports, in which case they are usually statically determinate. Such trusses can
be analysed manually by the method of joints or by the method of sections. Computer
programs are also available for the analysis of trusses. These programs are more useful in
the case of multi-span indeterminate trusses, as well as in the case of trusses in which the
joint rigidity has to be considered. The effect of joint rigidity is discussed later in greater
detail.
From the analysis based on pinned joint assumption, one obtains only the axial forces in
the different members of the trusses. However, in actual design, the members of the
trusses are joined together by more than one bolt or by welding, either directly or through
larger size end gussets. Further, some of the members, particularly chord members, may
be continuous over many nodes. Generally such joints enforce not only compatibility of
translation but also compatibility of rotation of members meeting at the joint. As a result,
the members of the trusses experience bending moment in addition to axial force. This
may not be negligible, particularly at the eaves points of pitched roof trusses, where the
depth is small and in trusses with members having a smaller slenderness ratio (i.e. stocky
members). Further, the loads may be applied in between the nodes of the trusses, causing
bending of the members. Such stresses are referred to as secondary stresses. The
secondary bending stresses can be caused also by the eccentric connection of members at
the joints. The analysis of trusses for the secondary moments and hence the secondary
stresses can be carried out by an indeterminate structural analysis, usually using computer
software.
The magnitude of the secondary stresses due to joint rigidity depends upon the stiffness of
the joint and the stiffness of the members meeting at the joint. Normally the secondary
stresses in roof trusses may be disregarded, if the slenderness ratio of the chord members
is greater than 50 and that of the web members is greater than 100. The secondary
stresses cannot be neglected when they are induced due to application of loads on
members in between nodes and when the members are joined eccentrically. Further the
secondary stresses due to the rigidity of the joints cannot be disregarded in the case of
bridge trusses due to the higher stiffness of the members and the effect of secondary
stresses on fatigue strength of members. In bridge trusses, often misfit is designed into the
fabrication of the joints to create prestress during fabrication opposite in nature to the
secondary stresses and thus help improve the fatigue performance of the truss members at
their joints.
3.0 CONFIGURATION OF TRUSSES
3.1 Pitched Roof Trusses
Most common types of roof trusses are pitched roof trusses wherein the top chord is
provided with a slope in order to facilitate natural drainage of rainwater and clearance of
dust/snow accumulation. These trusses have a greater depth at the mid-span. Due to this

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(a) Pratt Truss (b) Howe Truss
Fig. 2 Pitched Roof Trusses
(c) Fink Truss (d) Fan Truss
(e) Fink Fan Truss (f) Mansard Truss
even though the overall bending effect is larger at mid-span, the chord member and web
member stresses are smaller closer to the mid-span and larger closer to the supports. The
typical span to maximum depth ratios of pitched roof trusses are in the range of 4 to 8, the
larger ratio being economical in longer spans. Pitched roof trusses may have different
configurations. In Pratt trusses [Fig. 2(a)] web members are arranged in such a way that
under gravity load the longer diagonal members are under tension and the shorter vertical
members experience compression. This allows for efficient design, since the short
members are under compression. However, the wind uplift may cause reversal of stresses
in these members and nullify this benefit. The converse of the Pratt is the Howe truss [Fig.
2(b)]. This is commonly used in light roofing so that the longer diagonals experience
tension under reversal of stresses due to wind load.
Fink trusses [Fig. 2(c)] are used for longer spans having high pitch roof, since the web
members in such truss are sub-divided to obtain shorter members.
Fan trusses [Fig. 2(d)] are used when the rafter members of the roof trusses have to be
sub-divided into odd number of panels. A combination of fink and fan [Fig. 2(e)] can also
be used to some advantage in some specific situations requiring appropriate number of
panels.
Mansard trusses [Fig. 2(f)] are variation of fink trusses, which have shorter leading
diagonals even in very long span trusses, unlike the fink and fan type trusses.

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Fig. 3 Parallel Chord Trusses
(a) Floor Girder (b) Warren Truss
(d) K type Web
(c) Lattice Girder
(e) Diamond Type Web
The economical span lengths of the pitched roof trusses, excluding the Mansard trusses,
range from 6 m to 12 m. The Mansard trusses can be used in the span ranges of 12 m to
30 m.
3.2 Parallel Chord Trusses
The parallel chord trusses are used to support North Light roof trusses in industrial
buildings as well as in intermediate span bridges. Parallel chord trusses are also used as
pre-fabricated floor joists, beams and girders in multi-storey buildings [Fig. 3(a)]. Warren
configuration is frequently used [Figs. 3(b)] in the case of parallel chord trusses. The
advantage of parallel chord trusses is that they use webs of the same lengths and thus
reduce fabrication costs for very long spans. Modified Warren is used with additional
verticals, introduced in order to reduce the unsupported length of compression chord
members. The saw tooth north light roofing systems use parallel chord lattice girders [Fig.
3(c)] to support the north light trusses and transfer the load to the end columns.
The economical span to depth ratio of the parallel chord trusses is in the range of 12 to 24.
The total span is subdivided into a number of panels such that the individual panel lengths
are appropriate (6m to 9 m) for the stringer beams, transferring the carriage way load to
the nodes of the trusses and the inclination of the web members are around 45 degrees. In
the case of very deep and very shallow trusses it may become necessary to use K and
diamond patterns for web members to achieve appropriate inclination of the web members.
[Figs. 3(d), 3(e)]
3.3 Trapezoidal Trusses
In case of very long span length pitched roof, trusses having trapezoidal configuration,
with depth at the ends are used [Fig. 4(a)]. This configuration reduces the axial forces in
the chord members adjacent to the supports. The secondary bending effects in these

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Fig 4. Trapezoidal Trusses
(a)
(b)
members are also reduced. The trapezoidal configurations [Fig. 4(b)] having the sloping
bottom chord can be economical in very long span trusses (spans > 30 m), since they tend
to reduce the web member length and the chord members tend to have nearly constant
forces over the span length. It has been found that bottom chord slope equal to nearly half
as much as the rafter slope tends to give close to optimum design.
4.0 TRUSS MEMBERS
The members of trusses are made of either rolled steel sections or built-up sections
depending upon the span length, intensity of loading, etc. Rolled steel angles, tee sections,
hollow circular and rectangular structural tubes are used in the case of roof trusses in
industrial buildings [Fig. 5(a)]. In long span roof trusses and short span bridges heavier
rolled steel sections, such as channels, I sections are used [Fig. 5(b)]. Members built-up
using I sections, channels, angles and plates are used in the case of long span bridge
trusses [Fig. 5(c)]
Access to surface, for inspection, cleaning and repainting during service, are important
considerations in the choice of the built-up member configuration. Surfaces exposed to
the environments, but not accessible for maintenance are vulnerable to severe corrosion
during life, thus reducing the durability of the structure. In highly corrosive environments
fully closed welded box sections, and circular hollow sections are used to reduce the
maintenance cost and improve the durability of the structure.
(a) Light Section (b) Heavy Sections
Fig. 5 Cross Sections of Truss Members
(c) Built-up Sections

