Crsi Pilecap Aashto Limited Version

"CRSI-PILECAP (Bridges)" --- PILECAP DESIGN AND ANALYSIS SPREADSHEET
Program Description:
"CRSI-PILECAP (Bridges)" is a spreadsheet program written in MS-Excel for the design and analysis of rigid pile
caps. Pile caps are designed and analyzed in accordance with AASHTO (2014) with modifications as necessary
to account for beam shear and punching shear when piles are placed inside the AASHTO assumed shear failure plane.
Users input the pile type/material, pile shape, pile factored load, pile size, pile cap material properties,
pile cap depth, bar size, and minimum pile eccentrity (which can also be used to increase pile cap moment demand
if so desired). Design requires the user to simply yet interatively choose a minimum design depth that satisfies
all limit states required by AASHTO and those recommended by CRSI. The program is based on the LRFD
philosophy and the assumed strength resistance factor for shear is f = 0.9 for consistancy with AASHTO.
The user should note that although the intent of the spreadsheet is to meet pile cap design requirements
as found in AASHTO (2014), it is expected that all users will validate the solutions with hand calculations
and verify the applicability of the interpretation, assumptions, approach, and solution to their specific project.
This program is a workbook consisting of twenty seven (27) worksheets, described as follows:
Worksheet Name Description

Doc This documentation sheet
2 Pile Cap Pile Cap Design Spreadsheet for 2 Piles
3 Pile Cap Pile Cap Design Spreadsheet for 3 Piles (Round Column)
7 Pile Cap Pile Cap Design Spreadsheet for 7 Piles (Round Column)
Program Assumptions and Limitations:

General Note: This spreadsheet is part of the CRSI Pile Cap Design Guide for Bridges (2018). This program should only be used
in conjunction with the design requirements and recommendations as presented in the CRSI Pile Cap Design Guide for Bridges (2018).
For a complete list of terms used in this spreadsheet, please see the CRSI Pile Cap Design Guide for Bridges (2018).
1. The CRSI-PILECAP (Bridges) worksheets assume rigid cap load distribution to each pile (i.e., equal load distribution)
for a pile group.
2. The tabulated designs are based upon the use of a square reinforced concrete column of at least the minimum size indicated or
a structural steel column on a steel base plate such that the section half-way between the column face
and the edge of the base plate is equivalent to the size of the column.
3. The tabulated designs are adequate (and conservative) for rectangular columns or steel base
plates if the short side or section is equal to the minimum tabulated column size.
4. A minimum embedment of 6 inches has been established as good practice with structural
steel shapes to avoid use of cover plates for bearing. However, unless special exceptions apply, AASHTO requires a 12 in.
minimum embedment of all piles.
5. A minimum pile spacing equal to the maximum of (a) 3 ft o.c., (b) 3 times the pile dimension,
and (c) 2 ft clear between piles is used in the worksheets. The 3 times the pile dimension requirement is based on current
geotechnical recommendations that result in expected maximum pile capacities.
6. Note that some tabulated designs show dimensions on the line below. These dimensions are for use with “clipped corners”
to save concrete. In other words, the “clipped corner” dimensions are the truncated dimensions
of the pile cap with the corners removed.
7. Pile Factored Load is tabulated in tons as is usually done in accepted practice. The
range shown, 60 tons to 600 tons, covers the usual range for precast concrete or structural steel piles.
An additional load factor is provided by the user.
8. For pile caps with 2, 3, 4, 5, 6, 7, 8, or 9 piles, all reinforcing bars must be provided with standard
end hooks. For pile caps with 10, 11, or 12 piles, only the short reinforcing bars must be provided with
end hooks, and should be placed as the lower layer. As an alternative to hooked bars, the bar ends can be headed.
9. This program contains numerous “comment boxes” which contain a wide variety of information including
explanations of input or output items, equations used, data tables, etc. (Note: presence of a “comment box”
is denoted by a “red triangle” in the upper right-hand corner of a cell. Merely move the mouse pointer to the
desired cell to view the contents of that particular "comment box".)
10.The limit states listed in the output section of each worksheet are defined and shown graphically
in the CRSI Pile Cap Design Guide for Bridges (2018).
Terms Used in the Input Section of the Worksheets:

Add. Load. Fact. - an additional load factor if a larger factor of safety is desired or mandated
Pile Type - concrete, steel, or wood piling may be selected
Pile Shape - round or square piles may be selected; H-piles are considered square piles (see CRSI Pile Cap
Design Guide for Bridges); pipe piles are considered round piles (see CRSI Pile Cap Design Guide for Bridges)
f'c - 28 day strength of concrete (limited to 3,000 or 4,000 psi)
fy - yield strength of steel reinforcement; limited to 60 ksi
Accidential Pile Offset - pile location tolerence; increases shear and bending demands on pile cap;
minimum set as CRSI recommended value of 3 in.; maximum value arbitrarily set at 9 in.
Pile Factored Load (LRFD) - Ultimate compressive load on pile (used for structural design the pile). Ultimate pile
load (as determined by geotechnical engineers) may exceed this value.
Limited values between 60 tons and 600 tons may be selected.
Minimum Pile Embedment - the minimum pile embedment is set at 12 inches for all piles.
If less embedment is permitted, the overall cap thickness can be reduced by the desired reduction of embedment.
Clear Cover Over Top of Pile - the clear distance from the top of the pile to the first layer of reinforcement;
the minimum value is set as the CRSI recommended minimum of 3 in.; the maximum value is arbitrarily set at 5 in.
Minimum Pile Dimension dp - is the pile diameter for round columns and the pile side dimension for square piles
L - center to center pile spacing; minimum set by CRSI recommendations; see comment box
E - minimum edge distance from center of pile to edge of concrete; set by CRSI recommendations (see comment
box)
Dmin,recommended - minimum recommended pile cap thickness for start of iterative design
D - user selected pile cap thickness; all resulting limit state design checks must be "ok"
A - minimum pile cap width (long direction); automatically calculated based on pile spacing and edge distance
B - minimum pile cap width (short direction); automatically calculated based on pile spacing and edge distance
Short Bars Size (No.) - selected reinforcement size for bars spanning in the "B" direction
Long Bars Size (No.) - selected reinforcement size for bars spanning in the "A" direction
dshort bars - effective depth of concrete from top of cap to center of short bars (automatically calculated)
dlong bars - effective depth of concrete from top of cap to center of long bars (automatically calculated)
dshear checks - the effective depth of the concrete for shear limit states; based on the Minimum Pile Embedment
plus the Clear Cover Over Top of Pile plus 1 inch; this value is used for Limit States 4, 5, and 6
dv is used for Limit States 1, 2, and 3
Special Notes for Comparing Program Output with Tabulated Designs in Pile Cap Design Guide:
1. For the 2 Pile Cap, the previous CRSI Design Handbook (2008) added an extra degree of conservatism when
determining the one way shear nominal strength. Pile cap instability (weak axis) and one way action were
considerations. The CRSI Pile Cap Design Guide for Bridges (2018) and this software program also include this extra
conservatism. See 2 Pile Cap tab and Limit State 5 calculations for more information.
2. For the 3 Pile Cap, one way shear is possible on either side of the column (i.e., with 1 pile or 2 piles causing
demand and and reduced area towards the pile). The program considers both conditions and presents the worst
case results.
3. For pile caps with different distances w when considering orthogonal action and two way shear (e.g., 6 Pile Cap
and 8 Pile Cap), the program uses a slightly more conservative approach than the previous CRSI Design Handbook
(2008) for determining the nominal shear strength for Limit State 4. The previous approach used the average
w value (considering both directions) where this program uses the average demand capacity ratio for each w.
4. CRSI Limit State P4 is very restrictive for thicker pile caps and may not be entirely applicable in some cases.
The spreadsheet conservatively applies it to all caps with corner piles but assumes the one way shear strength
is limited to 2(fc')1/2 which is very conservative when the critical distance from the cap is limited to 13 in.
As a result, the design guide review committee has reviewed the results for the tabulated designs and believes that
increasing the one way shear strength to 3(f c')1/2 is still conservative for these configurations.
m should only be used
n Guide for Bridges (2018).
distribution)
nimum size indicated or
TO requires a 12 in.
based on current
h “clipped corners”
for Bridges)
e). Ultimate pile
on of embedment.
bitrarily set at 5 in.

