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Transcript of 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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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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Torsion Formula for Circular Section (contd)
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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Torsion Formula for Circular Section (contd)
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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Torsion Formula for Circular Section (contd)
J
Tr
max=
Shear stressdue to torsion
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Steel Structures
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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
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Pure Torsion For Non-Circular Section
Soap Film Analogy (contd)
Tmax
x
y
At themid ofLonger
sideSquare Cross Section
When material behavesin-elastically
Rectangular Cross Section
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Pure Torsion For Non-Circular Section
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
Pure Torsion For Non-Circular Section
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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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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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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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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Torsion in Hollow Tubes (contd)
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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Torsion in Hollow Tubes (contd)
=
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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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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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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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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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
P f D Z hid Ah d Siddi i d D A h S l
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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.
P f D Z hid Ah d Siddi i d D A h S l
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Example:
Locate the Shear Center for the given channel section.
bf
tw
tf
d
2
tbb wf=
ftdh =
Centerline Representation
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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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( )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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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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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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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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Differential Equation for Torsion of I-Shaped Section (contd)
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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Differential Equation for Torsion of I-Shaped Section (contd)
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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Differential Equation for Torsion of I-Shaped Section (contd)
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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q
Differential Equation for Torsion of I-Shaped Section (contd)
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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Steel Structures
q
Differential Equation for Torsion of I-Shaped Section (contd)
Total Solution
Ph += = Total Solution
h = Homogeneous SolutionP = Particular Solution
Homogeneous Equation
023
3
=dz
d
dz
d
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q
Differential Equation for Torsion of I-Shaped Section (contd)
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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Differential Equation for Torsion of I-Shaped Section (contd)
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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Differential Equation for Torsion of I-Shaped Section (contd)
( ) ( )[ ] ( ) ( )[ ] 321 sinhcoshcoshsinh AzzAzzAh +++=
( )( ) ( )( ) 32121 coshsinh AAAzAAzh +++=
( ) ( ) CzBzAh ++= coshsinh (13)
Homogeneous solution
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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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Differential Equation for Torsion of I-Shaped Section (contd)
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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Differential Equation for Torsion of I-Shaped Section (contd)
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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Differential Equation for Torsion of I-Shaped Section (contd)
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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Differential Equation for Torsion of I-Shaped Section (contd)
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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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
Prof. Dr. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
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7/24/2019 M.sc. Steel 03 Prof. Zahid Siddiqi
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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
Prof. Dr. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
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7/24/2019 M.sc. Steel 03 Prof. Zahid Siddiqi
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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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7/24/2019 M.sc. Steel 03 Prof. Zahid Siddiqi
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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
Prof. Dr. Zahid Ahmad Siddiqi and Dr. Azhar Saleem
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7/24/2019 M.sc. Steel 03 Prof. Zahid Siddiqi
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Steel Structures
ii) For
Fn = Fcr =
yy F
E
t
h
F
E07.345.2