[PDF] - Lecture 6 Lecture 6 Analysis and Analysis and Design for PDF Document

SLIDE 1

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

1 The Hashem ite University Departm ent of Civil Engineering

Lecture 6 Lecture 6 – – Analysis and Analysis and Design for Torsion Design for Torsion

Dr Hazim Dwairi Dr Hazim Dwairi

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Dr. Hazim Dwairi
Dr. Hazim Dwairi

Torsion in Pla in Concrete Torsion in Pla in Concrete Mem bers Mem bers

Torsion in circular members

Torsion in circular members

Distance Radial : Torque Applied :

p

T I T ρ ρ τ =

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Inertia

f

Moment Polar : Distance Radial :

p

I ρ

SLIDE 2

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

2

Torsion in Pla in Concrete Torsion in Pla in Concrete Mem bers Mem bers

Torsion in rectangular members

Torsion in rectangular members

The largest stress occurs at the middle of the wide face “a” The largest stress occurs at the middle of the wide face “a” – The largest stress occurs at the middle of the wide face a . The largest stress occurs at the middle of the wide face a . – The stress at the corners is zero. The stress at the corners is zero. – Stress distribution at any other location is less than that at the Stress distribution at any other location is less than that at the middle and middle and – greater than zero. greater than zero.

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Torsion in Pla in Concrete Torsion in Pla in Concrete Mem bers Mem bers

Torsion in rectangular members

Torsion in rectangular members

a b T

2 max

α τ =

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

a/b 1.0 1.5 2.0 3.0 5.0 α 0.208 0.219 0.246 0.267 0.290 1/3

∞

SLIDE 3

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

3

Cracking Strength Cracking Strength

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Thin Walled Tube Analogy Thin Walled Tube Analogy

The design for torsion is based on a thin walled

The design for torsion is based on a thin walled tube space truss analogy A beam subjected to tube space truss analogy A beam subjected to tube, space truss analogy. A beam subjected to tube, space truss analogy. A beam subjected to torsion is idealized as a thin torsion is idealized as a thin-

walled tube with the

walled tube with the core concrete cross section in a solid beam is core concrete cross section in a solid beam is neglected. neglected.

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 4

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

4

Thin Walled Tube Analogy Thin Walled Tube Analogy

V1=V =V3, V , V2

2=V

=V4

4

According to thin walled theory: According to thin walled theory: q = V q = V1/x /xo=V =V2

2/y

yo

=V

=V2/x /xo=V =V4/y yo q = shear force/unit length q = shear force/unit length q = shear flow = constant q = shear flow = constant

Reinforced Concrete II Reinforced Concrete II

q = q = τ. τ.t τ = shear stress = shear stress

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Thin Walled Tube Analogy Thin Walled Tube Analogy

Take moment about Take moment about centroid centroid: : T (V T (V V ) /2 (V /2 (V V ) /2 /2 T = (V T = (V1+V +V3

3)yo

/2 + (V

/2 + (V2+V +V4)x )xo/2 /2 T = 2V T = 2V1yo

/2 + 2V

/2 + 2V2xo/2 /2 Recall: V Recall: V1

1 =

= q.x q.xo

& V

& V2

2 =

= q.y q.yo T = 2V T = 2V1xo

yo
/2 + 2V

/2 + 2V2yoxo/2 /2 T = 2qA T = 2qAo

Reinforced Concrete II Reinforced Concrete II

q q

τ = T/(2A

= T/(2Ao

t)

t)

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 5

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

5

Threshold Torsion Threshold Torsion

Torques that do not exceed approximately one

Torques that do not exceed approximately one-

quarter of the cracking torque

quarter of the cracking torque T will not cause a will not cause a quarter of the cracking torque quarter of the cracking torque Tcr

cr will not cause a

will not cause a structurally significant reduction in either the structurally significant reduction in either the flexural or shear strength and can be ignored. flexural or shear strength and can be ignored.

Cracking is assumed to occur when the principal

Cracking is assumed to occur when the principal tensile stress reaches . In a tensile stress reaches . In a nonprestressed nonprestressed beam loaded with torsion alone the principal beam loaded with torsion alone the principal

'

33 .

c

f

Reinforced Concrete II Reinforced Concrete II

beam loaded with torsion alone, the principal beam loaded with torsion alone, the principal tensile stress is equal to the tensile stress is equal to the torsional torsional shear stress, shear stress, τ τ . Recall that: . Recall that:

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

t A T

2

= τ

Threshold Torsion Threshold Torsion

R11.6.1

ACI

to According A and

f

values ng Substituti ) ( 2 ; : where 4 3 ; 3 2

2

cp

cp cp cp cp

A

t A y x p xy A p A t A A ⎟ ⎞ ⎜ ⎛ + = = = =

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

4 en Torsion wh Neglect 33 .

