Beam Design Notes Nzs 3404

8/11/2019 Beam Design Notes Nzs 3404

1/24

Beam_Design_Notes.doc p1 JWB September 2005

DESIGN OF STEEL BEAMS

[Reading: Allen, H.G. and Bulson, P.S. Background to Buckling, McGraw-Hill, 1980; Trahair, N.S. and

Bradford, M.A. Behaviour and Design of Steel Structures, 2ndEd., Taylor & Francis, 1994; CIVIL 211:

Lecture Notes on Beam Behaviour, Plastic Analysis and Plate Buckling]

INTRODUCTION

The strength limit state requirement for beam design was stated earlier as (see plastic analysis notes)

n* MM (1)

where M*denotes the maximum bending moment due to the application of factored loads,

is the strength reduction factor (0.9), and

Mnis the nominal beam moment strength (MPor MRP)

1.2G+1.5Q

FactoredLoads

Analysis

Design Actions(bending moments)

(plasticor

elastic)

However, the assumption that the bending strength of the beam is MPis not always justified. Localor lateral buckling may occur before the full plastic moment capacity of the section is reached,and prevent it from being reached.

LOCAL BUCKLING AND SECTION MOMENT CAPACITY,MS

As discussed under plate buckling, parts of a

beam cross-section may buckle locally before the

fully plastic state is reached, or even before yield

stress is reached, depending on their slenderness

ratio. The high stresses in the compression flangetend to cause buckling as shown in the figure,

limiting the section moment capacity, Ms.

The effects of local buckling are accounted for in

the same way as used for column sections, using

an effective section property that depends on the

slenderness ratioof the web and flange.

Plate element slenderness ratio, e:

250tdor

250Tb y1y1

e= (2)

n* MM

b1d1

t

T


2/24


Plate element yield and plasticity limits, eyand ep:

eyand epare limiting values of slenderness ratio that determine the way a section will fail (for theorigins of eysee pages 9 and 10 of Plate Buckling Notes). The values are specified in Table 5.2 of theSteel Design Standard, NZS3404. Selected values are shown in Table 1 below.

Section slenderness ratio, s:

ey

ees largestwithelementplateof

= (3)

Section description: Hot-rolled

UB, UC

Heavily welded

BOX

Cold-formed

CHS

Cold-formed

RHS

Plate element widths: b1

d1

b1 b2 b1

d1

d0

b2

d1

ep 9 8Flange outstand b1

ey 16 14

ep 30 30Flange b2supportedalong both edges ey 35 40

ep 82 82 45Web d1supportedalong both edges ey 180 130 60

ep 50Diameter d0

ey 120

Table 1 Plasticity and yield slenderness limits (selected cases from Table 5.2, NZS3404).

Section moment capacity Msand effective section modulus Zeff

yeffs ZM = (4)

Beams are divided into three categories depending on how the section slenderness compares with the

plastic and yield slenderness limits, as shown in Table 2.


3/24


Slenderness range Behaviour Effective Section Modulus Classification

eps


4/24


Case 2 Slender section

Determine the classification, effective section modulus and moment capacity of a grade 350

400WC144 (welded column).

From section property tables, d1/t = 21.88, b1/T = 12, Zelastic=2550E3mm3, Zp= 2830E3mm

3and

y= 380MPa.

Flange slenderness: )14,8cf(8.1425038012

250Tb eyep

y1fl =====

Web slenderness )180,82cf(97.26250

38088.21

250T

deyep

y1web ====

=

Can see that flange is much more critical than web and so dictates the

Section slenderness: 8.14s =

Plasticity Limit: 8ep =

Yield Limit: 14ey =

And since eys > SLENDER section

Effective section modulus: 33

s

ey

elasticeff mm1024128.14

142550ZZ ==

=

Nominal section capacity kNm916Nm3802412ZM yeffs ===

NOTE

Most standard, hot-rolled UB and UC sections are proportioned such that they fall into the COMPACT

classification. A small number are NON-COMPACT and none are SLENDER.

