Design of Steel Beams

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Nov 29, 2013 (3 years and 6 months ago)

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Design of Steel Beams



Types of beams



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supported beams:


-

Overall behaviour


-

Design conside
rations:







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DEFINITIONS


Beam:


Structural member whic
h carries loads that are applied perpendicular to the
longitudinal axis of the member.

Girder:


A major, or deep, beam which often provides support to other beams.

Stringer:

A main longitudinal beam spanning between floor beams (bridge floors).

Floor beam:

A transverse beam in bridge floors.

Joist:


A light beam that supports a floor.

Lintel:


A beam spanning across an opening (door or window). Usually in masonry
construction.

Spandrel:

A beam on an outside perimeter of a building which supports, among oth
er loads, the
exterior wall.

Purlin:


A beam that supports a roof, and frames between or over supports such as roof
trusses or frames.

Girt:


A light beam that supports only the lightweight exterior siding of a building.

Steel Beams

2





Source: CISC Handbook of Steel C
onstruction





Steel Beams

3


Beams, along with columns, are the structural members used most frequently in structures. Its main
function is to carry gravity loads and transmit them to the supports.

Beams resist loads primarily through flexure.

Limitations in Civ
E
37
4:



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-

Laterally supported beams
: the compression flange is prevented from lateral movement by
bracing either continuously or at close intervals.



-

Laterally unsupported beams
: the c
ompression flange is laterally unsupported or is supported
at large intervals.



The member moves down with a lateral movement and twists (Lateral
-
Torsional Buckling)


Laterally Supported Beams

Cross
-
section behaviour


-

Plane sections

remain plane (


the strain distribution is linearly varying through the cross
-
section)

-

The moment corresponding to first yielding of the extreme fibers of the cross
-
section is called
the yield moment, M
y

=

y

S.

-

Gradual yielding of the cross
-
section progresses until the entire cross
-
section has yielded under
the plastic moment, M
p

=

y

Z.

Moment
-

Curvature Relationship

Residual Strains Considered
Strain Hardening
Considered
Residual Strains
Neglected
M
p
M
y
0.7 M
y
M
Curvature,

E I
1

Rotations are obtained by integrating the curvature (
M/EI) over the beam. Deflections are obtained
by integrating the rotations.

Load
-
Deflection Curve


M
M
p
M
y
Class 1
Class 2
Class 3
Class 4

P
Ideal Behaviour


beam deflects straight down

bracing offers no resistance to bending


Steel Beams

4


Because of strain
-
hardening, the moment capacity of a cross
-
section can exceed the theoretical
plastic moment given as M
p

=

y

Z.

Failure of the laterally supported beam will occur by local buckling.

Local Buckling

The compression flange and the portion of the web in compression can fail by instability (buckling).
If only a portion of the member buckles (i.e. portion of the fl
ange, or web) the buckling is called
local buckling. This is characterized by the deformation of the cross
-
section.


Steel sections are classified into four classes with respect to their local buckling capacity.



Class 1

(plastic design): The section is can reach the Plastic Moment, M
p
, and deform to allow
moment re
-
distribution before local buckling occurs.



Class 2

(compact section): The section can reach the plastic mo
ment, M
p
, before local buckling
occurs.



Class 3

(non
-
compact section): The section can reach the yield moment, M
y
, but cannot reach
the plastic moment before local buckling occurs.



Class
4

(slender section): The section local buckles elastically (i.e. at a moment less than the
yield moment).

Local buckling is essentially the buckling of a component plate of the cross
-
section. Most cross
-
sections are made of two different types of plate:


Flange Plates

Web Plates

Supported
Edge
Edge
Free


b
o


2
E
cr
2
b
2
o
12(1 )
t
k




 

 
 

2
E
2
h
2
12(1 )
w
k



 

 
 

By setting the critical stress to F
y
, the specified yield strength, and using the appropriate plate
buckling coefficient, k, the following table can be ob
tained.


Maximum Width
-
to
-
Thickness Ratios (CSA
-
S16
-
01)



From the above table, the following limitations for wide flange beams are obtained.

