mathcad - rc pier design

RC Pier Design.mcd Program

Reference:H:\References\Concrete\MathCad\Utilities.mcd(R)

A. INTRODUCTION

A.1 FOR UNDERGROUND PIPES, AT LOCATIONS WHERE THE PIPE EXITS FROM OR ENTERS TO THE GROUND, CONCRETEANCHOR BLOCKS SHALL BE PROVIDED IN ORDER O RESIT THER HORIZONTAL FORCE CAUSED BY THE PIPE.

A.2 THE HORIZONTAL FORE CAUSED BY THE PIPE IS RESISTED BY THE FOLLOWING:- FRICTION FORCE BETWEEN THE CONCRETE BLOCK AND THE GROUND.- PASSIVE EARTH PRESSURE AGAINST THE CONCRETE BLOCK.

A.3 ANCHOR BLOCKS USUALLY HAVE A RECTANGULAR PLAN BUT FOR LARGE FORCES H-SHAPE PLANS MAY BE USED IN ORDERTO INCREASE THE PASSIVE EARTH PRESSURE.

A.4 NO TENSION IS ALLOWED WHEN COMPUTING THE BEARING PRESSURE UNDER THE BLOCK.

A.5 THE FACTOR OF SAFETY AGAINST SLIDING SHALL NOT BE LESS THAN 1.20

A.6 THE FACTOR OF SAFETY AGAINST OVERTURNING SHALL NOT BE LESS THAN 1.50

C. INPUT PARAMETERS

MATERIAL PROPERTIES

CONCRETE COMPRESSIVE STRENGTH fc 4ksi:=

kcfkips

ft3:=

UNIT WEIGHT OF CONCRETE γc 0.15kcf:=

TENSILE STRENGTH OF REINFORCING BARS fy 60ksi:=

MODULUS OF ELASTICITY Es 29000ksi:=

CONCRETE COVER Cc 1.5in:=

PIER DIMENSIONS

PIER WIDTH bx 36in:=

PIER LENGTH by 36in:=

PIER HEIGHT hpier 4ft:=

MATH/FOOTING THICKNESS T 8ft:=

CONCRETE STRENGTH REDUCTION FACTORS [ACI 318-05, Sec. 9.3]

BENDING/TENSION ϕt 0.9:=

COMPRESSION ϕc 0.65:=

SHEAR ϕs 0.75:=

BEARING ϕb 0.65:=

MODIFICATION FACTOR λ 1.0:= NORMAL WEIGHT CONCRETE (ACI 318-05, Sec. 11.7.4.3)

B. DESIGN LOADS

MAX TENSION Tu 100− kip:=

MAX COMPRESSION Pu 1000kip:=

SHEAR ALONG X Vux 30kip:=

SHEAR ALONG Y Vuy 30kip:=

RC Pier Design.mcd LNT - of 7


TOTAL SHEARS VuTotal Vux2 Vuy

2+:= VuTotal 42.43kip=

MOMENT, Mux Mux 10kip ft:=

MOMENT, Muy Muy 20kip ft:=

C. RECTANGULAR PIER ANALYSIS AND DESIGN

FACTORED LOADS ACTING AT BOTTOM OF PIER

MAX COMPRESSION Pub Pu LF γc⋅ bx⋅ by⋅ hpier⋅+:= Pub 4481.85kN=

CONSERVATIVELY USE A LOAD FACTOR LF 1.4≡ FOR PIERSELFWEIGHT IN STRENGTH DESIGN

MOMENT ABOUT X Mux Mux Vuy hpier⋅+:= Mux 176.26kN m=

MOMENT ABOUT Y Muy Muy Vux hpier⋅+:= Muy 189.81kN m=

MAX SHEAR LIMIT BY CROSS SECTION PROVIDED

ASSUME TO USE A FACTOR 0.9 TO CALCULATE THE EFFECTIVE SECTION DEPTH. IT CAN BERE-EVALUATED AFTER REBAR DESIGN IF NECESSARY

dx 0.9bx:= dx 0.823m=

ϕVys.max ϕs 0.67⋅ fc MPa⋅⋅ dx⋅ by⋅:= ϕVys.max 446.431kip=

ϕVys.max >=? Vuy "YES!.. SATISFACTORY"=

dy 0.9by:= dy 0.823m=

ϕVxs.max ϕs 0.67⋅ fc MPa⋅⋅ dy⋅ bx⋅:= ϕVxs.max 446.431kip=

ϕVxs.max >=? Vux "YES!.. SATISFACTORY"=

CHECK MINIMUM/MAXIMUM REINFORCEMENT REQUIREMENTS

MINIMUM REINFORCEMENT RATIO ρmin 0.005:= [ACI 318-05, Sec. 10.8.4,10.9.1 & 15.8.2.1]

