3.0 Trusses
3.1 Introduction
How long can the largest sections in the table properties able to spans? Let say for a bridge, a
single longitudinal member is required to carry 100kN/m UDL along 40m span, can a single
914 × 419 × 398UB (the largest section in table of properties) adequate in shear capacity,
moment capacity and without excessive deflection, lateral torsional buckling? The use of a
single big section will be very costly, and may be infeasible in erection and fabrication.
Moreover, the bending moment capacity, which is governed by the depth of section, if
obtained by using a single cross-section, a large portion of the web actually is unused.
Therefore a truss system is suggested.
Trusses and lattice girders are fabricated from the various steel sections, jointed together by
welding or by bolting usually via gusset (connecting) plates. The joints could be pin or
continuous. Normally they are designed either acting in one plane or in three dimensions
(space frame). The Figure 3-1 is depicting some of them:
Figure 3-1 Examples of plane truss system
The members used in truss system normally are angles, double-angles, C-channels, double C-
channels, SHS, CHS, cold-formed steels etc. Some of them are depicted as follow:
Figure 3-2 Members used in truss system
Belgium
Pratt
Pratt
Warren
Fink HoweHowe
Curved
Angle C-channel Joist
CHS SHSCold-formed steel sections
1
3.2 Terminology of Truss
Simply labeled in Figure 3-3 is some basic terminology of a truss.
Figure 3-3 Terminology in truss system
3.3 Design of Roof Truss System
This session is purposely introducing the design of a simple plane roof truss system
(determinacy). The loading subjected by a truss is transferred through the purlins, either
directly onto the nodes or on the top members span. It is ideal if the loads can be transferred
to the truss at the node position, but commonly this is not possible. In roof truss design the
purlin positions may not be known initially, and allowing for the possibility of purlin changes
during future re-roofing, a random position for loads is often allowed. Therefore, the general
procedure is summarized in Figure 3-4 below:
Concentrated loads
Node
Vertical internal member
Sloped internal member
Top chord (rafter)
Bottom chordOverall span
Loading
Start
Analyses assuming all joints are pin-jointed and all loading on the nodes, therefore the out put will beTensile stress – normally occurs at bottom chord and sloped internal membersCompressive stress – occurs at top chord and the vertical internal members
Analyses of the load bearing member such as rafter as a continuous beam supported at the nodes and loaded by the purlins.
If the load positions are uncertain, the rafter moment may be taken as wL2/6 (cl. 4.10 d)) where L is the node-to-node length of the rafter and w is the total load per unit length applied perpendicular to the rafter.
Continue to next page
2
Figure 3-4 Design procedure of a roof truss system
3.4 Loading
The loading subjected to a truss system could be dead loads, live loads and wind load. For
roof truss system, the dead loads may be consisting of cladding, insulation, self-weight of
trusses and purlins, services etc. For live load, according to BS 6399-2 or CP3: Ch V: Part 2,
0.75kN/m2 may be used where the entrance to the roof is available only for services purpose.
Otherwise, 1.5kN/m2 may be used if the purpose is more than that. In local practice,
especially for buildings up to three storeys, no additional wind load is considered on the roof.
3.5 Purlin Design
Figure 3-5 Purlins
Assessment of stresses due to eccentricity of the connections
Assessment of the effects of joint rigidity and deflection
End
*Normally ignored
Continue from previous page
Sag rod
Sag rod
Rafters of roof truss
Purlin
ss
3
As depicted in Figure 3-5, purlins are those members in a truss system which carrying the roof
sheets and transferring the load to the rafters. It is normally placed perpendicular to the rafters
and sag rods may be added (in order to reduce the minimum size of purlins) (see Table 27).
The purlins are not necessary to be analyzed as complicated as the other structural members.