n for square piles
ns (see comment
edge distance
edge distance
Embedment
ervatism when
y action were
also include this extra
2 piles causing
d presents the worst
ar (e.g., 6 Pile Cap

SI Design Handbook
sed the average
ratio for each w.
n some cases.
ay shear strength
igns and believes that

2 PILE CAP (Traditional CRSI Approach)
for Vertical Loading
Per 2014 AASHTO LRFD
Job Name: Subject: ###
Job Number: Originator: Checker: ###
Input: Add. Load. Fact. 1 ###
Pile Type Steel ###
Pile Shape Round Bar No.
f'c= 4 ksi ###
fy= 60 ksi ###
Accidential Pile Offset 3 inches ###
Pile Factored Load (LRFD) 120 tons ###
Pile Factored Load (LRFD) 240 kips ###
Minimum Pile Embedment 12 inches ###
Clear Cover Over Top of Pile 3 inches ###
Note: Minimum cap dimensions
Minimum Pile Dimension dp= 12 inches shown above. Actual dimensions ###
L= 3 ft provided to the left referencing ###
E= 21 inches figure below. ###
Dmin,recommended= 38.6 inches Short hook add=
Dcap= 43 inches Long hook add=
A= 6.5 ft Short Bar Diamete
B= 3.5 ft Short Bar Area
Short Bars Size (No.) 4 Steel pile embedm
Long Bars Size (No.) 6
dshort bars = 27 inches
dlong bars = 27.625 inches
dshear checks = 27 inches If given pile is squa
Design Checks: Equivalent Pile Dia

Short bar size small enough to develop hooks? OK
Long bar size small enough to develop hooks? OK
CRSI Limit State 1 Adequate? NG d=
CRSI Limit State 2 Adequate? OK d/2=
CRSI Limit State 3 Adequate? Not applicable to this pilecap configuration Diameter for punc
CRSI Limit State 4 Adequate? Not applicable to this pilecap configuration Area in shear
CRSI Limit State 5 Adequate? OK 4(fc')0.5=
CRSI Limit State 6 Adequate? Not applicable to this pilecap configuration Vc=
Results: fVc=
Column Size Determination Vu=
Net Column Capacity Pu= 464.7148 kips (reduced by 1.25 x pile cap wt., column f'c = 4 ksi) Vu/fVc=
Minimum Column Dimension c= 10.77862 inches
Rounded Dimension c= 11 inches
Required Flexural Reinforcement - Short Bars

Mu = 0 k-in. Total moment on pile cap with offset
Mu = 0 k-in./ft 1 ft strip design
d= 27 in.
As,required,structural,per ft= 0.007943 in2/ft
As,required,structural,total= 0.051629 in2
(4/3)Mu= 0 k-in./ft
Mcr=1.2frSc 2104.939 k-in./ft
Mr = 0 k-in./ft
As, required, Mr = 0.007943 in2
As,required,Mr,total= 0.051629 in2
As,required= 0.051629 in2
Increase the calculated area of steel to maintain uniform spacing
b=A/B= 1.857143
gs= 0.7 % steel concentrated of width B of long side A
As,required,structural,total,modfied= 0.067117 in2
E'=E-Offset 18 in. Available for hook as past pile edge
Use 1.0 for no epoxy coating
ldh= 6.65 in. OK; sufficient length available to develop hook
Req. number of short bars 0.341836
Prov. number of short bars 8
Required Flexural Reinforcement - Long Bars

Mu = 4287.297 k-in. Total moment on pile cap with offset
Mu = 1224.942 k-in./ft 1 ft strip design
d= 27.625 in.
(4/3)Mu= 1633.256 k-in./ft
Mr = 1633.256 k-in./ft
Req. number of long bars 9.016762
Prov. number of long bars 8
Check two way shear at a distance d/2 from column - CRSI LIMIT STATE 1 -
d v= 24.3 in.
dv/2= 12.15 in. distance from colu
No. of piles outside d/2 from face of the column 2 distance from cent
4(fc') =
0.5
252.9822128135 psi distance from cent
b0= 84 in.
Vc= 516.3872927949 k based on Vc=4(fc')0.5b0d acting at d/2 from column face
fVc= 464.7485635154 k Limit State Allowab
Vu= 471.9579585938 k reduced by weight of concrete outside of critical section
Vu/fVc= 1.015512463393
Check deep beam (one way shear through short width) at a distance
d from column face - CRSI LIMIT STATE 2 -
d v= 24.3 in.
No. of piles outside d from face of the column 0 distance from colu
2(fc') =
0.5
bd= 1020.6 in.
Vc= 129.0968231987 k based on Vc=2(fc')0.5bd acting at d from column face Limit State Allowab
fVc= 116.1871408788 k
Vu= -1.2980625 k reduced by weight of concrete outside of critical section
Vu/fVc= -0.01117217009
Check deep beam (one way shear through short width) at face of column
(only applies when w/d<1.0) - CRSI LIMIT STATE 5 -
d= 27.625 in. d v= 24.8625 in.
w'= 15.5 in.
w'/d= 0.561085972851 OK, this limit state should be checked
Vu= 235.273359375 k
Mu = 3640.828769531 k-in.
Vud/Mu= 1.785150295209 Limit State Allowab
Mu/(Vud)= 0.560176923301
k= 7.777706855639
fVc= 462.2938990631 k
Vu/fVc= 0.508925944841
Conservatively assuming w'/d=1.0,
Vud/Mu= 1.785150295209
Mu/(Vud)= 0.560176923301
k= 4.363962217644
fVc= 259.3866220987 k
Vu/fVc= 0.907037369435
Quantities of Concrete and Reinforcing Bars

Concrete Summary:
3.0192901235 CY of Concrete
Short Steel Bar Summary:
8 Total No. 4 Bars of length 3 ft (per bar)
16 Total Hooks of length 0.56525 ft (per bar)
Weight = 22.07716394965 lbs
Long Steel Bar Summary:
8 Total No. 6 Bars of length 6 ft (per bar)
Weight = 91.04625352539 lbs
Total weight of all steel = 113.1234 lbs or 0.056562 tons

Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Clear Cover Over Top of Pile 3 inches Note: Minimum cap dimensions ###
shown above. Actual dimensions
Minimum Pile Dimension dp= 20 inches provided to the left referencing ###
L= 5 ft figure below. ###
E= 21 inches ###
Dmin,recommended= 38.6 inches A'= 2.020785 ft Short hook add=
Dcap= 66 inches B'= 2.218825 ft Long hook add=
Bar Size (No.) 11 Steel pile embedm
All Three Directions
dshort bars = 50.295 inches Pile Depth Recomm
dlong bars = 51 inches

Short bar size small enough to develop hooks? NG; select smaller bar size
CRSI Limit State 1 Adequate? OK d=
CRSI Limit State 2 Adequate? Not Applicable CRSI Limit State P1 Adequate? OK
CRSI Limit State 3 Adequate? OK CRSI Limit State P2 Adequate? OK
CRSI Limit State 4 Adequate? Not applicable CRSI Limit State P3 Adequate? OK
CRSI Limit State 5 Adequate? Not Applicable CRSI Limit State P4 Adequate? OK
CRSI Limit State 6 Adequate? OK Vc=
Results: fVc=
Required Flexural Reinforcement - Three Ways

d= 50.295 in.
(4/3)Mu= 3624.231 k-in./ft
Mr = 3624.231 k-in./ft
ldh= 18.753 in. NG; insufficient length available to develop hook
d v= 45 in.
4(fc')0.5= 252.9822128135 psi
b0= 188.49 in.
Vu/fVc= 0.356884196692
Check deep beam (one way shear through long width) at a distance
d v= 45 in.
2(fc')0.5= 126.4911064067 psi distance from cent
bd= 1254.278979187 in.
fVc= 142.7896222381 k
Vu= 0k reduced by weight of concrete outside of critical section Column center dist
Vu/fVc= 0 boriginal
bmodified
Check two way shear at column face - CRSI LIMIT STATE 4 -

d= 50 in. d v= 45 in.
d/2= 25 in. dv/2= 22.5 in.
No. of piles outside face of the column 3
wactual 30.14203233256 in. > d/2 Limit state not applicable
Vu 684.9912381836 k
4(fc')0.5= 252.9822128135 psi
32(fc')0.5= 2023.857702508 psi
b0= 204.1975 in.
bs= 47.1225 in. Limit State Allowab
k= 14.37638074806 based on k=2(d/w)(1+d/c) acting at column face
Vc= 1928.059351967 k based on Vc=k(fc')0.5bsd acting at column face
fVc= 1735.253416771 k
Vu/fVc= 0.394749972288
Check deep beam (one way shear through long width) at face of column
d= 48.885 in. d v= 43.9965 in. Column center dist
w'= 30.12 in. dv/2= 21.9983 in. boriginal
w'/d= 0.616139920221 OK, this limit state should be checked bmodified
Vu= 240 k batw'
Mu = 7228.8 k-in. Mu =
Mu/(Vud)= 0.616139920221 Mu/(Vud)=
k= 6.559205198628 k=
fVc= 1311.38841044 k fVc=
Vu/fVc= 0.183012140484 Above 0.18791 Below 0.18791 Controlling Vu/fVc=

