IS 800 - 1984

1. A 10 mm thick Gusset plate is connected to 6 mm angle

section by Lap Joint. Find the rivet value of 16 mm dia of

power drivern Rivets

Givert

Dia – 16mm

Dia of rivet hole – 16 + 1.5 = 17.5 mm

Permissible Stresses For Power driven rivet

(Table 8.1 Page 95. IS 800 – 1984)

vf = 100 N/mm2

bt = 300 N/mm2

(i) Strength in shearing = τ vf x πd2

4

= 100 x

π x 17 . 52

4

= 240252.82 N

= 24.052 KN.

(ii) Strength in bear = bft x d x t

= 300 x 17.5 x 10

= 52. 5 KN

(iii) Rivet Value (R) = Least of the strength in shearing

(or) bearing

Rivet value (R) = 24.052 KN

2. Find the value of the 16 mm power driven rivets

connected a pair of double angle section consisting of ISA

75 x 75 x 6 mm through 10 mm thick Gusset plate. Find

the Rivet value.

Given

(D) Dia = 16 mm

(d) Dia of rivet hole = 16 + 1.5 = 17.5 mm

vf = 100 N/mm2

bf = 300 N/m

[Refer Table 8.1 Page 95 – IS 800 – 1984]

(ii) Rivet value

(i) Strength of the rivet in double shearing = 2[τ vf x

πd2

4 ]

1

IS 800 - 1984

= 2[100 x

π x 17 . 52

4 ]= 48.104 KN

(ii) Strength of the rivet in bearing = bt x d x t

= 300 x 17.5 x 6

= 31.50 KN

(iii) Rivet value = 31.50 N (Least one)

3. Find the efficiency of the joint in a boiler. Shell

connected using 16 mm dia of the rivet at a pitch of 60

mm C/C in a single riveted Lap Joint thickness of the plate

is 8 mm. The rivets are power driven shop rivet

Given

(1) D = 16 mm

d = 16 + 1.5 mm = 17.5 mm

(2) Pitch Distance = 60 mm C/C

(3) PDS - Rivets - vs = 100 N/mm2

bt = 300 N/mm2

at = 100 N/mm2

(4) Thickness of plate = 8mm

(i) η = Least of the shearing , Bearing , Tearig

Strength of the solid plate

(i) Strength in shearing = vf x

πd2

4= 100 x

π x 17 .52

4

= 24.052 KN

(ii) Strength in bearing = bt x d x t = 300 x 17.5 x 8 mm

= 42.00 KN

(iii) Strength in Tearing = at (P- d) t = 100 (60 – 17.5) x

8

= 42.5 KN x 8

= 34.00 KN

η = 24 . 052Strength of solid plate

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

= 100 x 60 x 8

= 48 KN

(v) η = 24 .052

48x 100 = 50 .1%

2

IS 800 - 1984

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

tension of 40 KN is connected to a Gusset plate 10 mm

thick design the Joint & sketch the arrangement of rivet.

Given

Angle section = ISA 90 x 90 x 6 mm

Load (P) = 40 KN

Thickness of plate = 10 mm

Solution

Step 1 Assume dia of rivet

Assume Take Dia (D) = 12 mm

Dia of hole = 12 + 1.5 = 13.5 mm

Step 2 Find the value of rivet

Assume Hand driven rivet

vf = 80 N/mm2 Refer table 8.1

bf = 250 N/mm2 Page 95

tf = 80 N/mm2 IS 800 - 1984

(i) Strength in shearing = vf x

πd2

4

= 80 x

π x 13 .52

4

= 11.45 KN

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

= 250 x 13.5 x 10

= 33. 75 KN

(iii) Rivet value = Least value of shearing & bearing

Rivet value (R) = 11.45 KN

Step 3 Number of Rivet

No of Rivet =

P (6 load )R (Rivet value)

=

4011.45

= 3. 49 ≃ 4 Nos

Step 4 Arrangement of rivet

Edge distance (d)

(i) d = 13.5 => Edge Distance = 19 mm [Refer Table 8.2]

Pitch Distance

(ii) Min = 2.5 x 12 (D) = 300 mm ¿ 50 mm

3

IS 800 - 1984

(iii) Maxi = 16 t (or) 200 (whichever is less)

= 16 x 6 = 96 mm (or) 200 (Take whichever is less)

Maxi = 96 mm ¿ 100mm

Step 5:

η = Least of the Shearing , bearing ∧ bearingStrength of the solid plate

= P − dp

x 100

= 50 − 13 .550

x 100

= 73%

6. Two plates 6 mm tk are Jointed by 14 mm dia of the

rivet in a triple straggled rivet Lap Joint as Shown in

diagram in what way the Joint will failed. If allowable

tensile stress 150 mpa, Allowable shering stress 90 mpa,

Allowable bearing 270 mpa. Also find the efficiency of

Joint.

