shear analysis and design for shear - avant …avant-garde.engineering/concrete5_notes.pdf · shear...

Shear Analysis and Design for Shear

MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

LECTURE V

Dr. Jason E. Charalambides

Copyright J. CharalambidesCopyright J. Charalambides

Shear Vs Moment

! The main stresses applied on horizontal structural elements are shear and moment. Axial loads do exist but given the fact that gravity greatly surpasses the magnitude loads due to wind or the horizontal component of earthquakes (most of the times), we generally focus more on vertical than lateral loads on most common types of structures.

! Shear and moment go in parallel but the way they act is different; The change in moment (indicated by the slope of the moment diagram) equals the shear at that specific point.

! And, …so is different, the way that our structural members behave on these different types of stresses." Moment allows us to benefit from the ductility of the continuous rebars

within the structural elements. We can see the concrete beams sagging and cracking and we get a sign of failure.

" Shear failure is not that forgiving.! Flexural reinforcement can be considered essentially independent

of reinforcement for shear.


What Does Shear Do And How Do We Address It?

! So far we have seen that we can design an RC beam to perform in a ductile manner in terms of stress and strain.

! Therefore we want to make sure that nothing interferes with this nice behavior we design our beams for. One thing that can bother us in this manner is Shear!


! Stress due to M:

fM yI

Let's Take a Look at Stress


! Stress due to V:

Let's Take a Look at Stress


! Let’s take two points and find the principal stresses occurring:

Combined Stresses


Combined Stresses

! Tensile stresses, that cause cracking, result from combined shear and flexural stresses. Concrete will crack perpendicular to tensile stress trajectories.

! Note: Ask people with lots of field experience to prove that to you!


Where Does Cracking Occur?

! Let’s use this information to predict the cracking pattern of a RC beam w/ no shear reinforcement.


Flexure Theory for Reinforced Concrete! Applying flexural stress to a sample concrete beam:






How Do We Resist/Prevent Cracks?

! What is the resistance of a RC beam to these types of cracks?

Diagonal cracks: (Large V, Small M)" These will form when the shear V on a cross section exceeds about:

(f`c in psi)Flexural cracks developing into diagonal cracks" These will form when the shear V on a cross section exceeds about:

(f`c in psi)! Conclusion:

" Large bending moments can reduce the diagonal cracking resistance of a beam by about 50%.

V c=3.5∗bw∗d∗√ f ' c

V c=1.9∗bw∗d∗√ f ' c


What Happens in Between the Two Extreme Conditions?! Tests have proven that shear resistance of the concrete depends

on:

" the strength of concrete

" the ratio of steel within the concrete (test on t-beams also)

" (dimensionless ratio of shear to moment)

f ' c

ρw=Asbw∗d

V dowel

M


WHAT HAPPENS IN BETWEEN THE TWO EXTREME CONDITIONS?! Vc is nominal shear strength provided by concrete

V c=3.5∗bw∗d∗√ f ' c

V c=1.9∗bw∗d∗√ f ' c

V dowel

M

V c


How Does the ACI Address This?

! According to ACI 318-14, for non axially loaded, non prestressed members eq. 22.5.5.1 applies.

" Vu = factored shear force at section," Mu = factored moment at section

! both of the above occur simultaneously

" also, in eq. 11-6…

V c=[1.9 √ f ' c+2500⋅ρw⋅V u⋅d

M u ]bw⋅d≤3.5⋅bw⋅d⋅√ f ' c

V u⋅dM u

≤1.0

eq. 22.5.5.1

eq. 22.5.5.1


What Could Those Cracks Do to our Structural Element

! We allowed narrow flexural cracks to propagate to the flexural steel. Why do we need to worry about the diagonal cracks?

" The effect is different and we can not compare the two types of cracks. Flexural cracks are blocked by the flexural steel and they can only reach the compression zone anyway. It is the effect of concrete being in tension and cracking.

" Let’s take a look at what happens when we allow diagonal cracks to propagate to the NA.


