剪力與斜拉力 Shear and Diagonal Tension

- Shear Failure

- Shear Strength of Concrete

- Shear Reinforcement by Stirrup

- ACI Code Provision for Shear Reinforcement

44

鋼筋混凝土樑之抗剪機制及剪力強度預測式

• 鋼筋混凝土樑之抗剪機制• 剪力強度預測式

lR = w l2

w

R

dx剪力

y bAi

NA

T

V

C

dx

v

w

V- w dx

T + dT

C+dC

z

qmax =Vz

q = b v斷面 樑元素 撓曲應力 剪力流 剪應力

vmax

v =V Ai y

b I

剪應力之概念

主拉應力 :

主壓應力 :

45°

45°f1

f2

f1

f1

f2

f2

v

ff

vv

v

樑之主應力軌跡

The stress trajectories intersect the neutral axis at 45°. The principal tensile stresses become excessive due to cracking.

221 4

2

1vfff

222 4

2

1vfff

f

v2φ2tan

A

A

M

V

剪力與斜拉力 Shear and Diagonal Tension

A

A

Shear failure:

45o

Web-shear crack

Pure shear at neutral axis:

45o

(max)tf

tf90o

Below neutral axis:

tf tf

(max)tfmax

tf

tf

,tf

0,

(max)tf

max2

Combination of shear stress and tensile stress

2

2(max) max

1, tan 2

2 4 / 2t

t tt

ff f

f

Potential cracks

vag

Vcz

2VVd

1

T

C

x

無腹筋 RC 樑之抗剪機制

• 無腹筋 RC 樑開裂時，其主要的抗剪機制如下所列： – 抗壓區未開裂混凝土所提供的剪力抵抗 VCZ

– 裂縫兩側混凝土表面的骨材鎖結作用力 Vag之垂直分量 Vay

– 縱向鋼筋的合釘作用力 Vd

V

C

(2)

(1)

TVd

VayHa

G

Vcz

The equilibrium of the free body :

V = Vcz + Vay + Vd

Where :V = shear resistance in beam without

web reinforcementVcz = shear contribution of compression zoneVay = shear contribution of aggregate interlockVd = shear contribution of dowel actionx

Vcz

VVd

T

C

jd

G(1)

(2)

jd cot 12

M = x V = jd (T + Vd cot )

If the contribution of the dowel force isignored (particularly in the absence ofstirrups), then :

M = T jdWhere :M = moment resistance in beam without

web reinforcement

Equilibrium in the Shear Span of Beam

The Principal Mechanism of Shear Resistance

x

Vcz

VVd

T

C

jd

G(1)

(2)

jd cot 12

V = = (T jd) = jd + TdMdx

ddx

dTdx

d(jd)dx

If the Bond between Steel and Concrete is Good

In the elastic theory analysis of prismatic flexural members is assumedthat the internal lever arm remain constant, then d(jd)/dx = 0, the equation of perfect “beam action” is obtained :

Where :q = the bond force per unit length

= shear flow

qjddxdTjdV

V = = (T jd) = jd + TdMdx

ddx

dTdx

d(jd)dx

If the Bond between Steel and Concrete is Destroyed

* The tensile force T cannot change, hence dT/dx = 0.* The external shear can be resisted only by inclined internal compression.

This extreme case may be termed “arch action”. The shear resistanceis expressed by :

V = T = Cd(jd)dx

d(jd)dx

V = = (T jd) = jd + TdMdx

ddx

dTdx

d(jd)dx

拱作用 Arch Action in the Shear Span

jd

PLine of thrustCL

d

Pa

Slip

V = = (T jd) = jd + TdMdx

ddx

dTdx

d(jd)dx

* Second term of the equationsignifies that shear can be sustained by inclined compression in a beam.

* The shaded area indicates theextent of the compressed concrete outside which cracks can form.

剪跨 Shear Span (a = M /V )

P Pa a

Distance a over which the shear is constant

M = VaMomentDiagram +

V = +P

V = -P

ShearDiagram +

-

Variation in Shear Strength with a/d for rectangular beams

a/d0 1 2 3 4 5 6 7

Fai

lure

mom

ent

= V

a

Deepbeams

Shear-tension andshear-compressionfailures

Diagonal tensionfailures

Flexuralfailures

Inclined crackingstrength, Vc

Flexural momentstrengthShear-compression

strength

RC 樑之裂縫型態

撓曲裂縫及撓剪裂縫

腹剪裂縫 撓曲裂縫及撓剪裂縫 腹剪裂縫

RC 樑之剪力破壞模式

腹剪開裂

撓剪開裂

深樑之剪力破壞模式

拱作用

破壞之種類

1: 錨定破壞

2: 承壓破壞

3: 撓曲破壞

4 及 5: 拱肋破壞

RC 樑之剪力破壞模式

剪拉破壞模式

(shear-tension failure)

剪壓破壞模式

(shear-compression failure )

