Design Of Steel

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

the Rivet value.

vf

= 100 N/mm2

bt = 300 N/mm2

Given

(i) Strength in shearing =

(D) Dia = 16 mm πd 2 τ x vf 4

(d) Dia of rivet hole = 16 + 1.5 = 17.5 mm vf = 100 N/mm2 τ

=

π x 17.52 100 x 4

= 300 N/m

[Refer Table 8.1 Page 95 – IS 800 – 1984] (ii) Rivet value (i) Strength of the rivet in double shearing =

= 240252.82 N

πd 2   2 τ vf x 4  

= 24.052 KN. (ii) Strength in bear = τ

bf

bft

xdxt 1

IS 800 - 1984 =

(4) Thickness of plate = 8mm π x 17.5   2 100 x  4   2

(i) η =

= 48.104 KN (ii) Strength of the rivet in bearing = τ

bt

Least of the shearing , Bearing , Tearig Strength of the solid plate

(i) Strength in shearing = vf

xdxt

πd 2 π x 17.52 x = 100 x 4 4

= 300 x 17.5 x 6 = 31.50 KN (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 = bt x d x t = 300 x 17.5 x 8 mm

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

d = 16 + 1.5 mm = 17.5 mm η=

(2) Pitch Distance = 60 mm C/C (3) PDS - Rivets - vs = 100 N/mm2

24.052 Strength of solid plate

bt = 300 N/mm2

(iv) Strength of the solid plate = at x P x t

τ

= 100 x 60 x 8

at

= 100 N/mm2 2

IS 800 - 1984 = 48 KN

 tf = 80 N/mm2 IS 800 - 1984

(v)

(i) Strength in shearing = τ η=

vf

x

24.052 x 100 = 50.1% 48

πd 2 4

4. A tie member ISA 90 x 90 x 6 mm carning an axial

= 80 x

π x 13.52 4

tension of 40 KN is connected to a Gusset plate 10 mm thick design the Joint & sketch the arrangement of rivet. Given

= 11.45 KN

Angle section = ISA 90 x 90 x 6 mm

(ii) Strength in bearing = bf x d x t

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 = P (6 load) R (Rivet value)

Step 2 Find the value of rivet Assume Hand driven rivet vf = 80 N/mm2 τ

bf

Refer table 8.1

= 250 N/mm2 Page 95 3

IS 800 - 1984 =

=

40 = 3.49 ≅ 4 Nos 11.45

η = 73%

Step 4 Arrangement of rivet

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 (ii) Min = 2.5 x 12 (D) = 300 mm

≅

tensile stress 150 mpa, Allowable shering stress 90 mpa,

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

≅

50 − 13.5 x 100 50

Nominal dia (D) = 14 mm

100mm

Dia of rivet (d) = 15.5 mm Step 2 : Rivet value

Step 5:

(i) Strength in shearing =

Least of the Shearing , bearing & bearing η= Strength of the solid plate

τ vf =

P− d = x 100 p

π x 15.52 90 x 4 4

πd 2 x 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)

Failure of plate ‘A’ at section (1) – (1) = 89.10

Strength of plate ‘A’ (a) section (2) – (2)

KN

= Teaching strength (a) (2) – (2)

(iii)

+ Strength Rivet Sec (1) – (1)

Failure of plate ‘A’ at sec (2) – (2) = 109.114 KN

5

IS 800 - 1984 (iv)

bf = 300 Mpa

Failure of plate ‘A’ at sec (3) – (3) = 174. 01 KN

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 = πd 2   2  τvf x 4  

Strength of the solid plate = τ at x L x t (L = P) = 150 x 130 x 6 = 117.00 KN

= π(21.5) 2   2 100 x  4  

Efficiency = 89.1 x 100 = 76.15% 117.00

= 72.610 KN

Two plates 12 mm are joint by Double riveted double.

(ii) Strength of rivets in Bearing = bf. d.t

Cover bult joint as shown in dia. Using 20 mm dia of the

= 300 x 21.5 x 20

rivet design the pitch of the rivet. Take at = 150 Mpa

= 77.4 KN

also find the efficiency of the joint.

