Design of industrial roof truss

Analysis of panel point loads due to DL LL and WL and Calculation of member forces by method of joint.

1. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 1


















Unit No. 6
6a) Design of Gantry Girder: Selection of gantry girder, design of cross section,
Check for moment capacity, buckling resistance, bi-axial bending,
Deflection at working load and fatigue strength.

6b) Roof Truss: assessment of dead load, live load and wind load,
Design of purlin, design of members of a truss,
Detailing of typical joints and supports.

2. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 2

6b) Roof Truss: Assessment of dead load, live load and wind load, Design of purlin, Design
of members of a truss, Detailing of typical joints and supports.

6.1 Introduction:
Industrial buildings are low rise structures characterized by their low height, lack of
interior floors, walls or partitions. The roofing system for such buildings is truss with roof
covering material. Trusses are triangular formations of steel sections in which the
members are subjected to essentially axial forces due to externally applied load.


Figure 6.1Plane Truss
Trusses are frequently used to span long lengths in the place of solid web girders.
When the external load lie in the plane of truss it is termed as plane trusses (figure 6.1)
whereas when the loads may lie in any three dimensional plane then such trusses are
termed as space trusses (figure 6.2).


Figure 6.2 Space Truss
Steel members subjected to axial forces are generally more efficient than members
in flexure since the cross section is uniformly stressed. Trusses frequently consist of axially
loaded members, thus are very efficient in resisting these loads. They are extensively used,
especially to span large gaps. Usually trusses are adopted in roofs of single storey industrial
buildings, long span floors, to resist gravity loads. Trusses are also used in long span
bridges to carry gravity loads and lateral loads.
6.2 Components parts of roof truss:
Below given are the important component parts of industrial roof truss. (figure 6.3)

3. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 3

a) Principal Rafter (PR) - It is the top chord member of truss subjected to only
compressive force due to gravity load if the purlins are supported at nodes. If the
purlins are intermediate of nodes then the PR will be subjected to bending moment.
b) Principal Tie (PT) – The lower chord of truss is known as principal tie and carries
only tension due to gravity loads.
c) Strut – The members of roof truss other than PR and PT subjected to compressive
force are termed as strut.
d) Sling - The members of roof truss other than PR and PT subjected to tensile force
are termed as sling.
e) Purlin- These are the flexural members carrying the roof and roof covering loads
and distributing it over truss members.
f) Bracings- The member of truss which makes it stable for accidental loads, out of
plane loads or lateral loads is termed as bracing system.

Figure 6.3 Components of truss
6.3 Types of roof trusses:
Depending upon the span of truss, requirements of elegance, depending upon
demands of particular building and the ventilation requirements the types of roof trusses
are classified as below(figure 6.3)-
a) Pratt truss-
b) Howe truss-
c) Fink truss-
d) Fan truss-
e) Fink fan truss-
f) Mansard truss-
Strut
P. Tie

4. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 4


(Figure 6.3)- Types of trusses.
6.4 Loads on roof trusses:
Roof trusses are mainly subjected to Dead load, Live load and wind load.
1. Dead Load (DL) - The DL of the truss includes the weight of roofing material,
purlins, bracings, and truss load. The unit weight of different material are given in
IS875 part I. An empirical formula to calculate the approximate dead weight of truss
in N/m
2
is (


The weight of bracing may be assumed between 12 to 15
N/m
2
of the plan area. The design of purlin based on the roofing material load is
already done therefore the weight of purlin can be considered for the design
directly.
2. Live Load (LL) - IS 875 gives the live loads acting on inclined roof truss depending
upon the inclination of PR and access provided or not above the roof.
Table 6.1 Live load values
Roof Slope Access Live Load
≤ 10 ° Provided 1.5kN/m
2
of plan area
> 10 ° Not Provided 0.75 kN/m
2
of plan area
For roof membrane sheets or purlins the live load is to be
calculated by 750-20(θ-10°) in N/m
2

3. Wind Load (WL) – Wind load are most critical loads in design and analysis of
industrial roof truss. The design wind pressure for roofs or wall cladding must be

5. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 5

designed using the pressure difference between the opposite faces of such elements
to account for internal and external pressure exerted on the surface.
The wind force F on element is obtained by-
F= (cpe-cpi) APd
Where;
cpe= External pressure coefficient
cpi= Internal pressure coefficient
A= Inclined area of roof member (m
2
)
Pd= Design wind pressure (kN/m
2
)
a) External pressure coefficient (cpe) - The average external pressure coefficients and
pressure concentration coefficients for pitched roofs of rectangular clad building shall be as
given in Table 5of IS 875 Part 3. Where no pressure concentration coefficients are given,
the average coefficients shall apply.
Table 6.2a External pressure coefficients

6. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 6

Table 6.2b External pressure coefficients

b) Internal pressure coefficients (cpi) - Internal air pressure in a building depends upon
the degree of permeability of cladding to the flow of air. The internal air pressure may be
positive or negative depending on the direction of flow of air in relation to openings in the
buildings. In the case of buildings where the claddings permit the flow of air with openings
not more than about 5 percent of the wall area but where there are no large openings, it is
necessary to consider the possibility of the internal pressure being positive or negative.
Two design conditions shall be examined, one with an internal pressure coefficient of +0.2
and another with an internal pressure coefficient of -0.2.
c) Inclined Area (A) – Inclined area is calculated using spacing of truss multiplied by the
panel length of principal rafter.
d) Design wind pressure (Pd) - The design wind pressure at any height above mean
ground level shall be obtained by the following relationship between wind pressure and
wind velocity:
pz = 0.6 v z
2

where ;
pz = design wind pressure in N/m2 at height z, and
v= design wind velocity in m/s at height z.
e) Design wind speed (Vz)- The basic wind speed ( V) for any site shall be obtained from
Fig. 1 (IS 875 Part 3) and shall be modified to include the following effects to get design
wind velocity at any height ( Vz) for the chosen structure:
a) Risk level;
b) Terrain roughness, height and size of structure; and
c) Local topography.
It can be mathematically expressed as follows:
Vz= Vb k1 k2 k3
Where;
Vz = design wind speed at any height z in m/s; and for Vb refer below table;

7. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 7

Table 6.3 Basic wind speed

8. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 8

K1 = probability factor (risk coefficient) (see 5.3.1) (IS 875 Part 3) refer below table;
Table 6.4 Risk coefficients

9. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 9


K2 = terrain, height and structure size factor (see 5.3.2) (IS 875 Part 3)-
Terrain – Selection of terrain categories shall be made with due regard to the effect
of obstructions which constitute the ground surface roughness. The terrain category used
in the design of a structure may vary depending on the direction of wind under
consideration. Wherever sufficient meteorological information is available about the
nature of wind direction, the orientation of any building or structure may be suitably
planned. Terrain in which a specific structure stands shall be assessed as being one of the
following terrain categories:
Category 1 – Exposed open terrain with few or no obstructions and in which the
average height of any object surrounding the structure is less than 1.5 m.
Category 2 – Open terrain with well scattered obstructions having heights generally
between I.5 to 10 m.
Category 3 – Terrain with numerous closely spaced obstructions having the size of
building-structures up to 10 m in height with or without a few isolated tall structures.
Category 4 – Terrain with numerous large high closely spaced obstructions.
Table 6.5 k2 factor

The buildings/structures are classified into the following three different classes
depending upon their size:
Class A - Structures and/or their components such as cladding, glazing, roofing, etc.,
having maximum dimension (greatest horizontal or vertical dimension) less than 20 m.
Class B - Structures and/or their components such as cladding, glazing, roofing, etc.,
having maximum dimension(greatest horizontal or vertical dimension) between 20 to50m.
Class C - Structures and/or their components such as cladding, glazing, roofing, etc.,
having maximum dimension (greatest horizontal or vertical dimension) greater than 50 m.

10. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 10


K3 = topography factor (see 5.3.3) (IS 875 Part 3).
The basic wind speed Vb given in Fig. 1(IS 875 Part 3) takes account of the general
level of site above sea level. This does not allow for local topographic features such as hills,
valleys, cliffs, escarpments, or ridges which can significantly affect wind speed in their
vicinity. The effect of topography is to accelerate wind near the summits of hills or crest of
cliffs, escarpments or ridges and decelerate the wind in valleys or near the foot of cliff,
steep escarpments, or ridges.
The effect of topography will be significant at a site when the upwind slope (θ)
(figure 6.4) is greater than about 3°, and below that, the value of k3 may be taken to be
equal to 1. The value of k3 is confined in the range of 1 to 1.36 for slopes greater than 3°. A
method of evaluating the value of k3 for values greater than 1.0 is given in Appendix C (IS
875 Part 3). It may be noted that the value of k3 varies with height above ground level, at a
maximum near the ground, and reducing to 1.0 at higher levels. The topography factor k3 is
given by the following:
k3= 1+Cs
Where; Cs has the following values:
Table 6.6 K3 factor
Slope Cs
3°<θ<17° 1.2(z/L)
θ > 17° 0.36

Figure 6.4 Topographical Dimensions

11. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 11

Numerical 6.1) Determine the design forces and design the members L0U1 U1L1 L0L1 of an
truss where access is not provided and it is located in city area of Nashik. Assume c/c
spacing of truss 4m. Assume self-weight of purlin 100N/m, weight of bracing 80N/m
2
and
weight of AC sheets 130N/m
2
. Take Rise of truss 3m. The Length of shed is 38m and width
is 18m and consider design life of 50 years. Height of building up to eves is 10m.





L0 L1 L2 L3 L4 L5 L6

Solution:
1) Truss Geometry-
a) Length of principal rafter (L0U3) = √ [(L0L3)
2
+ (U3L3)
2
] = √ [9
2
+ 3
2
] = 9.4868m.
b) Length of each panel in sloping (L0U1, U1U2, U2U3)
= (L0U3/No. of panels) = (9.4868/3)
= 3.1622m
c) Inclination of principal rafter (θ) = tan
-1
(3/9) = 18.45°
d) Length of each panel in plan = 3.1622 cos 18.45° = 2.999 = 3m
e) Plan area = (plan length x spacing of truss) = 3 x 4 = 12 m
2

2) Panel point loads due to dead load-
Weight of AC sheets = 130N/m
2

Weight of bracing = 80N/m
2
Self-weight of truss = (


= (


…………………………………….………..(6.4 a)
= 110 N/m
2

Total area load = (130+80+110) = 320 N/m
2

Plan load = Total area load x Plan area = 320 x 4 x 3 = 3840 N
Weight of purlin = Self weight of purlin x Spacing of truss
= 100 x 4 = 400 N
Final load on all intermediate panels due to DL = 400+3840 = 4240 N = 4.3 kN.
Final load on end panels due to DL = (4.3/2) = 2.15 kN.

θ
U1
U2
U3
U4
U5
3m

12. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 12

3) Panel point loads due to live load-
As Inclination of principal rafter (θ) = tan
-1
(3/9) = 18.45° and the access is not provided to
roof then live load is calculated as;
Live load = 750-20(θ-10°) in N/m
2
……………………………………………………………..…………..(6.4 b)
= 750-20(18.45°-10°) in N/m
2

= 581 N/m
2

Final live load on each intermediate panel = Live load x Plan area
= 581

x 4 x 3 = 6972 N = 7kN.
Final live load on end panel = (7/2) = 3.5 kN.
4) Panel point loads due to wind load-
As the industrial shed is having design life of 50 years and it is located in city area of nashik
region, given data suggest the following conclusions,
Vb = 39 m/s ……………………………………………………………………………………………………..(Table 6.3)
K1 = 1.0 …………………………………………………………………………………………………….……..(Table 6.4)
K2 = 0.88 ………………………………………………………………………………………………..………..(Table 6.5)
K3 = 1.0 ……………………………………………………………………………………………………….…..(Table 6.6)
Vz= Vb k1 k2 k3 ……………………………………………………………………………………… ………………………… …………………………………….…. (6.4.3 e)
= 39x1.0x0.88x1.0 = 34.32 m/s
Design wind pressure = pz = 0.6 v z
2
……………………………………………………………………(6.4.3 d)
= 0.6x34.32
2
= 706.71 N/m
2

