M.sc rc-17 - [Download PDF]

1

Torsion Design

Slides prepared byAzhar

Lecturer of Civil Engineering

2

Torque:- (Twisting Moment)

Moment about longitudinal axis.Corresponding deformation produced is twist or torsion.

J = Polar M.O.IT = Torsional Shearing Stressr = Radial distance from center of shaft

T, Twisting Moment

T

JTr

=Τ

Valid only if no point on bar is stressed beyond proportional limit

3

Assumption : Plane section remains plane

JTr

=Τ

Torsional Shearing Stress Diagram

• Here the emphasis will be on design

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Reasons of Torsion:

1. Eccentric Loading

Resultant

Torsion Member

• Spandrel Beams

• Curved Stairs Slab

• Curved Beam

• Cantilever Slabs

5

T

mt

Torsion at a cantilever slab

T

mt

Torsion at an edge beam

A B

6

Reasons (contd…)

2. Different loadings and deformations of adjacent 2-D frames ( connecting members shall be subjected to torque)

3. If members are meeting at an angle in space then part of bending moment will become torque for other member.

7

Soap Film Analogy• Slope at any point is equal to shear stress at

that point• The volume between the

bubble and the original plane(by the analogy of governingdifferential equation) is proportional to the total torque resistance( applied). Steeper the slope of tangent at any point greater will bethe shear stress.

• SFA is more useful for noncircular and irregular section for which formulas are not available.

AIR

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SFA For Rectangular Section

Square Cross Section

x

y

At the mid of Longer

side

Tmax

When material behaves inelastically

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Shear Flow

Shear Flow = Shear Stress x Thickness= Shear force per unit length

Shear flow is opposite to applied shear

Shear Flow ( Resultant shear at a point)

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Shear Flow (contd...)

At one section , two directional shear flow, ONE CANCELS OTHER

BOX SECTION

At one section only one directional flow. So closed sections are very efficient in resisting torque

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Formula For Maximum Torsion

yxTx

= 3αΤ max

cTx=

Valid for Rectangular Section only (PCC)

x ( smaller side)

y

C, Torsion constant = α x3yα depends on x / y ratio.

1/3.290.267.246.219.208

∞5.03.02.01.51.0Y/x

α

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C For I-Sectionx

x

y

yC = ∑ α x3yPlastic Torsion• Whole the section will yield in torsion, T = Ty

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

Ty

Plastic analysis 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 sand

heap is constant everywhere as T=Ty throughout

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Sand Heap Analogy (contd…)

Volume under the sand heap is proportional to the torque.

yxTx

= 3p

maxα

T

αp = 0.33 for y/x = 1.0= 0.5 for y/x = ∞

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General Cracking Pattern

τ

T

Less inclined Almost Vertical

45o

Inclined cracks due to torsion which spiral around the member

Cracks due to Vu, Mu, and Tu

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General Cracking Pattern…

On one web face the torque shear adds ( & on the other face subtracts) relative to the vertical shear.

Section subjected to Shear

Section subjected to Torsion

• Cracks due to shear are parallel on both sides of the member butt due to torsion cracks are just like spiral around the member.

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Types of Torsion

There are two types of torsion depending upon its sources.

1- Equilibrium Torsion2- Compatibility Torsion

Equilibrium Torsion:Required for equilibrium & can’t be redistributed.

Pa

P x a, Can’t be redistributed,Must be here to maintain equilibrium

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P

Torque at B= P x L, it can’t be redistributedIf AC is not designed for torsion it will fail.

BC

LA

D

Compatibility Torsion

Compatibility means, compatibility of deformation b/w different members meeting together.

Redistribution/Adjustment is possible.

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Increased +veBending Moment

Decreased _ve B.M

Redistribution is allowed by the code

- +

PB

C• If AC is not designed for torque redistribution can be there, initially there will be some torque and if no resistance developed it will be redistributed and +ve BM in span BD will increase

A

D

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DesignACI CODE

Before 1995 1995 & Onwards

Design of shear and torsion was combined

Shear & Torsion design Separate. Reinforcement is combined

in the end.

