Module-Eccentric Connections

Unit No.II

Course-Design of Advanced Steel Structures Max. Marks. =25

Asst. Prof. Suryawanshi S.R.

In the frame of steel building, a beam may be attached to another beam or to a column. In such cases design of connections under system of loads on the elements depends on understanding of the behavior of the elementsThe beam column connections expected to resist transferred end reactions only are termed as shear connections or flexible connections ,permits free rotations of the beam end and do not have mmt restraint Other connection which do not permit any relative rotation between beam and column and expected to resist mmt in addition to end reaction are termed as mmt connections or rigid connections

Moment Diagram for Beams with different End Conditions

Types of Connections A. Shear connections (Flexible Connection)a)Riveted Shear connections 1. Bracket Connections–Type I &II 2. Seated Connections-Stiffened and Unstiffened 3.Framed Connectionsb)Welded Shear Connections 1. Bracket Connections–Type I &II 2. Seated Connections-Stiffened and Unstiffened 3.Framed Connections

B. Moment Resistant connections (Rigid Connections) a) Riveted mmt connection:- i.Heavy mmt connection ii. Light mmt Connectionsb) Welded mmt connections

C. Semi-Rigid Connections

WELDED SHEAR CONNECTIONS

• Welded connection are often subjected to shear & torsion or shear & bending

• Bracket type –I is an example of weld subjected to shear and torsion

• Bracket type –II is an example of weld subjected to shear and bending

• Seated and framed also be implemented depends on designer’s choices.

Bracket Connections 1.Bracket Connection Type -I

When twisting mmt acting is in the plane of weld i.e. C.G.of weld group lies in the plane of line of action of applied load, weld is subjected to shear and torsion.

Let ,M = Twisting mmt P = eccentric load acting over the jointe = eccentricity of load, t=effective throat thickness of the weldl = length of fillet weld (2a+d) as shown in fig. d = depth of the bracketr = the distance of extreme weld from CG of weld group, Ʈ = shear stress in wled , J = polar mmt. of Inertia

Weld group subjected to 1)Direct shear stress due to load P2) Shear stress due to twisting mmt.

For Direct shear stress Ʈ1 = Load/Effective area of weld = P/(2a+d)t------------(1)For shear stress due to twisting mmt.Can be computed using torsion equation.T/J= Ʈ/r Ʈ2= T.r/J= P.e.r/J-------------(2) The resultant Shear stress Ʈ can be given as

Ʈ= < 108 N/mm2 Allowable shear stress in weld

• The design of Connection is done as follows:-1. Assume overlap of bracket & work out the length of

fillet weld.2. Compute C. G. Of weld group (x)3. Compute polar Mmt. of Inertia (J)4. Calculate dist. of extreme weld from C.G. Of weld

group(r)5. Calculate Ʈ1, Ʈ2

6. Calculate Ʈ and equate it with Allowable shear stress.7. Weld size can be given as, t=0.7S, t is throat thk.

2.Bracket Connection Type –IIWhen moment is perpendicular to the weld i.e. C. G. of the weld group lies in a plane perpendicular to the plane of line of action of the applied load, the weld is subjected to shear and bending. The eccentric load can be made concentric along with a bending moment.The weld in such a case is subjected to---

a)Direct stress due to a concentric load P and b) Bending stress due to bending moment(P.e)

Either butt weld or a fillet weld can be used in such connections.

1. For Direct shear stress in weld :Ʈ1 = Load/(Eff.Area of weld)

For fillet weld Ʈ1 = P/2d.t

For butt weld Ʈ1 = P/d.t

2. For bending stress in the weld:Ϭb = moment/(section modulus)

For fillet Weld

Fillet weld when subjected to flexure the common relationship Ϭb = M.y/I ,used to evaluate stresses. Ϭb = {P.e(d/2)}/{2.td3/12} = P.e/(2td2/6) (Critical stresses exist on the throat of the weld at 450)The throat stresses then treated as a shear since,450 line of failure is indicative of a shear fractureƮ2 = P.e/(2td2/6) Combined stress in fillet weld

< 108N/mm2

For butt Weld

Ϭb = P.e/(td2/6)

Combined stresses and shear stress in butt weld may be checked by the interaction formula as,

< yield stress of steel used

The design of connection is done as follows:-Size of weld is assumed and the length is computed. If the length of the weld is more than twice the depth of the bracket, the size of weld is revised:-

1. Assume the size of the weld and compute the throat thickness. Calculate depth of the bracket from either of the following appropriate equations.

