1. 4 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

2. 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

3. 剪應力之概念 l 撓曲應力 剪應力 剪力流 剪力 斷面 樑元素 R = w l 2 w R dx y b Ai NA T V C dx v
V- w dx
T + dT
C+dC z
qmax =
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

4. 樑之主應力軌跡 The stress trajectories intersect the neutral
45° f1 f2   f1 f2 v f
The stress trajectories intersect the neutral
axis at 45°. The principal tensile stresses
become excessive due to cracking.
主拉應力 :
主壓應力 :

5. 剪力與斜拉力Shear and Diagonal Tension
Shear failure: A A M V
Pure shear at neutral axis:
45o
90o
45o
Web-shear crack

6. Below neutral axis: Potential cracks
Combination of shear stress and tensile stress
Potential cracks

7. 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
vag
Vcz 2 V Vd 1 T C x

8. The equilibrium of the free body : V = Vcz + Vay + Vd
Equilibrium in the Shear Span of Beam
The equilibrium of the free body :
V = Vcz + Vay + Vd V C
(2)
(1) T Vd
Vay Ha G
Vcz
Where :
V = shear resistance in beam without
web reinforcement
Vcz = shear contribution of compression zone
Vay = shear contribution of aggregate interlock
Vd = shear contribution of dowel action x
Vcz V Vd T C jd  G
(1)
(2)
jd cot  1 2
M = x V = jd (T + Vd cot )
If the contribution of the dowel force is
ignored (particularly in the absence of
stirrups), then :
M = T jd
Where :
M = moment resistance in beam without
web reinforcement

9. The Principal Mechanism of Shear Resistancex
Vcz V Vd T C jd  G
(1)
(2)
jd cot  1 2
V = = (T jd) = jd T dM dx d dT
d(jd)

10. If the Bond between Steel and Concrete is Good
In the elastic theory analysis of prismatic flexural members is assumed
that the internal lever arm remain constant, then d(jd)/dx = 0,
the equation of perfect “beam action” is obtained :
V = = (T jd) = jd T dM dx d dT
d(jd)
Where :
q = the bond force per unit length
= shear flow

11. 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 resistance
is expressed by :
V = = (T jd) = jd T dM dx d dT
d(jd)
V = T = C
d(jd) dx

12. 拱作用Arch Action in the Shear Spanjd P
Line of thrust C L d a
Slip
* Second term of the equation
signifies that shear can be
sustained by inclined
compression in a beam.
* The shaded area indicates the
extent of the compressed concrete
outside which cracks can form.
V = = (T jd) = jd T dM dx d dT
d(jd)

13. 剪跨Shear Span (a = M /V ) Distance a over which the shear is constant P
V = +P
V = -P
Shear
Diagram + -
M = Va
Moment
Diagram +

14. Variation in Shear Strength with a/d for rectangular beams
Flexural moment
strength
Shear-compression
strength
Inclined cracking
strength, Vc
Failure moment = Va
Shear-tension and
shear-compression
failures
Deep
beams
Flexural
failures
Diagonal tension
failures 1 2 3 4 5 6 7
a/d

15. RC樑之裂縫型態
撓曲裂縫及撓剪裂縫
腹剪裂縫

16. RC樑之剪力破壞模式
腹剪開裂
撓剪開裂

17. 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

18. 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 )

19. Crack Pattern in Several Lengths of Beam
Span
Mark (m) a/d 7/ 8/
10/ 9/ 1 2 3 4 5 6
7/1
8/1
10/1
9/1

20. SHEAR FAILURE MECHANISM
400
160
140
Theoretical flexural
strength of section Mu
350
Shear force corresponding
with the theoretical
flexural capacity Vu
300
120
Observed ultimate
moment
250
100
Observed ultimate shear
200 80
Moment, kNm
Shear force, kN
150 60
Shear corresponding
with “beam action”
Flexural capacity
corresponding with
“beam action”
in the shear span
100 40 50 20 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
Momen / Shear ratio = a d M
V d
Momen / Shear ratio = a d M
V 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)

21. ACI 318規範之剪力計算強度 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
精確剪力計算強度經驗公式
(U.S. customary units)
(SI units)
簡化剪力計算強度經驗公式
(SI units)
(U.S. customary units)

