5-shear, anchorage & hold down reinforcement

8/12/2019 5-Shear, Anchorage & Hold Down Reinforcement

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201- 1-

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201- 2-

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Shear Capacity Checkdp =+++tdeck thaunch ytop_nc em 59.989inde =dp 59.989in since As=0 - No mild rf provided1 0.85 for =f'cdeck 4ksi

k 0.28 for low relaxation strands Astr 0.217in 2b =wdeck 96in h =++hgirder tdeck thaunch 65.12in=c 8.517in =wweb 7.09in=a 7.239in Lbearing 10in

Shear depth @ all sections after

10ft. into the girder

dv =max

,,dp

a

2 0.9 de 0.72 h

4.697ft

=dv 56.369inShear width (min. web width at point of shear calculation)bv =wweb 7.09in(remains constant along entire length of the girder)

CRITICAL SECTION

fpo =0.7 fpu 189ksi 1st Iteration

Assume the length of critical section = =dv 4.697ftSince the shear and moment plots start from the end of the girder, and the cracks initiate from the edge ofthe bearing plate, the length of the bearing plate is added to the length of the critical section calculated

above to find the effective length of the critical section :

Lcr =+dv Lbearing 5.531ftMu@Lcr 2695.52kip ft

Vu@Lcr 357.96kipNu 0kipVp =ndrap Astr fpe slope 29.027kip

=Vu@Lcr Vp dv 1545.145 ip ft>Mu Vu@Lcr Vp dv OK.

Net tensile strain at centroid of tensile rf

s =

++Mu@Lcr

dv0.5 Nu Vu@Lcr Vp Aps fpo

+Aps Ep As Ep0.00308

Assume s 0shear crack angle cr =+29 g 3500 s g 29 g


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Lcr2 =+0.5 dv cot cr Lbearing 5.071ft 2nd Iteration

Using =Lcr2 5.071ftMu@Lcr2 2820.74kip ft Vu@Lcr2 359.72kip

s =

++Mu@Lcr2dv

0.5 Nu 0.5 Vu@Lcr2 Vp Aps fpo

+Aps Ep As Ep0.0036

Assume s 0cr =+29 g 3500 s g 29 g

Lcr3 =+0.5 dv cot cr Lbearing 5.071ft

Length of critical section Lcr =+max ,dv 0.5 dv cot cr Lbearing 5.531ftAngle of Shear crack @ Lcr =cr 29 g

Ultimate Shear at =Lcr 5.531ft Vu_Lcr 357.96kipUltimate Moment at =Lcr 5.531ft Mu_Lcr 2695.52kip ft1) Concrete Shear Capacity at Critical Section:

Assuming shear rf provided is more than min. reqd.

ecr 13.95in eccentricity of strands at critical lengthdp =+++tdeck thaunch ytop_nc ecr 52.78indv =dp

a

2 49.16in

=4.8

+1 750 s4.8

Vc_Lcr =0.0316 f'c ip

in 2 bv dv 149.532kip 0.9

= Vc_Lcr 134.579kip

Vs_reqd =Vu_Lcr +Vc_Lcr Vp

219.175kip

Vs Av

sfy dv cot cr

Using #4 bars - two legs, A4 0.2in 2 Av =2 A4 0.4in 2

s =Av fy dv cot cr

Vs_reqd9.711in

Use s 9in


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Vs_prov =Av fy dv cot cr

s236.5kip

Vn = ++Vs_prov Vc_Lcr Vp 373.553kip > =Vu_Lcr 357.96kip< =+0.25 f'c bv dv Vp 726.119kip

Minimum shear reinf required:

Av_min =0.0316 f'ckipin 2 bv

s

fy0.095in 2

Provided = =Av 0.4in 2 O.KMaximum Spacing of shear reinf.:

0.9

vu =Vu_Lcr Vp

bv dv1.058ksi =0.125 f'c 1ksi

< ,vu 0.125 f'c smax =0.8 dv 39.328insmax 24 in

=s 9in O.KProvide 2 legs of stirrups - #4 bars @ 9 in. c/c spacing upto 6ft into the girder.