In case of butt weld where Ϭb =165 N/mm2

In case of fillet weld where Ʈ1 = 108 N/mm22. Direct shear stress is computed in terms of throat thk. (Ʈ1)

3. the stress due to bending moment is computed I terms of throat thk. (Ʈ2 or Ϭb)

4. The equivalent stress is computed as appropriate

Welded Seated ConnectionsTwo types of seated connections used i.e.

1.Unstiffened Seat Connection 2. Stiffened Seat Connection In welded unstiffened seat connection two angles are used. one angle is welded to the column in the shop and forms the beam seat. A Cleat angle of nominal size (100x100x6 )mm is welded to the top of beam in the shop and to the column in the field.vertical welds are provided to connect the seat angle and these are turned at the ends.the resultant stress in the vertical weld is not uniform. resultant shear comprising of horizontal shear per mm and vertical shear per mm, is computed and should be less than allowable shear per mm of the weld.

Design steps:-A) A Seat angle is chosen suitably from following considerations 1. the seat angle is assumed to have a length B,equal to the width of the beam flange (B) 2. length of outstanding leg of seat angle b

The seat leg length is kept more than the calculated bearing length equation may yield negative value for large beam with small reaction therefore a minimum bearing length is specified as

where, R=end reaction in N ,

Ϭp=Permissible bearing stress in Mpa (0.75fy)

t= thk. Of the web of beam in mm,

h2=depth of the root of fillet weld from extreme fibre of flange for the beam in mm

3. thk. of seat angle is chosen such that the outstanding leg does not fail in bending on a section at the toe of the fillet (t)

bending stress

t2 = 6.R.e1 /B.Ϭbc

e1= the distance from the critical section to the reaction R acting

at the centre of the bearing length =10+0.5b-t-radious of root of fillet

M=mmt at critical section=R.e1

B=length of seat angle equal to width of flange

Ϭbc=bending stress in comp. assumed to bending stress in slab base

i.e.185Mpa

B)Calculate the vertical shear per mm Ʈ1 = R/2dt(t is unity) C)Calculate the eccentricity of the reaction and compute the

bending mm t.D)Calculate the horizontal Shear per mm due to the bending

mmt. Ʈ2 = M/(2.d2 .1÷6) (t is unity) d=length of weldE) Calculate resultant shear stress

F)This resultant shear per mm is equated to the strength of the weld per mm to find the size of the weld

Ʈ=0.7xSx1xƮ’ , Ʈ’=Allowable shear stress =108Mpa

Q. Design an unstiffened welded seat connection for a beam I. S. M. B. 500@86.9Kg/m, transmitting 170 Kn to the I. S. H. B. 350@367.4Kg/m (SRTMUN June’13)—8mks

Soln. the relevant properties of the section from IS Handbook No.1(SP 6(1964))

ISMB 500Beam

ISHB 350Column

B 180 250

tw 10.2 8.3

h2 37.95 27

tf 17.2 11.6

A) Design of seat angle 1. Length of seat angle B =Width of beam flange=180mm 2. for bearing length b

b=23.157 mm

b=45.045mm ,providing b=100mmProvide 10 mm clearance between beam and column Provide (ISA 200x100x12) mm r1 =12mm, t=12mm 3.for thickness of seat angle (t) t2 = 6.R.e1 /B.Ϭbc

e1 = 10+45.045/2-12-12=8.52mm

Try another section as (200x150x18)mmr1 =13.5mm t=18mmTherefore e1 = 10+45.045/2-18-13.5=1.0225mm

Therefore provide seat angle (200x100x18)mm

B) Vertical Shear /mmd=200mm i.e. length of weldƮ1 =R/2dt=170x103 / 2x200x1=425N/mmC)Horizontal Shear/mmƮ2 = M/(2.d2 .1÷6) (t is unity) =170x103x1.0225/(2x2002 ÷6)=13.036N/mmD) Resultant shear

E)For size of weldƮ=0.7xSx1xƮ’ Ʈ’=Allowable shear stress =108Mpa425.2=0.7xSx108S=5.62mm, therefore providing S=6mm

2. Stiffened Seat Connection

In the stiffened seat connection ,a T-section built up of two plates is used .

