A presentation onFlexural & Shear Design of Deep Beam Presented by Pragya N R P Roy (P t ) (Peter) Structural Engineer Contents • What is Deep Beam • Behavior of Deep Beam • Flexural Design of Deep Beam • Shear Design of Deep Beam g p • Examples p What is Deep Beam • Large depth/thickness ratio • And shear span depth ratio less than 2.5 for concentrated load and less than 5.0 for distributed load. • Where the shear span is the clear span of the beam for distributed load Application • • • • • Pile Cap Cap. Bridge Girder. Wall l b W ll slabs under vertical l d d ti l loads. Floors slabs under horizontal loads. Some shear walls. Behavior of Deep beam • Two-Dimensional Action. • Plane Section Do Not Remain Plane, in the deep beam design. • The strain distribution is no longer linear. • The shear deformation cannot be neglected as in the ordinary beam. • The stress distribution is not linear even in the elastic stage. • At the ultimate limit state the shape of concrete compressive stress bl k i not parabolic shape i block is b li h again. Behavior of Deep beam • The distribution of tensile stress at bottom fiber is constant over the span. • Tensile stress in the bottom fiber at support and at mid span is almost the same. • The tension reinforcement must be extended to the end of support. • Th maximum t The i tensile stress at th b tt il t t the bottom fiber is far exceed the magnitude of compressive stress. i t Design Criteria for flexure in deep Beams Simply Supported beams: • The ACI code does not specify a simplified g procedure design p • ACI procedure for flexural analysis and design of deep beams follows rigorous non linear approach • The simplified provisions presented in this section are based on the recommendations of the Euro International Concrete Committee Euro-International (CEB). Design Criteria for flexure in deep Beams • A schematic stress distribution in a homogeneous deep beam having a span/depth ratio Ln/h = 1 0 1.0 • It was experimentally observed that the moment lever arm does not change significantly even after initial cracking Design Criteria for flexure in deep Beams Nominal Resisting Moment, Mn = Asfy (moment arm jd) The reinforcement area ‘As’ for flexure is As = Mu/φfyjd ≥ 3√fc bd/fy ≥ 200bd/fy The lever arm as recommended by CEB jd= 0.2(l+2h) for j 0 2( 2 ) f 1 ≤ l/h <2 2 jd=0.6l for l/h <1 l= Effective span measured from c/c. The tension reinforcement has to be placed in the lower segment of beam height such that the g g segment height is Y=0.25h-0.05l < 0.20h Design Criteria for flexure in deep Beams SHEAR DESIGN OF DEEP BEAM • The shear design of deep beam is similar as shear design of ordinary beam. g y • The difference is the concrete shear strength. • Limitation of ultimate shear force. • Horizontal and vertical stirrups distribution. BASIC DESIGN EQUATION φVn ≥ Vu Vn = nominal shear strength φVn = design shear strength φ = strength reduction factor (0.85) g ( ) Vu = ultimate shear force, factored shear force • • • • SHEAR DESIGN OF DEEP BEAM • As shear design of ordinary beam, the shear force is resisted by the concrete component and by the shear reinforcement component, as follows : Vn = Vc + Vs where : • Vn = Nominal shear strength • Vc = Concrete shear strength without shear reinforcement • Vs = shear reinforcement (stirrup) shear strength ( p) g CONCRETE SHEAR STRENGTH The concrete Shear strength of deep beam is taken as where : Vc = concrete shear strength (N) Mu = ultimate fl lti t flexure moment (N t (Nmm) ) Vu = ultimate shear force (N) f’c = concrete cylinder strength (MPa) p d = effective depth bw = width of beam web ρw = longitudinal reinforcement ratio CONCRETE SHEAR STRENGTH Or ,the concrete shear strength can be determined as : The maximum limit of concrete shear strength is : s CONCRETE SHEAR STRENGTH The section must be enlarged if the ultimate shear force does not follow the condition below : Or, STIRRUP SHEAR STRENGTH The st e gt o horizontal a d vertical s ea e strength of o o ta and e t ca shear reinforcements is : LIMITS OF SHEAR REINFORCEMENT The minimum shear reinforcement area is : Av−min = 0.0015 (bsv) Avh−min = 0.0025 (b sh) where : Av-min = minimum vertical stirrups Avh-min = minimum horizontal stirrups h i b = width of beam sv = spacing of vertical stirrups i f ti l ti sh = spacing of horizontal stirrups MAXIMUM SPACING OF SHEAR REINFORCEMENT The maximum spacing of shear reinforcement is : p g CRITICAL SECTION IN DEEP BEAM The critical section to determines the ultimate shear force in the deep beam is : STEP – BY – STEP PROCEDURE The followings are the step – by – step procedure used in the shear design for deep beam, as follows : Determine the critical section to calculate the ultimate shear force Vu. STEP – BY – STEP PROCEDURE Check the ultimate shear force, enlarge the section if the condition is not achieved. STEP – BY – STEP PROCEDURE Calculate the concrete shear strength Vc: STEP – BY – STEP PROCEDURE If Vu < 0.5φVc then no shear reinforcements needed, but for practical reason provide minimum shear reinforcement: ti l id i i h i f t If Vu > φVc then provide the shear reinforcements. STEP – BY – STEP PROCEDURE Calculate the ultimate shear force carried by the stirrups Vs. Choose the vertical and horizontal stirrups until the condition Ch th ti l d h i t l ti til th diti achieved. STEP – BY – STEP PROCEDURE Check the spacing of shear reinforcement sv and sh. If necessary check the chosen shear reinforcements for the basic design equation for shear design. STEP – BY – STEP PROCEDURE The design procedure above is repeats until the basic design equation for shear design is achieved Example: Design Problem Flexural Design of a Simply Supported Beam Example: Design Problem • MATERIAL • Concrete strength = K – 300 • Steel grade = Grade 400 • Concrete cylinder strength = f'c = 0.83× 30 = 24.9 MPa • β1 = 0.85 • • • • DIMENSION b = 500 mm h = 2750 mm Concrete cover = 50 mm • d = 2700 mm Example: Design Problem • Design Force Mu=1.4((1/8)*qL^2 = 1 4((1/8)*6000*5^2 1.4((1/8)*6000*5^2 =26250 kgm Deep Beam Checking Ln/d = 4700/2700=1 74 4700/2700 1.74 Where, 1.0 ≤ 1.74 ≤ 5.0 Deep beam action Example: Design Problem • Lever Arm jd = 0.2(L + 2h)= 0.2(5000 + 2(2750)) =2100mm Positive Reinforcement Mu = 262500000Nmm Example: Design Problem Example: Design Problem • SHEAR DESIGN OF SIMPLY SUPPORTED DEEP BEAM Example: Design Problem • MATERIAL • Concrete strength = K – 300 • Steel grade = Grade 240 • Concrete cylinder strength = f'c = 0.83× 30 = 24.9 MPa • β1 = 0.85 • • • • DIMENSION b = 500 mm h = 2750 mm Concrete cover = 50 mm • d = 2700 mm Example: Design Problem • Design Force X =1.5Ln=0.15x4700 =705mm Vu =1 4(10770)= 15078 kg =1.4(10770)= Limitation Checking The Section is not enlarged. Example: Design Problem • Concrete Shear Strength Example: Design Problem • Design of Stirrups Vu= 150780 < 0 5*Φ*Vc =872161 0.5*Φ*V 872161 Provide minimum web reinforcement For horizontal and vertical stirrups we g choose 2 legs Φ10. Av=2((1/4)*π Φ2 =2((1/4)* π*102 =157mm2 Example: Design Problem Example: Design Problem • Reinforcement arrangement References • Reinforced Concrete by E G Nawy E. G. • http://www.asdipsoft.com/Deep-bm.htm THANK YOU