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Fig. 6 Typical Truss Joints
(a) Typical Roof Truss Joint (b) Typical Bridge Truss Joint
Gusset Plates
5.0 CONNECTIONS
Members of trusses can be joined by riveting, bolting or welding. Due to involved
procedure and highly skilled labour requirement, riveting is not common these days,
except in some railway bridges in India. In railway bridges riveting may be used due to
fatigue considerations. Even in such bridges, due to recent developments, high strength
friction grip (HSFG) bolting and welding have become more common. Shorter span
trusses are usually fabricated in shops and can be completely welded and transported to
site as one unit. Longer span trusses can be prefabricated in segments by welding in shop.
These segments can be assembled by bolting or welding at site. This results in a much
better quality of the fabricated structure. However, the higher cost of shop fabrication due
to excise duty in contrast to lower field labour cost frequently favour field fabrication in
India.
If the rafter and tie members are T sections, angle diagonals can be directly connected to
the web of T by welding or bolting. Frequently, the connections between the members of
the truss cannot be made directly, due to inadequate space to accommodate the joint
length. In such cases, gusset plates are used to accomplish such connections (Fig. 6). The
size, shape and the thickness of the gusset plate depend upon the size of the member being
joined, number and size of bolt or length of weld required, and the force to be transmitted.
The thickness of the gusset is in the range of 8 mm to 12 mm in the case of roof trusses
and it can be as high as 22 mm in the case of bridge trusses. The design of gussets is
usually by rule of thumb. In short span (8 – 12 m) roof trusses, the member forces are
smaller, hence the thickness of gussets are lesser (6 or 8 mm) and for longer span lengths

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(> 30 m) the thickness of gussets are larger (12 mm). The design of gusset connections are
discussed in a chapter on connections.
6.0 DESIGN OF TRUSSES
Factors that affect the design of members and the connections in trusses are discussed in
this section.
6.1 Instability Considerations
While trusses are stiff in their plane they are very weak out of plane. In order to stabilize
the trusses against out- of- plane buckling and to carry any accidental out of plane load, as
well as lateral loads such as wind/earthquake loads, the trusses are to be properly braced
out -of -plane. The instability of compression members, such as compression chord, which
have a long unsupported length out- of-plane of the truss, may also require lateral bracing.
Compression members of the trusses have to be checked for their buckling strength about
the critical axis of the member. This buckling may be in plane or out-of-plane of the truss
or about an oblique axis as in the case of single angle sections. All the members of a roof
truss usually do not reach their limit states of collapse simultaneously. Further, the
connections between the members usually have certain rigidity. Depending on the
restraint to the members under compression by the adjacent members and the rigidity of
the joint, the effective length of the member for calculating the buckling strength may be
less than the centre-to-centre length of the joints. The design codes suggest an effective
length factor between 0.7 and 1.0 for the in-plane buckling of the member depending upon
this restraint and 1.0 for the out of plane buckling.
In the case of roof trusses, a member normally under tension due to gravity loads (dead
and live loads) may experience stress reversal into compression due to dead load and wind
load combination. Similarly the web members of the bridge truss may undergo stress
reversal during the passage of the moving loads on the deck. Such stress reversals and the
instability due to the stress reversal should be considered in design.
The design standard (IS: 800) imposes restrictions on the maximum slenderness ratio,
( /r), as given below:
Member type Max /r limit
Members under compression under loads other than wind/ earthquake load 180
Tension members undergoing stress reversal due to loads other than wind load 180
or seismic forces
Members normally under tension but may have to resist compression under
wind load 250
Compression flange of a beam against lateral torsional buckling 300

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Members designed only for tension even though they may experience
stress reversal 350
Members always under tension (unless pre-tensioned to avoid sag) 400
These limits are imposed to ensure the following:
Too slender a member is avoided which may be damaged during transportation and
erection
Members do not sag excessively under self-weight during service causing excessive
deflection in truss.
Compression members do not sag greater than 1/1000th
of their length, which is
beyond the imperfection limit assumed in the compressive strength calculation.
It is a common practice to specify a minimum angle size of 50 X 50 X 6 in the case of roof
trusses.
7.0 ECONOMY OF TRUSSES
As already discussed trusses consume a lot less material compared to beams to span the
same length and transfer moderate to heavy loads. However, the labour requirement for
fabrication and erection of trusses is higher and hence the relative economy is dictated by
different factors. In India these considerations are likely to favour the trusses even more
because of the lower labour cost. In order to fully utilize the economy of the trusses the
designers should ascertain the following:
Method of fabrication and erection to be followed, facility for shop fabrication
available, transportation restrictions, field assembly facilities.
Preferred practices and past experience.
Availability of materials and sections to be used in fabrication.
Erection technique to be followed and erection stresses.
Method of connection preferred by the contractor and client (bolting, welding or
riveting).
Choice of as rolled or fabricated sections.
Simple design with maximum repetition and minimum inventory of material.
8.0 COMPOSITE TRUSSES
Trusses are efficient structural systems, since the members experience essentially axial
forces and hence the materials are fully utilized. Steel as a structural material is equally
strong both in tension and compression and hence steel trusses are more efficient. They
tend to be economical to support loads over larger span lengths. However, the members
in the compression chord of the simply supported steel truss (top chord) may prematurely
buckle before the stresses reach the material strength. In this context the concrete slab
acting in composite with the truss compression chord becomes useful (Fig. 6).

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Web Diagonals
Concrete deck slab
Steel truss
Top Chord
Bottom Chord
Fig. 6 Composite Truss
Shear studs
A reinforced concrete or composite deck floor is required in any case in building and other
structures to provide a flat surface. Using it as a part of the compression member in truss
system could be an economical proposition. Concrete has a lower strength compared with
steel and hence requires larger cross section to sustain a given compression.
Consequently, the concrete floor slab used as a part of the compression chord of the truss
is less vulnerable to buckling failure. Further, concrete can more economically carry
compression, whereas it is very weak in tension. In a composite truss system the relative
merits of steel and concrete as construction materials are fully exploited. It is one of the
most economical systems in longer span flooring construction. Thus composite truss
systems are structurally efficient and economical.
In multi-storey buildings, the composite truss systems also reduce the total height of the
building, by accommodating the services (heating, ventilation, lighting and
telecommunication ducts) within the depth of the truss, thus integrating structural,
mechanical and electrical systems within in the floor space. This minimises the inter-floor
height. Considering functional and structural efficiency and economy, it is only natural
that composite steel-concrete trusses are a popular choice for long span and high-rise
construction.
The composite truss usually consists of a parallel chord Warren truss, designed to resist,
the superimposed gravity load in conjunction with the reinforced composite concrete deck
slab attached to the truss through shear connectors (Fig. 6). The top and bottom chords
of the truss may be made of angles, T sections or rolled steel structural tubes (Fig. 7). The
Angles T - Sections RHS
Fig. 7 Types of Truss Members
Rod
s