LATERAL BUCKLING AND MEMBER MOMENT CAPACITY, Mb

Just as Euler buckling can limit the axial load capacity of a column so lateral

buckling can limit the bending capacity of a beam.

Since lateral buckling involves the entire beam, the associated moment

capacity is referred to as the membermoment capacity, Mb.

Consider a simply supported beam as shown, with the ends restrained against

twist. If a gradually increasing bending moment is applied to the beam it will

initially deflect in a plane perpendicular to the principal axis of bending(vertically), but will eventually become unstable and

buckle sideways. As buckling proceeds the cross-

section of the beam (at mid-span, for example), will

deflect laterally(out of plane) and also twistas shown.

The phenomenon is referred to as lateral-torsional or

flexural-torsional buckling, and the bending moment at

which buckling commences is known as the elastic

buckling moment Mcr.

Figure 1 Lateral torsional buckling


5/24


It can be shown (see limited derivation on p.7) that a uniform beam subject to a constant bending

moment about its major bending axis will buckle at a critical moment Mcr, defined by

+

=

2w

2

2

y2

crL

EIGJ

L

EIM (5)

where EIyis the minor axis flexural rigidity

GJ is the torsional rigidity

G is the shear modulus of elasticity

J is the torsion constant(a geometric property of the cross-section shape)

EIwis the warping rigidity

Iwis the warping constant(another geometric property of the cross-section).

The first term

2

y2

L

EI

, reflects the influence of the beams weak axis bendingresistance

The second term

+

2w

2

L

EIGJ , accounts for the torsional resistanceof the beam.

TORSIONAL RESISTANCE

Torsion is covered in greater depth later in the course. The notes below provide a brief overview of

some aspects that are relevant to the lateral buckling problem

UNIFORM OR ST VENANT TORSION

Illustrated in Fig. 2 is an I-beam, unrestrained against warping, subject to a constant torque T. The

effect is to cause a uniform rate of twist along the length of the member and the system of internal

shear stresses shown. The torque and twist are related by

)Ldz

dwith(

dz

dGJT =

= (6)

J is a relatively simple property of the cross-section shape. For the I-beam cross-

section shown

3/)tdBT(2J 313 + (7)

Open sections such as I beams are torsionally weak, with a low value of J and

correspondingly high internal shear stresses.

In contrast, closed sections such as RHS and CHS have a high value of J and a closed loop form of

stress distribution. Their torsional resistance is very high.

d1t

T

B


6/24


o

T

T

Torque, T,causing twist, o Plan view Shear stress distribution

in an I-beam due to torsion:OPEN section -low torsional resistance

Shear stress distributionin a box beam due to torsionCLOSED section -high torsional resistance

Figure 2 Uniform torsion

WARPING TORSION

Note in the plan view of Figure 2 how the ends of the I-beam displace out of their original plane as the

ends of the flanges rotate about the vertical (y) axis. The longitudinal movement is known as the

warping displacement. Provided warping is allowed to occur unrestrained it has little influence on the

torsional behaviour. However, if one end of the beam above was built in (rigidly fixed e.g. by casting

into concrete) the outcome would be different, as shown in Fig. 3.

warping prevented

Flanges forced to bend,

setting up flexuraland shear stresses

flange shear forces, S ,

generate warping torquef

T = S x dw f

Sf

Sfd

Tw

T (applied torque)

Figure 3 Torsion due to restrained warping

It can be seen in Fig. 3 that twisting the free end of the cantilever results in the flanges bending in a

horizontal plane. The resistance of the flanges to this horizontal bending sets up a system of equal and

opposite bending moments and shear forces acting in each flange. The pairs of equal and opposite

bending moments are known as bi-momentsand are a type of internal action (along with normal bending

moments, shear force, etc.). The associated horizontal shear forces, Sf, create an additional resisting

torque known as the warping torque. For thin-walled open sections, such as I-beams, the warping

torque is likely to be the main contributor to torsional resistance.

