CONNECTIONS:

1. Two plates are connected with G4.6 grade bolts of 20 mm diameter as shown in Figure. Find the efficiency of the joint. The yield and ultimate strength of plate are 250MPa and 410MPa respectively.

Solution:

Strength of a joint is the least of:

a. Strength of the plate in joint b. Strength of the bolt in the joint against Shear c. Strength of the bolt in the joint against bearing

a. Strength of the plate in joint

Tdn = 0.9 An fn /γmn An = Net area Available = 20 ( 180 – 3x22) = 2280 mm2

Therefore Tdn = 2280 x 0.9 x 410 / 1.25 = 673.056 kN

b. Strength of the bolt in the joint against single shear (Shear plane through unthreaded part)

43

2

∏=

dfV unsb

= (400/ √3)(22/7*1/4*202)

= 56.59 kN Strength of joint in shear due to all bolts = Ps= 6 x 56.59 = 339.54 kN

20 mm

180 mm

Page 1 of 16

c. Strength of the bolt in the joint against bearing

Vnpb = 2.5 kb d t fu

kb = e/3do = 40/(3x22) = 0.4545

(Assuming flame cutting, e=1.7 do = 1.7x22 = 37.4mm say 40 mm)

= 25.03

−od

p = 25.0223

60−

x= 0.66 (min. pitch = 2.5do =55mm, say 60mm)

= u

ub

ff =

410400 = 0.975

= 1

Vnpb = 2.5 x 0.454 x 20 x 20 x 410 = 186.35 kN

Strength of joint in bearing due to all bolts = Psb = 6 x 186.35 = 1118.1 kN

Strength of plate along 1-1 or 2-2 is given by

Pp= (b-nd)t {0.9 fu /γm1 } = ( 180 – 3 x 22) 20 x 0.9 x 410/1.25 x 10-3= 673 kN.

Strength of solid plate due to yielding = bt fy/γmo = 818.18 kN.

Efficiency = 10018.818

673 x = 82.25 %

2. Two plates of 16 mm thick are to be joined using M20 bolts of grade 4.6 by 1. Lap joint and 2. Butt joint using 10 mm, two cover plates. Determine the bolt value.

1. Lap joint :

Design shear strength of the bolt = 25.11 x

∏4

278.03

dfub

= 25.11 x

∏4

22078.03

400 = 45.27 kN

16 mm plates

Page 2 of 16

Design strength of bolt in bearing = Vnpb = 2.5 kb d t fu

Assuming flame cutting, e=1.7 do = 1.7x22 = 37.4mm say 40 mm

and minimum pitch = 2.5d =55mm, say 60mm

kb is the least of

= od

e3

= 40/ (3x22) = 0.606

= 25.03

−od

p = 25.0223

60−

x= 0.66

= u

ub

ff =

410400 = 0.975

= 1

Therefore Kb = 0.606

Design strength = 2.5 kb d t fu = 2.5 x 0.606 x 20 x 16 x 400 / 1.25 = 155.14kN

Bolt value = 45.27 kN

2. Butt Joint : 10 mm cover plate

Design shear strength of the bolt =

∏4

278.13

dfub / 1.25 = 103.31 kN

16mm

Page 3 of 16

(Since, one plate is connected in the threaded region where as another is through unthreaded. That is double shear.)

Design strength = 2.5 kb d t fu = 2.5 x 0.606 x 20 x 16 x 400 / 1.25 = 155.36 kN

(kb =0.606 as in the above ); (t is the minimum of (10+10) or 16 )

3. A member of a truss consists of 2 angles ISA(75x75x6)mm placed back to back. It carries a factored load of 200 kN and is connected to a gusset plate of 8 mm thick placed in between the angles. Determine the number of 16 mm dia 4.6 grade bolts required for the joint. Assume fu of the plates as 410 MPa.

