beams subject to torsion and bending

7/30/2019 Beams Subject to Torsion and Bending

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BEAMS SUBJECTED TO TORSION & BENDING-II

BEAMS SUBJECTED TO TORSION AND BENDING - II18

1.0 INTRODUCTION

In the previous chapter, the basic theory governing the behaviour of beams subjected totorsion was discussed. A member subjected to torsional moments would twist about a

longitudinal axis through the shear centre of the cross section. It was also pointed out

that when the resultant of applied forces passed through the longitudinal shear centre axisno torsion would occur. In general, torsional moments would cause twisting and warping

of the cross sections.

When the torsional rigidity (GJ) is very large compared with its warping rigidity (E),the section would effectively be in uniform torsion and warping moment would unlikely

to be significant from the designer's perspective. Examples of this behaviour are closed

hot-rolled sections (e.g. rectangular or square hollow sections) and rolled angles and

Tees. Note that warping moment is developed only if warping deformation is restrained.Warping deformation in angle and T-sections are not small, only warping moment would

be small. On the other hand, most thin walled open sections have much smaller torsional

rigidity (GJ) compared with warping rigidity (E) values and these sections will be

exhibiting significant warping moment. Hot rolled I sections and H sections wouldexhibit torsional behaviour in-between these two extremes and the applied loading is

resisted by a combination of uniform torsion and warping torsion.

2.0 DESIGNING FOR TORSION IN PRACTICE

Any structural arrangement in which the loads are transferred to an Ibeam by torsion isnot an efficient one for resisting loads. The message for the designers is "Avoid Torsion

- if you can ". In a very large number of practical designs, the loads are usually applied

in a such a manner that their resultant passes through the centroid. If the section isdoubly symmetric (such as IorHsections) this automatically eliminates torsion, as the

shear centre and centroid of the symmetric cross section coincide. Even otherwise load

transfer through connections may - in many cases - be regarded as ensuring that the loads

are effectively applied through the shear centre, thus eliminating the need for designingfor torsion. Furthermore, in situations where the floor slabs are supported on top flanges

of channel sections, the loads may effectively be regarded as being applied through the

shear centre since the flexural stiffness of the attached slab prevents torsion of the

channel.

Where significant eccentricity of loading (which would cause torsion) is unavoidable,alternative methods of resisting torsion efficiently should be investigated. These include

Copyright reserved

Version II 18 -1


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design using box sections, tubular (hollow) sections or lattice box girders which are fully

triangulated on all faces. All these are more efficient means of resisting torsionalmoments compared with I orH sections. Unless it is essential to utilise the torsional

resistance of an Isection, it is not necessary to take account of it. The likely torsional

effects due to a particular structural arrangement chosen should be considered in the early

stages of design, rather than left to the final stages, when perhaps an inappropriatemember has already been chosen.

3.0 PURE TORSION AND WARPING

In the previous chapter, the concepts of uniform torsion and warping torsion were

explained and the relevant equations derived.

When a torque is applied only at the ends of a member such that the ends are free to

warp, then the member would develop only pure torsion.

The total angle of twist () over a length ofzis given by

)1(JG

zTq

where Tq = applied torque

GJ = Torsional Rigidity

When a member is in non-uniform torsion, the rate of change of angle of twist will vary

along the length of the member. The warping shear stress (w) at a point is given by

)2(t

SE wmsw

where E = Modulus of elasticity

Swms = Warping statical moment at a particular point Schosen.

The warping normal stress (w) due to bending moment in-plane of flanges (bi-moment)

is given by

w = - E .Wnwfs . ''

where Wnwfs = Normalised warping function at the chosen point S.

4.0 COMBINED BENDING AND TORSION

There will be some interaction between the torsional and flexural effects, when a load

produces both bending and torsion. The angle of twist caused by torsion would be

amplified by bending moment, inducing additional warping moments and torsionalshears. The following analysis was proposed by Nethercot, Salter and Malik in reference

(2).

