Design of Steel Beams

A beam is a structural member subjected to transverse loads and negligible axial loads

Types of steel Beams

Types of Beams (Ref: www.esdep.org)

Behavior of Steel Beams The behavior is different for short (laterally

supported) and long beams (laterally unsupported)

Short beams exhibit 'plastic' behavior

When external loads are increased, the extreme fibers yield- yield stress fy

Any further increase in Load, results in yielding of the entire section.

A Plastic Hinge is formed at the maximum stressed location

Behavior of Short/Restrained beams

Behavior of Short Beams

The beam fails by a collapse mechanism after sufficient number of Plastic hinges are formed along the length.

Behavior of Beams Beams with restrained compression Flange reach Plastic

Moment Capacity

Check for Shear

Average Shear Stress

Design Shear Strength (Cl. 8.4.1)

wwav

dt

V

v

m

y

d Af

V30

Shear Areas

Other Failure Modes

Shear yielding near support

Web buckling Web Crushing Web crippling

450

d / 2

d / 2 b1 n1

Effective width for web buckling

Web Buckling (Clause 8.7.1.3)

t

d

td

r

L

t

t

t

A

I

r

d

r

L

y

E

y

yy

E

5.232

7.0

3212r

7.0

3

y

cwb ftnbP )( 11

Web Crippling (Clause 8.7.4)

b1 n2 1:2.5 slope

Root radius

Stiff bearing length

0/)( 21 mywcrip ftnbP

CHECK FOR DEFLECTION

CHECK FOR DEFLECTION

Deflection is checked at Working loads

Actual deflection > Allowable deflection (Table 6 of the Code)

Allowable deflection depends on cladding

Ranges between L/120 (elastic cladding) to L/300 (brittle cladding)

Only Live Load deflection is of concern in completed structure, since Dead Load deflection is often compensated by Cambering!

Behavior of Beams If the beams are long , then they fail by Lateral Torsional Buckling

Mode before attaining the Plastic Moment Capacity, Mp

Failure modes of beams include

Local buckling of

i) Flange in compression

ii) Web due to shear

iii) Web in compression due to concentrated loads

Design Strength in Bending or Flexure (Clause 8.2 )

The factored design moment, M at any section, in a beam due to external actions

shall satisfy

Laterally Supported Beam or short beam (Clause 8.2.1.2 )

For sections with stocky webs, d / tw 67 and V

DESIGN OF Laterally supported BEAMS

The steps to be followed in the design are:

1. Ascertain the loads acting on the beam. Using

appropriate partial load factors , calculate the

ultimate B.M. and S.F. and the required section

modulus

2 .Select the lightest section from the IS 808 or IS

Handbook No. 1. Use MB sections as they are

readily available in the market. Classify section

based on b/t ratios using Table 2 of code

DESIGN OF BEAMS (Cont.) 3. Add self-weight of designed section and check

design for Moment capacity-Clause 8.2.1.2

4. Check for shear as per Clause 8.4 of code.

5. Check for deflection, as per Table 6 of the code.

6. Check the web for buckling and crippling as per Clause 8.7.3 & 8.7.4

Beam Example: Compact & Braced

Design a simply supported beam of span 4 m subjected to uniformly distributed dead load of 20 kN/m and a uniformly distributed live load of 20 kN/m The dead load does not include the self-weight of the beam.

Step 1. Calculate the factored design loads (without self-weight).

wU = 1.5 wD + 1.5 wL = 60 kN/m

MU = wuL2/8 = 60 * 42/8 = 120 kNm

Beam Example (Cont.) Step 2. Select the lightest section from the IS 808 or IS

Handbook No. 1 ZP Req. = 120* 10

6 *1.1/250 = 528 * 103 mm3

Select ISMB 300 @ 0.442 kN/m with ZP = 651.74 * 103 mm3

Section classification

Hence the section is classified as plastic

;4.964.54.12

)2/140(

1250

250250

f

y

t

b

f 8496.325.7

2.247

wt

d

Beam Example (cont.) Step 3. Add self-weight of designed section and

check design wsw = 1.5*0.442 =0.663 kN/m Therefore, wu = 60.663 kN/m Mu = 60.663* 42/8 = 121.33 kNm. Req. Plastic section modulus = < ZP = 651.74 x 10

3 mm3

Thus the chosen section is adequate

336

10534250

1.1103.121mm

BEAM EXAMPLE (Cont.)

