ๅฐŠๆ•ฌ็š„ ๅพฎไฟกๆฑ‡็Ž‡๏ผš1ๅ†† โ‰ˆ 0.046239 ๅ…ƒ ๆ”ฏไป˜ๅฎๆฑ‡็Ž‡๏ผš1ๅ†† โ‰ˆ 0.04633ๅ…ƒ [้€€ๅ‡บ็™ปๅฝ•]
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DAYALBAGH
EDUCATIONAL
INSTITUTE
DESIGN OF
REINFORCED
CONCRETE
STRUCTURES
[CEM โ€“ 603]
(Additional Assignment)
TOPIC:
TORSION
REINFORCEMENT
IN R.C.C.
STRUCTURES
PRESENTED BY:
MUSKAN DEURA
B.TECH - III YEARโ€“ CIVIL
ROLL NO. - 1700655
Torsion
โ€ข The torsional moments produced in a beam are of two types:
(i) Primary torsion is the one which could be determined only
by using static equilibrium condition
(ii) Secondary or compatibility torsion is induced by rotation
applied to one or more points along the length of the
member through connected members. Twisting moment
induced here is proportional to the torsional stiffness of
member.
(i) (ii)
Effects of Torsional Moment
1. For this longitudinal reinforcements are provided which helps in reducing
the crack width through dowel action and stirrups crossing the cracks resist
shear due to vertical loads and torsion.
2. Where the torsional resistance or stiffness of members is taken into account
in the analysis , the members shall be designed for torsion.
โ€ข Because of torsion, beam fails in
diagonal tension forming spiral
cracks around the beam.
โ€ข Warping of the section does not
allow a plane section to remain
as plane after twisting.
Design Of Torsion Reinforcement:
Clause 41 of IS 456:2000 provides the provisions for
design of torsional reinforcements.
โ€ข The design rules for torsion are based on equivalent
moment.
โ€ข Types of reinforcements required to resist load:
Steps In Designing :-
1. Critical Section
2. Shear and Torsion
3. Reinforcements In
Members Subjected To
Torsion
oLongitudinal Reinforcement
o Transverse Reinforcement
4. Final designing for placing
of reinforcement in beam
1. Critical Section (Clause 41.2) :
โ€ข The details of this step is provided under the Clause
41.2 of IS 456:2000
โ€ข For design of torsion, section located at a distance less
than โ€˜dโ€™ from the face of support may be designed for
the same torsion as computed at a distance โ€˜dโ€™ (where
โ€˜dโ€™ is the effective depth).
2. Shear and Torsion (Clause 41.3) :
โ€ข The details of this step is provided under the Clause 41.3 of IS
456:2000
โ€ข The combined effect of torsion and shear is considered and is called
equivalent shear, which is calculated as:
๐‘ฝ ๐’† = ๐‘ฝ ๐’– + ๐Ÿ. ๐Ÿ”
๐‘ป ๐’–
๐’ƒ
where, ๐‘‰๐‘’= equivalent shear
๐‘‰๐‘ข= shear
๐‘‡๐‘ข= torsional moment
๐‘= breadth of beam
โœ“ The equivalent nominal shear, ๐‰ ๐’—๐’† =
๐‘ฝ ๐’†
๐’ƒโˆ—๐’…
โœ“ The maximum shear stress,
๐‰ ๐’Ž๐’‚๐’™= from Table-20 under Clause 40
โœ“ ๐‰ ๐’—๐’† should not be greater than ๐‰ ๐’Ž๐’‚๐’™
โœ“ ๐‰ ๐’„ is found from Table-19 under Clause 40 with respect to value of
๐Ÿ๐ŸŽ๐ŸŽ
๐‘จ ๐’”๐’•
๐’ƒโˆ—๐’…
and concrete grade.
โˆžIf (๐‰ ๐’—๐’† < ๐‰ ๐’„),
minimum reinforcement should be provided as per criteria
under Clause 26.5.1.6
๐ด ๐‘ ๐‘ฃ
๐‘ โˆ— ๐‘  ๐‘ฃ
โ‰ฅ
0.4
0.87 โˆ— ๐‘“๐‘ฆ
where, ๐ด ๐‘ ๐‘ฃ=total cross-sectional area of stirrup legs effective in shear
๐‘  ๐‘ฃ=stirrup spacing along the length of the member
๐‘=breadth of beam/web of flanged beam
๐‘“๐‘ฆ=characteristic strength of stirrup reinforcement
โˆžIf (๐‰ ๐’—๐’† > ๐‰ ๐’„),
both longitudinal and transverse reinforcement should be
provided in accordance with Clause 41.4
โˆžIf (๐‰ ๐’—๐’† > ๐‰ ๐’Ž๐’‚๐’™),
redesigning of the reinforcement should be done. Thus, this
case should never happen.