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web members, arranged in a Warren truss form, may be made of steel rods, single angle or
double angle sections, or structural tubing welded to the chord members either directly
(most common) or indirectly through gussets. The end nodes of the Warren truss are
usually arranged to coincide with the reaction points of the orthogonal flooring member.
The shear transfer between the steel truss and the concrete deck slab is mobilised usually
using shear studs. The deck slab may consist of cast in place concrete, either over
removable centering or left-in-place profiled sheeting. The profiles in the decking may run
either parallel or perpendicular to the truss in the orthogonal floor system.
The application of composites in construction is a mature technology in developed
countries, frequently chosen under competitive designs. Even in India, there are a few
interesting applications of the composite truss construction. Many [1,2,3] have reviewed
the progress of the technology since its inception.
Early applications relied essentially on the bond between concrete and steel to bring about
composite action. The requirement for efficient shear transfer led to mechanical shear
connectors in later applications. Angles, channels and several other proprietary shear
connectors were explored. Shear studs (straight round rods with upset head) evolved as
the standard, due to their ease of installation, labour reduction and cost efficiency.
The composite trusses consisting of readymade and made to order open-web joists/trusses
with cast-in-place concrete slab are most common. Instead of removable shuttering, left-
in-place permanent shuttering or steel profiled sheeting was subsequently used. These
evolved into composite decking slabs, wherein the profiled sheeting in addition to serving
as a shuttering for green concrete, also would act as tension reinforcement for hardened
concrete. The shear studs are welded to the compression chord of the truss through the
deck sheet, serving as a shear transfer unit both to the truss and profiled decking.
The World Trade Centre building in New York was one of the largest applications of the
composite, open-web joist system. Subsequent developments used cold-formed specially
shaped top chord members made of high strength steels. The profiled metal decking also
provide lateral support to the compression chord member until the concrete hardened. In
early 1970’s competitive, efficient systems were developed with wide concrete ribs,
requiring less number of shear studs. The volume of concrete in the deck was decreased
and the sprayed-on fire protection requirements were also decreased through field tests.
While the earlier studies concentrated on ultimate strength evaluation, the recent studies
have dealt with service load performance characteristics, such as creep and shrinkage
effects of concrete on deflection, connection detailing, improving the performance of shear
studs, slab crack control, member fatigue control, vibration and energy absorption
characteristics, and trusses continuous over many spans.
8.1 Stud shear connectors

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Stud shear connectors are commonly used to transfer shear between the steel compression
strut and concrete deck slab. These studs are welded through the metal decking on to the
compression chord of the truss in the case of composite deck slabs. The design of studs is
treated in greater detail in the chapter on composite beams.
The stud diameter should be limited to 2.5 times the thickness of the part to which it is
welded, in order to prevent the stud tearing out of the element. This could be a critical
requirement in the case of composite trusses, because of the thin chord members that may
be used.
8.2 Effective concrete slab
Due to shear lag, the entire width of the slab may not be fully stressed as per the simple
beam theory and for the purpose of composite action an effective section of concrete is
considered in stress, deflection and strength evaluation [4]. The equations for calculating
the effective width of the slab is given in the chapter on composite beams.
8.3 Design considerations
8.3.1 Preliminary Design
For the preliminary design of a composite truss the following data is needed:
The maximum bending moments and shear forces in the member
(a) at the construction stage (Ms, Vs),
(b) at the factored load acting at the limit state of collapse of the composite section
(Mc, Vc).
the concrete slab (regular or composite) sizes and
the truss spacing.
The following are the steps in the preliminary design:
1. Decide on the depth of the truss girder.
The span to depth ratio of a simply supported composite truss is normally 15 to
20.
2. Develop the web member layout, usually using Warren configuration.
Use a slope of 30 degrees to horizontal to increase the opening and reduce the
number of connections.
3. Design the top chord member.
Force in the top chord member at the construction load, Rt, is calculated from the
corresponding moment, Ms, and the lever arm between the chord members (Fig.
8).
Size of the member is based on the member strength as governed by lateral
buckling between the lateral supports to the top chord until the concrete hardens.

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Fig. 8 Moment Capacity of Steel and Composite Trusses
Xb
Xc/2
Xt
Rt
Rb
Rc
(a) Steel truss
(b) Composite truss
Ds
Dt
A minimum width of 120 mm for the top chord is usually acceptable to support the
decking in a stable manner during erection.
Minimum of 8 mm thickness of the leg of the compression chord is required to
weld the stud through the deck on to the leg.
Vertical leg of the member should be adequate to directly weld the web members.
Otherwise gusset may be required.
Local bending should be considered in between the nodal points in case of loading
between nodes
1. Design the bottom chord member.
Calculate the tension in the bottom chord, Rb, at the factored load moment using
the following equation.
Rb = Mc/(Dt + Ds - 0.5 Xc –Xb) (4)
where Xc = (Ds – Dp) Rb/Rc, , Dp = Depth of the profile, Rb, Rt, Rc are the forces in
the bottom chord, top chord of steel truss and the force in concrete slab,
respectively.
Area of the bottom chord and the bottom chord member shape may be designed
based on this force, Rb,, considering the yield strength of the member.
2. Check the slab capacity for the compression force at the limit state of collapse.
The slab capacity is given by
Rc = 0.45 fck beff (Ds – Dp) (5)
where fck = cube strength of concrete and beff is the effective width of the concrete
slab acting integral with the truss.
3. Design the web member.
The maximum force in the web member is calculated by setting the vertical
component of the member force equal to the maximum shear force in the truss.
The web member is designed to carry the force considering its yield strength in
tension and buckling strength in compression.