MINIMUM REINFORCEMENT As.min ρmin bx⋅ by⋅:=

As.min 6.48in2=

MINIMUM REINFORCEMENT RATIO ρmax 0.04:= [ACI 318-05, 10.9.1]

ALTHOUGH 0.08 IS ALLOWED BY THE CODE, PRACTICALLY0.04 MAY BE USED.

MAXIMUM REINFORCEMENT As.max ρmax bx⋅ by⋅:=

As.max 51.84 in2=



VERTICAL BAR SIZE bar 7:=

BAR DIAMETER dbbar 0.866 in=

BAR AREA Abbar 0.589 in2=

TIE BAR SIZE tie 4:=

TIE DIAMETER dbtie 0.472 in=

TIE BAR AREA Abtie 0.175 in2=

MINIMUM TIE SIZE dtie.min38

in bar 10≤if

48

in bar 10>if

:= dtie.min 0.375 in=

dbtie >=? dtie.min "YES!.. SATISFACTORY"=

BAR ARRANGEMENT (PER FACE)

NUMBER OF BARS IN A1 n1 7:= A1 n1 Abbar⋅:= A1 4.123in2=

NUMBER OF BARS IN A2 n2 5:= A2 n2 Abbar⋅:= A2 2.945in2=

NUMBER OF BARS n 2 n1 n2+( ):= n 24=

AREA OF STEEL PROVIDED As n Abbar⋅:= As 14.136in2=

As >=? As.min "YES!.. SATISFACTORY"=

As <=? As.max "YES!.. SATISFACTORY"=

D. CHECK PIER FOR APPLIED LOADS

DISTANCE TO CENTER OF TENSION REBARS

DISTANCE TO CENTER OF TENSION REBAR FROMTHE CONCRETE EDGE IN X-DIRECTION

dx bx Cc− dbtie−12

dbbar⋅−:= dx 33.594in=

DISTANCE TO CENTER OF TENSION REBAR FROMTHE CONCRETE EDGE IN X-DIRECTION

dy by Cc− dbtie−12

dbbar⋅−:= dy 33.594in=

PIER COMPRESSION CAPACITY

Fac minAs

2 As.min⋅1.0,

⎛⎜⎝

⎞⎟⎠

:= Fac 1= Fac REDUCES THE AREA OF CONCRETEWHE Pmin IS LESS THAN 0.01

[ACI 318-05 Sec 10.8.4 &10.9.1]

ϕPn 0.8 ϕc⋅ 0.85fc Fac⋅ bx by⋅ As−( )⋅⎡⎣ ⎤⎦⋅:= ϕPn 2266.34kip= [ACI 318-05 Eqn 10-2]

Pub <=? ϕPn "YES!.. SATISFACTORY"=



PIER BEARING CAPACITY (AT BASE OF PIER) PER ACI 318-05 Sec 10.17.1

ϕPbrg ϕb 0.85⋅ fc bx⋅ by⋅:= ϕPbrg 2864.16kip= [ACI 318-05 Eqn 10-2]

Pub <=? ϕPbrg "YES!.. SATISFACTORY"=

PIER TENSION CAPACITY (USE REINFORCING BARS AS DOWELS TO RESIST TENSION)

ϕTn ϕt As⋅ fy⋅:= ϕTn 763.35kip=

Tub Tu 0.9 γc⋅ bx⋅ by⋅ hpier⋅+:= Tub 95.14− kip=

Tub <=? ϕTn "YES!.. SATISFACTORY"=

AREA OF STEEL REQUIRED TO RESIST TENSION

Ast Asmax Tub− 0kips, ( )

ϕTn⋅:= Ast 1.762 in2=

REINFORCEMENT REQUIRED TO TRANSFER HORIZONTAL FORCE TO FOOTING (ACI 318-05 Sec 11.7.4)