It can be shown that the warping torque is given by

3

3

wwarpdz

dEIT = (8)

For a section of arbitrary shape the warping constant Iwis complicated to

calculate. For a doubly symmetric I-beam section such as that shown

4

)TD(II

2y

w

= (9)

T

B

D x

y


7/24


Iwhas units of length6, e.g. m6or mm6. Values of Iware included in the tables of section properties

that appear elsewhere in these notes.

Warping torque can still arise even if the beam ends are not restrained from warping. For example the

simply-supported beam of Fig. 1 would develop warping torques if subjected to an external torque at

mid-span (or as the result of twisting associated with lateral buckling).

Mcr

L

Mcr

flange bends in horizontal plane -warping stresses develop

warping unrestrainedat ends

PLAN VIEW

Figure 4 Non-uniform torsion warping restrained by continuity over central section

DERIVATION OF LATERAL BUCKLING MOMENT

The following brief derivation, for the case of equal applied end moments, is provided for the curious

and can be omitted without prejudice.

Mcr

L

x

y

u

Elevation Section

Plan at end

pins prevent end twistbut allow rotation aboutvertical axis (warping)

z

u dudz

Plan showing longitudinal axis

Mcr

o

Figure 5 Lateral buckling under equal end moments

As with column buckling we seek a deflected and twisted equilibrium position of the beam. The lowest

value of Mcrat which this is possible determines the required buckling moment.

The governing differential equation for lateral bending equilibrium is

= cr2

2

M

dz

udEI (10)

The left hand side defines the internal resisting moment and the right hand side is the horizontal

component of the applied moment.


8/24


The governing differential equation for torsional equilibrium is

dz

duM

dz

dEI

dz

dGJ cr3

3

w =

(11)

dz

dGJ

is the internal resistance to uniform torsion and3

3

w dz

dEI

is the internal resistance to warping

torsion. The right hand side defines the disturbing torque due to the applied moment.

When equations (10) and (11) are satisfied at all points along the beam the resulting deflected shape isone of equilibrium. Such a shape is defined by

L

zsin

L/EI

Mu

2y

2

cr == (12)

This buckled shape satisfies the boundary conditions of

zero lateral deflection ,0)L(u)0(u == (13)

zero twist at supports ,0)L()0( == (14)

and ends free to warp .0)L(dz

d)0(

dz

d

2

2

2

2

==

(15)

Eq. (12) also satisfies the equilibrium equations (10) and (11) provided

+

=

2w

2

2

y2

crL

EIGJ

L

EIM (5)

which defines the critical lateral-torsional buckling moment.

EFFECT OF BENDING MOMENT PATTERN

Equation (5) gives the critical moment for the standard case of uniform bm along the member. This is

the most severe loading case as it results in a constant, maximum compression force in the top flange.

Other loading cases allow the beam to reach higher buckling moments. A few cases are shown in the

table below.

Loading Case (and bms) Critical Moment Comment

McrThe standard case.

Value given in equation (5)

1.75Mcr

1.13Mcr

1.35Mcr

1.28Mcr


9/24


2.05Mcr

M

M

cr

cr2

M5.2

M)3.005.175.1(

++

Useful general rule for a beam segment

with linearly varying bm.11

Table 3 Variation of critical moment with bending moment pattern (Table 5.6.1 in NZS3404 lists

more cases see p.19)

EFFECT OF LOAD HEIGHTThe derivation of Mcrassumes that loading is

applied at the shear centre of the cross-

section (the point at which the resultant

internal shear force acts).

If the load is applied above this point it tends

to de-stabilise equilibrium and reduces the

critical moment. Conversely, applying the load

below the shear centre will increase Mcr.

CRITICAL FLANGEFor future reference we define the critical flangeas the flange which would deflect the furthest

laterally during buckling. Typically it is the compression flange, but in the case of a cantilever is the

tension flange.