3. Reinforcements In Members
Subjected To Torsion (Clause 41.4)
โ€ข The details of this step is provided under the Clause
41.4.2 of IS 456:2000
โ€ข The longitudinal reinforcement shall be designed to
resist an equivalent bending moment of
๐‘ด ๐’†๐Ÿ = ๐‘ด ๐’– + ๐‘ด ๐’•
where, ๐‘€ ๐‘ข= bending moment of cross section
and, ๐‘ด ๐’• = ๐‘ป ๐’–
๐Ÿ+
๐‘ซ
๐’ƒ
๐Ÿ.๐Ÿ•
where, ๐‘‡๐‘ข= torsional moment
๐ท= overall depth of beam
๐‘= breadth of beam
o LongitudinalReinforcement(Clause41.4.2)
In longitudinal reinforcement:-
โˆžIf (๐‘ด ๐’• > ๐‘ด ๐’–) ,
Reinforcement shall be provided on the flexural compression
face, to resist equivalent moment, given by
๐‘ด ๐’†๐Ÿ = ๐‘ด ๐’• โˆ’ ๐‘ด ๐’–
โˆžIf (๐‘ด ๐’• < ๐‘ด ๐’–) ,
No additional steel reinforcement is required on bending
compression side.
โ€ข The details of this step is provided under the Clause 41.4.3 of IS
456:2000
โ€ข Two-legged closed hoops enclosing the corner longitudinal bars
shall have an area of cross-section (stirrups),
๐ด ๐‘ ๐‘ฃ =
๐‘‡๐‘ข โˆ— ๐‘  ๐‘ฃ
{๐‘1 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ }
+
๐‘‰๐‘ข โˆ— ๐‘  ๐‘ฃ
{2.5 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ }
โ€ข But the total reinforcement should be greater than:
๐ด ๐‘ ๐‘ฃ โ‰ฅ
๐œ ๐‘ฃ๐‘’ โˆ’ ๐œ ๐‘ โˆ— ๐‘ โˆ— ๐‘  ๐‘ฃ
0.87 โˆ— ๐‘“๐‘ฆ
Where,
๐‘1 = center-to-center distance between corner bars in the direction of the
width
๐‘‘1 = center-to-center distance between corner bars in the direction of the
depth
o TransverseReinforcement(Clause41.4.3)
Since,
๐‘1 = ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ โˆ’ ๐‘ก๐‘œ โˆ’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘› ๐‘๐‘œ๐‘Ÿ๐‘›๐‘’๐‘Ÿ ๐‘๐‘Ž๐‘Ÿ๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘Ÿ๐‘’๐‘๐‘ก๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ค๐‘–๐‘‘๐‘กโ„Ž
๐‘1
= ๐‘ โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘ 
โˆ’ 2 โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ
2
And
๐‘‘1 = ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ โˆ’ ๐‘ก๐‘œ โˆ’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘› ๐‘๐‘œ๐‘Ÿ๐‘›๐‘’๐‘Ÿ ๐‘๐‘Ž๐‘Ÿ๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘Ÿ๐‘’๐‘๐‘ก๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘‘๐‘’๐‘๐‘กโ„Ž
๐‘‘1
= ๐ท โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘ 
โˆ’ 2 โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ
2 From figure
๐‘ฅ1 = ๐‘ โ„Ž๐‘œ๐‘Ÿ๐‘ก๐‘’๐‘Ÿ ๐‘‘๐‘–๐‘š๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘
๐‘ฅ1 = ๐‘ โˆ’ 2 โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘
2
And
๐‘ฆ1 = ๐‘™๐‘œ๐‘›๐‘”๐‘’๐‘Ÿ ๐‘‘๐‘–๐‘š๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘
๐‘ฆ1 = ๐ท โˆ’ 2 โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘
2
โ‘Longitudinal Reinforcement:
โ–ช Shall be placed as close as practicable to the corners of
the cross section and in all cases there shall be at least
one longitudinal bar at each corner of the ties.