LETS SEE THE EFFECT ONCE MORE






Notes

! Presence of shear permits cracks to propagate (slightly) into the compression block, making the state of stress of the element more critical.

! Due to inclined crack, the tension force in the longitudinal steel at a distance x1 from the support, must be great enough to resist the (larger) moment which would theoretically exist at a distance x2 from the support. (Increased steel stress)

! Dowel action increases tendency of concrete to split along longitudinal bars.


! If the beam has not been designed taking these issues into account, it will fail when diagonal cracks form. If it has been adequately designed, a brittle failure is still likely:

" The compressive block, subjected to shear and compression, will probably crash earlier than the same block subjected to compression alone.

" Dowel action reduces the chance that the longitudinal bars will develop their yield strength.

Notes


Conclusions

! A RC beam can not be designed for ductile, flexural behavior if diagonal cracks are permitted to propagate.

! Therefore RC beams should not be required to develop shear resistance in the concrete in excess of Vc (ACI 318-05 / 11-6)


Slabs w/out Shear Reinforcement

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

V n=Wu

2

V nc≥2∗bw∗d∗√ f ' c

V nc

bw∗d∗√ f ' c


The chart represents test data from slabs reinforced for flexure only. The abscissa is the ratio L/d and the ordinates are values of the ratio between the limit shear Vn and the result of (bw*d*√f'c). The chart indicates that there is more shear capacity for the shorter 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

Slabs w/out Shear Reinforcement

V nc≥2∗bw∗d∗√ f ' cV nc

bw∗d∗√ f ' c

V nc≥2∗√ f ' cbw∗d


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 will be equivalent to the beam depth “d” divided by the spacing “s”. Thus the strength of stirrups as shear reinforcement becomes:

The capacity safety factor given by ACI section 9.3.2.3 is 0.75. (Appendix C2 gives safety factor 0.85). The total shear strength is the sum of the concrete shear strength and the dowel strength:

Av≥2∗Abar.

V ns=Av∗ f y∗d

sV ns=

Av∗ f y∗ds

ΦV n=ΦV nc+ΦV ns


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

! The actual shear force that will cause the diagonal crack will not be the 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 their bottom side, may experience a crack no closer than distance “d” from the extreme fiber of that support.

! For ACI, maximum design shear force is the shear applied at that distance “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 the distance “d”.



! ACI requires that stirrups are used in all beams that experience shear values exceeding 0.5ΦVnc.– Beams in which bw exceeds 2.5d and joists (conforming to

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


V nc≥2∗√ f ' cbw∗d


! In this diagram, the abscissa represents the distance from the support along the length of the beam.

! The slope corresponds to the applied uniform load, and

! The ordinate represents the value of shear.

Design Stirrups for Shear Reinforcement


! ACI 318/11.2.4 requires that spacing of stirrups is no longer than d/2.! If the shear force that will be resisted by stirrups exceeds 2ΦVnc, or

Vu>6ΦVnc, the maximum distance between stirrups is reduced to d/4.

! The strength ordinate ΦVns2 represents the strength of stirrups at spacing d/2. That ordinate is added to the value of ΦVnc to indicate the total shear capacity ΦVn w/ stirrups at their widest spacing.

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

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


s≤Av∗ f y∗dV u−ΦV nc


! ACI 318-14 / 9.7.6.2.2 determines that maximum spacing between stirrups will be a minimum of 24” or d/2 if the maximum shear does not exceed 4bw*d*√f'c.

! If that maximum exceeds the (4bw*d*√f'c), the maximum spacing becomes the minimum of 12” or d/4


smax=if [(V d≤4⋅bw⋅d⋅√ f ' c) ,min(24in ,d2) ,min (12in , d

4)]

ACI-318-14 eq. 9.7.6.2.2


! ACI 318-14 / 9.6.3.3 determines that maximum spacing between stirrups will be given by the minimum value of the following:

! Where φv is 0.75.! This formula has been translated here to reflect spacing instead of

the way it is represented in the code as a division of Area of steel reinforcement divided by spacing.


smax=min [ Asv⋅f yφv⋅bw⋅√ f ' c

,Asv⋅f y50⋅bw ] ACI-318-14

eq. 9.6.3.3


! Finally, ACI 318-14 / 9.7.6.4.3 determines that maximum spacing between stirrups shall not exceed the minimum of the following three conditions:

" 16 times the largest flexural reinforcement's bar diameter," 48 times the shear reinforcement's bar diameter," Minimum dimension of the element.