MEMBERS,SEGMENTS,SUPPORTS AND RESTRAINTS

MEMBER

segmentsegmentsegment

support

twist restraintat support

restraint to main member providedby transverse secondary members(if suitably connected)

1. Membersspan between supportsor between a support and a free end.

2. Supports must provide FULL or PARTIAL restraint to a cross section (see definitions below).

3. Members are usually restrainedat discrete points between their supports.

4. The effect of the restraints is to divide the member into segments.Member moment capacity is

then determined on a segment by segment basis.

5. The member moment capacity needs to be determined only for x-axis bending. For y-axis

bending, only the sectionmoment capacity is required (a beam cannot buckle laterally when bent

about its minor axis).6. Restraints provide full, partial or lateral restraint to a cross section, depending on the nature of

the restraint and the location of the critical flange.

Load applied at shear centreNo change in Mcr

Load applied at top flangeDestabilising effectM reducedcr

load

beam resistance actsthrough shear centre

load

beams resisting forcein line with load


10/24


LATERAL AND TORSIONAL RESTRAINTS

The nature and positioning of the restraints on a beam are of profound importance. The otherwise

poor performance of an unrestrained beam can be greatly improved by the provision of restraints.

Restraint of some sort is always provided at the supportsof a beam, but frequently at other,

intermediate points along its length as well.

Restraints are classified as FULL, PARTIAL or LATERAL according to how well they restrain the

critical flange against lateral movement, and the whole cross-section against twisting.

A supportmust always provide FULL or PARTIAL restraint.

Type Description Examples (C denotes critical flange)

FULL (F)

Critical flange restrained against

lateral deflection and twist fully

restrained

webstiffener

C C

(Either flange could be critical)

PARTIAL (P)

Some point other than critical

flange restrained against lateral

deflection and twist partially

restrained, or

full twist restraint with partial or

no lateral deflection restraint

flexible

fly brace

CC

LATERAL (L)Critical flange restrained against

lateral deflection

C

UNRESTRAINED

(U)

No constraint to the critical

flange

C

ROTATIONAL RESTRAINT IN PLAN (WARPING RESTRAINT)

Figures 2 to 5 show warping rotations of the flanges about the minor y axis at beam ends. If these

minor axis warping rotations are restrained by a support, or other means, restraining moments and

shears develop adding significantly to the beams torsional resistance, raising its critical lateralbuckling moment. Fig. 6 shows four types of restraining moment that may be present at the ends of a

beam.

Mx Major axis bending moment which provides restraint about the major axis (e.g. at the fixed end

of a cantilever). Mxis determined by normal structural analysis (plastic or elastic) and depends

on the applied load, beam span, etc.

T Torque about the longitudinal z axis providing restraint against end twisting.

Mtop Top flange end moment providing restraint about the minor axis and against end warping.

Mbot Bottom flange end moment providing restraint about the minor axis and against end warping.

Rotational restraint about the y axis may for example, be provided by a continuation of the member

under consideration see Fig. 7


11/24


z

x

y

Mx

Mtop

Mbot

T

Figure 6 Types of restraining moment at a support

ELEVATION

PLAN VIEW

Effective length, Le

restrained segment

lateral rotationrestrained byadjacent segment

adjacent segment needs full lateralrestraint (or of shorter span)

Figure 7 Example of lateral rotation (warping) restraint provided by adjacent segments.

EFFECT OF IMPERFECTIONS

1.0

0

0.5

Full plasticity, MP

Elastic buckling, Mcr

NZS3404 Eq.5.6.1.1(3)

Dimensionlessm

omentcapacity,

M

/Mb

P

Slenderness ratio, L/ry

100 200 300

Figure 8 Moment capacity of imperfect beams

The inevitable presence of initial imperfections, such as residual stress and lack of straightness, will

influence the moment capacity of a beam just as they influenced the axial load capacity of a column.Short, stocky beams will reach a moment capacity of MPand long slender beams will fail by lateral

buckling at a moment close to the theoretical Mcr. In between these extremes failure will involve a


12/24


combination of inelastic material behaviour (yielding) and lateral buckling. Fig. 8 shows measured

failure moments for I-beams compared with the MPand Mcrlimits.