`

Double Shear Strength/ bolt = 25.11 x

∏4

278.13

dfub =

∏4

21678.13

400/ 1.25 = 66.12 kN

Design Strength of the bearing/ bolt = 2.5 kb d t fu

Assuming flame cutting, e=1.7 do = 1.7 x 18 = 30.6 mm say 35 mm

and minimum pitch = 2.5do = 45mm.

kb is the least of

= od

e3

= 35/ (3x18) = 0.648

= 25.03

−od

p = 25.0183

45−

x= 0.584

2-ISA 75x75x6 2-ISA 75x75x6

Gusset plate

Page 4 of 16

= u

ub

ff =

410400 = 0.975

= 1

Therefore kb = 0.584

Design strength = 2.5 kb d t fu = 2.5 x 0.584 x 16 x 8 x 410 / 1.25 x 10-3 = 57.52 kN

No of bolts required = 52.57

200 = 3.48 = Say 4 bolts.

5.An angle section of a truss with 8 mm thick carries 120kN to a gusset plate of 10 mm thick using lap joint by M20 bolt of 4.6 grade. Find number of bolts.

Design shear strength of the bolt = 25.11 x

∏4

278.03

dfub = 45.26 kN

Design strength of bolt in bearing / bolt = Vnpb = 2.5 kb d t fu

Assuming flame cutting, e=1.7 do = 1.7 x 22 = 37.4mm say 40 mm

& Min. pitch = 2.5d =55mm, say 60mm

kb is the least of

= e/3do = 40/ (3x22) = 0.606

= 25.03

−od

p = 25.0223

60−

x= 0.66

= u

ub

ff =

410400 = 0.975

= 1

Therefore Kb = 0.606

Design strength = 2.5 kb d t fu

= 2.5 x 0.606 x 20 x 8 x 410 / 1.25 = 77.57 kN

Bolt Value = 45.26 kN

Page 5 of 16

No of Bolts required = 26.45

120 =3 bolts should be provided.

6. A bracket connected to a column loaded as shown in fig. if M20 bolt of 4.6 grade are used. Determine the value of factored load ‘p’ which can be carried safely.

Given no. of bolts n = 5

For M20 bolts d = 20 mm

do = 22 mm

fbolt = 400 N/mm2

fu = 410 N/mm2

ISMC 300, thickness of flange = 13.6 mm

Tanα=80/60, α=53.13°, Cosα=0.6

Bolt value is given as,

Design strength of the bolt = 25.11 x

∏4

278.03

dfub = 45.26 kN

Design strength of bolt in bearing / bolt = Vnpb = 2.5 kb d t fu / 1.25

Assuming flame cutting, e=1.7 do = 1.7x22 = 37.4mm say 40 mm

& Min. pitch = 2.5d =55mm, say 60mm

kb is the least of

250mm

AA

10mm plate

ISMC 300 60 60

60

60

80

80

Page 6 of 16

= od

e3

= 40/ (3x22) = 0.606

= 25.03

−od

p = 25.0223

60−

x= 0.66

= u

ub

ff =

410400 = 0.975

= 1

Therefore Kb = 0.606

Design strength = 2.5 kb d t fu/1.25 = 2.5 x 0.606 x 20 x 10 x 410 / 1.25 x 10-3 = 99.384 kN

Direct shear force on the farthest bolt = F1= 5P = 0.2 P

Distance of farthest bolt = √602 + 802 = 100 mm

(since ∑𝑟2= 4x1002)

We know that F2 = ∑ 2rPer =

40000100250xPx = 0.625P

Considering bolt A-A,

F1 = 0.2 P

F2 = 0.625P

Therefore cosα = 0.6

Resultant force in the critical bolt is

FR = �𝐹12 + 𝐹22 + 2𝐹1𝐹2 cos𝛼

= √. 04𝑃2 + .39𝑃2 + 0.15𝑃2

Equating resultant force to bolt value, we have,

FR = 0.762P » 45.27 = 0.762P » P = 59.41 kN. (On one Bracket)

α

α

F2 F1

80mm

60mm CG

rmax

Page 7 of 16

7. A Bracket is bolted to a flange of a column ISHB 300 @ 577 N/m as shown in the figure. Determine the Maximum value of P (factored load) which can be carried safely.