Step 1 : Dia of rivets & holes

Nominal dia (D) = 14 mm

Dia of rivet (d) = 15.5 mm

Step 2 : Rivet value

(i) Strength in shearing = τ vf x

πd2

4

= 90 x

π x 15 . 52

4

= 16.982 KN

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

= 270 x 15.5 x 6

= 25.11 KN

(iii) Rivet value = 16.98 KN

(iv) Strength of Joint on the basis of rivet value = n x R.V

= 7 x 16.982

= 118.874 KN

Plate Failure (Consider Sec (1) – (1), (2) – (2), (3) – (3) for

plate A

Sec (3) – (3), (2) – (2), (3) – (3) for plate B)

at (P – d) t

4

IS 800 - 1984

Strength of the plate ‘A’ (a) section (1) – (1)

= at (L – 2d) t

= 150 (130 – 2 x 15.5) 6

= 89.1 KN

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

= Teaching strength (a) (2) – (2)

+ Strength Rivet Sec (1) – (1)

= 150 (130 – 3(15.5)) x 6 + 2(R.V)

= 150 [(130 – 3 (15.5)] x 6 + 2 (16, 982).

=109. 114 KN

Plate ‘A’ at section (2) – (2) can fail only it rivets at

section (1) – (1) also fail. In the strength of het rivet at sec

(1) – (1) will act along with the tearing of the plate (2) –

(2) section

Strength of the plat ‘A’ (a) sec (3) – (3)

= tearing strength of the section (2) – (2)

+ Rivet value of (1) – (1)

+ Rivet value of (2) – (2)

= 150 (130 – 2(15.5)) x 6 + 3 (16982) + 2(16982)

= 174.01 KN

Passable Failures

(i) Combined Failure of rivet = 118.87 KN

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

KN

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

KN

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

The weakness critical section is (1) – (1) of plate ‘A’

strength of the Joint = 89.1 KN

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

= 150 x 130 x 6

= 117.00 KN

Efficiency =

89 .1117.00

x 100 = 76 .15 %

Two plates 12 mm are joint by Double riveted double.

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

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

find the efficiency of the joint.

5

IS 800 - 1984

Given

at = 150 Mpa

Dia = 20 mm

Dia of the rivet hole = 20 + 1.5 = 21.5 mm

Thickness of plate = 12 mm

Assume PDS rivet, vf = 100 Mpa

bf = 300 Mpa

To find the rivet value

(i) Strength of rivets in double shear = 2[τvf x

πd2

4 ]=

2[100 xπ (21. 5 )2

4 ]= 72.610 KN

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

= 300 x 21.5 x 20

= 77.4 KN

(iii) Rivet value = least of shearing & bearing

= 72. 61 KN

For maximum efficiency of joint per pitch length,

Strength of plate per pitch = 2 x Rivet value

= 2 x 72.61

= 145. 22 KN

at (P – d) x t = 145.22KN

150 (P – 21.5) x 12 = 145. 22 KN (or) 145220

P = 102.178 mm.

Min pitch = 2.5 D = 2.5 x 20 = 50 mm

Provide Pitch = 100mm

η = P− dP

x 100

100 − 21 .5100

x 100 = 78 .5%

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

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

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

Find the efficiency

Take:

at = 150 N/mm2

6

IS 800 - 1984

vf = 80 N/mm2

bf = 250 N/mm2

Assume 20mm dia

Rivet hole = 20 + 1.5 = 21.5 mm

To find the rivet value

(i) Shearing = vf x d2/4 =

80 xπ x 21. 52

4= 29 . 04 KN

(ii) Bearing = bf x d x t = 250 x 21.5 x 10 = 53.75

KN

(iii) Rivet value = least of shearing (or) bearing =

29.04 KN

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

(1) – (1)

= at (P – d) t

= 1500 P – 32250 (1)

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

= at (P – 2d ) t + Rivet value

= 150 (P x 21.5)10 + 29044

= 1500 P – 35456 (2)

Sec (2) – (2) is weaker along which the strength of the

plate is 1500 P – 35456

For maximum efficiency the strength of the per

pitch length should be equal to strength of rivet per pitch

length.

1500P – 35456 = 4 x R.V

1500 P = (4 x 29044) + 35456

P = 101. 088mm

Min = 2.5D = 2.5 x 20 = 50mm

Max = 32t (or) 300 whichever is lesser.

= 32 (10) = 320 mm > 300 mm

Max = 300 mm

Outer row pitch = 120 mm

Inner row pitch = 60 mm

η =P − dd

=120 − 21 .5120

x 100 ( or )60 − 21 .560

x 100

(82% ) (64 % )

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

7

IS 800 - 1984

= 54, 544 N

42 = 4 x 29044N = 116, 76 N

Take which value is user so take pitch, efficiency = sec (2)

Design a bracket connection using two vertical lines of the

rivet load carried by each plate is 120 KN the bracket plate

of 10 mm tk are connected to 12mm tk flange plate.

Assume pitch of 10 cm and horizontal distance between

the vertical line is 12 cm. eccentriciting load is 25 cm.