It can be seen from Fig. 8 that the empirical beam design curve specified in NZS3404 provides a

transition between the full plasticity and elastic buckling curves that is close to the lower bound of

the experimental results shown. The equation of the beam design curve is

s

cr

s

2

cr

sb M

M

M3

M

M6.0M

+

= (16)

or ssb MM =

where

+

=

cr

s

2

cr

ss M

M3

M

M6.0 (17)

is the slenderness reduction factor,

Msis the section moment capacity defined earlier, and

Mcris the lateral buckling moment for a beam of effective length, Le.

+

= 2

e

w2

2e

y

2

crLEIGJ

LEIM (18)

(which is the same as equation (5) apart from L being replaced by Le, the effective length.)

STRENGTH LIMIT STATE DESIGN TO NZS3404

Beam strength must be checked for each segment of a beam (see p.9 for segment definition). The

segment requiring the greatest strength dictates the required section size.

If M*denotes the design bending moment in a beam determined by analysis using the factored loads,

then the beam must satisfy

s* MM

and b* MM

where is the strength reduction factor (0.9),Msis the nominal sectioncapacity in bending,

and Mbis the nominal membercapacity in bending

SECTION MOMENT CAPACITY

Depends on plate slenderness limits. Defined earlier as

yeffs ZM = (4)

MEMBER MOMENT CAPACITY

Depends on cross-section properties, bending moment pattern, degree of restraint and load height:

ssmb MM = (19)

where mis a moment modification factor that takes account of bending moment pattern

and sis the slenderness reduction factor defined in Eq. (17).


13/24


Moment modification factor, m

For each segment of a beam, mcan be determined from NZS3404s Table 5.6.1 (reproduced on p.19)which lists a range of common bending moment patterns.

For bending moment distributions which are not represented in Table 5.6.1 the following general rule

can be used:

( ) ( ) ( )5.2

MMM

M7.1

2*4

2*3

2*2

*max

m

++

= (20)

where signinpositiveastakensegment,theinmomentbendingdesignmaximumM*max =

segmenttheofpointsquartertheatmomentsbendingdesignM,M *4*2 =

segmenttheofpoint-midtheatmomentbendingdesignM*3 =

EXAMPLES

1. Consider the right hand span, BC, of the

continuous beam shown. The secondary beams

apply loads and also provide restraint, hence we

must consider the three segments shown.

Segment 1

72

-92

Use Case 1 from Table 5.6.1 (also shown as last row

of Table 3 on p.8).

M = 92, mM = 72. (Choose M as the larger of the end moments)

Hence m= 72/92 = 0.783, and

5.275.2)783.0(3.0)783.0(05.175.13.005.175.1 22mmm =++=++=

m= 2.5 (maximum value for this case)

Segment 2

72116

Using Case 1 again:

M=116, mM = -72, and m= -72/116 = -0.62.

)5.2(21.1)62.0(3.0)62.0(05.175.13.005.175.1 22mmm =++=++=

m= 1.21

Segment 3 (also segment AB)

116

Use Case 9 from Table 5.6.1 (also appears as row 2 of Table 3 on p.8)

m= 1.75.

Could also have used Case 1 from Table 5.6.1, with M=116, mM=0, and m=0 giving

)5.2(75.1)0.0(3.0)0.0(05.175.13.005.175.1 22mmm =++=++=

3m 2m 2m 2m

60kN 80kN

segment 1 segment 2 segment 3

-92

72116

restraintrestraint

A B C


14/24


2. Consider the right hand span, BC, of the

continuous beam shown. The secondary beams in

this case are assumed to provide no lateral

restraint and so BC must be treated as a single

segment.

Segment 1

Same bending moment pattern as segments 1 and 3

in previous example hence the samem= 1.75

Segment 2

M* =109max

M*=382

M* =1053 M* =934

Since there is no matching bm pattern in Table 5.6.1 we resort to the general rule (20)

( ) ( ) ( )5.2275.1

9310538

)109(7.1

MMM

M7.12222*

4

2*3

2*2

*max

m =++

=++

=

*maxM is the (positive) value of the absolute maximum moment anywhere in the segment i.e. the

109kNm moment at end B in this case.

