# 4 剪力與斜拉力Shear and Diagonal Tension - Shear Failure

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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

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

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 :

Shear failure: A A M V Pure shear at neutral axis: 45o 90o 45o Web-shear crack

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

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

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

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

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

jd 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)

V = +P V = -P Shear Diagram + - M = Va Moment Diagram +

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

RC樑之裂縫型態 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

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
Span Mark (m) a/d 7/ 8/ 10/ 9/ 1 2 3 4 5 6 7/1 8/1 10/1 9/1

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)

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)

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

. 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

from experiment vc

ACI 318規範對輕質骨材混凝土之 剪力計算強度修正

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）。

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)。

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:

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

Av = 2As d s Number of stirrup Shear strength provided by 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

(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)

(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

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

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

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

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

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

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

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

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