Welded Connection

2CL405 Design of Structures–III B. Tech. Civil Semester VII

Prof Urmil V Dave Institute of Technology, Nirma University

DESIGN OF STEEL ELEMENTS

1. Important Design Specifications for Fillet Welding as per IS 800:2007, IS 816:1969

and IS 9595:1996

Size of the weld, s ≥ 3mm

When a fillet weld is applied to a square edge of a part or section, s ≤ 1.5 mm less than

the edge thickness

When a fillet weld is applied to a rounded toe of a rolled section, s ≤ ¾ of the thickness

of the section at the toe

The minimum size of the fillet weld should be as given below to avoid cracking in the

absence of preheating

Thickness (t) of thicker part in mm S (mm)

t ≤ 10 3

10 < t ≤ 20 5

20 < t ≤ 32 6

32 < t ≤ 50 10 (8 mm for first run)

t >50 Special precaution like

pre-heating to be taken

Effective throat thickness: ≥ 3 mm, ≤ 0.7 × thickness of thinner part joined; For stress

calculations, effective throat thickness = Ks, where K depends on the angle between the

fusion faces. Usually, the angle is 90° for which K = 1/sqrt(2) = 0.7

Effective length of a fillet weld = (actual length – 2s) ≥ 4s

Length of end returns should not be less than 2s

In lap connections, the minimum length of weld should not be less than 4 times the

thickness of the thinner part joined or 40mm whichever is more. If only side fillets are

used, the length of the weld on either edge should not be less than the transverse

spacing between the welds.

The throat thickness of the end fillet weld, normal to direction of force, should not be

less than 0.5t where t is thickness of the part

For intermittent fillet welds,

Effective length (wl) ≥ 4s or 40 mm, whichever is greater

Clear spacing (uwl) ≥ 12t (for compression)

≤ 16t (for tension)

≤ 200 mm

Where t is the thickness of thinner part joined

Design shear strength of a fillet weld, fwd = fwn/γmw

where, fwn = nominal shear strength of fillet weld = fu/(sqrt3)

fu = ultimate strength of the weld or the parent metal

γmw = partial safety factor = 1.25 for shop welding,

= 1.5 for site welding

The design procedure for the fillet weld is as follows:

Assume the size of the weld based on the thickness of the members to be joined

By equating the deign strength of the weld to the external factored load, the effective



length of the weld to be provided either as longitudinal fillet welds (parallel to load

axis) or as transverse fillet welds (perpendicular to load axis) along with longitudinal

fillet welds. It is a common practice to treat both the welds as if they are stressed

equally. If the length exceeds 150te, reduce design capacity by a factor βlw as per

clause 10.5.7.3 of the code

If only the longitudinal fillet weld is provided, a check is made to see if the length of

each longitudinal fillet weld is more than the perpendicular distance between them

End returns of length equal to twice the size of the weld are provided at each end of

the longitudinal fillet weld

When subjected to combined tensile and shear stress, the equivalent stress, fe, should

satisfy,

fe = sqrt(fa2 + 3q

2) ≤ fu/(sqrt3 γmw) .............................................................(1)

where fa is normal stress due to axial force or bending moment, and q is shear stress

due to shear force or tension

2. Important Design Specifications for Bolted Connection as per IS 800:2007

Shear Capacity of a Bolt

The design strength of a bolt in shear Vdsb is given by Vdsb = Vnsb/ γmb .....................(2)

Where Vnsb is the nominal shear capacity of a bolt = (fub/sqrt3) × (nnAnb + nsAsb).....(3)

Bearing Capacity of a Bolt

The design strength of a bolt in bearing Vdpb is given by Vdpb = Vnpb/ γmb .....................(4)

Where Vnpb is the nominal bearing strength of a bolt = 2.5kb dtfu .................................(5)

kb is the smallest of e/3d0, [(p/3d0) – 0.25], fub/fu, 1

Tension Capacity of a Plate

The design strength of a plate in tension due to rupture at the net section

Tdn = 0.9 An fu/ γm1.....................................................................................................(6)