Concrete Summary: d=
9.8545732596 CY of Concrete d/2=
Diameter for punc
All Steel Bar Summary: Area in shear
36 Total No. 11 Bars of length 8 ft (per bar) 4(fc')0.5=
72 Total Hooks of length 1.72813 ft (per bar) Vc=
Weight = 2191.262178965 lbs fVc=
Vu=
Vu/fVc=
d=
Total weight of all steel = 2191.262 lbs or 1.095631 tons d/2=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
dshort bars = 49.4605 inches

Long bar size small enough to develop hooks? NG; select smaller bar size
Results: fVc=

d= 49.4605 in.
(4/3)Mu= 2108.864 k-in./ft
Mr = 2108.864 k-in./ft
b=A/B= 1
gs= 1 % steel concentrated of width B of long side A

d= 51.1535 in.
(4/3)Mu= 2108.864 k-in./ft
Mr = 2108.864 k-in./ft
d v= 45.9 in.
4(fc') =
0.5
b0= 243.6 in.
Vu/fVc= 0.363327336403
d v= 45.9 in.
2(fc') =
0.5
bd= 4681.8 in.
fVc= 532.9854557775 k
Vu/fVc= 0.002404146278
d v= 45.9 in.
2(fc') =
0.5
bd= 4681.8 in.
fVc= 532.9854557775 k
Vu/fVc= 0.002404146278

d= 51 in. d v= 45.9 in.
d/2= 25.5 in. dv/2= 22.95 in.
wactual 25.5 in. < d/2 This procedure applies
Vu 906.719296875 k
4(fc') =
0.5
252.9822128135 psi
32(fc')0.5= 2023.857702508 psi
b0= 264 in.
bs= 60 in. Limit State Allowab
fVc= 2758.98353579 k
Vu/fVc= 0.3286425182
d= 51.1535 in. d v= 46.0382 in.
w'= 25.5 in. dv/2= 23.0191 in.
Vu= 456.775078125 k
Mu = 11734.85794922 k-in.
Mu/(Vud)= 0.502227029132
k= 9.45099547696
fVc= 2526.202122986 k
Vu/fVc= 0.180814937162
d= 49.4605 in. d v= 44.5145 in.
w'= 25.5 in. dv/2= 22.2572 in.
Vu= 456.775078125 k
Mu = 11734.85794922 k-in.
Mu/(Vud)= 0.519417926117
k= 8.935080139347
fVc= 2309.256319286 k
Vu/fVc= 0.197801809314

Diameter for punc
Short Steel Bar Summary: Area in shear
Weight = 1023.746272746 lbs fVc=
Long Steel Bar Summary: Vu=
11 Total No. 14 Bars of length 8 ft (per bar) Vu/fVc=
Weight = 1023.746272746 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 61.4605 in.
(4/3)Mu= 2320.017 k-in./ft
Mr = 2320.017 k-in./ft
b=A/B= 1


d= 63.1535 in.
(4/3)Mu= 2320.017 k-in./ft
Mr = 2320.017 k-in./ft

d v= 56.7 in.
4(fc') =
0.5
b0= 294.8 in.
Vu/fVc= 0.234961228921
d v= 56.7 in.
2(fc') =
0.5
bd= 7192.5084 in.
fVc= 818.8095108202 k
Vu/fVc= 0.001696231709
d v= 56.7 in.
2(fc') =
0.5
bd= 7192.5084 in.
fVc= 818.8095108202 k
Vu/fVc= 0.001696231709

d= 63 in. d v= 56.7 in.
d/2= 31.5 in. dv/2= 28.35 in.
Vu 862.4693779363 k
4(fc') =
0.5
252.9822128135 psi
32(fc')0.5= 2023.857702508 psi
b0= 320 in.
fVc= 3524.064284902 k
Vu/fVc= 0.244737129692
d= 63.1535 in. d v= 56.8382 in.
w'= 36.926 in. dv/2= 28.4191 in.
Vu= 436.9976278744 k
Mu = 16543.50585431 k-in.
Mu/(Vud)= 0.599447351267
k= 7.07453320812
fVc= 2903.404511473 k
Vu/fVc= 0.15051214054
d= 61.4605 in. d v= 55.3145 in.
w'= 36.926 in. dv/2= 27.6572 in.
Vu= 436.9976278744 k
Mu = 16543.50585431 k-in.
Mu/(Vud)= 0.615959816439
k= 6.728281559188
fVc= 2687.277894284 k
Vu/fVc= 0.162617207846

Diameter for punc
12 Total No. 14 Bars of length 10.071 ft (per bar) 4(fc')0.5=
Weight = 1307.1785979 lbs fVc=
12 Total No. 14 Bars of length 10.071 ft (per bar) Vu/fVc=
Weight = 1307.1785979 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 41.436 in.
(4/3)Mu= 1599.582 k-in./ft
Mr = 1599.582 k-in./ft
b=A/B= 1.461538

d= 42.5 in.
(4/3)Mu= 3273.343 k-in./ft
Mr = 3273.343 k-in./ft
d v= 37.8 in.
4(fc') =
0.5
b0= 227.2 in.
Vu/fVc= 0.477822160336
d v= 37.8 in.
2(fc') =
0.5
bd= 2948.4 in.
fVc= 335.6517403167 k
Vu/fVc= -0.01021399069
d v= 37.8 in.
2(fc') =
0.5
bd= 4309.2 in.
fVc= 490.5679281551 k
Vu/fVc= 0.008739806465
Check two way shear at column face - CRSI LIMIT STATE 4 - waverage
d= 42 in. d v= 37.8 in. Make sure larger w
d/2= 21 in. dv/2= 18.9 in. OK
No. of piles outside face of the column 6 32(fc')0.5=
wactual 11.5 in. < d/2 This procedure applies wactual
Vu 1401.34390625 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 244 in. b0=
k= 23.45080091533 based on k=2(d/w)(1+d/c) acting at column face k=
Vc= 4260.818821708 k based on Vc=k(fc')0.5bsd acting at column face ###
fVc= 3834.736939537 k ###
Vu/fVc= 0.365434168848 ****COMBINED**** Vu/fVc= 0.65149 ###
d= 42.5 in. d v= 38.25 in.
w'= 29.5 in. dv/2= 19.125 in.
Vu= 463.21171875 k
Mu = 13761.27832031 k-in.
Mu/(Vud)= 0.699021140659
k= 5.15813099029
fVc= 875.9733911436 k
Vu/fVc= 0.528796563267
d= 41.436 in. d v= 37.2924 in.
w'= 11.5 in. dv/2= 18.6462 in.
Vu= 704.76140625 k
Mu = 8055.230742188 k-in.
Mu/(Vud)= 0.275840510212
k= 10
fVc= 2419.901508457 k
Vu/fVc= 0.291235574583

Diameter for punc
Weight = 268.2401512535 lbs fVc=
Weight = 385.4386160098 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

