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factored compressive load =335.175 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa
depending on .Assuming fcd=100 N/mm
2
Ac=
mm2
Since we will be using double angle on the same side of the gusset plate.
so area of each angle = 3351.75/2 = 1675.875 As this is the double angle rafter so the section must be joined on a single side of a gusset plate of
thickness 8mm and the dimension of the angle is 11011012 @ 19.7 kg/m. The angles are joinedby fillet welding.
As per IS 800:2007 SECTION 7 clause no. 7.5.1.2and 7.1.2.1 and table 5, 7, 10, 12
c=2 2
1 2 3vvk k k
2810
21.5 1.461 88.812
250
l
rvv
vvE
DESIGN OF MEMBER cd fG COMPRESSION DESIGN
TENSION DESIGN
Compressive load on CD and FG member
= 335.175 KN
Tensile load on the CD and FG member
=37.48 KN
DESIGN FOR COMPRESSIVE FORCE
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1 2 110 110
2 2 12 0.1031 88.812
250
b b
t
E
For hinge support we have ,k1=0.70, k2=0.60, k3=5.2 20.70 0.60 1.46 5 0.103 1.425
e
Again for fixed support we have, k1=0.20, k2=0.35, k3=20.
2 20.20 0.35 1.46 20 0.103 1.076e
It is given in the IS-800-2007 that
= 0.5[1+0.49(1.251-0.2) +1.251
2] = 1.539
2
2 2 0.5 N
2501.1
93.341.539 [1.539 1.251 ]
/mmcdf
Pc = cd cf A = 93.33 2 21.1100 = 393.852 KN > 335.175KN
Hence our design is safe.Here we are assuming that all the joints are partially hinged.
So for hinged joint K = 1
Again for fixed joint K = 0.625
So for our case K = 0.825
Hence K =0.825 is alright.
Now,0.825 2810
159.88 18014.5
KL
r
Hence ok.
Again we know that,
IS code specification
For partial restraint, the e can be interpolated between the eresults for fixed & hingedcases.
the interpolated result
e=1.251
According to IS-800-2007
In case for bolted, riveted & welded trusses the effective length,KL,of compression members shallbe taken as 0.7 to 1 times of distance between centres of commections depending upon degree ofend restraint provided.
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COMPRESSIONELEMENT
RATIO CONDITION RESULT
Double angle withcomponents separated,axial compression
Hence our assumed section is ok.
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.371
3 1.25
N/mm
3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 5 mm
So, taking weld size of 6 mm
Then t =0.7a = 0.74 = 4.2 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
2Lt189.371= 335.175103
L = 335.175 1000
210.72 4.2 189.371
mm
So we have to provide 225 mm length of welding.
.
Factored tensile load = 37.481.5=56.22 KN
According to IS-800-2007 table no-2
(The limiting width to thickness ratio )
clauses (3.7.2 & 3.7.4)
DESIGN OF WELDED JOINTS
DESIGN FOR TENSILESTRENGTH
TENSION
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25.1 250 100570.454
1.1
g y
dg
mo
dg
A fT
T KN
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 5 mm
So, taking weld size of 6 mm
Then t =0.7a = 0.76 = 4.2 mm
Now,Shearing area at the throat design shear strength of the weld = design load
2Lt189.371= 56.22103
L = 56.22 1000
35.342 4.2 189.371
mm
So as we have provided 225 mm length of welding , so it is ok.
Herebs= 110 mm, and Lc= 175 mm, W = 110 mm, t = 12 mm, fy =250N/mm2, and fu= 410 N/mm
2
0.7 =1.19
= 1.190.9 (110 12) 12 410 1.19 (110 12) 12 250
665.2091.25 1.1dnT KN
Design tensile strength=570.4542=1140.908KN > 56.22KN
As per IS 800: 2007 SECTION 6 clause 6.1, 6.2 and 6.3.3
Design strength due to Yielding of Gross Section,
Tdg = Ag fy/m0
Hence we can can find out the trial area for the design form this.
DESIGN OF WELDED JOINTS
Design strength due to Rupture of Critical Section
Tdn = 0.9Anc fu/m1+Ago fy/m0
Where, = 1.40.076 (w/t)(fy/fu) (bs/Lc)
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Hence our design is safe.