Pd = 0.706 kN/m
2

Height of the building is = 10m above the ground level and width of building is 18m.
(h/w) = (10/ 18) = 0.55

The external pressure coefficients (Cpe) for the condition








and θ=18.45° from Table
6.2a the coefficients can be estimated as;
Inclination of
principal
rafter(18.45°)
Wind angle θ= 0° Wind angle θ= 90°
EF
(windward)
GH
(lee ward)
EG
(windward)
FH
(lee ward)
10 -1.1 -0.6 -0.8 -0.6
20 -0.7 -0.5 -0.8 -0.6
18.45° -0.762 -0.515 -0.8 -0.6
Among the above calculated coefficients the wind load on panel point can be calculated
separately considering the wind ward and lee ward effect or else the maximum worse
effect(maximum coefficient) among all can be approximately used to calculate the
maximum force as given;
F= (cpe-cpi) APd ………………………………………………………………………………………………… ……(6.4.3)
Assuming normal permeability for the industrial building; cpi= ±0.2
A = 3.1622 x 4 = 12.65 m
2
cpe = -0.8

13. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 13

Pd = 0.706 kN/m
2

Total wind load on panel = (-0.8 - -0.2) 12.56 x 0.706 = -5.32 kN (Suction)
Total wind load on panel = (-0.8 - +0.2) 12.56 x 0.706 = -8.86 kN (Suction)
Final wind load on intermediate panel = -8.86 kN
Final wind load on end panel = - (8.86/2) = -4.43 kN











Figure 6.5 Final dead loads at panel points








Figure 6.7 Final live loads at panel points


7 kN
2.15 kN 2.15 kN
4.3kN
4.3kN
4.3kN
4.3kN
4.3kN
3.5 kN 3.5 kN
7 kN
7 kN
7 kN
7 kN

14. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 14











Figure 6.8 Final wind loads at panel points
5) Determining the member forces in L0U1 U1L1 L0L1 by method of joints for all types of
loadings.
a) Member forces due to dead load-
∑Fy=0
RL0 + RL6 = (2.15X2 + 4.3X5) = 25.8 kN
RL0 = RL6 = (25.8/2) = 12.9kN
Joint L0:
∑Fy=0
12.9-2.15+ FL0U1 sin 18.45°=0
FL0U1= -34kN (Compressive)
∑Fx=0
F L0L1 + FL0U1 cos18.45°=0
FL0L1 = (34 cos18.45°) = 32.25 kN (Tensile)
Joint L1:
∑Fy=0
FL1U1=0 …Zero force member

∑Fx=0
- FL0L1 + FL1L2 = 0
FL1L2 = 32.25 kN (Tensile)

FL1U1=0
FL0L1= 32.25kN FL1L2= 32.25kN
L0
12.9 kN
F L0L1
F L0U1
2.15 kN
18.45°
8.86kN
8.86kN
8.86kN
8.86kN
4.43 kN 4.43 kN
4.43 kN
4.43 kN

15. SRES’ Sanjivani College of Engineering, Kopargaon.

Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune) Page 15

b) Member forces due to Live load- Similarly performing calculations for live load
the member forces are calculated as;
FL0U1= -55.296kN (Compressive)
FL0L1 = 52.45 kN (Tensile)
FL1L2 = 52.45 kN (Tensile)
FL1U1=0
c) Member forces due to Wind load- Similarly performing calculations for wind load
the member forces are calculated as;
FL0U1= 66.09 kN (Tensile)
FL0L1 = -61.25 kN (Compressive)
FL1L2 = -61.25 kN (Compressive)
FL1U1=0
Design force table
Members
Member force (kN) due to Design Forces (kN)
DL LL WL
1.5
(DL +LL)
1.5
(DL +WL)
1.2
(DL + LL+WL)
FL0U1 -34.00 -55.29 66.09 -133.93 +48.13 -27.840
FL0L1 32.25 52.45 -61.25 +127.05 -43.50 +28.14
FL1U1 0.00 0.00 0.00 0.00 0.00 0.00
FL1L2 32.25 52.45 -61.25 +127.05 -43.50 +28.14

By considering above forces and their nature i.e. negative force as compression
member whereas positive force as tension member can be designed as per given in Unit
No1 and Unit No2.