! Where " dbmax is the largest diameter of flexural bars." dbv is the bar diameter of the shear reinforcement" d is the effective depth of the beam and bw is the width.


smax=min [16⋅d bmax ,48⋅dbv ,min(d , bw) ]ACI-318-14 eq. 9.7.6.4.3


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 in tension. At the same time, concrete acts as a compressive element along those diagonal lines. ACI 318-11, 11.5.6.8 limits the Vsn to 4Vcn, the stress value under which concrete diagonal struts fail in C. No additional stirrup reinforcement shall increase the shear strength after concrete struts fail.



! The designer should keep in mind the following index values of shear strength:

Stirrups must be used whenever Vu exceeds (Φ√f`c)(bw*d)/2Maximum spacing of stirrups is d/2 unless Vu exceeds (6Φ√f`c)(bw*d),

above which maximum stirrup spacing is set to d/4The section itself must be made larger if Vu exceeds (10Φ√f`c)(bw*d)

! According to ACI318/11.5.5.3, where stirrups are required, the quantity Av/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 shear crack 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.



! Deep and thin beams may not develop their potential.! For non prestressed beams that are not subjected to axial stress,

the nominal factored shear capacity is given by the following formula:

! Because of the above mentioned condition about thin beams, beams deeper than four times their length should contain horizontal shear reinforcement in addition to vertical bars.


ΦV nc=φ⋅2⋅√ f ' c⋅bw

f yACI-318-14 eq. 22.5.5.1


! The design of shear reinforcement requires selection of stirrup size and determination of spacing needed to resist shear. Stirrups size is related to beam size.

" Per ACI-318 14, (9.7.6.4.2), #3 dowels shall suffice for enclosure of #10 or smaller longitudinal bars. Dowels of #4 or #5 should be used for larger longitudinal bars.

! To determine stirrups as shear reinforcement, it is recommended that a diagram is constructed. With abscissa as length and ordinates as shear strength, horizontal lines at the values of 0.5ΦVnc, ΦVnc, ΦVd, and a vertical line at distance “d” will aid in the process.



! Summarizing ACI requirements in formulae:



In Class Example Attn: Solved with previous code

jcharalambides

Line

jcharalambides

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RC Beam Shear Design with recap on flexural designBar Designation

NumberWeight per foot (lbf) Diameter db Area As Perimeter

3 0.376 0.375 0.11 1.1784 0.668 0.500 0.20 1.5715 1.043 0.625 0.31 1.9636 1.502 0.750 0.44 2.3567 2.044 0.875 0.60 2.7498 2.670 1.000 0.79 3.1429 3.400 1.128 1.00 3.54410 4.318 1.270 1.27 3.99011 5.304 1.410 1.56 4.43014 7.650 1.693 2.25 5.31918 13.600 2.257 4.00 7.091

Steel Grade is 60, f`c=4ksi, bw=12in, h=22in, w=4k/ft, L=26ft.Select dowel types for shear and determine appropriate spacing.

wu 4kipft

:=fy 60ksi:= f'c 4ksi:= bw 12in:= h 22in≡ cover 2.5in≡

L 26ft:= d h cover− 19.5 in⋅=:= ϕv .75:= Es 29000ksi:=

Muwu L2

⋅

8:= Mu 338 k'⋅=

Estimating area of steel:

AsMu4d

= As_est 4.33 in2⋅= Abar1 .79in2

≡ Abar2 1.56in2≡

Select Number & size of rebars: nbar1 0:= nbar2 3:=Take two #8 and two #11 bars for tensile

bar1dia 1in≡ bar2dia 1.41in≡As Abar1 nbar1⋅( ) Abar2 nbar2⋅( )+:= As 4.68 in2

⋅=

Largest bar diameter selected:

db_max max bar1dianbar1nbar1

⋅ bar2dianbar2nbar2

⋅,

:= db_max 1.41 in⋅=

sidecover 1.5in:=

sbw nbar1 bar1dia⋅ nbar2 bar2dia⋅+( )− 2 sidecover⋅−

nbar1 nbar2+ 1−( ):= s 2.385 in⋅=

Condition_smin. if s max 1in bar1dia, bar2dia, ( )< "redesign", "Ok", ( ):=

ρAs

bw d⋅:= ρ 0.02= Condition_smin. "Ok"=

ρmin max200psi

fy

3 f'cfy

,

= ρmin 0.0033=

ρmin_condition if ρmin ρ< "OK", "Not acceptable", ( ):= ρmin_condition "OK"=

Determining maximum tension for rebars:T As fy⋅:= T 280.8 kip⋅=

Determining concrete's a:

aT

.85 f'c⋅ bw⋅:= a 6.88 in⋅=

ad

0.35= This is too high. We areaiming for 19.6%, not 35%!

caβ1

:= β1 0.85=

c 8.1 in⋅=

εt .003dc

1−

⋅:= εt 0.0042= εtyfyEs

:= εty 0.00207=

This is a dangerous design. The Φ factor is definitely not 0.9. For the sake of argument we can calculate the Φfactor and calculate what the ΦMn would be anyway. The option of using 0.65 is always available, but let's seewhat happens with the calculated value of Φ

ϕ 0.9 .65 0.25εt εty−( )

0.005 εty−( )⋅+≥ 0.65≥= ϕ 0.8339=

Determining Maximum Moment: note: Make sure that ΦMn>Mu initialΦMn ϕ T⋅ d

a2

−

⋅:= ΦMn 313.359 k'⋅=

Strength if ΦMn Mu≥ "OK", "Not acceptable", ( ):= Strength "Not acceptable"=

Efficiency if ΦMn Mu 1.1⋅≥ "Overdesigned", "Not Overdesigned", ( ):= Efficiency "Not Overdesigned"=

Essentially that Φ factor reduced the capacity of the beam below the Mu. It is prudent to use compressive rebars. Let's proceedwith compressive steel and the process in which the flexural stress f's equals the yield strength fy, omitting the scenario where the f'sdoes not equal fy for reasons of simplicity.

Proceeding with compressive steel design:

ρ10.196 f'c⋅ β1⋅

fy:= ρ10 0.0111=

A's_ini As ρ10 bw⋅ d⋅−:= A's_ini 2.08 in2⋅=

We can assume that two #9 rebars can satisfy our needs for compressive steel:

A's 2 1⋅ in2:= A's 2 in2

⋅=

As1 As A's−:= As1 2.68 in2⋅=

T1 As1 fy⋅:= T1 160.8 kip⋅=

a1T1

0.85 f'c⋅ bw⋅:= a1 3.941 in⋅=

c1a1β1

:= c1 4.63668 in⋅=

εt1 .003dc1

1−

⋅:= εt1 0.0096= Fair enough!

ΦMn1 .9T1 da12

−

⋅:= ΦMn1 2536.856 k''⋅= Or ΦMn1 211.405 k'⋅=

T2 A's fy⋅:= T2 120 kip⋅=

ΦMn2 .9T2 h 2cover−( )⋅:= ΦMn2 1836 k''⋅= Or ΦMn2 153 k'⋅=

ΦMn ΦMn1 ΦMn2+:= ΦMn 4372.856 k''⋅= Or ΦMn 364.405 k'⋅=

Strength if ΦMn Mu≥ "OK", "Not acceptable", ( ):= Strength "OK"=

Efficiency if ΦMn Mu 1.1⋅≥ "Overdesigned", "Not Overdesigned", ( ):= Efficiency "Not Overdesigned"=

