Chapter 31 Trusses

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


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

[email protected]

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


Since max = 149.25 and py = 275 N/mm2, pc = 68N/mm2 and the


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


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)


Since max = 88.08 and py = 275 N/mm2, pc = 146N/mm2 and the

Table 23

Table 24 c)

17


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


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


Since max = 88.27 and py = 275 N/mm2, pc = 145N/mm2


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


Tension Capacity


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


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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Documents

Chapter 31 Trusses