M.sc. Steel 03 Prof. Zahid Siddiqi

7/24/2019 M.sc. Steel 03 Prof. Zahid Siddiqi

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Steel Structures

SE-505

Lecture # 3

Design for Torsion

M.Sc. Structural Engineering


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Steel Structures

Prof. Dr. Zahid Ahmad Siddiqi and Dr. Azhar Saleem

TorqueMoment about longitudinal axis

Corresponding deformation produced is twist ortorsion.

T, Twisting Moment

Torque can be resisted in two different ways

1. Pure Torsion (St. Venant Torsion)

2. Warping Torsion


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Pure TorsionIn this case the various cross-sections along the length of the

member rotate relative to each other causing twist of the member.

Any particular cross section twists as a wholeTypical example is the torque applied on a circular rod.

Warping TorsionThe whole cross-sections do not rotate as a whole

z


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C

C

T

T

Under the Action of Torque

Plane section do not remain plane inwarping torsion


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Torsion Formula for Circular Section(Pure Torsion)

1. Plane section remains plane.

2. Radial lines remain straight.

3. Moment is applied along longitudinal axis.

4. Material remains elastic.


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Torsion Formula for Circular Section (contd)

Ldz

TrT

A

C

B

O

Helix, deformed positionof line AB after twist

= total rotation of any

section w.r.t. the referencepoint. = change of angle per unitlength

= /L for linear increase = d/dz in general= radial distance up to anypoint where stresses are to becalculated.

= shear stress at any point= shear strain at any point


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Shear stress due to pure torsion is always

perpendicular to the redial distance at thatpoint.

r

Reference

pointd

C

B

dz

dz'C'B= dz

d=

dz

d

= =


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dAdT =( ) AdG= G=Q

( ) AGd =

AGd2 =

==A

r ATT Gd2

= A AT dG 2

GJ=TGJ = torsional rigidity

yx IIJ += For circular section

G=Now

GGJT

=

JT =


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Steel Structures



J

Tr

max=

Shear stressdue to torsion


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Steel Structures


Pure Torsion For Non-Circular Section

(Experimental Method)

Soap Film Analogy

AIR

Slope at any point is equal to shearstress at that point

The volume between the bubbleand the original plane (by the

analogy of governing differentialequation) is proportional to thetotal torque resistance (applied).Steeper the slope of tangent at any

point greater will be the shearstress.SFA is more useful for noncircularand irregular section for which

formulas are difficult to derive.


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Steel Structures



Soap Film Analogy (contd)

Tmax

x

y

At themid ofLonger

sideSquare Cross Section

When material behavesin-elastically

Rectangular Cross Section


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Soap Film Analogy (contd)

At one section , twodirectional shear ispresent, GIVING

RESISTINGTORQUE

Soap Film


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C, Torsion constant

depends on b/t ratio.

3bt

Tt

max

=

CTt=

Valid for Rectangular Section only

b

t ( smaller side)

1/3.290.267.246.219.208

5.03.02.01.51.0b/t


By Timoshenko

3

bt 3=

3

btC

3

=

For section consisting ofmore than one rectangular

= 1/3 for practical sectionwith large b/t ratio.


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Steel Structures


Plastic Torsion

Whole the section will yield in torsion, y

Plastic analysis assumes uniform shear intensity allaround the surface and all around the cross section.

y

Plastic torsions can be envisioned in

terms of SAND HEAP ANALOGY


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Steel Structures


Sand Heap Analogy

Put sand on a plate having a shape same as that of cross section(Circular, Rectangular, Irregular)Slope of sandheap is constanteverywhere as y

throughout


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Volume under the sand heap is proportional to the torque.

