ALL ABOUT CIVIL & MECHANICAL ENGINEERING: 2016

DESIGN OF PRATT TRUSS

CALCULATION OF LOADS

Span of truss = 10m

Slope of the sheeting = 30⁰

Number of panels = 3 (on both sides)

Spacing of truss = 1/3^rd – 1/5^thof span

≈ 4m

3.1 LIVE LOAD

Imposed load,

= 0.75 – (0.02 x 20)

= 0.35 KN/m².

As per the IS code imposed load should not be less than 0.4 KN/m.

≈ 0.4 KN/m².

Imposed load in horizontal for trusses,

= 2/3^rd of imposed load

= 2/3 x 0.4

= 0.267 KN/m².

3.2 WIND LOAD

Basic wind speed at10m height for the region “Coimbatore”,

V_b = 9m/sec

The design wind speed,

V_z = V_b x k₁ x k₂ x k₃

Where,

K₁= 1.0

K₂ = 0.91

K₃ = 1

V_z = 39 x 1 x 0.91 x 1

= 35.49 m/sec

Design wind pressure,

P_z = 0.6 x V_z²

= 0.6 x ( 35.49)²

= 755.72 N/m².

Wind pressure on roof ,

F = ( Cpe - Cpi )Pz

WIND NORMAL TO THE RIDGE	TOTAL PRESSURE, F = ( Cpe - Cpi )Pz
Windward cpe = -0.2	Cpi= + 0.5	Cpe = - 0.5
Windward cpe = -0.2	- 529.025	- 226.716
Leeward = -0.5	- 755.72	0
WIND PARALLEL TO RIDGE
Windward cpe = -0.8	- 982.436	-227.716
Leeward = -0.8	- 982.436	-227.716

Maximum wind load = - 982.436 N/m²uplift on both slopes.

3.3 DEAD LOAD

Dead load of sheeting = 112.7 N/m^.

DESIGN OF PURLIN

Spacing of Purlin = 1.92m

Weight of 20 gauge (GI sheets) = 112.7 N/m²

Load on purlin per meter length

1. Weight on sheeting = 112.7 x 1.9

= 216.38 N/m

2. Weighting on purlin (assume) = 100 N/m

Total dead load = 316.38N/m

Imposed load = 400 x 1.92 Cos 30 ̊

= 665.11 N/m

Wind load = - 982.436 x 1.92

= 1886.27 N/m

Dead load + Imposed load = 316.38 + 665.11

= 981.48N/m

Dead load + wind load = 316.38 – 1886.27

= 1569.89 N/m

Since increase in permissible stress is 33.33%, when wind load is considered, D.L + W.L may consider 33.33% less effective.

1886.27/1.33 = 1418.25 N/m> 981.48 N/m

Maximum bending moment = w l ²/10

= (141.25 x 4 ²)/10

= 2269.2Nm

= 2269.2KNmm

For angle purlin Zx required = (2269.2 x 10³⁾/1.65

= 13752.72mm³

= 13.752 x 10³ mm³

Trial section;

Minimum depth = L/45

= 4000/45

= 88.89 mm

Minimum width = L/60

= 4000/60

= 66.67 mm

Providing the section ISA 90 x 90 x 8 mm @ 108 N

Zx provided = Zx required

16.0 x 10³ mm³ = 13.752 x 10³ mm³

Hence it is safe

LOADS ON TRUSSES

1. Dead load (assumed to be acting on top panel)

GI sheets = 112,7 x 4 x (10/cos30 ̊)

= 5205.4N

Purlin = 108 x 4 x 8

= 3456N

Truss (assume 100 N/m²)

= 100 x 4 x 10

= 4000N

Total dead load = 12661.4N

Dead load on end panel points.

= 2110.23/2

= 1055.11N

2. Imposed load:

For truss on top panel point

= 267 x 4 x 10

= 10680N

Imposed load per panel = 10680/6

= 1780N

Imposed load at two ends = 890N

3.Wind load on each of the top panel

Uplift force:

= - 982.436 x 1.92 x 4

= - 7545.11N

Download force = o

The force due to imposed load are calculating by force due imposed load by dead load .

= imposed load / dead load

= 1780 / 2110.23

Effective length of Prismatic compression member

Column pin-ended at Both Ends l_e = 1.0l

Column pin-ended at One End and fixed at the other le = 0.8l

Column Fixed at Both Ends le = 0.65l

Column Fixed at One End and on roller support at the other,le = 1.2l

Column Fixed at One End and Free at the other le = 2l

Using limit state method the design force are calculated as follows are

Principle rafter:

1.5 x – 19.44 = - 29.16KN (compression)

1.5x – 32.74 = + 49.11KN (tension)

Main tie:

1.5x – 11.281 = - 16.92 (compression)

1.5x – 16.829 = +25.24 (tension)

Main sling:

1.5x – 11.281 = -16.92KN (compression)

1.5 x +6.806 = +10.209KN (tension)

Main strut:

1.5 x -5.894 = - 8.841KN (compression)

1.5 x +10.066 = +15.099 (tension)

Minor sling:

1.5 x – 8.671 = - 13KN (compression)

1.5 x +5.075 = + 7.612KN (tension)

Minor strut:

1.5 x – 3.888 = - 5.832KN (compression)

1.5 x + 6.643 = + 7.612KN (tension)

1.Principle rafter:

Cross area of angles, Ag = 2 x 568

= 1136mm²

Strength governed by yielding:

(Ag.fy)/ϒ_mo = (1136 x 250)/1.1

= 258.18KN

Compression strength:

Effective length(KL) = 0.7 x L

= 0.7 x 5.8

= 4060mm

Iy = 2[12.9 x 10⁴ + 568(14.5 + 4)²

= 646796mm⁴

r_y = √(646796/1136)