M20, bolts of 4.6 Grade are used,

Solution:

Given number of bolts n = 24

For M20 bolts d = 20 mm

do = 22 mm

fub = 400 N/mm2

fu = 410 N/mm2

ISHB 300, thickness of flange = 13.6 mm

350 mm

8 mm plate

90 80 80

ISHB 300

50

25

1

2

4

3

5

6

Page 8 of 16

Bolt value is given as,

Design strength of the bolt = 25.11 x

∏4

278.03

dfub = 45.26 kN

Design strength of bolt in bearing / bolt = Vnpb = 2.5 kb d t fu / 1.25

kb is the least of (from the figure, pitch and edge distance are taken)

= od

e3

= 25/ (3x22) = 0.378

= 25.03

−od

p = 25.0223

80−

x= 0.507

= u

ub

ff =

410400 = 0.97

= 1

Therefore Kb = 0.378

Design strength = 2.5 kb d t fu/1.25 = 2.5 x 0.378 x 20 x 8 x 410 / 1.25 = 48.38 kN

Direct shear force on the farthest bolt (A) = F1= 24P = 0.042 P

Distance of farthest bolt = √2752 + 702 = 283.76 mm

∑𝑟2= 4[283.762+235.632+188.482+143.262+102.592+74.332] = 4[208150] = 832600 mm2

Therefore F2 = ∑ 2r

Per = 832600

76.283350xPx = 0.12p

Cosα = 76.283

70

FR = �𝐹12 + 𝐹22 + 2𝐹1𝐹2 cos𝛼

FR = 0.135P

» P = 333.8 kN.

Page 9 of 16

8. A bracket is bolted to flange of a column as shown using 8mm thick gusset plate. Use M20 of grade 4.6 design the connections.

Solution:

For M20 bolts d = 20 mm

do = 22 mm

fub = 400 N/mm2

fu = 410 N/mm2

e = 40mm say

p = 60mm say

Design strength of the bolt = 25.11 x

∏4

278.03

dfub = 45.26 kN

Design strength of bolt in bearing per bolt = Vnpb = 2.5 kb d t fu / 1.25

kb is the least of

= od

e3

= 40/ (3x22) = 0.606

300 kN 350 mm

Page 10 of 16

= 25.03

−od

p = 25.0223

60−

x= 0.66

= u

ub

ff =

410400 = 0.975

= 1

Therefore Kb = 0.606, assuming the thickness of bracket as 8mm less than tf.

Design strength = 2.5 kb d t fu/1.25 = 2.5 x 0.606 x 20 x 8 x 410 / 1.25 = 79.51 kN

We know that e = �6𝑀2𝑣𝑝

= �6 𝑥 350 𝑥 3002 𝑥 45.26 𝑥 60

= 10.76. Say 12 numbers in each line.

Distance of extreme bolts from CG = r = √702 + 3302 = 337.34 mm

∑𝑟2 = 4[3302+2702+2102+1502+902+302+6(702)]

∑𝑟2= 1147200 mm2

Force in the extreme bolt due to moment = ∑ 2rPer =

114720034.337350300 xx = 30.87 kN

And also F1 = nP =

24300 = 12.5

Cosα = 34.337

70

FR = �𝐹12 + 𝐹22 + 2𝐹1𝐹2 cos𝛼

FR = 36.65 kN (which is less than bolt value 45.27kN )

Therefore Ok.

Page 11 of 16

9. Determine the shear capacity of the bolt used in connecting 2 plates as shown. Slip resistance is designed at the service loads. Use HSFG bolts of 8.8 grade µf =0.3, d= 20mm.

For HSLG bolts of 8.8 Grade with 20mm dia,

Fub = 800MPa

Fy = 0.8x800 = 640MPa

Design Strength of HSFG bolt = 1.1

1 x µf x ne x kn x fo

Here, µf = 0.3

ne = 2

Kn = 1 (Assumed)

Fo = Anb x fo

= 0.78 x 4∏ d2 x 0.7 x 800 = 137.2 kN

Therefore strength = 1.1

1 x 0.3 x 2 x 1x 137.2 = 74.85 kN

Therefore strength of the joint = 2x74.85 = 149.7kN

10. Design the bolted connection between the flange of a column ISHB 450 at 907 N/m and a bracket plate 15mm thick. The bracket plate is supporting a load of 150 kN at an eccentricity of 350mm. Adopt 20 dia HSFG bolts of grade 8.8 consider µf =0.3

40 40 40 40 65 65

Page 12 of 16

From the steel table, for ISHB 450,

Width of the flange = b = 250mm

Thickness = tf = 13.7 mm

Strength of the bolt = 1.1

1 x µf x ne x kn x fo

Here, µf = 0.3

ne = 1

Kn = 1 (Assumed)

Fo = Anb x fo

= 0.78 x 4∏ d2 x 0.7 x 800 = 137.2

Therefore strength = 1.1

1 x 0.3 x 1 x 1x 137.2 = 37.425 kN

No. of bolts required in each row is

= �6𝑀2𝑣𝑝

= �6 𝑥 150 𝑥 3502 𝑥 37.42 𝑥 60

= 8.375. Therefore 9 numbers in each line.

