IS 800 - 1984
1. A 10 mm thick Gusset plate is connected to 6 mm angle
= 300 x 17.5 x 10
section by Lap Joint. Find the rivet value of 16 mm dia of
= 52. 5 KN
power drivern Rivets
(iii) Rivet Value (R) = Least of the strength in shearing
Givert
(or) bearing
Dia – 16mm
Rivet value (R) = 24.052 KN
Dia of rivet hole – 16 + 1.5 = 17.5 mm
2. Find the value of the 16 mm power driven rivets
Permissible Stresses For Power driven rivet
connected a pair of double angle section consisting of ISA
(Table 8.1 Page 95. IS 800 – 1984)
75 x 75 x 6 mm through 10 mm thick Gusset plate. Find
τ
= 100 N/mm2
the Rivet value.
= 300 N/mm2
Given
vf
bt
(i) Strength in shearing =
(D) Dia = 16 mm τ
vf
x
πd 2
(d) Dia of rivet hole = 16 + 1.5 = 17.5 mm
4
=
π x 17.5
2
100 x
vf
= 100 N/mm2
τ
bf
= 300 N/m
[Refer Table 8.1 Page 95 – IS 800 – 1984]
4
(ii) Rivet value (i) Strength of the rivet in double shearing =
= 240252.82 N
= 24.052 KN. (ii) Strength in bear =
τ
2 τ vf x
bft x d x t
1
πd 2 4
IS 800 - 1984
=
(4) Thickness of plate = 8mm
2 100 x
π x 17.5 4
2
(i)
η =
Least of the shearing , Bearing , Tearig Strength of the solid plate
= 48.104 KN (ii) Strength of the rivet in bearing =
τ
bt
(i) Strength in shearing =
xdxt
vf
= 300 x 17.5 x 6
x
= 31.50 KN
πd 2 4
= 100 x
π x 17.52 4
(iii) Rivet value = 31.50 N (Least one)
= 24.052 KN
3. Find the efficiency of the joint in a boiler. Shell
(ii) Strength in bearing =
connected using 16 mm dia of the rivet at a pitch of 60
= 42.00 KN
mm C/C in a single riveted Lap Joint thickness of the plate
(iii) Strength in Tearing = σ at (P- d) t = 100 (60 – 17.5) x
is 8 mm. The rivets are power driven shop rivet
8
Given
= 42.5 KN x 8
(1) D = 16 mm
= 34.00 KN
bt
x d x t = 300 x 17.5 x 8 mm
d = 16 + 1.5 mm = 17.5 mm (2) Pitch Distance = 60 mm C/C (3) PDS - Rivets -
vs
η
= 100 N/mm2
=
24.052 Strength of solid plate
bt
= 300 N/mm2
(iv) Strength of the solid plate =
τ
at
= 100 N/mm2
= 100 x 60 x 8 2
at
xPxt
IS 800 - 1984
= 48 KN
(v)
(i) Strength in shearing = η=
24.052 48
x 100
tf = 80 N/mm2 N/mm2 IS 800 800 - 1984 1984
τ
x
vf
πd 2
= 50.1%
4
4. A tie member ISA 90 x 90 x 6 mm carning an axial
= 80 x
tension of 40 KN is connected to a Gusset plate 10 mm
π x 13.52
thick design the Joint & sketch the arrangement of rivet.
4
Given
= 11.45 KN
Angle section = ISA 90 x 90 x 6 mm
(ii) Strength in bearing =
Load (P) = 40 KN
= 250 x 13.5 x 10
Thickness of plate = 10 mm
= 33. 75 KN
Solution
(iii) Rivet value = Least value of shearing & bearing
Step 1 Assume dia of rivet
Rivet value (R) = 11.45 KN
Assume Take Dia (D) = 12 mm
Step 3 Number of Rivet
Dia of hole = 12 + 1.5 = 13.5 mm
No of Rivet =
= 80 N/mm2
τ
bf
= 250 N/mm2 Page 95
xdxt
R (Rivet value)
Assume Hand driven rivet vf
bf
P (6 load)
Step 2 Find the value of rivet
Refer table 8.1
3
IS 800 - 1984
= 40 11.45
=
3.49
≅
=
4 Nos
η
Step 4 Arrangement of rivet
50
− 13.5 50
x 100
= 73%
6. Two plates 6 mm tk are Jointed by 14 mm dia of the
Edge distance (d)
rivet in a triple straggled rivet Lap Joint as Shown in
(i) d = 13.5 => Edge Distance = 19 mm [Refer Table 8.2]
diagram in what way the Joint will failed. If allowable
Pitch Distance
tensile stress 150 mpa, Allowable shering stress 90 mpa,
(ii) Min = 2.5 x 12 (D) = 300 mm
≅
50 mm
Allowable bearing 270 mpa. Also find the efficiency of Joint.
(iii) Maxi = 16 t (or) 200 (whichever is less)
Step 1 : Dia of rivets & holes
= 16 x 6 = 96 mm (or) 200 (Take whichever is less) Maxi = 96 mm
≅
Nominal dia (D) = 14 mm
100mm
Dia of rivet (d) = 15.5 mm Step 2 : Rivet value
Step 5:
η=
(i) Strength in shearing =
Least of the Shearing , bearing & bearing
τ vf x
Strength of the solid plate
= − P
p
d
= x 100
90 x
4
π x 15.52 4
πd 2 4
IS 800 - 1984
= 16.982 KN (ii) Strength in bearing = τ
= 150 (130 – 3(15.5)) x 6 + 2(R.V) bf
xdxt
= 150 [(130 – 3 (15.5)] x 6 + 2 (16, 982).