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4.1 Maximum Stress Check or "Capacity check"

The maximum stress at the most highly stressed cross section is limited to the design

strength (fy /m). Assuming elastic behaviour and assuming that the loads produce bending

about the major axis in addition to torsion, the longitudinal direct stresses will be due to

three causes.

)3(

.. ''

nwfsw

y

yt

byt

x

xbx

WE

Z

M

Z

M

byt is dependent onMyt, which itself is dependent on the major axis moment Mx and the

twist .

Myt = Mx (4)

Thus the "capacity check" for major axis bending becomes:

bx + byt+w fy /m. (5)

Methods of evaluating , , and for various conditions of loading and boundaryconditions are given in reference (2).

4.2 Buckling Check

Whenever lateral torsional buckling governs the design (i.e. when pb is less than fy) the

values ofw and byt will be amplified. Nethercot, Salter and Malik have suggested a

simple "buckling check" along lines similar toBS 5950, part 1

)6(/

1M

M0.51

fM

M

b

x

my

wbyt

b

x

where , equivalent uniform moment = mx MxxM

21

2pEBB

pE

MM

MM

and Mb, the buckling resistance moment =

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in which

2

M1M ELTpB

MP , the plastic moment capacity = fy . Zp /m

Zp = the plastic section modulus

ME , the elastic critical moment =

where LT is the equivalent slenderness. m

y2LT

2p

f

EM

4.3 Applied loading having both Major axis and Minor axis moments

When the applied loading produces both major axis and minor axis moments, the"capacity checks" and the "buckling checks" are modified as follows:

Capacity check:

bx + byt+w + by fy/m (7)

Buckling check:

yybyt

yyy

b

x

my

wbyt

myy

y

b

x

ZM

MmMwhere

M

M0.51

fZf

M

M

M

/

)8(1//

4.4 Torsional Shear Stress

Torsional shear stresses and warping shear stresses should also be amplified in a similar

manner:

)9(

b

x

wtvtM

M0.51

This shear stress should be added to the shear stresses due to bending in checking the

adequacy of the section.

5.0 DESIGN METHOD FOR LATERAL TORSIONAL BUCKLING

The analysis for the lateral torsional buckling is very complex because of the different

types of structural actions involved. Also the basic theory of elastic lateral stability

cannot be directly used for the design purpose because

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the formulae for elastic critical momentMEare too complex for routine use and there are limitations to their extension in the ultimate rangeA simple method of computing the buckling resistance of beams is given below. In a

manner analogous to the Perry-Robertson Method for columns, the buckling resistance

moment,Mb, is obtained as the smaller root of the equation

(ME- Mb) (Mp - Mb) = LT. MEMb (10)

As explained in page 3,Mbis given by,

21

2pEBB

pE

MM

MM

Mb =

2

M1M ELTpB

where

[ As defined above,ME = Elastic critcal momentMp = fy . Zp /m

LT = Perry coefficient, similar to column buckling coefficient

Zp = Plastic section modulus]

In order to simplify the analysis, BS5950: Part 1 uses a curve based on the above

concept (Fig. 1 ) (similar to column curves) in which the bending strength of the beam is

expressed as a function of its slenderness (LT). The design method is explained below.

The buckling resistance momentMb is given by

Mb= pb .Zp (11)

wherepb = bending strength allowing for susceptibility to lateral -torsional buckling.

Zp = plastic section modulus.

It should be noted thatpb = fy for low values of slenderness of beams and the value ofpbdrops, as the beam becomes longer and the beam slenderness, calculated as given below,increases. This behaviour is analogous to columns.