โ–ช When the cross-sectional dimensions (either b or D) of
the member exceeds 450 mm, additional longitudinal
bars shall be provided along the two faces. The total area
of such reinforcement shall not exceed 0.1% of the web
area and shall be distributed equally on two faces at a
spacing not exceeding 300 mm or web thickness
(whichever is less).
โ–ช Minimum diameter of this bar shall not be less than 10
mm.
4. Final designing for placing of
reinforcement in beam
โ‘Transverse Reinforcement:
โ–ช This reinforcement for torsion shall be rectangular closed
stirrups placed perpendicular to the axis of the member.
โ–ช The spacing of the stirrups shall not exceed the least of
๐’” ๐’— ๐’„๐’‚๐’๐’„๐’–๐’๐’‚๐’•๐’†๐’… , ๐’™ ๐Ÿ ,
๐’™ ๐Ÿ+๐’š ๐Ÿ
๐Ÿ’
, ๐Ÿ‘๐ŸŽ๐ŸŽ ๐’Ž๐’Ž
QUESTION
Determine the reinforcement required of a
beam of b = 400 mm, d = 650 mm, D = 700 mm
and subjected to factored bending moment (๐‘€ ๐‘ข)
= 200 kN-m, factored torsional moment (๐‘‡๐‘ข) =
50 kN-m and factored shear (๐‘‰๐‘ข) = 100 kN. Use
M-20 and Fe 415 for the design.
Given:
โ€ข Factored bending moment (๐‘€ ๐‘ข) = 200 kN-m
โ€ข Factored torsional moment (๐‘‡๐‘ข) = 50 kN-m
โ€ข Factored shear (๐‘‰๐‘ข) = 100 kN
โ€ข M- 20 grade concrete
โ€ข Fe-415 steel
hence, ๐‘“๐‘ฆ=250 N/sq.mm
โ€ข Breadth(b) = 400 mm
โ€ข Depth(D) = 700 mm
โ€ข Effective depth(d) = 650 mm
Solution:
Equivalent shear,
๐‘‰๐‘’ = ๐‘‰๐‘ข + 1.6
๐‘‡๐‘ข
๐‘
= 100 + 1.6
50
0.4
= ๐Ÿ‘๐ŸŽ๐ŸŽ kN
The equivalent nominal shear,
๐œ ๐‘ฃ๐‘’ =
๐‘‰๐‘’
๐‘โˆ—๐‘‘
=
300
0.4โˆ—0.65
= ๐Ÿ. ๐Ÿ๐Ÿ“๐Ÿ’ N/sq.mm
The maximum shear stress(from table-20),
๐œ ๐‘š๐‘Ž๐‘ฅ = ๐Ÿ. ๐Ÿ– N/sq.mm
Since,
๐‰ ๐’—๐’† < ๐‰ ๐’Ž๐’‚๐’™
Hence, the section is OK and does not require redesigning.
1. CheckforDepth
Assuming, 100
๐ด ๐‘ ๐‘ก
๐‘โˆ—๐‘‘
= 50%
๐œ ๐‘(found from Table-19) = 0.48 N/sq.mm
Since, ๐‰ ๐’„ < ๐‰ ๐’—๐’†
So, both longitudinal and transverse reinforcement
should be provided.
2. Checkforshearreinforcement
โ€ข Equivalent Bending Moment,
๐‘€๐‘’1 = ๐‘€ ๐‘ข + ๐‘€๐‘ก
Here,
๐‘ด ๐’• = ๐‘‡๐‘ข
1+
๐ท
๐‘
1.7
= 50
1+
0.7
0.4
1.7
= ๐Ÿ–๐ŸŽ. ๐Ÿ–๐Ÿ– kN-m
Hence,
๐‘€๐‘’1 = ๐‘€ ๐‘ข + ๐‘€๐‘ก = 200 + 80.88 = ๐Ÿ๐Ÿ–๐ŸŽ. ๐Ÿ–๐Ÿ–kN-m
Since (๐‘ด ๐’• < ๐‘ด ๐’–) ,
No additional steel reinforcement is required on bending
compression side.