Steel Beams

5



Flange

Web

Class 1:

y
o
F
145
t
b


y
F
1100
w
h



Class 2:

y
o
F
170
t
b


y
F
1700
w
h


C
lass 3:

y
o
F
200
t
b


y
F
1900
w
h


Note that these limitations can be more conveniently rewritten as:

200 1900
o
y y
b
h
F or F
t w
 


for example

Flexural Strength

For a rectangular cross
-
section,

d
b
C
T
2/3 d

y

y
M = M
y
C
T

y

y
M = M
p
d/2
(a)
(b)
M

From (a),



y
y
y
S
2
/
d
I
M




; where
6
d
b
S
2


(NOTE: This equation
for S applies to rectangular cross
-
sections only)

From (b)



;
Z
M
y
p




where
4
d
b
Z
2


(for rectangular sections)

Factored Bending Resistance

Class 1 and 2:

M
r

=


M
p

=


F
y

Z

Class 3 :

M
r

=


M
y

=


F
y

S

Class 4 :

local buckling governs (refer to CSA
-
S136 standard)

In the above equations


= 0.9

Example 1


-

Laterally supported beam, W360x64

-

G40.21M
-

300W steel

-

Neg
lect self weight of beam

-

Find the maximum factored load P that can be
safely applied on the beam.


From tables of section properties,

S
x

= 1030x10
3

mm
3
; Z
x

= 1140x10
3

mm
3

; d = 347 mm

b = 203 mm; t = 13.5 mm; w = 7.7 mm

Check Class of section

203
300 130 2 145
2 13 5
o
y
b
F.
t.
  



(Class 1 flange)

h = d
-

2t = 347
-

2x13.5 = 320 mm

320
300 720 1100
7 7
y
h
F
w.
  


(Class 1 web)

Therefore, the section is a Class 1 section and,

M
r

=


M
p

= 0.9 F
y
Z
x

= 0.9 x 300 x 1140x10
3


M
r

= 307.8 kN
.
m


M
f

= 2P
f

P
f



154 kN

Check the effect of neglecting the weight of the beam:

For a W360x64 mass = 64 kg/m; w
D

= 0.627 kN/m

Moment due to weight =
8
L
w
2
D

= 4.8 kN∙m

M
r

= 307.8 kN∙
m


2P
f

+ 4.8 (1.25)

2 m
3 m
2 m
2P
M
P
P

Steel Beams

6



P
f



151 kN (The previous solution was approximately 1.6% unconservative).


Shear in Beams

Although theoretically shear should reduce the flexural strength of beams, tests on standard rolled
sections indicate that beams subjected t
o high shear forces can carry significantly higher moments
than beams with no shear.

The interaction between shear and moment is usually neglected for standard rolled I
-
sections. The
problem is to prevent yield of the web under shear forces. Using von
Mises yield criterion, the yield
stress in shear is given as:



y
y
y
577
.
0
3






Note: the following diagram of shear stress distribution in a I
-
shaped cross
-
section indicates that
most of the shear force is carried by the web.




=
V Q
I t
V/A
w



Shear Stress Distribution in a W
-
Shape

Since the major portion of the shear in a wide flange member is carried by the web, Clause
13.4.1

of
S16 expresses the shear resistance as:



V
r

=


A
w

F
s

where A
w

is the shear area given as
d x w

for r
olled members and
h x w

for welded built
-
up
members, and F
s

is the ultimate shear strength taken as 0.66 F
y
. This is greater than the shear yield
strength based on von Mises criterion to take into account the effect of strain hardening.

If the web is rela
tively slender, the web can fail in shear buckling. For this reason, S16.1 places a
limitation on the web slenderness for which the full shear yield capacity can be developed
(F
s

=

0.66

F
y
).



y
v
F
k
439
w
h


where k
v

is the shear buckling coefficien
t given as:


2
v
h
a
34
.
5
4
k









when a/h < 1.0


2
v
h
a
4
34
.
5
k









when a/h


1.0

for an unstiffened web a/h




and k
v

= 5.34.

For design, V
r



V
f


Example 1

(Check shear strength)

6
.
41
7
.
7
5
.
13
x
2
347
w
t
2
d
w
h






The web is unstiffened,


k
v

= 5.34

6
.
58
300
34
.
5
439
F
k
439
y
v



,


F
s

= 0.66 F
y

A
w

= d w = 347 x 7.7 = 2672 mm
2



V
r

=


A
w

F
s

= 0.9 (2672) (198) =
476 kN

V
f

= P
f

= 151.5 kN << V
r



shear strength O.K.

Serviceability Limit State

Deflections must be checked under service (specified) (unfactor
ed) live loads.