The beam slenderness (LT) is given by,

(12)LT

y

LTf

E 2

EM

MpLT where

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Beam fails by yield

Beam buckling

50

0

100

pbN/mm2

200

300

150 200 250100LT

Fig.1 Bending strength for rolled sections of design strength

275 N/mm2

according to BS 5950

Fig. 2 is plotted in a non-dimensional form comparing the observed test data with the twotheoretical values of upper bounds, viz. Mp and ME. The test data were obtained from a

typical set of lateral torsional buckling data, using hot-rolled sections. In Fig. 2 three

distinct regions of behaviour can be observed:-

stocky beams which are able to attain the plastic moment Mp, for values of belowabout 0.4.

LT

Slender beams which fail at moments close toME, for values of above about 1.2LT

beams of intermediate slenderness which fail to reach eitherMp orME . In this case0.4


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For more slender beams,pb is a function ofLTwhich is given by ,

u is called the buckling parameter andx, the torsional index.

For flanged sections symmetrical about the minor axis,

and

For flanged sections symmetrical about the major axis

and

In the aboveZp = plastic modulus about the major axis

A = cross sectional area of the member

)13(y

LTr

uv

0.2 0.4 0.6 0.8 1.0 1.2 1.40

0.2

0.4

0.6

0.8

1.0

stocky intermediate slender

ME/ MP

Plastic yield

M / Mp

EM

PM

LT

Fig.2 Comparison of test data (mostly I sections) with theoretical elastic critical moments

41

s

pu

2hA

Z4

2

2 2

1

JAh0.566x s

21

41

2

2

u

A

ZI py

JI1.132x

A

y

x

y

I

I1

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322311

32

3121

2

btbt12

bbtths

= torsional warping constant

J = the torsion constant

hs = the distance between the shear centres of the flangest1, t2 = flange thicknesses

b1, b2 = flange widths

We can assume

u = 0.9 for rolled UBs, UCs, RSJs and channels

= 1.0 for all other sections.

is given in Table 14 ofBS5950: Part I

r

functionav

y

xof ,

(for a preliminary assessment v = 1)

x = D/Tproviding the above values ofu are used.

5.1 Unequal flanged sections

For unequal flanged sections, eqn. 11 is used for finding the buckling moment of

resistance. The value of LT is determined by eqn.13 using the appropriate section

properties. In that equation u may be taken as 1.0 and v includes an allowance for thedegree of monosymmetry through the parameterN = Ic / (Ic + It ) . Table 14 ofBS5950:

Part1 must now be entered with (E/ry )/x andN .

5.2 Evaluation of differential equations

For a member subjected to concentrated torque with torsion fixed and warping free

condition at the ends ( torque applied at varying values ofL), the values of and

its differentials are given by

(1-

Tq

azsinh

acosh

atanh

asinhaz1

GJaTq

.For 0 z ,

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a

zcosh

acosh

atanh

asinh

1GJ

Tq

a

zsinh

acosh

atanh

asinh

aJG

Tq

a

zcosh

acosh

atanh

asinh

aJG

Tq

2

Similar equations are available for different loading cases and for different values of.Readers may wish to refer Ref. (2) for more details. We are unable to reproduce these on

account of copyright restrictions.

6.0 SUMMARY

This chapter is aimed at explaining a simple method of evaluating torsional effects and toverify the adequacy of a chosen cross section when subjected to torsional moments. The

method recommended is consistent with BS 5950: Part 1.

7.0 REFERENCES

(1) British Standards Institution, BS 5950: Part 1: 1985. Structural use of steelwork in

Building part 1: Code of Practice for design in simple and continuousconstruction: hot rolled sections. BSI, 1985.

(2) Nethercot, D. A., Salter, P. R., and Malik, A. S. Design of Members Subject toCombined Bending and Torsion , The Steel construction Institute , 1989.

(3) Steelwork design guide to BS 5950: Part 1 1985, Volume 1 Section properties and

member capacities. The Steel Construction Institute, 1985.(4) Introduction to Steelwork Design to BS 5950: Part 1,

The Steel Construction Institute, 1988.

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Job No. Sheet 1 of 14 Rev.