(iii) Rivet value = least of shearing & bearing

Given

= 72. 61 KN

at = 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

= 145. 22 KN

Assume PDS rivet, τ

vf

= 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.

π x 21.52 80 x = 29.04 KN 4

Min pitch = 2.5 D = 2.5 x 20 = 50 mm Provide Pitch = 100mm

KN (iii)

100 - 21.5 x 100 = 78.5% 100

Strength of the plate (thinner) per pitch length along sec (1) – (1)

lab joint, In which the pitch of the centrel row of the rivet

=σ

is half the pitch of rivet in outer row. Design the Joint &



vf

= 80 N/mm2

(P – d) t

Strength of plate per pitch length along sec (2) – (2)

Take: = 150 N/mm2

at

= 1500 P – 32250 (1)

Find the efficiency

at

Rivet value = least of shearing (or) bearing = 29.04 KN

Two plates 12mm & 10 mm tk are jointed by trible riveted



Bearing = bf x d x t = 250 x 21.5 x 10 = 53.75

(ii)

P −d η = x 100 P

= σ at (P – 2d ) t + Rivet value = 150 (P x 21.5)10 + 29044 = 1500 P – 35456  (2)

bf = 250 N/mm2

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

P − d 120 − 21.5 60 − 21.5 η= = x 100 ( or ) x 100 d 120 60 (82%) (64%)

Soln:-

Note:- Strength of the plate = 1500 (60) = 35456

d = 21.5 mm

= 54, 544 N

Step 2:

Step:

Assume 20 mm dia of rivet (PDS)

D = 20mm

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 = Tvf x

(i)

Step 4: Step 5: Check for the safety fo the joint q1 = qv1 + qv2

πd 2 π x 21.52 = 100 x 4 4

(ii)

P M + .x n ε r2

= 36305.03 N.

Strength in bearing = bf x d x t qn =

= 300 x 21.5 x 10

 (2) M .y ε r2

= 64500 N Rivet value R = 36/305 KN Step 3:

 (1)

qv =

r2 = Σ x2 + Σ y2

To find the no of the rivets vertical line

= 0.(6)2 + 4 (102 + 202)

Vertical line = 2 (given)

r2 = 2360 cm

M=PxQ

qv =

= 120 x 25 = 3000 KN.cm

120 3000 + .6 10 2360

M1 = M 3000 = = 150 KN.cm no . of. vertical rivet 2

qv = 19.627 qn =

n1= 6M1 6 x 1500 = R.P 36305 x 10

M . y max εr 2

= 4.979 ≅ 5 nos

9

IS 800 - 1984 =

ft max = M . y max εy 2

3000 x 20 2320 qn = 25.42

M = P x R = 80 x 16 = 1280 KN. Cm

q=

Ymax = 6 + 6 + 6 + 6 = 24 cm qn + qv = 19.627 + 25.42 2

2

2

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 x 24 2160

Check the safety of the joint as shown in diagram Step 1:-

Ftmax = 14.22KN

Assume diameter = 16 mm

Step 4

Using PDS dia of the rivet hole = 16 + 1.5

Find vf (cal) =

= 17.5 mm

Q πd 2 / 4

Step 2: Shear stress due to

=

Direct laod (Q) = p/n

8 πx17.5 2 / 4

= 80/10 Q = 8 KN

= 0.083

Step 3: Find ft max

= 83. 26 N/mm2 10

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 / 4

Step 1 Assume 20mm dia of the rivet

=

Using PDS rivet 14.22 π17.52 / 4

= 59.11 N/mm

Dia of the rivet hoel = 20 + 1.5 = 21.5 mm 2

Minimum pitch distance (D) = 2.5

Step 6

= 2.5 x 20 = 50 mm

Check

Maximum pitch distance = 32t

τvf (cal ) σ (cal ) + tf ≤ 1.4 τvf σ tf

= 32 x 21.5 = 688

≅

690 mm

Adopt pitch distance to 100 mm

33.26 59.11 + ≤ 1.4 0.4fy 0.6fy

Step 2: To find rivet value Strength in shearing = vf x π d2 / 4 = 100 x

33.26 0.4 x 4

π x 21.52 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

IS 800 - 1984 Strength is bearing = σ

bt

xdxt

n1 = 0.8 6 x 1000 36.3 x 10

= 300 x 21.5 x 10 = 64.5 KN 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 RP