Appendix D of S16 gives allowable deflections under live load and wind load.


Industrial buildings:



Simple span floors,

L.L.



L/300



Simple span roofs with inelastic covering,

L.L.



L/240


All other buildings:



Simple span (flo
ors and roofs supporting finishes susceptible to cracking),

L.L.



L/360



Simple span (floors and roofs supporting finishes
not
susceptible to cracking),

L.L
.



L/300

Steel Beams

7


Example 1

Assume the beam is part of the floor s
ystem of an industrial building
(

L.L.



L/300).

Let P
f

= 154 kN and assume ½ D.L. and ½ L.L.

The service live load =
5
.
1
1
x
2
154

= 51.3 kN

from CISC Handbook p. 5
-
134




max

=
)
a
4
L
3
(
I
E
24
a
P
2
2


= 15.7 mm


L/300 = 7000/300 = 23.33 mm

> 15.7 mm




the deflection is satisfactory

Note: in some instances the deflections under unfactored dead loads should be considered. (e.g. if
a significant portion of the dead load is applied after the elements susceptible to cracking or other
servi
ceability limit state are installed, that portion of the dead load should be included in the
calculations).


Laterally unsupported beams

Members loaded in bending can fail by:


1)

Flexural yielding;


2)

Local buckling;


3)

Shear yielding or buckling; or


4)

Lateral
-
torsional buckling.

Lateral
-
torsional buckling occurs if sufficient lateral bracing is
not

provided to the compression
flange. Out
-
of
-
plane bending and twisting of the cross
-
section will occur when the applied l
oads
reach a certain limit.

The value of the lateral
-
torsional buckling load depends on:


-

shape of the cross
-
section;


-

the unbraced length of the beam (short members behave as fully supported beams
while slender beams buckle laterally when the section
is still elastic);


-

the support conditions of the beam;


-

type and position of the applied loads along the member axis;

-

location of the applied loads w.r.t. centroidal axis of the beam.

The resistance to lateral
-
torsional buckling depends on the section
's ability to resist torsion. The
resistance to torsion is made up of two components:

1)

St. Venant torsion;

2)

warping restraint torsion.

St. Venant Torsion

From strength of materials,


GJ
T
L
sv



where,


/L

= angle of twist per unit length

T
sv


=

St. Venant torsion

G

=

shear modulus

J

=

torsional constant of the cross
-
section


Upon rearranging the above equation and expressing the twist per unit length in terms of rate of
twist, the following expression for the St. Venant torsion is obtained:

dz
d
GJ
T
sv



For thin
-
walled, open sections made up of slender component plates,
with midline length b and thickness t, the torsional constant can be
obtained from:






n
1
i
3
i
i
3
t
b
J

2 m
3 m
2 m
P
P
a
a
L


St. Venant shear stress
distribution in an I-section
Steel Beams

8


Warping Restraint Torsion

When a beam of thin
-
walled open cros
s section is subjected to torsion the cross section will warp.
During warping plane sections of the cross section no longer remain plane. When a beam is free to
warp no resistance to torsion is developed.

When warping is restrained at some sections of t
he beam, warping restraint torsion will develop in
the cross
-
section.

top flange
bottom flange
T
V
f
V
f

Because of the warping restraint, the top and bottom flange curve. This curvature results in bending
moments in the plane of the

flanges, which in turn

give rise to shear forces V
f

in both flanges. The
resulting force couple provides the torsion resistance.



3
3
w
f
w
dz
d
C
E
h
V
T





where EC
w

is the warping rigidity and C
w

is called the warping constant. For a wide flange shape,



4
)
t
d
(
I
C
2
y
w



Later
al
-
Torsional Moment Resistance



Uniform moment and elastic material



w
y
2
y
u
C
I
L
E
J
G
I
E
L
M












2
w
2
y
L
C
E
J
G
I
E
L







For other moment distributions



w
y
2
y
2
u
C
I
L
E
J
G
I
E
L
M














2

= equivalent moment factor


= 1.75 + 1.05


+ 0.3

2



2.5


1
2
M
M



where
1
2
M
M




is therefore positive for double curvature and negative for
single curvature

Notes
:




If the bending moment anywhere within the unbraced length is
greater than M
1
,

2

is taken as 1.0 (refer to Cl. 13.6).



If there is no
effective lateral support for the compression
flange at one of the ends of the unsupported length

2

is taken
as 1.0.