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8.3.2 Detailed Analysis and Design
The composite truss thus evolved may be analysed in an exact fashion using a more
accurate truss analysis following either manual or computer method. The methods of
modeling for computer analysis are presented in reference 3.
The composite truss should be checked for (a) limit state at construction load, (b) limit
state at service load and (c) limit state of collapse.
a) The limit state at construction load
During the construction the truss has to carry all the superimposed loads until concrete
sets. The top chord of the truss at this stage can fail either by reaching the material
strength or lateral buckling strength, the lateral buckling being the more vulnerable mode
of failure. In order to improve the lateral buckling strength, the top chord may be laterally
braced in between supports either temporarily or permanently. In case the truss supports
composite deck slab, once the profiled metal decking is attached to the top chord by the
welding of studs to the top chord through the deck metal, it may be assumed to provide
adequate lateral support to the compression chord.
Until the green concrete hardens, the steel section alone has to support all the dead weight
and construction live load. Hence, the failure mode of the truss can be due to yielding /
buckling (lateral or in-plane) of the top chord in the plane of truss due to compression,
failure of the web member by yielding / buckling. In order to reduce the forces in
members during this stage, propping of the truss from below at one or more points can be
done.
b) The limit state at service load
Strength: Before the concrete hardens, the members of the truss experience forces due to
its self-weight and the weight of composite deck profile, green concrete and
reinforcements. The composite truss resists the loads applied after the concrete hardens
(the super imposed dead load, and floor live loads). These loads cause axial forces in all
the members due to truss action. Furthermore, the top chord is subjected to bending
moment due to UDL / concentrated load between the nodes of the truss, which is resisted
by the steel alone before concrete hardens and by the composite section after the concrete
hardens.
In the allowable stress method, the members have to be checked for stresses at this service
load to ensure adequate factor of safety in addition to deflection. If the construction is
shored, then the stresses have to be calculated for the entire dead load acting on the
composite section. If the construction is un-shored, the stresses due to the self weight
including green concrete, sustained by steel section acting alone, have to be superposed on
stresses due to super imposed dead load and live load acting on the composite member. In

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Version II 27 - 16
the case of cyclically loaded composite trusses, as in composite bridges, the stress range at
the service load has to be calculated and checked for fatigue. Further, deflection at this
service load is to be checked, as discussed in the following section.
Deflection: The deflection of the steel truss alone due to construction load has to be
checked before concrete hardens and that of the composite truss for the full dead and
service live load as given below.
At the time of concreting, the deflection of the truss system could cause ponding of
concrete leading to a larger slab thickness while leveling concrete. In order to overcome
this, pre-camber is specified for the truss, particularly in the unshored construction. If the
calculated deflection of the steel truss alone under the construction load (dead load and
construction live load) is less than 20mm no cambering is necessary. If the deflection is
greater than 20mm camber is provided in the top chord of the truss to an extent slightly
less than the calculated deflection. This is to account for moment restraint provide by
even simple connections at the ends of the truss, the stiffness of the supporting member,
non-hinged nature of the truss joints, all of which reduce the actual deflection to a value
below the theoretical value.
The deflection under the full dead load and live load is calculated, considering the
composite action under super imposed dead load and live load and simple steel truss
action for dead load until the concrete hardens after accounting for camber given in the
top chord. The deflection calculation should include the instantaneous deflection, creep
effect and shrinkage effect. The shrinkage effect can be accounted for by calculating the
deflection due to net restrained shrinkage strain of around 200 microns at the slab level.
The creep deflection is calculated for the sustained load corresponding to the total dead
load and sustained live load in the case of shored construction and only superimposed
dead load and sustained live load in the case of unshored construction. For this purpose,
the transformed area of concrete is calculated using the modular ratio corresponding to the
creep modulus of concrete. The instantaneous deflection is calculated using the
transformed section arrived at using the elastic modular ratio.
Span to depth ratio limitation can be effective to prevent excessive deflection and vibration
under moving loads. The span to depth ratio of 20 for steel truss alone and 25 for the
composite truss would be usually adequate for buildings. Slightly reduced values would
be appropriate (15 to 20 respectively) in bridge trusses. The vibration control could be
achieved by ensuring that any applied vibration frequency of any machinery is not close to
the natural frequency of the composite flooring and ensuring the natural frequency is
above 4 cycles per second. There is also a strong correlation between deflection control
and vibration control, so much so that usually strict deflection control under loads would
also ensure satisfactory vibration performance.
c) The limit state of collapse

TRUSSES
Version II 27 - 17
Cs Cs+ Cs
Cs,bTs,b
(a) Due to Composite Truss
(b) Due to Composite Bending of Top Chord
Fig. 9 Forces in Studs at Limit State of
Collapse
At the limit state of collapse the sequence of loading and the corresponding non-
composite / composite member behaviour is immaterial. The composite member resists
the total factored load. The different members of the composite truss are checked for
their limit state of collapse under factored loads as given below:
Ultimate tensile strength of bottom chord as governed by yield strength of the gross
area or ultimate strength of net effective area.
Ultimate tensile / compressive strength of the web members, depending upon the type
of axial force under factored loading.
Ultimate strength of the composite compression chord under combined bending (at
nodes and in between nodes) due to load in between nodes and compression.
8.3.3 Design of Studs
The shear studs within a panel of a truss have to transfer the shear between the slab and
top chord. This is due to overall composite truss action and the additional shear due to
the bending of top chord between panel points, caused by the UDL/concentrated load
between the panel points of the truss.
In the composite truss action, the forces in the composite top chord would be due to full
load in the case of shored construction and due to super imposed dead and live load only
in the case of un-shored construction. The unbalanced component of the compressive
load on the concrete slab ( Ct) causes shear in the studs. The bending moments at the
nodal point are calculated, only due to super imposed dead and live load. Due to this
bending, the shear in the stud over half the span is calculated as (Tsb + Csb ) as shown in
Fig. 9. The studs have to resist these combined forces due to local bending between nodes
and overall truss action, at ultimate load, assuming the shear to be uniformly shared by the
studs in the region.
8.3.4 Partial Shear Connection
In the elastic range, the actual shear
force in shear connections over the span length varies according to the variation of the
shear diagram. At the ultimate load, redistribution of the shear force among shear
connectors takes place due to the ductility of stud shear connector and the slip between

TRUSSES
Version II 27 - 18
the steel and concrete. Hence, the shear in the shear connectors in a shear span is assumed
as uniform, at the ultimate load.
The shear connectors in bridges are spaced according to the elastic theory to avoid stress
concentration and fatigue failure at service load and a limit of 55% of the shear stud
capacity is imposed at the service load limit state. In buildings the shear connectors are
spaced uniformly over the length.
The number of shear connectors as required by the elastic design may be very high. In
such cases partial shear connection (50 – 70% of full shear connection) may be used. In
such a case, the shear capacity of the shear connections and hence the effectiveness of the
concrete in compression is to be reduced accordingly, some times leading to increase in
the size of steel chord members. The use of partial shear connection also leads to slight
increase in the service load deflection. However, considering the large area of concrete in
compression, partial shear connections usually do not cause any appreciable changes in the
final design.
8.3.5 Concrete Cracking
The deck slab may have a tendency to crack, especially at the interior supports of
continuous composite beams. In order to minimise the cracking, the steel reinforcement is
employed in the direction perpendicular to the potential cracks at supports.
8.3.6 Practical Considerations
Ductile failure can be obtained, provided the design is governed by the ultimate
strength of the tension chord member and the strength of top chord, web and stud
connectors are large enough to preclude their premature failure.
To facilitate stud welding, the top chord made of T or tubular section with a minimum
width of 50mm is preferred, instead of smaller single or double angles.
8.4 Cost implications
The steel weight savings, the change of ratio in labour content to weight of the structure
and the reduction in time for the completion of the work are three important factors that
contribute to the cost reduction of the composite truss design. The project analysis
division of the Canadian Institute of Steel Construction carried out a review of a number
of design examples, covering steel framed buildings with braced steel core, gravity steel
framing with concrete core(s). The total building costs including the deck slab and fire
protection were considered. The results are tabulated in ref [2]. The summary of the
findings of this study is as follows:
The material savings in composite construction can be as high as 20 to 40 percent
compared to non-composite steel construction, in the case of girder flooring. Further