.
factored compressive load =231.5=34.5 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa
depending on .Assuming fcd=100 N/mm
2
DESIGN OF MEMBER aj mi COMPRESSION DESIGN
TENSION DESIGN
Compressive load on AJ and MI member
= 23 KN
Tensile load on the AJ and MI member
=133.305 KN
DESIGN FOR COMPRESSIVE FORCE
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Ac=
mm2
Since we will be using double angle so area of each angle = 345/2 = 172.5 As this is the double angle rafter so the section must
be joined on a single side of a gusset plate ofthickness 8mm and the dimension of the angle is
70706 @ 6.3 kg/m. The angles are joined byfillet welding
As per IS 800:2007 SECTION 7 clause no. 7.5.1.2and 7.1.2.1 and table 5, 7, 10, 12
c=2 2
1 2 3vvk k k
2865
13.6 2.371 88.812
250
l
rvv
vvE
1 2 70 70
2 2 6 0.1311 88.812
250
b b
t
E
For hinge support we have ,k1=0.70, k2=0.60, k3=5.
2 20.70 0.60 2.37 5 0.131 2.038e
Again for fixed support we have, k1=0.20, k2=0.35, k3=20.
2 20.20 0.35 2.37 20 0.131 1.584e
It is given in the IS-800-2007 that
= 0.5[1+0.49(1.811-0.2) +1.811
2] = 2.534
2
2 2 0.5 N
2501.1
52.8542.534 [2.534 1.811 ]
/mmcdf
Pc = cd cf A = 52.854 2 8.06100 = 85.2 KN > 34.5KN
Hence our design is safe.
Here we are assuming that all the joints are partially hinged.
So for hinged joint K = 1
Again for fixed joint K = 0.65
So for our case K = 0.825
IS code specification
For partial restraint, the e can be interpolated between the eresults for fixed & hingedcases.
the interpolated result
e=1.811
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Hence K =0.825 is alright.
Now,0.825 2865
168.53 18013.6
KL
r
Hence ok.
Again we know that,
COMPRESSIONELEMENT
RATIO CONDITION RESULT
Double angle withcomponents separated,axial compression
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 3 mm
So, taking weld size of 4 mm
Then t =0.7a = 0.74 = 2.8 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
4Lt189.371= 28.8103
L = 28.8 1000
38.024 2.8 189.371
mm
So we have to provide 40 mm length of welding.(as per IS Code).
According to IS-800-2007
In case for bolted, riveted & welded trusses the effective length,KL,of compression members shallbe taken as 0.7 to 1 times of distance between centres of commections depending upon degree ofend restraint provided.
According to IS-800-2007 table no-2
(The limiting width to thickness ratio )
clauses (3.7.2 & 3.7.4)
DESIGN OF WELDED JOINTS
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Factored tensile load = 133.305 KN
8.06 100 250183.181
1.1
g y
dg
mo
dg
A fT
T KN
.
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 3 mm
So, taking weld size of 4 mm
Then t =0.7a = 0.74 = 2.8 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
4Lt189.371= 133.305103
L =133.305 1000
62.854 2.8 189.371
mm
So we have to provide 75 mm length of welding.
DESIGN FOR TENSILESTRENGTH
TENSION
As per IS 800: 2007 SECTION 6 clause 6.1, 6.2 and 6.3.3
Design strength due to Yielding of Gross Section,
Tdg = Ag fy/m0
Hence we can can find out the trial area for the design form this.
DESIGN OF WELDED JOINTS
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Herebs= 70 mm, and Lc= 75 mm, W = 70 mm, t = 6 mm, fy =250N/mm2, and fu= 410 N/mm
2
0.7 = .89 (fum0/fym1)=1.44
= 0.890.9 (70 6) 6 410 0.89 (70 6) 6 250
191.021.25 1.1
dnT KN
Design tensile strength=183.1812=366.362KN > 133.305KNHence our design is safe.