Check for Design Capacity in Bending (Cl.8.2.1.2):

kNmfZ

kNm

kNmM

m

ye

d

4.156101.1

250106.5732.12.112.148

12.148101.1

2501074.6510.1

6

3

0

6

3

Design capacity > Maximum bending moment

Md =148.12 kNm > 121.33 kNm

Beam Example (cont.)

Step 4: Check for Shear

Design shear force, V =

Design shear strength, Vd=

kNwl

33.1212

4663.60

2

3

0

105.730031.1

250

3

w

m

yth

f

= 295.2 kN > 121.33 kN

Hence safe to carry the SF.

Also 0.6Vd = 177; the design shear force V< 0.6Vd

Beam Example (cont.)

Step 5- check for Deflection

Actual deflection

mm

EI

wl7.3

108990102384

4000205

384

545

44

Allowable maximum deflection

mmmmL

7.333.13300

4000

300max

Hence OK.

Effect of Holes

Ideal Location for Holes: In Webs: At sections of low shear In Flanges: At sections of low B.M. Holes in Tension Flange of Beams should be checked.

Clause 8.2.1.4: Holes have no effect on Md when

(Anf / Agf) (fy/fu) (m1 / m0 ) / 0.9

When the holes do not satisfy the above condition,

If V exceeds 0.6Vd For plastic and compact sections

Mdv = Md (Md-Mfd) 1.2 Zefy/m0 For Semi-compact sections

Mdv = Md (1-) Zefy/m0 Where, = [2V/Vd -1 ]

2 Mdv= Design bending strength under high shear

Design Bending Strength with High Shear Force

(Clauses 8.2.1.3 and 9.2.2)

Laterally Unsupported Beams (Clause 8.2.2)

The design bending strength of laterally unsupported beam

is given the code as:

Md = b Zp fbd fbd = design stress in bending, obtained as ,fbd = LT fy /m0

LT = reduction factor to account for lateral torsional buckling

LT = 0.21 for rolled section,

LT = 0.49 for welded section

Cont

Elastic Lateral Torsional Buckling Moment (Clause 8.2.2.1)

For simply supported, prismatic members with symmetric cross-

section,

For standard rolled or welded doubly symmetric I- sections, the

above equation is simplified as

Annex E gives equation for Elastic Critical Moment of a Section

Symmetrical about Minor Axis

Design of of Laterally Unsupported Beams

Design procedure is same as for a laterally supported Beams.

Only difference is in step 3, the design strength, Md, is computed based on Clause 8.2.2 instead of Clause 8.2.1.2) and compared with Mu

The design is a trial and error process

DESIGN OF PURLINS(Cont.)

It is a trial and error procedure; the various steps are:

The components of the applied loads in the direction perpendicular and parallel to the sheeting are determined.

The factored BMs(Mz and My) and SFs (Fz and Fy) are determined. Choose a trial section

The required plastic section modulus is Zpz = 1.1 Mz gmo / fy + 2.5 (d/b) [1.1 My / fy]

Where, d and b are depth and breadth of trial section

Mz, My = factored bending moments about Z and Y axis; fy = yield stress of steel.

DESIGN OF PURLINS(Cont.) Check for section classification (Table 2 of the

code)

Check for the shear capacity of the section for both Z and Y axis . The shear capacity in Z and Y axis is taken as

Vdz = fy / (3m0) Avz Vdy = fy / (3 m0 ) Avy where Avz = htw ; Avy = 2 bf tf ; h , tw = height and

thickness of web, bf , tf = breadth and thickness of flange .

Compute the design capacity of the section in both the axes. Mdz = Zpz fy / mo 1.2 Zez fy / mo

Mdy = Zpy fy / mo f Zey fy / mo

DESIGN OF PURLINS(Cont.)