TRUSSES
Version II 27 - 19
material savings of about 20 percent is possible if composite trusses are used instead of
composite girders.
The cost saving of composite girders is smaller (between 15 and 30%) compared to
weight savings, due to the cost of studs and additional labour associated with
composite construction. Further, cost savings of about 15% is possible by using
composite trusses instead of composite girders.
In the case of composite construction in India, the difference between the percentage of
weight saving and cost saving should be lesser due to the lower labour cost. Consequently
the composite construction, particularly use of composite trusses in long span structure,
could mean considerable economy as realised in U.K., New Zealand, South Africa,
Australia and Singapore, in the past two decades.
9.0 SUMMARY
In this chapter, initially the behaviour and design of steel trusses were dealt with.
Important aspects of truss systems such as the systems, their economy, their connections
were discussed. Then the use of steel truss and reinforced concrete slab acting together as
a composite truss was discussed in this chapter. After a brief introduction, the historical
evolution of the system was discussed. The background information for the design of the
composite trusses was presented. The economy of the system, particularly in the Indian
context was evaluated. The discussions indicate that there is great potential for the use of
the system in the Indian context.
10.0 REFERENCES
1. Anon, “Design of Composite Trusses”, Steel Construction Institute”, Ascot, 1992.
2. Anon “ Constructional Steel Design: An International Guide”, Elsevier, London, 1993.
3. Chien, E.Y.L. and Ritchie, J.K, “Composite Floor Systems – A Mature Design
Option”, Journal of Constructional Steel Research, V. 25, 1993, pp107-139.
4. Vallenilla, C.R., and Bjorhovde, R., “Effective Width Criteria for Composite Beams”,
Engineering Journal, AISC, Fourth Quarter, 1985, pp. 169-175.

TRUSSES
Version II 27 - 20
Job No: Sheet 1 of 13 Rev
Job Title: ROOF TRUSS
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
PROBLEM 1:
Design a roof truss for an industrial building with 25 m span and 120 m
long. The roofing is galvanized iron sheeting. The basic wind speed is 50
m/s and terrain is open industrial area and building is class A building. The
building clear height at the eaves is 9 m.
Structural form:
For the purpose of this design example a trapezoidal truss is adopted with a
roof slope of 1 to 5 and end depth of 1 m. For this span range the
trapezoidal trusses would be normally efficient and economical.
Economical span to depth ratio is around 10.
Then, Span/depth = 25/3.5 = 7.1
Hence, depth is acceptable.
Truss spacing:
Truss spacing should be in the region of 1/4th
to 1/5th
of the span length.
For 6 m spacing,
Spacing/span = 6/25 = 1/4.17 (acceptable)
Then, number of bays = 120/6 = 20
120 m, trusses@ 6 m
spacing
25 m
Plan

TRUSSES
Version II 27 - 21
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Truss configuration:
Loading:
kN/m2
Dead load: GI sheeting = 0.085
Fixings = 0.025
Services = 0.100
Total load = 0.210
For 6 m bays,
Roof dead load = 0.21 * 25 * 6 = 31.5 kN
Weight of purlin = 0.07 * 6 * 25 = 10.5 kN
(Assuming 70 N/m2
)
*
Self-weight of truss = 0.133 * 6 * 25 = 20.0 kN
Total dead load = 62.0 kN
*
[For welded sheeted roof trusses, the self-weight is given approximately by
w = (1/100) (5.37 + 0.053A) kN/m2
= (5.37 + 0.053 * 6 * 25) = 0.133 kN/m2
2.5 m
9.0 m
1.0 m
25 m
Elevation
A B C D E F
G
H I
J K
a b c
f g
d e h
Truss
25 m = 20 @ 1.25 m

TRUSSES
Version II 27 - 22
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Dead Loads
Intermediate nodal dead load (W1) = 62.0/20 = 3.1 kN
Dead load at end nodes (W1 / 2) = 3.1/2 = 1.55 kN
(Acts vertically downwards at all nodes)
Wind load (IS: 875-1987):
Basic wind speed = 50 m/s
Wind load F on a roof truss by static wind method is given by
F = (Cpe - Cpi) * A * pd
where, Cpe, Cpi are force co-efficient for exterior and interior of the
building.
Value of Cpi:
Assume wall openings between 5-20% of wall area.
Then, Cpi = ± 0.5
Value of Cpe:
Roof angle =
Height of the building to eaves, h = 9 m
Lesser dimension of the building in plan, w = 25 m
Building height to width ratio is given by,
01
3.11
5
1
tan
5.036.0
25
9
w
h

TRUSSES
Version II 27 - 23
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Roof angle Wind angle Wind angleh/w
Windward
side
Leeward
side
Windward
side
Leeward
side
100
- 1.2 - 0.4 - 0.8 - 0.8
200
- 0.4 - 0.4 - 0.7 - 0.7
Here, = 11.30
, then by interpolation we get
0.36
11.30
- 1.1 - 0.4 - 0.79 - 0.79
Risk Co-efficient, k1 = 1.0
(Assuming the industrial building as general building and its probable life
about 50 years)
Terrain, height, structure size factor, k2:
Roof elevation - 9 m to 12.5 m.
Height (m) Terrain category and class of building
10 0.91
15 0.97
For 12.5 m, k2 = 0.94
Assume, topography factor = k3 = 1.0
9 m
3.5 m

TRUSSES
Version II 27 - 24
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Wind pressure:
Total height of the building = 12.5 m
Basic wind speed, vb = 50 m/s
Design wind speed vZ is given by,
vZ = k1* k2* k3 * vb.
k1 = 1.0
k2 = 0.94
k3 = 1.0
vZ = 0.94 * 1 * 1 * 50 = 47 m/s
Design wind pressure (pd) = 0.6 vZ
2
= 0.6 * (47)2
= 1325 N/m2
= 1.325 kN/m2
Tributary area for each node of the truss:
Length of each panel along sloping roof
Spacing of trusses = 6m
Tributary area for each node of the truss = 6 * 1.27 = 7.62 m2
m1.4m1.27
11.3cos
1.25
0
6 m
1.27 m
Node
Area