Design Strength due to Block Shear

The block shear strength Tdb of the bolted connection is the least of

Tdb = (Avg fy/γmo sqrt3) + 0.9Atnfu/γm1 .................................................................(7)

OR

Tdb = (0.9 Avn fu/γm1 sqrt3) + Atg fy/γm0 .......................................................................(8)

Tension Capacity of a Bolt



The design strength of a bolt in tension Tdb is the least of

(i) the design strength of the bolt due to the yielding of the gross section (i.e. the shank)

Tdbg = fyb Asb/ γm0 .............................................................................................(9)

(ii) the design strength of the bolt due to the rupture at the net section (i.e. at the root of

the threads)

Tdbn = 0.9 fub Anb/ γmb ....................................................................................(10)

Bolt subjected to Combined Shear and Torsion

A bolt subjected to shear and torsion simultaneously should satisfy the condition

(Vsb/Vdb)2 + (Tb/Tdb)

2 ≤ 1 ..............................................................................(11)

Other Specifications

The diameter of the hole should be the nominal diameter of the bolt plus the

clearance as given below:

Nominal diameter of bolt (mm) 12-14 16-24 24

Standard Clearance 1.0 2.0 3.0

Minimum pitch: the distance between the centres of the bolts in the direction of

stress should not be less than 2.5 times the nominal diameter of the bolt

Maximum pitch:

(i) 32t or 300 mm, whichever is less for the bolts in members including the tacking

bolts,

(ii) 16t or 200 mm, whichever is less for the bolts in tension members

(iii)12t or 200 mm, whichever is less for the bolts in compression members

Where t is the thickness of thinner plate

Edge and End distance

(i) Minimum edge and end distances from the centre of any hole to the nearest edge

of a plate should not be less than 1.7 times the hole diameter for sheared or hand-

flame cut edges; and 1.5 times the hole diameter for rolled, machine-flame cut,

sawn and planned edges

(ii) The maximum edge distance from the centre of hole to the nearest edge should not

exceed 12tε, where ε = sqrt(250/fy) and t is the thickness of the thinner outer plate.

Tacking Bolts

These are the additional bolts provided other than strength consideration. The

maximum pitch of these bolts should be 32t or 300 mm, whichever is less, where t is

the thickness of the thinner plate. If the members are exposed to weather, the pitch

should not exceed 16 times the thickness of the outside plate or 200 mm, whichever is

less.



3. Tension Members

The designer has to arrive at the type and size of member, based on tensile force acting on

the member. The type of member is chosen based on the type of the structure and location

of the member. The design is iterative, involving a choice of a trial section and an

analysis of its capacity. The various steps are as follows:

Step-I The net area required An to carry the designed load T is obtained by the equation,

An = Tu/(fu/γm1) ...............................................................................................(12)

Step-II From the required net area, the gross area may be computed by increasing the net area

by about 25% to 40%. The required gross area may also be checked against that

required from the yield strength of the gross section as follows:

Ag = Tu/(fu/γm0) ..............................................................................................(13)

A suitable trial section may be chosen from steel tables to meet the required gross

area.

Step-III The number of bolts or welding required for the connections is calculated. They are

arranged in a suitable pattern and the net area of the chosen section is calculated. The

design strength of the trial section is evaluated using equations (12) to (15) for plates

and threaded bars and additionally equations (16) to (18) in case of angles.

Tdg = 0.909 fyAg ............................................................................................(14)

where Ag is the gross area of cross-section in mm2, fy is the yield strength of the

material (MPa)

Tdn = 0.72 fuAn ...............................................................................................(15)

where An is the net effective area of cross-section in mm2, fu is the ultimate strength of

the material (MPa)

Plates: The block shear strength Tdb of the connection is taken as the smaller of

Tdb1 = 0.525 Avgfy + 0.72 fuAtn ....................................................................(16)

Tdb2 = 0.416 fuAvn +0.909 fyAtg ....................................................................(17)

where Avg and Avn are the minimum gross and net area in shear along a line of

transmitted force, respectively. Atg and Atn are the minimum gross and net area in

tension from the hole to the toe of the angle or next last row of bolt in plates,

perpendicular to the line of force, respectively and fu and fy are the ultimate and yield

stress of the material (MPa), respectively.