The satisfaction of purlins is approached by the empirical rules suggested in cl. 4.12.4.3 as:
a) The slope of the roof should less than 30˚ from the horizontal.
b) The loading on the purlin should be substantially uniformly distributed. Not more than
10% of the total roof load on the member should be due to other types of load.
c) The limitations of section modulus Z about its axis parallel to the plane of the
cladding, member dimensions D perpendicular to the plane of cladding, B parallel to
the plane of cladding are given in Table 27 as shown below:
Table 3-1 Empirical values for purlins (Table 27)
Purlin section Zp (cm3) Zq (cm3) D (mm) B (mm)
Wind load from
BS 6399-2
Wind load from
CP3:Ch V:Part2
Angle WpL/1800 WqL/2250 WqL/1800 L/45 L/60
CHS WpL/2000 WqL/2500 WqL/2000 L/65 -
RHS WpL/1800 WqL/2250 WqL/1800 L/70 L/150
NOTE 1 Wp and Wq are the total unfactored loads (in kN) on the span of the purlin, acting perpendicular to the plane of the
cladding, due to (dead plus imposed) and (wind minus dead) loading respectively.
NOTE 2 L is the span of the purlin (in mm) centre-to-centre if main vertical supports. However, if properly supported sag
rod used, L may be taken as the sag rod spacing in determining B only.
3.6 Worked Example for Purlin Design and Loading Transfer
A plane truss (as shown below) is arranged where all purlins on its nodes. Design the purlins
using single angle sections, with the following data:
- Spacing between trusses = 5m
- Weight of roof sheet, insulation and purlins (on slope) = 0.35kN/m2
- Self-weight of truss (on slope) = 0.20kN/m2
- Imposed load (on plan) = 0.75kN/m2
4
Figure 3-6
Solution
Purlin Design
Dead load = 0.35kN/m2 (on slope)
Imposed load = 0.75 × 6 / 6.324
= 0.71kN/m2 (on slope)
Spacing of purlins = 6.324 / 3
= 2.11m
Wp = (0.35 + 0.71) × 2.11 × 5
= 11.18kN
Roof slope = tan-1 (2/6)
= 18.4˚ < 30˚
Use angles for purlin, therefore adopt the limitation:
Zp = WpL/1800 = 11.18 × 5000 / 1800
= 31.06cm3
D = L/45 = 5000 / 45
= 111.11mm
Assume sag rod are assigned on the middle of purlins between two trusses
B = (L/2)/60 = (5000/2) / 60
= 41.67mm
Therefore, use single angle 125 × 75 × 10L (Zx = 36.5cm3)
Table 27
Loading Transferred to the Trusses (on nodes)
Dead load = 0.35 + 0.2
= 0.55kN/m2 (on slope)
Imposed load = 0.71kN/m2 (on slope)
Total dead load Gk = 0.55 × 2.11 × 5
= 5.80kN
Total imposed load Qk= 0.71 × 2.11 × 5
12m
2m
6.324m
5
= 7.49kN
Design load P = 1.4Gk + 1.6Qk = 1.4 × 5.8 + 1.6 × 7.49
= 20.10kN
Figure 3-7
3.7 Tension Members
3.7.1 General
The specifications of tension members are given in cl. 4.6.1. Generally the tension capacity Pt
is given by:
Pt = py Ae
where Ae is the sum of the effective areas ae of all the elements of the cross-section. It should
be less than 1.2 times the total net area An.
The above formula is based on the assumption that the member is loaded on its axis. If
members are connected eccentric to their axes, the reduction of tensile capacity may be
limited by using the tensile-moment expression, (cl. 4.8.2). However,
angles, channels and T-sections could be treated as axially loaded by using reduced tension
capacity as follows (cl. 4.6.2):
For simple tied (no moment along the member) single angle consisting of a single angle
connected only through one leg only; single channel connected only through one web, or a T-
section connected only through the flange,
- for bolted connections: Pt = py (Ae – 0.5a2)
- for welded connections: Pt = py (Ag – 0.3a2)
where
a2 = Ag – a1
10.05kN20.1kN
20.1kN20.1kN
20.1kN20.1kN
10.05kN
6
a1 is the gross area of the connected element, taken as the product of its thickness
and the overall leg width for an angle, the overall depth for a channel or the flange
width for a T-section.