TRUSSES
Version II 27 - 25
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Wind load on roof truss:
Pressure co-efficient (Cpe-Cpi) Wind load F
(kN)
Cpe
Wind
angle
Wind
ward
Lee
ward
Cpi Wind
ward
Lee
ward
A pd
(kN)
Wind
ward
Lee
ward
00
- 1.10 - 0.4 0.5
- 0.5
-1.6
- 0.6
- 0.9
0.1
10.1
10.1
- 16.2
- 6.1
- 9.1
1.0
900
- 0.79 - 0.79 0.5
- 0.5
- 1.29
- 0.29
- 1.29
- 0.29
10.1
10.1
- 13.0
- 2.9
- 13.0
- 2.9
Maximum Cpe – Cpi:
Critical wind loads to be considered for analysis:
Wind ward side (W3) Lee ward side (W4)Wind
angle
Intermediate
nodes W3
End and
apex nodes
W3 /2
Intermediate
nodes W4
End and
apex nodes
W4 / 2
00
- 16.2 - 8.1 - 9.1 - 4.55
900
- 13.0 - 6.5 - 13.0 - 6.5
*Loads in kN
- 0.9- 1.6
Wind on side
Wind angle = 00
- 1.29- 1.29
Wind on end
Wind angle = 900
120 m, trusses@ 6 m
25 m

TRUSSES
Version II 27 - 26
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Imposed load:
Live load = 0.35 kN/m2
[From IS: 875 – 1964]
Load at intermediate nodes, W2 = 0.35 * 6 * 1.25
= 2.63 kN
Load at intermediate nodes, W2 / 2 = 1.32 kN
(Acts vertically downwards)
Loading pattern:
W1 /2 W1
W1W1
W1
W1
W1
W1
W1
W1
W1 /2
W1
W1
W1
W1
W1
W1
W1
W1
W1
W1
(a) Dead load
W2 /2 W2
W2W2
W2
W2
W2
W2
W2
W2
W2 /2
W2
W2
W2
W2
W2
W2
W2
W2
W2
W2
(b) Live load
W4
W3/2
W3
W3
W3
W3
W3
W3
W3W3
W3
W3/2
W4
W4
W4
W4
W4
W4 /2
W4/2
W4
W4
W4
(c) Wind load

TRUSSES
Version II 27 - 27
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Forces in the members:
The truss has been modeled as a pin jointed plane truss and analysed using
SAP90 software. The analysis results are tabulated below.
[See truss configuration for member ID]
Member Forces (kN)Member
Dead load Live load Wind on side Wind on end
A-B 0 0 1.6 1.3
B-C -47.4 -40.2 214.9 172.5
C-D -47.4 -40.2 218.1 175.0
D-E -63.2 -53.6 284.3 228.1
E-F -63.2 -53.6 287.5 230.7
F-G -66.4 -56.3 294.8 236.6
G-H -66.4 -56.3 298 239.1
H-I -63.2 -53.6 276 221.5
I-J -64.5 -54.8 286.2 229.7
J-K -64.5 -54.8 289.4 232.2
a-A -1.6 -1.3 8.3 6.7
a-B -41.6 -35.3 186.5 149.7
a-b 29.5 25 -131.8 -105.8
b-B 24.1 20.5 -104.8 -84.1
b-C -3.1 -2.6 16.5 13.2
b-D -17.1 -14.5 70.8 56.8
b-c 56.5 47.9 -247.1 -198.3
c-D 9.5 8.1 -35.4 -28.4
c-E -3.1 -2.6 16.5 13.2
c-F -5.3 -4.5 14 11.2
c-d 64.6 54.8 -274.5 -220.3
d-F 1 0.9 5.8 4.7
d-G -3.1 -2.63 16.5 13.2
d-H 2.4 2 -23.7 -19.0
d-e 64.1 54.4 -262 -210.2
e-H -5.1 -4.3 36.4 29.2
e-I -4.6 -3.9 24.8 19.9
e-f 11.4 9.7 -71.1 -57.1
e-h 55.4 47 -205.5 -164.9
f-I 1.8 1.6 -9.7 -7.8
f-J -3.1 -2.6 16.5 13.2
f-K 13.6 11.6 -83 -66.6

TRUSSES
Version II 27 - 28
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Load factors and combinations:
For dead + imposed
1.5*DL + 1.5*LL
For dead + wind
1.5*DL + 1.5*LL
or
0.9*DL + 1.5*LL
For dead + imposed + wind
Not critical as wind loads act in opposite direction to dead and imposed
loads
Member Forces under Factored loads in kN:

TRUSSES
Version II 27 - 29
Worked Example - 1
Made by
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Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet Checked by
PU
Date 16-08-00

TRUSSES
Version II 27 - 30
Member Design Forces
(kN)
Member
DL + WL DL + LL
A-B 2.4 0
B-C 251.25 -131.4
C-D 256.05 -131.4
D-E 331.65 -175.2
E-F 336.45 -175.2
F-G 342.6 -184.05
G-H 347.4 -184.05
H-I 319.2 -175.2
I-J 332.55 -178.95
J-K 337.35 -178.95
a-A 10.05 -4.35
a-B 217.35 -115.35
a-b -153.45 81.75
Top Chord Design:(G-H)
Maximum compressive force = 174.1 kN
Maximum tensile force = 357.4 kN
Trying ISNT 150 X 150 X 10 mm @ 0. 228 kN/ m
Sectional Properties:
Area of Cross section = At = 2908 mm2
Membe
r
DL + WL DL + LL
b-B -121.05 66.9
b-C 20.1 -8.55
b-D 80.55 -47.4
b-c -285.9 156.6
c-D -38.85 26.4
c-E 20.1 -8.55
c-F 13.05 -14.7
c-d -314.85 179.1
d-F 10.2 2.85
d-G 20.1 -8.595
d-H -31.95 6.6
d-e -296.85 177.75
e-H 46.95 -14.1
e-I 30.3 -12.75
e-f -89.55 31.65
e-h -225.15 153.6
f-I -11.85 5.1
f-J 20.1 -8.55
f-K -104.1 37.8

TRUSSES
Version II 27 - 31
Worked Example - 1
Made by
SSSR
Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00

TRUSSES
Version II 27 - 32
Section classification:
=(250/fy)0.5
= (250/250)1/2
= 1.0
Flange:
B/T =75/10 = 7.5 8.9 (Flange is plastic)
Web:
d/t = 140/10 = 14 [> 9.975 and 19.95 ]
(Web is semi-compact)
As no member in the section is slender, the full section is effective and there
is no need to adopt reduction factor.
Maximum unrestrained length = y = 3810 mm
(Assuming every two alternative nodes are restrained)
ryy = 30.3 mm
y = 3810/30.3 = 125.7
Then, c = 84.3 N/mm2
Hence, section is safe against axial compression
Axial tension capacity of the section = 2908 * 250/1.15 = 632 kN > 357.4 kN
Hence, section is safe in tension.
Bottom chord design:(c-d)
Maximum tensile force = 169.4 kN [Try same section as top chord]
Axial tension capacity of the selected section = 2908 * 250/1.15 = 632 kN
Hence, section is safe in tension.
Axial capacity = (84.3/1.10)*2908/1000 = 222.86 kN > 184.05 kN
(Assuming every node is restrained by longitudinal tie runner)
ryy = 30.3m
y = 2500/30.3 = 82.5
Then, c = 145.5 N/mm2
Axial capacity = (145.5/1.15)*2908/1000 = 368 kN > 314.85 kN
Hence, section is safe against axial compression also.

