Design of Bottom Chord Bracing & Tie Runner :
1.Analysis:
Portion of wind load from gable end along the ridge will be transferred as axial load to
tie runners provided along the length of the building at tie level.
i. Calculation of Loads
Design wind pressure = 1.176 kN/m2
Maximum Wind Co-efficient = 1.2
2.062 4.126 2.062
Tie Runner Truss
A B C D
2
5.000 Tie Bracing
3
Truss
Plan
1.4
Cladding
0.000
8.25
Elevation
Load on Node 'A & D' = 1.2 x 1.176 x 1.031 x 0.000
= 0.000 kN
Load on Node 'B & C' = 1.2 x 1.176 x 3.094 x 0.700
= 3.056 kN
ii. Calculation of Forces in Members
0.000 3.056 3.056 0.000
2.062 4.126 2.062
A B C D
5.000
F H
E G
3.056 3.056
Consider Node "A" :
0.000
A AB
67.589 Deg.
AF
AE
FAE = 0.000 + FAF * Sin ( 67.589 )
= 0.000 + FAF * 0.924 1
FAB = FAF * Cos ( 67.589 )
= FAF * 0.381 2
Consider Node "E" :
AE
EB
67.589 Deg.
E AB
3.056
FAB = FEB * Cos ( 67.589 )
= FEB * 0.381 3
FAE = 3.056 - FEB * Sin ( 67.589 )
= 3.056 - FEB * 0.924 4
From Eq. 1
0.000 + FAF * 0.924 = 3.056 - FEB * 0.924
Assume FAF = FEB
1.849 * FEB = 3.056
FEB = 1.653 kN C
FAF = 1.653 kN T
FAB = 0.630 kN C
FAE = 1.528 kN C
2. Design :
i. Design of Bracing :
a. Design as a Comp. Member
Total Length of Bracing ( l ) = 6.483 m
Half the Length = 3.241 m
Since, these are tension members, ( l/r) max. = 350 ( IS:800, Table No.3.1)
rmin. required = (3.242*10^3)/350
= 9.261 mm
Provide ISA50 x 50 x 6
rmin. provided = 15.10 mm
Hence OK
b. Design as a Tension Member :
Permissible stress in Tension = 0.6 * fy
Permissible stress = 150 N/mm2
Required area of member = (1.653*10^3)/150
= 11.018 mm2
Angles are connected to a gusset plate and In welded construction net area is same as gross area.
Area of angle = 568 mm2
Effective area (Ae) = A1 + k * A2
A1 = 284 mm2
A2 = 284 mm2
k = 0.75 , for angles connected on one side of a gusset.
Effective area (Ae) = 497 mm2
Load carrying capacity = (150 * 497) / 1000
= 74.550 kN
Hence OK
ii. Design of Tie runner:
a. As per ' l/r ' ratio :
Length = 5.000 m
Since, these are tension members, ( l/r) max. = 350 ( IS:800, Table No.3.1)
rmin. required = (5*10^3)/350
= 14.286 mm
Provide 2 ISA 50 x 50 x 6 Star Angles.
rmin. provided = 18.23 mm
Hence OK
b. Design as a Tension Member :
Permissible stress in Tension = 0.6 * fy
Permissible stress = 150 N/mm2
Required area of member = (497*10^3)/150
= 3313.333 mm2
Angles are connected to a gusset plate and In welded construction net area is same as gross area.
Area of angle = 1014 mm2
Effective area (Ae) = A1 + k * A2
A1 = 507 mm2
A2 = 507 mm2
k = 0.75 , for angles connected on one side of a gusset.
Effective area (Ae) = 887.25 mm2
Load carrying capacity = (150 * 887.25) / 1000
= 133.088 kN
Hence OK
Design of Rafter Bracing :
1. Analysis :
For the Design of Rafter Bracing, IS:800-1984 (cl.6.6.6.1) recomends the following.
In the case of series of latticed beams, girders or roof truss which are connected together by
the same system of restraint members, the sum of the restraining forces required shall be taken as
2.5% of the maximum compression flange plus 1.25% of this this force for every member of the
series other than first upto maximum total of 7.5%.
Rafter bracing shall be provided in end panels as shown in fig. below. The Purlins shall be
assumed to give lateral support to other trusses.
