Chapter18 Beam Subjected to Torsion & Bending II

BEAMS SUBJECTED TO TORSION & BENDING-II

BEAMS SUBJECTED TO TORSION AND BENDING - II

18

1.0 INTRODUCTION In the previous chapter, the basic theory governing the behaviour of beams subjected to torsion 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 axis no 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 (EΓ), 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 (EΓ) values and these sections will be exhibiting significant warping moment. Hot rolled I sections and H sections would exhibit 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 I beam by torsion is not 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 is doubly symmetric (such as I or H sections) 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 designing for 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


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 torsional moments compared with I or H sections. Unless it is essential to utilise the torsional resistance of an I section, 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 inappropriate member 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 (φ ) over a length of z is given by

)1(JGzTq ⋅

= φ

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 (τw) at a point is given by

)2(t

SE wmsw

φτ′′′

−=

where E = Modulus of elasticity Swms = Warping statical moment at a particular point S chosen. The warping normal stress (σw) due to bending moment in-plane of flanges (bi-moment) is given by σw = - E .Wnwfs . φ'' 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 φ caused by torsion would be amplified by bending moment, inducing additional warping moments and torsional shears. The following analysis was proposed by Nethercot, Salter and Malik in reference (2).

Version II 18 -2


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 /γm). 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(

.. ''⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=

=

=

φσ

σ

σ

nwfsw

y

ytbyt

x

xbx

WE

ZMZM

σbyt is dependent on Myt, which itself is dependent on the major axis moment Mx and the twist φ. Myt = φ Mx (4) Thus the "capacity check" for major axis bending becomes: σbx + σbyt +σw ≤ fy /γm. (5) Methods of evaluating φ, φ′, φ′′ and φ′′′ for various conditions of loading and boundary conditions 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 of σw and σbyt will be amplified. Nethercot, Salter and Malik have suggested a simple "buckling check" along lines similar to BS 5950, part 1

( )( ) )6(

/1

MM0.51

fMM

b

x

my

wbyt

b

x ≤⎥⎦

⎤⎢⎣

⎡+

++

γσσ

where , equivalent uniform moment = mx Mx xM

( ) 212

pEBB

pE

MM

MM

−+ φφand Mb , the buckling resistance moment =

Version II 18 -3


in which ( )

2M1M ELTp

B++

=η

φ MP , the plastic moment capacity = fy . Zp / γm

Zp = the plastic section modulus ME , the elastic critical moment = where λLT is the equivalent slenderness. m

y2LT

2p

fEM

γλ

π

⋅

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: σbx + σbyt +σw + σby ≤ fy/γm (7) Buckling check:

( )( )

yybyt

yyy

b

x

my

wbyt

myy

y

b

x

ZM

MmMwhere

MM0.51

fZfM

MM

/

)8(1//

=

=

≤⎥⎦

⎤⎢⎣

⎡+

+++

σ

γσσ

γ

4.4 Torsional Shear Stress Torsional shear stresses and warping shear stresses should also be amplified in a similar manner:

( ) )9(⎟⎟⎠

⎞⎜⎜⎝

⎛++=

b

xwtvt M

M0.51τττ

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

Version II 18 -4


• the formulae for elastic critical moment ME are too complex for routine use and • there are limitations to their extension in the ultimate range A 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) = ηLT. ME Mb (10) As explained in page 3, Mb is given by,

( ) 212

pEBB

pE

MM

MM

−+ φφ Mb = ( )

2M1M ELTp

B++

=η

φwhere [ As defined above, ME = Elastic critcal moment Mp = fy . Zp / γm ηLT = 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 (λLT ). The design method is explained below. The buckling resistance moment Mb is given by Mb= pb .Zp (11) where pb = bending strength allowing for susceptibility to lateral -torsional buckling. Zp = plastic section modulus. It should be noted that pb = fy for low values of slenderness of beams and the value of pb drops, as the beam becomes longer and the beam slenderness, calculated as given below, increases. This behaviour is analogous to columns. The beam slenderness (λLT) is given by, (12) LT

yLT f

E λπλ ⋅= 2

EM

M pLT =λwhere

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Beam fails by yield

Beam buckling

50 0

100

pbN/mm2

200

300

150 200 250 100 λLT 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 two theoretical 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 below

about 0.4.