= 270 x 15.5 x 6
=109. 114 KN
= 25.11 KN
Plate ‘A’ at section (2) – (2) can fail only it rivets at
(iii) Rivet value = 16.98 KN
section (1) – (1) also fail. In the strength of het rivet at sec
(iv) Strength of Joint on the basis of rivet value = n x R.V
(1) – (1) will act along with the tearing of the plate (2) –
= 7 x 16.982
(2) section
= 118.874 KN
Strength of the plat ‘A’ (a) sec (3) – (3)
Plate Failure (Consider Sec (1) – (1), (2) – (2), (3) – (3) for
= tearing strength of the section (2) – (2)
plate A
+ Rivet value of (1) – (1)
Sec (3) – (3), (2) – (2), (3) – (3) for plate B)
+ Rivet value of (2) – (2)
at (P – d) t
= 150 (130 – 2(15.5)) x 6 + 3 (16982) + 2(16982)
Strength of the plate ‘A’ (a) section (1) – (1)
= 174.01 KN
= σ at (L – 2d) t
Passable Failures
= 150 (130 – 2 x 15.5) 6
(i)
Combined Failure of rivet = 118.87 KN
= 89.1 KN
(ii) (ii)
Failur Failuree of plate plate ‘A’ ‘A’ at at secti section on (1) (1) – (1) = 89.1 89.10 0
Strength of plate plate ‘A’ (a) section (2) (2) – (2)
KN
= Teaching strength (a) (2) – (2)
(iii) (iii)
+ Strength Rivet Sec (1) – (1)
Failur Failuree of plat platee ‘A’ ‘A’ at sec sec (2) (2) – (2) (2) = 109.1 109.114 14 KN
5
IS 800 - 1984
(iv) (iv)
Failur Failuree of plat platee ‘A’ ‘A’ at sec sec (3) (3) – (3) (3) = 174. 174. 01 KN
bf
= 300 Mpa
The weakness critical section is (1) – (1) of plate ‘A’
To find the rivet value
strength of the Joint = 89.1 KN
(i) Strength of rivets in double shear =
Strength of the solid plate =
τ
πd 2 τvf x 4 2
at x L x t (L = P)
= 150 x 130 x 6 = 117.00 KN
=
2 100
Efficiency = 89.1 117.00
x 100 = 76.15%
rivet design the pitch of the rivet. Take
. d.t
bf
= 300 x 21.5 x 20
at = 150 Mpa
= 77.4 KN
also find the efficiency of the joint.
(iii) Rivet value = least of shearing & bearing
Given
= 72. 61 KN
= 150 Mpa
For maximum efficiency of joint per pitch length,
Dia = 20 mm
Strength of plate per pitch = 2 x Rivet value
Dia of the rivet hole = 20 + 1.5 = 21.5 mm
= 2 x 72.61
Thickness of plate = 12 mm Assume PDS rivet,
(ii) Strength of rivets in Bearing =
Cover bult joint as shown in dia. Using 20 mm dia of the
at
4
= 72.610 KN
Two plates 12 mm are joint by Double riveted double.
x
π(21.5)2
τ
vf
= 145. 22 KN
= 100 Mpa
σ 6
at
(P – d) x t = 145.22KN
IS 800 - 1984
150 (P – 21.5) x 12 = 145. 22 KN (or) 145220
Shearing = τ vf x π d2/4 =
(i)
P = 102.178 mm. Min pitch = 2.5 D = 2.5 x 20 = 50 mm
80 x
π x 21.5 4
Provide Pitch = 100mm
=
η
P −d P
100 - 21.5 100
(iii) (iii) x 100
=
78.5%
(1) – (1) =σ
bf
(P – d) t (1)
Strength of plate per pitch length along sec (2) – (2)
Take: = 80 N/mm2
at
= 1500 P – 32250
Find the efficiency
vf
Rivet Rivet value value = leas leastt of shea shearin ring g (or) (or) bearin bearing g=
Strength of the plate (thinner) per pitch length along sec
is half the pitch of rivet in outer row. Design the Joint &
= 150 N/mm2
bf x d x t = 250 x 21.5 x 10 = 53.75
29.04 KN
lab joint, In which the pitch of the centrel row of the rivet
at
= 29.04 KN
KN
Two plates 12mm & 10 mm tk are jointed by trible riveted
Bearing =
(ii)
x 100
2
= σ at (P – 2d ) t + Rivet value = 150 (P x 21.5)10 + 29044 = 1500 P – 35456
= 250 N/mm2
(2)
Sec (2) – (2) is weaker along which the strength of the
Assume 20mm dia
plate is 1500 P – 35456
Rivet hole = 20 + 1.5 = 21.5 mm To find the rivet value 7
IS 800 - 1984
For maximum efficiency the strength of the per
Design a bracket connection using two vertical lines of the
pitch length should be equal to strength of rivet per pitch
rivet load carried by each plate is 120 KN the bracket plate
length.
of 10 mm tk are connected to 12mm tk flange plate.
1500P – 35456 = 4 x R.V
Assume pitch of 10 cm and horizontal distance between
1500 P = (4 x 29044) + 35456
the vertical line is 12 cm. eccentriciting load is 25 cm.
P = 101. 088mm
Given
Min = 2.5D = 2.5 x 20 = 50mm
Load (P = 120 KN
Max = 32t (or) 300 whichever is lesser.