Given

Load (P = 120 KN

Thickness of the plate = 12 mm

Thickness of the Flange = 12 mm

Pitch (P) = 10 cm

Gauge (G) = 12 cm

Eccentricity (e) = 25 cm

Soln:-

Step: Assume 20 mm dia of rivet (PDS)

D = 20mm

d = 21.5 mm

Step 2: Find the rivet value

(i) Strength in shear = Tvf x

πd2

4= 100 x

π x 21. 52

4

(ii) = 36305.03 N.

Strength in bearing = bf x d x t

= 300 x 21.5 x 10

= 64500 N

Rivet value R = 36/305 KN

Step 3: To find the no of the rivets vertical line

Vertical line = 2 (given)

M = P x Q

= 120 x 25 = 3000 KN.cm

M1 =

Mno . of . vertical rivet

= 30002

= 150 KN.cm

n1= √6M1

R . P= √ 6 x 1500

36305 x 10= 4 .979 ≃ 5 nos

Step 4:

8

IS 800 - 1984

Step 5: Check for the safety fo the joint

q1 = qv1 + qv2

qv =

Pn

+ M

ε r2. x

(1)

qn =

M

ε r2. y

(2)

r2 = x2 + y2

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

r2 = 2360 cm

qv =

12010

+ 30002360

.6

qv = 19.627

qn =

M

εr2. ymax

=

30002320

x 20

qn = 25.42

q = √qn2+qv2 = √19 .6272 + 25 .422

q = 32.118 KN < R = 36.305 K

q < R hence safe

Check the safety of the joint as shown in diagram

Step 1:-

Assume diameter = 16 mm

Using PDS dia of the rivet hole = 16 + 1.5

= 17.5 mm

Step 2:

Shear stress due to

Direct laod (Q) = p/n

= 80/10

Q = 8 KN

Step 3: Find ft max

ft max =

M

εy2. ymax

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

Ymax = 6 + 6 + 6 + 6 = 24 cm

y2 = 2(62 + 122 + 182 + 242)

y2 = 2160 cm

9

IS 800 - 1984

Ft max =

12802160

x 24

Ftmax = 14.22KN

Step 4

Find vf (cal) =

Q

πd2 /4

=

8

πx 17 .52/4

= 0.083

= 83. 26 N/mm2

Step 5

To find tf (cal) =

f t max

πd2 /4

=

14 . 22

π 17 . 52/ 4

= 59.11 N/mm2

Step 6

Check

τvf (cal )τvf

+σ tf (cal )

σ tf

≤ 1 . 4

33 . 260 .4 fy

+ 59. 110 . 6 fy

≤ 1 . 4

33 . 260 .4 x 4

A bracket plate of 10 mm thick is to be connected to the

base of the flange using angles the load is applying

through the bracket to the column. E = 10cm, P = 200KN.

Design the connection between the angle & column

Step 1

Assume 20mm dia of the rivet

Using PDS rivet

Dia of the rivet hoel = 20 + 1.5

= 21.5 mm

Minimum pitch distance (D) = 2.5

= 2.5 x 20 = 50 mm

Maximum pitch distance = 32t

= 32 x 21.5 = 688 ¿ 690 mm

10

IS 800 - 1984

Adopt pitch distance to 100 mm

Step 2: To find rivet value

Strength in shearing = vf x d2 / 4

= 100 x

π x 21 .52

4

= 36.30 KN

Strength is bearing = bt x d x t

= 300 x 21.5 x 10

= 64.5 KN

Rivet value = 36.3 KN

Step 3

No of rivet (n1) = 0.8 √ 6mRP

= 0.8 √ 6 x M36 .3 x 10

M = P x e

= 200 x 10

= 2000 KN.cm

M1 =

mno of rivet line

= 2000/2

M1 = 1000 KN.cm

n1 = 0.8 √ 6 x 100036 .3 x 10

= 3.25 4 Nos

Adopt 4 nos of rivet each row

Step 4 Arrangement of the rivet

Step v To find ft max

Ft max =

M

∑ y2x ymax

M = 2000 KN.cm

y2 = 2(102 + 202 + 302) = 2800

Ymax = 10 + 10 + 10 = 30 cm

Ft (max) =

20002800

x 30

Ft max = 21.42 KN

11

IS 800 - 1984

Q =

Pn

= 2008

Q = 25 KN

Step VI

vt (cal) =

Q

πd2 /4

=

25

π x 21 .52 /4

= 0.0688 KN/mm2 = 68.8 N/mm2

tf (cal)=

F t max

πd2 /4

=

21 . 42

π x 21. 52 /4

= 0.05

tf (cat) = 59 N/mm2

Check

τ vf (cal )τ vf

+σ tf (cal )

σ tf

≤ 1.4

68 . 80 .4 fy

+ 5 . 90 .6 fy

≤ 1. 4

68 . 80 .4 x ?

+ 590 . 6 x ?

≤ 1 . 4

Design the riveted connection between the column ISMB

300 & beam ISMB 350 transmitting the load of 35 KN/m

over a span of 9m. Assume 20mm dia PDS rivet

Given data:-

Load = 35 KN/m

Span l = 9m

Solution:

Step 1

The beam is connected to the column using angle.