SHEAR FRICTION COEFFICIENT μ 1.0 λ⋅:= (ASSUMED ROUGHENEDSURFACE)

[ACI 318-05 Sec 11.7.4.3]

AvfVuTotal

ϕs μ⋅ fy⋅:= Avf 0.943 in2=

Avf Ast+( ) <=? As "YES!.. SATISFACTORY"=

PIER BENDING CAPACITY

CALCULATE REMAINING REINFORCEMENT AREA AVAILABLE FOR BENDING

Ax.net 1Ast Avf+

As−

⎛⎜⎝

⎞⎟⎠A1:= Ay.net 1

Ast Avf+

As−

⎛⎜⎝

⎞⎟⎠A2:=

Ax.net 3.334 in2= Ay.net 2.382 in2=

axAx.net fy⋅

0.85fc by⋅:= ax 1.634in= ay

Ay.net fy⋅

0.85fc bx⋅:= ay 1.167in=

BENDING CAPACITY ABOUT Y-AXIS

ϕMny ϕb Ax.net⋅ fy⋅ dxax2

−⎛⎜⎝

⎞⎟⎠

⋅:= ϕMny 355.17kip ft=

Muy <=? ϕMny "YES!.. SATISFACTORY"=

BENDING CAPACITY ABOUT X-AXIS

ϕMnx ϕb Ay.net⋅ fy⋅ dyay2

−⎛⎜⎝

⎞⎟⎠

⋅:= ϕMnx 255.50kip ft=

Mux <=? ϕMnx "YES!.. SATISFACTORY"=



HORIZONTAL TIES [ACI 318-05 Sec 7.10.5 &11.5.5]

PROVIDE THE TIE BAR DIAMETER

TIE SIZE tie 4=

TIE DIAMETER dbtie 0.472 in=

TIE BAR AREA Abtie 0.175 in2=

Se 0:=

[ACI 318-05 Sec 7.10.5]

Smax

min 16dbbar 48dbtie, bx, by, ( )min 8dbbar 24dbtie,

min bx by, ( )2

, 12in, ⎛⎜⎝

⎞⎟⎠

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=[ACI 318-05 Sec 21.12.5.2]

SmaxSe 14in=

PROVIDE THE TIE VERTICAL SPACING

Spacingtie 12in:=

Spacingtie <=? SmaxSe "YES!.. SATISFACTORY"=

PIER SHEAR CAPACITY

ϕVnc1 ϕs 0.17⋅ fc MPa⋅⋅ 1Pu

min bx by, ( ) min dx dy, ( )⋅ 14⋅ MPa+

⎛⎜⎝

⎞⎟⎠

⋅ min bx by, ( )⋅ min dx dy, ( )⋅:= [ACI 318-05 Eqn 11-4]

ϕVnc1 165.276kip=

ϕVnc2 ϕs 0.17⋅ fc MPa⋅⋅ 1Tub

min bx by, ( ) min dx dy, ( )⋅ 3.5⋅ MPa+

⎛⎜⎝

⎞⎟⎠

⋅ min bx by, ( )⋅ min dx dy, ( )⋅:= [ACI 318-05 Eqn 11-8]

ϕVnc2 99.249kip=

ϕVnc max min ϕVnc1 ϕVnc2, ( ) 0kN, ( ):= ϕVnc 99.249kip=

12

ϕVnc⎛⎜⎝

⎞⎟⎠

>=? VuTotal "YES!.. SATISFACTORY"=

kcs if12

ϕVnc⎛⎜⎝

⎞⎟⎠

>=? VuTotal "YES!.. SATISFACTORY"≠ 1, 0, ⎡⎢⎣

⎤⎥⎦

:= kcs 0=

SHRINKAGE REINFORCEMENT DESIGN

SHEAR AREA OF SINGLE TIE Abtie 0.175 in2=



NUMBER OF LEGS NoLeg 3:=

Av NoLeg Abtie×:= Av 0.525in2=SHEAR REINFORCEMENT AREA

Av.min max 0.062 fc MPamax bx by, ( ) Spacingtie⋅

fy⋅

0.345MPamax bx by, ( ) Spacingtie⋅

fy,

⎛⎜⎝

⎞⎟⎠

:= [ACI 318-05 Sec 11.5.6.3]