TRUSSES
Version II 27 - 33
Worked Example - 1
Made by
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Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Web member design:(b-B)
Maximum tensile force = 66.9 kN
Try – ISA 80 X 80 X 8.0
A = 1221 mm2
rxx = 24.4 mm
ruu = 30.8 mm
b/t = 80/8 = 10.0 <14.0
Hence, the section is not slender
Length of member = (12502
+ 12502
)0.5
= 1767.5 mm
Slenderness ratio is taken as the greater of
0.85 * 1767.5/24.4 = 61.6
1.0 * 1767/ 30.8 = 57.4

TRUSSES
Version II 27 - 34
Worked Example - 1
Made by
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Date 9-2-2000
Structural Steel
Design Project
Calculation Sheet
Checked by
PU
Date 16-08-00
Then, c = 182.1 N/mm2
Design compressive strength = 1221 * (182.1/1.10)/1000
=202.13 kN > 121.05 kN
Hence, safe in compression.
Tensile capacity of the section = (250/1.10)*1221/1000
= 277.5 kN > 66.9kN
Hence ISA 80 X 80 X 8.0 is adequate for the web member
(The web members away from the support would have lesser axial force
but longer and can be redesigned, if so desired)

TRUSSES
Version II 27 - 35
Job Title: COMPOSITE TRUSS
Worked Example - 2
Made by
SSSR
Date 17-10 -99
Structural Steel
Design Project
Calculation Sheet
Checked by
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Date 16-08-00
PROBLEM 2:
Design a composite truss of span 10.0 m with following data:
DATA:
Span = = 10.0 m
Truss spacing = 3.0 m
Slab thickness = Ds = 150 mm
Profile depth = Dp = 75.0 mm
Self weight of deck slab = 2.80 kN/m2
Maximum laterally un-restrained length in top chord is 1.5 m.
Grade of concrete, M20 =(fck)cu=20 MPa
Composite
Trusses
ELEVATION
10 m
10 m
3.0 m
FLOOR
PLAN
TRUSS

TRUSSES
Version II 27 - 36
Worked Example - 2
Made by
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Date 17-10 -99
Structural Steel
Design Project
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Checked by
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Date 16-08-00
Loading:
kN/m2
Factored Load
(kN/m2
)
Deck slab weight 2.8 2.8*1.5 = 4.20
Truss weight (assumed) 0.4 0.4*1.5 = 0.60
Ceiling, floor finish and
Services 1.0 1.0*1.5 = 1.5
Construction Load 1.0 1.0*1.5 = 1.5
Superimposed live load 5.0 5.0*1.5 = 7.5
PRE-COMPOSITE STAGE:
Loading kN/m2
Factored Load
(kN/m2
)
Truss weight 0.4 0.4*1.5 = 0.60
Construction load 1.0 1.0*1.5 = 1.5
Total factored load = 6.30 kN/m2
Choose depth of truss = Span/20 = 10000/20
= 500 mm
Total factored load = 6.30 * 3 = 18.9 kN/m
Maximum bending moment = w 2
/8 = 18.9 *102
/8 = 240.98 kN-m
Maximum shear = w /2 = 18.9*10/2 = 94.50 kN
Depth of truss (centre to centre distance of chords) = 0.5 m
Maximum axial compressive force in top chord = 240.98/0.5 =481.96 kN

TRUSSES
Version II 27 - 37
Worked Example - 2
Made by
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Date 17-10 -99
Structural Steel
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Dt = 500 + xt + xb
Truss configuration: Choose the following truss configuration
Top chord design:
Try ISNT 150 X 150 X 10 mm @ 0.228 kN/m
Sectional properties:
Area of cross-section = At = 2908 mm2
Depth of section = 150 mm
Width of section, b = 2b1
= 150 mm
Thickness of flange = T = 10.0 mm
Thickness of web = t = 10.0 mm
Centre of gravity = xt = 39.5 mm
rxx = 45.6 mm
ryy = 30.3 mm
T
X
Y
t
B
xt
D = 500 mm
xb
xt
Ds = 150 mm
Dp=75 mm
500 mm
1500 mm 1500 mm 1500 mm500
CA E G
D F HB

TRUSSES
Version II 27 - 38
Worked Example - 2
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=(250/fy)0.5
= (250/250)1/2
= 1.0
Flange:
b1
/T = 75/10 = 7.5 8.9 Flange is plastic
Web:
d/t = 140/10 = 14 ( >9.98 and 19.95 ) Web is semi-compact
As no member in the section is slender, there is no need of adopting
reduction factor (Yielding govern).
Given, maximum un-restrained length of top chord is 1.5 m during
construction stage.
x = 0.85*1500 = 1275 mm
rxx = 45.6 mm
ryy = 30.3 mm
x = 1275/45.6 = 28
y = 1500/30.3 = 49.5
Then, c = 202.8 N/mm2
[From Table - 3 of Chapter on axially compressed
Columns]
Axial capacity = (202.8/1.15)*2908/1000 = 512.8 kN > 437.4 kN
Axial capacity = (202.8/1.1)*2908/1000 =536.13 kN > 481.96 kN
Hence, section is safe against axial compression at construction stage.
[Other member design is governed by composite loading]

TRUSSES
Version II 27 - 39
Worked Example - 2
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COMPOSITE STATE:
kN/m2
Factored Load (kN/m2
)
Truss weight (assumed) 0.4 0.4*1.5 = 0.60
Ceiling, floor finish and Services 1.0 1.0*1.5 = 1.5
Superimposed live load 5.0 5.0*1.5 = 7.5
Total factored load = (4.2+0.6+1.5+7.5)* 3
= 13.8*3 = 41.40 kN/m
Maximum bending moment (Mc) = w 2
/8 = 41.4*102
/8 = 527.85 kN-m
Maximum shear = w /2 = 41.4*10/2 = 207 kN
Bottom chord design:
Force in bottom chord, Rb,req is given by: [See Fig. of the text]
Rb,req {D + xt + Ds-(Ds-Dp)/2} = Mc
[Assume NA is in the concrete slab]
Rb,req (500+39.5+(150-37.5))/1000 = 527.85
Rb,req (652/1000) = 527.85 kN-m
Rb,req = 527.85/0.652 = 809.59 kN
Area required = 809.59*1000/(fy/1.1)
=809.59*1000/(250/1.1) = 3562.2 mm2
Trial-1 Trying ISHT 150 @ 0.294kN/m
Sectional properties:
A = 3742 mm2
; xb = Centre of gravity = 26.6 mm
Width of the section, b = 2b1 = 250 mm

