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![Design of Shear Links
Sv Sv
Sv
V =resistance due to concrete +Resistance due to reinforcements
= (concrete stress x area) + (Steel stress x area of links)
The force in each link = [
𝑓𝑦𝑣
ɣ𝑚
Asv]= 0.87fyv.Asv
number of links cut by 45o plane = d/sv
Total force in the cut links = 0.87fyv.Asv. d/sv
V=vcbvd+ 0.87fyv.Asv. d/sv
0.87fyv.Asv. = [V-vcbvd].sv/d=[vbvd- vcbvd] sv/d=[bvsv(v-vc)]
as V=vbvd
Hence
x
x
d
d
V
bv
d
Asv
As
Asv ≥
𝑏𝑣
𝑠𝑣
(𝑣−𝑣𝑐)
0.87𝑓𝑦𝑣](https://image.slidesharecdn.com/4sheardesign-231018100356-32087d4c/85/4-shear-design-pptx-13-320.jpg)
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4 shear design.pptx
- 1. Design of Reinforced ConcreteStructures to BS8110 Design for Shear Eng. Sunil Jayawardena BSc Eng (Hons)., PGDip(CPM)., Ceng.,MIE(SL)
- 2. Shear Stress inConcrete beams at ULS Compression- takes the form of arch Tension- takes the form of suspended chain at mid-span bending is high and stresses are more parallel to the beam axis Towards support shear force is higher. principal stresses are at steeper angle and tensile stresses are causing diagonal cracks (if tensile stress exceed the tensile strength of the section) If the diagonal tension exceed tensile strength of concrete then shear reinforcements shall be provided (links, bent up bars and links)
- 3. • As indicatedabove RC beams are subjected to diagonal tension as a result of shear forces. A D C B Lets say AB=CD=L After deformation AB=L+δ CD=L-δ Hence we can say that tensile forces in AB direction cause deformation of length AB If this tensile stress exceed the tensile strength of concrete it will induce diagonal cracking of concrete perpendicular to AB. Experiment evidence shows that these shear diagonal cracks occurs at 45o to the Longitudinal Axis of the beam.
- 4. P R d 45o σt If beam widthis b PR surface area = 2 𝑏𝑑 Total tensile force on face PR, T= σt 2 𝑏𝑑 If shear force acting on the plane is V For the equilibrium of the section V=(T/√2) V=σt 2 𝑏𝑑/√2 σt = 𝑉 𝑏𝑑 in code above is given as v= 𝑉 𝑏𝑣 𝑑
- 5. Shear Resistance ofconcrete Section The shear of a section without shear reinforcements are taken mainly by three components. • Concrete in compression zone (20-40%) • Dowelling action of tensile r/f (15-25%) • Aggregate interlocking of flexural cracks.(33-50%) The shear resistance of a section (vc) is depend on; • Concrete grade (fcu) • Area of steel in tension (100As/bd) • Effective depth of section (d) Note Values of vc for grade 25 concrete is given in table 3.8 of the code.
- 7. Shear Resistance ofconcrete Section Alternatively we can use following equation; 𝑣𝑐 = 0.79 100𝐴𝑠 𝑏𝑣𝑑 1 3 1.25 400 𝑑 1 4 ( 𝑓𝑐𝑢 25 ) 1 3 Ɣm - partial safety factor 100𝐴𝑠 𝑏𝑣𝑑 ≤ 3 400 𝑑 ≥ 1 fcu max 40
- 8. Enhance Shear Resistancenear Supports X av X Ө • Failure is caused by loads to RHS of x-x • For beams without rlf Ө≈30o (av/d≈2) (experimentally verified) • Hence for a section closer to the support than 2d, the concrete shear strength will be vc( 2𝑑 𝑎𝑣 ) 𝑣𝑟𝑒𝑠 𝑣𝑐 𝑎𝑣 𝑑 1 2
- 9. Enhance Shear Resistancenear Supports Note 1. However the maximum shear stress of the section should not exceed the lesser of 0.8√𝑓𝑐𝑢 or 5N/mm2 (cl 3.4.5.2) 2. For most beams, design for shear is carried out at a distance ‘d’ from the face of the support and without considering shear enhancement near support, shear reinforcement provided up to the support (cl 3.4.5.10) vc v= 𝑉 𝑏𝑣 𝑑 d 2d Shear stress taken for design Shear resistance taken for design
- 10. Enhance Shear Resistancenear Supports Note 3. The shortfall of (v-vc) has to be met by shear reinforcements. 4. In order to longitudinal steel to function as a dowel, it has to anchored at least distance ‘d’ from the section considered. 5. If bottom steel is not properly anchored at the support, top steel shall be used for 100As/bd calculation (cl 3.4.5.4).