' cr u cp cp c cr

T T p A f T φ ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ∴

SLIDE 6

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

6

Torsion in Torsion in Reinforced Reinforced Concrete Concrete Mem bers Mem bers Mem bers Mem bers

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

stirrup

f

CL to CL from distance &

y

x

Reinforcem ent Requirem ent for Reinforcem ent Requirem ent for RC Mem bers in Torsion RC Mem bers in Torsion

Reinforcement is determined using space truss

Reinforcement is determined using space truss analogy analogy analogy. analogy.

In space truss analogy, the concrete compression

In space truss analogy, the concrete compression diagonals (struts), vertical stirrups in tension (ties), diagonals (struts), vertical stirrups in tension (ties), and longitudinal reinforcement (tension chords) and longitudinal reinforcement (tension chords) act together as shown in figure on the next slide. act together as shown in figure on the next slide.

The analogy derives that

The analogy derives that torsional torsional shear stress will shear stress will

Reinforced Concrete II Reinforced Concrete II

The analogy derives that

The analogy derives that torsional torsional shear stress will shear stress will be resisted by the vertical stirrups as well as by be resisted by the vertical stirrups as well as by the longitudinal steel the longitudinal steel

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 7

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

7

Reinforcem ent Requirem ent for Reinforcem ent Requirem ent for RC Mem bers in Torsion RC Mem bers in Torsion

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Vertical Stirrup Reinforcem ent Vertical Stirrup Reinforcem ent

θ τ cot 2 2 : Recall

4 yv t

yv

t

f

A s y f nA y A T qy V A T t q = = = = = =

crossing stirrups

f

number = n

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

θ θ θ cot 2 cot 2 2 cot 2

h

yv t yv t

n
yv

t

A

s f A s f A y x T y A f A s y T s A = = =

ent reinforcem shear

f

strength Yield stirrup leg

ne
f

Area crack diagonal g p = =

yv t

f A

SLIDE 8

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

8

Vertical Stirrup Reinforcem ent Vertical Stirrup Reinforcem ent

For

For No failure No failure, i.e., torsion capacity greater than , i.e., torsion capacity greater than

f equal to torsion demand:
f equal to torsion demand:
f equal to torsion demand:
f equal to torsion demand:
For torsion capacity equal to or greater than

For torsion capacity equal to or greater than torsion demand, we have at the limit state: torsion demand, we have at the limit state:

θ φ cot 2

h

yv t u

A f A T = 75 . ; = > φ φ

u n

T T

Reinforced Concrete II Reinforced Concrete II

Therefore steel area in one leg stirrup is:

Therefore steel area in one leg stirrup is:

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

φ

h

u

s θ φ cot 2

h

yv u t

A f s T A =

Vertical Stirrup Reinforcem ent Vertical Stirrup Reinforcem ent

ACI 6.3.6 assumes

ACI 6.3.6 assumes θ θ = 45 = 45o for for nonprestressed nonprestressed members and replaces members and replaces A by by A where: where: members and replaces members and replaces Aoh

h by

by Ao where: where: Ao = 0.85A = 0.85Aoh

h.
Therefore:

Therefore:

yv

u t

A f s T A φ 2 =

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

yv

f φ

SLIDE 9

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

9

Longitudinal Steel Reinforcem ent Longitudinal Steel Reinforcem ent

Diagonal Compression Struts Diagonal Compression Struts

2 4 4

cot cot : Torsion to due Force Axial f y s A V N

yv

t

= = Δ θ θ

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

3 1 2 1 1 2 4

cot cot : Similarly N N f x s A V N N N

yv

t

Δ = Δ = = Δ Δ = Δ θ θ

Longitudinal Steel Reinforcem ent Longitudinal Steel Reinforcem ent

Total axial force is: Total axial force is:

4 3 2 1 total

N N N N N Δ + Δ + Δ + Δ = θ θ θ θ

2 2 2 2

cot cot ) ( 2 cot 2 cot 2

yv h t yv t

total

yv

t

yv

t

total

f p s A f s A y x N f x s A f y s A N = + = + =

Longitudinal Steel Force: Longitudinal Steel Force:

θ

2

t

f A f A N stirrup

f

permimeter =

h

p

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

θ

2

cot

yv h t y l total

f p s f A N = = ent reinforcem al longitudin

f

strength Yield sion resist tor ent to reinforcem al longitudin

f

area Total = =

y l

f A

θ

2

cot

h y yv t l

p f f s A A =

SLIDE 10

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

10

Longitudinal Steel Reinforcem ent Longitudinal Steel Reinforcem ent

h yl yv t l

p f f s A A θ : Recall cot2 =

yv

yv

u l t

yv

u t

f A f s T A A A A f s T A φ φ φ 2 45 for and , in Substitute 2 : Recall

=

=

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

h y yv

yv

l

p f f s f A φ =

y

h

u l

f A p T A φ 2 =

Com bined Shear and Torsion Com bined Shear and Torsion

t

w v

A t A A t A T d b V / 85 ti k d F ) 2 /( : Stress Torsional / : Stress Shear = = τ τ

h

h
h
p

A t A A / , 85 . : section cracked For = =

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

2

7 . 1

h

h w t v

A Tp d b V + = + = τ τ τ

2 2 2

) 7 . 1 ( ) (

h

h w

A Tp d b V + = τ

SLIDE 11

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

11

Equilibrium and Com patibility Equilibrium and Com patibility Torsion Torsion

Equilibrium Torsion

Equilibrium Torsion: : torsion moment is required for

torsion moment is required for equilibrium of the structure (cannot be reduced by equilibrium of the structure (cannot be reduced by equilibrium of the structure (cannot be reduced by equilibrium of the structure (cannot be reduced by internal forces redistribution). internal forces redistribution).

Compatibility Torsion

Compatibility Torsion: : torsional

torsional moment results moment results from the compatibility of deformations between members from the compatibility of deformations between members meeting at a joint ( meeting at a joint (torsional torsional moment can be reduced by moment can be reduced by redistribution of internal forces after cracking if the redistribution of internal forces after cracking if the torsion arises from the member twisting to maintain torsion arises from the member twisting to maintain

Reinforced Concrete II Reinforced Concrete II

torsion arises from the member twisting to maintain torsion arises from the member twisting to maintain compatibility of deformations). The reduction in compatibility of deformations). The reduction in Tu is of is of the magnitude: the magnitude:

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Equilibrium Torsion Equilibrium Torsion

The torsion in the beams The torsion in the beams in Figs (a) & (b) must be in Figs (a) & (b) must be in Figs (a) & (b) must be in Figs (a) & (b) must be resisted by the structural resisted by the structural system if the beam is to system if the beam is to remain in equilibrium. If remain in equilibrium. If the applied torsion is not the applied torsion is not resisted, the beam will resisted, the beam will rotate about its axis until rotate about its axis until

Reinforced Concrete II Reinforced Concrete II

the structure collapses. the structure collapses.

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 12

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

12

Equilibrium Torsion Equilibrium Torsion

A B

Reinforced Concrete II Reinforced Concrete II

The canopy applies a The canopy applies a torsional torsional moment to the beam moment to the beam A-

B

B. . For this structure to stand, the beam must resist the For this structure to stand, the beam must resist the torsional torsional moment, and the columns must resist the moment, and the columns must resist the resulting bending moments. resulting bending moments.

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Com patibility Torsion Com patibility Torsion

If joint If joint A is monolithically built is monolithically built with cross beam with cross beam C-D then then with cross beam with cross beam C D then then beam beam A can only develop end can only develop end slope at A if beam slope at A if beam C-

D

D twists twists about its own axis. If ends about its own axis. If ends C and and D are restrained against are restrained against rotation, a rotation, a torsional torsional moment moment T will be applied to beam will be applied to beam C-

D
D. if

. if

Reinforced Concrete II Reinforced Concrete II

C and and D are free to rotate are free to rotate about axis about axis C-

D

D then then T would would be zero. be zero.

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 13

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

13

Com patibility Torsion Com patibility Torsion

A B

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Hinge

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

Equilibrium Torsion Equilibrium Torsion: : design for full design for full Tu Compatibility Torsion Compatibility Torsion: : reduce reduce Tu to the following to the following

Nonprestressed

Nonprestressed member member without without axial force: axial force:

Reinforced Concrete II Reinforced Concrete II

Nonprestressed

Nonprestressed member member with with an axial force: an axial force:

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 14

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

14

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

It shall be permitted to It shall be permitted to neglect torsion neglect torsion effects if effects if the factored the factored torsional torsional moment moment T is less than: is less than: the factored the factored torsional torsional moment moment Tu is less than: is less than:

Nonprestressed

Nonprestressed members members without without axial force: axial force:

Reinforced Concrete II Reinforced Concrete II

Nonprestressed

Nonprestressed members members with with an axial force: an axial force:

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

The cross

The cross-

sectional dimensions shall be such

sectional dimensions shall be such that: that: that: that:

τmax

Reinforced Concrete II Reinforced Concrete II

If

If NOT NOT, , increase section dimensions increase section dimensions.

