rc slab shearing_d

8/21/2019 RC Slab Shearing_D

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Design For Shear

UNIVERSITY OF WISCONSIN STOUTCOLLEGE OF SCIENCE, TECHNOLOGY, ENGINEERING, AND MATHEMATICS

LECTURE VII

Dr. Jason E. Charalambides

Slabs Without Shear Reinforcement

! Shear reinforcement is a rarity for slabs. Here we use a slab toillustrate the capacity of concrete to sustain shear w/out shearreinforcement.


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Slabs Without Shear Reinforcement

The chart represents test data from slabs reinforced for flexure only. The abscissa is theratio L/d and the ordinates are values of the ratio between the limit shear Vn and theresult of (bw*d*!f`c). The chart indicates that there is more shear capacity for theshorter rather than the longer spans of an element with same thickness and width. If spans are more than 4 to 6 times the depth of the slab, nominal shear strength of concrete can be given by the following formula:

" Note: ACI318 (11.3.2.1) can yield a value gain of 5%-15% in regions near supports,but due to its relative complexity and the limited gain, it is not used frequently

Design For Shear Reinforcement! When a shear failure mechanism is taken

along a crack at 45˚, the number of stirrups that will intercept the crack willbe equivalent to the beam depth “d”divided by the spacing “s”. Thus thestrength of stirrups as shearreinforcement becomes:

! The capacity safety factor given by ACIsection 9.3.2.3 is 0.75. (Appendix C2gives safety factor 0.85). The total shear

strength is the sum of the concrete shearstrength and the dowel strength:


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Design For Shear Reinforcement!

Shear reaches maximum at the points of support. If we can assumethat the support provides compression at the bottom of the beam,diagonal cracks will begin as flexural cracks (remember last lecture) atdistance “d” or further from the points of support.

! The actual shear force that will cause the diagonal crack will not bethe shear at the face of the column, but the shear at that distance “d”.

! Slabs and beams that are supported by elements deep enough(columns or deep beams) that can apply a compression force at theirbottom side, may experience a crack no closer than distance “d” fromthe extreme fiber of that support.

! For ACI, maximum design shear force is the shear applied at thatdistance “d”, where a 45˚ crack may lead toward the top of the beam.Stirrups need to be placed at the face of the support through thedistance “d”.

Design For Shear Reinforcement

! ACI requires that stirrups are used in all beams that

experience shear values exceeding 0.5"Vnc." Beams in which bw exceeds 2.5d and joists (conforming to

ACI318/8.11 definition of joist) are to be exempted from thisbasic design requirement for shear. In fact a 10% gain can be givento the strength of joists over the basic equation:


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Design Stirrups For Shear Reinforcement

! In this diagram,the abscissarepresents thedistance fromthe supportalong the lengthof the beam.

! The slopecorresponds tothe applieduniform load,and

! The ordinaterepresents the

value of shear.

Design Stirrups For Shear Reinforcement! ACI318/11.2.4 requires that spacing of stirrups is no longer than d/2.

! If the shear force that will be resisted by stirrups exceeds 2"Vnc, orVu>6"Vnc, the maximum distance between stirrups is reduced tod/4.

! The strength ordinate "Vns2 represents the strength of stirrups atspacing d/2. That ordinate is added to the value of "Vnc to indicatethe total shear capacity "Vn w/ stirrups at their widest spacing.

! Stirrups should be placed at closer intervals from the face of thesupport to the point where maximum-spaced stirrups are adequate.

! Stirrup spacing at any location where Vu is known is given by thefollowing formula:


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Design Stirrups For Shear Reinforcement!

Let’s recall these stresses from the previous session:

! Take a look at points 1 & 2. We can picture the action of the stirrups being intension. At the same time, concrete acts as a compressive element along thosediagonal lines. ACI318/11.5.6.8 limits the Vsn to 4Vcn, the stress value underwhich concrete diagonal struts fail in C. No additional stirrup reinforcementshall increase the shear strength after concrete struts fail.

Design Stirrups For Shear Reinforcement! The designer should keep in mind the following index values of shear

strength:

Stirrups must be used whenever Vu exceeds ("!f`c)(bw*d)/2

Maximum spacing of stirrups is d/2 unless Vu exceeds (6"!f`c)(bw*d), above whichmaximum stirrup spacing is set to d/4

The section itself must be made larger if Vu exceeds (10"!f`c)(bw*d)

According to ACI318/11.5.5.3, where stirrups are required, the quantityAv/s must be greater than (.75!f`c)*(bw/fy)>50(bw/fy).

!

This clause that is intended to assure stirrups shall not yield after a shearcrack develops as they pin together, and it may reduce the spacing “s” of stirrups to less than d/2 if beams are large or wide.


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Design Stirrups For Shear Reinforcement!

Deep and thin beams may not develop their potential Vcn=(2ef`c)(bw/fy).

! Beams deeper than four times their length should contain horizontalshear reinforcement in addition to vertical bars.

! The design of shear reinforcement requires selection of stirrup sizeand determination of spacing needed to resist shear. Stirrups size isrelated to beam size. If bw*d


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In Class Example (intro):

! Steel Grade is 60, f`c=4ksi, bw=20in,d=19.5", L=26`, Construct a Shearstrength diagram.

In Class Example cont:


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Deep Beams

! The definition of “deep beams” refers to beams that have a length/depth ratiohigher than 4 or they assume a significant concentrated load at a location twotimes the member’s depth from the support.

! For deep beams shear reinforcement must incorporate horizontal as well asvertical re-bars.

! Concrete compressed as a shear strut must be “confined” laterally byreinforcement with a density Avi*(sin#)/bsi in each direction such that thesum of both densities exceeds the quantity 0.003. The diagonal angle # istaken w/ respect to the bar (horizontal or vertical).

! The compression strut area Acs has a width b and a depth that may beconsidered to increase at a rate equal to the distance along the strut fromthe center of the nodal point.

! The strength of the concrete struts is 0.85*f`c*Acs.! Also, CCC nodal points at intersection of 3 struts must possess the same

compression strength limit of 80% or 0.65f`cAcs. The value $=0.75 appliesfor shear.

Deep Beams(ACI Definitions)! 11.8.1. – Deep beams are all beams loaded on one face and supported on

the opposite face and for which 4dPln or for which a significantconcentration force acts within 2d of a support.

! 11.8.2. – Deep beams shall be designed using Appendix A. (Strut & Tie Model)! 11.8.3. – Vn0.0025*bw*s with s0.0015*bw*s2 with s2


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In Class Example:

In Class Example:


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Corbel Design

!

Not covered in class.

Shear Combined WithTorsion

! Not covered in class.


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! Reading:" Required:

! Furlong Chapter 6" Recommended:

! McCormac & Nelson, Chapter 8 for this week’s lectures.

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Assignments will be received at the beginning of class period.

! Assignment 6 is due next session (not in a week for this time.)


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Clarification On Textbook !

Q: You state that d/2 is the maximum spacing and itbecomes d/4 if our Vu is larger than6*"*!f`c*(bw*d). Could you please verify this?

! A: The text you quoted is correct. When stress V/bw*d exceeds 4!fc', spacing must be less than d/2.Thus when Vu /(bw*d) exceeds Vc plus 4 Vs itexceeds 6!fc'.

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