Step v To find ft max Ft max = M x y max ∑ y2

= 0.8 6xM 36.3 x 10

M = 2000 KN.cm

M=Pxe

y2 = 2(102 + 202 + 302) = 2800

= 200 x 10

Ymax = 10 + 10 + 10 = 30 cm

= 2000 KN.cm

Ft (max) = 2000 x 30 2800

M1 = m no of rivet line

Ft max = 21.42 KN

= 2000/2 M1 = 1000 KN.cm 12

IS 800 - 1984 Q=

Check P 200 = n 8

τ vf (cal) σ tf (cal ) + ≤ 1.4 τ vf σ tf

Q = 25 KN 68.8 5.9 + ≤ 1.4 0.4 fy 0.6 fy

Step VI τ

vt

(cal) = Q πd 2 / 4

68.8 59 + ≤ 1 .4 0 .4 x ? 0 .6 x ?

=

Design the riveted connection between the column ISMB

25 π x 21.52 / 4

300 & beam ISMB 350 transmitting the load of 35 KN/m over a span of 9m. Assume 20mm dia PDS rivet

= 0.0688 KN/mm2 = 68.8 N/mm2

Given data:-

tf (cal)=

Load = 35 KN/m

Ft max πd 2 / 4

Span l = 9m Solution:

=

Step 1

21.42 π x 21.52 / 4

The beam is connected to the column using angle. The size of the angle should not be less than 3d.

= 0.05 

tf

(cat) = 59 N/mm2 13

IS 800 - 1984

∴

length of the angle = 3 x 21.5

Rivet value = 52.24 KN Number of rivets (n) = P R .v

= 64.5 mm 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 35 x 9 = =157.5 2 2

line Angle & flange of column line

n =

To find the rivet value: Strength of rivet in double shearing = 2 x τ

vf

157.5 52.24

x

πd 2 4

n=3 Step 3

= 2 x 100 x

π x 21.52 4

Connection between flange of the column of angle. To find the rivet value Strength in single shearing = τ

= 7261 KN.

vf

x

πd 2 4

Bearing for web of ISMB 350 = bE. d x t = 300 x 21.5 x 8.1 = 52. 24 KN 14

IS 800 - 1984 = 100 x

Maximum pitch = 32 t (or) 300

π x 21.5 4

2

= 32 x 8.1 = 259.2 < 300

= 36. 30 KN

Max pitch distance = 260 mm

Bearing for flange of ISMB = σ

bf

xdxt

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

∴

Number of rivets n =

A tie bar 100 mm x 16 mm is to be welded to another plate P R .V

150 mm x 16 mm. find the minimum overlab length required if 8 mm fillet weld of used. Take σ N/mm2. σ

= 157.5 36.3 n = 4.33

bt

= 165 N/mm2, σ

vf

= 100 N/mm2

Given data:at = 150 N/mm2

≅

5 nos

bt = 165 N/mm2 vf = 100 N/mm2

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 = at x b x t

= 50 15

at

= 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/mm2. 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/mm2.

=

Given data:-

2400 x 10 560

3

σ

at

= 150 N/mm2

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 = σ

Length is to be provided by side

= 130 x 150 x 16

Fillet = 430 – 200

= 360 KN

Side = 230 mm

Strength of the Joint = vf. K.S.L 16

at

xbxt

IS 800 - 1984 L = 50 + 2

L= Load Value of the weld

502 + 802 ]2 L = 477.38 mm

Value of the weld = vf. K.S

Strength of the Joint = 100 x 0.707 x 8 x 4m

= 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 103 424.2

Load 4 strength 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/mm2.

equation to find the either x1 (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 = x1 + x2

17

IS 800 - 1984 Load acting at a distance lxx = 44.6 mm ( from steel

An =



W 150 x 103 = = 1000 mm 2 σ at 0.6 x 250

table, Pg.) Moment of the load at top = 200 x 103 x 44.6

Step 2: choose 70 x 70 x 10 mm in steel table

= 8.92 x 10 N/mm (1)