Say 18 bolts of totally

∑𝑟2 = {(702+2402) + (702+1802) + (702+1202) + (702+602) +702= 132500mm2.

∑ 2rPer =

132500250350150 xx = 99.05 kN

15mm bracket plate

350

140 45 45

60 40

1

2

4 3

Page 13 of 16

11. Design a suitable longitudinal fillet weld to connect 2 plates as shown in figure for the full strength of smaller plate.

Solution.

Minimum size of weld = S = 1/3 x 16 = 5.3mm say 6mm.

Maximum size of weld = (12-1.5) = 10.5 mm say 10mm.

So, let the size of the weld be S= 10mm

fu= 410 MPa, fy= 240 Mpa

Strength of the plate (due to shear) = m

yg fAγ

. = 1.1

240 x 12 x 120 = 314.18 kN ------- i

Let effective length of the weld = Lw

Let the normal throat thickness of weld = 0.7(size of weld, S) = 0.7x10 = 7mm

Design Strength of the weld = 3uf x

25.11 x Area of weld ------ ii

From i and ii we get

= 3

410 x25.11 x Lw x 7 = 314180

That is Lw= 237 mm

Therefore provide 10mm thick weld of length 120mm on either side as shown.

Note: The lap can be reduced by providing welds on 3 or 4 sides. For four side welding using S=6 or 7 mm, suitable lap can be provided.

12mm thick

120mm

16mmthick plate

S=10mm

180mm 120mm

P

Page 14 of 16

12. A tie member of a roof truss 2ISA 100x75x8 mm has to resist a factored force of 300kN. Design the welded connection if 8mm thick gusset plate is used for connection. The connection is made at the fabricators shop.

Sol:

Size of the weld,

Near the curved edge, size of weld = 0.75 x 8 = 6 mm

Near the square edge 1.5 mm less than 8mm = 8-1.5 = 6.5mm

Minimum size of weld = 8/3 = 2.66, say 3 mm.

So adopt 6mm size weld.

Strength of weld per unit length = 3uf x

25.11

= 3

410 x25.11 = 189.37 N

Equating strength to the given force we get, 300x103 = Lw x 6 x 189.37

Therefore Lw = 264mm

Actual length = 264 x 100/70 = 377mm say 380mm

L1 x 31 = 69L2

L1= 31

69 2L and L1 + L2 = 190mm

On solving, L1 = 131mm and L2 = 59mm

2-ISA 100x75x8

Gusset plate

2-ISA 100x75x8

135 mm

59 mm

Page 15 of 16

10 mm

Load carried P = 0.7 x 6 x 190 x 2 x 410/(√3 x1.25) = 302.94 kN

13. Design a suitable Plug weld to connect 2 plates as shown in figure for 100% efficiency.

200 mm

Strength of the plate by yielding = mo

yg fAγ

. = 1.1

20 x 10 x 200= 454.54 kN

Minimum size = 1/3 x10 = 3.33 mm say 4mm.

Maximum = 10 – 1.5 = 8.5mm

So let the size of the weld = 8mm

Total length of the weld available = 200+200=400mm

Effective length of the weld = 2(200-2x8) = 368mm

Load taken by the weld = 0.7 S L x 3uf x

25.11

= 5.6 x 368 x 3

410 x25.11 = 390.25 kN

To achieve 100% efficiency, slot welds are to be provided to resist a force of (454.54 – 390.25) That is 64.29 kN.

Strength of the weld/mm2 = = 3uf x

25.11 =

3410 x

25.11 = 189.37 N/mm2

Area of slot = 37.1891029.64 3x = 339.49 = 340 mm2

So let the slot size be 10x20 mm2 of 2 slots as shown in figure.

20 10

20 10

Page 16 of 16