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 15

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

15

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

Reinforcement for torsion

Reinforcement for torsion R ll ACI E (11 R ll ACI E (11 21) 21) Recall ACI Eq. (11 Recall ACI Eq. (11-21) 21)

Combined shear and torsion reinforcement

Combined shear and torsion reinforcement

(for closed stirrup) (for closed stirrup)

yv

u t

A f T s A 60 3 ; cot 2 ≤ ≤ = θ θ φ

Reinforced Concrete II Reinforced Concrete II

(for closed stirrup) (for closed stirrup)

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

s A s A s A

t v t v

2 + =

+

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

Maximum spacing of torsion reinforcement

Maximum spacing of torsion reinforcement ⎧

Spacing is limited to ensure the development of

Spacing is limited to ensure the development of the ultimate the ultimate torsional torsional strength of the beam, to strength of the beam, to ⎪ ⎩ ⎪ ⎨ ⎧ = mm p s

h

300 8

f

smaller

max

Reinforced Concrete II Reinforced Concrete II

prevent excessive loss of prevent excessive loss of torsional torsional stiffness after stiffness after cracking, and to control crack widths. cracking, and to control crack widths.

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 16

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

16

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

Minimum area of closed stirrups

Minimum area of closed stirrups

⎧ s b 35

Minimum area of longitudinal

Minimum area of longitudinal torsional torsional reinforcement reinforcement

⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = +

yv w c yv w t v

f s b f f s b A A

'

062 . 35 .

f

larger ) 2 (

1- shall be distributed around the perimeter f th l d ti ith i

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

yv w t y vt h t y cp c l

f b s A f f p s A f A f A 175 . : where 42 .

' min ,

≥ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =

f the closed stirrups with a maximum

spacing of 300 mm. 2- The longitudinal bars shall be inside the stirrups. 3- shall have a diameter at least 0.042 times the stirrup spacing, but not less than φ10

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

Torsion reinforcement shall be provided for a

Torsion reinforcement shall be provided for a distance of at least ( distance of at least (b + d) beyond the point + d) beyond the point distance of at least ( distance of at least (bw

w + d) beyond the point

+ d) beyond the point where where T Tu

u is less than

is less than ΦT ΦTcr

cr/4

4. .

Stirrup Detailing

Stirrup Detailing

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 17

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

17

ACI Requirem ents for Torsion ACI Requirem ents for Torsion Design Design

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

A cantilever beam supports its own weight plus a A cantilever beam supports its own weight plus a concentrated load. The beam is concentrated load. The beam is 1400mm 1400mm long, long, g g and the concentrated load acts at and the concentrated load acts at 150mm 150mm from the from the end of the beam and end of the beam and 150mm 150mm away from the away from the centroidal centroidal axis of the beam. axis of the beam. The The unfactored unfactored concentrated load consists of a concentrated load consists of a 90 90 kN kN dead and dead and 90 90 kN kN live load. The beam also live load. The beam also

Reinforced Concrete II Reinforced Concrete II

supports an supports an unfactored unfactored axial compression dead axial compression dead load of load of 180 180 kN kN. Use normal weight concrete with Use normal weight concrete with f’ f’c = 21 = 21 MPa MPa and and both both fy and and fvy

vy = 420

= 420 MPa MPa. .

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

SLIDE 18

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

18

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

m 80 kN 180 kN 600 mm 350 mm

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University mm

d mm mm and b h mm l h 525 2 / 25 10 50 600 350 600 USE 175 8 1400 8 9.5(a) Table

9.5.2.1

ACI to According

min

≈ − − − = = = = = =

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

( )

LL DL m kN 6 1 2 1 load d concetrate Factored / 3 . 6 25 10 600 350 1.2 weight self beam Factored

6

+ = = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = kN V kN kN LL DL

u

82 . 232 : to leads Analysis Structural 216 180 2 . 1 ) (N load axial Factored 224 80 6 . 1 80 2 . 1 6 . 1 2 . 1 load d concetrate Factored

u

= = × = = × + × = + =

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

kN N m kN T m kN M kN d V

u u u u

216 . 6 . 33 . 17 . 286 51 . 229 525 . 3 . 6 82 . 232 @ = = = = × − =

SLIDE 19

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

19

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

441 1000 ) 60 . 35 . ( 21 1 . 1 . 216 1 . if check

' '

kN A f kN N A f N

g c u g c u

= × × × × = = ≥ n. interactio load axial and bending for designed be shall member Otherwise, design. flexure in neglected be can affect force axial Therefore 1 . ) (