22. Shear strength with axial load:
Simple formula:
Shear strength with axial load:
Compression:
Tension:

23. . Comparison of Simplified Expression with Experimental Result 0.2 0.4
=  3.5 v
f’c
2500 w Vd
M f’c
= 2
0.2
0.4
0.6
0.8
1.0
1.5
2.0 
6.0
5.0
4.0
3.0
bwdf’c V
= , psi
1000 wVd

24. 剪力強度Shear Strength of Rectangular Concrete Section
from experiment vc

25. 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。

26. 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）。

27. 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)。

28. Example: Determine shear strength of the concrete section.
As = 18 cm2
25 cm
45 cm
5 cm
Vu = 9 ton, Mu = 4 t-m
Use simple formula:

29. 剪力鋼筋Shear Reinforcement Requirement
No need for shear reinforcement only when:
1) Vu < f Vc /2 ; f = 0.85 for shear
2) Shallow beam (slab, footing)
- d ฃ 25 cm
- d ฃ 2.5 tf
- d ฃ 1/2 bw

30. 腹筋所提供之剪力強度Shear Strength Provided by Stirrup
Av = 2As d s
Number of stirrup
Shear strength provided by stirrup

31. 含腹筋粱之剪力強度Shear Strength of RC Beam with Stirrup
Vu ฃ f Vn where f = 0.85 for shear
Vn = Vc + Vs ; Shear strength from concrete and steel
Vu ฃ f (Vc + Vs)
腹筋之最大間距Maximum Stirrup Spacing (smax)
smax ฃ d/2 or 60 cm when
smax ฃ d/4 or 30 cm when

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

33. 剪力設計之分類Shear Design Categories
(1) Vu ฃ 0.5 f Vc No shear reinforcement
(2) 0.5 f Vc < Vu ฃ f Vc Min. shear reinforcement
Min Vs = 3.5 bw d
Max s ฃ d/2 ฃ 60 cm
(3) f Vc < Vu ฃ f (Vc + min Vs) same as (2)

34. (4) f (Vc + min Vs) < Vu ฃ f (Vc + 1.1 bwd)
Max s ฃ d/2 ฃ 60 cm
(5) f (Vc bwd) < Vu ฃ f (Vc bwd)
Vs = Vu /f - Vc
Max s ฃ d/4 ฃ 30 cm

35. 剪力設計之步驟Design Procedure for Shear
STEP 1
Compute Vu at critical section d from face of support h d L w
d+h/2
wL/2
L/2
Critical section for shear

36. STEP 2
Compute shear strength Vc of concrete or
STEP 3
Design shear reinforcement
Where is Vn = Vu / f ?
0.5 Vc
Vc + min Vs

37. 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=55cm
b=40cm
wu=1.4(2)+1.7(5) = 11.3 t/m
Vu = 11.3(6)/2-(.55+.2)11.3 = 25.4 ton
[Vu=25.4 ton] > [fVc=0.85(18.1)=15.4 ton]
f (Vc + min Vs)= 0.85( ด40ด55/1,000) = 21.9 ton
21.9 ton < Vu=25.4 ton < 47.3 ton ฎ Category 4
Required Vs = Vu /f - Vc = 25.4/ = 11.8 ton

38. 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 ฃ d/2 = 55/2 = ฃ 60 cm
USE Stirrup 0.14 m Ans

39. Variation of Shear Capacity
Mid span
Support d
critical section
wuL/2
f Vc
f Vc/2 wu
f Vn

40. Example: Redesign the beam in Ex
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 = ton
[f Vc=15.4 t]< Vu<[fVc+min fVs =22.0 t] USE Min stirrup
USE RB9: Av= 2(0.636) = 1.27 cm2
smax ฃ 55/2 = 27.5 ฃ 60 cm
USE Stirrup 0.25 m Ans

41. Shear reinforcement detailing of beam in Example
0.40
1.30
3.00
1.30
6.00

42. 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.
40 cm
d = 53 cm
Section A-A
PL = 5 tons
PD = 2 tons
wL = 3 t/m
wD = 2 t/m
2.5 m
4.0 m A