LTλ

• Slender beams which fail at moments close to ME, for values of above about 1.2

LTλ

• beams of intermediate slenderness which fail to reach either Mp or ME . In this case 0.4 < < 1.2

LTλ

Beams having short spans usually fail by yielding. So lateral stability does not influence their design. Beams having long spans would fail by lateral buckling and these are termed "slender". For the practical beams which are in the intermediate range without lateral restraint, design must be based on considerations of inelastic buckling. In the absence of instability, eqn. 11 permits that the value of fy can be adopted for the full plastic moment capacity pb for λLT < 0.4 . This corresponds to λLT values of around 37 (for steels having fy= 275 N/mm2) below which the lateral instability is NOT of concern.

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For more slender beams, pb is a function of λLT which is given by ,

u is called the buckling parameter and x, the torsional index. For flanged sections symmetrical about the minor axis, and For flanged sections symmetrical about the major axis and In the above Zp = plastic modulus about the major axis A = cross sectional area of the member

)13(y

LT ruv λ

=λ

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

PMLT =λ

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

41

s

pu⎟⎟

⎜⎜= 2hA

Z4

⎠

⎞

⎝

⎛2

2 γ ( ) 21

JAh0.566x s=

21

41

2

2

⎟⎟

⎜⎜=u

⎠

⎞

⎝

⎛ ⋅

ΓAZI py γ

⎟⎟

⎜⎜=

JI1.132x

⎠

⎞

⎝

⎛ ΓA

y

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

x

y

II

1γ

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( )322

311

32

3121

2

btbt12bbtths

+≅ Γ = torsional warping constant

J = the torsion constant hs = the distance between the shear centres of the flanges t1, 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 of BS5950: Part I

⎟⎟⎠⎝ r⎞⎛

= functionavy

λ⎜⎜ xof ,

(for a preliminary assessment v = 1)

x = D/T providing the above values of u 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 λLT 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 the degree of monosymmetry through the parameter N = Ic / (Ic + It ) . Table 14 of BS5950: Part 1 must now be entered with (λE /ry )/x and N . 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 of αL), the values of φ and its differentials are given by

λ (1-α

Tq

( )⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+−=azsinh

acosh

atanh

asinh

az1

GJaTq λ

λ

λα

α

αφ.

For 0 ≤ z ≤ α λ,

Version II 18 -8


( )⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+−=′azcosh

acosh

atanh

asinh

1GJTq λ

λ

λα

α

αφ

azsinh

acosh

atanh

asinh

aJGTq

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=′′ λλ

λα

α

φ

azcosh

acosh

atanh

asinh

aJGTq

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=′′′ λλ

λα

α

φ 2

Similar equations are available for different loading cases and for different values of αλ. 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 to verify 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 continuous construction: hot rolled sections. BSI, 1985.

(2) Nethercot, D. A., Salter, P. R., and Malik, A. S. Design of Members Subject to Combined 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.

Version II 18 -9


Job No. Sheet 1 of 14 Rev. Job title: Design of members subjected to bending and torsion Worked Example. Flexural member

Made by RSP Date Jan. 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 applied to 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 against torsion 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 load applied through the shear centre and a torsional moment as shown below.

A

λ = 4000 mm

W = 100 kN

B

W = 100 kNe = 75 mm

Stiffener to prevent flange and web buckling

W = 100 kNe = 75 mm

2000mm

=

negative angle of twist due to Tq

φ

W

Tq = W.e

W

e

Version II 18 -10






Loadings due to plane bending and torsion are shown below. Loading (Note: These are factored loads and are not to be multiplied by γf) 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 this example, 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

α

Tq

λ

(ii) Torsional

λ

W

(i) Plane

+

Version II 18 -11





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, = = 1.76 × 1012 mm6

Shear modulus, = Torsional bending constant, =

14 7 mm

9.9 mm

B = 250 mm

D = 500 mm

( )[ ]3t2TD2BT1−3

3+

( )[ ]1 33 9.914.7250014.725023

×−+××

4hI y

2⋅=Γ

( )4

14.7500102988 × 24 −×

( )υ+=

12EG

( )2/ mmkN76.9

0.312102 5

=+

×

21

⎟⎟⎠

⎞⎜⎜⎝

⎛=

JGΓEa

mm2591106821076.9

101.76102 21

33

125=⎟⎟

⎠

⎞⎜⎜⎝

⎛

××××××

Version II 18 -12






Normalized warping function, = 30331 mm2

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 = 1166 × 103 mm3

( )4

25014.75004BhWnwfs

×−=

=

1614.7250485.3

16TBhS

2

wms

××=

=2

235.39.925014.72

+⎦⎣

235.3235.39.9242.725014.7

××

⎥⎤

⎢⎡ ××+××

Version II 18 -13




Structural Steel Design Project

CALCULATION SHEET

Checked by RN Date Jan. 2000 Material Properties Shear modulus, G = 76.9 kN/mm2

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

Check for Combined bending and torsion (i) Buckling check ( at Ultimate Limit State) Effective length λE = 1.0 L λE = 4000 mm The buckling resistance moment, where ME = elastic critical moment Mp = plastic moment capacity = fy.Zp / γm =