The quarter point and mid-span moments can be read directly from the bending moment diagram.

Slenderness reduction factor, s

scan be calculated from Equations (17) and (18):

+

=

cr

s

2

cr

ss M

M3

M

M6.0 (17)

+

=

2e

w2

2e

y2

crL

EIGJ

L

EIM (18)

Ms, the section moment capacity and Le, the effective length, are needed first. However, the

calculation is tedious and it is usual to look up pre-calculated values of sin tables.

Table A1 sets out values of sfor grade 300 UB sections with effective lengths ranging up to 10m.

Table A2 covers grade 300 UC sections.

Table A3 lists grade 300 PFC (Parallel Flange Channel) sections with effective lengths up to 7m.

Similar tables are available for most grades of standard steel sections.

Effective Length, Le

The effective length concept is used in the same way and for the same reasons as in the case ofcolumn buckling. A single buckling moment equation can then be used, with different support

conditions, etc, accommodated by altering the effective length.

3m 2m

-109

no restraintat loads

A B C

2m 2m 2m

40kN 40kN 40kN

38

105 93

segment 2segment 1


15/24


Effective length will depend on lateral-torsional (twist) restraint, y-axis rotational restraint (warping)

and load height relative to the shear centre.

LkkkL rLte = (21)

where kt = a twist restraint factor given in Table 5.6.3(1),

kL = a load height factor given in Table 5.6.3(2),

kr = a rotation restraint factor, typically = 1.0, otherwise as given in Tables 5.6.3(3).

In Tables 5.6.3(1), 5.6.3(2) and 5.6.3(3)

d = overall depth of section

L = segment length

nw = number of webs (e.g. 1 for I-beam, 2 for single-cell box)

tf = thickness of critical flange

tw = thickness of web

F = fully restrained

L = laterally restrained

P = partially restrained

U = unrestrained

and two of the symbols F, L, P, U are used to indicate the restraint conditions at the two ends,

e.g. FF.

Note For loads applied to a member along a principal (y) axis passing through the shear centre andthe centroid (such as an I-beam), the classification of load height applied through the top

flange need be applied only when the load itself or the structural system transferring the load

to the segment is laterally unrestrained.


16/24


Note

Restraint of minor axis rotation requires a support of significant stiffness. Typically this is not

available and the factor is left as kr= 1. NZS3404 states the following conditions:

1. krshall only be taken as less than unity when effective rotational restraints act at one or both endsof a segment which is restrained (torsionally) at both ends. krshall be taken as unity for all

segments which are unrestrained at one end.

2. A rotational restraint at a cross section which is fully, partially or laterally restrained may be

considered to provide restraint against lateral rotation of the critical flange about the minor y-

axis, providing that its flexural stiffness in the plane of rotation is comparable with the

corresponding stiffness of the restrained member, EIy.

3. A segment which has full lateral restraint may be deemed to provide rotational restraint to an

adjacent segment which is part of the same member (.e. the member containing both segments is

continuous through the point of restraint). See Fig.7.

4. A segment which does not have full lateral restraint shall be assumed to be unable to providerotational restraint to an adjacent segment which is part of the same member.


17/24


18/24


4. Moment modification factor, 13.1m = (Table 5.6.1, case 6 or 7, with 0m = )

5. kt= 1.0 (FF in Table 5.6.3.1)

kL= 1.4 (FF in Table 5.6.3.2, load within segment)

kr= 1.0 (FF, but no minor axis rotation restraint Table 5.6.3.1)m72.68.40.14.10.1LkkkL rLte ===

6. Slenderness reduction factor, 295.0

0.60.7

0.672.6)281.0333.0(333.0s =

= (Table A1).