For simple tied double angle consisting of a single angle connected only through one leg
only; two channels connected only through one web, or two T-sections connected only
through the flange, see Figure 5-8 below:
Figure 3-8 Pt for double angle, channel or T-section members
3.4 Section Properties3.4.1 Gross cross-sectionGross cross-section properties should be determined from the specified shape and nominal dimensions of the member or element. Holes for bolts should not be deducted, but due allowance should be made for larger openings. Material used solely in splices or as battens should not be included.
3.4.2 Net areaThe effective area of a cross-section or an element of a cross-section should be taken as its gross area, less the deduction for bolt holes given in 3.4.4
3.4.3 Effective net areaThe effective net area ae of each element of a cross-section with bolt holes should be determined from ae = Kean but ae < ag
in which the effective net area coefficient Ke is given by:- for grade S 275: Ke = 1.2- for grade S 355: Ke = 1.1- for grade S 460: Ke = 1.0- for other steel grades: Ke = (Us/1.2)/py
whereag is the gross area of the element;an is the net area of the element;py is the design strength;Us is the specified minimum tensile strength
3.4.4 Deductions for bolts holes3.4.4.2 Holes not staggeredProvided that the bolt holes are not staggered, the area to be deducted should be the sum of the sectional areas of the bolt holes in a cross-section perpendicular to the member axis or direction of direct stress.
Figure 3-9 Definition of some terms about the sectional area
Gusset or other section
Two angles are longitudinal parallel to the other end
- for bolted connections: Pt = py (Ae – 0.25a2)
- for welded connections: Pt = py (Ag – 0.15a2)
a)
b)
- design as two separated members
7
3.8 Work Example for Tension Members
An internal member of a truss system is subjects to tensile force 260kN from truss analysis (as
shown in the Figure 3-10 below). Propose a suitable cross-section for it if
a) the end connections are welded;
b) the end connections are bolted (24mm bolt)
Figure 5-10
Solution
a) Welded end
Preliminary Sizing
Ft = 260kN
py = 275 N/mm2
Area needed = 260 × 103 / 275
= 945.45mm2
Try angle 100 × 65 × 7L where Ag = 11.2cm2
Tension Capacity
Assume the longer leg of the section welded to gusset; therefore the neutral axis
is eccentric away.
a1 = 100 × 7
= 700mm2
a2 = Ag - a1 = 1120 - 700
= 420mm2
Pt = py (Ag – 0.3a2)
= 275 × (1120 – 0.3 × 420) × 10-3
= 275.35kN > Ft = 260kN Ok
cl. 4.6.3.1
26
0kN
8
65mm
100m
m a1
a2
7mm
7mm
Figure 510 a1 and a2
a) Bolted end
Preliminary Sizing
Ft = 260kN
py = 275 N/mm2
Assume the bolt size D = 24mm
Bolt hole size = D + Allowance = D + 3 = 24 + 3
= 27mm
Assume the section thickness is 8mm
Area needed = 260 × 103 / 275 + (27 × 8)
= 1161.45mm2
Try angle 100 × 65 × 8L where Ag = 12.7cm2
Tension Capacity
Since the member is grade S 275, Ke = 1.2
An = Ag – Ah = 1270 – (27 × 8)
= 1054mm2
Ae = Kean = 1.2An = 1.2 × 1054
= 1264.8mm2
a1 = 100 × 8 – 27 x 8
= 800mm2
a2 = Ag - a1 = 1270 - 800
= 470mm2
Pt = py (Ag – 0.5a2)
= 275 × (1264.8 – 0.5 × 470) × 10-3
= 283.19kN > Ft = 260kN Ok
cl. 3.4.3
cl. 4.6.3.1
3.9 Compression Members
The design basis of a compression member in truss is generally similar to a column. To
simplify the design procedure, the angles, channels and T-sections are allowed to be designed
ignoring the effect of end-connection eccentricity; through the empirical-based limitations
given in cl. 4.7.10. The limitations are describing the slenderness of the member in terms of
end connection, effective length and different axes. The critical (largest value) is determined
9
from the rest (e.g. xx, yy, and vv) therefore results the compression capacity through the
formula Pc = Ag pc.