Thickness of the plate = 12 mm
= 32 (10) = 320 mm > 300 mm
Thickness of the Flange = 12 mm
Max = 300 mm
Pitch (P) = 10 cm
Outer row pitch = 120 mm
Gauge (G) = 12 cm
Inner row pitch = 60 mm
Eccentricity (e) = 25 cm
η=
P
−d d
=
120 − 21.5 120 (82%)
x 100 (or )
60
− 21.5 60 (64%)
Soln:x 100
Step:
Assume 20 mm dia of rivet (PDS)
D = 20mm
Note:- Strength of the plate = 1500 (60) = 35456
d = 21.5 mm
= 54, 544 N
Step 2:
42 = 4 x 29044N = 116, 76 N Take which value is user so take pitch, efficiency = sec (2) 8
Find the rivet value
IS 800 - 1984
Strength in shear = T vf x
(i)
Step 4: Step 5: Check for the safety fo the joint
πd 2 4
(ii)
= 100
q1 = qv1 + qv2
π x 21.52
x
qv =
4
P
= 36305.03 N.
Strength in bearing =
n
ε r 2
qn = M
.y
ε r 2
Rivet value R = 36/305 KN
r 2 = Σ x2 + Σ y2
To find find the the no no of of the the rive rivets ts vert vertic ical al line line
= 0.(6)2 + 4 (102 + 202)
Vertical line = 2 (given)
r 2 = 2360 cm
M=PxQ
qv =
= 120 x 25 = 3000 KN.cm
120
M1 = M no . of. vertical rivet
=
3000 2
10
= 150 KN.cm
+
3000 2360
qv = 19.627 qn =
n1= R.P
=
(2)
.x
bf x d x t
= 64500 N
6M1
(1)
= 300 x 21.5 x 10
Step Step 3:
M
+
6 x 1500 36305 x 10
M
=
εr
4.979 ≅ 5 nos
2
9
. y max
.6
IS 800 - 1984
=
f t max = 3000 2320
M
x 20
εy
2
. y max
qn = 25.42
M = P x R = 80 x 16 = 1280 KN. Cm
q=
Ymax = 6 + 6 + 6 + 6 = 24 cm qn 2
+ qv = 2
19.6272 + 25.42 2
y2 = 2(62 + 122 + 182 + 242)
y2 = 2160 cm
q = 32.118 KN < R = 36.305 K
Ft max =
q < R hence safe
1280
Check the safety of the joint as shown in diagram
2160
x 24
Step 1:Ftmax = 14.22KN
Assume diameter = 16 mm
Step 4
Using PDS dia of the rivet hole = 16 + 1.5
Find
= 17.5 mm
vf
(cal) = Q
Step 2:
πd
Shear stress due to
=
Direct laod (Q) = p/n
8
= 80/10
πx17.5
Q = 8 KN
2
/4
= 0.083
Step 3: Find ft max
= 83. 26 N/mm 2 10
2
/4
IS 800 - 1984
Step 5
through the bracket to the column. E = 10cm, P = 200KN.
To find
tf
(cal) =
Design the connection between the angle & column f t max
πd
2
Step 1
/4
Assume 20mm dia of the rivet
=
Using PDS rivet 14.22
π17.5
2
Dia of the rivet hoel = 20 + 1.5
/4
= 21.5 mm
= 59.11 N/mm
2
Minimum pitch distance (D) = 2.5
Step 6
= 2.5 x 20 = 50 mm
Check
Maximum pitch distance = 32t
τ
vf
(cal cal )
τ
+
σ
tf
vf
(cal cal )
σ
= 32 x 21.5 = 688
≤ 1.4
690 mm ≅
tf
Adopt pitch distance to 100 mm 33.26 0.4fy
+ 59.11 ≤ 1.4
Step 2: To find rivet value
0.6fy
Strength in shearing = = 100 x
33.26
π x 21.5
0.4 x 4
4
A bracket plate of 10 mm thick is to be connected to the = 36.30 KN
base of the flange using angles the load is applying 11
2
vf
x π d2 / 4
IS 800 - 1984
Strength is bearing =
σ bt x d x t
n1 = 0.8
= 300 x 21.5 x 10
6 x 1000
= 64.5 KN
36.3 x 10
Rivet value = 36.3 KN
= 3.25 4 Nos
Step 3
Adopt 4 nos of rivet each row
No of rivet (n1) = 0.8
Step 4 Arrangement of the rivet 6m
Step v To find ft max
RP
Ft max = M
= 0.8
x y max
6xM
∑y
36.3 x 10
M = 2000 KN.cm
2
y2 = 2(102 + 202 + 302) = 2800
M=Pxe
= 200 x 10
Ymax = 10 + 10 + 10 = 30 cm
= 2000 KN.cm
Ft (max) = 2000
M1 =
2800
m no of rivet line
x 30
Ft max = 21.42 KN
= 2000/2 M1 = 1000 KN.cm 12
IS 800 - 1984
Q=
Check P n
=
200
τ
8
vf
+σ
(cal)
τ
vf
tf
(cal cal )
σ
≤ 1 .4
tf
Q = 25 KN Step VI
68.8
τ
0.4 fy
vt
(cal) = Q
πd
2
68.8
/4
0.4 x ?
= 2
tf
0.6 fy
+
≤
59 0.6 x ?
1.4
≤ 1.4
300 & beam ISMB 350 transmitting the load of 35 KN/m
/4
over a span of 9m. Assume 20mm dia PDS rivet
= 0.0688 KN/mm 2 = 68.8 N/mm2
5 .9
Design the riveted connection between the column ISMB
25
π x 21.5
+
Given data:-
(cal)=
Load = 35 KN/m
Ft
πd
max 2
Span l = 9m
/4
Solution:
=
Step 1
21.42
π x 21.5
2
The beam is connected to the column using angle.