Note that in the second equation 1.2 is replaced by f. It is because, in the Y direction, the shape factor Zp / Ze will be

greater than 1.2.

Check for local capacity by using the interaction equation

(Mz / Mdz) + (My / Mdy) 1.0

Check for deflection (Table 6 of the code)

DESIGN OF PURLINS(Cont.)

Under wind section (combined with dead load), the lateral-torsional buckling capacity of the section is calculated.

The overall member buckling check is done using

(Mz / Mdz) +(My / Mdy) 1.0

where Mdz is the lateral-torsional buckling strength of the member.

Empirical Design of Purlins

General requirements as per BS 5950 -1:2000 :

Unfactored loads are used in the design

The span should not exceed 6.5 m

If the purlin spans one bay only, and connected at the ends by at least two bolts.

If the purlins are continuous over two or three spans, with staggered joints, at least one end of any single bay member should be connected by not less than two bolts.

Empirical Design of Purlins

The roof slope should not exceed 30.

The loading :substantially uniformly distributed.

The elastic modulus about the axis parallel to the plane of cladding >WpL / 1800 mm

3 where Wp is the total unfactored load, and L is the span in mm.

Dimension D perpendicular to the plane of the cladding >L/45 and dimension B parallel to the plane of cladding >L/60.

SUMMARY

Beams are structural members that support loads that are applied transverse to their longitudinal axis

They resist the load primarily by bending and shear.

The sections are classified as plastic, compact, semi-compact and slender depending on w/t ratios of

the individual elements.

Local buckling can be prevented by limiting the width-to-thickness ratios.

SUMMARY (cont)

Beams attain their full plastic moment capacity, when the compression flange

is restrained by roofing, and the

sections chosen are plastic or

compact.

When the beam is long, lateral-torsional buckling occurs. This is

similar to that of Euler buckling of

columns.

A simplified design approach has been presented

SUMMARY (Cont)

Factors that affect the behaviour of beams include: type of cross section, type of

loading, support conditions, restraint from

other members, level of application of

transverse load, effects of plasticity, residual

stresses and imperfections.

Single effective length factor only is specified for restraints against lateral

bending, and increased by 20% when

restraints are not provided for warping.

SUMMARY (Cont) The stiffness and strength required for

the bracings (which are used to alter

the lateral torsional buckling

behaviour) are also provided .

The effect of type of loading can be included by specifying an equivalent uniform moment

factor.

The code gives expressions for the elastic critical moment for lateral torsional buckling,

Mcr for symmetric and mono symmetric

beams only. Refer AISC Codes for other

sections

SUMMARY (Cont)

Shear forces may control short beams which carry heavy concentrated loads.

The beams should not excessively deflect or vibrate during the service life

of the structure

Web buckling, web crippling, web and flange holes, purlins, and biaxial

bending are also included.

Lateral buckling of a cantilever

Lateral Torsional Buckling of a Beam Girder

Stiffeners are secondary plates or sections which are attached to beam webs or flanges to stiffen them against out of plane deformations. Almost all main bridge beams will have stiffeners. However, most will only have transverse web stiffeners, i.e. vertical stiffeners attached to the web. Deep beams sometimes also have longitudinal web stiffeners. Flange stiffeners may be used on large span box girder bridges but are unlikely to be encountered elsewhere.

What are stiffeners for? Stiffeners have one or both of the following functions: Controlling local buckling Connecting bracing or transverse beams Controlling local buckling Local buckling occurs when a cross section is slender enough for buckling to occur within the cross section, due either to compression or shear. The webs of bridge beams are usually vulnerable to local buckling, but flanges are usually much thicker and inherently more resistant to buckling. Local buckling can occur due to transverse compression load e.g. a web subjected to a bearing reaction, longitudinal compression load e.g. from bending, or from shear. In all cases the addition of a relatively small stiffener to a slender plate can increase the resistance to local buckling substantially.

Connecting bracing or transverse beams The easiest way to brace steel beams together is by fixing the bracing to transverse

stiffeners.Thus stiffener positions almost always coincide with bracing positions.