• For shear design shear strength of concrete is considered.

• For torsion design shear strength of concrete is considered zero. Compressive strength is considered to an extent.

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Reinforcements

Two types of reinforcements are required to resist torque.i) Transverse reinforcement in the form of stirrups

( closed loop)ii) Extra reinforcement in longitudinal directions

specially in corners and around perimeter.

Open stirrups are for Shear not for Torsion

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Design (contd…)

If section is solid consider it hollow. Consider some outer surface to resist torque.

Hollow

t

Space Truss

Analogy

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Space Truss AnalogyShear stresses are considered constant over a finite thickness t around the periphery of member, allowing the beam to be represented by an equivalent tube.Within the walls of the tube torque is resisted by the shear flow q. in the analogy q is treated as constant around the perimeter.

Not in center but close to edge

t

Tt = thickness of assumed / actual hollow section yo

xo

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Space Truss Analogy (contd…)

xo & yo are measured from the center of wall.we are neglecting the internal area and using shear flow calculated from average shear stress, instead of maximum, so these two things are balancing each other. MAKING THE PROBLEM SIMPLER.

T

Aoxo

Ao = Area enclosed by the shear flow path,

Ao = xoyo

t

yo

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Space Truss Analogy (contd…)

T = Average shear stress in the wall thicknessq = shear flow = T x t

Moment about center due to shear flow must be equal to applied torque T, so

o

oo

oooo

oo

A2T

=q

qA2=Tyqx2=T

yqx+yqx=T

2×]2y

×x×q+2x

×y×q[=TAs t×=q T

tq

=T

tA2T

=o

T

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ACI 11.6.1 CommentaryPrior to cracking due to torsion the thickness we

consider is

cp

PA

75.0=t

Area enclosed by the wall center line, Ao

≈

Tensile strength of concrete is considered =( which is generally )

oA cpA32

'fc31

'fc21

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Cracking Torque

Torque at which cracking starts.For pure shear case principal tensile stress is equal to Torsional shear stress.

T

T f1 = T

cp

2cp

cp

cpcp

o

PA

'fc31

=Tcr

PA

43

×A32

×'fc32

=Tcr

t×A×'fc32

=Tcr

'fc31

=Aot2T

=T

Tcr = Torque at which cracking starts

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ACI 11.6.1Neglect torsion effect if

ACI 11.6.2.2Compatibility torque is allowed to be reduced to

cp

2cp

PA

'fc12Φ

<=Tu

75.0=Φ4TcrΦ

<=Tu

cp

2cp

PA

×'fc3Φ Value of moment and shear

in adjoining member must be consider

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ACI 11.6.2.3Unless determined by more exact analysis it shall be permitted to take the Torsional loading from slab as uniformly distributed along the member (beam).

w Tu

LL

WL/2

dDesign Torque

TuL/2

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ACI 11.6.2.4

Face of the support can be considered as critical section if there is some point momentwithin d distance from face of support.

Face of support d

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Space Truss Analogy

N

NN

N

32

Aoh = Area enclosed by the centerline of the outermost closed transverse reinforcement

N = Tensile force in longitudinal bars

Aoh = Shaded area

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fcd = Stress acting over compression diagonalN2 = longitudinal force required for equilibrium

at face 2

Face 2

θ

fcd Ni/2 = N2/2

V2yo

Total Area

Ni/2 = N2/2

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o2

2

A2Ty

=V

y×q=VFor Face - 2

Vi is resolved in Di & Ni

Di

θ

θCosθtSinA2T

=f

θCosθtSinyy×A2

T=f

t×θCosy1

×θSin

V=f

AreaDi

=f

ocd

o

oocd

o

icd

cdVi

θSinVi

=Di

θSin=DiVi

Ni

θθ

y

y cos θArea of this face is (y Cos θ) t

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Face - 2At = Area of one leg of

closed stirrup.n = number of stirrups

intercepted by Torsional crack

θ

Atfyv

V2 yo

sθCoty

=n o2

sFor vertical equilibrium yo Cot θ Tn = Nominal Torsional

moment strength

Ao ≈ 0.85Aoh, or get by exact analysis

θ = 30o to 60o

= 45o better for non-prestressed members

θCot×s

fAA2=Tn

A2yT

=fA×s

θCotyV=fA×n

yvto

o

onyvt

o

2yvt2

ACI 11.6.3.6

36

Face – 2 (contd…)