3max bt

Tt

pp =

p = 0.33 for b/t = 1.0= 0.5 for b/t =

Sand Heap Analogy (contd)


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Torsion in Hollow Tubes

T

AB

DC

r

z

ds

A B

D C

2

1

3

4 VBC

VDC

VDA

VAB

dz

t1

t2

1

2

1=2


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Steel Structures


Torsion in Hollow TubesV = Resultant shear force at a face

1

remains constant throughout the length

dzVAB = 11 t

dzVCD = 24 tCDABz VVF == 0 To maintain equilibrium

2411 tt =For equilibrium of infinitesimal element at corner B, 1 = 2

Similarly, at corner C, 3 = 4


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Torsion in Hollow Tubes (contd)

2312 tt =Shear stress is more in the portion where thickness is less butx tremains constant

= 2411 tt

The product t is referred to as the shear flow,q having units of N/mm. The shear flow remains

constant around the perimeter of the tube.

This term comes from an analogy to water flowing

in a loop of pipes having different diameters,

where the total discharge remains the same.


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CB qq =

(Shear flow)qt =

In general shear flow is same throughout the cross section.

=

P

dsqrT

Torsional shear force acting on ds length of wall = qds

Resisting moment of this force = rqds

Integrating this differential resisting torque around the perimetergives the total resisting torque.


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=

P

dsrqTr

r

ds

=P 2

dsrq2T

oAq2T =

Ao = Area enclosed by shear flow path

oAt2T =

t2A

T

o= For hollow closed tube


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Steel Structures


Shear center is defined as the point in the cross-sectional plane of abeam through which the transverse loads must pass so that the beam

bends without twisting.

ePT =

Shear Center

Pe

S.C.e is from Shear Center

In other words, loads applied through the shear center will cause notorsional stresses to develop.

0)(0

= dsrtn


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Shear Center (contd)r

ds

Closed Thin Walled Section

( )

==

n

0

0dsrtT

Open Thin Walled Section

=

s

0

s

0

xyy2

xyyx

yxtdsIytdsI

III

Vq

I

VQq=

If we put Ixy = 0, we will get (1)

(1) Valid for sections having Ixy = 0

I is about the axis of bending

Magnitude of Shear Flow for Transverse LoadsThrough Shear Center


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Steel Structures


Rules For Plotting Shear Flow Diagram

1. The shear flow in the part of element parallel

to the applied shear is always in a directionopposite to this applied shear.

2. Shear flow due to direct shear occurs in onedirection through thin walls of open sections.

3. At junction of elements, incoming shear flowis equal to outgoing shear flow.

Shear Flow In Thin Walled Open SectionsDue to Applied Shear Force


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Rules For Plotting Shear Flow Diagram (contd)

4. The value of shear flow is zero at free tips ofthe element and more shear flow is generatedas more area is added.

5. Shear flow is assumed to be generated on one side ofthe neutral axis and consumed/absorbed on the otherside.

6. Shear flow generated is proportional to the firstmoment of the area added.


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7. Shear flow increases linearly for the elementsperpendicular to the load and parabolically for

the elements parallel to the load.

8. Shear flow is considered zero for elements which

have insignificant contribution in correspondingI value.


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Steel Structures


Rules For Plotting Shear Flow Diagram (contd)

q1 q2

q3

q3 = q1 + q2

Ix is very small so this portioncan be neglected

x

Applied Load

P f D Z hid Ah d Siddi i d D A h S l


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Steel Structures


General Rules For Locating Shear Center

1. Shear center always lie on axis of symmetry.

2. If two axes of symmetry exist for a section, S.C. will beat the intersection of these two axis.

3. If the centerlines of all the elements of a sectionintersect at a single point this is the shear center.

4. Shear center of Z section is at the centroid.

Shear Center

Shear Center



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Steel Structures


Procedure to Locate Shear Center

1. To find horizontal location (ex) apply vertical load (V)

at ex from reference point.