Design strength due to Rupture of Critical Section
Tdn = 0.9Anc fu/m1+Ago fy/m0
Where, = 1.40.076 (w/t)(fy/fu) (bs/Lc)
DESIGN OF MEMBER kl COMPRESSION DESIGNTENSION DESIGN
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.
factored compressive load =11.31.5=16.95 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa
depending on .Assuming fcd=100 N/mm
2
Ac=
mm2
Since we will be using double angle so area of each angle = 169.5/2 = 84.75 As this is the double angle rafter so the section mustbe joined on a single side of a gusset plate of
thickness 8mm and the dimension of the angle is
20020020 @ 4.5 kg/m. The angles are joined
by fillet welding
As per IS 800:2007 SECTION 7 clause no. 7.5.1.2and 7.1.2.1 and table 5, 7, 10, 12
c=2 2
1 2 3vvk k k
9542
39.3 2.7341 88.812
250
l
rvv
vvE
1 2 200 200
2 2 20 0.1121 88.812
250
b b
t
E
For hinge support we have ,k1=0.70, k2=0.60, k3=5.
2 20.70 0.60 2.734 5 0.112 1.789e
Again for fixed support we have, k1=0.20, k2=0.35, k3=20.
2 20.20 0.35 2.734 20 0.112 1.751e
Compressive load on KL member
= 11.3 KN
Tensile load on the KL member
=65.16 KN
DESIGN FOR COMPRESSIVE FORCE
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It is given in the IS-800-2007 that
= 0.5[1+0.49(1.811-0.2) +1.811
2] = 2.451
2
2 2 0.5 N
2501.1
54.822.451 [2.451 1.77 ]
/mmcdf
Pc = cd cf A = 54.82 2 76.4100 = 837.649 KN > 16.95KN
Hence our design is safe.Here we are assuming that all the joints are partially hinged.
So for hinged joint K = 1
Again for fixed joint K = 0.65
So for our case K = 0.825
Hence K =0.825 is alright.
Now,0.825 9542
194.24 25039.3
KL
r
Hence ok.
Again we know that,
COMPRESSIONELEMENT
RATIO CONDITION RESULT
Double angle with
components separated,axial compression
IS code specification
For partial restraint, the e can be interpolated between the eresults for fixed & hingedcases.
the interpolated result
e=1.77
According to IS-800-2007
In case for bolted, riveted & welded trusses the effective length,KL,of compression members shallbe taken as 0.7 to 1 times of distance between centres of commections depending upon degree ofend restraint provided.
According to IS-800-2007 table no-2
(The limiting width to thickness ratio )
clauses (3.7.2 & 3.7.4)
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From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 5 mm
So, taking weld size of 6 mm
Then t =0.7a = 0.76 = 4.2 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
4Lt189.371= 28.8103
L = 16.95 1000
3.894 4.2 189.371
mm
So we have to provide 40 mm length of welding.(as per IS Code).
Factored tensile load = 65.16 KN
76.4 100 2501736.36
1.1
g y
dg
mo
dg
A fT
T KN
.
DESIGN OF WELDED JOINTS
DESIGN FOR TENSILESTRENGTH
TENSION
As per IS 800: 2007 SECTION 6 clause 6.1, 6.2 and 6.3.3
Design strength due to Yielding of Gross Section,
Tdg = Ag fy/m0
Hence we can can find out the trial area for the design form this.
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From IS 800:2007 clause 10.5.7.1.1
2
1
410 189.3713 1.25 N/mm3
uwd
mff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 5 mm
So, taking weld size of 6 mm
Then t =0.7a = 0.76 = 4.2 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
4Lt189.371= 65.16103
L = 65.16 1000
40.962 4.2 189.371
mm
So we have to provide 350 mm length of welding.
Herebs= 200 mm, and Lc= 350 mm, W = 200 mm, t = 20 mm, fy =250N/mm2, and fu= 410 N/mm
2
0.7 = 1.135 (fum0/fym1)
= 1.1350.9 (200 20) 20 410 1.135 (200 20) 20 250
1991.351.25 1.1
dnT KN
Design tensile strength=1736.362=3472.72KN > > 65.16KNHence our design is safe.