BS 5950: Part I App.B.2

kNm5074470.6240.1

4500250

15.1250 22

=⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−

××

( )1

MM0.51fM

M

b

x

m

y

wbyt

b

x ≤⎥⎦

⎤⎢⎣

⎡+

++

γ

σσ

kNm102M1.0M

1.0mMmM

xBx

xBx

=×=∴

=×=

( ) 212

pEBB

pEb

MM

MMM =

−+ φφ

( )2

M1M ELTpB

++=

ηφ

Version II 18 -14





CALCULATION SHEET

Checked by RN Date Jan. 2000 Elastic critical moment, λLT = the equivalent slenderness = nuv λ λ = the minor axis slenderness = λE / ry = 4000 / 49.6 = 80.7 n = 0.86, u = 0.9 v = slenderness factor (according to N and λ/x) = 0.5 ( for equal flanged sections) λ / x = 80.7 / 36.6 = 2.2 v = 0.948 λLT = nuvλ = 0.86 × 0.9 × 0.948 × 80.7 = 59.2

BS 5950: Part I App.B.2.2 BS 5950: Part I Table 14 BS 5950: Part 1 App.B.2.5

y2

LT

2p

Ep

EMM

⋅=

λ

π

tfcf

cf

IIN

+=

I

36.6310681.6102988

101.76121221.132

J.IΓA1.132x

21

34

12

21

y

=⎟⎟⎠

⎞⎜⎜⎝

⎛

×××××

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

kNm114317259.2

102π10583M

2

526

E

=×

××××=

Version II 18 -15





CALCULATION SHEET

Checked by RN Date Jan. 2000 The Perry coefficient, ηLT = αb ( λLT - λLO ) Limiting equivalent slenderness, ηLT = 0.007 ( 59.2 – 38.2 ) = 0.15 Myt = Mx . φ To calculate φ λ / a = 4000 / 2591 = 1.54 z = α λ , α = 0.5 = 0.5 × 4000 = 2000 α λ / a = 0.77

BS 5950: Part 1 App.B.2.3

( )2

M1M ELTpB

++=

ηφ

2.83172

102π0.4

pEπ

0.4

21

52

21

y

2

LO

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ××=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=λ

( )kNm911

2143110.15075

B =×++

=∴ φ

( )

( )kNm114

0751431911911

0751431

MM

MMM

212

21

pE2

BB

pEb

=

×−+

×=

−+

=∴

φφ

Version II 18 -16





CALCULATION SHEET

Checked by RN Date Jan. 2000 Myt = 102 × 0.023 = 2.36 kNm σw = E . Wnwfs .φ′′ To calculate φ′′ σw = 2 × 105 × 30331 × 1.8 × 10-8 = 109 N / mm2

Ref. 2.0 App. B Ref. 2.0 App. B

( )

( )

rads0.023

sinh0.77cosh0.77tanh1.54sinh0.770.770.51

10681.61076.9

2591107.5

azsinh

acosh

atanh

asinh

az1

GJ

a.T

33

6

q

−=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ −+−

×××

××−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+−=λ

λ

λα

α

αφ

2/ mmN9.8910239102.36

ZM

3

6

y

ytbyt =

××

==σ

8

33

6

q

101.8

sinh0.770.77coshtanh1.54sinh0.77

259110681.61076.9

107.5

azsinh

acosh

atanh

asinh

aJG

T

−×=

⎥⎦⎤

⎢⎣⎡ −

××××

×−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=′′λ

λ

λα

α

φ

Version II 18 -17





CALCULATION SHEET

Checked by RN Date Jan. 2000 ∴ Buckling is O. K (i) Local "capacity" check σbx + σbyt + σw ≤ fy / γm σbx = 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

MM0.51fM

M

γ

σσ

( )( ) 160.8

10114

101020.51

1.15250

109.29.9

10114

101026

6

6

6<=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

×

××+

++

×

×

2/

..