CRSI Limit State 4 Adequate? NA CRSI Limit State P3 Adequate? NA
CRSI Limit State 5 Adequate? OK CRSI Limit State P4 Adequate? NA
Results: fVc=
Rounded Dimension c= 23 inches DIAMETER FOR 7 PILE CAP

d= 31.875 in.
(4/3)Mu= 1850.586 k-in./ft
Mr = 1850.586 k-in./ft
b=A/B= 1.092381

d= 32.625 in.
(4/3)Mu= 2266.947 k-in./ft
Mr = 2266.947 k-in./ft
d v= 28.8 in.
4(fc') =
0.5
252.9822128135 psi
b0= 162.734362 in.
Vu/fVc= 1.31505183633
d v= 28.8 in.
2(fc') =
0.5
bd= 3005.54496 in. distance from cent
fVc= 342.157236611 k
Vu/fVc= -0.01910085248
d v= 28.8 in.
2(fc') =
0.5
bd= 3283.2 in.
fVc= 373.7660404991 k
Vu/fVc= -0.01358745432

d= 32 in. d v= 28.8 in.
d/2= 16 in. dv/2= 14.4 in.
Vu 1397.370192 k
4(fc') =
0.5
252.9822128135 psi
32(fc')0.5= 2023.857702508 psi
b0= 172.7825 in.
bs= 72.2545 in. Limit State Allowab
fVc= 659.1921494613 k
Vu/fVc= 2.119822259931
d= 32.625 in. d v= 29.3625 in.
w'= 9.5 in. dv/2= 14.6813 in.
Vu= 702.1937115 k
Mu = 10754.90693663 k-in.
Mu/(Vud)= 0.469460655226
k= 10
fVc= 1744.199975693 k
Vu/fVc= 0.402587846168
d= 31.875 in. d v= 28.6875 in.
w'= 22.6796 in. dv/2= 14.3438 in.
Vu= 462.609471 k
Mu = 10532.48811825 k-in.
Mu/(Vud)= 0.714276339985
k= 4.915124291553
fVc= 914.9651737867 k
Vu/fVc= 0.505603365301

Diameter for punc
Weight = 205.5596991909 lbs fVc=
Weight = 222.4677307788 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

CRSI Limit State 3 Adequate? NG CRSI Limit State P2 Adequate? OK
Results: fVc=

d= 34.436 in.
(4/3)Mu= 3527.976 k-in./ft
Mr = 2961.03 k-in./ft
b=A/B= 1.115654

d= 35.5 in.
(4/3)Mu= 3745.002 k-in./ft
Mr = 2961.03 k-in./ft
d v= 31.5 in.
4(fc') =
0.5
b0= 210 in. distance from cent
Vu/fVc= 1.200048099819
d v= 31.5 in.
2(fc') =
0.5
fVc= 597.8559946881 k
Vu/fVc= 0.746203607706
d v= 31.5 in.
2(fc') =
0.5
bd= 5859 in.
fVc= 667.0002531934 k
Vu/fVc= 1.033506287403
d/2= 17.5 in. dv/2= 15.75 in. OK
wactual 28.5 in. > d/2 Limit state not applicable wactual
Vu 1798.203145125 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 224 in. b0=
Vc= 1096.078724148 k based on Vc=k(fc')0.5bsd acting at column face Vc=
fVc= 986.4708517329 k fVc=
Vu/fVc= 1.822864955377 ****COMBINED**** Vu/fVc= 2.6656 Vu/fVc=
d= 35.5 in. d v= 31.95 in.
w'= 28.5 in. dv/2= 15.975 in.
Vu= 665.1978384375 k
Mu = 35539.41083555 k-in.
Mu/(Vud)= 1.504981084625
k= -0.6428602694
fVc= -194.914203375 k
Vu/fVc= -3.41277252719
d= 34.436 in. d v= 30.9924 in.
w'= 54.852 in. dv/2= 15.4962 in.
w'/d= 1.592867928912 Limit state not applicable
Vu= 666.00450225 k
Mu = 37526.40561517 k-in.
Mu/(Vud)= 1.636240372675
k= -0.72713923565
fVc= -238.593296306 k
Vu/fVc= -2.79137977706

Diameter for punc
Weight = 824.1634815372 lbs fVc=
Weight = 639.574094485 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 72.4605 in.
(4/3)Mu= 3873.852 k-in./ft
Mr = 3873.852 k-in./ft
b=A/B= 1

d= 74.1535 in.
(4/3)Mu= 3873.852 k-in./ft
Mr = 3873.852 k-in./ft
d v= 66.6 in.
4(fc') =
0.5
b0= 354.4 in.
Vu/fVc= 0.322279648843
d v= 66.6 in.
2(fc') =
0.5
bd= 12387.6 in.
fVc= 1410.229106752 k
Vu/fVc= -0.014281589
d v= 66.6 in.
2(fc') =
0.5
bd= 12387.6 in.
fVc= 1410.229106752 k
Vu/fVc= -0.014281589

d= 74 in. d v= 66.6 in.
d/2= 37 in. dv/2= 33.3 in.
wactual 64 in. > d/2 Limit state not applicable
Vu 1680.15 k
4(fc') =
0.5
252.9822128135 psi
32(fc')0.5= 2023.857702508 psi
b0= 384 in.
fVc= 3366.353351601 k
Vu/fVc= 0.499100903713
d= 74.1535 in. d v= 66.7381 in.
w'= 64 in. dv/2= 33.3691 in.
Vu= 612.759375 k
Mu = 41683.134375 k-in.
Mu/(Vud)= 0.917357782313
k= 2.808657736075
fVc= 1984.533489831 k
Vu/fVc= 0.308767465069
d= 72.4605 in. d v= 65.2145 in.
w'= 64 in. dv/2= 32.6072 in.
Vu= 612.759375 k
Mu = 41683.134375 k-in.
Mu/(Vud)= 0.938791345778
k= 2.619402812577
fVc= 1808.554392427 k
Vu/fVc= 0.338811692679

Diameter for punc
Weight = 2640.3718558 lbs fVc=
Weight = 2640.3718558 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Long bar size small enough to develop straight bars? OK
Results: fVc=

d= 63.4605 in.
(4/3)Mu= 2356.375 k-in./ft
Mr = 2356.375 k-in./ft
b=A/B= 1.54752

d= 65.1535 in.
As,required,structural,per ft= 1.233151 in2/ft c. to c. spacing of long ba
As,required,structural,total= 17.13242 in2 concrete cover =

(4/3)Mu= 5611.077 k-in./ft cb=
Mr = 5611.077 k-in./ft yt=
As, required, Mr = 1.645489 in 2

ye=
As,required,Mr,total= 22.86111 in2 ys=
As,required= 22.86111 in 2
(cb+Ktr)/db=
ld=
ld = 67.53241 in. lavailable=
lavailable= 114.5 in. OK; sufficient length available to develop bar

d v= 58.5 in.
4(fc') =
0.5
Vc= 4824.623780566 k based on Vc=4(fc')0.5b0d acting at d/2 from column face distance from cent
Vu/fVc= 0.389170045488
d v= 58.5 in.
2(fc') =
0.5
Vc= 1233.67110015 k based on Vc=2(fc')0.5bd acting at d from column face distance from cent
fVc= 1110.303990135 k
Vu= 657.754122375 k reduced by weight of concrete outside of critical section Limit State Allowab
Vu/fVc= 0.592409041325
d v= 58.5 in.
2(fc') =
0.5
bd= 15093 in.
fVc= 1718.217242097 k
Vu/fVc= -0.01269396025
d/2= 32.5 in. dv/2= 29.25 in. NG, but answer co
Vu 2131.153740375 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 352 in. b0=
fVc= 5540.917617926 k fVc=
d= 65.1535 in. d v= 58.6381 in.
w'= 27.5 in. dv/2= 29.3191 in.
Vu= 836.0357521875 k
Mu = 53677.10044102 k-in.
Mu/(Vud)= 0.985431514756
k= 4.914646153787
fVc= 2734.818781284 k
Vu/fVc= 0.305700603605
d= 63.4605 in. d v= 57.1145 in.
w'= 53.852 in. dv/2= 28.5572 in.
Vu= 602.678628 k
Mu = 34558.13003259 k-in.
Mu/(Vud)= 0.903568233103
k= 2.940644510174
fVc= 2466.497823504 k
Vu/fVc= 0.244345898973

Diameter for punc
Weight = 2821.966479069 lbs fVc=
Weight = 4021.461175899 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

CRSI Limit State 4 Adequate? NA CRSI Limit State P3 Adequate? OK
CRSI Limit State 6 Adequate? NG Vc=
Results: fVc=

d= 45.885 in.
(4/3)Mu= 3284.942 k-in./ft
Mr = 3284.942 k-in./ft
b=A/B= 1.54752

d= 47.295 in.