(From steel table) L1 = L2 = 70mm, t = 10mm, d = 20 + 1.5

Moment resistance of the bottom = 424.4 x2 x 130

= 21.5

6

3

= 63.66 x 10 x 2

A = 1302 mm2

x2 =

Anet = A1 + A2

6

8.92 x 10 63.66 x 103

A1 = (L1 – t/2) t – d x t = [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

3A 1 3A 1 + 3A 2

KN. Assume Fy. 250 N/mm2 and dia of the rivet is 20mm. step 1

18

IS 800 - 1984 =

A2 = 700 mm2 3 x 55 (3 x 455) + ( 650)

K= 3A 1 3 x 735 = = 0.76 3A1 + A 2 (3 x 735) + 700

K = 0.67 Anet = 870.50 mm2

Anet = 735 + (0.76 x 700)

Step 3

Anet = 1266.32 mm2

Load = Anet x σ

Load = Anet x σ

at

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

tension of 250 KN using double angle section back to back

Gross Area A = 1650 mm2

of the Gusset plate (Opp side) dia of rivet is 20mm.

L1 = 100, L2 – 75, t = 10

Step 1

Anet = A1 + KA2

An =

W 250 x 103 = σ at 150

A1 = [L1 – t/2] t – (d x t) = [100 – 10/2]10 (21.5 x 10) 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 mm2 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 Anet

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 = σ

Ag = 1.35 Anet

Bending stress due to moment = bt (cal) = M/I. y

(b) Angles on either side of the gusset plate

The section is safe the following intraction formula is

Ag = 1.25 Anet

satisfied.

at

(cal) = W/An

(iii) (a) For chain riveting in plate section

for uniaxial

(b) for zig – zag riveting (or) staggered riveting

σat (cal ) σ bt ( cal ) + ≤ 1 for uniaxial bending σat σ bt

(i) Anet = t [(b – nd) 4 + m [s2/4g]

bending

Anet = t (b – nd)

(ii) Anet = t [(b – nd) + s2/4g1 + s22 4g2)] 20

IS 800 - 1984 Sectional properties

σ (cal ) σat (cal ) σ btx (cal ) + + bty ≤1 0.6 fy 0.66fy 0.66 fy

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

= 11846.4 mm2

= 30% to 40% of the preliminary stress

at (cal)

at = (0.3 to 0.4) of 0.6 fy

W 450 x 103 = = 37.99N / mm 2 An 11846.4

= 0.3 x 0.6 x 250 = 45

bt (cal) =

Area required = W 450 x 1000 = 0.3σat 0.3 x 0.6 x 250

M 19000 x 1000 400 .y = 3 x 15082x 109 x = 12.60 N / mm 2 I 2

= 10000 mm2 This is offered by two channel section

Check for Intraction formula

Area = 10000/2 = 5000 mm2

σat ( cal ) σ bt (cal ) + ≤ 1 0.6 x fy 0.6 x fy

Choose ISMC 400 21

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 leff = 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

σ

Aread =

at

= 150 N/mm2

ac

= 80 N/mm2

Load 600 x 103 = = 7500mm 2 σ ac 80

Step 1 Anet = A1 + KA2

Choose ISHB 300 (1) 63.0 kg/m

K= 5A l 5 A1 + A 2

Area = 80.25 cm2 = 8025 mm2 rxx = 12.70 m ryy = 5.29 cm

K = 0.714 Anet = 650 + (0.714 x 1300) = 1578.2 mm2 Strength = Anet x σ

at

22

IS 800 - 1984 σ

Slanderness ratio λ =

bc

(act) =

L eff 3.5 = rmin 0.052

W 600 x 103 = = 74.76 area 8025

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  Pg 46]

bc

permissible

Fy = 250 (assume)