โ€ข From Table 2 of SP-16,
For
๐‘€ ๐‘ข
๐‘โˆ—๐‘‘2 = ๐Ÿ. ๐Ÿ”๐Ÿ”N/sq.mm
3. LongitudinalReinforcement:
โ€ข By linear interpolation
๐ด ๐‘ ๐‘ก
๐‘โˆ—๐‘‘
= 0.5156
๐ด ๐‘ ๐‘ก =
0.5156โˆ—400โˆ—650
100
= ๐Ÿ๐Ÿ‘๐Ÿ’๐ŸŽ. ๐Ÿ“๐Ÿ” ๐’”๐’’. ๐’Ž๐’Ž
โ€ข If we provide 2-25T and 2-16T = 981 + 402 = 1383 sq.mm
This gives
๐ด ๐‘ ๐‘ก
๐‘โˆ—๐‘‘
= 0.532
for which,๐œ ๐‘= 0.488 N/sq.mm
โ€ข Minimum percentage of tension reinforcement =
0.85
๐‘“๐‘ฆ
โˆ— 100 = 0.205
โ€ข Maximum percentage of tension reinforcement is 4.0.
โ€ข So, 2-25T and 2-16T bars satisfy the requirements
According to dimensions,
We have ๐‘ซ > ๐Ÿ’๐Ÿ“๐ŸŽ๐’Ž๐’Ž
Hence, side face reinforcement should be provided
Let us provide 2-10 mm diameter bars at the mid-
depth of the beam and one on each face
Area of bars = 157 sq.mm
The total area required
=
0.1โˆ—400โˆ—300
100
= 120 sq.mm < 157 sq.mm
Hence, OK
Using twoโ€“legged 10 mm - diameter bars for stirrups
๐ด ๐‘ ๐‘ฃ = 2 โˆ—
๐œ‹โˆ—102
4
= ๐Ÿ๐Ÿ“๐Ÿ•. ๐ŸŽ๐Ÿ– sq.mm
And,
๐‘1 = ๐‘ โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘  โˆ’ แ‰„
แ‰…
2
โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ
2
= ๐Ÿ‘๐ŸŽ๐Ÿ“ mm
๐‘‘1 = ๐ท โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘  โˆ’ แ‰„
แ‰…
2
โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ
2
= ๐Ÿ”๐ŸŽ๐ŸŽ mm
๐‘ฅ1 = ๐‘ โˆ’ 2 โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘
2
= ๐Ÿ‘๐Ÿ’๐ŸŽmm
๐‘ฆ1 = ๐ท โˆ’ 2 โˆ—
๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘
2
= ๐Ÿ”๐Ÿ๐Ÿ–mm
4. TransverseReinforcement:
Also, we know that
๐ด ๐‘ ๐‘ฃ =
๐‘‡๐‘ข โˆ— ๐‘  ๐‘ฃ
{๐‘1 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ }
+
๐‘‰๐‘ข โˆ— ๐‘  ๐‘ฃ
{2.5 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ }
157.08 =
50 โˆ— 106
โˆ— ๐‘  ๐‘ฃ
{305 โˆ— 600 โˆ— 0.87 โˆ— 250 }
+
100 โˆ— 103
โˆ— ๐‘  ๐‘ฃ
{2.5 โˆ— 600 โˆ— 0.87 โˆ— 250 }
๐‘  ๐‘ฃ(๐‘๐‘Ž๐‘™๐‘๐‘ข๐‘™๐‘Ž๐‘ก๐‘’๐‘‘) = ๐Ÿ๐Ÿ”๐ŸŽ ๐’Ž๐’Ž
And,
๐‘  ๐‘ฃ = least of ๐‘  ๐‘ฃ ๐‘๐‘Ž๐‘™๐‘๐‘ข๐‘™๐‘Ž๐‘ก๐‘’๐‘‘ , ๐‘ฅ1 ,
๐‘ฅ1+๐‘ฆ1
4
, 300 ๐‘š๐‘š
๐‘  ๐‘ฃ = least of 160, 340, 242, 300 ๐‘š๐‘š
๐‘  ๐‘ฃ = 160 mm
Area check,
๐ด ๐‘ ๐‘ฃ โˆ—
0.87โˆ—๐‘“๐‘ฆ
๐‘  ๐‘ฃ
โ‰ฅ ๐œ ๐‘ฃ๐‘’ โˆ’ ๐œ ๐‘ โˆ— ๐‘
๐Ÿ‘๐Ÿ‘๐Ÿ—. ๐Ÿ— > ๐Ÿ๐Ÿ”๐Ÿ—. ๐Ÿ”
Hence, OK
โ€ข Final Diagram will look like:
 Design of torsion reinforcement

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Design of torsion reinforcement

  • 3. TOPIC: TORSION REINFORCEMENT IN R.C.C. STRUCTURES PRESENTED BY: MUSKAN DEURA B.TECH - III YEARโ€“ CIVIL ROLL NO. - 1700655
  • 4. Torsion โ€ข The torsional moments produced in a beam are of two types: (i) Primary torsion is the one which could be determined only by using static equilibrium condition (ii) Secondary or compatibility torsion is induced by rotation applied to one or more points along the length of the member through connected members. Twisting moment induced here is proportional to the torsional stiffness of member. (i) (ii)
  • 5. Effects of Torsional Moment 1. For this longitudinal reinforcements are provided which helps in reducing the crack width through dowel action and stirrups crossing the cracks resist shear due to vertical loads and torsion. 2. Where the torsional resistance or stiffness of members is taken into account in the analysis , the members shall be designed for torsion. โ€ข Because of torsion, beam fails in diagonal tension forming spiral cracks around the beam. โ€ข Warping of the section does not allow a plane section to remain as plane after twisting.