(4/3)Mu= 7324.257 k-in./ft cb=
Mr = 4518.389 k-in./ft yt=

ye=
(cb+Ktr)/db=
ld=

d v= 42.3 in.
4(fc') =
0.5
b0= 269.2 in. distance from cent
Vu/fVc= 0.852627922368
d v= 42.3 in.
2(fc') =
0.5
fVc= 802.8351928669 k
Vu/fVc= 0.82098284328
d v= 42.3 in.
2(fc') =
0.5
bd= 10913.4 in.
fVc= 1242.403236593 k
Vu/fVc= 0.743506014125
d/2= 23.5 in. dv/2= 21.15 in. OK
Vu 2191.369853625 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 288 in. b0=
fVc= 2459.729444278 k fVc=
d= 47.295 in. d v= 42.5655 in.
w'= 26.5 in. dv/2= 21.2828 in.
Vu= 1104.403929938 k
Mu = 69431.52891886 k-in.
Mu/(Vud)= 1.329271078035
k= 0.62333842666
fVc= 251.7895365866 k
Vu/fVc= 4.386218525636
d= 45.885 in. d v= 41.2965 in.
w'= 52.8592 in. dv/2= 20.6483 in.
Vu= 870.01988775 k
Mu = 47556.87261503 k-in.
Mu/(Vud)= 1.191278684326
k= 0.898641029651
fVc= 544.9939736527 k
Vu/fVc= 1.596384418563

Diameter for punc
Weight = 1790.461277893 lbs fVc=
Weight = 2454.657263953 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 31.355 in.
(4/3)Mu= 3853.672 k-in./ft
Mr = 2622.92 k-in./ft
b=A/B= 1.387097

d= 32.365 in.

(4/3)Mu= 8034.974 k-in./ft cb=
Mr = 2622.92 k-in./ft yt=

ye=
(cb+Ktr)/db=
ld=
lavailable= 113 in. OK; sufficient length available to develop bar

d v= 28.8 in.
4(fc') =
0.5
Vu/fVc= 1.886309991798
d v= 28.8 in.
2(fc') =
0.5
fVc= 609.8288029196 k
Vu/fVc= 1.080923034209
d v= 28.8 in.
2(fc') =
0.5
bd= 7430.4 in.
fVc= 845.8915653401 k
Vu/fVc= 1.076336539228
d/2= 16 in. dv/2= 14.4 in. OK
wactual 26 in. > d/2 Limit state not applicable wactual
Vu 2702.58 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 232 in. b0=
fVc= 936.1805278899 k ###
d= 32.365 in. d v= 29.1285 in.
w'= 26 in. dv/2= 14.5643 in.
Vu= 1359.09 k
Mu = 84587.22 k-in.
Mu/(Vud)= 1.923007258618
k= -3.17709514434
fVc= -979.79177849 k
Vu/fVc= -1.38712125355
d= 31.355 in. d v= 28.2195 in.
w'= 62 in. dv/2= 14.1098 in.
Vu= 882.6 k
Mu = 56424 k-in.
Mu/(Vud)= 2.038886933378
k= -1.57435004724
fVc= -652.443377306 k
Vu/fVc= -1.35276106816

Diameter for punc
Weight = 772.2128896375 lbs fVc=
Weight = 1448.296162542 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

Short bar size small enough to develop straight bars? OK
Results: fVc=

d= 41.23 in.
(4/3)Mu= 3278.349 k-in./ft
Mr = 3278.349 k-in./ft
b=A/B= 1.56671
gs= 0.779208 % steel concentrated of width B of long side A c. to c. spacing of long ba
As,required,structural,total,modfied= 33.80569 in 2
concrete cover =
cb=
ld = 23.56125 in. yt=
lavailable= 76.5 in. OK; sufficient length available to develop bar ye=
Req. number of short bars 47.21695 ys=
Prov. number of short bars 29 (cb+Ktr)/db=
ld=
Required Flexural Reinforcement - Long Bars lavailable=

d= 42.365 in.
As,required,structural,total= 49.8045 in 2
concrete cover =
(4/3)Mu= 9211.117 k-in./ft cb=
Mr = 3829.645 k-in./ft yt=

ye=
As,required,Mr,total= 26.97001 in 2
ys=
As,required= 49.8045 in2 (cb+Ktr)/db=
ld=

d v= 37.8 in.
4(fc') =
0.5
fVc= 2230.793104874 k
Vu/fVc= 1.189453468008
d v= 37.8 in.
Vc= 889.3336709245 k based on Vc=2(fc')0.5bd acting at d from column face
fVc= 800.400303832 k
Vu/fVc= 1.399843457562
d v= 37.8 in.
2(fc') =
0.5
fVc= 1253.994901823 k
Vu/fVc= 0.530255383641
d/2= 21 in. dv/2= 18.9 in. OK
Vu 2637.70097625 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 276 in. b0=
fVc= 1956.196567999 k ###
d= 42.365 in. d v= 38.1285 in.
w'= 51.852 in. dv/2= 19.0643 in.
Vu= 1088.57681625 k
Mu = 99750.54470776 k-in.
Mu/(Vud)= 2.162962472105
k= -3.03305084545
fVc= -1224.37614732 k
Vu/fVc= -0.88908691878
d= 41.23 in. d v= 37.107 in.
w'= 25.5 in. dv/2= 18.5535 in.
Vu= 1095.02482125 k
Mu = 52347.23664469 k-in.
Mu/(Vud)= 1.159461846717
k= 1.931213860388
fVc= 1188.666954555 k
Vu/fVc= 0.921220882816

Diameter for punc
Weight = 1162.518619792 lbs fVc=
Weight = 2665.485637432 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 53.166 in.
(4/3)Mu= 5589.374 k-in./ft
Mr = 5578.258 k-in./ft
b=A/B= 1.080837
concrete cover =
cb=
ld = 29.97896 in. yt=
ld=

d= 54.365 in.
concrete cover =
(4/3)Mu= 5933.726 k-in./ft cb=
Mr = 5578.258 k-in./ft yt=

ye=
ys=
ld=

d v= 48.6 in.
4(fc') =
0.5
fVc= 3346.189657311 k distance from cent
Vu/fVc= 0.912817192931 Limit State Allowab
d v= 48.6 in.
Vu/fVc= 0.660770333817
d v= 48.6 in.
2(fc') =
0.5
fVc= 1427.442016511 k
Vu/fVc= 0.447808736787

d= 54 in. d v= 48.6 in.
d/2= 27 in. dv/2= 24.3 in.
Vu 3027.19029375 k
4(fc') =
0.5
252.9822128135 psi
32(fc')0.5= 2023.857702508 psi
b0= 324 in.
fVc= 3796.097510394 k
Vu/fVc= 0.79744798058
d= 54.365 in. d v= 48.9285 in.
w'= 25.5 in. dv/2= 24.4643 in.
Vu= 1289.22485625 k
Mu = 79852.73544844 k-in.
Mu/(Vud)= 1.13930954407
k= 2.761919531394
fVc= 1836.137180112 k
Vu/fVc= 0.702139725841
d= 53.166 in. d v= 47.8494 in.
w'= 25.5 in. dv/2= 23.9247 in.
Vu= 1530.64944375 k
Mu = 80399.15245991 k-in.
Mu/(Vud)= 0.987965423714
k= 4.297955384363
fVc= 3020.164583032 k
Vu/fVc= 0.506809944183

Diameter for punc
Weight = 2044.151859449 lbs fVc=
Weight = 2172.444243812 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
dlong bars = 32.295 inches x=
Design Checks: ###

Results: fVc=

d= 30.885 in.
(4/3)Mu= 5886.572 k-in./ft
Mr = 2622.92 k-in./ft
b=A/B= 1.129488
concrete cover =
cb=
ld = 46.84212 in. yt=
ld=

d= 32.295 in.
concrete cover =
(4/3)Mu= 6566.77 k-in./ft cb=
Mr = 2622.92 k-in./ft yt=

ye=
ys=
ld=

d v= 28.8 in.
4(fc') =
0.5
Vu/fVc= 2.057752322163
d v= 28.8 in.
fVc= 845.8915653401 k
Vu/fVc= 1.600678989944 boriginal
bmodified
d v= 28.8 in.
2(fc') =
0.5
fVc= 955.4246871033 k
bmodified
d/2= 16 in. dv/2= 14.4 in. OK
Vu 3107.369962822 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 244 in. b0=
fVc= 1044.885515294 k ###
d= 32.295 in. d v= 29.0655 in.
w'= 50.852 in. dv/2= 14.5328 in.
Vu= 1326.136856411 k
Mu = 96394.90707315 k-in.
Mu/(Vud)= 2.25076656364
k= -2.62645419273
fVc= -1121.08835165 k
Vu/fVc= -1.18290128914
d= 30.885 in. d v= 27.7965 in.
w'= 24.5 in. dv/2= 13.8983 in.
Vu= 1567.953416411 k
Mu = 95337.78025327 k-in.
Mu/(Vud)= 1.968721456821
k= -3.4964928224
fVc= -1612.11764524 k
Vu/fVc= -0.97260483504