Horizontal Shear at Critical Section

The nominal shear resistance of the interfaceplane shall be taken as:

Vn +c Acv +Avf fy Pc Fy 60ksiShear at the abutment due to strength loading:

Vu_Lcr 295.38kipEffective shear depth:

=dv 49.16inInterface Shear Force:

Vui =Vu_Lcr

dv72.102

ipft

Area of interface between girder and deck per foot.

Acv =whaunch 1ft 566.88in2

vui =Vuift

Acv0.127ksi


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Permanent Net Compressive force Pc 0kipcohesion factor: (assumed concrete placed against clean, hardened concrete withsurface not intentionally roughened (smooth)).

c 0.28ksifrom AAHTO 5.8.4.3

friction factor: (for normal weight concrete)

1

use above equation to solve for required area of shear reinforcement:

Avf_reqd =

Vni_reqd c Acv ft 1

Fy1.31

n2ft

Avf_reqd_min =0.05 Acv

Fyksi 1 ft0.472

n2ft

To satisfy the minimum requirement criteria, use the same reinforcement as shear with a horizontal

spacing , extending every alternate bar into the deck.sh

=s 2 18inAvf_prov =

A4 dvshft 0.546

in 2ft

Vni_prov =+c Acvft 1 +Avf_prov Fy Pcft 1 191.5kipft2) Longitudinal reinf check

Transfer length lt =60 db 3ft strands have developed at critical sectionfpe

fps_t =fpe 167.205ksi As_top 0in 2Tn =+Aps fps_t As_top fy 1523.906kipf 1.0 N 0.9 v 0.9

T =++Mu_Lcr

dv f0.5

Nu

N

|

|

Vu_Lcr

vVp

|

|0.5 Vs_prov

cot cr 984.369kip

T

Tn No longitudinal rf reqd.


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3) Anchorage Zone Transverse Reinforcement:

Ldisturbed =hgirder 4.593ftLdisturbed.reinf =0.25 hgirder 13.78in

=Pi 1691.215kip force due to strands before lossesPr.min =0.04 Pi 67.649kip

Web Reinforcement Resistance required :>Pr 0.04 Pi

fs 20ksi maximum allowed

area of web rf reqd

As.reqd =Pr.min

fs3.382in 2

Using #6 bars - 2 legs, A6 0.44in 2As =2 A6 0.88in 2

spacing =As

As.reqdLdisturbed.reinf 3.585in

use spacing 3.5inAs.provided =

As Ldisturbed.reinfspacing

3.465in 2

Web Resistance >Pr =As.provided fs 69.294kip Prmin Provide 2 legs of #6 bars spaced at 3.5 inches c/c upto a length of 14 inches into the

girder, (i.e 4 pairs of #6 bars) followed by the shear reinforcement as stated above

for the length of critical section, i.e 2 legs of stirrups - #4 bars @ 9 in. c/c spacing

upto 12ft into the girder.

4) Shear Reinforcement along beam length (every 12 ft. (L/10))

Since the moment and shear values have the highest range along the halves of the girders connected to

the pier, the shear reinforcement has been designed just for one of the two halves stated above, and the

same shear reinforcement design has been used symmetrically.

Strength 1 MOMENTS: Strength 1 SHEAR:

Mu_12 3251.85kip ft Vu_12 333.20kipMu_24 5659.38kip ft Vu_24 278.76kipMu_36 7244.32kip ft Vu_36 219.89kipMu_48 8063.91kip ft Vu_48 168.06kipMu_60 8134.17kip ft Vu_60 110.39kip

Nu 0kip throughout the girderVp =ndrap Astr fpe slope 29.027kip upto 37.6ft after which Vp=0


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Shear Reinforcement at 12 ft.

e12 15.4in e24 18.1indp =+++tdeck thaunch ytop_nc e12 54.23indv =dp

a

2 50.61in

=Vu_12 Vp dv 1282.858 ip ft>Mu_12 Vu_12 Vp dv OK.


s =

++Mu_12

dv0.5 Nu Vu_12 Vp Aps fpo


Assume s 0 only when has a negative value, as we go ahead along the length of the girder,sbecomes positive at about 36ft. for which the positive value should bes

considered, which results in the crack angle increasing as we near midspan.(sample calculation for this shown later).

shear crack angle =+29 g 3500 s g 29 gAssuming shear rf provided is more than min. reqd.