TRUSSES
Version II 27 - 40
Worked Example - 2
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Axial tension capacity of the selected section (Rb):
Rb = (250/1.1)* 3742/1000 = 850.46kN > 809.59 kN
Hence, O.K.
Capacity of Composite Section in Compression:
Capacity of concrete slab, Rc, is given by
Rc = 0.45 (fck)cu*beff*(Ds – Dp)
Effective width of the slab, beff: [See the chapter Composite beams – II]
beff /4 = 10000/4 = 2500 mm
Therefore, beff = 2500 mm
Rc = 0.45*20*2500*75/1000 {fck = 20 N/mm2
}
= 1687.5 kN > Rb (tension governs)
Neutral axis depth :
xc = (Ds – Dp)*850.46/1687.5 = 75*850.46/1687.5 = 37.8 mm
Dt = 0.5+0.0266+0.0395 = 0.566 mm
Then, maximum moment it can carry
Mu, design = 850.46(0.566+0.15-0.5*0.0378-0.0266)
= 570.23 kN-m > 527.85 kN-m
Hence, the slab and chord members are designed.

TRUSSES
Version II 27 - 41
Worked Example - 2
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Date 17-10 -99
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Web members:
FAB = V(1.414) =207(1.414) = 292.7kN (tension)
FBC = (V-0.5* 39.5) (5002
+7502
)0.5
/500 =335.86kN (compression)
FCD =(V – 2.0*39.5) (5002
+7502
)0.5
/500 = 223.91kN (tension)
Hence, maximum tensile force in bracing members =292.7kN
Maximum compressive force in bracing members =335.86 kN
Design of tension members:
Trial gross area required =292.7*103
/(250/1.1) =1287.88 mm2
Trying 2 – ISA 70 X70 X6 .0 @ 0.126 kN/m
Agross provided =2*806 =1612 mm2
Effective area:
(Assume, angle is welded to T- section)
A net effective = 1612 mm2
Axial tension capacity = Ae*(fy/ m)
= 1612*250/1.1
= 366.36 kN 292.7 kN
Hence, 2 – ISA 70 X 70 X 6.0 are adequate
V = 207 kN
A
500 mm
B D
C
1500 mm
41.4 kN/m2

TRUSSES
Version II 27 - 42
Worked Example - 2
Made by
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Date 17-10 -99
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Date 16-08-00
Design of compression member:
Maximum compressive load = 320 kN
Trying 2 – ISA 80 X 80 X 6.0 @ 0.146 kN/m
A = 1858 mm2
rxx = 24.6 mm
ruu = 34.9 mm
b/t = 80/6 = 13.3 <15.75
Hence, the section is not slender and no need to apply any reduction factor.
Slenderness ratio is taken as the greater of
Length of member = (7502
+ 5002
)0.5
= 901 mm
xx = 0.85 * 901/24.6 = 31.1
xx =1.0 * 901/ 34.9 =25.8
Design buckling strength = c = 231.2 Mpa
[Table – 3 of chapter on axially compressed columns]
Design compressive strength = 1858* (231.2/1.1)/103
=390.52 kN > 335.86
kN
Hence the 2 – ISA 80 X 80 X 6.0 are adequate for the web members
(The web members away from the support would have lesser axial force
and can be redesigned, if so desired. Preferably use the same section for all
web members)
X
ISA 80 X 80 X 6 mm
X
Y
Y

TRUSSES
Version II 27 - 43
Worked Example - 2
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Weight Schedule:
Description Section
mm X mm X mm
Weight
kN/m
Number Length
(m)
Total
Length (m)
Weight
kN
Top Chord
Bottom
Chord
Bracing
Members
Tension
Members
Compression
Members
ISNT 150 X150X10
ISHT 150
2-ISA 70 X 70 X 6
2-ISA 70 X 70 X 6
2-ISA 80 X 80 X 6
0.228
0.294
0.126
0.126
0.146
1
1
2
6
6
10.0
10.0
0.71
0.9
0.9
10.0
10.0
1.42
5.4
5.4
2.28
2.94
0.18
0.68
0.79
6.87
Allow 2 ½ % Extras 0.17
7.04
Average weight per unit area of floor
= 7.04 = 0.23 kN/m2
0.4 kN/m2
(Assumed)
10*3
Hence, O.K.

TRUSSES
Version II 27 - 44
Worked Example - 2
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Date 17-10 -99
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Deflection:
Pre-composite stage:
The second moment of area of the steel truss, It can be calculated from the
following equation.
Where,
Ab - Cross-sectional area of bottom chord.
At - Cross-sectional area of top chord.
In this problem,
Ab = 3742 mm2
xb = 26.6 mm
At = 2908 mm2
xt = 39.5 mm
Dt = 566 mm
Loading:
kN/m2
Deck slab weight 2.80
Truss weight 0.23
Construction load 1.00
--------
4.03
Total Load = 4.03*3*10=121 kN
2
)(
tbt
tb
tb
t xxD
AA
AA
I
46
2
t
mm10940
39.526.6656
29083742
29083742
I

TRUSSES
Version II 27 - 45
Worked Example - 11
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Date 17-10 -99
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Deflection at pre composite state is given by
0 = (5*121*100003
)/(384*200*409*106
) = 19.3 mm
Deflection at composite state due to dead load = 1 = (3.03/4.03)*19.3
= 14.5 mm
[For composite stage construction load has to removed for calculating
deflections]
Deflection - Composite stage:
The second moment of area, Ic, of a composite truss can be calculated from
the following equation
Where,
Ac = Cross-sectional area of the concrete in the effective breadth of slab
= (Ds - Dp)beff
m = modular ratio
In this problem,
Ab = 3742 mm2
; beff = 2500 mm
Ac = (150 - 75)* 2500 = 1875*102
mm2
m = 15(light weight concrete)
Dt = 566 mm
xb = 26.6 mm
2
2/)(
/
/
bpst
cb
cb
c xDDD
mAA
mAA
I
46
2
2
2
c
mm104221
26.6
2
225
656
15/1087513742
15/1018753742
I

TRUSSES
Version II 27 - 46
Worked Example - 12
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Date 17-10 -99
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Date 16-08-00
Loading:
Super Imposed load = 5.0 kN/m2
Total Load = 5.0*3*10=150 kN
Deflection at composite state due to superimposed load is given by
2 = (5*150*100003
)/(384* 200* 1224*106
) = 8.0 mm
10% allowance is given
Then, 2 = 8.8 mm < /360 = 10000/360 = 28 mm
Total deflection = 1 + 2 =14.5 + 8.8 = 23.3 mm ( /429) < ( /325)
Hence, design is O.K.

Chapter27

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Chapter27