DESIGN OF WELDED JOINTS
Design strength due to Rupture of Critical Section
Tdn = 0.9Anc fu/m1+Ago fy/m0
Where, = 1.40.076 (w/t)(fy/fu) (bs/Lc)
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.
factored compressive load =103.631 KNThe anglebelongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa
depending on .Assuming fcd=100 N/mm
2
Ac=
mm2
Since we will be using double angle so area of each angle = 1036.31/2 = 518.155
As this is the double angle rafter so the section must
be joined on a single side of a gusset plate ofthickness 8mm and the dimension of the angle is
70708 @ 8.3 kg/m. The angles are joined byfillet welding
As per IS 800:2007 SECTION 7 clause no. 7.5.1.2and 7.1.2.1 and table 5, 7, 10, 12
c=2 2
1 2 3vvk k k
DESIGN OF MEMBER ck lg COMPRESSION DESIGN
TENSION DESIGN
Compressive load on CK and LG member = 103.631 KN
Tensile load on the CK and LG member
=8.97 KN
DESIGN FOR COMPRESSIVE FORCE
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2084
13.5 1.781 88.812
250
l
rvv
vvE
1 2 710 70
2 2 8 0.0981 88.812
250
b b
t
E
For hinge support we have ,k1=0.70, k2=0.60, k3=5.
2 20.70 0.60 1.78 5 0.098 1.627e
Again for fixed support we have, k1=0.20, k2=0.35, k3=20.
2 20.20 0.35 1.78 20 0.098 1.225e
It is given in the IS-800-2007 that
= 0.5[1+0.49(1.426-0.2) +1.426
2] = 1.817
2
2 2 0.5 N
2501.1
77.221.817 [1.817 1.426 ]
/mmcdf
Pc = cd cf A = 77.22 2 10.6100 = 163.766 KN > 103.631KN
Hence our design is safe.
Here we are assuming that all the joints are partially hinged.
So for hinged joint K = 1
Again for fixed joint K = 0.65
So for our case K = 0.825
Hence K =0.825 is alright.
Now,0.825 2084
123.49 18013.5
KL
r
Hence ok.
Again we know that,
IS code specification
For partial restraint, the e can be interpolated between the eresults for fixed & hingedcases.
the interpolated result
e=1.426
According to IS-800-2007
In case for bolted, riveted & welded trusses the effective length,KL,of compression members shallbe taken as 0.7 to 1 times of distance between centres of commections depending upon degree ofend restraint provided.
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COMPRESSIONELEMENT
RATIO CONDITION RESULT
Double angle withcomponents separated,axial compression
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 5 mm
So, taking weld size of 6 mm
Then t =0.7a = 0.76 = 4.2 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
2Lt189.371= 103.631103
L = 103.631 1000
65.172 4.2 189.371
mm
So we have to provide 75 mm length of welding.
Factored tensile load = 8.971.5=13.455KN
According to IS-800-2007 table no-2
(The limiting width to thickness ratio )
clauses (3.7.2 & 3.7.4)
DESIGN OF WELDED JOINTS
DESIGN FOR TENSILESTRENGTH
TENSION
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10.6 100 250240.9
1.1
g y
dg
mo
dg
A fT
T KN
.
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 3 mm
So, taking weld size of 4 mm
Then t =0.7a = 0.74 = 2.8 mmNow,
Shearing area at the throat design shear strength of the weld = design load
4Lt189.371= 13.455103
L = 13.455 1000
6.3434 2.8 189.371
mm
So we have to provide 75 mm length of welding.
Herebs= 70 mm, and Lc= 75 mm, W = 70 mm, t = 8 mm, fy =250N/mm2, and fu= 410 N/mm
2
0.7 =1.021 (fum0/fym1)
= 1.0210.9 (70 8) 8 410 1.021 (70 8) 8 250 261.513
1.25 1.1dn
T KN
As per IS 800: 2007 SECTION 6 clause 6.1, 6.2 and 6.3.3
Design strength due to Yielding of Gross Section,
Tdg = Ag fy/m0
Hence we can can find out the trial area for the design form this.
DESIGN OF WELDED JOINTS
Design strength due to Rupture of Critical Section
Tdn = 0.9Anc fu/m1+Ago fy/m0
Where, = 1.40.076 (w/t)(fy/fu) (bs/Lc)
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Design tensile strength=240.92=481.8KN > 13.455 KN
Hence our design is safe.
.
factored compressive load =19.821.5=29.73 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa
depending on .