mmN11.7

9.910522911011661052

tIQF

4

33

x

wVAbw

=

×××××

==τ

Version II 18 -18





CALCULATION SHEET

Checked by RN Date Jan. 2000 In flange, At midspan :- In web, τbw = 11.3 N / mm2

In flange, τbf = 2.8 N / mm2

Shear stresses due to torsion ( at Ultimate Limit state ) Stress due to pure torsion, τt = G.t.φ′ Stress due to warping, To calculate φ′ and φ′′′ At α = 0.5,

Ref. 2.0 App.B

2/

..

mmN2.9

14.7105229110428.21052

TIQF

4

33

x

fVAbf

=

×××××

==τ

tSE wms

wφτ

′′′−=

..

( )⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+−=′azcosh

acosh

atanh

asinh

1JG

Tq λλ

λα

α

αφ

azcosh

acosh

atanh

asinh

aJG

T2

q

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=′′′ λλ

λα

α

φ

0.913a

tanh1.313,a

cosh,0.851a

sinh

77.02591

40000.5a

===

=×

=

λλλ

λ

αα

α

Version II 18 -19





CALCULATION SHEET

Checked by RN Date Jan. 2000 At support, z = 0 At support Stresses due to pure torsion. In web, τtw = G.t.φ′ τtw = 76.9 × 103× 9.9 × (-1.7 × 10-5 ) = - 12.95 N / mm2

In flange, τtf = G. T . φ′ τtf = 76.9 × 103 × 14.7 × (-1.7 × 10-5) = - 19.22 N / mm2

1.313 cosh(0.77)azcosh

2000 z midspan,

1.0cosh(0)azcosh

==

=

==

At

11

233

6

100.812

11.3130.9130.851

259110681.61076.9107.5

−×=′′′∴

×⎥⎦⎤

⎢⎣⎡ −

×××××−

=′′′

φ

φ

( )

5

33

6

101.7

11.3130.9130.8510.51

10681.61076.9107.5

−×−=

⎥⎦

⎤⎢⎣

⎡×⎥⎦

⎤⎢⎣⎡ −+−

××××−

=′φ

Version II 18 -20





CALCULATION SHEET

Checked by RN Date Jan. 2000 Stresses due to warping in flange, At midspan φ′ = 0 Stresses due to pure torsion, In web, τtw = G.t.φ′ = 0 In flange, τtf = G.T.φ′ = 0 Stresses due to warping in flange, By inspection the maximum combined shear stresses occur at the support.

2/

..

mmN3.114.7

100.812102787102T

SE

1145

wf

wmswf

−=×××××−

=

′′′−=

−

τ

φτ

11

233

6

101.06

1.3131.3130.9130.851

259110681.61076.9107.5

−×=

×⎥⎦⎤

⎢⎣⎡ −

×××××−

=′′′φ

2/

..

mmN4.0214.7

101.06102787102T

SE

1145

wf

wmswf

−=×××××−

=

′′′−=

−

τ

φτ

Version II 18 -21





CALCULATION SHEET

Checked by RN Date Jan. 2000 At support This must be added to the shear stresses due to plane bending. τ = τbw + τvt τ = ± 11.7 - 14.6 = - 26.3 N / mm2( acting downwards) In the top flange at 1, τtf = - 19.2 N / mm2

τ wf = - 3.1 N / mm2

τ = τbf + τvt = - 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

MM

0.51

−=⎟⎠⎞

⎜⎝⎛ ×+−=∴

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

τ

τ

τττ

( ) 2vt mm/N1.52

1141020.513.119.2 −=⎟

⎠⎞

⎜⎝⎛ +−−=∴ τ

Version II 18 -22





CALCULATION SHEET

Checked by RN Date Jan. 2000 Shear strength, fv = 0.6 fy / γm = 0.6 × 250 /1.15 = 130 N / mm2

Since τ < fv 27.9 < 130 N / mm2

Section is adequate for shear Referring back to the determination of the maximum angle of twist φ, in order to obtain the value at working load it is sufficient to replace the value of torque Tq

with the working load value as φ is linearly dependent on Tq. Since Tq is due to solely the imposed point load W, dividing by the appropriate value of γf will give :- ∴ Working load value of Tq is the corresponding value of φ 0.93° On the assumption that a maximum twist of 2° is acceptable at working load, in this instance the beam is satisfactory.