Job title:

Design of members subjected to bending and torsion

Worked Example. Flexural member

Made by RSP DateJan. 2000

Structural Steel

Design Project

CALCULATION SHEET Checked by RN Date Jan. 2000

Example 1

The beam shown below is unrestrained along its length. An eccentric load is appliedto the bottom flange at the centre of the span in such a way that it does not provide

any lateral restraint to the member.

The end conditions are assumed to be simply supported for bending and fixed againsttorsion but free for warping. For the factored loads shown, check the adequacy of the

trial section.

Replace the actual loading by an equivalent arrangement, comprising a vertical loadapplied through the shear centre and a torsional moment as shown below.

= 4000 mm

W = 100 kN

B

W = 100 kN

e = 75 mm

Stiffener to

prevent flange

and web

buckling

W = 100 kN

e = 75 mm

2000mm

=

negative angle of

twist due to Tq

W

Tq = W.e

W

e

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Job No. Sheet 2 of14 Rev.

Job title:



Made by RSP Date Jan. 2000

Structural Steel

Design Project


Loadings due to plane bending and

torsion are shown below.

Loading (Note: These are factored loads and are not to be multiplied by f)

Point load, W = 100 kN

Distributed load (self weight), w = 1 kN/m (say)Eccentricity, e = 75 mm

Bending effects ( at U.L.S)

Moment at B, MxB = 102 kNm

Shear at A, FvA = 52 kN

Shear at B, FvB = 50 kN

Torsional effects ( at U.L.S)

Torsional moment, Tq = W.e

Tq = 100 75 10-3

= 7.5 kNm

This acts in a negative sense, Tq = -7.5 kNm

Generally wide flange sections are preferable to deal with significant torsion. In thisexample, however, an ISWB section will be tried.

Try ISWB 500 250 @ 95.2 kg/m

Section properties from steel tables.

Depth of section D = 500 mm

Width of section B = 250 mm

z

Tq

(ii) Torsional

W

i Plane

+

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Job title:




Structural Steel

Design Project

CALCULATION SHEET

Checked by RN Date Jan. 2000

Web thickness t = 9.9 mm

Flange thickness T = 14.7 mm

Moment of inertia Ixx = 52291 cm4

Moment of inertia Iyy = 2988 cm4

Radius of gyration ry = 49.6 mm

Elastic modulus Zx = 2092 cm3

Elastic modulus Zy = 239 cm3

Cross sectional area A = 121.2 cm2

Additional properties

Torsional constant, J =

= = 682 103

mm4

Warping constant,

= = 1.761012

mm6

Shear modulus,

=

Torsional bending constant,

=

9.9 mm

B = 250 mm

D = 500 mm

3t2TD2BT1 33

1 33 9.914.7250014.725023

4

hIy2

4

14.7500102988 24

12

EG

2/ mmkN76.9

0.312

102 5

21

JG

Ea

mm2591106821076.9

101.76102 21

33

125

Version II 18 -12


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Job title:




Structural Steel

Design Project


Normalized warping function,

= 30331mm2

Warping statical moment,

=2787 104

mm4

Statical moment for flange, Qf = Af . yf

= ( 120.05 14.7) 242.7

= 428.2 103

mm3

Statical moment for web, Qw = (A/2) yw

yw = = 194.2 mm

Qw = 6061 194.2

= 1166103 mm3

4

25014.7500

4

BhWnwfs

16

14.7250485.3

16

TBhS

2

wms

2

235.39.925014.7

2

235.3235.39.9242.725014.7

Version II 18 -13


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Job title:




Structural Steel

Design Project


Material Properties

Shear modulus, G = 76.9 kN/mm2

Design strength, py = 250 /m = 250 / 1.15 = 217 N/mm2

Check for Combined bending and torsion

(i) Buckling check ( at Ultimate Limit State)

Effective length E = 1.0 L E = 4000 mm

The buckling resistance moment,

where

ME = elastic critical momentMp = plastic moment capacity

= fy.Zp /m =

BS 5950:

Part IApp.B.2

kNm5074

470.6240.1

4

500250

15.1

250 22

1

M

M0.51

fM

M

b

x

m

y

wbyt

b

x

kNm102M1.0M

1.0m

MmM

xBx

xBx

21

2

pEBB

pE

b

MM

MMM

2

M1M ELTpB

Version II 18 -14


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Job No. Sheet 6of 14 Rev.