Diameter for punc
Weight = 1896.780613055 lbs fVc=
Weight = 1642.771404234 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=
Net Column Capacity Pu= 3698.398 kips (reduced by 1.25 x pile cap wt., column f c' = 4 ksi) Vu/fVc=

d= 40.885 in.
(4/3)Mu= 6236.057 k-in./ft
Mr = 3829.645 k-in./ft
b=A/B= 1
gs= 1 % steel concentrated of width B of long side A c. to c. spacing of long ba
concrete cover =
cb=
ld = 46.84212 in. yt=
ld=

d= 42.295 in.
concrete cover =
(4/3)Mu= 6236.057 k-in./ft cb=
Mr = 3829.645 k-in./ft yt=

ye=
ys=
ld=

d v= 37.8 in.
4(fc') =
0.5
Vu/fVc= 1.181971522165
d v= 37.8 in.
bd= 5670 in. distance from cent
fVc= 645.4841159936 k
Vu/fVc= 1.464405951795
d= 37.8 in.
fVc= 645.4841159936 k
Vu/fVc= 1.464405951795

d= 42 in. d v= 37.8 in.
d/2= 21 in. dv/2= 18.9 in.
Vu 3742.40140625 k
4(fc') =
0.5
252.9822128135 psi
32(fc')0.5= 2023.857702508 psi
b0= 292 in.
k= 32 based on k=2(d/w)(1+d/c) acting at column face
fVc= 8537.603240875 k
Vu/fVc= 0.438343326653
d= 42.295 in. d v= 38.0655 in.
w'= 5.5 in. dv/2= 19.0328 in.
Vu= 1879.55859375 k
Mu = 43916.86816406 k-in.
Mu/(Vud)= 0.55244171324
k= 10
fVc= 3250.089367375 k
Vu/fVc= 0.578309818991
d= 40.885 in. d v= 36.7965 in.
w'= 5.5 in. dv/2= 18.3983 in.
Vu= 1879.55859375 k
Mu = 43916.86816406 k-in.
Mu/(Vud)= 0.57149375716
k= 10
fVc= 3141.740247904 k
Vu/fVc= 0.598253975644

Diameter for punc
Weight = 701.3306468438 lbs fVc=
Weight = 701.3306468438 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

Results: fVc=

d= 33.955 in.
(4/3)Mu= 5821.648 k-in./ft
Mr = 2961.03 k-in./ft
b=A/B= 1.129488
concrete cover =
cb=
ld = 38.00194 in. yt=
ld=

d= 35.295 in.
concrete cover =
(4/3)Mu= 8282.015 k-in./ft cb=
Mr = 2961.03 k-in./ft yt=

ye=
ys=
ld=

d v= 31.5 in.
4(fc') =
0.5
Vu/fVc= 1.998756742081
d v= 31.5 in.
fVc= 925.1938995908 k
bmodified
d v= 31.5 in.
2(fc') =
0.5
fVc= 1044.995751519 k
bmodified
d/2= 17.5 in. dv/2= 15.75 in. OK
Vu 3572.058710499 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 264 in. b0=
fVc= 1409.993671816 k ###
d= 35.295 in. d v= 31.7655 in.
w'= 49.852 in. dv/2= 15.8828 in.
Vu= 1560.048378687 k
Mu = 121595.6683927 k-in.
Mu/(Vud)= 2.208344640728
k= -2.78323524759
fVc= -1298.36805782 k
Vu/fVc= -1.20154556275
d= 33.955 in. d v= 30.5595 in.
w'= 23.5 in. dv/2= 15.2798 in.
Vu= 1562.111583687 k
Mu = 94065.86534469 k-in.
Mu/(Vud)= 1.773439051122
k= -2.63906761357
fVc= -1337.73699216 k
Vu/fVc= -1.16772698433

Diameter for punc
Weight = 1448.296162542 lbs fVc=
Weight = 1895.505466424 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

Results: fVc=

d= 41.166 in.
(4/3)Mu= 6532.351 k-in./ft
Mr = 3829.645 k-in./ft
b=A/B= 1.129488
concrete cover =
cb=
ld = 29.97896 in. yt=
lavailable= 110 in. OK; sufficient length available to develop bar ye=
ld=

d= 42.365 in.
concrete cover =
(4/3)Mu= 9000.011 k-in./ft cb=
Mr = 3829.645 k-in./ft yt=

ye=
ys=
ld=

d v= 37.8 in.
4(fc') =
0.5
Vu/fVc= 1.665223394304
d v= 37.8 in.
fVc= 1110.232679509 k
Vu/fVc= 1.416422439209
d v= 37.8 in.
2(fc') =
0.5
fVc= 1253.994901823 k
Vu/fVc= 0.87775693697
d/2= 21 in. dv/2= 18.9 in. OK
Vu 3983.965835 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 296 in. b0=
fVc= 2325.987979719 k ###
d= 42.365 in. d v= 38.1285 in.
w'= 49.352 in. dv/2= 19.0643 in.
Vu= 1528.3679175 k
Mu = 132935.6361857 k-in.
Mu/(Vud)= 2.053081997391
k= -2.73122058985
fVc= -1529.32120046 k
Vu/fVc= -0.99937666269
d= 41.166 in. d v= 37.0494 in.
w'= 23 in. dv/2= 18.5247 in.
Vu= 1770.7899975 k
Mu = 104849.6348588 k-in.
Mu/(Vud)= 1.438338788574
k= -0.33787079125
fVc= -207.637505801 k
Vu/fVc= -8.52827619303

Diameter for punc
Weight = 1285.34780367 lbs fVc=
Weight = 1537.780175442 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

Results: fVc=

d= 36.955 in.
(4/3)Mu= 6958.361 k-in./ft
Mr = 3319.633 k-in./ft
b=A/B= 1.204279
concrete cover =
cb=
ld = 38.00194 in. yt=
ld=

d= 38.295 in.
concrete cover =
(4/3)Mu= 10382.6 k-in./ft cb=
Mr = 3319.633 k-in./ft yt=

ye=
ys=
ld=

d v= 34.2 in.
4(fc') =
0.5
Vu/fVc= 1.94006682031
d v= 34.2 in.
fVc= 1004.496233841 k
Vu/fVc= 1.797466788994
d v= 34.2 in.
2(fc') =
0.5
fVc= 1209.693790075 k
Vu/fVc= 0.906601611083
d/2= 19 in. dv/2= 17.1 in. OK
Vu 3986.138115 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 280 in. b0=
fVc= 1801.118963275 k ###
d= 38.295 in. d v= 34.4655 in.
w'= 49.352 in. dv/2= 17.2328 in.
Vu= 1768.3240575 k
Mu = 153307.6670304 k-in.
Mu/(Vud)= 2.263914106167
k= -3.25823508193
fVc= -1649.14639539 k
Vu/fVc= -1.07226627208
d= 36.955 in. d v= 33.2595 in.
w'= 23 in. dv/2= 16.6298 in.
Vu= 2011.8815775 k
Mu = 119071.3091288 k-in.
Mu/(Vud)= 1.60151683395
k= -1.58852215433
fVc= -934.390343697 k
Vu/fVc= -2.15314893938