/4
The size of the angle should not be less than 3d.
= 0.05 tf
(cat) = 59 N/mm 2 13
IS 800 - 1984
length of the angle = 3 x 21.5
Rivet value = 52.24 KN
∴
Number of of rivets (n) = P
= 64.5 mm
R .v
Choose ISA 75 x 75 x 10mm Angles Step 2
Load at the Joint (P) = Reaction from the beam Connection between the angle & web of the beam
= WL
line
2
Angle & flange of column line
=
35 x 9 2
=157.5
n =
To find the rivet value: Strength of rivet in double shearing = 2 x
τ
vf
157.5
x
πd
52.24
2
4
n=3 Step 3
= 2 x 100 x
π x 21.5
2
Connection between flange of the column of angle.
4
To find the rivet value Strength in single shearing =
= 7261 KN. Bearing for web of ISMB 350 =
bE
τ
vf
x
πd
.d xt
4
= 300 x 21.5 x 8.1 = 52. 24 KN
14
2
IS 800 - 1984
= 100 x
Maximum pitch = 32 t (or) 300
π x 21.5
2
= 32 x 8.1
4
= 259.2 < 300
= 36. 30 KN
Max pitch distance = 260 mm
Bearing for flange of ISMB = σ bf x d x t
Minimum edge distance = 29 mm [From IS 800 – 1984.
= 300 x 21.5 x 10
Pg
= 64.5 KN
Provide 30 mm edge distance ∴
Rivet value = 36.3 KN
A tie bar 100 mm x 16 mm is to be welded to another plate
Number of rivets n = P
150 mm x 16 mm. find the minimum overlab length
R .V
required if 8 mm fillet weld of used. Take N/mm2. σ bt = 165 N/mm2, σ
= 157.5
vf
= 150 N/mm2
= 165 N/mm2
= 100 N/mm2
at
n = 4.33
5 nos
bt
≅ vf
Step 4 Arrangement of rivet
Size of the fillet welt = 8mm
Minimum pitch = 2.50
Load = the strength of the smaller plate
= 2.5 x 20
Strength of the smaller plate =
= 50 15
at
= 100 N/mm2
Given data:-
36.3
σ
at
xbxt
= 150
IS 800 - 1984
= 150 x 100 x 16mm
The length shared by two side
P = 240 KN
Length of the onside = 230 / 2 = 115 mm
The value of the weld = ks. Fs
The weld Lab Joint is to be provided to connect two tie bar
= 0.78 x 8 x 100
150 x 16 mm stress in the plate is 150 N/mm 2. To check
R = 560 N/mm
the design if the size of the weld is 8mm & shear stress is
Length of the Weld = P/R
taken as 100 N/mm 2.
=
Given data:-
2400 x 10
3
σ
at
= 150 N/mm2
560
b = 150 mm
L = 428. 57 mm
t = 16 mm
L
τ
430 mm ≅
vf
= 100 N/mm2
S = 8 mm
For minimum over lab in of the plate both end fillet weld
Solution
& side fillet weld are provided.
To check the safety of the Joint should not be more
The length of the end fillet = 2 x 100
than load at the joint.
= 200 mm
Load at the Joint =
Length is to be provided by side
= 130 x 150 x 16
Fillet = 430 – 200
= 360 KN
Side = 230 mm
Strength of the Joint = 16
σ
at
xbxt
. K.S.L
vf
IS 800 - 1984
L = 50 + 2
L=
502
+ 80
2
Load
]2
Value of the weld
L = 477.38 mm Value of the weld =
Strength of the Joint = 100 x 0.707 x 8 x 4m
. K.S
vf
= 100 x 0.707 x 6 mm
= 266. 61 KN
= 424.2 N/mm2
Hence the design is unsafe
Load at the Joint P = 200 KN
Load = 360
L=
Strength = 267 KN
200 x 103
Load 4 strength
424.2
A 150 mm x 115 mm x 8mm angle section carries a tensile L= 471. 47 mm
load of 200 KN it is to be connected gusset plate using 6
x1 + x2 – 150 = 471.48 – 150 = 321.48 mm
mm fillet weld at the extreame of the longer length (leg)
Two unknowns S1 one equation to create another
Design the Joint along the shear stress 100 N/mm 2.
equation to find the either x 1 (or) x2.
Angle section is unequal. The load is acting excentricity.
Moment of the at the top = Moment of resistance of het
We have to adopt
bottom weld at top.
Let x1 be the length of the weld at tob
Unequal section = 150 x 115 x 8 mm
X2 be the length of the weld at bottom Total length = x 1 + x2
17
IS 800 - 1984
Load acting at a distance lxx = 44.6 mm (
from steel
An = W
σ
table, Pg.) Moment of the load at top = 200 x 103 x 44.6 6
= 8.92 x 10 N/mm
at
=
150 x 10 3 0.6 x 250
= 1000 mm
2
Step 2: choose 70 x 70 x 10 mm in steel table
(1)
(From steel table) L 1 = L2 = 70mm, t = 10mm, d = 20 + 1.5
Moment resistance of the bottom = 424.4 x2 x 130
= 21.5
3
= 63.66 x 10 x 2
A = 1302 mm 2
x2 =
Anet = A1 + A2 8.92 x 106
A1 = (L1 – t/2) t – d x t
63.66 x 103
= [70 – 10/2] 10 – 25 x 10
x2 = 140.119 mm
A1 = 435 mm2
x1 + x2 = 321.48
A2 = [L2 – t/2]t
x1 + 140.119 = 321.48
= [70 – 10/2] 10
x1 = 321.48 – 140.119
A2 = 650 mm2
x1 = 181.36 mm
K =
Design a single angle section carring a axial load of 150
3A1 3A 1 + 3A 2
KN. Assume Fy. 250 N/mm 2 and dia of the rivet is 20mm. step 1
18
IS 800 - 1984
=
A2 = 700 mm2 3 x 55
K=
(3 x 455) + (650)
3A 1 3A 1
K = 0.67
+A
= 2
3 x 735 (3 x 735) + 700
Anet = 870.50 mm2
Anet = 735 + (0.76 x 700)
Step 3
Anet = 1266.32 mm 2
Load = Anet x σ
Load = Anet x σ
at
= 0.76
at
= 870.50 x 150
= 1266.32 x (0.6 x 250)
= 130.575 KN < 150 KN
Load (w) = 189.948 KN
So unsafe
Design a tension member of roof truss carrings a axial
Hence trial section choose ISA 100 x 75 x 10 Gross Area A = 1650 mm
tension of 250 KN using double angle section back to back
2
of the Gusset plate (Opp side) dia of rivet is 20mm.