The size of the angle should not be less than 3d.

∴ length of the angle = 3 x 21.5

= 64.5 mm

Choose ISA 75 x 75 x 10mm Angles

Step 2

12

IS 800 - 1984

Connection between the angle & web of the beam

line

Angle & flange of column line

To find the rivet value:

Strength of rivet in double shearing = 2 x vf x

πd2

4

= 2 x 100 x

π x 21 .52

4

= 7261 KN.

Bearing for web of ISMB 350 = bE. d x t

= 300 x 21.5 x 8.1

= 52. 24 KN

Rivet value = 52.24 KN

Number of rivets (n) =

PR . v

Load at the Joint (P) = Reaction from the beam

=

WL2

= 35 x 92

= 157 .5

n =

157 .552 .24

n = 3

Step 3

Connection between flange of the column of angle.

To find the rivet value

Strength in single shearing = vf x

πd2

4

= 100 x

π x 21 .52

4

= 36. 30 KN

Bearing for flange of ISMB = bf x d x t

= 300 x 21.5 x 10

= 64.5 KN

Rivet value = 36.3 KN

Number of rivets n =

PR .V

=

157 . 536 . 3

13

IS 800 - 1984

n = 4.33 ¿ 5 nos

Step 4 Arrangement of rivet

Minimum pitch = 2.50

= 2.5 x 20

= 50

Maximum pitch = 32 t (or) 300

= 32 x 8.1

= 259.2 < 300

Max pitch distance = 260 mm

Minimum edge distance = 29 mm [From IS 800 – 1984.

Pg

∴Provide 30 mm edge distance

A tie bar 100 mm x 16 mm is to be welded to another plate

150 mm x 16 mm. find the minimum overlab length

required if 8 mm fillet weld of used. Take at = 150

N/mm2. bt = 165 N/mm2, vf = 100 N/mm2

Given data:-

at = 150 N/mm2

bt = 165 N/mm2

vf = 100 N/mm2

Size of the fillet welt = 8mm

Load = the strength of the smaller plate

Strength of the smaller plate = at x b x t

= 150 x 100 x 16mm

P = 240 KN

The value of the weld = ks. Fs

= 0.78 x 8 x 100

R = 560 N/mm

Length of the Weld = P/R

=

2400 x 103

560

L = 428. 57 mm

L ¿ 430 mm

For minimum over lab in of the plate both end fillet weld

& side fillet weld are provided.

The length of the end fillet = 2 x 100

= 200 mm

Length is to be provided by side

14

IS 800 - 1984

Fillet = 430 – 200

Side = 230 mm

The length shared by two side

Length of the onside = 230 / 2 = 115 mm

The weld Lab Joint is to be provided to connect two tie bar

150 x 16 mm stress in the plate is 150 N/mm2. To check

the design if the size of the weld is 8mm & shear stress is

taken as 100 N/mm2.

Given data:-

at = 150 N/mm2

b = 150 mm

t = 16 mm

vf = 100 N/mm2

S = 8 mm

Solution

To check the safety of the Joint should not be more

than load at the joint.

Load at the Joint = at x b x t

= 130 x 150 x 16

= 360 KN

Strength of the Joint = vf. K.S.L

L = 50 + 2 √502+802 ]2

L = 477.38 mm

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

= 266. 61 KN

Hence the design is unsafe

Load = 360

Strength = 267 KN

Load 4 strength

A 150 mm x 115 mm x 8mm angle section carries a tensile

load of 200 KN it is to be connected gusset plate using 6

mm fillet weld at the extreame of the longer length (leg)

Design the Joint along the shear stress 100 N/mm2.

Angle section is unequal. The load is acting excentricity.

We have to adopt

Let x1 be the length of the weld at tob

X2 be the length of the weld at bottom

15

IS 800 - 1984

Total length = x1 + x2

L =

LoadValue of the weld

Value of the weld = vf. K.S

= 100 x 0.707 x 6 mm

= 424.2 N/mm2

Load at the Joint P = 200 KN

L =

200 x 103

424 .2

L= 471. 47 mm

x1 + x2 – 150 = 471.48 – 150 = 321.48 mm

Two unknowns S1 one equation to create another

equation to find the either x1 (or) x2.

Moment of the at the top = Moment of resistance of het

bottom weld at top.

Unequal section = 150 x 115 x 8 mm

Load acting at a distance lxx = 44.6 mm (∵ from steel

table, Pg.)