Figure 2
-
17 in CISC handbook presents some examples of

2

for various loading and lateral support conditions.


Calculation of

2

for ge
neral moment diagrams

An approximate formula has been proposed by Kirby and Nethercot (“Design for Structural
Stability”, 1979) to calculate the equivalent moment factor for general bending moment diagrams.

max
3
2
1
max
2
M
2
M
3
M
4
M
3
M
12






where the absolute values

of bending moment are used.


The above equations for M
u

apply for beams that remain elastic.

M
M
Unbraced
length (L)

M
1
M
2
Single Curvature
M
1
M
2
Double Curvature
Steel Beams

9


S16 uses the following formulae for inelastic LTB (Cl. 13.6)

p
u
p
p
cr
M
M
M
28
.
0
1
M
15
.
1
M












for Class 1 & 2 sections


y
u
y
y
cr
M
M
M
28
.
0
1
M
15
.
1
M












for Class 3 sections


Design equations

For Class 1 & 2 sections

If
p
3
2
u
M
M


, M
r

=


M
u

otherwise,

p
u
p
p
r
M
M
M
28
.
0
1
M
15
.
1
M














For Class 3 sections

If
y
3
2
u
M
M


, M
r

=


M
u

otherwise,

y
u
y
y
r
M
M
M
28
.
0
1
M
15
.
1
M














Design procedu
re for a laterally unsupported beam

1.

Choose a trial section based on the unbraced length and factored moment.

2.

Check class of the section selected.

3.

Calculate M
u

based on the unsupported length and the moment diagram.

4.

Calculate the factored moment resistance
, M
r
, considering inelastic LTB if necessary.

5.

If the factored resistance is less than the maximum factored moment within the unsupported
length go back to step 1 and select a stronger section.

6.

Check shear capacity.

7.

Check deflections.


Note: Sections with
plate components all intersecting at one point (angle, tee, cruciform) do not
warp. This is also the case for axisymmetric (circular tube or bar). For all these sections
C
w

= 0. Warping for narrow rectangular sections and box sections is usually very sm
all and
C
w



0.


Example 2



Laterally Unsupported Beam

The same as example 1 except that:



(a) no lateral supports
between end supports


(b) with lateral supports at load points


(a)

No lateral supports between ends

L = 7000 mm,

2

= 1.0 (the maximum moment wi
thin the unsupported length exceeds the
moments at the supports).

W360 x 64, Class 2 section, Z
x

= 1140 x 10
3

mm
3

I
y

= 18.8 x 10
6

mm
4
; J = 438 x 10
3

mm
4
; C
w

= 524 x 10
9

mm
6

w
y
2
y
2
u
C
I
L
E
J
G
I
E
L
M






















7000
1
0
200
10
18
8
10
77
10
438
10
200
10
7000
18
8
10
524
10
3
6
3
3
3
2
6
9
x
x
x
x
x
x
x
x
.
(
.
)(
)(
)
(
)
(
.
)(
)

M
u

= 203.8x10
6

N∙mm = 203.8

kN∙m

2/3 M
p

= 2/3 Z
x

F
y

= 2/3 (1140x10
3
) 300 = 228 kN∙m > M
u

Elastic LTB takes place

M
r

=


M
u

= 0.9 x 203.8 =
183.4 kN∙m

2 m
3 m
2 m
2P
M
P
P
Steel Beams

10


P
f

= M
r
/2 =
91.7 kN

(compared to 154 kN for the laterally supported case)

(b) With lateral supports at load points

We have to consider two unbraced lengths

2000
3000
2P
2P
(1)
(2)



2

= 1.75


2

= 1.0

By inspe
ction, it is obvious that Case (2) will govern the design.

For L = 3000 mm and

2

= 1.0,

w
y
2
y
2
u
C
I
L
E
J
G
I
E
L
M











)
10
x
524
)(
10
x
8
.
18
(
3000
)
10
x
200
(
)
10
x
438
(
)
10
x
77
(
)
10
x
8
.
18
(
10
x
200
0
.
1
x
3000
9
6
2
3
3
3
6
3













M
u

= 782.9x10
6

N∙mm = 782.9 kN∙m > 2/3 M
p

= 228 kN∙m

The beam will buckle inelastically

M
r

= 1.15


M
p

(1
-

0.28 M
p
/M
u
)

where M
p

= Z
x

F
y

= 342 kN∙m

M
r

= 1.15 x 0.9 x 342 (1
-

0.28 x (342 / 782.9)) = 310.7 kN∙m



M
p

= 0.9 x 342 = 307.8 kN∙m



Yielding of the cross
-
section controls (
Mr = 308 kN∙m
)



P
f

= 154 kN (This is the same as in example 1)


We can c
alculate the maximum unbraced length for which the maximum capacity (M
p

or M
y
) of the
section can be reached (L
u
).