2. Plot shear flow diagram due to applied load.

3. Find the internal shear force in each element.4. Apply M = 0 at convenient location and find ex

5. Similarly apply horizontal load at a vertical distance

ey from reference point (say centroid) and repeat theabove procedure to calculate ey

6. The distances exand ey locate the shear center.



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Steel Structures


Example:

Locate the Shear Center for the given channel section.

bf

tw

tf

d

2

tbb wf=

ftdh =

Centerline Representation

Prof Dr Zahid Ahmad Siddiqi and Dr Azhar Saleem


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Steel Structures


Solution

I

VQq=

V

ex

A

P

V

By symmetry about z-axis, the shear center must lie athalf the depth. Only horizontal location is to be found.

qA

qP



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Steel Structures


( )2

htbQ f =

Point A

( )2

htb

I

Vq f

x

A =

Point P

42

ht

h

I

Vqq w

x

AP

+=

+= wf

x

P thh

btI

Vq

82

2



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Steel Structures


Solution

bhbtIVV

f

x

f =22

1

Shear force in flange

4

2 htb

IVV f

x

f =

Shear force in web

+= hhtI

Vhhbt

IVV w

x

f

x

w83

22

2

+= 122

32 hthbt

I

V

V wf

x

w

V

ex

P

Vf

Vw

Vf



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Solution

0= PM0

22=

hV

hVeV ffx

V

ex

P

Vf

Vw

Vf

hVeV fx =

= 4

2 htb

I

V

V

h

e

f

xx

x

f

x I

htbe

4

22

= Positive means on theassumed left side.



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Solution

V P

VfVw

Vf

2hey=

For vertical location of shear center.

N.A.

Vw

ey

Applied Torque = Load x Perpendicular distance from S.C.



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Steel Structures


Differential Equation for Torsion of I-Shaped Sections

y

2

hu f ft

fV

fV

2

h

2

h

fb

x

uf = lateral deflection of one of theflanges

= twist angle at the selected section

Vf = Shear force in flange due totorsion. (internal force developed)

is smaller and is in radians, so

(1)

uf



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Steel Structures


Differential Equation for Torsion of I-Shaped Section (contd)

f

ff

EI

M

dz

ud=

2

2

The lateral curvature relationship of one flange alone is:

(2)

Mf = Lateral Bending moment on one flange

If = Moment of inertia of one flange about y-axis of beam

12

3

ff

f

btI =



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dz

dM

Vdz

dMV

f

f==Differentiating (2) and using (3):

f

ff

EI

V

dz

ud =3

3

(4)

(3)

3

3

dz

ud

EIV f

ff =

( )3

32

dz

dhEIV

ff =

3

3

2dz

dhEIV

ff = (5)



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Steel Structures



hVM fw =

Torsion resistance due to warping

hdz

dh

EIM fw = 33

2

3

32

2 dz

dhEIf =

3

3

dz

dECM ww = (6)

2

2hIC fw=

Warping Constant

212

2

3

hbtC ff

w =

22

2hI

C y

w =

4

2hI

C

y

w=



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Steel Structures



Torsion resistance due to Pure torsion

dz

dGJMs

=

dz

dGCMs

=OR

For Circular Section For Non-Circular Section

wsz MMM +=

3

3

dz

dEC

dz

dGCM wz =

(8)

Total Torque Applied



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Steel Structures

q


Dividing by -ECw

w

z

w EC

M

dz

d

EC

GJ

dz

d=

3

3

w

z

EC

M

dz

d

dz

d= 2

3

3

(9)

(10)

wEC

GC=2

where

wEC

GC= (11)

2 = Ratio of pure torsion rigidity to warping torsion rigidity

Non homogeneous differential equation



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q


Total Solution

Ph += = Total Solution

h = Homogeneous SolutionP = Particular Solution

Homogeneous Equation

023

3

=dz

d

dz

d



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Steel Structures

q


Trial Function

mzh Ae =

mzeAmdz

d 33

3

=

A, m are constants. z is independent variable

023

= mzmz

AmeeAm

023 = mmAemz

0AFor non-trivial solution



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023

= mm022 =mm

Possible Solutions: m = 0, m = +, m = -Sum of all solutions is total homogeneous solution

ozz

h eAeAeA 321 ++=

321 AeAeA zz ++=

( ) ( ) xexx =+ coshsinh ( ) ( ) xexx = coshsinhand

We know



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Steel Structures


( ) ( )[ ] ( ) ( )[ ] 321 sinhcoshcoshsinh AzzAzzAh +++=

( )( ) ( )( ) 32121 coshsinh AAAzAAzh +++=

( ) ( ) CzBzAh ++= coshsinh (13)

