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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.

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Presentation on theme: "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."— Presentation transcript:

1 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 4

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 l R = w l 2 w R dx 剪力 y b AiAi NA T V C dx v w V- w dx T + dT C+dC z q max = VzVz q = b v 斷面 樑元素 撓曲應力 剪力流 剪應力 v max v = V A i y b I 剪應力之概念

4 主拉應力 : 主壓應力 : 45° f1f1 f2f2   f1f1 f1f1 f2f2 f2f2 v f f v v 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.

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

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

7 v ag V cz 2 V VdVd 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 V CZ – 裂縫兩側混凝土表面的骨材鎖結作用力 V ag 之垂直分量 V ay – 縱向鋼筋的合釘作用力 V d

8 V C (2) (1) T VdVd V ay HaHa G V cz The equilibrium of the free body : V = V cz + V ay + V d Where : V = shear resistance in beam without web reinforcement V cz = shear contribution of compression zone V ay = shear contribution of aggregate interlock V d = shear contribution of dowel action x V cz V VdVd T C jd  G (1) (2) jd cot  1 2 M = x V = jd (T + V d 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 Equilibrium in the Shear Span of Beam

9 The Principal Mechanism of Shear Resistance x V cz V VdVd T C jd  G (1) (2) jd cot  1 2 V = = (T jd) = jd + T dM dx d dx dT dx d(jd) dx

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 : Where : q= the bond force per unit length = shear flow V = = (T jd) = jd + T dM dx d dx dT dx d(jd) dx

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 = C d(jd) dx d(jd) dx V = = (T jd) = jd + T dM dx d dx dT dx d(jd) dx

12 拱作用 Arch Action in the Shear Span jd P Line of thrust C L d P a Slip V = = (T jd) = jd + T dM dx d dx dT dx d(jd) dx *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.

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

14 Variation in Shear Strength with a/d for rectangular beams a/d Failure moment = Va Deep beams Shear-tension and shear-compression failures Diagonal tension failures Flexural failures Inclined cracking strength, V c Flexural moment strength Shear-compression strength

15 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

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 / / / / /1 8/1 10/1 9/1

20 Shear force corresponding with the theoretical flexural capacity V u Observed ultimate shear Shear corresponding with “beam action” Shear force, kN Momen / Shear ratio = adad M V d Theoretical flexural strength of section M u Observed ultimate moment Flexural capacity corresponding with “beam action” in the shear span Moment, kNm Momen / Shear ratio = adad 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 影響剪力計算強度之主要因子 : 混凝土強度 (  f c ’) 拉力鋼筋比 (  w = A s / b w d) 跨深比 (Shear span to depth ratio) M/Vd ACI 318 規範之剪力計算強度 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) (U.S. customary units)

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

23 =  3.5 v  f’ c 2500  w Vd M  f’ c v  f’ c =  v  f’ c V b w d  f’ c =, psi 1000  w Vd M  f’ c Comparison of Simplified Expression with Experimental Result

24 剪力強度 Shear Strength of Rectangular Concrete Section Shear strength: from experiment vcvc

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 f ct 已予以規定時, f c '1/2 須以 f ct /8 替代修正之,但所用之 f ct /8 值不得超過 f c '1/2 。 當輕質骨材混凝土之平均開裂抗拉強度 f ct 未予規定時, f c '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 A s /(bd) 與 0.02 中之小值  Rd = 0.25f ctk0.05 /  c (其中:  c =1.5 ; f ctk0.05 = 0.7f ctm ; f ctm =0.3 f c '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. V u = 9 ton, M u = 4 t-m A s = 18 cm 2 25 cm 45 cm 5 cm Use simple formula:

29 剪力鋼筋 Shear Reinforcement Requirement 2) Shallow beam (slab, footing) - d ฃ  25 cm - d ฃ 2.5 t f - d ฃ 1/2 b w No need for shear reinforcement only when: 1) V u <  V c /2 ;  = 0.85 for shear

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

31 含腹筋粱之剪力強度 Shear Strength of RC Beam with Stirrup V u ฃ  V n where  = 0.85 for shear V n = V c + V s ; Shear strength from concrete and steel V u ฃ  V c + V s ) 腹筋之最大間距 Maximum Stirrup Spacing (s max ) s max ฃ  d/2 or 60 cm when s max ฃ  d/4 or 30 cm when

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

33 剪力設計之分類 Shear Design Categories (1) V u ฃ  0.5  V c No shear reinforcement (2) 0.5  V c < V u ฃ  V c Min. shear reinforcement Min V s = 3.5 b w d Max s ฃ  d/2 ฃ  60 cm (3)  V c < V u ฃ  V c + min V s ) same as (2)

34 Max s ฃ  d/2 ฃ  60 cm (4)  V c + min V s ) < V u ฃ  V c b w d) V s = V u /  - V c (5)  V c b w d) < V u ฃ  V c b w d) Max s ฃ  d/4 ฃ  30 cm V s = V u /  - V c

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

36 STEP 2 Compute shear strength V c of concrete or STEP 3 Design shear reinforcement Where is V n = V u /  ? 0.5 V c V c + min V s

37 Example: The simply supported beam spans 6 m, support width 40 cm, w DL = 2 t/m, w LL = 4 t/m, f y = 2,400 ksc and f’ c =240 ksc. Design a vertical stirrup. d=55cm b=40cm V u = 11.3(6)/2-(.55+.2)11.3 = 25.4 ton [V u =25.4 ton] > [  V c =0.85(18.1)=15.4 ton] w u =1.4(2)+1.7(5) = 11.3 t/m  V c + min V s )= 0.85(  ด  ด 55/1,000) = 21.9 ton 21.9 ton < V u =25.4 ton < 47.3 ton ฎ Category 4 Required V s = V u /  - V c = 25.4/ = 11.8 ton

38 Select stirrup RB9, A v = 2(0.636) = 1.27 cm 2 Required spacing s = A v f y d / V s = 1.27(2,400)(55)/11,800 = 14.2 cm USE 14 cm s max ฃ  d/2  55/2 = 27.5 ฃ  60 cm USE Stirrup 0.14 m Ans

39 Variation of Shear Capacity Mid spanSupport d critical section w u L/2  V c  V c /2 wuwu  V n

40 Example: Redesign the beam in Ex. for stirrup at the middle half of span at x = L/4; V u = wL/4 = 11.3(6)/4 = ton [  V c =15.4 t]< V u <[  V c +min  V s =22.0 t] USE Min stirrup USE RB9: A v = 2(0.636) = 1.27 cm 2 s max ฃ  55/2 = 27.5 ฃ  60 cm USE Stirrup 0.25 m Ans

41 Shear reinforcement detailing of beam in Example

42 Home work: Select the stirrup spacing for the beam shown below. = 280 ksc, and f y = 4,000 ksc Use DB10 stirrups. Show your results on a scaled sketch. P L = 5 tons P D = 2 tons P L = 5 tons P D = 2 tons w L = 3 t/m w D = 2 t/m 2.5 m 4.0 m A A 40 cm d = 53 cm Section A-A


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