For a Class 2 section,

M
r

= 1.15


M
p

(1
-

0.28 M
p
/M
u
) =


M
p



(1
-

0.28 M
p
/M
u
) = 1.0

Solving for M
u
, we obtain M
u

= 2.15 M
p



p
w
y
2
y
2
M
15
.
2
C
I
L
E
J
G
I
E
L












for

2

= 1.0, L
u

is listed in beam tables

For example, for the W360x64, L
u

= 3110 mm > 3000 mm



M
r

=


M
p

= 308 kN∙m


Lateral Bracing

Spacing (L):

When L


L
u
,

M
r

=


M
p

or


M
y

when L > L
u
,

M
r

=


M
u

when M
u



2/3 M
p

or 2/3 M
y


M
r

= 1.15


M
p

(1
-

0.28 M
p
/M
u
) when M
u



2/3 M
p


M
r

= 1.15


M
y

(1
-

0.28 M
y
/M
u
) when M
u



2/3 M
y

Bracing force

Refer to Clause 9.2.5 in S16

The bracing (bracing member and connections) to resist a force equal to 2% of the force in the

compression flange at the location of the brace.

From part (b) of Example 2,

M
r

= 307.8 kN∙m = M
f

kN
887
mm
347
mm
kN
10
x
8
.
307
C
3
f





F
b

= 0.02 x 887 kN = 17.8 kN

Each brace must be designed for a force of
2 m
3 m
2 m
P
f
P
f
= 154 kN
Steel Beams

11


17.8 kN.



Concentrated Loads and Reactions

R
N
N
P

Failu
re Modes


1)

Compressive yielding


2)

Web buckling (crippling)

Design provisions
: Both criteria are based on empirical results.

1)

Compressive yielding

a)

Interior loading:



r bi y
B w F
N+10 t


b)

End loading:



r be y
B w F
N+4 t


bi
0 80
.
 
,
be
0 75
.
 

2)

Web buckling (crippling)


a)

Interior loading: Clause 14.3.2 (a)



2
r bi y
B 1 45 w F E
.
 


b)

End loading: Clause 15.9 (b) (ii)



2
r be y
B 0 6 w F E
.
 

Example

(refer to Example 1)

W360x64 beam

d =
347 mm; t = 13.5 mm

w = 7.7 mm







The factored reaction, R
f
, is 154 kN.



The bearing length, N, would have to be determined either to prevent crushing of the support
(e.g. this could be a concrete wall) or to prevent yielding and crippling of the web.



Fo
r this example assume that N = 150 mm for both the interior and end loads.


Interior loading

a)

Compressive yielding




r bi y
B w F
N+10 t



= 0.80 x 7.7 (150 + 10 x 13.5) x 300


= 527 kN >> 154 kN


O.K.


b)

Web buckling


2
r bi y
B 1 45 w F E
.
 


= 1.45

x 0.80 x 7.7
2

300 200000



= 533 kN >> 154 kN


O.K.


The bearing capacity is adequate for the interior loads.


End loads

a)

Compressive yielding




r be y
B w F
N+4 t



= 0.75 x 7.7 (150 + 4 x 13.5) x 300


=

353 kN >> 154 kN


O.K.


b)

Web buckling


2
r be y
B 0 6 w F E
.
 


= 0.6 x 0.75 x 7.7
2

300 200000




= 207 kN > 154 kN


O.K.


The bearing capacity is adequate at the end reactions.


Design of Bearing Stiffeners

If P
f

(or

R
f
) > B
r
, then

1)

increase the bearing length, N, or,




2)

increase the web thickness, or





3)

provide bearing stiffeners.