kNm4.71.67.5

=

=== rads0.0160.0261.6

Version II 18 -23


Job No. Sheet 1 of 6 Rev. Job title: Design of members subjected to bending and torsion Worked Example.Flexural member

Made by RSP Date Jan 2000


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. S Section properties. Depth of section D = 300 mm Width of section B = 200 mm Web 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) = 56064 mm2 The mean perimeter, h = 2[(B - t) + ( D - T)] = 2[( 200 - 8) + ( 300 - 8)] = 968 mm

8 mm

8 mmD = 300

200

h

3AK2

3htJ ⋅+=

Version II 18 -24





CALCULATION SHEET

Checked by RN Date Jan 2000 ∴ Torsional constant, Torsional modulus constant, Material properties Shear modulus, Design strength, py = 250 / γm = 250 / 1.15 = 217 N / mm2

Check for combined bending and torsion (i) Buckling check Since slenderness ratio (λE / 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..

2mm927968

8560642h

tA2K h =××

=⋅

=

4mm10104

56064927239688

6

3

×=

××+×

=J

3mm1084089278

10104t

KtJC

36

×=+

×=

+=

( ) ( )2/9. mmkN76

0.312102

12E 5

=+

×=

+=

υG

( )1

MM0.51fM

M

b

x

m

y

wbyt

b

x ≤⎥⎦

⎤⎢⎣

⎡+

++

γ

σσ

⎟⎟⎠

⎞⎜⎜⎝

⎛=×× 385

f250350 250275

y

Version II 18 -25





CALCULATION SHEET

Checked by RN Date Jan 2000 Hence Mb = Mcx

Shear capacity Pv = 0.6 fy / γm . Av Shear area Av = ∴ Pv = 0.6 × (250 /1.15) × 46.3 × 102 × 10-3 = 604.3 kN Since FVB < 0.6 Pv 50 < 363 Mcx = fy. Zp / γm ≤ 1.2 fy / γm. Zx ( for plastic sections) ∴ Mcx = 1.2 × (250 /1.15) × 653× 10-3 = 170 kNm To calculate φ The 100 kN eccentric load gives a value of Tq = 100 × 0.75 = 7.5 kNm

BS 5950: Part 1 4.2.5 BS 5950: Part 1 4.3.7.2 table 13

2cm46.377.1200300

300ABD

D=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

kNm1021021.0M

1.0mMmM xB

=×=

=⋅=

100 kN

L

T0 = Tq / 2

T0 = Tq / 2

75 mm 100 kN

Version II 18 -26





CALCULATION SHEET

Checked by RN Date Jan 2000 At centre of span, z = λ / 2 = 2000 mm Myt = φ . MxB = 0.001 × 102 = 0.102 kNm Warping stresses ( σw ) are insignificant due to the type of section employed. Check becomes ∴ O. K

(ii ) Local capacity check σbx + σbyt + σw ≤ fy /γm σbx = MxB / Zy

kNm3.752

7.52T

T

zJG

T

q ===

⋅=

0

0φ

radians0.001101041076.9

2000103.7563

6=

×××××

=φ

2/ mmN0.19510522100.102

ZM

3

6

y

ytbyt =

××

==σ

160.701

1020.511.15

2500.195

701102

1MM

0.51fM

M

b

x

my

byt

b

x

<=⎥⎦⎤

⎢⎣⎡ ×++

≤⎥⎥⎦

⎤

⎢⎢⎣

⎡++

γ

σ

Version II 18 -27





CALCULATION SHEET

Checked by RN Date Jan 2000 196 + 0.195 + 0 = 196.2 < 217 N / mm2

∴ O. K

Shear 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 =×

×=σ

1..tIQF

x

wVAbw =τ

21

1

AA

yAQw

=

⋅=

2

31

/ mmN13.182109798

103951052

cm395A

395AQ

cmA

39510A

146818422

1508150y

4

33

bw

1w

1

3

1

=×××

×××=

=×=∴

=×××+×××

= −

τ

y8150

200

20 /2

mmN4.5108372

107.5C

TCT

3

6q

t =××

×===τ

Version II 18 -28





CALCULATION SHEET

Checked by RN Date Jan 2000 Total shear stress ( at Ultimate Limit State ) τ = 13.1 + 5.9 = 19 N / mm2

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

Since τ < 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

MM

0.51ττ

=⎟⎠⎞

⎜⎝⎛ ×++=

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

+=

τ

τττ

Version II 18 -29

Documents

Chapter18 Beam Subjected to Torsion & Bending II