L1 = 100, L2 – 75, t = 10
Step 1
Anet = A1 + KA2
An =
A1 = [L1 – t/2] t – (d x t)
W
= [100 – 10/2]10 (21.5 x 10)
σ
at
=
250 x 103 150
A1 = 735 mm2
An = 1666.66 mm2
A2 = [L2 – t/2] t
Step 2: Selected a section whose Gross area is
= [75 – 10/2] x 10
1.5 x An area = 1.5 x 1666.66 = 2500 mm 2 19
IS 800 - 1984
Take section ISA 150 x 115 x 12 mm
Where
L1 = 150 L2 = 115 t = 12 mm A = 3038mm2
b – breadth
Anet = Ag – Area of Rivet holes.
n – no of rivets
= 3038 – 2(21.5 x 10)
d – dia of nivet hole
Anet = 2608 mm2
m – no of zig. Zag line along the failure line
Load = 2606 x 0.6 x 250
s – Pitch
Load = 391.2 KN
g – guage
Note: (i) for single angle section
Member under axial load and moment
Ag = 1.35 to 1.5 times of A net
There will be axial tension due to axial force and
(ii) For Double angle section
bending stress due to bending moment.
(a) angles on some side of the gusset plate
Direct stress due to axial tension =
Ag = 1.35 Anet
Bending stress due to moment =
(b) Angles on either side of the gusset plate
The section is safe the following intraction formula is
Ag = 1.25 Anet
satisfied.
σ
bt
at
(cal) = W/An
(cal) = M/I. y
(iii) (a) For chain riveting in plate section
for uniaxial
σ
Anet = t (b – nd)
at
(cal cal )
σ
(b) for zig – zag riveting (or) staggered riveting (i) Anet = t [(b – nd) 4 + m [s 2/4g]
at
bending
(ii) Anet = t [(b – nd) + s 2/4g1 + s22 4g2)] 20
+
σ
bt
(cal cal )
σ
bt
≤ 1 for uniaxial
bending
IS 800 - 1984
σ
at
(cal cal )
0.6 fy
+
σ
btx
(cal cal )
0.66fy
+
σ
bty
(cal cal )
0.66 fy
Sectional properties
≤1
Area = 6293 mm2 Ixx = 15082 cm4
For biaxial bending
tw = 8.6 mm
A tension member made of two channels placed back to
Adopt 20mm dia
back carries a moment of 1900 N.m in addition to a direct
Rivet for the connection
tension of 450 KN. Design the section assume that f y =
An = Gross Area – area of Rivet hole
250 N/mm2
= 2 x 6293 – 4 (21.5 x 8.6)
For the selection of the section assume that
σ
at
at
= 11846.4 mm 2
= 30% to 40% of the preliminary stress
at
= (0.3 to 0.4) of 0.6 fy
(cal)
W
= 0.3 x 0.6 x 250
=
An
450 x 103 11846.4
= 37.99 N / mm
2
= 45
bt
Area required = W 0.3σat
=
(cal) =
450 x 1000 0.3 x 0.6 x 250
M I
= 10000 mm2
.y
=
19000 x 1000
9
3 x 15082x 10 x
Check for Intraction formula
This is offered by two channel section Area = 10000/2 = 5000 mm 2
σ
Choose ISMC 400
0.6 x fy 21
at
(cal cal )
+
σ
bt
(cal cal )
0.6 x fy
≤
1
400 2
= 12.60 N / mm
2
IS 800 - 1984
37.99 0.6 x 250 +
12.60 0.66 x 250
= 1578.2 x 150
≤1
= 236730 N (or) 236.730 KN. A rolled steel is used as a column of height 5.5 m both
0.253 + 0.07 < 1
ends are hinged. Design the column to carried axial toad
0.33 < 1
of 600 KN.
Hence the section is safe
Solution:
A tie of roof truss consist of double angles ISA 100 x 75
Both ends are hinged l eff = L
x10 mm with it’s short leg back. To back and long leg
Leff = 3.5 m
connected to the same side of the gusset plate with 16 mm
Load (P) = 600 KN
dia of the rivet determine the strength of the member take
Rolled steel section
σ
Aread =
at
= 150 N/mm2
Load
Step 1
σ
Anet = A1 + KA2
ac
=
σ
ac
= 80 N/mm2
600 x 103 80
= 7500mm
Choose ISHB 300 (1) 63.0 kg/m
K= 5A l 5 A1
+A
Area = 80.25 cm2 = 8025 mm2 r xx xx = 12.70 m
2
r yy yy = 5.29 cm
K = 0.714 Anet = 650 + (0.714 x 1300) = 1578.2 mm 2 Strength = Anet x σ
at
22
2
IS 800 - 1984
σ bc (act)
Slanderness ratio λ = L eff r min
=
=
3.5
W
0.052
area
=
600 x 10 3 8025
= 74.76
λ = 66.16
Design a single angle discontinuous structs connected by 2
IS 800 – 1984 Table 3.5 page 38
rivets to a gusset plate length 2.5m, applied load 150 KN.