Moment of the load at top = 200 x 103 x 44.6

= 8.92 x 106 N/mm (1)

Moment resistance of the bottom = 424.4 x2 x 130

= 63.66 x 103 x 2

x2 =

8 . 92 x 106

63 . 66 x 103

x2 = 140.119 mm

x1 + x2 = 321.48

x1 + 140.119 = 321.48

x1 = 321.48 – 140.119

x1 = 181.36 mm

Design a single angle section carring a axial load of 150

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

step 1

An =

Wσat

= 150 x 103

0.6 x 250= 1000 mm2

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

16

IS 800 - 1984

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

= 21.5

A = 1302 mm2

Anet = A1 + A2

A1 = (L1 – t/2) t – d x t

= [70 – 10/2] 10 – 25 x 10

A1 = 435 mm2

A2 = [L2 – t/2]t

= [70 – 10/2] 10

A2 = 650 mm2

K =

3 A1

3 A1+3 A2

=

3 x 55(3 x 455 ) + (650)

K = 0.67

Anet = 870.50 mm2

Step 3

Load = Anet x at

= 870.50 x 150

= 130.575 KN < 150 KN

So unsafe

Hence trial section choose ISA 100 x 75 x 10

Gross Area A = 1650 mm2

L1 = 100, L2 – 75, t = 10

Anet = A1 + KA2

A1 = [L1 – t/2] t – (d x t)

= [100 – 10/2]10 (21.5 x 10)

A1 = 735 mm2

A2 = [L2 – t/2] t

= [75 – 10/2] x 10

A2 = 700 mm2

K =

3 A1

3 A1 + A2

= 3 x 735(3 x 735 ) + 700

= 0 .76

Anet = 735 + (0.76 x 700)

Anet = 1266.32 mm2

Load = Anet x at

17

IS 800 - 1984

= 1266.32 x (0.6 x 250)

Load (w) = 189.948 KN

Design a tension member of roof truss carrings a axial

tension of 250 KN using double angle section back to back

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

Step 1

An =

Wσat

= 250 x 103

150

An = 1666.66 mm2

Step 2: Selected a section whose Gross area is

1.5 x An area = 1.5 x 1666.66 = 2500 mm2

Take section ISA 150 x 115 x 12 mm

L1 = 150 L2 = 115 t = 12 mm A = 3038mm2

Anet = Ag – Area of Rivet holes.

= 3038 – 2(21.5 x 10)

Anet = 2608 mm2

Load = 2606 x 0.6 x 250

Load = 391.2 KN

Note: (i) for single angle section

Ag = 1.35 to 1.5 times of Anet

(ii) For Double angle section

(a) angles on some side of the gusset plate

Ag = 1.35 Anet

(b) Angles on either side of the gusset plate

Ag = 1.25 Anet

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

Anet = t (b – nd)

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

(i) Anet = t [(b – nd) 4 + m [s2/4g]

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

Where

b – breadth

n – no of rivets

d – dia of nivet hole

m – no of zig. Zag line along the failure line

s – Pitch

g – guage

18

IS 800 - 1984

Member under axial load and moment

There will be axial tension due to axial force and

bending stress due to bending moment.

Direct stress due to axial tension = at (cal) = W/An

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

The section is safe the following intraction formula is

satisfied.

σat (cal)σ at

+σbt (cal )

σbt

≤ 1 for uniaxial bendingfor uniaxial

bending

σat (cal)0. 6 fy

+σbtx (cal)0 . 66 fy

+σ bty( cal)0 .66 fy

≤ 1

For biaxial bending

A tension member made of two channels placed back to

back carries a moment of 1900 N.m in addition to a direct

tension of 450 KN. Design the section assume that f y =

250 N/mm2

For the selection of the section assume that

at = 30% to 40% of the preliminary stress

at = (0.3 to 0.4) of 0.6 fy

= 0.3 x 0.6 x 250

= 45

Area required =

W0 .3σ at

= 450 x 10000 .3 x 0 . 6 x 250

= 10000 mm2

This is offered by two channel section

Area = 10000/2 = 5000 mm2

Choose ISMC 400

Sectional properties

Area = 6293 mm2

Ixx = 15082 cm4

tw = 8.6 mm

Adopt 20mm dia

Rivet for the connection

An = Gross Area – area of Rivet hole

= 2 x 6293 – 4 (21.5 x 8.6)

= 11846.4 mm2

19

IS 800 - 1984

at (cal)

WAn

= 450 x 103

11846.4= 37 .99N /mm2

bt (cal) =

MI

. y = 19000 x 10003 x 15082 x 109 x

4002

= 12. 60 N /mm2

Check for Intraction formula

σ at (cal )0 .6 x fy

+σbt (cal )0 .6 x fy

≤ 1

37 . 99

0 .6 x 250 +12. 600. 66 x 250

≤ 1

0.253 + 0.07 < 1

0.33 < 1

Hence the section is safe

A tie of roof truss consist of double angles ISA 100 x 75

x10 mm with it’s short leg back. To back and long leg

connected to the same side of the gusset plate with 16 mm

dia of the rivet determine the strength of the member take

at = 150 N/mm2

Step 1

Anet = A1 + KA2

K =

5 A l

5 A1 + A2

K = 0.714

Anet = 650 + (0.714 x 1300) = 1578.2 mm2

Strength = Anet x at

= 1578.2 x 150

= 236730 N (or) 236.730 KN.