Diameter for punc
Weight = 1538.814672701 lbs fVc=
Weight = 1753.836675762 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 52.026 in.
(4/3)Mu= 7138.486 k-in./ft
Mr = 5420.017 k-in./ft
b=A/B= 1.27907
concrete cover =
cb=
ld = 29.97896 in. yt=
ld=

d= 53.295 in.
concrete cover =
(4/3)Mu= 11174.81 k-in./ft cb=
Mr = 5420.017 k-in./ft yt=

ye=
ys=
ld=

d v= 47.7 in.
4(fc') =
0.5
Vu/fVc= 1.111350652291
d v= 47.7 in.
fVc= 1401.007905095 k
Vu/fVc= 1.270377735577
d v= 47.7 in.
2(fc') =
0.5
fVc= 1791.986855354 k
Vu/fVc= 0.605320776634
d/2= 26.5 in. dv/2= 23.85 in. OK
Vu 4296.56953125 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 344 in. b0=
fVc= 4400.20260563 k ###
d= 53.295 in. d v= 47.9655 in.
w'= 58.5 in. dv/2= 23.9828 in.
Vu= 1713.468984375 k
Mu = 166105.0720898 k-in.
Mu/(Vud)= 1.818947565196
k= -1.86539968043
fVc= -1313.99310121 k
Vu/fVc= -1.30401672794
d= 52.026 in. d v= 46.8234 in.
w'= 22.5 in. dv/2= 23.4117 in.
Vu= 2199.873046875 k
Mu = 129142.8588867 k-in.
Mu/(Vud)= 1.128372104422
k= 3.122516976354
fVc= 2746.339347741 k
Vu/fVc= 0.801020110164

Diameter for punc
Weight = 2570.69560734 lbs fVc=
Weight = 3442.895902688 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

Results: fVc=

d= 42.885 in.
(4/3)Mu= 8495.245 k-in./ft
Mr = 4098.312 k-in./ft
b=A/B= 1
gs= 1 % steel concentrated of width B of long side A c. to c. spacing of long ba
concrete cover =
cb=
ld = 46.84212 in. yt=
ld=

d= 44.295 in.
concrete cover =
(4/3)Mu= 8883.15 k-in./ft cb=
Mr = 4098.312 k-in./ft yt=

ye=
ys=
ld=

d v= 39.6 in.
4(fc') =
0.5
Vu= 4407.005680379 k reduced by weight of concrete outside of critical section distance from cent
Vu/fVc= 1.660267712805 distance from cent
Check deep beam (one way shear through short width) at a distance Limit State Allowab
d v= 39.6 in.
fVc= 1357.716594825 k
bmodified
d v= 39.6 in.
2(fc') =
0.5
bmodified

d= 44 in. d v= 39.6 in.
d/2= 22 in. dv/2= 19.8 in.
wactual 22 in. < d/2 This procedure applies ###
Vu 4387.032430379 k ###
4(fc') =
0.5
252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 312 in. b0=
fVc= 2813.081252178 k ###
Vu/fVc= 1.559511452783 ###
d= 44.295 in. d v= 39.8655 in. Limit State Allowab
w'= 22 in. dv/2= 19.9328 in. w'=
w'/d= 0.496670053053 OK, this limit state should be checked w'/d=
Vu= 1975.56606519 k Vu=
Mu = 155907.2439015 k-in. Mu =
Vud/Mu= 0.561280519543 Vud/Mu=
Mu/(Vud)= 1.781640312074 Mu/(Vud)=
k= -3.75771256801 k=
fVc= -2649.35479776 k fVc=
Vu/fVc= -0.74567818054 **CONTROLLING** Vu/fVc=
w'= 22 in. dv/2= 19.2983 in. w'=
Vu= 1975.56606519 k Vu=
Mu = 148374.5124478 k-in. Mu =
Vud/Mu= 0.57100204953 Vud/Mu=
Mu/(Vud)= 1.751307199026 Mu/(Vud)=
k= -3.35060208322 k=
fVc= -2287.12619251 k fVc=
Vu/fVc= -0.86377659075 **CONTROLLING** Vu/fVc= -0.86378 Vu/fVc=

Diameter for punc
Weight = 2293.478729842 lbs fVc=
Weight = 2293.478729842 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

Results: fVc=

d= 44.885 in.
(4/3)Mu= 8201.293 k-in./ft
Mr = 4376.086 k-in./ft
b=A/B= 1.132433
concrete cover =
cb=
ld = 46.84212 in. yt=
ld=

d= 46.295 in.
concrete cover =
(4/3)Mu= 9820.708 k-in./ft cb=
Mr = 4376.086 k-in./ft yt=

ye=
ys=
ld=

d v= 41.4 in.
4(fc') =
0.5
Vu/fVc= 1.692782033596
d v= 41.4 in.
bmodified
d v= 41.4 in.
2(fc') =
0.5
fVc= 1555.309346156 k
bmodified
d/2= 23 in. dv/2= 20.7 in. OK
wactual 21.5 in. < d/2 This procedure applies ###
Vu 4853.925023458 k ###
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 324 in. b0=
fVc= 3267.136081058 k ###
w'= 21.5 in. dv/2= 20.8328 in. w'=
Vu= 2208.697096104 k Vu=
Mu = 159740.3151395 k-in. Mu =
Vud/Mu= 0.640111621007 Vud/Mu=
Mu/(Vud)= 1.562227535295 Mu/(Vud)=
k= -1.71515810106 k=
fVc= -1185.37218483 k fVc=
w'= 47.852 in. dv/2= 20.1983 in. w'=
w'/d= 1.066102261335 Limit state not applicable w'/d=
Vu= 1971.968371104 k Vu=
Mu = 150871.2579488 k-in. Mu =
Vud/Mu= 0.58667105677 Vud/Mu=
Mu/(Vud)= 1.704532699305 Mu/(Vud)=
k= -1.39873584919 k=
fVc= -1061.36775608 k fVc=

Diameter for punc
Weight = 2274.606559709 lbs fVc=
Weight = 2438.71793107 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

Results: fVc=

d= 40.885 in.
(4/3)Mu= 9136.957 k-in./ft
Mr = 3829.645 k-in./ft
b=A/B= 1.062104
concrete cover =
cb=
ld = 46.84212 in. yt=
ld=

d= 42.295 in.
concrete cover =
(4/3)Mu= 10066.44 k-in./ft cb=
Mr = 3829.645 k-in./ft yt=

ye=
ys=
ld=

d v= 37.8 in.
4(fc') =
0.5
Vu/fVc= 1.917515438193
d v= 37.8 in.
bmodified
d v= 37.8 in.
2(fc') =
0.5
fVc= 1420.065055186 k
bmodified

d= 42 in. d v= 37.8 in.
d/2= 21 in. dv/2= 18.9 in.
wactual 57 in. > d/2 Limit state not applicable ###
Vu 4852.881900372 k ###
4(fc') =
0.5
252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 312 in. b0=
fVc= 989.2893398807 k ###
Vu/fVc= 4.90542221041 ###
w'= 21 in. dv/2= 19.0328 in. w'=
Vu= 2208.846495186 k Vu=
Mu = 175799.9177037 k-in. Mu =
Vud/Mu= 0.531417555447 Vud/Mu=
Mu/(Vud)= 1.881759437094 Mu/(Vud)=
k= -4.73776743732 k=
fVc= -3189.51483843 k fVc=
w'= 57 in. dv/2= 18.3983 in. w'=
Vu= 1970.420325186 k Vu=
Mu = 169578.8459558 k-in. Mu =
Vud/Mu= 0.475062998225 Vud/Mu=
Mu/(Vud)= 2.104983978413 Mu/(Vud)=
k= -2.4619956063 k=
fVc= -1701.68915102 k fVc=

Diameter for punc
Weight = 2293.478729842 lbs fVc=
Weight = 2438.71793107 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 48.955 in.
(4/3)Mu= 9862.885 k-in./ft
Mr = 4958.957 k-in./ft
b=A/B= 1.062104
concrete cover =
cb=
ld = 38.00194 in. yt=
ld=

d= 50.295 in.
concrete cover =
(4/3)Mu= 10712.5 k-in./ft cb=
Mr = 4958.957 k-in./ft yt=

ye=
ys=
ld=

d v= 45 in.
4(fc') =
0.5
Vu/fVc= 1.441469121083
Limit State Allowab
d v= 45 in.
d v= 45 in.
2(fc') =
0.5
fVc= 1690.553637126 k
Vu/fVc= 1.326699875558
d= 50 in. d v= 45 in. Make sure larger w
d/2= 25 in. dv/2= 22.5 in. OK
wactual 21 in. < d/2 This procedure applies wactual
Vu 5238.000225 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 344 in. b0=
fVc= 4195.890701092 k ###
d= 50.295 in. d v= 45.2655 in.
w'= 21 in. dv/2= 22.6328 in.
Vu= 2164.4960775 k
Mu = 188610.4616963 k-in.
Mu/(Vud)= 1.732543560066
k= -3.89802245615
fVc= -3120.54901235 k
Vu/fVc= -0.69362668842
d= 48.955 in. d v= 44.0595 in.
w'= 47.352 in. dv/2= 22.0298 in.
Vu= 2166.2869875 k
Mu = 183994.3251536 k-in.
Mu/(Vud)= 1.734967654753
k= -1.69486061369
fVc= -1402.68449595 k
Vu/fVc= -1.54438649159