Crack Pattern in Several Lengths of Beam SpanMark (m) a/d1 0.90 1.02 1.15 1.53 1.45 2.04 1.70 2.55 1.95 3.06 2.35 4.07/1 3.10 5.08/1 3.60 6.010/1 4.70 8.09/1 5.80 7.0

1

234

5

6

7/1

8/1

10/1

9/1

400

350

300

250

200

150

100

50

00 1 2 3 4 5 6 7 8

Shear force correspondingwith the theoreticalflexural capacity Vu

Observed ultimate shear

Shear correspondingwith “beam action”Sh

ear f

orce

, kN

Momen / Shear ratio = ad

MV d

160

140

120

100

80

Theoretical flexuralstrength of section Mu

Observed ultimatemoment

Flexural capacitycorresponding with“beam action”in the shear span

60

40

20

00 1 2 3 4 5 6 7 8

Mom

ent,

kNm

Momen / Shear ratio = ad

MV d

a/d < 2.5 (Crushing / Splitting of concrete - Type 3)

2 < a/d < 3 (Shear compression failure - Type 2)

3 < a/d < 7 (Flexural tension failure - Type 1)

影響剪力計算強度之主要因子 :•混凝土強度 ( fc’)•拉力鋼筋比 (w = As / bwd)•跨深比 (Shear span to depth ratio) M/Vd

ACI 318 規範之剪力計算強度

dbf.dbM

dVf.V w

'cw

u

uw

'cc 3017160

dbfdbM

dVfV wcw

u

uwcc

'' 5.3ρ25009.1

精確剪力計算強度經驗公式

簡化剪力計算強度經驗公式

(U.S. customary units)

(SI units)

dbf

V wc

c

6

'

(SI units) dbfV wcc'2 (U.S. customary units)

Simple formula:

20.53 kg/cmc cv f

0.53 kgc c wV f b d

Shear strength with axial load:

Compression:

Tension:

0.53 1 0.0071 kguc c w

g

NV f b d

A

0.53 1 0.0029 kguc c w

g

NV f b d

A

. . .

.. . . .... .. ... .. .. .......

... . ... ... .... .....

...... .....

. . ......... ...... ..... ..

.. ...

. .. .

..

. . .. ..

.. .. .

. . ...

... ... . . . .. .....

..

. ..

.. . .

..

..

. . ..

. . ... .. .

= 1.9 + 3.5vf’c

2500 w VdM f’c

vf’c

= 2

0 0.2 0.4 0.6 0.8 1.0 1.5 2.0

6.0

5.0

4.0

3.0

2.0

1.0

v f’ c

Vb w

df’ c

=

,

psi

1000 wVdM f’c

Comparison of Simplified Expression with Experimental Result

剪力強度 Shear Strength of Rectangular Concrete Section

Shear strength: cc

w

Vv

b d

from experiment 20.50 176 0.93 kg/cmuc c w c

u

V dv f f

M

vc

0.50 176 uc c w

u

V dv f

M

0.93 cf

/w u uV d M

1.0u

u

V d

M

sw

w

A

b d

ACI 318 規範對輕質骨材混凝土之剪力計算強度修正

• 當其平均開裂抗拉強度 fct已予以規定時， fc'1/2須以 fct/8

替代修正之，但所用之 fct/8 值不得超過 fc'1/2 。

• 當輕質骨材混凝土之平均開裂抗拉強度 fct 未予規定時

， fc'1/2 值對全輕質骨材混凝土須乘以 0.75 ；對於砂輕

質骨材混凝土須乘以 0.85 ；介於以上兩者間之含有部分輕質骨之輕質骨材混凝土可採內插法決定之。

Eurocode 規範之剪力計算強度

式中：• k = 1.6-d 1 （ d 之單位為 m ） = 1 （當 a/d > 2.5 ）或 = 2.5d/a 5 （當 a/d < 2.5 ） 為 As/(bd) 與 0.02 中之小值

Rd = 0.25fctk0.05/c（其中： c =1.5 ； fctk0.05 = 0.7fctm ； fctm

=0.3 fc'2/3）。

db.kV wRdc 4021

Zsutty 之剪力計算強度

• 當 a/d < 2.5 時，則上式右邊必須乘以 2.5(d/a) 。

dba

df.V w

'cc

31

172

Example: Determine shear strength of the concrete section.