Let us look in the expression of . Consider the single angle strut with double-bolt fixing:
Figure 3-12 for a double-bolted member
There are given:
v = 0.85Lv/rv but > 0.7Lv/rv + 15
a = 1.0La/ra but > 0.7La/ra + 30
b = 0.85Lb/rb but > 0.7Lb/rb + 30
where
ra is the radius of gyration about an axis through the centriod of the angle parallel
to the gusset, so on for rb.
rv is the minimum radius of gyration.
Recall the coefficient of effective length LE in column design, 0.85 is indicating the element
which is partially restraint at the end and 1.0 is indicating the pin-joint. From this view, the
expressions a = 1.0La/ra and b = 0.85Lb/rb is similar to the = LE/r in column design. The
next expressions which contain slenderness factor of 0.7 purposely for indicating the
allowance of eccentricity. 0.7 is pessimistically indicates the effective length of the member
and the additional constant (15 and 30) are assigned to indicate the allowance of eccentricity.
For compression member subjected to bending moment, it should satisfy the three moment-
axial compression interaction limitations which expressed in flexural expression in cl. 4.8.3.
See Chapter 3 part 4.3.7 to 4.3.9. An alternative of checking all of this is to use the simplified
method provided in I.4.3, where as: .
(Double-bolt connection)
y y
a
a
v
v
x x
a
a
v
v
Axes
10
3.10 Work Example for Compression Members
Design the compression member for the truss shown in Figure 3-13 below. Use different
section for the top chord and internal vertical members. Use equal angle grade S 275 and
double-bolted connection.
Figure 3-13
Solution
a) Internal vertical member
Preliminary sizing
Fc = 46kN
py = 275 N/mm2
Since the member is bolted, assume D = 24mm
Assume the section thickness t = 8mm
Ah = (24 + 3) × 8
= 216mm2
Assume pc = 0.4py
Ag = Fc/(0.4 py) + 216 = 46 × 103 / (0.4 × 275) + 216
= 634mm2
Try 60 × 60 × 6L (Ag = 6.91cm2)
ra = rb = ry = rx = 1.82cm, rv = 1.17cm
Compression capacity
Since max = 145.30 and py = 275 N/mm2, pc = 72N/mm2
Therefore the compression capacity of the angle (class 3)
Pc = Ag pc
= 691 × 72 × 10-3
= 49.75kN > Fc = 46kN Ok
Table 23
Table 24 c)
cl. 4.7.4
90kN
2m
46kN
11
a) Top Chord
Preliminary sizing
Fc = 90kN
py = 275 N/mm2
Ah = 2 × (24 + 3) × 8
= 432mm2
To use a double-angle, assume pc = 0.4py
Ag = Fc/(0.4 py) + 432 = 90 × 103 / (0.4 × 275) + 432
= 1250mm2
Try 2/65 × 50 × 8L (Ag = 16.5cm2)
rx = 2.01cm, ry = 2.28cm, rv = 1.05cm
Section Classification (consider a single angle)
= (275 / 275)0.5
= 1.0
b/t = 50 / 8
= 6.25 < 15 = 15
d/t = 65 / 8
= 8.13 < 15 = 15
(b + d)/t = (50 + 65)/8
= 14.38 < 24 = 24 Class 3 semi-compact
Table 11
Slenderness
x
1.0Lx/rx = 1.0 × 3000 / 20.1 = 149.25
0.7Lx/rx + 30 = 0.7 × 3000 / 20.1 + 30 = 134.48< 149.25 use 149.25
y
Assume the back-to-back struts are interconnected each 500mm spacing using
bolts, as shown in Figure below:
Figure 3-14 Interconnection between two angles
Table 25
cl. 4.7.10.3a)
End connection End connection
4 @ 500mm c/c interconnection between two members
12
c = Lv/rv = 500 / 10.5
= 47.62
[(0.85Ly/ry)2 + c2]0.5 = [(0.85 × 3000 / 22.8)2 + 47.6232]0.5 = 121.56
1.4c = 1.4 × 47.62 = 66.67 < 121.56 use 121.56
max = 149.25
Compression capacity
Since max = 149.25 and py = 275 N/mm2, pc = 68N/mm2 and the
Therefore the compression capacity of the angle (class 3)
Pc = Ag pc
= 1650 × 68 × 10-3
= 112.20kN > Fc = 90kN Ok
Table 23
Table 24 c)
cl. 4.7.4
3.11 Worked Example of Designing a Truss
Design a roof truss (Pratt) for a factory which covers an area of 40 × 12m. Details of the truss
are shown below. Use mild steel for all members and apply welding to all connections.