'A

f N f

g c u g c

≥ ⇒

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

. . . 5 . 342 1960mm 25 4 use 1610mm A error and By trial : Flexure for Design

2 2 s

K O M m kN M

u n

> = = = φ φ

600 mm 350 mm 4φ25

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

mm h b A

w cp

000 , 210 600 350 : Torsion and Shear Both for Design

2

= × = = mm y mm x mm h b p

w

cp w cp

490 55 2 600 240 55 2 350 55mm 5 50 stirrup

f

center cover to Assume 900 , 1 600 2 350 2 2 2 , = × − = = × − = = + = = × + × = + =

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

mm y x p mm A A mm y x A

h
h
h

460 , 1 ) ( 2 960 , 99 85 . 600 , 117 490 240

2 2

= + = = = = × = =

SLIDE 20

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

20

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

: beam

f

size for Check

( )

21 66 . 21 17 . 75 . 117600 1460 10 6 . 33 525 350 10 5 . 229 66 . 7 . 1

2 2 6 2 3 ' 2 2 2

+ ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

c w c

h

h u w u

f d b V A p T d b V φ

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

kay

is size Beam 853 . 2 260 . 2 ⇒ ≤

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

N A f T

u cp

1 083 : T torsion critical for Check

2 ' c

+ ⎟ ⎟ ⎞ ⎜ ⎜ ⎛ φ φ φ T m kN T T f A p f T

u c c c g u cp p c c

considered be must Torsion . 640 . 9 28 600 350 33 . 10 216 1 1900 210000 21 083 . 75 . 33 . 1 083 .

3 2 '

⇒ < = × × × + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × = + ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ = φ φ φ φ

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

mm mm f A T s A

yv

u

t

/ 534 . 420 99960 75 . 2 10 6 . 33 2 ent Reinforcem Torsion (a)

2 6

= × × × × = = φ

SLIDE 21

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

21

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

d b f A N V

w c u c

14 1 17 . ent Reinforcem Shear (b)

'

⎟ ⎟ ⎞ ⎜ ⎜ ⎛ + = φ kN V V V kN V f A

c u s c w c g c

3 . 152 7 . 153 75 . 51 . 229 7 . 153 525 350 21 ) 600 350 ( 14 10 216 1 17 . 14

3 3

= − = − = = × × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × + = ⎟ ⎠ ⎜ ⎝ φ φ φ

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

mm mm d f V s A

yv s v

/ 691 . 525 420 10 3 . 152

2 3

= × × = =

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

/ 759 . 1 2 stirrups select and ent reinforcem shear Add (c)

2

= + =

+

mm mm s A s A s A

t v t v

226 : stirrups 12 Assume 237 . 420 350 21 062 . 292 . 420 350 35 .

f

larger stirrups minimum Check

2

φ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = × = × ≥

+

mm A s A

t v

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Stirrups Closed 12@125mm USE OK 5 . 182 8 / 1460 6 . 128 759 . 1 / 226 226 : stirrups 12 Assume

max

φ φ ⇒ = = ⇒ = = =

+

mm s mm s mm A

t v

SLIDE 22

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi

22

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

( )( )

780 420 420 1460 534 . n for torsio ent reinforcem al longitudin Design (d)

2

= = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = mm f f p s A A

vt h t l

( )( )

3 . 182 420 420 1460 534 . 420 210000 21 42 . 42 . 420

2 2 min , ' min ,

= − × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎠ ⎝ mm A f f p s A f A f A f s

l y vt h t y cp c l y

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

mm 10

r

5.25mm 125 0.042 diameter bar min. bottom at 2 and middle at 2 at top, 2 : bars 6 use 300mm spacing Max. 780

2

= × = ⇒ = = ⇒ mm Al

Exam ple: Equilibrium Torsion Exam ple: Equilibrium Torsion

mm 64 1 616

780

ent reinforcem flexural to add beam the

f

half bottom in mm 16 6 14 4 Use

2 2

φ = = 25 4 1774mm 64 1 1610 ent Reinforcem Flexural

2

φ < = + = ⇒

600 mm 4φ25 2φ14 φ12@125mm

Reinforced Concrete II Reinforced Concrete II

Dr. Hazim Dwairi
Dr. Hazim Dwairi

The Hashemite University The Hashemite University

6 350 mm 2φ14