Job title:




Structural Steel

Design Project


Elastic critical moment,

LT = the equivalent slenderness = nuv

= the minor axis slenderness = E/ ry = 4000 / 49.6 = 80.7

n = 0.86, u = 0.9

v = slenderness factor (according to N and/x)

= 0.5 ( for equal flanged sections)

/ x = 80.7 / 36.6 = 2.2

v = 0.948

LT = nuv

= 0.860.9 0.948 80.7 = 59.2

BS 5950:

Part IApp.B.2.2

BS 5950:Part I

Table 14

BS 5950:

Part 1

App.B.2.5

y2

LT

2p

Ep

EMM

tfcf

cf

IIN

I

36.6310681.6102988

101.7612122

1.132

J.I

A1.132x

21

34

12

21

y

kNm1143

17259.210210583M

2

526

E

Version II 18 -15


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Job No. Sheet 7of14 Rev.

Job title:




Structural Steel

Design Project


The Perry coefficient, LT = b (LT - LO)

Limiting equivalent slenderness,

LT = 0.007 ( 59.2 38.2 ) = 0.15

Myt = Mx .

To calculate

/ a = 4000 / 2591 = 1.54z = , = 0.5

= 0.5 4000 = 2000

/ a = 0.77

BS 5950:

Part 1

App.B.2.3

2

M1M ELTpB

2.83172

1020.4

pE

0.4

21

52

21

y

2

LO

kNm911

2

143110.15075B

kNm114

0751431911911

0751431

MM

MMM

21

2

21

pE2

BB

pEb

Version II 18 -16


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Job title:




Structural Steel

Design Project


Myt = 102 0.023 = 2.36 kNm

w = E . Wnwfs .

To calculate

w = 2 10530331 1.8 10

-8= 109 N / mm

2

Ref. 2.0

App. B

Ref. 2.0

App. B

rads0.023

sinh0.77cosh0.77tanh1.54

sinh0.770.770.51

10681.61076.9

2591107.5

a

zsinh

acosh

atanh

asinh

a

z1

GJ

a.T

33

6

q

2/ mmN9.8910239

102.36

Z

M3

6

y

yt

byt

8

33

6

q

101.8

sinh0.770.77coshtanh1.54

sinh0.77

259110681.61076.9

107.5

a

zsinh

acosh

atanh

asinh

aJG

T

Version II 18 -17


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Job title:




Structural Steel

Design Project


Buckling is O. K

(i) Local "capacity" check

bx + byt + w fy /m

bx = Mx / Zx = 102 106

/ 2092 103

= 48.8 N / mm2

48.8 + 9.9 + 109.2 = 168 N / mm2

< 217 N / mm2

O. K

Strictly the shear stresses due to combined bending and torsion should be checked,

although these will seldom be critical.

Shear stresses due to bending (at Ultimate Limit state)

At support:-

In web,

1

b

x

m

y

wbyt

b

x

M

M0.51

fM

M

160.8

10114

101020.51

1.15250

109.29.9

10114

10102

6

6

6

6

2/

.

.

mmN11.79.91052291

1011661052

tI

QF4

33

x

wVAbw

Version II 18 -18


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Job title:




Structural Steel

Design Project


In flange,

At midspan :-

In web, bw = 11.3 N / mm2

In flange, bf = 2.8 N / mm2

Shear stresses due to torsion ( at Ultimate Limit state )

Stress due to pure torsion, t = G.t.

Stress due to warping,

To calculate and

At = 0.5,

Ref. 2.0App.B

2/

.