Figure 3-15 Truss 40m × 12m
A A
5.0
5.0
Purlin
Truss
6 @ 2.0m = 12m
2.0
1.5
0.3
1.0
Purlins
Truss span = 12 m
Truss spacing = 5 m
Truss depth = 1.3 m
Roof slope, = tan-1 (0.3 / 6)
= 2.86º > 2º
- Water would not stay on the roof
Nodes spacing of bottom chord = 2 m
Nodes spacing of top chord, L2 = 2 m × cos2.86º
= 2 m
Purlin spacing, L1 = 1.5 m
13
Solution
Loading
Dead load (on slope)
7mm thick steel sheeting = 0.1kN/m2
Insulation and lamps = 0.15kN/m2
Self-weight of purlins = 0.05kN/m2
Self-weight of trusses = 0.1kN/m2
Total dead load, Gk = 0.4kN/m2
Live load:
For services
Live load on plan = 0.75kN/m2
Live load on slope = 0.75 × 6 / 6.008
Total live load, Qk = 0.75kN/m2
Design load, q = 1.4Gk + 1.6Qk
= 1.4 × 0.4 + 1.6 × 0.75
= 1.76kN/m2
Concentrated load on nodes, P = q × trusses spacing × nodes spacing
= 1.76 × 5 ×1.5
= 13.2kN
P/2 = 6.6kN
Analysis
On-node analysis
Apply truss analysis based on assumption that the loads are concentrated on
nodes:
14
Figure 3-16 On-node analysis
Table 3-2 Analysis results of the truss
Member Node Axial load (factored) (kN)(+) Tension(–) Compression
Bottom chord 1 – 22 – 33 – 4
0.078.0115.5
Internal members 1 – 82 – 82 – 93 – 93 – 104 – 104 – 11
-52.887.2-39.042.8-20.6-6.87.0
Top chord 8 – 99 – 1010 – 11
-78.3-116.0-122.0
Bending analysis
The top chord members are subjected to the transverse loading due to the purlins
loads. Therefore a moment analysis which treating the member as continuous
member should be carried on:
Cmax
= -52.8
Tmax = 87.2
Tmax
= 115.5
Cmax
= -122
Extreme reactions of the members
1 2 3 4 5 6 7
8
9
10 11 12
13
14
Loading diagram
[email protected] m = 12m
13.2kN
13.2kN 13.2kN
13.2kN
6.6kN
0.3
1.0
13.2kNP/2 =6.6kN
15
Figure 5-17 Bending analysis
Resulted from the computer analysis (plane frame analysis), the maximum
moment Mmax is 3.8kNm.
To allow future re-roofing, we may determine the maximum moment as:
Mmax = WL/6 = 13.2 × 2 / 6
= 4.4kNm
a) Top Chord Design
For this case, the top chord should be checked on
- its compression capacity assuming only axial load applied
- interaction check
Compression resistance
Preliminary sizing
Fc = 112kN
py = 275 N/mm2
Due to existence of moment, Assume pc = 0.15 py
Ag = Fc/(0.15py) = 112 × 103 / (0.15 × 275)
2.0m 2.0m 2.0m
13.2 13.2 13.2
6.6kN 6.6kN
[email protected] = 4.5m 1.4m0.1m
Analysis assuming the thop chord is continuous
2.8
3.8kNm
3.4
2.0
Bending moment of top chord
2.1
2.0
16
= 2715mm2
It is recommended to use equal angle in strut subjected to moment, because the
Mb formula is available in BS. The larger Zx is more beneficial to the Mb.
