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Transcript of design of Reinforce concrete Torsion in Beams Slides
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CHAPTER 2
TORSION IN RC BEAMS
Prepared By Mesfin D.
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Torsion when encountered in reinforced concrete members
usually occurs in combination with flexure and transverseshear. Due to this Torsion in its pure form (generally
associated with metal shafts) is rarely encountered in
reinforced concrete.
Equi libr ium tors ion and compatib i li ty tors ion This is associated with twisting moments that are developed
in a structural member to maintain static equilibrium with the
external loads.
are independent of the torsional stiffness of the member. Such torsion must be necessarily considered in design.
INTRODUCTION
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Such type of torsion is induced in beams supporting lateral
overhanging projections, in beams curved in plansubjected to gravity loads, and in beams where the
transverse loads are eccentric with respect to the shear
centre of the cross-section.
fig.1a). Example of
Equilibrium torision
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This is the name given to the type of torsion induced in a
structural member by rotations (twists) applied at one ormore points along the length of the member.
The twisting moments induced are directly dependent on
the torsional stiffness of the member.
These moments are generally statically indeterminate.
Their analysis necessarily involves (rotational)
compatibility conditions; hence the name compatibility
torsion.
Torsions due to compatibility torsion are not necessaryfor equilibrium and may be neglected in ultimate limit
state calculations and Design.
Compatib i li ty Tors ion
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Members subjected to a torsional moment develop shear
stresses.
In general, these tend to increase in magnitude from the
longitudinal axis of the member to its surface.
If the shear stresses are sufficiently large, cracks will
propagate through the member and, if torsion
reinforcement is not provided, the member will collapsesuddenly.
circu lar member sub jected to a torque T
Torsional s tresses
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a). Circular member b). Member in Torsion c . Shear distribution
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shear stress at a given cross-section varies linearly from the
longitudinal axis of the member to a maximum value, max, atthe periphery of the section.
The stress at any distance r from the longitudinal axis of a
circular member is given by:
The maximum shear stress, maxis found by setting r= /2 in
equation (1).
7
34max
16
32/
2
TT
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But for non-circular member, the distribution of shearstress is not straightforward.
rectangular member sub jected to a torque T
Contd
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Contd
9
Unlike in the circular member, the stress distribution in arectangular member is non-linear.
The maximum shear stress occurs at the mid-point of the
longer side.
Shear stress at the corners of the section is zero indicatingthat the corners of the section are not distorted under torsion.
Analytical studies have shown that the maximum shear stress,
max, in rectangular section is given by:
The parameter depends on the relative values of x and y.
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The stress distribution in thin-walled hollow members is
much easier to determine than for solid non-circularmembers.
The shear stress in the walls is reasonably constant and is
given by:
Thin-wal led ho l low member sub jected to a
torque T
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Where t is the thickness of the wall of the member and
Aois the area within a perimeter bounded by the centreline of the wall.
On a given section, the shear stress is maximum where
the thickness of the wall is minimum.
Fai lure of conc rete members w i th tors ion Consider a rectangular member subjected to a torque T
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Consider small elements on each face of the member, as
illustrated in the figure.As for members with applied shear, shear stresses act on
the sides of each element in the directions.
The equivalent principal stresses are inclined at an angle
of 450to the horizontal. In the same way as for shear, the principal tensile stresses
cause the development of cracks inclined at an angle of
450.
However, in the case of torsion, they form a spiral crack allaround the member
Since the shear stresses in members with torsion are
greatest at the surface, these cracks develop inwards from
the surface of the member.
(Contd...)
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For members with no form of reinforcement to preventthe opening of Torsional cracks, failure of the member
will occur almost as soon as the cracking begins.
Therefore, torsional failure of a member without
reinforcement is prevented only if the shear strength of
the concrete exceeds the shear stress due to applied
torsion.
The torsional strength of a concrete member can be
significantly increased by providing suitable torsion
reinforcement across the cracks.
This is usually provided in the form of closedfour-sided
stirrups, in combination with longitudinal bars distributed
around the periphery of the section.
(Contd...)
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This reinforcement controls the propagation of cracks and
ensures that when failure occurs due to yielding of the
reinforcement, it is not sudden. To quantify the behavior of members with such torsional
reinforcement, an equivalent space truss model, similar to
the plane truss model for shear, can be used.
(Contd...)
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This theory, developed by Lampert and Collins(1972),
assumes that solid members can be designed as equivalentthin walled hollow members .
The thickness of the wall, t, is commonly taken as:
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The concrete cracks are inclined at an angle which,
like shear, varies in the range 220< < 680.
The truss dimensions,xoand yo, are measured from
centre to centre of the notional thin walls.
(Contd...)
