rc16_torsion2

11
Thin-walled Tube Combined Shear & Torsion Space Truss Analogy Torsion Design by ACI318 Compatibility Torsion Mongkol JIRAVACHARADET S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING Reinforced Concrete Design Reinforced Concrete Design Torsion 2 Torsion 2 T กกก Solid Hollow 0 1 2 3 Percent of torsional reinforcement 0 T n T cr solid section T cr hollow section Torsional Strength of Reinforced Concrete SDM กก กก กก กกก

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Page 1: RC16_Torsion2

� Thin-walled Tube

� Combined Shear & Torsion

� Space Truss Analogy

� Torsion Design by ACI318

� Compatibility Torsion

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Reinforced Concrete DesignReinforced Concrete Design

Torsion 2Torsion 2

T

�����������ก�� ����� ����������������������ก�� �����������ก�������������ก����������������ก���ก������������� ��

Solid Hollow

0 1 2 3Percent of torsional reinforcement

0

Tn

Tcr solid section

Tcr hollow section

Torsional Strength of Reinforced Concrete SDM

����� �����!�����ก����� ������ก�� "���ก�� ก�� ����������� ������������ก�� ��#����ก��������������������ก��������ก��

�� ��$�ก�� �����ก����"% �"ก�����ก�� ���������������ก�����

Page 2: RC16_Torsion2

Shear Stress in Thin-walled Tube

Cracking Torque (Tcr)

SDM

#������ ���"�&'ก#������(����)�� � ��������ก�ก� #��ก���

A0

t

Shear flow (q) Shear flow:0

Tq kg/cm

2A=

��#�# �������������'����ก� #����)�� ���

������� �*+��"กก���� τt = q / t :

t0

T2A t

τ =

ก���ก����ก���%$���+�� τ ��#�&% c1.06 f ′ crcr c

0

T1.06 f

2A t′τ = =

�� ��$� �����!�����ก��� ( )cr c 0T 1.06 f 2 A t′=

SDMACI ก���,�����#�#����)�� ���� t = 0.75Acp/pcp ���

-+$����������� A0 = 2Acp/3 ��+��

pcp #+���������'�.���ก�� ������#��ก���

Acp #+�-+$����/%� ������� �� pcp

�� ��$� �����!�����ก��� cp cpcr c

cp

2A 0.75AT 1.06 f 2

3 p

′= × ×

2cp

cr ccp

AT 1.06 f

p

′=

������ #��)��� ก������+�� 4/TT cru φ≤

���#'���ก�� �����ก���� φ = 0.85

( )2cp

ccp

A0.265 f

p

′≤ φ

Page 3: RC16_Torsion2

ตัวอยางที ่7.1 คานยื่นรับน้ําหนักบรรทกุประลัย 3 ตันที่มุมหนาตัดหางจาก

ศูนยกลางหนาตัด 15 ซม. ตรวจสอบดูวาจําเปนตองคิดผลของการบิดในการออกแบบ

หรือไม กําหนด f’c = 240 กก./ซม.2

��������'������� pcp = 2(60+30) = 180 /�.

-+$���������� Acp = (60)(30) = 1,800 /�.2

������������� ���������� ������

���"ก�� �����!��� φTcr/4 ( )2cp

ccp

A0.265 f

p

′= φ

218000.85 0.265 240

180

= × ×

= 62,812 กก.-/�. = 0.63 ���-����

�����!�������ก��� Tu = (3)(0.15) = 0.45 ���-���� OK< 0.63 ���-����

SDM

3 ton

30 cm

15 cm

60 cm

tA2T

0t =τ

Combined Shear and Torsion

Torsionalstresses

Shearstresses

Hollow section

Torsionalstresses

Shearstresses

Solid section

SDM

������� �*+��"กก��*+��:db

V

wv =τ

������� �*+��"กก����:��+�� t = 0.75Acp/pcp

��� A0 = 2Acp/3

�����������������ก�����:

2h0

hu

w

u

A7.1

pTdb

V+

2

2h0

hu2

w

u

A7.1

pTdb

V

+

Page 4: RC16_Torsion2

Torsional Geometric Parameters

Gross area A0h = x0 y0

Shear perimeter ph = 2(x0 + y0)

x0 , y0 = Distance from center to

center of stirrup

SDM

y0 h

x0

b

�����������ก�� &�)�� ��������ก��,��,5�#�����,��������� �ก����6� ���&�)�� ��#���� t < A0h/ph ,��������� �*+��"กก������(�

2h0

hu

A7.1

pTtA7.1

T

h0

u

Maximum Shear + Torsion Stress SDM

��������� ������-�� -����"��,��������� �*+������ก�����#�����ก�����"ก��

(a) For solid sections

2h0

hu

w

u

A7.1

pTdb

V+

2

2h0

hu2

w

u

A7.1

pTdb

V

+

′+φ≤ c

w

c f12.2db

V

(b) For hollow sections

′+φ≤ c

w

c f12.2db

V

For reinforced concrete cc f53.0V ′=

Page 5: RC16_Torsion2

Torsional Crack

ก���ก���"กก��������ก7����(��)������

���6����#��ก����������� �% �������ก�����#�ก

ก�� �����ก��������,�8��"ก����ก���ก�9� ����ก��������� ����)��#��ก������� ����� ���

#��ก�����ก����ก���ก���#�����)����ก���� �� ��$���� ก���ก���"�,5�-+$����A0h /%� &'ก������� ������ก���ก

����ก���ก�9�

A0h = -+$�������

T

45o

Torsional crack

Txo

yo

����ก���ก�����ก���

θV1V2

V3

V4

����ก���������

�)��#��ก������� ����� ���

Space Truss Analogy

ก�����*+�� q ,�)�� ���"�&'ก��� ��ก��(��� �*+�� V1 V2 V3 ��� V4 �������

����� #� &�ก�������� ,��'�

Page 6: RC16_Torsion2

! "��#����������������$%�: V4

V4

x0

V4y0

s

At fyv

At fyv

At fyv

y0 cot θ �����!���"ก�� �*+�� V4 :

4 04

V xT

2=

�� �*+�� V4 = n At fyv

"������ก,������*+��:

o0 0y cot 45 y

ns s

= =

t yv 04

A f yV

s=

t yv 0 04

A f x yT

2s= 1 2 3T T T= = =

�����!�����$ ��� : n 1 2 3 4T T T T T= + + +

t yv 0 0n

2 A f x yT

s= t yv 0h2A f A

s=

! "��#������������������&��

V4y0

∆N4/2

∆N4/2θθθθ

V4

θθθθ

D4

N4

�� �% ,�����ก���������: ∆N4 = V4 cot θ

t yv 04

A f yN

s∆ = 2N= ∆

t yv 01 3

A f xN N

s∆ = ∆ =

perimeter of stirrup2(xo+yo)

y0

x0�� �% ��$ ���: ∆N = ∆N1 + ∆N2 + ∆N3 + ∆N4

t yvy 0 0

A fA f 2(x y )

s= +

ℓ ℓ

t vy hA f p

s=

Page 7: RC16_Torsion2

Torsion Design by ACI318-08 SDM

��+�� �����!��� Tu ��#��ก�� φφφφTcr/4 ,����ก������ �����!����-+��,��

φφφφTn ≥≥≥≥ Tu

,�ก�#��� Tn �����!�����$ ���"�&'ก��� ��

����ก���ก�������ก��������� ��,�� Tc = 0

���������ก���ก#��� �� 0 t yvn

2A A fT

s= ��+�� A0 = 0.85A0h

t n

0 yv

A Ts 2A f

= u

0 yv

T2 A f

-+$��������ก��������������� ,���-����%$��-+�������ก���� yvth

y

fAA p

s f

=

���������ก���ก������ ก�������� �*+����� �����!���

v t tvA AATotal 2

s s s+ = +

2 legs Av

1 leg At

SDMMinimum Torsion Reinforcement

���������ก���ก��������6� ( ) wv t c

yv

b sA 2A 0.199 f

f′+ = w

yv

b s3.5

f≥

������ ����ก���ก��#�����ก��#��������ก���� ph/8 ��+� 30 /�.

-+$��������ก�����������������6� yvc t,min cp h

y y

f1.33 f AA A p

f s f

′= −

ℓ ℓ

����� At/s "���� �������ก�� 1.8 bw/fyv

���� ���������ก�������ก����ก��"� �����.�,�����ก���ก ���������� �ก����6� 30 /�.

���� ���������ก������� ������%� �������������6��� ����ก���ก

���� ,5�����ก��������� ∅∅∅∅ ≥≥≥≥ 1/24 ��������� ����ก���ก ≥≥≥≥ 10 �.�.

Page 8: RC16_Torsion2

Example 14-1: The 8-m span beam carries a cantilever slab 1.5 m. The beam supports a live load of 1.2 t/m along the beam centerline plus 200 kg/m2 over the slab surface. The effective depth of beam is 54 cm, and the distance from the

surface to the stirrup is 4 cm. f’c= 280 kg/cm2, fy = 4,000 kg/cm2

8 m 1.5 m

60 cm

30 cm15 cm

Load from slab:

wu = 1.4(0.15)(2,400)(1.5)+1.7(200)(1.5) = 1,266 kg/m

Eccentricity = 1.5/2 = 0.75 m

Load from beam:

wu = 1.4(0.6)(0.3)(2,400)+1.7(1,200) = 2,645 kg/m

2 2(1,266 2,645)(8.0)25.0 t-m

10 10 1,000u

w LM

+= = =

×

�� �*+��������: uu

w L (1,266 2,645)(8.0)V /1,000 15.6 ton

2 2+

= = =

�����!���������: tuu

w L 1,266 0.75 8.0T /1,000 3.80 t-m

2 2× ×

= = =

�����!���������:

Flexural Design

22 2

25(1,000)(100)31.8 kg/cm

0.9(30)(54)u

n

MR

bdφ= = =

ρmin = 0.0035, ρmax = 0.0229

0.85 21 1 0.0086

0.85c n

y c

f Rf f

ρ ′

= − − = ′ OK

Al,flexure = ρ b d = 0.0086(30)(54) = 13.9 cm2

Page 9: RC16_Torsion2

Shear Design

Cracking Torque

60cm

15 cm

30 cm

Acp = 30 × 60 = 1,800 cm2 pcp = 2(30 + 60) = 180 cm

( )2cp

cr ccp

AT / 4 0.265 f

p

′φ = φ

21,8000.85 0.265 280

180= × ×

= 67,845 kg-cm = 0.68 t-m < Tu = 3.80 t-m

∴∴∴∴ ����ก������'ก�(� ����)����*� �

sv

y

VAs f d

=

cV 0.53 280 30 54 /1,000= × × = 14.37 ton

Vu / φ = 15.6 / 0.85 = 18.35 ton

Vs = Vu / φ – Vc = 18.35 – 14.37 = 3.98 ton

3.984.0 54

= 0.0184 cm2 / cm / two legs

bwd = (30)(54) = 1,620 cm2

xo = 30 - 2(4) = 22 cm

yo = 60 - 2(4) = 52 cm

Aoh = xoyo = (22)(52) = 1,144 cm2

Ao = 0.85(1,144) = 972.4 cm2

ph = 2(22+52) = 148 cm

Torsional Geometric Parameters

30 cm

60 c

m

Check adequacy of section

���"����'�������������,�8��-�� -���+����?

2

2h0

hu2

w

u

A7.1

pTdb

V

+

′+φ≤ c

w

c f12.2db

V

2 2

2

15.6 3.80 100 14830 54 1.7 1,144

× × + × × ( )0.85 0.53 2.12 280 /1,000+

0.0271 t/cm2 0.0377 t/cm2< ������&�����!&��!�

Page 10: RC16_Torsion2

Torsional reinforcement

Combined shear & torsion stirrup

t u

0 yv

A Ts 2 A f

3.80 1002 0.85 972.4 4.0

×=

× × ×= 0.0575 cm2 / cm / one leg

v t tvA AATotal 2

s s s+ = +

= 0.0184 + 2 × 0.0575 = 0.1334 cm2 / cm

two legs

��+�ก����ก���ก�9� DB12 (-+$������ � 2(1.13) = 2.26 /�.2)

������ ����ก���ก������ ก� s = 2.26 / 0.1334 = 16.9 cm

������ ����ก���ก�ก����6� smax = ph / 8 = 148 / 8 = 18.5 cm < 30 cm

Use closed stirrup DB12 @ 0.16 m (Av+t/s = 2.26/16 = 0.141 cm2 / cm)

Torsion longitudinal steel

Minimum torsion reinforcement

v t w wc

y y

A b bmin 0.199 f 3.5

s f f+ ′= ≥ f’c > 309 ksc

control

303.5

4,000= × = 0.0263 cm2 / cm < (Av+t/s = 0.141 cm2 / cm) OK

yvth

y

fAA p

s f

=

= 0.0575 × 148 = 8.51 cm2

yvc t,min cp h

y y

f1.33 f AA A p

f s f

′= −

ℓ ℓ

1.33 2801,800 8.51

4,000= × − = 1.51 cm2

t w

yv

A b 300.0575 1.8 1.8 0.0135

s f 4,000 = > = =

OK

Page 11: RC16_Torsion2

Total Longitudinal Steel

Bending: Al = 13.9 cm2 (Top reinforcement)

Torsion: Al = 8.51 cm2 (Distributed along perimeter)

Provide 4DB16 in the bottom half of beam

60 cm

30 cm

To satisfy 30 cm max. spacing of Al

and add 8.51 – 4(2.01) = 0.47 cm2 to flexural steel

Steel area required at top = 13.9 + 0.47 = 14.37 cm2

USE TOP 3DB25 (As = 14.73 cm2)

Section @ Supports

60 cm

3DB25

2DB25

DB12 @ 0.16 m

30 cm

2DB16

2 2(1,266 2,645)(8.0)17.87 t-m

14 14 1,000u

w LM

+= = =

×At midspan:

Required As = 9.68 cm2 (2DB25)