  • 6. Design Of Torsion Reinforcement: Clause 41 of IS 456:2000 provides the provisions for design of torsional reinforcements. โ€ข The design rules for torsion are based on equivalent moment. โ€ข Types of reinforcements required to resist load:
  • 7. Steps In Designing :- 1. Critical Section 2. Shear and Torsion 3. Reinforcements In Members Subjected To Torsion oLongitudinal Reinforcement o Transverse Reinforcement 4. Final designing for placing of reinforcement in beam
  • 8. 1. Critical Section (Clause 41.2) : โ€ข The details of this step is provided under the Clause 41.2 of IS 456:2000 โ€ข For design of torsion, section located at a distance less than โ€˜dโ€™ from the face of support may be designed for the same torsion as computed at a distance โ€˜dโ€™ (where โ€˜dโ€™ is the effective depth).
  • 9. 2. Shear and Torsion (Clause 41.3) : โ€ข The details of this step is provided under the Clause 41.3 of IS 456:2000 โ€ข The combined effect of torsion and shear is considered and is called equivalent shear, which is calculated as: ๐‘ฝ ๐’† = ๐‘ฝ ๐’– + ๐Ÿ. ๐Ÿ” ๐‘ป ๐’– ๐’ƒ where, ๐‘‰๐‘’= equivalent shear ๐‘‰๐‘ข= shear ๐‘‡๐‘ข= torsional moment ๐‘= breadth of beam โœ“ The equivalent nominal shear, ๐‰ ๐’—๐’† = ๐‘ฝ ๐’† ๐’ƒโˆ—๐’… โœ“ The maximum shear stress, ๐‰ ๐’Ž๐’‚๐’™= from Table-20 under Clause 40 โœ“ ๐‰ ๐’—๐’† should not be greater than ๐‰ ๐’Ž๐’‚๐’™ โœ“ ๐‰ ๐’„ is found from Table-19 under Clause 40 with respect to value of ๐Ÿ๐ŸŽ๐ŸŽ ๐‘จ ๐’”๐’• ๐’ƒโˆ—๐’… and concrete grade.
  • 10. โˆžIf (๐‰ ๐’—๐’† < ๐‰ ๐’„), minimum reinforcement should be provided as per criteria under Clause 26.5.1.6 ๐ด ๐‘ ๐‘ฃ ๐‘ โˆ— ๐‘  ๐‘ฃ โ‰ฅ 0.4 0.87 โˆ— ๐‘“๐‘ฆ where, ๐ด ๐‘ ๐‘ฃ=total cross-sectional area of stirrup legs effective in shear ๐‘  ๐‘ฃ=stirrup spacing along the length of the member ๐‘=breadth of beam/web of flanged beam ๐‘“๐‘ฆ=characteristic strength of stirrup reinforcement โˆžIf (๐‰ ๐’—๐’† > ๐‰ ๐’„), both longitudinal and transverse reinforcement should be provided in accordance with Clause 41.4 โˆžIf (๐‰ ๐’—๐’† > ๐‰ ๐’Ž๐’‚๐’™), redesigning of the reinforcement should be done. Thus, this case should never happen.
