Bolt Connection

BOLTED CONNECTION LAP JOINT Desi gn a l ap joi nt conn ectin g two plates plates 120mm 120mm x 8mm to tr ansmi ansmi t a 1. Desi f actore actor ed load of 120 kN.Use kN .Use 12mm diame di ameter ter bolts bol ts of gr ade 4.6 and and plate pl ates s of of grade gr ade 410.

Sln: Gi ven n data  1) Give a) Type of joint = lap joint  b) Plate thickness = t = 8 mm c) Bolt diameter = 12mm 2 d) Ultimate strength of bolt = f u bolt = 4 x 100 = 400 N/mm 2 e) Yield stress of bolt material = f y = 0.6 x 400 = 240 N/mm . 2 f) Ultimate strength of plate = f u plate = 410 N/mm 2 g) Yield stress for plate = f y = 250 N/mm . 2) Str Str ength of the bol bol t 

Minimum edge distance (For 12mm bolts)= 20 mm e =20mm Minimum pitch = 2.5 x bolt diameter = 2.5 x 12 = 30 mm P = 30mm Gross diameter = d0 = 12+1 = 13 mm (i) Design shear strength of one bolt   Vdsb =    √    nn = 1 ; ns = 0    = 16298 N Vdsb =

[         ]

 [ √   

(a) Design bearing strength of one bolt Vdpb =





[ 2.5 K  b dt f u]

 K  b is the least of the following (i) (ii) (iii) (iv)

   =    = 0.513    –  0.25  0.25 =  0.25 = 0.519     –  0.25    =  = 0.98 1





 ]

K  b = 0.513

Vdpb =





[ 2.5 K  b dt f u] =

  [ 2.5 x 0.513 x 12x8 x 400] = 39398.4 N

Lesser design strength of one bolt = 16298 N 3) No. of bolts

Bolt value = 16298 N  Number of bolts required =

     = 7.4

4) Strength of the plate

(a) Design tensile strength of plate ( Yielding consideration) Tdg =



 [ Ag f y]  =  [ 120 x 8 x 250] = 218182 N

(b) Design tensile strength of plate ( Rupture consideration) Tdn =

  =

[0.9 An f u]

 

[0.9 (120 –  2x13) x 8 x 410] = 221990 N

Lesser will be the Design strength of joint = 218182 N > 120 kN Hence design is safe.

2. Two plates 12mm X 60mm are conn ected in a l ap j oin t with 4 bolts of 16mm diameter i n shown i n figur e below.Determi ne the str ength of the 2 

 j oint.U l ti mate str ength of bolt = 400 N/mm . Ul tim ate str ength f or pl ate 2 

= 410 N/mm .

Sln: Step : 1- Gi ven data

a)  b) c) d) e)

Type of joint = lap joint Plate thickness = t = 12mm Bolt diameter = 16mm Ultimate strength of bolt = f u bolt = 400 N/mm 2 Ultimate strength of plate = f u plate = 410 N/mm 2.

Step : 2- Str ength of th e bolt

Gross diameter

= d0 = 16+2 = 18 mm

(b) Design shear strength of one bolt   Vdsb =  √   Vdsb =

 

[         ]   [ √       ] = 28975 N

(c) Design bearing strength of one bolt Vdpb =





[ 2.5 K  b dt f u]

 K  b is the least of the following

(v) (vi) (vii) (viii)

   =    = 0.56    –  0.25 =     –  0.25 = 0.49    =  = 0.98 1

 K  b = 0.49 Vdpb =





[ 2.5 K  b dt f u] =

  [ 2.5 x 0.49 x 16x12 x 400 = 75264 N

Lesser design strength of one bolt = 28975 N Total design strength of four bolt = 4 x 28975 = 115900 N Step : 3- Str ength of the plate

(a) Design tensile strength of plate ( Yielding consideration) Tdg =



 [ Ag f y]  =  [ 60 x 12 x 250] = 163636 N

(b) Design tensile strength of plate ( Rupture consideration) Tdn =

  =

[0.9 An f u]

 

[0.9 (60 –  18) x 12 x 410] = 148780 N

Lesser will be the Design strength of joint = 115900 N

BUTT JOINT

Find the efficiency of the butt joint as shown in fig below. Bolts are 16mm diameter of grade 4.6.Cover plates are 8mm thick.