.

mmN2.9

14.71052291

10428.21052

TI

QF

4

33

x

fVA

bf

t

SE wmsw

..

a

zcosh

acosh

atanh

asinh

1JG

Tq

a

zcosh

acosh

atanh

asinh

aJG

T

2

q

0.913a

tanh1.313,a

cosh,0.851a

sinh

77.02591

40000.5

a

Version II 18 -19


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Job title:




Structural Steel

Design Project


At support, z = 0

At support

Stresses due to pure torsion.

In web, tw = G.t.

tw = 76.9 1039.9 (-1.710

-5)

= - 12.95 N / mm2

In flange, tf = G. T .

tf = 76.9 103 14.7 (-1.710-5)

= - 19.22 N / mm2

1.313cosh(0.77)a

zcosh

2000zmidspan,

1.0cosh(0)a

zcosh

At

11

233

6

100.812

11.3130.913

0.851

259110681.61076.9

107.5

5

33

6

101.7

11.3130.913

0.8510.51

10681.61076.9

107.5

Version II 18 -20


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Job title:




Structural Steel

Design Project


Stresses due to warping in flange,

At midspan

= 0

Stresses due to pure torsion,

In web, tw = G.t. = 0

In flange, tf = G.T. = 0

Stresses due to warping in flange,

By inspection the maximum combined shear stresses occur at the support.

2/

..

mmN3.114.7

100.812102787102

T

SE

1145

wf

wmswf

11

233

6

101.06

1.3131.3130.913

0.851

259110681.61076.9

107.5

2/

..

mmN4.0214.7

101.06102787102

T

SE

1145

wf

wmswf

Version II 18 -21


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Job title:




Structural Steel

Design Project


At support

This must be added to the shear stresses due to plane bending.

= bw + vt

= 11.7 - 14.6 = - 26.3 N / mm2( acting downwards)

In the top flange at 1, tf = - 19.2 N / mm2

wf = - 3.1 N / mm2

= bf + vt = - 27.9 N / mm2

( acting left to right)

3

0 1 2

2vt

2tw

b

xwtvt

/mmN614.114

1020.5112.95

mm/N12.95,3atwebIn

M

M0.51

2vt mm/N1.52114

1020.513.119.2

Version II 18 -22


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Job title:




Structural Steel

Design Project


Shear strength, fv = 0.6 fy/m = 0.6250 /1.15 = 130 N / mm2

Since < fv 27.9 < 130 N / mm2

Section is adequate for shear

Referring back to the determination of the maximum angle of twist, in order to

obtain the value at working load it is sufficient to replace the value of torque Tqwith the working load value as is linearly dependent on Tq. Since Tq is due to solely

the imposed point load W, dividing by the appropriate value offwill give :-

Working load value of Tq is

the corresponding value of 0.93

On the assumption that a maximum twist of 2is acceptable at working load, in thisinstance the beam is satisfactory.

kNm4.71.6

7.5

rads0.0160.026

1.6

Version II 18 -23


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Job title:


Worked Example.Flexural member

Made by RSP DateJan 2000

Structural Steel

Design Project

CALCULATION SHEET Checked by RN Date Jan 2000

Example 2

Redesign the member shown in example 1, using a rectangular hollow section.

Try 300 200 8 @ 60.5 kg / m R. H. SSection properties.