Homogeneous solution



46/94

Steel StructuresDifferential Equation for Torsion of I-Shaped Section (contd)

Particular solution

ConsiderMz

to be constant or linearly varying along the length

Mz =f(z) [Constant or function of first degree]. p may assumed tobe a polynomial of degree up to 2, as twist due to pure torque is firstintegral of moment.

Uniform torque

Let

( )zfP 1=

Polynomial of second order. One

order higher that applied torque.

e.g. ( ) FEzDzzf ++= 21

(14)



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Steel Structures


Try this particular integral in (10)

( ) ( )( )zf

ECdz

zdf

dz

zfd

w

1 12

3

1

3

= ( )zfMz=As

Polynomial of Ist order

( ) ( )zfECdz

zdf

w

112 =

Boundary conditions1- Torsionally Simply Supported

weld

0= 022

=dzd

0dzd

Flanges can bend laterally

(15)



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Steel StructuresDifferential Equation for Torsion of I-Shaped Section (contd)

2- Torsionally Fixed End

0= 02

2

dz

d0=

dz

d weld

Both Flanges and Web areconnected

The constant of integration will beevaluated for individual cases.

This is equivalent to deflection and moment made equal to zero forsimple support for bending. Change of twist d / dz may have any

value at the end.Flange may displace at the end but web is held at its position.



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After getting the value of constants and full solution for , the

stresses may be evaluated as follows:Pure Torsional Shear Stress

C

Trs=

dz

dGCT=where

dz

dGts= (16)



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Steel Structures


Warping Shear Stress

ff

ff

w

tI

QV=

No stress in the web

( )ff

ff

ff

w tI

bbt

dz

dhEI

= 422 3

3

mag.max.

From (5)

( )3

32

mag.max.

16 dz

dhbE

f

w = (16)



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Normal Warping Stress

f

f

bwI

xMf =

(in the flanges)

( ) 2

2

dz

udEIM f

fmagf = ( )

2

2

dz

d

2

hEIM fmagf =

( ) 22

dzd

hCEM wmagf =

( )f

f

f

bwI

b

dz

dhEI

f 22

2

2

max

= ( ) 22

max4 dz

dhb

Ef f

bw =

(fbw)max is at

the tips offlange



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Steel Structures

DESIGN AND ALLOWABLE

TORSION STRENGTHS

The design and allowable torsion strengths are below:

Design torsional strength in LRFD = tTn

Allowable torsional strength in ASD = Tn / t

Resistance factor for torsion in LRFD = t = 0.9

Safety factor for torsion in ASD = t = 1.67



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Steel Structures

The nominal torsional strength (Tn) according to

the limit states of torsional yielding and torsionalbuckling is:

Tn = FnC

The following nomenclature may be used in the

further discussion:

C = torsion constant

= 2(Bt)(H t) 4.5(4 )t3 forrectangular HSS



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Steel Structures

C = for round HSS

B = overall width of rectangular HSS

H = overall height of HSS

h = clear distance between the flanges

less the inside corner radius on each

side

D = outside diameter of round HSS

L = length of the member

2

)( 2 ttD



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Steel Structures

Fn For Round HSS

Fn = Fcr = larger of and

but the value should not exceed 0.6Fy

4/5

23.1

t

D

D

L

E2/3

60.0

t

D

E

Fn For Rectangular HSS

i) For Fn = Fcr = 0.6FyyF

E

t

h45.2



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ii) For

Fn = Fcr =

yy F

E

t

h

F

E07.345.2

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M.sc. Steel 03 Prof. Zahid Siddiqi