In Table A1: Le= 6m: s= 0.333Le= 7m: s= 0.281

(result obtained by interpolation see alternative direct calculation below*)

7. kNM0.38114295.013.1MM sxsmbx ===

8. Check: bx*x MM

OK0.386.34

9. Try a smaller size, 250 UB 25:

kNm92Msx =m= 1.13 (no change)

Le= 6.72m (no change)

s= 0.238 (interpolation, Table A1)kNM7.2492238.013.1MM sxsmbx ===

bx*x MM

NG7.246.34

Smaller size inadequate USE 250 UB 31

* Direct calculation of sProperties needed for calculation:

E kPa10200 6 (material property)

G kPa1080 6 (material property)

y kPa10320 3 (from Table of UB section properties)Iy

46m1047.4 (from Table of UB section properties)

Iw69m109.65 (from Table of UB section properties)

J 49m103.89 (from Table of UB section properties)Zelastic

36 m10354 (from Table of UB section properties)

Zp

36

m10397

(from Table of UB section properties)Ze

46 m10395 (from Table of UB section properties, or by using

36elasticp

epey

sey

elasticeeyeps m10395)ZZ(ZZ,16,9,2.9 =

+====

Le 6.72m

Msx kNm4.1261032010395Z36

ye ==

kNm3.44L

EIGJ

L

EIM

2e

w2

2e

y2

cr =

+

= (Eq. 18)

290.0M

M

3M

M

6.0 cr

s

2

cr

s

s =

+

=

(c.f. 0.295 from Table A1)


19/24



20/24



21/24


EXAMPLE 2

Member:

250UB31 Gr 300 beam, simply supported over a span of 6.0m.

Loading:

Factored loading amounts to 12kN/m applied to the top flange

plus a moment of 40 kNm at the left-hand end. The load is not

laterally restrained.

Restraints:

LEFT: Both flanges fully restrained laterally, and against

minor axis rotation (warping restraint).

RIGHT: Bottom flange restrained against rotation and lateral displacement, top flange unrestrained.

DESIGN CHECK

1. Design moment:

From BM diagram kNm0.40M*x = .

2. Segments:

Restraints at ends only beam consists of a single, full length segment (6.0 m long).

3. Section moment capacity kNm114Msx= (Table A1)

4. Moment modification factor, m:

Table 5.6.1, case 6: 74.08/612

40

8/wL

4022m

===

34.15.325.1 mm =+=

5. Effective length:

Segment has Full restraint at one end, Partial at the other

End restraint = FP

kt= 015.11

1

1.62

6.8

6000

2521

n

1

t2

t

L

d1

3

w

3

w

f =

+=

+ (FP in Table 5.6.3.1)

kL= 1.4 (Table 5.6.3.2: FP, load within segment, Top Flange)

kr= 1.0 (FP, but no minor axis rotation restraint Table 5.6.3.1)m25.70.685.04.1015.1LkkkL rLte ===

6. Slenderness reduction factor, sIn Table A1: Le= 7m: s= 0.281Le= 8m: s= 0.243Le= 7.25m: s= 0.272 (linear interpolation)

7. Member capacitykNM5.41114272.034.1MM sxsmbx ===

8. Check: bx*x MM

OK5.410.40

6.0m

w = 12kN/m

250 UB 31

40kNm

40 kNm


22/24


44.1

(e)

177.9

EXAMPLE 3

4m

A B C

2.5m 2.5m 4m

D

2.5m 2.5m

160kN

The continuous beam ABCD is to be designed to carry load-factored midspan live loads of 1.5Q=160kN

that may act on any or all spans. The task is to select a suitable grade 300 UB section and suggestappropriate lateral restraint locations.

Restraints:

All supports are assumed to provide full torsional restraint. Loads provide no restraint and are applied

to the top flange.

Analysis:

We will ignore self-weight and hope to show that its

effect is negligible. This leaves 5 load cases to consider,

as shown.

The results could be obtained from just two analyses (bymoment distribution, for example):

(a) Load on AB

(b) Load on BC

All other cases can be obtained as combinations of (a) and

(b). For example case (c) is simply (a)+(b), and case (d) is

(a)+(b)+mirror image of (a).