- 11. Shear & AxialCompression Clause 3.4.5.12 v’c= vc+0.6 𝑵 𝑨𝒄 𝑽𝒉 𝑴 Shear resistance of concrete is improved by presence of an axial force. N- axial load Ac-Area of section V-Shear force M-moment 𝑉ℎ 𝑀 should not be taken as greater than 1.0
- 12. Design of ShearLinks 45o Sv Sv Sv • If Sv=d then 45o diagonal crack can just developed between two adjacent links and such links ineffective. Therefore to prevent above distance between links are limited to a maximum value of 0.75d (cl 3.4.5.5) • Closer the link load carry by each link is the smaller.
- 13. Design of ShearLinks Sv Sv Sv V =resistance due to concrete +Resistance due to reinforcements = (concrete stress x area) + (Steel stress x area of links) The force in each link = [ 𝑓𝑦𝑣 ɣ𝑚 Asv]= 0.87fyv.Asv number of links cut by 45o plane = d/sv Total force in the cut links = 0.87fyv.Asv. d/sv V=vcbvd+ 0.87fyv.Asv. d/sv 0.87fyv.Asv. = [V-vcbvd].sv/d=[vbvd- vcbvd] sv/d=[bvsv(v-vc)] as V=vbvd Hence x x d d V bv d Asv As Asv ≥ 𝑏𝑣 𝑠𝑣 (𝑣−𝑣𝑐) 0.87𝑓𝑦𝑣
- 14. Bent up Bars Shearresistance of single bar is given by V=0.87fyvAsbsinα For multiple bars shear resistance increased proportionately Hence V=0.87fyvAsbsinα (d-d’)(cotα+cotβ)/sb In general code requires sb maximum value of 1.5d where β=α=45o Thus V=1.23fyvAsb β α x x Asb Asb Anchorage length d’ d =(d-d’)(cotα+cotβ) sb Note: • Only 50% of shear reinforcements can be used as bent up bars.
- 15. Design of ShearLinks It is important to provide minimum steel in concrete section to prevent thermal & thermal cracks and this minimum steel provide resistance of 0.4 N/mm2. Consequently design links required when v>(vc+0.4) Nominal link required when 0.5vc<v<(vc+0.4) Nominal links are given by Minor important section such as lintel beam where v<05vc link may be omitted. Asv ≥ 0.4 𝑏𝑣𝑠𝑣 0.87𝑓𝑦𝑣 Asv Designed links required vc (vc+0.4) Nominal link provided No links required in members of minor importance v 0.5vc
- 16. Notes Shear reinforcements • Carriesshear stress (vc) • Improve shear strength of concrete (links improve the dowel action of longitudinal bars, bent up bars prevent opening up diagonal cracks) Links also contain the compression reinforcement preventing buckling. Links confines the concrete improving its strength. • Code specify link spacing should not exceed 0.75d to be effective in preventing diagonal cracks. • Leg spacing; any tension bar should not away more than 150mm or d(whichever lesser) to ensure dowel action of longitudinal bars.