BAU-RCIII-Lecture 1
S h e a r Fr Fr i c t i o n
GENERAL CONSIDERATIONS
Reinforcement details for precast concrete structures, and other situations where it is considered appropriate to investigate shear transfer across a plane in structural concrete.
An interface between concrete and steel
Shear friction concept applications
Where concrete is cast against concrete not intentionally roughened
Provisions of the shear-friction are to be applied where it is appropriate to consider shear transfer across a given plane, such as: an existing or potential crack, an interface between dissimilar materials, or an interface between two concretes cast at different times.
The concept is simple to apply and allows the designer to visualize the structural action within the member or joint. The approach is to assume that a crack has formed at an expected location, as illustrated in Fig. 1-1. As slip begins to occur along the crack, the roughness of the crack surface forces the opposing faces of the crack to separate. This separation is resisted by reinforcement (Avf) across the assumed crack.
Nu
Vu
Vu = RusinαF + TucosαF Nu = TusinαF - RucosαF
Shear-friction design is to be used where direct shear is being transferred across a given plane. Situations where shear-friction design is appropriate include the interface between concretes cast at different times, an interface between concrete and steel, and connections of precast constructions, etc. Example locations of direct shear transfer and potential cracks for application of the shear-friction concept are shown in Fig. 16-2 for several types of members.
Fig. 1-1, Applications of the Shear-Friction Concept and Potential Crack Locations
Required shear-transfer strength = Design shear-transfer strength The required area of shear-friction reinforcement, can be computed directly from ɸ = 0.75
1-2
For shear reinforcement inclined to the crack, the required area of shear-friction reinforcement can be computed directly from
Ac=bwd=concrete section resisting shear transfer
The shear friction equations assume that there are no forces other than shear acting on the shear plane. A certain amount of moment is almost always present in brackets, corbels, and other connections due to eccentricity of loads or applied moments at connections. Therefore, it is recommended, although not generally required, that the member be designed for a minimum direct tensile force of at least 0.2Ru in addition to the shear. Assuming the direct tension perpendicular to the assumed crack (shear plane) then the , Sin( f ) α
Tu
0.2Ru
For the beam support shown, design for shear transfer across the potential crack plane. Assume a crack at an angle of about 20 degrees to the vertical, as shown below. Beam reactions are DL = 110 kN, LL= 130 kN. Use T = 90 kN as an estimate of shrinkage and temperature change effects. Concrete was cast monolithically. f’c=25MPa (Normal weight) and fy=400Mpa 200mm 100mm 25mm
400mm
Required: Design for shear transfer across the potential crack plane. i.e required reinforcement and checking SOLUTION MATERIAL Concrete cylinder strength = 25Mpa Steel grade = Grade 400Mpa DIMENSION b = 400 mm FACTORED LOADS
LIMITATION CHECK Potential crack length=125/sin20=365.5mm Ac=(l)(b)=(365.5)(400)=146000mm2 1) 0.2f’c=5Mpa (Controls) 2) 3.3+0.08f ’c=5.3Mpa 3) 11Mpa then øVn=(0.75)(0.2f’c) (Ac)=547.5kN øVn>Vu, Design can be proceeded in the next slide
Ru=1.2DL+1.6LL=1.2(110)+1.6(130)=340kN Tu due to temperature and shrinkage effects, Tu=1.6(90)=144kN (controls) Tu should not be less than 0.2Ru=0.2(340)=68kN EVALUATE FORCE CONDITION ALONG POTENTIAL CRACK PLAN
Direct shear transfer force along shear plan, Vu=Ru sinα+T =340 sin(70)+144 (70)=368.75kN, see next figure
Net tension or compression across shear plan, Nu=Tu sinα- Ru cosα=144 sin (70)-340 cos(70)= 19kN, SHEAR FRICTION REINFORCEMENT TO RESIST DIRECT SHEAR TRANSFER
μ=1.4λ=1.4
Therefore Avf =741.5mm2 REINFORCEMENT TO RESIST NET TENSION
Therefore An=63.3mm2 Sin( ) αf
67.4
TOTAL AREA OF REQUIRED REINFORCEMENT 809
As=Avf +An=804.8mm2 Distribute the above reinforcement uniformly along the potential crack plane Required number of ties=804.8/[(2(71)]=5.66, say 6 ties So use No. 10 closed ties ( 2 legs per tie) 5.7 Ties should be distributed along length of potential crack plane;
αf = α
CHECK REINFORCEMENT REQUIREMENTS FOR DEAD LOAD ONLY PLUS SHRINKAGE AND TEMPERATURE EFFECTS TO MAXIMIZE NET TENSION ACROSS SHEAR PLANE.
75mm
Use 0.9 load factor for DL Ru=0.9DL=0.9(110)=99kN Vu=99 sin(70)+144 cos(70)=142.3kN Nu=144 sin (70)-99 cos(70)= 101.5kN 360 Therefore, Avf =310.3mm2 and An=338.2mm2
Then, As=Avf +An=648.5mm2<804.8mm2 OK 670
m m
5 . 3 4 3
6 No. 10 closed ties spaced at 60 mm o.c
