Design Of Industrial Roof Truss And Analysis Of Member Forces 564a1i

SRES’ Sanjivani College of Engineering, Kopargaon.

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 of a truss, Detailing of typical ts and s.

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

6b) Roof Truss: Assessment of dead load, live load and wind load, Design of purlin, Design of of a truss, Detailing of typical ts and s. 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 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 subjected to axial forces are generally more efficient than in flexure since the cross section is uniformly stressed. Trusses frequently consist of axially loaded , 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)

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

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 ed 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 of roof truss other than PR and PT subjected to compressive force are termed as strut. d) Sling - The of roof truss other than PR and PT subjected to tensile force are termed as sling. e) Purlin- These are the flexural carrying the roof and roof covering loads and distributing it over truss . 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.

Strut P. Tie

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 trussb) Howe trussc) Fink trussd) Fan trusse) Fink fan trussf) Mansard truss-

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

(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/m2 is ( The weight of bracing may be assumed between 12 to 15 N/m2 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/m2 of plan area

> 10 °

Not Provided

0.75 kN/m2 of plan area

For roof membrane sheets or purlins the live load is to be calculated by 750-20(θ-10°) in N/m2 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 Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune)

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SRES’ Sanjivani College of Engineering, Kopargaon.

designed using the pressure difference between the opposite faces of such elements to for internal and external pressure exerted on the surface. The wind force F on element is obtained by-

F= (e-i) APd Where; e= External pressure coefficient i= Internal pressure coefficient A= Inclined area of roof member (m2) Pd= Design wind pressure (kN/m2) a) External pressure coefficient (e) - 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

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

Table 6.2b External pressure coefficients

b) Internal pressure coefficients (i) - 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 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;

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

Table 6.3 Basic wind speed

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

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

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

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.

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

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 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: Where; Cs has the following values:

k3= 1+Cs

Table 6.6 K3 factor Slope Cs 3°<θ<17° 1.2(z/L) θ > 17° 0.36

Figure 6.4 Topographical Dimensions

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

Numerical 6.1) Determine the design forces and design the 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/m2 and weight of AC sheets 130N/m2. 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. U3

U4

U2

3m U5

U1 θ

L0

L1

L2

L3

L4

L5

L6

Solution: 1) Truss Geometrya) Length of principal rafter (L0U3) = √ [(L0L3)2 + (U3L3)2] = √ [92 + 32] = 9.4868m. b) Length of each in sloping (L0U1, U1U2, U2U3) = (L0U3/No. of s) = (9.4868/3) = 3.1622m c) Inclination of principal rafter (θ) = tan -1 (3/9) = 18.45° d) Length of each in plan = 3.1622 cos 18.45° = 2.999 = 3m e) Plan area = (plan length x spacing of truss) = 3 x 4 = 12 m2 2) point loads due to dead loadWeight of AC sheets = 130N/m2 Weight of bracing = 80N/m2 Self-weight of truss = (

= (

…………………………………….………..(6.4 a)

= 110 N/m2 Total area load = (130+80+110) = 320 N/m2 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 s due to DL = 400+3840 = 4240 N = 4.3 kN. Final load on end s due to DL = (4.3/2) = 2.15 kN.

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

3) point loads due to live loadAs 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/m2 ……………………………………………………………..…………..(6.4 b) = 750-20(18.45°-10°) in N/m2 = 581 N/m2 Final live load on each intermediate = Live load x Plan area = 581 x 4 x 3 = 6972 N = 7kN. Final live load on end = (7/2) = 3.5 kN. 4) point loads due to wind loadAs 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) = 0.6x34.322 = 706.71 N/m2 Pd = 0.706 kN/m2 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 (e) for the condition

and θ=18.45° from Table

6.2a the coefficients can be estimated as; Inclination of Wind angle θ= 0° Wind angle θ= 90° principal EF GH EG FH rafter(18.45°) (windward) (lee ward) (windward) (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 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= (e-i) APd ………………………………………………………………………………………………………(6.4.3) Assuming normal permeability for the industrial building; i= ±0.2 A = 3.1622 x 4 = 12.65 m2 e = -0.8 Structural Design I (Steel Structures) Prof. Gayake Sudhir B. (SPPU, Pune)

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SRES’ Sanjivani College of Engineering, Kopargaon.

Pd = 0.706 kN/m2 Total wind load on = (-0.8 - -0.2) 12.56 x 0.706 = -5.32 kN (Suction) Total wind load on = (-0.8 - +0.2) 12.56 x 0.706 = -8.86 kN (Suction) Final wind load on intermediate = -8.86 kN Final wind load on end = - (8.86/2) = -4.43 kN

4.3kN

4.3kN

4.3kN 4.3kN

4.3kN

2.15 kN

2.15 kN

Figure 6.5 Final dead loads at points 7 kN 7 kN

7 kN

7 kN

7 kN 3.5 kN

3.5 kN

Figure 6.7 Final live loads at points

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

4.43 kN

4.43 kN

8.86kN

8.86kN 8.86kN

8.86kN 4.43 kN

4.43 kN

Figure 6.8 Final wind loads at points 5) Determining the member forces in L0U1 U1L1 L0L1 by method of ts 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 t 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) t L1: ∑Fy=0 FL1U1=0 …Zero force member

2.15 kN

L0

F L0U1

18.45°

F L0L1

12.9 kN

FL0L1= 32.25kN

FL1U1=0

FL1L2= 32.25kN

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

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

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SRES’ Sanjivani College of Engineering, Kopargaon.

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 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

0.00

0.00

0.00

0.00

0.00

0.00

32.25

52.45

-61.25

+127.05

-43.50

+28.14

FL1U1 FL1L2

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.

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

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