Maximum design moment (all cases) = 207.1kNm.

First approximation:Assume a compact section, and take Ze= Zp.

For ye* ZM <

Require 3336p mm10767m10767000,3009.0

1.207Z ==

>

Try 360UB44.7, Zp=777x103mm3(isnt quite compact, but has only slightly smaller Ze=770 x10

3mm3,

more than compensated for by y=320MPa, so that kNm222Z ye = , > 207.1)

Check end spans AB, CD

Top flange

Maximum M*(+ve)=178kNm - load case (e) - top flange critical (comprn).

FF segment, 5m long. kt=1, kL=1.4, kr=1, Le= 1.4*5 = 7.0m

s= 0.293 (Table A1)

294.051603

161.44m =

= (Table 5.6.1, case 4)

39.129.015.035.1m =+=

NG)kNm178(MkNm4.90222293.039.1M *maxb


23/24


This creates two segments, 2.5m long. Considering the left hand segment:

FL segment, 2.5m long. kt=1, kL=1 (load outside seg), kr=1, Le= 1.0*2.5 = 2.5m

s= 0.778 (Table A1).

m= 1.75 (Table 5.6.1, case 1, m= 0, or case 9).

kNm302222778.075.1Mb == ,

However OK)kNm178(MkNm222MbemustM*

maxsb >=

Right segment certain to be less critical as bending moment pattern will give rise

to a higher m:

25.0178

1.44m == (Table 5.6.1, case 1).

(max)5.278.3)25(.3.0)25.0(05.175.1 2m ==++= .

OKkNm222MstillMso,1 sbsm ==>

Bottom flange

Top flange restraint does not restrain bottom flange. For loading case

(b) bottom flange is critical so spans AB and BC revert to 5m segments.

Maximum M*(-ve)=113kNm

FF segment, 5m long. kt=1, kL=1, kr=1, Le= 5.0m

s= 0.436 (Table A1)

m= 1.75 (Table 5.6.1, case 1, m= 0).

OK)kNm113(MkNm169222436.075.1M *maxb >==

Check middle span, BC

Top flange

Top flange is critical flange for load cases (b), (c) and (d).

Maximum M*(+ve)=207kNm - load case (b) - top flange critical.

FF segment, 8m long. kt=1, kL=1.4, kr=1, Le= 1.4*8 = 11.2m

s< 0.190 (Table A1; s= 0.19 is for Le=10m)

706.08160

8113;113

8

FLmm =

== (Table 5.6.1, case 4)

60.1706.036.035.1m =+=

NG)kNm207(MkNm5.6722219.06.1M *maxb ===

177.9

F L

A

2.5m

177.9

44.1

F

L

B

5m

113

(b)F F

113

207

113

FF FF

B C

207

F L

B

4m

113


24/24

Bottom flange

Predominantly ve bm in loading cases (e) and (a) results in critical bottom flange. Since bottom flange

is unrestrained it becomes an 8m segment.

Load case (e):

Maximum M*(-ve)=44 kNm


s= 0.248 (Table A1)m= 1.0 (Table 5.6.1, case 8)

OK)kNm44(MkNm55222248.00.1M *maxb >==

Load case (a):

Maximum M*(-ve)=64 kNm


s= 0.248 (Table A1)

31.07.63

6.19

m

== (Table 5.6.1, case 1).

1.2)31(.3.0)31.0(05.175.1 2m =++=

OK)kNm64(MkNm115222248.01.2M *maxb >==

Check that dead load IS negligible:

Dead load is 44.7 kg/m giving m/kN53.045.02.1wdead == . Applying a uniform spread load of

0.53kN/m gives the bms shown below.

2.47

1.75

0.64

2.47

0.64

These are insignificant compared with the 1.6Q (live load) bms which ranged up to 207 kNm.

Final configuration:

4m

A B C

2.5m 2.5m 4m

D

2.5m 2.5m

F F F FL L L

= additional restraint to top flange

C

360 UB 45

Documents

Beam Design Notes Nzs 3404