1.4
0.00 Cladding
8.25
Elevation
4.356 4.356 Truss
Rafter Bracing
5.000 Ridge Purlin
Purlin
Truss
Plan of Top Chord Bracing
i. Calculation of Loads
Design wind pressure = 1.176 kN/m2
Maximum wind pressure co-efficient = 1.2
Load on Node 'A & C' = 1.2 x 1.176 x 2.178 x 0.370
= 1.136 kN
Load on Node 'B' = 1.2 x 1.176 x 4.356 x 1.400
= 8.606 kN
1.136 8.606 1.136
4.356 4.356
A
B C
5.000
D E F
5.439 5.439
ii. Calculation of Forces in Members
Consider Node "A" :
1.136
A AB
48.938 Deg.
AE
AD
ΣV = 0:
FAD = 1.136 + FAE * Sin ( 48.938 )
= 1.136 + FAE * 0.754 1
ΣH = 0:
FAB = FAE * Cos ( 48.938 )
= FAE * 0.657 2
Consider Node "D" :
DA
DB
D 48.938 Deg.
DE
5.439
ΣH = 0:
FDE = FDB * Cos ( 48.938 )
3
= FDB * 0.657 3
ΣV = 0:
FDA = 5.439 - FDB * Sin ( 48.938 )
= 5.439 - FDB * 0.754 4
From Eq. 1
1.136 + FAE * 0.754 = 5.439 - FDB * 0.754
Assume FAE = FDB
1.508 * FEB = 4.303
FDB = 2.853 kN C
FAE = 2.853 kN T
FAB = 1.874 kN C
FAD = 3.288 kN C
iii. Shear due to Rafter :
Maximum Comp. Force in Rafter = 57.03 kN (From STAAD)
Shear in rafter Bracing = (2.5 + 2*1.25) x 57.03
100
= 2.8515 kN
Total Shear = i + ii = 2.853 + 2.852
= 5.705 kN
2. Design :
i. As per 'l/r' ratio :
Length of bracing member = 4.356 ^2 + 5.000 ^2
= 6.631 m
rxx required = 6.631 x 1000.0
350.0
= 18.947 mm
Ptovid 65 x 65 x 6
rxx provided = 19.400 mm
Hence OK
ii. Design as a Tension Member :
Permissible stress in Tension = 0.6 * fy
Permissible stress = 150 N/mm2
Required area of member = (5.705*10^3)/150
= 38.033 mm2
Angles are connected to a gusset plate and In welded construction net area is same as gross area.
Area of angle = 744 mm2
Effective area (Ae) = A1 + k * A2
A1 = 372 mm2
A2 = 372 mm2
k = 0.75 , for angles connected on one side of a gusset.
Effective area (Ae) = 651 mm2
Load carrying capacity = (150 * 651) / 1000
= 97.650 kN
Hence OK
Design of Welds :
Member 50 x 50 x 6
Area of the angle = 568 mm2
Permissible tensile stress = 150 N/mm2
Load carrying capacity of angle = 85.20 kN
85.200
35.50
14.5
L1
L2
50
Maximum size of weld = 0.75 * Thickness of member
Required Max.size of weld = 4.5 mm
Provided size of weld = 6 mm
Permissible shear stress in the weld = 110 N/mm2
Actual shear capacity of weld = 0.707*6*110
= 0.467 kN/mm
Effective Length of weld required = 85.2 / 0.467
= 182.590 mm
Let us distribute the welds in such a way that the c.g. of the welds coinsides with
the c.g. of the angle section.
50 + L1 + L2 = 182.590
Taking moments about ' L1 ' edge
0.467 x L2 x 50 + 50 x 25 = 1235
x x
L2 x 50 + 50 25 = 2647.550
50 x L2 = 1397.550
L2 = 27.951 mm
L1 = 105 mm
Total Length of weld = Effective length + 2*size of weld
Total length of weld (L 1) = 116.64 mm Say 180
Total length of weld (L 2) = 39.95 mm Say 100
Design of Welds :
Member 65 x 65 x 6
Area of the angle = 744 mm2
Permissible tensile stress = 150 N/mm2
Load carrying capacity of angle = 111.6 kN
111.600
46.90
18.10
L1
L2
65
Maximum size of weld = 0.75 * Thickness of member
Required Max.size of weld = 4.5 mm
Provided size of weld = 6 mm
Permissible shear stress in the weld = 110 N/mm2
Actual shear capacity of weld = 0.707*6*110
= 0.467 kN/mm
Effective Length of weld required = 111.6 / 0.467
= 239.167 mm
Let us distribute the welds in such a way that the c.g. of the welds coincides with
the c.g. of the angle section.
65 + L1 + L2 = 239.167
Taking moments about ' L1 ' edge
0.467 x L2 x 65 + 65 x 32.5 = 2019.96
x x
L2 x 65 + 65 x 32.5 = 4328.919
65 x L2 x = 2216.419
L2 = 34.099 mm
L1 = 140 mm
Total Length of weld = Effective length + 2*size of weld
Total length of weld (L 1) = 152.07 mm Say 180
Total length of weld (L 2) = 46.10 mm Say 80