To find the
[Refer Is 800 – 1984 -> CL 5.2
bc
permissible
Pg 46]
Fy = 250 (assume)
Effective length = 0.85L = 2.125 m
λ = 66.16
σ
60
Areqd =
70
122
ac
= 60 N/mm2 W
112
σ
x = 122 –
122 − 112 (66.16 − 60) 60 − 70
=
r yy yy = 46.3mm λ
(assume) = 80 N/mm 2
bc
2
A = 2903 mm 2
σ bc permissible = 115.84 N/mm 2 σ bc Permissible >
= 2500mm
Choose the ISA 150 x 150 x 20
r xx xx = 46.3mm
bc
60
ac
x = 115.84
150 x 10 3
assume => Hence safe
=
L eff r min
=
2.125 x 103 46.3
= 45.80
[Refer IS 800 – 1984 Table 5.1 Pg.30] 23
IS 800 - 1984
f y = 250
Solution
λ = 45.89
Leff = 0.85l
40 139
= 0.85 x 3m
50 132
Leff = 2.55 m
45.89 134.88
Assume double angle
σ bc Permissible = 134.88 N/mm 2 σ bc (assume) < bc permissible
Area required =
σ bc = 80 N/mm2
W
σ
Hence safe
bc
Where
=
250 x 103
σ bc (act) = =
W area
80
3
= 150 x 10 = 51.67 2903
= 3125 mm2
Design a double angle strut continuous to a load of 250
Single angle area = 3125 / 2
KN/m3 length 3m.
= 1562.5 mm 2
Given
Select ISA 90 x 90 x 100 mm @ 13.41 kg/m
Load = 250 KN
A = 1703 mm 2
L = 3,
Ixx = 126.7 x 10 4 mm4
If Double angle continuous member
Iyy = 126.7 x 10 4 mm4
Leff = 0.7L to L
Lyy = 25.9 mm (lyy – centrid distance) 24
IS 800 - 1984
To calculate r min min:
= 2.55 x 103
r min min =
R xx
=
I xx
=
A
27.27
2L xx
=
2a
λ = 93.50 To find the σ bc (Permissible)
126.7 x 10 4
[Refer IS 800 – 1984
1703
Pg 39
Table 5.1]
90 90
r xx xx = 27.27 mm
100 80
Ryy 2iyy
+ a ( iyy + t / 2)
93.5 86.5
2
bc
2a
(Permissible) = 86.5
σ bc (act) =
=
250 x 103 2
126.7 x 10 + 1703 126.7 x 104 + 10 2 2 2 2 4
2(1703)
2
2x
bc
1703
(act) = 73.4
Design a compression member consist of two channels
2
placed with toes facing each other subjected to load of
R yy 447.9 x 103 mm yy = 447.9
1300 KN. Eff ht of the column is 8m. Design the comp.
λ= l eff
member and also design a lacing system
r min
Solu: 25
IS 800 - 1984
Assume
ac
= 110 N/mm2
R xx xx = r xx xx = 154.8 mm
Areq =
r yy yy = K 1
σ
=
1300 x 10
3
= 11818.18mm
110
ac
Iyy
2
A
399.074 x 10 6 2 x 6293
= 178.07 mm
(or )
This is offered two channel. Therefore
Iyy / 2a
Area of single channel = 11818.18 2
= 5909.09
λ=
L eff r min
Select ISMC 400 @ 494 N/m Area = 6293 mm2
=
8 x 103
= 51.68
154.8
λ = 51.68 4
4
ixx = 15082.8 x 10 mm 4
=
[Refer Table 5.1 Pg. 39 Is 800 = 1984
4
iyy = 504.8 x 10 mm
50 132
r xx xx = 154.8 mm
60 122
r yy yy = 28.3 mm
51.68 130.32
cyy = 24.2 mm
bc 6
4
Ixx = 2ixx = 39.656 x 10 mm
= 130.32
σ bc (dct) =
2
Iyy = 2[iyy + a (S – (yy) ]
Load
= 2 [504.8 x 10 4 + (6293) [200 – 24.2) 2]
Area
= 399.074 x 106 mm4
σ 26
ac
=
(act) = 103.29
1300 x 10 3 2 x 6293
= 103.29
IS 800 - 1984
bc
(per) = 130.32 > σ bc (act) = 103.29
Assume 20mm dia of rivet for connections.
