RISA Foundation Verification Problems

RISAFoundation Rapid Interactive Structural Analysis – Foundations Verification Problems

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Table of Contents

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Table of Contents Verification Problem 1: Strip Footing Design .............................................................................................................. 3 Verification Problem 2: Square Spread Footing #1 ................................................................................................... 5 Verification Problem 3: Rectangular Spread Foot #1.............................................................................................. 7 Verification Problem 4: Pile Cap Shear .......................................................................................................................... 9 Verification Problem 5: Eccentrically Loaded Footing ......................................................................................... 13 Verification Problem 6: Cantilever Retaining Wall #1 .......................................................................................... 15 Verification Problem 7: Cantilever Retaining Wall #2 ........................................................................................... 17 Verification Problem 8: Rectangular Footing #2 ...................................................................................................... 19 Verification Problem 9: Square Footing #2 ................................................................................................................ 21 Verification Problem 10: Cantilever Retaining Wall #3 ........................................................................................ 23 Verification Problem 11: Pile Cap Design Example ................................................................................................. 25

Appendices Appendix A10: Cantilever Retaining Wall #3 Calculations ....................................................................... A10.1 Appendix A11: Pile Cap Design Example Calculations ................................................................................ A11.1

Verification Overview

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Verification Overview

Verification Methods We at RISA Technologies maintain a library of hundreds of test problems used to validate the computational aspects of RISA programs. In this verification package we will present a representative sample of these test problems for your review and compare RISAFoundation to textbook examples listed within each problem. The input for these test problems was formulated to test RISAFoundation’s performance, not necessarily to show how certain structures should be modeled and in some cases the input and assumptions we use in the test problems may not match what a design engineer would do in a “real world” application. The data for each of these verification problems is provided. The files where these RISAFoundation problems are located is in the C:\RISA\Examples directory and they are called Verification Problem 1.fnd (2, 3, etc). Verification Version This document contains problems that have been verified in RISAFoundation version 5.0.

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Verification Problem 1: Strip Footing Design

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Design of a Wall Footing This problem represents a typical design of a wall footing. The hand verification of this problem can be taken directly from the 4th edition of Macgregor and Wight’s, Reinforced Concrete Mechanics and Design (Example 16-1, p.802-805). Description/Problem Statement A 12 in. thick concrete wall carries service dead and live loads of 10 kips per foot and 12.5 kips per foot, respectively. The allowable soil pressure, qa, is 5 ksf at the level of the base of the footing, which is 5 ft below the final ground surface. The wall footing has a strength of 3 ksi and fy = 60 ksi. The density of the soil is 120 lb/ft3. Note that the text does not account for the self-weight of the footing. Therefore, the RISA model has the density of the concrete material set to zero.

Figure 1.1 – RISAFoundation Model View Comparison Comparison of Results (Units Specified Individually) Value RISAFoundation Text Value % DifferenceFactored Net Pressure, qnu (ksf) 6.19* 6.19 0 Vu (k/ft) 7.87** 8.51*** 7.52

φ*Vc (k/ft) 9.613 9.37**** 2.59 Mu (k*ft/ft) 13.455 13.4 0.41 φ*Mn (k*ft/ft) 14.268 14.0 1.91As min (in^2) 1.451 1.45 0.07Table 1.1 – Results Comparison


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*The detail report for LC2 shows a Loading Diagram with 6.2 ksf on the toe end and 6.18 ksf on the heel. The average of these values is used in the above table. ** The detail report shows a Vu Toe = 7.88 k/ft and a Vu Heel = 7.86 k/ft. The average of these values is used in the above table. ***The value from the text is using a d = 8.5”. RISAFoundation is being more exact and using d = 13 – 3 – 0.5/2 = 9.75”. This produces a Vu = (1/12)*(25-9.75)*6.19 = 7.87 k/ft ****The value from the text is using d = 9.5” where RISAFoundation is being more exact and is using d = 9.75”. (9.75/9.5)*9.37 = 9.617 k/ft. Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design examples except in instances which are explained above.

Verification Problem 2: Square Spread Footing #1

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Design of a Square Spread Footing This problem represents a typical design of a square spread footing. The hand verification of this problem can be taken directly from the 4th edition of Macgregor and Wight’s Reinforced Concrete Mechanics and Design (Example 16-2, p.805-810). Description/Problem Statement A square spread footing supports an 18 in. square column supporting service dead and live loads of 400 kips and 270 kips, respectively. The column is built of 5 ksi concrete and has eight No. 9 longitudinal bars with fy = 60 ksi. The footing has concrete of strength 3 ksi and Grade-60 bars. The top of the footing is covered with 6 in. of fill with a density of 120 lb/ft3 and a 6 in. basement floor. The basement floor loading is 0.1 ksf. The allowable bearing pressure on the soil is 6 ksf. Load and resistance factors are taken from ACI sections 9.2 and 9.3.