Estimating area of steel for Shear:

Vu .5wu L⋅:= Vu 52 kip⋅=

Asv if db_max 1.27in≤ .22in2, .4in2

,

:= Asv 0.4 in2

⋅=

dbv if Asv 0.22in2= 0.375in, .5in,

:= dbv 0.5 in⋅=

Vd Vu d wu⋅−:= Vd 45.5 kip⋅=

ΦVc ϕv 2⋅ bw⋅ d⋅ f'c⋅= ΦVc 22.2 kip⋅=ΦVc

211.1 kip⋅= Vc

ΦVcϕv

29.6 kip⋅=:=

Vstir Vd ΦVc−:= Vstir 23.3 kip⋅=

Establish need for reinforcement and that other ACI codes are met on side within supports :

CondACI_318_14_9631 if .5ΦVc Vd( )< "Reinf. reqd", "no dowels", := CondACI_318_14_9631 "Reinf. reqd"=

Calculate maximum spacing allowed by codes in region within supports:

ACI 318-14, 9.7.6.2.2

smax9622 if Vd 4 bw⋅ d⋅ f'c( )⋅≤ min 24ind2

,

, min 12ind4

,

,

= smax9622 9.8 in⋅=

ACI 318-14, 9.6.3.3

smax9633 minAsv fy⋅( )

ϕv f'c( )⋅ bw⋅

Asv fy⋅( )50 bw⋅

,

=smax9633 40 in⋅=

ACI 318-14, 9.7.6.4.3

smax97643 min 16 db_max⋅ 48 dbv⋅, min d bw, ( ), ( ):= smax97643 12 in⋅=

smax trunc min smax9622 smax9633, smax97643, ( )( )= smax 9 in⋅=

ACI 318-14, 22.5.1.2

Condition22512 if Vd ϕv Vc 8 10⋅ bw d⋅ f'c( )⋅+ ⋅> "Resize element", "Proceed with design", =

Condition22512 "Proceed with design"=

Establish values of shear capacity based on concrete capacity in conjunction with steel and varying spacing:

ΦVsmin Asv ϕv⋅ fy⋅d

smax⋅:= ΦVsmin 39 kip⋅= Using the largest allowable spacing

ΦVnmin ΦVsmin ΦVc+:= ΦVnmin 61.2 kip⋅=

ΦVn9 Asv ϕv⋅ fy⋅d

9 in⋅⋅

ΦVc+:= ΦVn9 61.2 kip⋅= Vd 45.5 kip⋅=

ΦVn8 Asv ϕv⋅ fy⋅d

8 in⋅⋅

ΦVc+:= ΦVn8 66.1 kip⋅= No need for this. These values are simply added but given that the d/2relation established a maximum spacing that already satisfies themaximum shear, it is not necessary to use any other spacing. There willbe a ontinuous spacing at 9 inches. ΦVn6 Asv ϕv⋅ fy⋅

d6 in⋅

⋅

ΦVc+:= ΦVn6 80.7 kip⋅=

Estimating number of stirrups: Note that these are not ACI formulae. These are simple mathematicalrelations that were assembled by the instructor to calculate distances andnumber of stirrups.

ReinfspanVu .5ΦVc−( )

wu:= Reinfspan 122.7 in⋅= Reinfspan 10.23 ft⋅=

nstir roundReinfspan trunc

smax2in

in−

smax

1+:= nstir 14=

Initially we added an extra space and then we subtracted a truncated value of half a spacing from the total distance that needed to becovered with stirrups. That was done to establish that the first stirrup would be placed at a half spacing distance and then we wouldcontinue with the regular spacing. Then the remaining length that would be covered with stirrups was divided by the spacing, and thevalue was rounded to an integer. It is easier to do this calculation with a calculator than the above process because rounding is not ideal, but a formula including "if"conditions could become somewhat complicated to visualize.

shear analysis and design for shear - avant …avant-garde.engineering/concrete5_notes.pdf · shear...

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