Effective length = 0.85L = 2.125 m

λ = 66.16

σ

60  122

Areqd =

ac

= 60 N/mm2

W 150 x 103 = = 2500mm 2 σ ac 60

70  112 x = 122 –  122 − 112   ( 66 . 16 − 60 )    60 − 70    

Choose the ISA 150 x 150 x 20 A = 2903 mm2

x = 115.84

rxx = 46.3mm

σ

ryy = 46.3mm

bc

permissible = 115.84 N/mm2

λ

bc (assume) = 80 N/mm2 σ

bc

L eff 2.125 x 103 = = = 45.80 rmin 46.3

Permissible > bc assume => Hence safe

[Refer IS 800 – 1984 Table 5.1 Pg.30] 23

IS 800 - 1984 fy = 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

σ

bc

Permissible = 134.88 N/mm2

Area required =

σ

bc

(assume) < bc permissible

Where bc

= 80 N/mm2

W σ bc

Hence safe σ

bc

=

250 x 103 80

(act) =

W 150 x 103 = = = 51.67 area 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 mm2

Given

Select ISA 90 x 90 x 100 mm @ 13.41 kg/m

Load = 250 KN

A = 1703 mm2

L = 3,

Ixx = 126.7 x 104 mm4

If Double angle continuous member

Iyy = 126.7 x 104 mm4

Leff = 0.7L to L

Lyy = 25.9 mm (lyy – centrid distance) 24

IS 800 - 1984 To calculate rmin:

=

2.55 x 103 27.27

rmin =

R xx =

I xx = A

2L xx 2a

λ = 93.50 To find the σ

126.7 x 104 = 1703

(Permissible)

[Refer IS 800 – 1984  Pg 39  Table 5.1] 90  90

rxx = 27.27 mm

100  80

Ryy

93.5  86.5

2 iyy + a ( iyy + t / 2 ) 2 2a

bc (Permissible) = 86.5 σ

=

 126.7 x 104  1703  126.7 x 104 10  2 +  +  2 2  2 2   1703 2x 2

bc

bc

(act) = 250 x 103 2(1703)

2

bc (act) = 73.4 Design a compression member consist of two channels placed with toes facing each other subjected to load of

Ryy = 447.9 x 103 mm

1300 KN. Eff ht of the column is 8m. Design the comp.

λ=

member and also design a lacing system

l eff rmin

Solu: 25

IS 800 - 1984 Assume ac = 110 N/mm2

Rxx = rxx = 154.8 mm

Areq =

ryy =

K 1 1300 x 103 = = 11818.18mm 2 σ ac 110

Iyy 399.074 x 106 = = 178.07mm A 2 x 6293 (or )

This is offered two channel. Therefore

Iyy / 2a

Area of single channel = 11818.18 = 5909.09 2

λ=

L eff 8 x 103 = = 51.68 rmin 154.8

Select ISMC 400 @ 494 N/m Area = 6293 mm2

λ = 51.68 4

ixx = 15082.8 x 10 mm 4

iyy = 504.8 x 10 mm

4

[Refer Table 5.1 Pg. 39 Is 800 = 1984

4

50  132

rxx = 154.8 mm

60  122

ryy = 28.3 mm

51.68  130.32

cyy = 24.2 mm 6

Ixx = 2ixx = 39.656 x 10 mm

4



bc

= 130.32

σ

bc

(dct) =

2

Iyy = 2[iyy + a (S – (yy) ]

Load 1300 x 103 = = 103.29 Area 2 x 6293

= 2 [504.8 x 104 + (6293) [200 – 24.2)2] = 399.074 x 106 mm4

σ 26

ac

(act) = 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 = 3 x 20 = 60 mm

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 + 3002

Connection are at 50mm from the edge. Distance C/C of rivet across = 400 – 50 – 50 = 300 mm

l1 = 424.26 mm

Assume the angle of inclination of lacing bar  = 450

‘t’ =

C = 2 x 300 = 600 mm [For angle 450 = 25 desare equal]

424.26 = 10.61mm 40

Check < 0.7 λ < 50

T

C rmin

≅

12 mm

Size of the bar = 60 x 12 mm < 0.7 x 51.68 < 50

Check for:-

600 28.3

(i) Slenderness ratio λ > 145

21.20 < 36.176 < 50 Hence ‘C’ is ok. Size of the Lacing bar:27

IS 800 - 1984 rL =

P 22.98 x 103 σ bc (cal ) = = = 31.92 N / mm 2 A 60 x 12

 60 x 123    IL 12   = = 3.46 AL (60 x 12)

bc (Per) [For lacing bar] => λ = 122.62 fy = 250

122.62 > 145

120  64

Hence O.K

130  57 from Table 5.1

(ii) Check for compressive stress

122.62  62.17 in IS 800 – 1984

Compression leading in the lacing bar =

bc (Per) = 62.17 N/mm2

V n Sin Q

(62.17) σ

bc

(Per) > σ

bc

(cal) (31.92)