  • 11. 3. Reinforcements In Members Subjected To Torsion (Clause 41.4) โ€ข The details of this step is provided under the Clause 41.4.2 of IS 456:2000 โ€ข The longitudinal reinforcement shall be designed to resist an equivalent bending moment of ๐‘ด ๐’†๐Ÿ = ๐‘ด ๐’– + ๐‘ด ๐’• where, ๐‘€ ๐‘ข= bending moment of cross section and, ๐‘ด ๐’• = ๐‘ป ๐’– ๐Ÿ+ ๐‘ซ ๐’ƒ ๐Ÿ.๐Ÿ• where, ๐‘‡๐‘ข= torsional moment ๐ท= overall depth of beam ๐‘= breadth of beam o LongitudinalReinforcement(Clause41.4.2)
  • 12. In longitudinal reinforcement:- โˆžIf (๐‘ด ๐’• > ๐‘ด ๐’–) , Reinforcement shall be provided on the flexural compression face, to resist equivalent moment, given by ๐‘ด ๐’†๐Ÿ = ๐‘ด ๐’• โˆ’ ๐‘ด ๐’– โˆžIf (๐‘ด ๐’• < ๐‘ด ๐’–) , No additional steel reinforcement is required on bending compression side.
  • 13. โ€ข The details of this step is provided under the Clause 41.4.3 of IS 456:2000 โ€ข Two-legged closed hoops enclosing the corner longitudinal bars shall have an area of cross-section (stirrups), ๐ด ๐‘ ๐‘ฃ = ๐‘‡๐‘ข โˆ— ๐‘  ๐‘ฃ {๐‘1 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ } + ๐‘‰๐‘ข โˆ— ๐‘  ๐‘ฃ {2.5 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ } โ€ข But the total reinforcement should be greater than: ๐ด ๐‘ ๐‘ฃ โ‰ฅ ๐œ ๐‘ฃ๐‘’ โˆ’ ๐œ ๐‘ โˆ— ๐‘ โˆ— ๐‘  ๐‘ฃ 0.87 โˆ— ๐‘“๐‘ฆ Where, ๐‘1 = center-to-center distance between corner bars in the direction of the width ๐‘‘1 = center-to-center distance between corner bars in the direction of the depth o TransverseReinforcement(Clause41.4.3)
  • 14. Since, ๐‘1 = ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ โˆ’ ๐‘ก๐‘œ โˆ’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘› ๐‘๐‘œ๐‘Ÿ๐‘›๐‘’๐‘Ÿ ๐‘๐‘Ž๐‘Ÿ๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘Ÿ๐‘’๐‘๐‘ก๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ค๐‘–๐‘‘๐‘กโ„Ž ๐‘1 = ๐‘ โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘  โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ 2 And ๐‘‘1 = ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ โˆ’ ๐‘ก๐‘œ โˆ’ ๐‘๐‘’๐‘›๐‘ก๐‘Ÿ๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘› ๐‘๐‘œ๐‘Ÿ๐‘›๐‘’๐‘Ÿ ๐‘๐‘Ž๐‘Ÿ๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘Ÿ๐‘’๐‘๐‘ก๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘‘๐‘’๐‘๐‘กโ„Ž ๐‘‘1 = ๐ท โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘  โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ 2 From figure ๐‘ฅ1 = ๐‘ โ„Ž๐‘œ๐‘Ÿ๐‘ก๐‘’๐‘Ÿ ๐‘‘๐‘–๐‘š๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘ ๐‘ฅ1 = ๐‘ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘ 2 And ๐‘ฆ1 = ๐‘™๐‘œ๐‘›๐‘”๐‘’๐‘Ÿ ๐‘‘๐‘–๐‘š๐‘’๐‘›๐‘ ๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘ ๐‘ฆ1 = ๐ท โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘ 2
  • 15. โ‘Longitudinal Reinforcement: โ–ช Shall be placed as close as practicable to the corners of the cross section and in all cases there shall be at least one longitudinal bar at each corner of the ties. โ–ช When the cross-sectional dimensions (either b or D) of the member exceeds 450 mm, additional longitudinal bars shall be provided along the two faces. The total area of such reinforcement shall not exceed 0.1% of the web area and shall be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness (whichever is less). โ–ช Minimum diameter of this bar shall not be less than 10 mm. 4. Final designing for placing of reinforcement in beam
  • 16. โ‘Transverse Reinforcement: โ–ช This reinforcement for torsion shall be rectangular closed stirrups placed perpendicular to the axis of the member. โ–ช The spacing of the stirrups shall not exceed the least of ๐’” ๐’— ๐’„๐’‚๐’๐’„๐’–๐’๐’‚๐’•๐’†๐’… , ๐’™ ๐Ÿ , ๐’™ ๐Ÿ+๐’š ๐Ÿ ๐Ÿ’ , ๐Ÿ‘๐ŸŽ๐ŸŽ ๐’Ž๐’Ž
  • 17.