The design is safe
Width of the bar = 3
Design of Lacing
Thickness of bar ‘t’ = l1 / 40 for single Lacing
Assume that the connection to the lacking bar in mode at
l1 = length of the lacking bar
the centre of the flange width
l1 = 3002
Connection are at 50mm from the edge. Distance C/C of rivet across = 400 – 50 – 50 = 300 mm Assume the angle of inclination of lacing bar
+ 300
= 3 x 20 = 60 mm
2
l1 = 424.26 mm
= 450
‘t’ =
C = 2 x 300 = 600 mm [For angle 450 = 25 desare equal]
424.26
Check
40
< 0.7 λ < 50
T
C
= 10.61mm
12 mm ≅
r min
Size of the bar = 60 x 12 mm < 0.7 x 51.68 < 50
Check for:-
600
(i) Slenderness ratio
28.3
λ > 145
21.20 < 36.176 < 50 Hence ‘C’ is ok. Size of the Lacing bar:27
IS 800 - 1984
r L =
σ
60 x 12 12 = = 3.46 3
IL AL
bc
bc
(60 x 12 )
( cal cal ) =
P A
=
22.98 x 103 60 x 12
= 31.92 N / mm
(Per) [For lacing bar] => λ = 122.62
fy = 250 122.62 > 145
120
Hence O.K
64
130 57 from Table 5.1
(ii) Check for compressive compressive stress
122.62 62.17 in IS 800 – 1984
Compression Compression leading in the lacing bar =
bc
V
(Per) = 62.17 N/mm2
(62.17) σ bc (Per) > σ bc (cal) (31.92)
n Sin Q
Hence safe V = 25% of the load
(iii) Check for tensile stresses:-
= 2.5 100
P= V
x 1300
n Sin Q
V = 32.5 KN
σ
Comp. Load =
at
= 22.98 KN
(cal) = P
32.5 x 103 2 x sin 450
=
A net
22.98 KN
Anet = Agross – Area of rivet hole = (60 x 12) – (12 x 21.5) 28
2
IS 800 - 1984
= 462 mm 2
Assume square base length of one side (l)
σ
L=
at
(cal) = 22.98 462
at
l
= 49.74
at
(Per) >
at
A
=
187.5 x 103
= 433.012mm
Provide 450 x 450 mm
(Per) = 0.6 fy = 0.6 x 250 = 150 KN
[150]
=
Thickness
(Cal)
t=
Hence safe in tensile stress.
3w
Design a simple slab base resting on a concrete slab for
σ
bs
(a 2
− b
2
/ 4)
the following data a=
Load from the column = 750 KN
450
Size of the column = ISHB 400
σ
cc
− 250 2
= 4 N/mm2
b =
SBC = 100 KN/m 2
450 − 400
Design the slab base.
2
Soln:-
Load Area
σ
cc
= 2500
W= Bearing Area =
Load
= 100 mm
=
750 x 10 3 4
= 187.5 x 10
3
mm 2
29
=
750 x 10 3 450
2
= 3.7 N / mm
2
IS 800 - 1984
t=
Provide 3m x 3m of the pedestil
3 x (3.7)
(100 2
185 t = 24.3 mm
−
50 4
2
Area of pedestile = 3 x 3 = 9m 2
)
Depth of the pedestile Assume 450 despersion projection of the pedestile beyond
25 mm ≅
the base plate =
Design of pedestile
3 − 0.45
Size of the pedestil is design such that pressure on
2
the soil is with in the safe bearing capacity of soil.
= 1.275 ≅ 1.3m
Add 10% of the self wt
Adopt depth = 1.3 m
Total Load =
Size of the pedestal = 3 x 3 x 1.3 m
750 + 10 x 750 100
Size of the base plate = 450 x 450 x 25 mm Design of gusseted plate A builtup steel column compressing 2ISWB 400
Base area of the pedestil:-
RSJ section with their webs spaced at 325 mm and
Area = Load SBC
=
825 100
= 8.25m
connections by 10mm thick battens. It transmit and axial
2
load of 2000KN. SBC of soil at site is 300 KN/m 2. The
Adopt square base, length of the one side (1)
safe permissible stresses of concrete base is 4 N/mm 2.
L=
Design the gusseted base. Grillage foundation.
A
=
8.25
= 8.25m
2
Load = 2000 KN
30
IS 800 - 1984
SBC = 300 KN/m 2
= 714.29 mm
Area required =
Provide square plate = 750 mm x 750 mm Load con. permissibl e
2000 x 10
=
3
Cantilever projection of the plate from face of the gusset
4
angle is checked for bending stress due to the concrete
= 500 x 10 3 mm2
below. Intensity of pressure below the plate =
It is shared by two angle =
500 x 10
load
3
2
mm 2
Area W=
Adopt angle section 150 x 75 x 12 mm gusset angles on
2000 x 103
flange side width 75 mm long horizontal
750 x 750
Allow 30 mm projection on either side in the direction of
= 3.56 N / mm
2
Moment in the cantilever portion:
parallel to web. Length base plate parallel to the web
M
L reqd = 400 x 2(10) + 2(12) + 2(75) + 2(30) = 654 mm
=
Wl 2 2
Where l = [750 – 400 – 2(10) – 2(12)] / 2
Provide length of base plate = 700 mm
l = 153 mm
Breath of the plate = A reqd L reqd
=
w = load per ‘m’ length = 3.56 N/mm for 1 mm width
500 x 103
bs
700
31
= M/Z
IS 800 - 1984
185 =
Using 20mm 41.67 x 10
T
bt / 12 3
(or ) Y 185 =