Di

θ

Vi

Ni

ACI 11.6.3.7Al = Total area of longitudinal steel to resist torsion

1

( )

θ=

×θ

=×

+θ

=

×θ=

θθ=

θ=

2

y

yvh

t

h

2yvt

y

oo

2yvt

i

oyvt

2

o

oyvto2

oo

2

22

Cotff

Ps

AA

PsCotfA

fA

y2x2sCotfA

N

yCotsfA

N

A2Coty

]s

CotfAA2[N

CotyA2Tn

N

CotVN

2

ll

∑

37

θ=

CotfA2T

sA

yvo

ntGet and put in Eq-1

• Smaller θ value we use, lesser shall be the stirrups but more longitudinal reinforcement & vice versa. θ Cot θ

Total Reinforcement (Shear + Torsion)

sA2

sA

sA tvtv +=+

For closed loop stirrup

2 is not here, because we use twice the area of shear reinforcement

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ACI 11.6.5.3Min longitudinal steel

]ff

[Ps

A_

fA

12'fc5

Ay

yvh

t

y

cpmin

lll ×=

y

w

fb

61

sAt

>=

ACI 11.6.5.2

y

w

y

w

tv

fb

31

)ii

fb

'fc483

)i

)s

A( min

+is larger of

Where

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ACI 11.6.6 (Spacing requirement)

Transverse stirrups spacing should not be more thani) Ph / 8ii) 300 mm

Longitudinal bar spacing should not be more than 300 mm.

Dia of longitudinal bar should not be less thani) s / 24ii) 10 mm

•Distributed around the perimeter of closed stirrup.•At least one in each corner.

s = spacing of shear reinforcement

40

ACI 11.6.6.3Torsional reinforcement shall be provided for a distance of at least ( bt + d) beyond the point theoretically required

bt = width of torsion section

Check for the x-sectional dimensions for combined shear and torsion

ACI 11.6.3.1

)'fc32

dbV

()A7.1pT

()db

V(

w

c22

oh

hu2

w

u +Φ<=+

If this equations does not satisfy increase x-sectional dimensions

41

Design Procedure1. Plot SF & BM diagram, design for flexure, do partial detailing.2. Draw factored torque diagram, get torque at different sections ad

also get critical torque.3. Neglect torsion if

4. If compatibility torsion it can be reduced to

5. Check the x-sectional dimensions for combined actions of shear and torque, if not ok increase dimensions.

6. Design for shear and calculate Av/s.7. Compute the torsion transverse reinforcement At/s.8. Calculate total transverse reinforcement

4T

TucrΦ

<=

cp

2cp

PA

×'fc3Φ

sA2

sA

sA tvtv +=+

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Design Procedure (contd…)

9. Calculate extra longitudinal reinforcement

10. Combine with flexural reinforcement and decide placing according to the code requirement.

θ= 2

y

yvh

t Cotff

Ps

AA

ll

]ff

[Ps

A_

fA

12'fc5

Ay

yvh

t

y

cp

minll

l ×=&

If depth of beam is >0.9m provide skin reinforcement.

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ExamplePanel C.L

450

525175

133

d

100.3

65

25

fc’ = 25Mpa

fy = 420Mpa

d = 450mm

Cover up to center of stirrup =45 mm

Cover up to center of stirrup for flange (slab part) = 25mm

150 3000

118 113.6

150

SFD (kN)

Torque D (kN-m) 18

110 Moment Envelop(kN-m)

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Statement : Design part of beam closer to junction and mid span section for combined action of shear, bending and torque.