Try 100 × 100 × 15L (Ag = 35.6cm2)
rb = rx = ra = ry = 2.98, rv = 1.93cm, Zx = 35.6cm3
Section Classification
= (275 / 275)0.5
= 1.0
b/t = d/t = 100 / 15
= 6.67 < 15 = 15
(b + d)/t = (100 + 100)/15
= 13.33 < 24 = 24 Class 3 semi-compact
Table 11
Slenderness
Lb = Lx = L2 = 2.0m (top chord nodes spacing)
La = Ly = L1 = 1.5m (purlins spacing)
Lv = L1 = 2.0m
v
0.85Lv/rv = 0.85 × 2000 / 19.3 = 88.08
0.7Lv/rv + 15 = 0.7 × 2000 / 19.3 + 15 = 87.54 < 88.08 use 88.08
a
1.0La/ra = 1.0 × 1500 / 29.8 = 50.34
0.7La/ra + 30 = 0.7 × 1500 / 29.8 + 30 = 65.23 > 50.34 use 65.23
b
1.0Lb/rb = 0.85 × 2000 / 29.8 = 57.05
0.7Lb/rb + 30 = 0.7 × 2000 / 29.8 + 30 = 76.98 > 57.05 use 76.36
max = 88.08
Table 25
cl. 4.7.10.2a)
Compression capacity
Since max = 88.08 and py = 275 N/mm2, pc = 146N/mm2 and the
Table 23
Table 24 c)
17
Therefore the compression capacity of the angle (class 3)
Pc = Ag pc
= 3560 × 146 × 10-3
= 519.76kN > Fc = 112kN Ok
cl. 4.7.4
Linear Interaction Checking
- Heel of angle in compression:
Mb = 0.8pyZx = 0.8 × 275 × 35.6 × 10-3
= 7.83kNm
cl. 4.3.8.3
The column interaction expression
= = 0.777 < 1 Okcl. 4.7.7
b) Internal Vertical Member
Preliminary sizing
Fc = 52.8kN
py = 275 N/mm2
Assume pc = 0.4py
Ag = Fc/(0.4 py) = 52.8 × 103 / (0.4 × 275)
= 480mm2
Try 50 × 50 × 6L (Ag = 5.69cm2)
ra = rb = ry = rx = 1.5cm, rv = 0.963cm
cl. 4.5.2
cl. 4.8.3.2
Section Classification
= (275 / 275)0.5
= 1.0
b/t = d/t = 50 / 6
= 8.33 < 15 = 15
(b + d)/t = (50 + 50)/6
= 16.67 < 24 = 24 Class 3 semi-compact
Table 26
Slenderness
La = Lb = Lv = 1.0m
v
0.85Lv/rv = 0.85 × 1000 / 9.63 = 88.27
0.7Lv/rv + 15 = 0.7 × 1000 / 9.63 + 15 = 87.69 < 88.27 use 88.27
Table 18
18
Since one leg of the angle is bolted
a
1.0La/ra = 1.0 × 1000 / 15 = 66.67
0.7La/ra + 30 = 0.7 × 1000 / 15 + 30 = 76.67 > 66.67 use 76.67
b
0.85Lb/rb = 0.85 × 1000 / 15 = 56.67
0.7Lb/rb + 30 = 0.7 × 1000 / 15 + 30 = 76.67 > 93.41 use 76.67
max = 88.27
Compression capacity
Since max = 88.27 and py = 275 N/mm2, pc = 145N/mm2
Therefore the compression capacity of the angle (class 3)
Pc = Ag pc
= 569 × 145 × 10-3
= 82.50kN > Fc = 52.8kN Ok
c) Internal Sloped Member Design
Preliminary Sizing
Ft = 87.2kN
py = 275 N/mm2
Area needed = 87.2 × 103 / 275
= 317mm2
Try angle 45 × 45 × 5L where Ag = 4.3cm2
Tension Capacity
Assume the longer leg of the section welded to gusset; therefore the neutral axis
is eccentric away.
a1 = 45 × 5
= 225mm2
a2 = Ag - a1 = 430 - 225
= 205mm2
Pt = py (Ag – 0.3a2)
Table 11
19
= 275 × (430 – 0.3 × 205) × 10-3
= 101.33kN > Ft = 87.2kN Ok
c) Bottom Chord
Preliminary Sizing
Ft = 115.5kN
py = 275 N/mm2
Area needed = 115.5 × 103 / 275
= 420mm2
Although 40 × 40 × 5L has adequate cross-sectional area. To ease the erection,
we may use angle 50 × 50 × 6L where Ag = 5.69cm2 so that only one type of
internal members section being used.