3
41.67 x 10
t/2
=
41.67 x 10
3
rivet (DDS)
To find the rivet value
1x t 3
/ ( t / 2) 12
Strength in shearing = v
πxd x
bc
xbxt
τ
2
4
3
=
t2 / 6
185t2 = 41.67 x 10 3
100 x
π x 20.5
t = 36.76 mm
2
4
= 36.305 x 10 3
thickness of the base = 36.76 mm – 12 (thickness of the
Strength in bearing =
angle leg)
= 300 x 21.5 x 10
= 24.76 mm
= 64.5 x 103
Provide 25 mm plate thickness, size of Gusset base plate =
Rivet value = 30.305 KN
750 x 750 x 25 mm
No of rivet =
CONNECTIONS:
Load
Outstanding Outstanding length of the each side =
R .V
750 − 400
=
467.250 36305
= 12.87
= 13 nos (or) 14 nos
2
Pitch:
Load on each connection = 3.56 x 750 x 175
Min pitch = 2.5 x D = 2.5 x 20 = 40 mm
= 467.250 KN 32
IS 800 - 1984
Max pitch = 12 x t = 12 x 10 = 120 mm
Depth = 325 mm
Provide 60 mm pitch edge distance code book = 30 mm
bf = 165 mm
A beam supporting a floor glab carries a distributed load
tf = 9.8 mm
of 20 KN/m span for the beam is 6m design suitable I –
tw = 7.0 mm
section for the beam
Zxx = 607.7 x 10 3 mm3
Step I
Ixx = 9874.6 x 10 4 mm4 Iyy = 510.8 x 10 4 x mm4
Assume 3% adding as a selt wb of section Total load = 20 +
= 20.6
Step 3 check for shear
20 x 3 100
Shear is calculated at a distance of ‘d’ from the support V = w (l/2 – d) W = adi + self wt (of section)
B.M = 2
WL r
= 20 + 0.481
= 92.7 KN m
W = 20.481 V = 20.431 (6/2 = 0.2)
Step 2:
V = 54.65 KN
Z= M
σ
bt
=
0.27 x 10 10
5
6
= 561.81 x 103mm
τ
3
av
(cal) = V Area of web portion
Choose ISLB 325 @ 431 N/m Area = 5490 m2 33
IS 800 - 1984
=
Hence the section is safe in deflection. 54.65 [32 5 − 2( 007] x 7
τ
= 25.56 N / mm 2
A s/s beam of span 6m carring a point lead low Joist at Mid span and at support load applied at Midspan is 150
av
(Per) = 0.45 fy = 0.45 x 250 = 112.5 N/mm
av
(Per) > τ
av
2
KN Design the beam, assuming fy = 250 N/mm 2 the beam
(cal)
developes B.M, S.F and check for shear and deflection
2
112.5 N/mm > 25.56 N/mm
2
Step 1
Step 4
Assuming 3% adding as a self wt of the section
Check for deflection
Total load = 150 + (150 x 3/100) = 154.5 mm
Ymax =
B.M = 5 WL2
154.5 x 6
384 E I xx
4
Z = M/σ
= 4
at
= 231.75 KN.m
=
5 x 20.43 x 6000
231.75 x 106
384 x ( 2.1x 105 ) x 9874.6 x 10 4
0.66 fy
= 1.404 x 106
= 1404.55 x 10 6 mm 3
Ymax = 16.62 mm Step 2
Permissible deflection =
λ 325
=
6000 325
Take ISLB 500 at 750 N/m
= 18.46 mm
W = 750
Ixx = 38570 x 10 4 mm4
Ymax < y per
A = 9550
Iyy = 1063.9 x 10 4 mm4
16.62 < 18.46
D = 500
Zxx = 1543.2 x 10 3 mm3
34
IS 800 - 1984
bf = 180
=r 33.4 mm
yy
Ymax = WL3
tf = 14.1
48EI
tw = 9.2
=
154.5 x 10 3 x 60003 48 x 2.1 x10 5 x 38579 x 10 4
Step 3 Check for shear
Y per = L/325 = 18.46 mm
Shear is calculated at a distance of ‘d’ from the support
Ymax < y per
V=
Hence in deflection W
In the above problem the beam is laterly un support
2
between the own beam
W = P.L + Selt wt
Assume
= 150 + (0.750 x 6)
M = 231.75 KN.m
τ
av
(cal) dw.tw
=
77.25 x 103 [500
− 2(14.1)] (9.2 )
120
= 120 N/mm2 (120 to 130 N/mm2)
(cal) < τ
av
= 1.93125m 3
Z = 1031.25 mm 3
= 112.5 N/mm2 av
bc
231.75
= 17.80 N / mm 2
av (Per) = 0.45 fy = 0.45 x 250
Z= V
τ
= 3.58mm
Choose ISLB 550 at 863 N/m A = 12669 mm 2
(Per)
Hence safe in shearing
Zxx = 1933.2 x 10 3 mm4
Step 4 Check for deflection
Ixx = 53161.6 x 104 mm4 tf = 15 mm tw 9.9 mm 35
IS 800 - 1984
r yy yy = 34.8 mm
86.21
130.01
129.71 129.04
To find τ
90
127
120
bc
permissible for the selected section effective
length of the compressive flange distance between the
131 − 12 7 = 13 1− x 1.2 85 N 80 = 13 0.04
cross beam. L = 6/2 = 3m ∴
l
3000
=
ry
D1 T
34.8
d
=
dw tw
t f
=
=
130 N 126 = 130 − x 1.2 80 N 85 = 129.04
= 86.207
550 15
550 − 2(15) 99
130.04 N 129.09 = 130.04 − 35 − 40 = 129.71
= 36.67
bc
= 52.53
(cal) =
=
[Refer IS 800 – 1984 => Table 6.1 B => Page 58
T t
=
tf tw
=
15 9.9
bc
= 1.5
Z xx
=
23 1.75 x 10 6 1933.2 x 10 3
(Per) = 129.71
Hence the section is safe B5
85
M
131
36.67
bc
40
(Per) >
bc
Check for shear:
130
36
(cal)
= 119.88 N / mm 2
IS 800 - 1984
Max shear at the support V =
= =
150 2
+
W 2
+
WL 2
(or )
W 2
+
1 2
(0.863 x 6) 2
V = 75.59 KN.
37