Solution:M-ve = 133 kN-m

Asmin =

= 675.0 mm2

As = 815 mm2

M +ve = 110 kN-mAs = Asmin = 675.0 mm2

dbf wy

4.1dbf wy

4.1l

25x450x45010x85

'fcbdMu

'fcR

0506.0fy

'fc85.0s

2

6

2==

==

45

b is lesser of :i) bw + 4hf = 450 + 700ii) bw + hw = 450 +350 = 800 mm

So b = 800 mmTu 65 kN-m

b =800 mm

525 hw = 350175

450

cp

2cp

PA

'fc12

4/TcrΦ

=

Acp = 450 x 525 + 175 x 350 = 297500 mm 2

Pcp = ( 800 + 525 ) x 2 = 2650 mm

kNm44.1010

12650

29750025

1275.0

PA

'fc12

4/Tcr 6

2

cp

2cp

=××=Φ

= < Tu

We cannot redistribute because we don't know about the other members. Their design will change if we redistribute.

46

730

Aoh = 210050 mm2

Ph = 2370 mm 125525–25-45= 455Step # 6

Check for the x-sectional dimensions 330

2)A7.1pT

()db

V(LS 2

oh

hu2

w

u +=

450 – 90 = 360Mpa

xxx

xxLS 113.3)

2100507.123701065()

4504501030.100( 2

2

62

3

=+=

For right hand side we need Vc

RSLS

Mpa125.3'fc65

x75.0'fc65

xRS

]'fc32

'fc61

[RS

db'fc61

V wc

<

==Φ=

+Φ=

=

So dimensions are O.K.

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Step # 7 (independent design for shear)

(Vu)max = 100.3 kN

0s

A0 V

2

3.632

56.1262

56.12675.16875.075.168

1000/4504502561

'61

v

s

=

Φ<<Φ

==Φ

==Φ=

=

cuc

wc

VVV

kNVckNxVc

kNVc

xxVc

dbfcV

So

For the actually applied shear force. However, min shear reinforcement is required for combined shear and torsion. To be calculated latter.

All shear is taken by concrete

48

Step # 8 (Transverse reinforcement required for torque)

legmmmmxxxx

xsAt

xfAxxTu

sA

AA

yvoh

t

oho

o

//2578.042021005085.0275.0

1065

85.02

85.045

6

==

Φ=

==θ

Step # 9 (Total Transverse reinforcement)

mmdmms

mms

legmmmmxsx

sA

sA tvtv

2252

4502

120

123

//156.1578.020712

2

==<=

=

=+=

+=+

Area of #10 bar

Check is required because Vu>ΦVc/2

49

Minimum shear + Torsional reinforcement

14.1375.0420450

31min)(

5.28',31)( min

<==+

<=+

stAv

Mpafcfb

sA

yv

wtv

O.K.

Transverse stirrup spacing should be less than:

1. Ph / 8 = 2370 / 8 = 296 mm

2. 300 mm O.K.

50

Step # 10 (Longitudinal Reinforcement)

θ= 2

y

yvh

t Cotff

Ps

AA

ll

22 1370112370578.0 mmxxxCotff

sAA

y

yvh

t ==

= θρ

ll

][_12

'5min

lll

y

yvh

t

y

cp

ff

sA

fAfc

A ρ×=

210434204502370157.0

42029750025

125][_

12'5

min mmxxxff

PsA

fAfc

Ay

yvh

t

y

cp ==×=ll

l -

Replaced by bw/(3fy)Al = 1370 mm2

51

(Longitudinal Reinforcement)

Maximum spacing = 300 mmMinimum dia = 10 mmFor three layers, Al / 3 = 1370 / 3 = 457 mm2

Top Flexural + Torsional Steel = 815 + 457 = 1272 mm2

Bottom Flexural + Torsional Steel = 1/4A+ + 457 = 626 mm2

5 # 19 for top layer

4 # 19 + 1 # 13 for bottom layer

For middle layer = 457 mm2 2 # 19

52

5 # 19

2 # 19# 10 @ 120 mm c / c

4 # 19 Support Section

1 #13

53

Concluded

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