A rolled steel is used as a column of height 5.5 m both

ends are hinged. Design the column to carried axial toad of

600 KN.

Solution:

Both ends are hinged leff = L

Leff = 3.5 m

Load (P) = 600 KN

Rolled steel section ac = 80 N/mm2

20

IS 800 - 1984

Aread =

Loadσ ac

= 600 x 103

80= 7500mm2

Choose ISHB 300 (1) 63.0 kg/m

Area = 80.25 cm2 = 8025 mm2

rxx = 12.70 m

ryy = 5.29 cm

Slanderness ratio λ =

Leff

rmin

= 3 .50 .052

λ = 66.16

IS 800 – 1984 Table 3.5 page 38

To find the bc permissible

Fy = 250 (assume)

λ = 66.16

60 122

70 112

x = 122 – [(122−112

60−70 ) (66 . 16 − 60 )]x = 115.84

bc permissible = 115.84 N/mm2

bc (assume) = 80 N/mm2

bc Permissible > bc assume => Hence safe

bc (act) =

Warea

= 600 x 103

8025= 74 .76

Design a single angle discontinuous structs connected by 2

rivets to a gusset plate length 2.5m, applied load 150 KN.

[Refer Is 800 – 1984 -> CL 5.2 Pg 46]

Effective length = 0.85L = 2.125 m

ac = 60 N/mm2

Areqd =

Wσac

= 150 x 103

60= 2500mm2

Choose the ISA 150 x 150 x 20

A = 2903 mm2

rxx = 46.3mm

ryy = 46.3mm

λ =

Leff

rmin

= 2 .125 x 103

46 . 3= 45 . 80

21

IS 800 - 1984

[Refer IS 800 – 1984 Table 5.1 Pg.30]

fy = 250

λ = 45.89

40 139

50 132

45.89 134.88

bc Permissible = 134.88 N/mm2

bc (assume) < bc permissible

Hence safe

Where

bc (act) = = W

area= 150 x 103

2903= 51. 67

Design a double angle strut continuous to a load of 250

KN/m3 length 3m.

Given

Load = 250 KN

L = 3,

If Double angle continuous member

Leff = 0.7L to L

Solution

Leff = 0.85l

= 0.85 x 3m

Leff = 2.55 m

Assume double angle bc = 80 N/mm2

Area required =

Wσbc

=

250 x 103

80

= 3125 mm2

Single angle area = 3125 / 2

= 1562.5 mm2

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

A = 1703 mm2

Ixx = 126.7 x 104 mm4

Iyy = 126.7 x 104 mm4

Lyy = 25.9 mm (lyy – centrid distance)

22

IS 800 - 1984

To calculate rmin:

rmin = Rxx = √ I xx

A= √ 2 Lxx

2a

= 126 . 7 x 104

1703

rxx = 27.27 mm

Ryy √ 2 iyy + a ( iyy+t /2)2

2a

= √ 2(126 .7 x 104

2 ) + 17032 (126 .7 x 104

2+ 10

2 )2

2 x1703

2

Ryy = 447.9 x 103 mm

λ =

leffrmin

=

2. 55 x 103

27 . 27

λ = 93.50

To find the bc (Permissible)

[Refer IS 800 – 1984 Pg 39 Table 5.1]

90 90

100 80

93.5 86.5

bc (Permissible) = 86.5

bc (act) =

250 x 103

2(1703 )

bc (act) = 73.4

Design a compression member consist of two channels

placed with toes facing each other subjected to load of

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

member and also design a lacing system

Solu:

Assume ac = 110 N/mm2

Areq =

K 1

σac

= 1300 x 103

110= 11818. 18mm2

This is offered two channel. Therefore

23

IS 800 - 1984

Area of single channel =

11818. 182

= 5909 . 09

Select ISMC 400 @ 494 N/m

Area = 6293 mm2

ixx = 15082.8 x 104 mm4

iyy = 504.8 x 104 mm4

rxx = 154.8 mm

ryy = 28.3 mm

cyy = 24.2 mm

Ixx = 2ixx = 39.656 x 106 mm4

Iyy = 2[iyy + a (S – (yy)2]

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

= 399.074 x 106 mm4

Rxx = rxx = 154.8 mm

ryy =

√IyyA = √399 .074 x 106

2 x 6293= 178. 07mm

(or )√Iyy /2a

λ =

Leff

rmin

= 8 x 103

154 . 8= 51. 68

λ = 51.68

[Refer Table 5.1 Pg. 39 Is 800 = 1984

50 132

60 122

51.68 130.32

bc = 130.32

bc (dct) =

LoadArea

= 1300 x 103

2 x 6293= 103 . 29

ac (act) = 103.29

bc (per) = 130.32 > bc (act) = 103.29

The design is safe

Design of Lacing

Assume that the connection to the lacking bar in mode at

the centre of the flange width

Connection are at 50mm from the edge.

Distance C/C of rivet across = 400 – 50 – 50 = 300 mm

24

IS 800 - 1984

Assume the angle of inclination of lacing bar = 450

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

Check

Crmin < 0.7 λ < 50

60028 . 3 < 0.7 x 51.68 < 50

21.20 < 36.176 < 50

Hence ‘C’ is ok.