  • 18. QUESTION Determine the reinforcement required of a beam of b = 400 mm, d = 650 mm, D = 700 mm and subjected to factored bending moment (๐‘€ ๐‘ข) = 200 kN-m, factored torsional moment (๐‘‡๐‘ข) = 50 kN-m and factored shear (๐‘‰๐‘ข) = 100 kN. Use M-20 and Fe 415 for the design.
  • 19. Given: โ€ข Factored bending moment (๐‘€ ๐‘ข) = 200 kN-m โ€ข Factored torsional moment (๐‘‡๐‘ข) = 50 kN-m โ€ข Factored shear (๐‘‰๐‘ข) = 100 kN โ€ข M- 20 grade concrete โ€ข Fe-415 steel hence, ๐‘“๐‘ฆ=250 N/sq.mm โ€ข Breadth(b) = 400 mm โ€ข Depth(D) = 700 mm โ€ข Effective depth(d) = 650 mm
  • 20. Solution: Equivalent shear, ๐‘‰๐‘’ = ๐‘‰๐‘ข + 1.6 ๐‘‡๐‘ข ๐‘ = 100 + 1.6 50 0.4 = ๐Ÿ‘๐ŸŽ๐ŸŽ kN The equivalent nominal shear, ๐œ ๐‘ฃ๐‘’ = ๐‘‰๐‘’ ๐‘โˆ—๐‘‘ = 300 0.4โˆ—0.65 = ๐Ÿ. ๐Ÿ๐Ÿ“๐Ÿ’ N/sq.mm The maximum shear stress(from table-20), ๐œ ๐‘š๐‘Ž๐‘ฅ = ๐Ÿ. ๐Ÿ– N/sq.mm Since, ๐‰ ๐’—๐’† < ๐‰ ๐’Ž๐’‚๐’™ Hence, the section is OK and does not require redesigning. 1. CheckforDepth
  • 21. Assuming, 100 ๐ด ๐‘ ๐‘ก ๐‘โˆ—๐‘‘ = 50% ๐œ ๐‘(found from Table-19) = 0.48 N/sq.mm Since, ๐‰ ๐’„ < ๐‰ ๐’—๐’† So, both longitudinal and transverse reinforcement should be provided. 2. Checkforshearreinforcement
  • 22. โ€ข Equivalent Bending Moment, ๐‘€๐‘’1 = ๐‘€ ๐‘ข + ๐‘€๐‘ก Here, ๐‘ด ๐’• = ๐‘‡๐‘ข 1+ ๐ท ๐‘ 1.7 = 50 1+ 0.7 0.4 1.7 = ๐Ÿ–๐ŸŽ. ๐Ÿ–๐Ÿ– kN-m Hence, ๐‘€๐‘’1 = ๐‘€ ๐‘ข + ๐‘€๐‘ก = 200 + 80.88 = ๐Ÿ๐Ÿ–๐ŸŽ. ๐Ÿ–๐Ÿ–kN-m Since (๐‘ด ๐’• < ๐‘ด ๐’–) , No additional steel reinforcement is required on bending compression side. โ€ข From Table 2 of SP-16, For ๐‘€ ๐‘ข ๐‘โˆ—๐‘‘2 = ๐Ÿ. ๐Ÿ”๐Ÿ”N/sq.mm 3. LongitudinalReinforcement:
  • 23. โ€ข By linear interpolation ๐ด ๐‘ ๐‘ก ๐‘โˆ—๐‘‘ = 0.5156 ๐ด ๐‘ ๐‘ก = 0.5156โˆ—400โˆ—650 100 = ๐Ÿ๐Ÿ‘๐Ÿ’๐ŸŽ. ๐Ÿ“๐Ÿ” ๐’”๐’’. ๐’Ž๐’Ž โ€ข If we provide 2-25T and 2-16T = 981 + 402 = 1383 sq.mm This gives ๐ด ๐‘ ๐‘ก ๐‘โˆ—๐‘‘ = 0.532 for which,๐œ ๐‘= 0.488 N/sq.mm โ€ข Minimum percentage of tension reinforcement = 0.85 ๐‘“๐‘ฆ โˆ— 100 = 0.205 โ€ข Maximum percentage of tension reinforcement is 4.0. โ€ข So, 2-25T and 2-16T bars satisfy the requirements