Depth of section D = 300 mm

Width of section B = 200 mmWeb thickness t = 8 mm

Flange thickness T = 8 mm

Area of section A = 77.1 cm2

Moment of inertia Ix = 9798 cm4

Radius of gyration ry = 8.23 cm

Elastic modulus Zx = 653 cm3

Elastic modulus Zy = 522 cm3

Plastic modulus Zp = 785 cm3

Additional properties

Torsional constant

Area enclosed by the mean perimeter of the section, Ah = (B - t ) (D - T)

(neglecting the corner radii)

= ( 200 - 8 )(300 - 8)

= 56064mm2

The mean perimeter, h = 2[(B - t) + ( D - T)]

= 2[( 200 - 8) + ( 300 - 8)] = 968 mm

8 mm

8 mmD = 300

200

h

3

AK23

htJ

Version II 18 -24


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Job title:




Structural Steel

Design Project


Torsional constant,

Torsional modulus constant,

Material properties

Shear modulus,

Design strength, py = 250 /m = 250 / 1.15 = 217 N / mm

2

Check for combined bending and torsion

(i) Buckling check

Since slenderness ratio (E/ ry = 4000 / 82.3 = 48.6) is less than the limiting value

given in BS 5950 Part 1, table 38, lateral torsional

buckling need not be considered..

2mm927

968

8560642

h

tA2K h

4mm10104

5606492723

9688

6

3

J

3mm1084089278

10104

tKt

J

C

36

2/9. mmkN76

0.312

102

12

E 5

G

1

M

M0.51

fM

M

b

x

m

y

wbyt

b

x

385

f250350 250275

y

Version II 18 -25


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Job title:




Structural Steel

Design Project


Hence Mb = Mcx

Shear capacity Pv = 0.6 fy /m . Av

Shear area Av =

Pv = 0.6 (250 /1.15) 46.3 10210

-3= 604.3 kN

Since FVB < 0.6 Pv 50 < 363

Mcx = fy. Zp /m 1.2 fy /m. Zx ( for plastic sections)

Mcx = 1.2 (250 /1.15) 65310-3

= 170 kNm

To calculate

The 100 kN eccentric load gives a value of Tq = 100 0.75 = 7.5 kNm

BS 5950:

Part 14.2.5

BS 5950:

Part 1

4.3.7.2table 13

2cm46.377.1200300

300A

BD

D

kNm1021021.0M

1.0m

MmM xB

100 kN

L

T0 = Tq / 2

T0 = Tq / 2

75 mm100 kN

Version II 18 -26


27/29



Job title:




Structural Steel

Design Project


At centre of span, z = / 2 = 2000 mm

Myt = . MxB = 0.001 102 = 0.102 kNm

Warping stresses (w) are insignificant due to the type of section employed.

Check becomes

O. K(ii ) Local capacity check

bx + byt + w fy/m

bx = MxB / Zy

kNm3.752

7.5

2

TT

zJG

T

q

0

0

radians0.001101041076.9

2000103.75

63

6

2/ mmN0.19510522

100.102

Z

M3

6

y

yt

byt

160.701

1020.51

1.15250

0.195

701

102

1M

M0.51

fM

M

b

x

m

y

byt

b

x

Version II 18 -27


28/29



Job title:




Structural Steel

Design Project


196 + 0.195 + 0 = 196.2 < 217 N / mm2

O. KShear stresses due to bending ( at Ultimate Limit state)

Maximum value occurs in the web at the support.

Shear stresses due to torsion ( at Ultimate limit State)

2/ mmN19610522

101023

6

bx

1.

.

tI

QF

x

wVAbw

21

1

AA

yAQw

2

31

/ mmN13.182109798

103951052

cm395A

395AQ

cmA

39510

A

146818422

1508150

y

4

33

bw

1

w

1

3

1

y8

150

200

20 /2

mmN4.5108372

107.5C

T

CT 3

6

qt

Version II 18 -28


29/29


Job No. Sheet 6of 6 Rev.

Job title:




Structural Steel

Design Project


Total shear stress ( at Ultimate Limit State )

= 13.1 + 5.9 = 19 N / mm2

Shear strength pv = 0.6 fy /m = 0.6 250 /1.15 = 130 N /mm2

Since < pv 19 < 130 N / mm2

the section is adequate for shear.

BS 5950:

Part 1

4. 2. 3

2

b

xwtvt

vtbw

mm/N9.5701

1020.5104.5

M

M0.51

beams subject to torsion and bending

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