Diameter for punc
Weight = 1751.196959965 lbs fVc=
Weight = 2295.263935125 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###
Design Checks: ###

CRSI Limit State 6 Adequate? NG Vc=
Results: fVc=

d= 50.885 in.
(4/3)Mu= 8950.083 k-in./ft
Mr = 5264.054 k-in./ft
b=A/B= 1.101236
concrete cover =
cb=
ld = 46.84212 in. yt=
ld=

d= 52.295 in.
concrete cover =
(4/3)Mu= 12808.49 k-in./ft cb=
Mr = 5264.054 k-in./ft yt=

ye=
ys=
ld=

d v= 46.8 in.
4(fc') =
0.5
d v= 46.8 in.
fVc= 1758.175782611 k
bmodified
d v= 46.8 in.
2(fc') =
0.5
bmodified

d= 52 in. d v= 46.8 in.
d/2= 26 in. dv/2= 23.4 in.
wactual 20.5 in. < d/2 This procedure applies ###
Vu 5647.223887331 k ###
4(fc') =
0.5
252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 356 in. b0=
fVc= 4811.138224345 k ###
Vu/fVc= 1.173781260068 ###
w'= 20.5 in. dv/2= 23.5328 in. w'=
Vu= 2852.408349916 k Vu=
Mu = 238036.7491949 k-in. Mu =
Vud/Mu= 0.626654057255 Vud/Mu=
Mu/(Vud)= 1.595776790116 Mu/(Vud)=
k= -2.45049480548 k=
fVc= -2166.42125528 k fVc=
w'= 20.5 in. dv/2= 22.8983 in. w'=
Vu= 2615.691729916 k Vu=
Mu = 179342.0410774 k-in. Mu =
Vud/Mu= 0.742154337473 Vud/Mu=
Mu/(Vud)= 1.34742862705 Mu/(Vud)=
k= 0.644050302264 k=
fVc= 610.1253437423 k fVc=
Vu/fVc= 4.287138301569 **CONTROLLING** Vu/fVc= 4.28714 Vu/fVc=

Diameter for punc
Weight = 3155.987910797 lbs fVc=
Weight = 3481.405330932 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 50.955 in.
(4/3)Mu= 10292.51 k-in./ft
Mr = 5264.054 k-in./ft
b=A/B= 1.101236
concrete cover =
cb=
ld = 38.00194 in. yt=
ld=

d= 52.295 in.
concrete cover =
(4/3)Mu= 13666.15 k-in./ft cb=
Mr = 5264.054 k-in./ft yt=

ye=
ys=
ld=

d v= 46.8 in.
4(fc') =
0.5
d v= 46.8 in.
fVc= 1758.175782611 k
d v= 46.8 in.
2(fc') =
0.5
d/2= 26 in. dv/2= 23.4 in. NG, but answer co
Vu 6090.9806625 k Vu
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 364 in. b0=
fVc= 5171.523110064 k ###
Vu/fVc= 1.177792409096 ###
Check deep beam (one way shear through short width) at face of column ###
w'= 19.5 in. dv/2= 23.5328 in. w'=
Vu= 3075.6361125 k Vu=
Mu = 251960.239996 k-in. Mu =
Vud/Mu= 0.638356236308 Vud/Mu=
Mu/(Vud)= 1.566523428647 Mu/(Vud)=
k= -2.19253251528 k=
fVc= -1938.36323724 k fVc=
Vu/fVc= -1.58671814106 Vu/fVc=
w'= 19.5 in. dv/2= 22.9298 in. w'=
Vu= 2599.0969725 k Vu=
Mu = 208524.3047494 k-in. Mu =
Vud/Mu= 0.635115347311 Vud/Mu=
Mu/(Vud)= 1.574517139656 Mu/(Vud)=
k= -2.23853426463 k=
fVc= -2123.53806588 k fVc=
Vu/fVc= -1.22394649489 Vu/fVc=

Diameter for punc
Weight = 3258.666365719 lbs fVc=
Weight = 3639.651027793 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=
Pile Type Steel ###
f'c= 4 ksi ###
fy= 60 ksi ###

Results: fVc=

d= 41.885 in.
(4/3)Mu= 10691.39 k-in./ft
Mr = 3962.84 k-in./ft
b=A/B= 1.218182
concrete cover =
cb=
ld = 46.84212 in. yt=
ld=

d= 43.295 in.
concrete cover =
(4/3)Mu= 16283.85 k-in./ft cb=
Mr = 3962.84 k-in./ft yt=

ye=
ys=
ld=

d v= 38.7 in.
4(fc') =
0.5
d v= 38.7 in.
fVc= 1453.876127928 k
d v= 38.7 in.
2(fc') =
0.5
fVc= 1771.085464931 k
d/2= 21.5 in. dv/2= 19.35 in. OK
wactual 19 in. < d/2 This procedure applies ###
Vu 6595.8953125 k ###
4(fc')0.5= 252.9822128135 psi 4(fc')0.5=
32(fc')0.5= 2023.857702508 psi 32(fc')0.5=
b0= 332 in. b0=
fVc= 3310.292759538 k ###
Check deep beam (one way shear through short width) at face of column ###
w'= 19 in. dv/2= 19.4828 in. w'=
Vu= 3324.68203125 k Vu=
Mu = 302683.7238281 k-in. Mu =
Vud/Mu= 0.475552853396 Vud/Mu=
Mu/(Vud)= 2.102815686751 Mu/(Vud)=
k= -7.79750016969 k=
fVc= -5707.18684932 k fVc=
w'= 55 in. dv/2= 18.8483 in. w'=
Vu= 2611.31953125 k Vu=
Mu = 242600.6660156 k-in. Mu =
Vud/Mu= 0.450844263385 Vud/Mu=
Mu/(Vud)= 2.218060827686 Mu/(Vud)=
k= -3.02942282193 k=
fVc= -2613.12076003 k fVc=

Diameter for punc
Weight = 4733.981866195 lbs fVc=
Weight = 4558.649204484 lbs
d=
Straight dimension
Round dimension (
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
d/2=
Length dimension
Area in shear
4(fc')0.5=
Vc=
fVc=
Vu=
Vu/fVc=
d=
dcritical=
diagonal =
Area in shear
2(fc')0.5=
Vc=

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"CRSI-PILECAP (Bridges)" --- PILECAP DESIGN AND ANALYSIS SPREADSHEET

Worksheet Name Description

Program Assumptions and Limitations:

Terms Used in the Input Section of the Worksheets:

nimum size indicated or

e). Ultimate pile

bitrarily set at 5 in.

ar (e.g., 6 Pile Cap

igns and believes that

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars

Quantities of Concrete and Reinforcing Bars

Total weight of all steel = 113.1234 lbs or 0.056562 tons

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Three Ways

Check two way shear at column face - CRSI LIMIT STATE 4 -

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars

Check two way shear at column face - CRSI LIMIT STATE 4 -

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

ldh= 22.5169 in. NG; insufficient length available to develop hook

Required Flexural Reinforcement - Long Bars

ldh= 22.5169 in. NG; insufficient length available to develop hook

Check two way shear at column face - CRSI LIMIT STATE 4 -

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars

Check two way shear at column face - CRSI LIMIT STATE 4 -

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars

Check two way shear at column face - CRSI LIMIT STATE 4 -

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars

As,required,structural,total= 17.13242 in2 concrete cover =

As, required, Mr = 1.645489 in 2

As,required,Mr,total= 22.86111 in2 ys=

ld = 67.53241 in. lavailable=

lavailable= 114.5 in. OK; sufficient length available to develop bar

Quantities of Concrete and Reinforcing Bars

Design Checks: Equivalent Pile Dia

Required Flexural Reinforcement - Short Bars

Required Flexural Reinforcement - Long Bars