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Since the horizontal projection of the crack is yo(cot ),
the number of stirrups crossing the crack is yo(cot )/swhere sis the stirrup spacing.
Hence, assuming that reinforcement has yielded, the
total vertical shear force transmitted across the cracks by
the stirrups on one side face is:
Similarly the shear force transferred across the top or
bottom face is:
(Contd...)
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The shear force on the side wall of a thin walled member is
Substituting for from equation (3) to equation (7) gives:
Similarly it can be shown that:
(Contd...)
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Substituting from equation (5) into equation (9)
Thus the stirrup reinforcement required to resist torsion
of Tis:
In addition to the stirrup reinforcement, longitudinal
reinforcement is required to resist torsion.
(Contd...)
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From equilibrium of forces, the compressive force in the
diagonal members, C, must equal V1/sin.
And the force in the longitudinal members is
N = Ccos = V1cot.
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Substituting for V1
The total force in the longitudinal members from all four
joints at a given cross section is:
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cot2cot2cot2 2121 VVVVNTotal
cot2 21 VVf
ANTotals
y
long
cotcot22
200
00
00 yx
yxTy
T
x
TfAs
y
long
cot
00
00
sy
long
fyx
yxTA
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Alternatively, it can be rearranged to give the maximum
torsion carried without leading to yielding of thelongitudinal reinforcement:
Regardless of how much stirrup and longitudinal
reinforcement is provided, the torsion must not be of
such magnitude as to cause crushing of the concrete in
the diagonal struts.
This force is resisted by stresses in the concrete
between the diagonal cracks.
22
cot00
00
yx
fyxAT
sy
long
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The surface area of concrete to which this is applied is,
from Fig. yo(cos ) t, Hence the stress in the struts is:
This must not exceed the compressive strength of
concrete, fck/c, where is the effectiveness factor for
torsion
cossincos
sin
0
1
0
1
ty
V
ty
V
strutinstress
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Substitution for V1 from equation (7) gives:
From which the torsion which could cause crushing of
the concrete struts is Tw, is:
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cckf
tyV
cossin0
1
cckf
tyxT
cossin2
0
0
c
ck
w
ftyxT
cossin200
Members with combined actions
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For members with combined M and T, the longitudinal
reinforcement required for torsion should be provided in
addition to the amount required for moment.
For members with combined T and V, the ultimate
torque, T, and the ultimate shear force, V, should satisfy
the condition:
Members with combined actions
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1
22
ww VV
TT
Design o f members for tors ion
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Design o f members for tors ion
in acco rdance w ith EBCS2
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The torsional resistance of members is calculated
on the basis of an equivalent thin-walled section (i.e.the truss analogy) as describe above.
Wall thickness t=teff=hef A/u the actual wall
thickness
For sections of complex shape, such as T-sections,
the torsional resistance can be calculated by
dividing the section into individual elements of
simple, say rectangular, shape.
The torsional resistance of the section is equal to
the sum of the capacities of the individual elements,
each modeled as an equivalent thin-walled section.
Contd.
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The critical section for torsion is at the face of supports.
Tsd does not exceed the torsional capacity, TRd whereTRd= 0.80fcdAefhef Aef= xoyo
The torsional resistance of concrete Tcshall be taken as:
The torsional resistance of the reinforcement Teffis given
by:
Members with pure torsion
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350mmor8
S
effU
Spacing
29
yk
wf
4.0min,
Members w i th combined
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Members subjected to combined moment and tors ion
The longitudinal reinforcement required for torsion andFlexure should be calculated separately and hence total
area of reinforcement to be provided should be the sum
of the two.
Members sub jected to combined shear and to rs ion
TRd and VRd(TRd= 0.80fcdAefhef and VRd= 0.25fcdbwd),
respectively, multiplied by the following reduction factors
tand v.
Members w i th combined
act ions
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2
1
1
RdSd
RdSd
t
TT
VV
2
1
1
RdSd
RdSd
v
VV
TT
RdVvRdVRdTtRdT 'and'
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Tcand Vc(Tc= 1.2fctdAefhef and Vc= 0.25fctdk1k2bwd),
respectively, must be multiplied by the reduction factorstcand vc.
The calculations for the design of stirrups may be made
separately for torsion and shear and the total stirrup
required for shear and torsion are, of course, additive.
31
2
1
1
cSd
cSd
tc
TT
VV
2
1
1
cSd
cSd
vc
VV
TT
cVvcVTtcT ccc 'and'
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.
.
.
32
S
A
S
A
S
Atvtv
2
torsionandshearforstirrupleg2Total
ydeff
effst
fA
T
S
A
*2
)(
leg1 torsionforStirrup
df
V
S
A
yd
sdv
*
hear2 SforStirrupleg
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THE END