Angles: Strength as governed by block shear failure in angle end connection is

calculated using equations (5) and (6). The design strength is governed by

the tearing of the net section is given in cl. 6.3.3 of the code. Substituting the

values of γm1 = 1.25 and γm0 = 1.1, we get

Tdn = 0.72fuAnc + 0.909 β fy Ago .................................................................(18)

and

β = 1.4 – 0.076[(bs/Lc)(w/t)(fy/fu)] ≤ 0.88 fy/fu ≥ 0.7..................................(19)

where, w and t are the size and thickness of the outstanding leg, respectively, bs is the

shear distance from the edge of the outstanding leg to the nearest line of fasteners,

measured along the centre line of the legs in the cross-section, Lc is the length of the

end connection measured from the centre of the first bolt hole to the centre of the last

bolt hole in the end connection or length of the weld along load direction and, Anc is

the net area of connected leg at the critical cross-section, computed after deducting the

diameter of hole (the diameter of the holes should be taken as 2-mm larger than the

nominal size in case of punched holes), and Ago is the gross area of the outstanding

leg. Alternatively, the equation suggested by code, for preliminary design, with the

partial safety factor for material γm1 = 1.25, we get

Tdn = 0.8αA fu................................... ............................................................(20)

with α = 0.6 for one or two bolts, 0.7 for three bolts, and 0.8 for four or more bolts in

end connections or equivalent weld length.

Step-IV If the design strength is either small or too large compared to the design force, a new

trial section is chosen and step-III.

Step-V The slenderness ratio of the member is checked as per Table 3 of code.

4. Compression Members

The strength of a compression member is based on its gross area Ag (for slender cross-

section, Aeff should be used). The strength is always a function of the effective slenderness

ratio KL/r, and for short columns the yield stress fy of the steel. Since the radius of

gyration r depends on the section selected, the design of compression members is an

iterative process. The design procedure involves the following steps.

Step-I: The axial force in the member is determined by a rationale frame analysis, or by

statics for statically determinate structures. The factored load Pu is determined by

summing up the specified loads multiplied by the appropriate partial load factors γf.

Step-II: Select a trial section. Note that the width/thickness limitations as given in Table 2 of

code to prevent local buckling must be satisfied (most of the rolled section satisfy the

width-to-thickness ratios specified in Table 2). If it is not satisfied and a slender section is

chosen, the reduced effective area Aeff should be used in the calculation. The trial section

may be chosen by making initial guesses for Aeff/A, and fcd (between 0.4-0.6fy) and

calculating the target area A. Following member sizes may be used as a trial section:



(a) Single angle size 1/30 of the length of compression member

(b) Double angle size 1/35 of the length of compression member

(c) Circular hollow section diameter = 1/40 of length

The slenderness ratios as given in the Table 1 will help the designer to choose the trial

sections.

Table 1 Slenderness ratios to be assumed while selecting the trial sections

Type of Member Slenderness ratio(L/r)

Single angles 100-150

CHS, SHS, RHS 90-110

Single channels 90-150

Double angles 80-120

Double channels 40-80

Single I-section 80-150

Double I-section 30-60

Step-III: Compute KL/r for the section selected. The computed value of KL/r should be

within the maximum limiting value given in Table 2. Using Fig 8 and Tables 9 and 10 of

code, compute fcd and the design strength Pd = Afcd.

Table 2 Maximum Slenderness ratio of compression members

Type of Member KL/r

Carrying loads resulting from dead loads and superimposed loads 180

Carrying loads resulting from wind and seismic loads only, provided

the deformation of such a member does not adversely affect the stress

in any part of the structure

250

Normally acting as a tie in a roof truss or a bracing system but subject

to possible reversal of stress resulting from the action of wind or

seismic forces

350

Lacing bars in columns 145

Elements (components) in built-up sections 50

Step-IV: Compare Pd with Pu. When the strength provided does not exceed the strength

required by more than a few percentages, the design would be acceptable: otherwise

repeat steps 2 through 4.