Table 25
cl. 4.7.10.2a)
Tension Capacity
Assume the longer leg of the section welded to gusset; therefore the neutral axis
is eccentric away.
a1 = 50 × 6
= 300mm2
a2 = Ag - a1 = 569 - 300
= 269mm2
Pt = py (Ag – 0.3a2)
= 275 × (569 – 0.3 × 269) × 10-3
= 134.28kN > Ft = 115.5kN Ok
Table 23
Table 24 c)
cl. 4.7.4
20
Summary
Figure 3-18
To simplify the erection work, the 45 × 45 × 5L may be replaced with 50 × 50
×6L. Therefore the truss is using 100 × 100 × 100L for top chord, and 50 × 50 ×
6L for the rest remained.
cl. 4.6.3.1
cl. 4.6.3.1
Top chord 100 × 100 × 15L
Internal vertical members50 × 50 × 6L
Bottom chord 50 × 50 × 6L
Internal sloped members45 × 45 × 5L
21
3.12 Problems
1. Design a computer program that able to calculate the tension and compression
capacity using spreadsheets. The users are only required to enter the member length,
section properties and the end condition.
2. A tie member in a roof truss is acted with ultimate tensile force 1000kN. Choose the
minimum size equal angle (S 275) to resist the force.
3. A grade S 275 double-angle member 2/150 × 100 × 8L is connected back to back.
Two bolt holes with diameter 22mm are drilled through the longer heel at the ends.
Determine ultimate tensile capacity of the member.
4. A member in heavy truss is subjected to ultimate axial load 2000kN and ultimate
moment of 500kNm. Design the member using a mild steel universal beam.
5. A tie member is subjected to axial load and biaxial moment. The ultimate tensile force
is 3000kN; moment about major axis is 160kNm and moment about minor axis is
90kNm. Investigate whether 305 × 305 × 158UC (S 275) adequate.
6. An 18m-spanned flat roof is supported by trusses which have a height of 1.5m and
spaced 4m to each other. The purlins are located 1.5m spaced to each other too.
Assume the dead load is 0.7kN/m2 and imposed load is 0.75kN/m2,
a. Analysis the truss using method of joints
b. Design the truss using angles. Assume the connections between members are
to be welded.
7. A roof truss is shown in Figure 5-19 below. The trusses are placed 6m adjacent to each
other. The building height is 5m (to the eave) and overall length is 36m. The loadings
on roof are:
Dead load of roof = 0.4kN/m2 (on slope)
Imposed load = 0.75kN/m2 (on plan)
Estimate the wind load using CP3:Ch V:Part 2.
The building is located at country side and the basic wind speed is 45m/s.
a. Design the purlins
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b. Analyze the forces in the truss.
i. Design the trusses
Figure 5-11
8. Part of a building is shown in Figure 5-20 below. The trusses are supported by column
A and B; and part of the truss is cantilevering to the front. A roller door is placed
under the end of cantilever. The trusses are placed at a spacing of 6m to each other.
The overall length of the building is 48m.
The loadings on roof are:
Dead load = 0.45kN/m2 (on slope)
Imposed load = 0.75kN/m2 (on plan)
The building is located at country side and the basic wind speed is 45m/s.
Consider the truss will be subjected to the uplift force and gravity force. Analyze and
propose adequate members for the truss.
Figure 5-12
3 @ 2000 = 6000mm
2000mm
Purlin spacing 4 @ 1550 =6200mm
12m 6m
5m3m
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3.13 References
1. L. J. Morris, D. R. Plum (1988), Structural Steelwork Design to BS 5950, Longman
Scientific & Technical, UK.
2. BSI (2000), BS 5950-1:2000 Guide to Amendments, SCI, UK.
3. F. Arbabi (1991), Structural Analysis and Behavior, McGraw-Hill, Inc. USA.
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