Assuming fcd=100 N/mm2
DESIGN OF MEMBER CN GO COMPRESSION DESIGN
TENSION DESIGN
Compressive load on CN and OG member
= 19.82 KN
Tensile load on the CN and OG member
=275.925 KN
DESIGN FOR COMPRESSIVE FORCE
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Ac=
mm2
Since we will be using double angle so area of each angle = 345/2 = 172.5 As this is the double angle rafter so the section mustbe joined on a single side of a gusset plate ofthickness 8mm and the dimension of the angle is
70706 @ 6.3 kg/m. The angles are joined byfillet welding
As per IS 800:2007 SECTION 7 clause no. 7.5.1.2and 7.1.2.1 and table 5, 7, 10, 12
c=2 2
1 2 3vvk k k
2865
13.6 2.371 88.812
250
l
rvv
vvE
1 2 70 70
2 2 6 0.1311 88.812
250
b b
t
E
For hinge support we have ,k1=0.70, k2=0.60, k3=5.
2 20.70 0.60 2.37 5 0.131 2.038e
Again for fixed support we have, k1=0.20, k2=0.35, k3=20.
2 20.20 0.35 2.37 20 0.131 1.584e
It is given in the IS-800-2007 that
= 0.5[1+0.49(1.811-0.2) +1.811
2] = 2.534
2
2 2 0.5 N
2501.1
52.8542.534 [2.534 1.811 ]
/mmcdf
Pc = cd cf A = 52.854 2 8.06100 = 85.2 KN > 29.732KN
Hence our design is safe.
Here we are assuming that all the joints are partially hinged.
So for hinged joint K = 1
Again for fixed joint K = 0.65
So for our case K = 0.825
IS code specification
For partial restraint, the e can be interpolated between the eresults for fixed & hingedcases.
the interpolated result
e=1.811
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Hence K =0.825 is alright.
Now,0.825 2865
168.53 18013.6
KL
r
Hence ok.
Again we know that,
COMPRESSIONELEMENT
RATIO CONDITION RESULT
Double angle withcomponents separated,axial compression
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 3 mm
So, taking weld size of 4 mm
Then t =0.7a = 0.74 = 2.8 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
2Lt189.371= 19.82103
L = 19.83 1000
18.682 2.8 189.371
mm
So we have to provide 40 mm length of welding.
According to IS-800-2007
In case for bolted, riveted & welded trusses the effective length,KL,of compression members shallbe taken as 0.7 to 1 times of distance between centres of commections depending upon degree ofend restraint provided.
According to IS-800-2007 table no-2
(The limiting width to thickness ratio )
clauses (3.7.2 & 3.7.4)
DESIGN OF WELDED JOINTS
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Factored tensile load = 275.925KN
8.06 100 250183.181
1.1
g y
dg
mo
dg
A fT
T KN
.
From IS 800:2007 clause 10.5.7.1.1
2
1
410
189.3713 1.25
N/mm3
uwd
m
ff
Maximum size of welding (a) = 81.5 = 6.5 mm
Minimum size of welding (a) = 3 mm
So, taking weld size of 4 mm
Then t =0.7a = 0.74 = 2.8 mm
Now,
Shearing area at the throat design shear strength of the weld = design load
2Lt189.371= 275.925103
L =275.925 1000
230.22 2.8 189.371
mm
So we have to provide 250 mm length of welding.
DESIGN FOR TENSILESTRENGTH
TENSION
As per IS 800: 2007 SECTION 6 clause 6.1, 6.2 and 6.3.3
Design strength due to Yielding of Gross Section,
Tdg = Ag fy/m0
Hence we can can find out the trial area for the design form this.
DESIGN OF WELDED JOINTS
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Herebs= 70 mm, and Lc= 250 mm, W = 70 mm, t = 8 mm, fy =250N/mm2, and fu= 410 N/mm
2
0.7 =1.21 (fum0/fym1)
= 1.210.9 (70 8) 8 410 1.21 (70 8) 8 250
282.8191.25 1.1
dnT KN
Design tensile strength=183.1812=366.362 > 275.925
Hence our design is safe.
.