Size of the Lacing bar:-

Assume 20mm dia of rivet for connections.

Width of the bar = 3 = 3 x 20 = 60 mm

Thickness of bar ‘t’ = l1 / 40 for single Lacing

l1 = length of the lacking bar

l1 = √3002 + 3002

l1 = 424.26 mm

‘t’ =

424 .2640

= 10 .61 mm

T ¿ 12 mm

Size of the bar = 60 x 12 mm

Check for:-

(i) Slenderness ratio

λ > 145

rL = √ ILAL

= √ (60 x 123

12 )(60 x 12 )

= 3 .46

122.62 > 145

Hence O.K

(ii) Check for compressive stress

Compression leading in the lacing bar =

Vn Sin Q

V = 25% of the load

=

2 .5100

x 1300

V = 32.5 KN

25

IS 800 - 1984

Comp. Load =

32 .5 x 103

2 x sin 450= 22. 98 KN

σ bc (cal ) = PA

= 22. 98 x 103

60 x 12= 31. 92 N /mm2

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

fy = 250

120 64

130 57 from Table 5.1

122.62 62.17 in IS 800 – 1984

bc (Per) = 62.17 N/mm2

(62.17) bc (Per) > bc (cal) (31.92)

Hence safe

(iii) Check for tensile stresses:-

P =

Vn Sin Q

= 22. 98 KN

at (cal) =

PAnet

Anet = Agross – Area of rivet hole

= (60 x 12) – (12 x 21.5)

= 462 mm2

at (cal) =

22 . 98462

= 49 .74

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

[150] at (Per) > at (Cal)

Hence safe in tensile stress.

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

following data

Load from the column = 750 KN

Size of the column = ISHB 400

cc = 4 N/mm2

SBC = 100 KN/m2

Design the slab base.

Soln:-

Bearing Area =

Loadσ cc

= 750 x 103

4= 187 .5 x 103 mm2

Assume square base length of one side (l)

26

IS 800 - 1984

L = l = √A = √187 .5 x 103 = 433 . 012mm

Provide 450 x 450 mm

Thickness

t = √ 3 wσbs

(a2 − b2/ 4 )

a =

450 − 2502

= 100 mm

b =

450 − 4002

= 2500

W =

LoadArea

= 750 x 103

4502= 3 .7 N /mm2

t = √ 3 x (3 . 7 )185

(1002 − 502

4)

t = 24.3 mm ¿ 25 mm

Design of pedestile

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

Total Load = (750 + 10

100x 750)

Base area of the pedestil:-

Area =

LoadSBC

= 825100

= 8 .25m2

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

L = √A = √8 .25 = 8.25m2

Provide 3m x 3m of the pedestil

Area of pedestile = 3 x 3 = 9m2

Depth of the pedestile

Assume 450 despersion projection of the pedestile beyond

the base plate =

3−0 . 452

= 1. 275 ≃ 1 .3m

Adopt depth = 1.3 m

Size of the pedestal = 3 x 3 x 1.3 m

Size of the base plate = 450 x 450 x 25 mm

Design of gusseted plate

27

IS 800 - 1984

A builtup steel column compressing 2ISWB 400

RSJ section with their webs spaced at 325 mm and

connections by 10mm thick battens. It transmit and axial

load of 2000KN. SBC of soil at site is 300 KN/m2. The

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

Design the gusseted base. Grillage foundation.

Load = 2000 KN

SBC = 300 KN/m2

Area required =

Loadcon . permissible

= 2000 x 103

4

= 500 x 103 mm2

It is shared by two angle =

500 x 103

2mm2

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

flange side width 75 mm long horizontal

Allow 30 mm projection on either side in the direction of

parallel to web.

Length base plate parallel to the web

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

= 654 mm

Provide length of base plate = 700 mm

Breath of the plate =

A reqd

Lreqd

= 500 x 103

700

= 714.29 mm

Provide square plate = 750 mm x 750 mm

Cantilever projection of the plate from face of the gusset

angle is checked for bending stress due to the concrete

below.

Intensity of pressure below the plate =

loadArea

W =

2000 x 103

750 x 750= 3. 56 N /mm2

Moment in the cantilever portion:

M = Wl2

2

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

l = 153 mm

28

IS 800 - 1984

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

bs = M/Z

185 =

41 . 67 x 103

(TY ) (or ) [ bt 3 /12t /2 ]

= 41 .67 x 103

( 1 x t3

12 ) / ( t /2)

185 = 41. 67 x 103

t2/6

185t2 = 41.67 x 103

t = 36.76 mm

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

angle leg)

= 24.76 mm

Provide 25 mm plate thickness, size of Gusset base plate =

750 x 750 x 25 mm

CONNECTIONS:

Outstanding length of the each side =

750 − 4002

Load on each connection = 3.56 x 750 x 175

= 467.250 KN

Using 20mm rivet (DDS)