= 23.8mm

KL/r = 4060/23.8

= 170.58

For KL/r = 170 ; fcd = 48.1 N/mm²

For KL/r = 180 ; fcd = 43.9 N/mm²

For KL/r = 170.58

Fcd = 48.1 –[ ((170.58 – 170)/10)(48.1 – 43.9)]

= 47.85 N/mm²

Pd = 2 x 568 x 47.85

= 54.35 KN

2. Main tie :

Design load = - 16.92KN(compression)

= + 25.24KN(tension)

Tension strength:

(Ag.fy)/ϒ_mo = (1136 x 250)/1.1

= 258.18KN

Compression strength:

Effective length = 0.7 x 5000

KL = 3500mm

Iy = 2[12.9 x 10⁴ + 568(14.5 + 4)²

= 646796mm⁴

r_y = √(646796/1136)

= 23.8mm

KL/r = 3500/23.8

= 147.05

KL/r = 140 ; fcd = 66.2

KL/r = 150 ; fcd = 59.2

For KL/r = 147.05

Fcd = 66.2 – [((147.05-140)/10) x (66.2 – 59.2)]

= 61.26 N/mm²

Pd = 2 x 568 x 61.26

= 60.59KN.

3. Main sling:

Design load = - 16.92KN(compression)

= +10.209KN(tension)

Try the section ISA 50 x 50 x 6mm

Tension strength = 258.18KN

Compression strength:

Effective length = 0.7 x 3330

KL = 2310mm

r_y = √I/A

= 23.8mm

KL/r = 2310/23.8

= 97.05 N/mm²

KL/r = 90 ; fcd = 121 N/mm²

KL/r = 100 ; fcd = 107 N/mm²

For KL/r = 97.05

Fcd = 121 – [((97.05 – 90)/10) x (121 – 107)

= 111.13 N/mm²

Pd = 2 x 568 x 111.13

= 126.24KN

4.Main strut:

Design loads = - 8.84KN

= + 15.099KN

Try the section ISA 40 x 40 x 6mm

Tension strength:

(Ag.fy)/ϒ_mo

Ag = 2 x 447

= 894mm²

= (894 x 250)/1.1

= 203.18KN

Tension strength = 203.18KN

Compression strength:

ffective length KL = 0.7 x 1920

= 1344mm

From steel table r_y = 11.9mm

KL/r = 1344/11.9

= 112.94

KL/r = 110 ; fcd = 94.6 N/mm²

KL/r = 120 ; fcd = 83.7 N/mm²

For KL/r = 112.94

Fcd = 94.6 – [((112.94 – 110)/10) x (94.6 – 83.7)]

= 91.39 N/mm²

Pd = 894 x 91.39

= 81.70KN

5. Minor sling:

Design loads = - 8,671KN

= + 5.075KN

Tension strength = 203.18KN

Compressive strength:

Effective length KL = 0.7 x 2560

= 1792mm

From steel table r_y = 11.9mm

KL/r = 1792/11.9

= 150.58

KL/r = 150 ; fcd = 59.3 N/mm²

KL/r = 160 ; fcd = 53.3 N/mm²

For KL/r = 150.58

Fcd = 59.2 – [((150.88 – 150)/10) x (59.2 – 53.3)]

= 58.68 N/mm²

Pd = 58.68 x 894

= 52.46KN

Since the other members are not severely loaded, a single ISA 40 x 40 x 6mm will be sufficient for minor sling and minor strut.

DESIGN OF JOINTS

Design loads = 32.74KN

Factored load = 1.5 x 32.74

= 49.11KN

Thickness of weld:

1. Size of the weld = ¾ x 8

= 6mm

2. At top thickness should not exceed

S = t – 1.5

= 8 – 1.5

= 6.5mm

Size of weld = 6mm

Each load carry a factored pull of

= 49.11/2

= 24.55KN

Let l_w be the total length of the weld required assuming normal weld

t = 0.7 x 6

design strength of weld = L_w x t x (fu/√3) x (1/1.25)

= L_wx 0.7 x 6 x (410/√3) x (1/1.25)

Equating it to the factored load. We get,

24.55 x 10³ = L_wx 0.7 x 6 x (410/√3) x (1/1.25)

L_w = 30.86mm

Centre of gravity of the section is at a distance 15mm from top

L₁x 1.5 = L₂ x (50 – 15)

L₁ = 35/15 x L₂

L₁+ L₂ = 30.86mm

L₂ + (35/15) x L₂ = 30.86

L₂ = 9.26mm

≈ 10mm

L₁ = 21.6mm

≈ 25mm

REFERENCE

1. S.S. BHAVIKATTI ‘Design Of Steel Structures’2^nd Edition, I.K. International Publishing House Pvt.Ltd.

2. L.S.NEGI ‘Design Of Steel Structures’2^nd Edition, Tata McGraw-Hill Publishing Company Limited, New Delhi.

3. Dr.B.C.PUNMIA, Ashok Kumar Jain, Arun Kumar Jain ‘Design Of Steel Structures’2^nd Edition, Lakshmi publications (P) Ltd, New Delhi.

4. IS 800-2007, Indian Standard General Construction In Steel(Code Of Practice)Third Revision, Bureau Indian Standards, New Delhi .

5. IS 875-1987 PART – I, Dead Load ( Code of practice for design loads (other than earthquake) for buildings and structures, Second Revision, Bureau Indian Standards, New Delhi.

6. IS 875-1987 PART – II Live Load (Code of practice for design loads (other than earthquake) for buildings and structures), Second Revision, Bureau Indian Standards, New Delhi.

7. IS 875-1987 PART - III Wind Load (Code of practice for design loads (other than earthquake) for buildings and structures), Second Revision, Bureau Indian Standards, New Delhi.

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