=4.8

+1 750 s4.8

Vc =0.0316 f'c ip

in 2 bv dv 153.942kip

0.9= Vc 138.548kip

Vs_12.reqd =Vu_12 +Vc Vp

187.253kip

Vs Av

sfy dv cot


s =Av fy dv cot

Vs_12.reqd11.702in

Use s 10inVs_12 =

Av fy dv cot

s219.128kip

Vn = ++Vs_12 Vc Vp 361.888kip > =Vu_12 333.2kip< =+0.25 f'c bv dv Vp 746.68kip


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Av_min =0.0316

f'ckipin 2 bv

s

fy0.106in 2

Provided = =Av 0.4in 2 O.KMaximum Spacing of shear reinf.:

0.9

vu =Vu_12 Vp



=s 10in O.KHorizontal Shear at 12 ft.





Vui =Vu_12

dv79.004

ipft


Acv =whaunch 1ft 566.88in 2

vui =Vuift

Acv0.139ksi


8/11

use Vni equation to solve for required area of shear reinforcement:

Avf_reqd =

Vni_reqd c Acv ft 1

Fy1.182

in 2ft

Avf_reqd_min =

0.05 Acv

Fyksi 1 ft 0.472in 2ft

To satisfy the minimum requirement criteria, use the same reinforcement as shear with a horizontal

spacing , extending every alternate bar into the deck.sh =s 2 20inAvf_prov =

A4 dvshft 0.506

n2ft

Vni_prov =+c Acvft 1 +Avf_prov Fy Pcft 1 189.093ipft

Shear Reinforcement at 48 ft.e48 =em 21.159in

dp =+++tdeck thaunch ytop_nc e48 59.989indv =dp

a

2 56.369in

Vp 0kip=Vu_48 Vp dv 789.452 ip ft

>Mu_48 Vu_48 Vp dv OK.


s =

++Mu_48

dv0.5 Nu Vu_48 Aps fpo


shear crack angle =+29 g 3500 s g 30.993 gAssuming shear rf provided is more than min. reqd.

=4.8

+1 750 s3.364

Vc =0.0316 f'c ipin 2 bv dv 120.152kip 0.9

= Vc 108.137kip Vs_48.reqd =

Vu_48 +Vc Vp

66.582kip

Vs Av

sfy dv cot


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s =Av fy dv cot

Vs_48.reqd33.826in

Use s 24inVs_48 =

Av fy dv cot

s93.841kip

Vn = ++Vs_48 Vc Vp 192.593kip > =Vu_48 168.06kip< =+0.25 f'c bv dv Vp 799.317kip


Av_min =0.0316 f'c ip

in 2 bvs

fy0.253in 2

Provided = =Av

0.4in 2 O.KMaximum Spacing of shear reinf.:

0.9

vu =Vu_48 Vp



=s 24in O.KProvide 2 legs of stirrups - #4 bars @ 10 inch. c/c spacing after 12ft upto 24ft. into the girder.

Using the same procedure and taking Vp=0 after 37.6 ft., the shear reinforcement is determined as:

2 legs of stirrups - #4 bars @ 18 inch. c/c spacing after 24ft upto 36ft. into the girder.

2 legs of stirrups - #4 bars @ 24 inch. c/c spacing after 36ft upto 60ft. into the girder.

Beyond this, the shear reinforcement has a symmetrical layout till the end of the girder, and the

same for the second girder.

Horizontal Shear at 48 ft.





Vui =Vu_48

dv35.777

kipft


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Acv =whaunch 1ft 566.88in 2

vui =Vuift

Acv0.063ksi


11/11

Hold down point force and rebar

=slope 5.73 g angle of draped strands =slope 5.73 gthe hold down points need to be designed for the jacking force

Pdraped =fpj ndrap Astr 351.54kip force along the draped strandsPdraped.normal =Pdraped sin 35.095kip resultant normal force of the draped strands

Phold.down =Pdraped.normal 35.095kiptotal hold down reinforcement area reqd

Ahold.down.rebar =Phold.down

fy0.585in 2

hold down reinf area for each ps strand

Aindividual =Ahold.down.rebar

ndrap

0.073in 2 A3 0.11in 2 area of #3 bar

Provide a hooked/bent down #3 bar at the hold down point for each draped strand

5-shear, anchorage & hold down reinforcement

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