To find the rivet value

Strength in shearing = τ v x

π x d2

4

= 100 x

π x 20 .52

4

= 36.305 x 103

Strength in bearing = bc x b x t

= 300 x 21.5 x 10

= 64.5 x 103

Rivet value = 30.305 KN

No of rivet =

LoadR .V

= 467 .25036305

= 12 . 87

= 13 nos (or) 14 nos

Pitch:

Min pitch = 2.5 x D = 2.5 x 20 = 40 mm

Max pitch = 12 x t = 12 x 10 = 120 mm

Provide 60 mm pitch edge distance code book = 30 mm

29

IS 800 - 1984

A beam supporting a floor glab carries a distributed load

of 20 KN/m span for the beam is 6m design suitable I –

section for the beam

Step I

Assume 3% adding as a selt wb of section

Total load = 20 + (20 x

3100 )

= 20.6

B.M =

WL2

r= 92 .7 KN m

Step 2:

Z =

Mσbt

= 0 .27 x 106

105= 561.81 x 103mm3

Choose ISLB 325 @ 431 N/m

Area = 5490 m2

Depth = 325 mm

bf = 165 mm

tf = 9.8 mm

tw = 7.0 mm

Zxx = 607.7 x 103 mm3

Ixx = 9874.6 x 104 mm4

Iyy = 510.8 x 104 x mm4

Step 3 check for shear

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

V = w (l/2 – d)

W = adi + self wt (of section)

= 20 + 0.481

W = 20.481

V = 20.431 (6/2 = 0.2)

V = 54.65 KN

av (cal) =

VArea of web portion

=

54 . 65[ 325 − 2(007 ] x 7

= 25. 56 N /mm2

av (Per) = 0.45 fy = 0.45 x 250 = 112.5 N/mm2

av (Per) > av (cal)

112.5 N/mm2 > 25.56 N/mm2

Step 4

30

IS 800 - 1984

Check for deflection

Ymax =

5WL2

384 E I xx

=

5 x 20 . 43 x 60004

384 x (2 .1 x 105 ) x 9874 .6 x 104

Ymax = 16.62 mm

Permissible deflection =

λ325

= 6000325

= 18 .46 mm

Ymax < yper

16.62 < 18.46

Hence the section is safe in deflection.

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

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

developes B.M, S.F and check for shear and deflection

Step 1

Assuming 3% adding as a self wt of the section

Total load = 150 + (150 x 3/100) = 154.5 mm

B.M =

154 . 5 x 64

= 231 .75 KN .m

Z = M/at =

231 .75 x 106

0 .66 fy= 1 . 404 x 106

¿ 1404 .55 x 106 mm3

Step 2

Take ISLB 500 at 750 N/m

W = 750 Ixx = 38570 x 104 mm4

A = 9550 Iyy = 1063.9 x 104 mm4

D = 500 Zxx = 1543.2 x 103 mm3

bf = 180 ryy = 33.4 mm

tf = 14.1

tw = 9.2

Step 3 Check for shear

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

V =

W2

W = P.L + Selt wt

= 150 + (0.750 x 6)

31

IS 800 - 1984

av (cal)

Vdw . tw

= 77 . 25 x 103

[ 500 − 2(14 .1) ] (9 .2 )= 17 .80 N /mm2

av (Per) = 0.45 fy = 0.45 x 250

= 112.5 N/mm2

av (cal) < av (Per)

Hence safe in shearing

Step 4 Check for deflection

Ymax =

WL3

48 EI= 154 . 5 x 103 x 60003

48 x 2 .1 x105 x 38579 x 104= 3 .58mm

Yper = L/325 = 18.46 mm

Ymax < y per

Hence in deflection

In the above problem the beam is laterly un support

between the own beam

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

M = 231.75 KN.m

Z =

231 .75120

= 1. 93125m3

Z = 1031.25 mm3

Choose ISLB 550 at 863 N/m

A = 12669 mm2

Zxx = 1933.2 x 103 mm4

Ixx = 53161.6 x 104 mm4

tf = 15 mm

tw 9.9 mm

ryy = 34.8 mm

To find bc permissible for the selected section effective

length of the compressive flange distance between the

cross beam.

∴L = 6/2 = 3m

lry

= 300034 .8

= 86 .207

D1

T= d

t f= 550

15= 36 .67

dwtw

=550−2(15 )99

= 52 .53

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

Tt

= tftw

= 159. 9

= 1. 5

32

IS 800 - 1984

B5 36.67 40

85 131 130

86.21 130.01 129.71 129.04

90 127 120

= 131 − (131−12785 N 80 ) x 1 .2

= 130 .04

= 130 − (130 N 12680 N 85 ) x 1 . 2

= 129 .04

= 130 .04 − (130 .04 N 129 .0935−40 )

= 129 .71

bc (cal) = = M

Z xx

= 231 .75 x 106

1933 . 2 x 103= 119. 88 N /mm2

bc (Per) = 129.71

Hence the section is safe

bc (Per) > bc (cal)

Check for shear:

Max shear at the support V = = W

2+ WL

2( or ) W

2+ 1

2

= 1502

+(0 .863 x 6 )

2

V = 75.59 KN.

33