Vu = 9 ton, Mu = 4 t-m

9(45)1.01 1.0

4(100)u

u

V d

M

180.016

(25)(45)s

ww

A

b d

As = 18 cm2

25 cm

45 cm

5 cm

240 ksccf

0.93 0.93 240(25)(45) /1,000 16.2 tonc wf b d

(25)(45)0.50 240 176(0.016)(1.0)

1,000

11.9 ton < 16.2 ton

cV

OK

Use simple formula:(25)(45)

0.53 240 9.24 ton1,000cV

剪力鋼筋 Shear Reinforcement Requirement

2) Shallow beam (slab, footing)

- d ฃ25 cm

- d ฃ 2.5 tf

- d ฃ 1/2 bw

No need for shear reinforcement only when:

1) Vu < Vc /2 ; = 0.85 for shear

腹筋所提供之剪力強度 Shear Strength Provided by Stirrup

d

d

s s s/n d s

Number of stirrup

v ys v y

A f dV A f n

s

Shear strength provided by stirrup

Av = 2As

含腹筋粱之剪力強度 Shear Strength of RC Beam with Stirrup

Vu ฃVn where = 0.85 for shear

Vn = Vc + Vs ; Shear strength from concrete and steel

Vu ฃVc + Vs)

腹筋之最大間距 Maximum Stirrup Spacing (smax)

smax ฃd/2 or 60 cm when 1.1s c wV f b d

smax ฃd/4 or 30 cm when 1.1 2.1c w s c wf b d V f b d

腹筋之最大與最小用量 Maximum & Minimum Shear Reinforcement

,max

,max

2.1

2.1

s c w

c wv

y

V f b d

f b dA

f

Maximum:

Minimum: ,min

max

3.5

3.5

wv

y

v y

w

b sA

f

A fs

b

剪力設計之分類 Shear Design Categories

(1) Vu ฃ0.5 Vc No shear reinforcement

(2) 0.5 Vc < Vu ฃVc Min. shear reinforcement

Min Vs = 3.5 bw d

Max s ฃd/2 ฃ60 cm

(3) Vc < Vu ฃVc + min Vs) same as (2)

Max s ฃd/2 ฃ60 cm

cf (4) Vc + min Vs) < Vu ฃVc + 1.1 bwd)

Vs = Vu / - Vc

v y

s

A f ds

V

(5) Vc + 1.1 bwd) < Vu ฃVc + 2.1 bwd)cf cf

Max s ฃd/4 ฃ30 cm

Vs = Vu / - Vc

v y

s

A f ds

V

剪力設計之步驟 Design Procedure for Shear

STEP 1 Compute Vu at critical section d from face of support

/ 2

2 / 2 2

2 2

u

wL wL hV d

L

wL hd w

h

d

L

w

d

d+h/2

wL/2

L/2

Critical section for shear

STEP 2 Compute shear strength Vc of concrete

0.50 176 0.93

0.53

uc c w w c w

u

c c w

V dV f b d f b d

M

V f b d

or

STEP 3 Design shear reinforcement

Where is Vn = Vu / ?

0.5 Vc Vc + min Vs 1.1c cV f b d 2.1c cV f b d

Example: The simply supported beam spans 6 m, support width 40 cm,

wDL= 2 t/m, wLL= 4 t/m, fy = 2,400 ksc and f’c=240 ksc. Design a vertical stirrup.

d=55

cm

b=40cm

Vu = 11.3(6)/2-(.55+.2)11.3 = 25.4 ton

0.53 240(40)(55) /1,000 18.1 toncV

[Vu=25.4 ton] > [Vc=0.85(18.1)=15.4 ton]

wu=1.4(2)+1.7(5) = 11.3 t/m

Vc + min Vs)= 0.85(18.1+3.ฃฃ55/1,000) = 21.9 ton

21.9 ton < Vu=25.4 ton < 47.3 ton ฃ Category 4

1.1 0.85(18.1 1.1 240 40 55 /1,000) 47.3 tonc cV f b d

Required Vs = Vu / - Vc = 25.4/0.85 - 18.1 = 11.8 ton

Select stirrup RB9, Av = 2(0.636) = 1.27 cm2

Required spacing s = Av fy d / Vs

= 1.27(2,400)(55)/11,800

= 14.2 cm USE 14 cm

smax ฃd/255/2 = 27.5 ฃ60 cm

USE Stirrup RB9 @ 0.14 m Ans

Variation of Shear Capacity

Mid spanSupport

d critical sectionwuL/2

Vc Vc/2

wu

Vn

Example: Redesign the beam in Ex. for stirrup at the middle half of span

at x = L/4; Vu = wL/4 = 11.3(6)/4 = 16.95 ton

[Vc=15.4 t]< Vu<[Vc+min Vs =22.0 t] USE Min stirrup

USE RB9: Av= 2(0.636) = 1.27 cm2 1.27 2,400

50.9 cm3.5 3.5 40

v yA fs

b

smax ฃ55/2 = 27.5 ฃ60 cm

USE Stirrup RB9 @ 0.25 m Ans

6.00

1.30 1.303.00

0.40

RB9@0.14 RB9@0.14RB9@0.25

Shear reinforcement detailing of beam in Example

Home work: Select the stirrup spacing for the beam shown below.

= 280 ksc, and fy = 4,000 ksc Use DB10 stirrups.

Show your results on a scaled sketch.

'cf

PL = 5 tons

PD = 2 tons

PL = 5 tons

PD = 2 tons wL = 3 t/m

wD = 2 t/m

2.5 m 2.5 m4.0 m

A

A

40 cm

d = 53 cm

Section A-A