1 DESIGN FOR TORSION LECTURE 2 By Dr. Rashid (MSc Structural Engineering -2012) Reinforced Concrete Structures

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DESIGN FOR TORSION

LECTURE 2 By Dr. Rashid

(MSc Structural Engineering -2012)

Reinforced Concrete Structures

2

Design for TorsionReinforced Concrete Structures

MSc in Structural Engineering

Design before 1995 Design for shear and torsion was combined

Design after 1995 Design for shear, flexure and torsion are carried out separately

and then the transverse steel for shear and torsion are added and also the longitudinal steel of flexure and torsion are combined at the time of detailing.

Concrete shear and compressive strength For shear design, full shear strength of concrete is considered

while no shear strength of concrete is considered in torsion design. However, comp. Strength of concrete is indirectly used as concrete compression diagonals in the space truss analogy.

DESIGN

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Types of Reinforcement Required to resist Torque

Transverse Reinforcement in the form of stirrups (closed loops)Note: open stirrups are for shear not for torsion

Longitudinal reinforcement in addition to steel required for flexure, specially in cornersand around perimeter

DESIGN

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Concrete section for Design The design of a cracked reinforced concrete section is performed

considering it as an equivalent hollow tubular section

This is done because…..

Experimental and theoratical evidence show that, after cracking, the concrete in the central portion of a solid member has little effect on the torsional strength of the member

DESIGN

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Concrete section for Design

Shear flow, q

t

(A solid cross-section of beam subjected to torsion is idealized as a thin walled tube with core concrete neglected)

DESIGN

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Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY

Shear flow, q

t

Shear stresses are considered constant over a finite thickness around the periphery of the member allowing the beam to be presented by an equivalent tube.

Applied torque is resisted by the shear flow, q which is taken as constant around the perimeter

DESIGN

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After cracking the tube is idealized as a hollow truss consisting of closed stirrups, longitudinal bars in the corners, and compression diagonals. The diagonals are idealized as being between the cracks that are at angle , generally taken as 45 degrees for RC.

DESIGN

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Shear flow, q

t

Y0

X0A0

Ao = Xo x Yo , Xo and Yo are measured from the center of the wall

Internal area is neglected and shear flow is calculated from average shear stress instead of maximum.

t q FlowShear

DESIGN

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X0

Y0

Relationship between applied torque and shear flow for a tube section

22

qy2

qxT oo

oo xy

oo yqx2T

oA

T

2q

1

2

3

DESIGN

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X0

Y0

Relationship between applied torque and shear flow for a tube section

tA

T

o2 4

avg. shear stress due to torsion at any point along the perimeter of the tube

gross area enclosed by the shear flow path.

oA

DESIGN

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area enclosed by the outside perimeter of concrete section or gross area of the section

cpA

outside perimeter of the concrete sectioncpPcpA

As per ACI Code

Prior to cracking, approximate values of thickness (t) of hollow section and Ao

cp

cp

P

At 75.0 5

cpo AA3

2 6

DESIGN

Let

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Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Cracking of Concrete

Cracking of concrete occurs when the principal tensile stress of the concrete exceeds the tensile strength of the concrete

DESIGN

Concrete Cracking

Tensile strength of the concrete is appx. = '5.0 fc

ACI Code considers a conservative value of '33.0 fc

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Cracking TorqueThe twisting moment at which cracking starts is cracking torque

and is denoted by Tcr

DESIGN

In pure torsion, the principal tensile stress is equal to the torsional shear stress, hence

'33.0 fc 7

'33.02

fctA

T

o

cr 8

cp

cpcr P

AfcT

2'33.0 9

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According to ACI 11.6.1

Torsional effects may be neglected when the factored torsional moment is less than one-fourth the cracking torque along with the Φ factor

DESIGN

cru TT 4

10

cp

cpu P

AfcT

2

'083.0 11

or

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Critical Section for Design Torque

DESIGN

In non-prestressed members, the design torque is calculated at d distance from the face of the support

Critical Section

d

As per ACI 11.6.2.4, If a concentrated torque acts within d distance from the edge of support, the critical section for design must be taken at the face of the support

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Basic Derivation for Space Truss Analogy

DESIGN

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DESIGN

A space truss consisting of longitudinal bars in the corners, closed stirrups and diagonal concrete compression member between the cracks.

The height of the truss between centers of bars is taken equal to yo and the width between centers is taken equal to xo. The angle of crack is close to 45° but actually it varies and become lesser at high torques.

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DESIGN

Face-1

Face-2

Face-3

Face-4

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DESIGN

The shear force force acting over the wall in cross-section for each face is denoted by Vi.

The tensile force required in the longitudinal bars required per face may be denoted by Ni.

The symbol used for diagonal compressive force required per face is Di and the corresponding diagonal truss is fcd

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DESIGN

X0

Y0

Shear flow =oA

T

2

Shear flow is converted into total shear force per face by multiplying with the length of the face.

oxA

TVV

031 2

, 12

oyA

TVV

042 2

, 13

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DESIGN

By taking the sum of moments produced by the shears on each face , the resisting torque (Tn) is found and is proved that it is equal to the applied torque.

X0

Y0V2

V1

V4

V3

2)(

2)( 4231

oon

xVV

yVVT

Values of V1, V2, V3 and V4 from equations 12 and 13

222

222 o

oo

oo

on

xy

A

Tyx

A

TT

14

15

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DESIGN

X0

Y0V2

V1

V4

V3

ooo

n yxA

TT

17

16

) (As ooon AyxTT

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DESIGN

Compressive Stress, fcd

Considering face 2, the force V2 is resolved into two component; a diagonal compressive force D2 and axial tensile force N2.

V2

D2

N2

sin2

2

VD 18

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DESIGN

Compressive Stress, fcd

The force D2 acts over a width equal to and thickness equal to t and the resulting compressive stress then becomes;

cos

sin/2

ocd yt

Vf

19

cosoycdf

cossin2 tA

Tf

o

ncd 20

Same stresses are developed in all four walls of the tube and, combined with the stress due to direct shear force, must not exceed the crushing strength of concrete.

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Required Transverse Reinforcement

DESIGN

2

cotV

s

yfA oytt

21

s

fAAT ytton

cot2 22

tan2 yto

nt

fA

T

s

A 23

Discussion about Ao Discussion about angle

Minimum amount of transverse reinforcement combined with that of for the shear

yt

w

yt

wtv f

sb

f

sbfcAA 35.0 and '062.0 oflarger )2( min

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Required Longitudinal Reinforcement

DESIGN

2cotyl

yt

f

f

ht

l ps

AA 24

When

Minimum Longitudinal Steel

cp

cpu p

AfcT

2

'083.0

yl

yth

t

y

cpl f

fp

s

A

f

AfcA

'42.0min,

)175.0 than less benot must (yt

wt

f

b

s

Awhere

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1 DESIGN FOR TORSION LECTURE 2 By Dr. Rashid (MSc Structural Engineering -2012) Reinforced Concrete Structures