Av.min 0.36in2=

Av >=? Av.min "YES!.. SATISFACTORY"=

dp min bx by, ( ) Cc− dbtie−12

dbbar−:= dp 33.594in=

SHEAR BY TIES PROVIDED

ϕVsprovkcs ϕs⋅ Av⋅ fy⋅ dp⋅

Spacingtie:= ϕVsprov 0kip= [ACI 318-05 Sec 11.5.7.2]

ϕVsprov <=? min ϕVxs.max ϕVys.max, ( ) "YES!.. SATISFACTORY"=

ϕVsprov ϕVnc+( ) >=? VuTotal "YES!.. SATISFACTORY"=

E. DEVELOPMENT LENGTH OF REBARS (ACI 318-05 SECTION 12.2)

CENTER TO CENTER BAR SPACING Sp

bx 2Cc−

2 n2+( ) 1−

by 2Cc−

n1( ) 1−

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

:= Sp5.5

5.5

⎛⎜⎝

⎞⎟⎠

in=

REINFORCEMENT LOCATION FACTOR ψt 1:= (FOR VERTICAL DOWELS)

REINFORCEMENT LOCATION FACTOR ψe 1:= (FOR UNCOATED REBARS)

CONCRETE TYPE FACTOR λd 1:= (FOR NORMAL WEIGHT OF CONCRETE)

cb min Cc dbtie( )+dbbar

2+

min Sp( )2

, ⎡⎢⎣

⎤⎥⎦

:= cb 2.406in=

TRANSVERSE REINFORCMENT INDEX Ktr 0in:= AS A DESIGN SIMPLIPICATION, IT IS CONSERVATIVE TOASSUME Ktr=0, EVEN IF TRANSVERSE REINFORCEMENT ISPRESENT (PER ACI 318-05 SECTION 12.2.3)

DEVELOPMENT LENGTH OF STRAIGHT BAR

Ld340

ψe⋅ ψt⋅ 0.80⋅ 12⋅λd fy⋅

fc MPa⋅ mincb Ktr+

dbbar2.5,

⎛⎜⎝

⎞⎟⎠

⋅

⋅ dbbar⋅ bar 6≤if

340

ψe⋅ ψt⋅ 1.0⋅ 12⋅λd fy⋅

fc MPa⋅ mincb Ktr+

dbbar2.5,

⎛⎜⎝

⎞⎟⎠

⋅

⋅ dbbar⋅ bar 7≥if

:= [ACI 318-05 Sec 12.2.3, Eqn12-1]



Ld 24.562in=

DEVELOPMENT LENGTH OF BAR WITH STANDARD HOOK

Ldh max 0.24 ψe⋅λd fy⋅

fc MPa⋅⋅ dbbar⋅ 150mm, 8 dbbar,

⎛⎜⎝

⎞⎟⎠

:= Ldh 16.375in= [ACI 318-05 Sec 12.5.2]

CHECK THE DEVELOPMENT LENGTH OF THE STRAIGHT BAR INTO THE PIER

REDUCTION FACTOR PER ACI 318-05 SECTION 12.2.5 & 12.5.3.D

IR maxMuy

ϕMny

MuxϕMnx

, TubϕTn

, ⎛⎜⎝

⎞⎟⎠

:= IR 0.509=

IR <=? 1.0 "YES!.. SATISFACTORY"=

max Ldh IR⋅ 12in, ( ) <=? hpier 2in−( ) "YES!.. SATISFACTORY"=

CHECK THE DEVELOPMENT LENGTH OF THE HOOKED BAR INTO THE BASE MAT

max Ldh IR⋅ 6in, 8 dbbar, ( ) <=? T 3in−( ) "YES!.. SATISFACTORY"=1

0.08312.048=

F. PIER DESIGN SUMMARY 12 0.02⋅ 0.24=

PIER WIDTH bx 36in=

PIER LENGTH by 36in=

VERTICAL BAR SIZE bar 7=

NUMBER OF VERTICAL BARS n 24=

BAR ARRANGEMENT

NUMBER OF BARS IN A1 n1 7=

NUMBER OF BARS IN A2 n2 5=

TIE BAR SIZE (2 LEGS) tie 4=

TIE SPACING Spacingtie 12in= NOTE:TIE ARRANGEMENT DETAILS SHOULD ALSOBE PROVIDED


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