  • 24. According to dimensions, We have ๐‘ซ > ๐Ÿ’๐Ÿ“๐ŸŽ๐’Ž๐’Ž Hence, side face reinforcement should be provided Let us provide 2-10 mm diameter bars at the mid- depth of the beam and one on each face Area of bars = 157 sq.mm The total area required = 0.1โˆ—400โˆ—300 100 = 120 sq.mm < 157 sq.mm Hence, OK
  • 25. Using twoโ€“legged 10 mm - diameter bars for stirrups ๐ด ๐‘ ๐‘ฃ = 2 โˆ— ๐œ‹โˆ—102 4 = ๐Ÿ๐Ÿ“๐Ÿ•. ๐ŸŽ๐Ÿ– sq.mm And, ๐‘1 = ๐‘ โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘  โˆ’ แ‰„ แ‰… 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ 2 = ๐Ÿ‘๐ŸŽ๐Ÿ“ mm ๐‘‘1 = ๐ท โˆ’ ๐‘๐‘™๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘œ๐‘ฃ๐‘’๐‘Ÿ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘๐‘  โˆ’ แ‰„ แ‰… 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“๐‘™๐‘œ๐‘›๐‘”๐‘–๐‘ก๐‘ข๐‘‘๐‘–๐‘›๐‘Ž๐‘™ ๐‘๐‘Ž๐‘Ÿ 2 = ๐Ÿ”๐ŸŽ๐ŸŽ mm ๐‘ฅ1 = ๐‘ โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘ 2 = ๐Ÿ‘๐Ÿ’๐ŸŽmm ๐‘ฆ1 = ๐ท โˆ’ 2 โˆ— ๐‘‘๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘–๐‘Ÿ๐‘Ÿ๐‘ข๐‘ 2 = ๐Ÿ”๐Ÿ๐Ÿ–mm 4. TransverseReinforcement:
  • 26. Also, we know that ๐ด ๐‘ ๐‘ฃ = ๐‘‡๐‘ข โˆ— ๐‘  ๐‘ฃ {๐‘1 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ } + ๐‘‰๐‘ข โˆ— ๐‘  ๐‘ฃ {2.5 โˆ— ๐‘‘1 โˆ— 0.87 โˆ— ๐‘“๐‘ฆ } 157.08 = 50 โˆ— 106 โˆ— ๐‘  ๐‘ฃ {305 โˆ— 600 โˆ— 0.87 โˆ— 250 } + 100 โˆ— 103 โˆ— ๐‘  ๐‘ฃ {2.5 โˆ— 600 โˆ— 0.87 โˆ— 250 } ๐‘  ๐‘ฃ(๐‘๐‘Ž๐‘™๐‘๐‘ข๐‘™๐‘Ž๐‘ก๐‘’๐‘‘) = ๐Ÿ๐Ÿ”๐ŸŽ ๐’Ž๐’Ž And, ๐‘  ๐‘ฃ = least of ๐‘  ๐‘ฃ ๐‘๐‘Ž๐‘™๐‘๐‘ข๐‘™๐‘Ž๐‘ก๐‘’๐‘‘ , ๐‘ฅ1 , ๐‘ฅ1+๐‘ฆ1 4 , 300 ๐‘š๐‘š ๐‘  ๐‘ฃ = least of 160, 340, 242, 300 ๐‘š๐‘š ๐‘  ๐‘ฃ = 160 mm Area check, ๐ด ๐‘ ๐‘ฃ โˆ— 0.87โˆ—๐‘“๐‘ฆ ๐‘  ๐‘ฃ โ‰ฅ ๐œ ๐‘ฃ๐‘’ โˆ’ ๐œ ๐‘ โˆ— ๐‘ ๐Ÿ‘๐Ÿ‘๐Ÿ—. ๐Ÿ— > ๐Ÿ๐Ÿ”๐Ÿ—. ๐Ÿ” Hence, OK
  • 27. โ€ข Final Diagram will look like:
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