5. Beams

5.1 Laterally Supported Beams

The design of laterally supported beams consists of selecting a section on the basis of the

modulus of section and checking it for shear, deflection, and web crippling. The steps to

be followed are a follows:

Step-I: The loads that may be acting on the beam are ascertained. The design loads are

obtained by summing up the loads multiplied by the appropriate partial load factors as

given in Table 4 of the code.



Step-II: A trial beam section is assumed and the distribution of the bending moment along the

length of the beam is determined by an elastic analysis (if the beam is statically

indeterminate) or by statics (if the beam is statically determinate). The maximum

bending moment and shear force are calculated.

Step-III: The required section modulus may be determined by

Md = 0.909βbZpfy ≤ 1.09Zefy ≤ 1.36 Zefy (for cantilever), when V < Vd .... (21)

Md is design bending strength, V is design shear force, Vd is the design shear strength,

βb is 1 for plastic and semi-compact sections and βb is Ze/Zp, for semi-compact

sections

Md = 0.909Zefy for class 3 semi-compact sections .......................................(22)

Md = Zefy’ for class 4 slender sections ..........................................................(23)

Zp and Ze are the plastic and elastic section moduli of the cross-section, respectively.,

and fy is the yield stress of the material.

Step-IV: Select a suitable section from steel table, which has a section modulus equal to or

more than the calculated section modulus.

Step-V: Evaluate whether the selected section falls in the category of compact, semi-compact

or plastic section. Check the adequacy of the selected section including self weight of the

beam to satisfy for required plastic section modulus

Step-VI: The selected beam is to be checked for shear. Calculate the factored design shear

force V. The V in a beam due to external actions should satisfy,

V ≤ Vd .........................................................................................................(24)

Where Vd, the design strength is given by

Vd = 0.909 Vn ...............................................................................................(25)

Vn is nominal shear strength of a cross-section and may be governed by plastic shear

resistance or the strength of the web governed by shear buckling.

Step-VII: The design capacity of the member is to be evaluated and should be more than the

maximum bending moment calculated.

Step-VIII: The beam is checked for deflection, as per Table 6 of the code. Actual deflection

should be less than allowable maximum deflection.

Step-IX: The beam is checked for web buckling. Strength of the section against web buckling

Fwb should be more than the calculated shear force.



Fwb = (b1 + n1) tfc ..........................................................................................(26)

Where, b1 is the stiff bearing length, n1 is the dispersion of 45° line at the mid depth of

the section, t is the web thickness and fc is the allowable compressive stress

corresponding to the assumed web strut. The effective length of the strut is taken as LE

as 0.7d, where d is the depth of the strut in between the flanges. The slenderness ratio λ

of the idealized strut is 2.5d/t.

Step-X: The beam is checked for web crippling/ web bearing. Strength of the section against

web buckling Fcrip should be more than the calculated shear force. The crippling strength

of the web (also called as the web bearing capacity) at supports is calculated as

Fcrip = (b1 + n2) tfyw ......................................................................................(27)

Where, n2 is the length obtained by dispersion through the flange, to the flange to web

connection (web toes of fillets), at a slope of 1:2.5 to the plane of flange (i.e. n2 = 2.5d1).

fyw is the design yield strength of the web. At an interior panel where concentrated load is

acting, the crippling strength is given by,

Fcrip = (b1 + 2n1) tfyw.......................................................................................(28)

If the above bearing capacity or crippling strength of the beam web is exceeded,

stiffeners must be provided to carry the load.

5.2 Laterally Unsupported Beams

When the compression flange of the beam is laterally unsupported, lateral torsional

buckling may take place, leading to failure at or below the elastic critical moment. The

design of laterally unsupported beams with equal flanges (e.g., I-sections and channel

sections) or mono symmetric beams is essentially a trial and error process, since the

section dimensions are not initially known. Hence, we have to assume a section in order

to compute the strength of that section. The design procedure is essentially the same as

described for a laterally supported section, except that in Step-III, the design strength Md

is computed based on provisions given in section 8.2.2 as well as E1.2 of the code for

mono symmetric section and compared with the factored design moment M. If it is not

equal to or greater than the factored design moment, next higher section is chosen and

the process is repeated till a section which satisfies the condition M ≤ Md is found.

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Welded Connection