Bearing stiffeners are designed in accordance to Clause 15.6 of S16

Interior
Exterior
x
x
Steel Beams

12


Bearing stiffeners are also required at unframed ends of sing
le web girders for which
y
F
1100
w
h




b
eff

=

25w for interior loading


=

12w for end loading


Failure modes:

1)

Local buckling of the stiffeners


y
b
b
F
200
t
b


2)

Bearing
-

High contact stresses


P
f




B
r

= 1.5


A F
y


where A = contact area

The contact area, A = 2 x (b
b

-

cope width) x t
b





cope width = k
1

-

w/2

3) Overall buckling of the stiffeners

Calculate
x
b
r
L
k

where k is taken as 0.75 and r
x

is calculated for the cross
-
sec
tion shown.

The column buckling capacity of the stiffeners is calculated using the column curve presented in
Clause 13.3.1 of S16.1.


Example 3



Design of Bearing Stiffeners

Check the beam from the previous example for end reactions of
350

kN

(a)

Yiel
ding B
r

= 353 kN > 350 kN

(b)

Web buckling,


B
r

= 207 kN < 350 kN


Bearing stiffeners are required at the supports.

Try a pair of 160x16 mm stiffeners



1)

Local Buckling


b/t = 160/16 = 10 <
y
F
200

= 11.5



O.K.

2)

Contact St
resses


Contact area, A = 2 x (b
b

-

cope width) x t
b


cope width =
1
2
k w/


= 24
-

7.7/2 = 20.2 say 22 mm


A = 2 x (160
-

22) x 16 = 4416 mm
2


B
r

= 1.5


A F
y

= 1.5 x 0.9 x 4416 x 300 = 1788 kN



>> 350 kN




O.K.


3
)

Column Buckling

I
x

=
3 3
16 (327 7) (92 4 16) (7 7)
12 12
...




= 46.9 X 10
6

mm
4

A

= 327.7 x 16 + (92.4
-

16) x 7.7


= 5832 mm
2

r
x

=
A
I
x

= 89.7 mm; length of column = d
-

2t = 320 mm

effective length, kL = 0.75 L
b

= 240 mm

033
.
0
000
,
200
300
90
240
E
F
r
L
k
2
2
y









1/n
2n
r y
C AF
1

 
 

= 0.9 x 5832 x 300


1/1.34
2 1 34
1 0 033
.
.




= 1575 kN >> 350 kN

Therefore, a pair of 160x16 stiffeners is satisfactory.


b
b

cope
width


b
b
t
b
L
b
b
eff
k
1
350

kN
150
347
7.7
W360x64
Steel Beams

13


Welding of stiffeners



The weld between the stiffeners and the web of the beam are designed
to carry the full
concentrated load/reaction (350

kN).



The minimum weld size is usually sufficient.



Fillet welds are used at four locations.

For t
b

= 16 mm and w = 7.7 mm , minimum weld size = 6 mm (refer to Table 4
-
4 of W59
-
M1989)

For E480 electrode and

6

mm weld

Base metal:


V
r

= 0.67


A
m

F
y


= 0.67 x 0.9 x (6 x 1) x 300


= 1085 N/mm

Weld metal:


V
r

= 0.67

w

A
w

X
u

= 0.67 x 0.67 x (6 x 0.707 x 1) x 480


= 914 N/mm


governs


Length of weld required =
mm
/
N
914
N
10
x
350
3

= 383 mm

If the st
iffeners are welded continuously on four sides, the provided length of weld is


1200 mm.
This is more than sufficient.


Design of Beams

(summary)

1.

Perform a structural analysis (beam sizes may have to be assumed).

2.

Choose a trial section (may use the bea
m selection tables of CISC Hdbk)

3.

Determine the class of the cross
-
section (Table 1 of S16.1). The maximum capacity for
compact sections is M
p

and for non
-
compact sections is M
y
.

4.

Calculate the bending resistance, M
r

a)

Laterally supported (Clause 13.5)


Compac
t sections M
r

=


Z
x

F
y


Non
-
compact sections M
r

=


S
x

F
y

b)

Laterally unsupported


w
y
2
y
2
u
C
I
L
E
J
G
I
E
L
M














2
= 1.75 + 1.05


+ 0.3

2



2.5




For Compact sections


If
p
3
2
u
M
M


, M
r

=


M
u


otherwise,


p
u
p
p
r
M
M
M
28
.
0
1
M
15
.
1
M















For
non
-
compact sections

replace M
p

by M
y

5.

Compare factored moment capacity with the factored applied moment (M
r



M
f
).

6.

Check shear strength according to Clause 13.4 of S16.

7.

Check deflections (Appendix D of S16).


grind to bear
gap