The bearing length of the seat plate is calculated as in case of the unstiffened seat connection.

thk. Of seat plate is kept equal to the thk, of flange of beam thk. Of stiffening plate is kept equal to thk. Of web of the

beam. The depth of the stiffening plate is decided depending upon

the length of the vertical weld required. Seat plate kept wider than flange of the beam by atleast

twice of the weld on each side of the beam flange to facilitate welding

• Design steps:-1. Calculate width of seating plate

2. thk. of the seat plate is assumed equal to the thk. of beam flange.

3. thk. of stiffening plate is assumed equal to the thickness of the web of the beam

4. The eccentricity of load & mmt. due to it are calculated 5. vertical and horizontal shear per mm length of weld Ʈ1 &

Ʈ2 are computed 6. the resultant shear per mm of weld length Ʈ is computed by

the vector sum of Ʈ1 & Ʈ2

7. calculate size of weld from Ʈ = 0.7xsx1x Ʈ’

Welded Double Plate Framed Connections

• One plate is shop welded to the supported beam while other plate is welded to plate element of supporting structure /girder/column.

• In field beam is properly aligned, both plates are brought in position and erection bolts are inserted to hold them together ,welding is then completed in field.

• There are two line of weld (weld A&B) provided for design purposes.

• For Weld A:-designed for vertical shear and mmt.• For Weld B:- designed for only vertical shear • Check for vertical shear stress in column flange • Check for stress in web.

• Design steps:-A)For weld A:-1. Assume width of plate as 50 mm wide so e=50mm2. Calculate length of weld or depth of plate as:-

Where M=R.e & t=0.7x s

3. Calculate vertical shear/mm Ʈ1 = P/2d.t t is unity 4. Calculate horizontal shear/mm Ʈ2 = R.e/(2td2/6) 5. Calculate resultant

6. Strength of weld=0.7xSx1xƮ’ > Resultant shear ….hence ok

B)For weld B:- Vertical shearƮ1 = R/2d For field weld Ʈper=allowable stress=0.8 Ʈ’=0.8x108 Equate vertical shear to strength of weld as Ʈ1 = 0.7xsx0.8xƮ’ Calculate size of weld

C)Check for vertical shear stress in column flange:-

Ʈ=V/2tf .d <0.4 fy tf = thk.of flange of beam

D)Check for stress in web:-Ʈ= V/tw.d <0.4 fy tw =thk. Of web of column

Welded double angle framed connections

• Beams are connected to columns or to the other beams by means of framing angles or plates.

• A pair of angles or plates are placed one on each side of the web of the beam to be connected.

• These angles or plates used to transmit shear and mmt.• One of the leg of angle is welded to the beam in the shop

and the other leg is welded to the column in the field.• Seats are only provided for erection and can be removed

afterwards.• Seat angle is set back from the beam web by 10-12 mm

as shown in fig.• Welds connecting the angles to the column are subjected

to vertical shear and to bending mmt.

A)For weld connecting the framing angle leg to column Calculate V=R/2 For vertical shear per mm Ʈ1= V/d For horizontal shear per mm Ʈ2 = 1.8.R.e1/d2

For Resultant shear

For size of weld Ʈ=0.7xSx1xƮ’ , Ʈ’=Allowable shear stress =108Mpa

B)Weld connecting the angle with web of the beam Max.horizontal shear in field weld occurs at the bottom of the angle section.N.A. passes at distance of d/6 from top of the angle ,d-depth of section

Vertical shear =R/2, bending mmt.=R.e2/2 Find out CG of weld group to connect the angle section to web of

beam. Calculate polar mmt. of Inertia. Compute twisting mmt. For vertical shear Ʈ1 = V/Effective area of weld=V/(2a+d)t For horizontal shear Ʈ2 = T.r/J=V.e2.r /J=R.e2.r/2J

For Resultant shear

For size of weld Ʈ=0.7xSx1xƮ’ , Ʈ’=Allowable shear stress=108Mpa