Hence safe V = 25% of the load

(iii) Check for tensile stresses:-

=

P=

2 .5 x 1300 100

V = 22.98 KN n Sin Q

V = 32.5 KN

σ

Comp. Load =

at

(cal) = P A net

3

32.5 x 10 = 22.98 KN 2 x sin 450

Anet = Agross – Area of rivet hole = (60 x 12) – (12 x 21.5) 28

IS 800 - 1984 = 462 mm2

Assume square base length of one side (l)

σ

L=

at

(cal) = 22.98 = 49.74 462

l = A = 187.5 x 103 = 433.012mm Provide 450 x 450 mm

at (Per) = 0.6 fy = 0.6 x 250 = 150 KN

Thickness

[150] at (Per) > at (Cal)

t=

Hence safe in tensile stress.

3w 2 (a − b 2 / 4 ) bs σ

Design a simple slab base resting on a concrete slab for the following data

a=

Load from the column = 750 KN

450 − 250 = 100 mm 2

Size of the column = ISHB 400 σ

cc

= 4 N/mm2

b=

SBC = 100 KN/m2

450 − 400 = 2500 2

Design the slab base. Soln:-

W=

Bearing Area =

Load 750 x 103 = = 3.7 N / mm 2 2 Area 450

Load 750 x 103 = = 187.5 x 103 mm 2 σcc 4 29

IS 800 - 1984 t=

Provide 3m x 3m of the pedestil

3 x (3.7) 502 (1002 − ) 185 4 t = 24.3 mm

≅

Area of pedestile = 3 x 3 = 9m2 Depth of the pedestile Assume 450 despersion projection of the pedestile beyond

25 mm

the base plate =

Design of pedestile

3 − 0.45 = 1.275 ≅ 1.3m 2

Size of the pedestil is design such that pressure on the soil is with in the safe bearing capacity of soil. Add 10% of the self wt

Adopt depth = 1.3 m

Total Load =

Size of the pedestal = 3 x 3 x 1.3 m 10   x 750   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 825 = = 8.25m 2 SBC 100

connections by 10mm thick battens. It transmit and axial load of 2000KN. SBC of soil at site is 300 KN/m2. The

Adopt square base, length of the one side (1)

safe permissible stresses of concrete base is 4 N/mm2.

L=

Design the gusseted base. Grillage foundation.

A = 8.25 = 8.25m 2

Load = 2000 KN 30

IS 800 - 1984 SBC = 300 KN/m2

= 714.29 mm

Area required =

Provide square plate = 750 mm x 750 mm Load 2000 x 103 = con . permissibl e 4

Cantilever projection of the plate from face of the gusset angle is checked for bending stress due to the concrete below.

= 500 x 103 mm2

Intensity of pressure below the plate =

It is shared by two angle =

load Area

500 x 10 mm 2 2 3

W=

Adopt angle section 150 x 75 x 12 mm gusset angles on

2000 x 103 = 3.56 N / mm 2 750 x 750

flange side width 75 mm long horizontal Allow 30 mm projection on either side in the direction of parallel to web.

Moment in the cantilever portion:

Length base plate parallel to the web

Wl 2 M = 2

L reqd = 400 x 2(10) + 2(12) + 2(75) + 2(30) = 654 mm

Where l = [750 – 400 – 2(10) – 2(12)] / 2

Provide length of base plate = 700 mm

l = 153 mm

Breath of the plate =

w = load per ‘m’ length = 3.56 N/mm for 1 mm width

A reqd 500 x 103 = L reqd 700



31

bs

= M/Z

IS 800 - 1984 Using 20mm  rivet (DDS)

185 = 41.67 x 10  bt 3 / 12  T   ( or )   Y  t/2  3

185 =

=

41.67 x 10  1x t 3    / ( t / 2)  12  3

To find the rivet value Strength in shearing =

π x d2 τv x 4

41.67 x 103 t2 / 6

2

185t = 41.67 x 10

=

π x 20.52 100 x 4

3

t = 36.76 mm

= 36.305 x 103

thickness of the base = 36.76 mm – 12 (thickness of the

Strength in bearing = bc x b x t

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 467.250 = = 12.87 R .V 36305

