rciii 02 shear friction

7/24/2019 RCIII 02 Shear Friction

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Shear Fr ic t ion

BAU-RCIII-Lecture 1

Dr. Zaher Abou SalehBAU-Dept. of Civil

Engineering-DebbiehCampus-RCIII-Lecture 1 1


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Reinforcement details for precast

concrete structures, and other

situations where it is consideredappropriate to investigate shear

transfer across a plane in structural

concrete.

Where concrete is

cast against concrete

not intentionally

roughened

An interface

between concrete

and steel

Shear friction

concept

applications

GENERAL CONSIDERATIONS

2

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.


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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 thecrack to separate. This separation is resisted by reinforcement (Avf) across the

assumed crack.

Fig. 1-1, Idealization of the Shear-Friction Concept

Vu= RusinF+ TucosF

Nu= T

usin

F- R

ucos

F

Vu

Nu


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Required shear-transfer strength = Design shear-transfer strength

= 0.75

The required area of shear-friction

reinforcement, can be computed directly from

For shear reinforcement inclined to the crack,

the required area of shear-friction

reinforcement can be computed directly from

Fig. 1-2

1-2


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Ac=bwd=concrete section resisting shear transfer


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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)

Tu 0.2Ru


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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 andtemperature change effects. Concrete was cast monolithically.

fc=25MPa (Normal weight) and fy=400Mpa

25mm

200mm

100mm

400mm


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Required:

Design for shear transfer across the potential crack plane. i.e required reinforcement

and checking

SOLUTION

MATERIAL LIMITATION CHECK

Concrete cylinder strength = 25Mpa Potential crack length=125/sin20=365.5mm

Steel grade = Grade 400Mpa Ac=(l)(b)=(365.5)(400)=146000mm2

1) 0.2fc=5Mpa (Controls)

DIMENSION 2) 3.3+0.08fc=5.3Mpa

b = 400 mm 3) 11Mpathen Vn=(0.75)(0.2fc) (Ac)=547.5kN

FACTORED LOADS Vn>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 sin+Tu cos =340 sin(70)+144 cos(70)=368.75kN, see next figure


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Net tension or compression across shear plan,

Nu=Tu sin- Ru cos=144 sin (70)-340 cos(70)= 19kN,

SHEAR FRICTION REINFORCEMENT TO RESIST

DIRECT SHEAR TRANSFER

f= =1.4=1.4

Therefore Avf=741.5mm2

REINFORCEMENT TO RESIST NET TENSION

Therefore An=63.3mm2

TOTAL AREA OF REQUIRED REINFORCEMENT

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)

Ties should be distributed along length of potential crack plane;

approximate length=125/tan20=343.5mm, see next figure

Sin(f)67.4

809

5.7


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6 No. 10 closed

ties spaced at

60 mm o.c

75mm

343.

5mm

CHECK REINFORCEMENT REQUIREMENTS FOR

DEAD LOAD ONLY PLUS SHRINKAGE AND

TEMPERATURE EFFECTS TO MAXIMIZE NET

TENSION ACROSS SHEAR PLANE.

Dr. Zaher Abou SalehBAU-Dept. of Civil

Engineering-DebbiehCampus-RCIII-Lecture 1

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

Therefore,

Avf=310.3mm2 and An=338.2mm

2

Then, As=Avf+An=648.5mm2

rciii 02 shear friction

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