Here we are supposed to find out the maximum load at minor & major axis. To do so we
need to do the load combinations for various condition. They are explained below
LOAD COMBINATIONS
Load Types Major axis MAX. LOADING IN KN Type
1.5(DL+LL) 1.5{(0.588+1.03)+5.12} cos = 9.44 Tension
1.5(DL+WL 1) 1.5{(0.588+1.03) cos 1.68} = 0.25 Compression
1.5(DL+WL 2) 1.5{(0.588+1.03) cos 0.6} = 1.35 Tension
1.5(DL+WL 3) 1.5{(0.588+1.03) cos 2.68} = 1.75 Compression
Design strength due to Rupture of Critical Section
Tdn = 0.9Anc fu/m1+Ago fy/m0
Where, = 1.40.076 (w/t)(fy/fu) (bs/Lc)
DESIGN OF PURLIN
LO D COMBIN TION
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1.5(DL+WL 4) 1.5{(0.588+1.03) cos 1.5} = 0.018 Tension
1.5(DL+WL 5) 1.5{(0.588+1.03) cos 2.38} = 1.30 Compression
1.5(DL+WL 6) 1.5{(0.588+1.03) cos 1.19} =0.255 Compression
1.2(DL+LL+WL 1) 1.2[{(0.588+1.03)+5.12} cos 1.68] = 5.26 Tension
1.2(DL+LL+WL 2) 1.2[{(0.588+1.03)+5.12} cos 0.6] = 6.836 Tension
1.2(DL+LL+WL 3) 1.2[{(0.588+1.03)+5.12} cos 2.68] = 4.34 Tension
1.2(DL+LL+WL 4) 1.2[{(0.588+1.03)+5.12} cos 1.5] = 5.76 Tension
1.2(DL+LL+WL 5) 1.2[{(0.588+1.03)+5.12} cos 2.38] = 4.7 Tension
1.2(DL+LL+WL 6) 1.2[{(0.588+1.03+5.12} cos 1.19] = 6.13 Tension
Load Types Minor axis MAX. LOADING IN KN Type
1.5(DL+LL) 1.5{(0.588+1.03)+5.12} sin = 3.59 Tension
1.5(DL+WL 1) 1.5{(0.588+1.03) sin } = 0.864 Tension
1.5(DL+WL 2) 1.5{(0.588+1.03) sin } = 0.864 Tension
1.5(DL+WL 3) 1.5{(0.588+1.03) sin } = 0.864 Tension
1.5(DL+WL 4) 1.5{(0.588+1.03) sin } = 0.864 Tension
1.5(DL+WL 5) 1.5{(0.588+1.03) sin } = 0.864 Tension
1.5(DL+WL 6) 1.5{(0.588+1.03) sin } = 0.864 Tension
1.2(DL+LL+WL 1) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension
1.2(DL+LL+WL 2) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension1.2(DL+LL+WL 3) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension
1.2(DL+LL+WL 4) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension
1.2(DL+LL+WL 5) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension
1.2(DL+LL+WL 6) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension
Maximum design load on plane of major axis = 9.44 KN Maximum design load on plane of minor axis = 3.59 KN
Let assume length of purlin = 3 m
Now factor applied moment about the minor axis, Mz=2 2
1 WL 1 (3.59) 30.41
10 8 10 8
KNm
Now factor applied moment about the major axis, My=2 2
1 WL 1 (9.54) 31.07
10 8 10 8
KN-m
Assuming fcd =100 N/mm2
DESIGN
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According to IS 808: 1989,
Try with ISMC 100,
Area (a) = 1220 mm2, b = 50 mm, tw = 5 mm, tf = 7.7 mm,
Zpz = 43.83cm3
According to IS 800: 2007,
Now, b/ tf= 50/7.7 =6.493 < 9.4
And d/tw= (100-27.7)/5 = 16.92 < 42
So the section is plastic.
Now, Zpy=31220 50
152502 2 2 2
a bmm
Design strength under corresponding moment
Mndz= (b Zpz fy)/ m0
and Mndy = (b Zpy fy)/ m0
For plastic section b= 1,
So, Mdz = (1 43.83 103 250) / 1.1
= 9.9613106N-mm = 9.9613 KN-m
Mdy= (1 15250 250) / 1.1
= 3.465106N-mm = 3.465 KN-m
Now,
1 2
1y z
ndy ndz
M M
M M
1 2
0.8619 0.30640.2497 1
3.465 9.9613
So our design is safe
CHECK FOR DEFLECTION
DEEFLECTION OF THE PURLI N:-
=
4
4
5 4
9.445 3000
5 1.5 3 5.762384 384 2 10 192 10
WL mmEI
Whereas allowable deflection is3000
20150
So it is also safe from deflection.
CHECK FOR DEFLECTION
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