Outstanding length of the each side = 750 − 400 2

= 13 nos (or) 14 nos 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 103 mm3

Step I

Ixx = 9874.6 x 104 mm4 Iyy = 510.8 x 104 x mm4

Assume 3% adding as a selt wb of section Total load = 20 +

= 20.6

Step 3 check for shear

3    20 x  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 = 92.7 KN m r

= 20 + 0.481 W = 20.481

Step 2:

V = 20.431 (6/2 = 0.2)

Z=

V = 54.65 KN

M 0.27 x 106 = = 561.81 x 103mm 3 5 σ bt 10

τ

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 = 25.56 N / mm 2 [325 − 2(007] x 7

τ

av

(Per) = 0.45 fy = 0.45 x 250 = 112.5 N/mm



av

(Per) > τ

av

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 2

KN Design the beam, assuming fy = 250 N/mm2 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 =

2

5 WL 384 E I xx

154.5 x 6 = 231.75 KN.m 4

=

Z = M/σ 4

at

= 231.75 x 106 = 1.404 x 106 0.66 fy

5 x 20.43 x 6000 384 x ( 2.1x 105 ) x 9874.6 x 104

= 1404.55 x 106 mm 3

Ymax = 16.62 mm Step 2

Permissible deflection = λ 6000 = = 18.46 mm 325 325

Take ISLB 500 at 750 N/m W = 750

Ixx = 38570 x 104 mm4

Ymax < yper

A = 9550

Iyy = 1063.9 x 104 mm4

16.62 < 18.46

D = 500

Zxx = 1543.2 x 103 mm3

34

IS 800 - 1984 bf = 180

ryy = 33.4 mm

Ymax = WL3 154.5 x 103 x 60003 = = 3.58mm 48EI 48 x 2.1 x105 x 38579 x 104

tf = 14.1 tw = 9.2 Step 3 Check for shear

Yper = L/325 = 18.46 mm

Shear is calculated at a distance of ‘d’ from the support

Ymax < y per

V=

Hence in deflection W 2

In the above problem the beam is laterly un support between the own beam

W = P.L + Selt wt

Assume bc = 120 N/mm2 (120 to 130 N/mm2)

= 150 + (0.750 x 6)

M = 231.75 KN.m

τ

av

(cal)

Z= 3

V 77.25 x 10 = = 17.80 N / mm 2 dw.tw [500 − 2(14.1)] (9.2 )

231.75 = 1.93125m 3 120

τ av (Per) = 0.45 fy = 0.45 x 250

Z = 1031.25 mm3

= 112.5 N/mm2

Choose ISLB 550 at 863 N/m

av (cal) < τ

av

A = 12669 mm2

(Per)

Hence safe in shearing

Zxx = 1933.2 x 103 mm4

Step 4 Check for deflection

Ixx = 53161.6 x 104 mm4 tf = 15 mm tw 9.9 mm 35

IS 800 - 1984 ryy = 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− 127   x 1.2 = 131−   85 N 80  = 130.04

cross beam. ∴

L = 6/2 = 3m

l 3000 = = 86.207 ry 34.8

 130 N 126   x 1.2 = 130 −   80 N 85  = 129.04

D1 d 550 = = = 36.67 T tf 15

 130.04 N 129.09  = 130.04 −   35 − 40   = 129.71

dw 550 − 2(15) = = 52.53 tw 99



(cal) = M 231.75 x 106 = = = 119.88 N / mm 2 3 Z xx 1933.2 x 10

[Refer IS 800 – 1984 => Table 6.1 B => Page 58

bc (Per) = 129.71

T tf 15 = = = 1 .5 t tw 9.9

Hence the section is safe B5

85

bc

131

36.67

bc (Per) > bc (cal)

40

Check for shear:

130 36

IS 800 - 1984 Max shear at the support V = =

=

W WL W 1 + (or) + 2 2 2 2

150 (0.863 x 6) + 2 2

V = 75.59 KN.

37