Sln: 1) Given data 

h) Type of joint = Butt joint i) Plate thickness = t = 12 mm  j) Bolt diameter = 16 mm 2 k) Ultimate strength of bolt = f u bolt = 4 x 100 = 400 N/mm 2 l) Yield stress of bolt material = f y = 0.6 x 400 = 240 N/mm . 2 m) Ultimate strength of plate = f u plate = 410 N/mm 2 n) Yield stress for plate = f y = 250 N/mm . 2) Str ength of pl ate 

(a) Design tensile strength of plate ( Yielding consideration) Tdg =



 [ Ag f y]  =  [ 190 x 12 x 250] = 436360 N

(b) Design tensile strength of plate ( Rupture consideration) Tdn =

  =

[0.9 An f u]

 

[0.9 (160 – 2x 18) x 12 x 410] = 439260N

3) Str ength of bolt 

Gross diameter

= d0 = 16+2 = 18 mm

(d) Design shearing strength of one bolt   Vdsb =2x  √   Vdsb =2x

[         ]     [ √       ] = 57950N

(e) Design bearing strength of one bolt Vdpb =





[ 2.5 K  b dt f u]

 K  b is the least of the following (ix) (x) (xi) (xii)

   =    = 0.56    –  0.25 =     –  0.25 = 0.676    =  = 0.98 1

 K  b = 0.56 Vdpb =





[ 2.5 K  b dt f u] =

  [ 2.5 x 0.56x 16x12 x 400 = 85250 N

Lesser design strength of one bolt = 57950 N Total design strength of four bolt = 4 x 57950 = 231800 N

Efficiency of the joint =

             =  x 100 = 53.12%

ECCENTRIC BOLTED CONNECTIONS

There are two types of bracket connections 1. Bolted connection subjected to moments in the plane of the connection. 2. Bolted connection subjected to moments in a plane normal to the plane of the connection.

Type -1 A wor ki ng l oad of 150 kN i s appl ied to a bracket plate at an eccentr i city of 300mm. Si xteen bolts of 20mm diameter are arr anged in two r ows with 8 bol ts per r ow. The rows are 200mm apar t. The pitch of bolts i n each ver ti cal r ow is 80mm.The thi ckness of the bracket plate is 12.5mm.I nvesti gate the safety of th e desi gn.

With respect to G. 2

2

2

2

2

2

2

2

∑x +∑y = 16 (100) +4 [280 + 200 + 120 + 40 ] mm

= 697600 mm

2

Consider the bolt marked A (i)

Resistance against translation =

(ii)

Torsional shear = Sa = K r a K=

(iii)

(iv)

      = 

    =   = 93575 N

= 64.5 N/mm

Sa = K r a = 64.5 x r a Total vertical component V= 9375 + 64.5 r a sinϴ V= 9375 + 64.5 x 100 = 15825 N Horizontal component H= 64.5 r a Cos ϴ H = 64.5 x 280 = 18060 N

(v)

Resultant resistance

√    = √  

 R= R = 24010 N

Factored shear load on the bolt A = 1.5 x 24010 = 36015 N

Bolt diameter = do = 20 mm Gross diameter

= d0 = 20+2 = 22 mm

1. Design shearing strength of one bolt

         ] [    √        =  [ √       ] = 45274 N

Vdsb =



Vdsb

2. Design bearing strength of one bolt Vdpb =





[ 2.5 K  b dt f u]

 K  b is the least of the following

   =    = 0.606   –    –  (xiv)  0.25 =     0.25 = 0.962   (xv)   =   = 0.975 (xiii)

(xvi) 1  K  b = 0.606

Vdpb = =





[ 2.5 K  b dt f u]

  [ 2.5 x 0.606x 22x12.5 x 400 = 121200 N

Bolt Value = 45274 N But factored load on the bolt = 36015 N

The design is safe.

Type -2

I nvesti gate the safety of the bolts connecting the angl es and the colu mn f lange in the ar r angement shown in F igur e. Th e bolts are 20mm in diameter . The factored load on th e bracket i s 180 kN at an eccentr icity of 250mm fr om the column flange.

Sln. Bolt diameter = d = 20mm Design shearing strength of one bolt

 [  √          ]  [     ] = 45274 N  =  √  

Vdsb =



Vdsb

Design tensile strength of one bolt

  Tdb =0.9x f u x 0.78x    Tdb =0.9x 400 x 0.78x  = 88216 N h= height of the uppermost bolts from the lower edge of the  bracket h = 4 x 80 + 40 = 360mm Height of neutral axis above the lower edge of the bracket

=

 =  = 51.43 mm  

There are 4 pairs of the bolts above the neutral axis. ∑y= 68.75 + 148.57 + 228.57 + 308.57 = 754.28mm 2

2

2

2

2

∑y = 68.75 +148.57 + 228.57 + 308.57 = 174234.58

       K=     

 K 

2



mm

K=

= 224.89 N/mm

Maximum tension in the uppermost bolt Tmax = K ymax = 224.89 x 308.57 = 69394 N

   Shear force on the bolt = V b =  = 18000 N      [ ]  + [   ] < 1     [ ]  + [ ] = 0.158 + 0.619 = 0.777 <1 Hence the design is safe.