Figure 2.1 – RISAFoundation Model View Solve the model and look at the detail report for the footing. Note that the text uses the net soil bearing to calculate the size of footing. This size is used directly in RISAFoundation and thus the soil overburden and self-weight are set to zero.


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Comparison Comparison of Results (Units Specified Individually) Value RISAFoundation Text Value % Difference Soil Pressure, qu (ksf) 7.31* 7.31 11.6 Vu Punching (k) 804.591 804 0.07 φ*Vc Punching (k) 0.75*1128.747= 846.56** 846 0.07 Vu One-Way (k) 204.254 204 0.12 φ*Vc One-Way (k) 0.75*411.134 = 308.35** 308 0.11 Mu (k*ft) 954.34 954 0.04 As Required (in^2) 7.763 8.41 7.7*** Table 2.1 – Results Comparison *To actually see this value, check the "Service" checkbox for LC 2 and solve the model. Then look at the detail report in the Soil Bearing section. When viewing the rest of the results, uncheck this checkbox and re-solve. **In RISAFoundation the Vc value is reported without the φ value. If the Vc value is multiplied by the text φ then there is agreement. ***If you use RISA’s value of As Required and calculate a new “a”, you will get a φ*Mn = 954.3 k*ft. This value exceeds Mu. The As required by the text is using a back of the envelope calculation to come up with As that is conservative in this case. When it comes to the calculation of φ*Mn RISA is following ACI 318-11 Section 10.5.3 in providing (4/3)*As required, whereas the text is not.

Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design examples except in instances which are explained above.

Verification Problem 3: Rectangular Spread Foot #1

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Design of a Rectangular Spread Footing This problem represents a typical design of a rectangular spread footing. The hand verification of this problem can be taken directly from the 4th edition of Macgregor and Wight’s Reinforced Concrete Mechanics and Design (Example 16-3, p.810-812). Description/Problem Statement Note that the text uses the net soil bearing to calculate the size of footing. This size is used directly in RISAFoundation and thus the soil overburden and self-weight are set to zero. This footing has been designed assuming that the maximum width is 9 ft. Following the hand calculation from the textbook the footing is found to be 9’ wide by 13’ 8” long by 32” thick. The example assumes the same net soil pressure of 7.31 ksf for both 16-2 and 16-3. However, (11.17 ft)2 = 124.77 ft2 and 13.666 ft * 9 ft = 123 ft2. Thus, the smaller footing in this example produces a slightly higher soil pressure than the text.

Figure 3.1 – RISAFoundation Detail Report View The text example uses #8 bars in one direction and #5 bars in the other for the bottom steel. In RISAFoundation this is not possible, so two footings have been created to verify the calculations. Node N1 is using the #8 bars and node N2 is using #5 bars. When viewing the results in RISAFoundation use the footing node numbers given in Table 3.1 below.


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Comparison Comparison of Results (Units Specified Individually) Value RISAFoundation Text Value % DifferenceVu One-Way (k) - N1 250.23 247 1.31 φ*Vc One-Way (k) - N1 0.75*331.263 = 248.45 248 0.18 Mu Long (k*ft) - N1 1234.69 1217 1.45 As Min Long (in2) - N1 6.221 6.22 0.02 As Provided Long (in2) - N1 10.21 in2 (13- #8 bars) 11.1 in2 (14-#8 bars)* 8.02 Mu Short (k*ft) - N1 712.5 702 1.5 As Min Short (in2) - N2 9.446 9.45 0.4

As Provided Short (in2) - N2 10.12 in2 (33 - #5 bars; 25 are banded)** 9.61 in2 (31-#5 bars; 25 are banded) 5.3 Table 3.1 – Results Comparison *In the text approximate methods are used to determine As Req’d. We can see that the ф*Mn = 1330 k*ft. RISAFoundation is able to remove a bar and still produce a ф*Mn greater than Mu. **In RISAFoundation the program is adding one extra bar to each side of the unbanded region. Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design examples, except in the instances explained above.

Verification Problem 4: Pile Cap Shear

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Design for Depth of Footing on Piles This problem represents the design for a footing supported on piles. The hand verification of this problem can be taken directly from PCA’s Notes on ACI 318-05 Building Code Requirements for Structural Concrete (Example 22.7, p.22-20). Description/Problem Statement Footing Size = 8.5’ x 8.5’ Column Size = 16” x 16” Pile Diameter = 12 in. f’c = 4000 psi Load per Pile: PD = 20 kips PL = 10 kips

Figure 4.1 – RISAFoundation Detail Report View Note that RISAFoundation will not place top steel reinforcement in a pile cap unless there is tension in the top face of the pile cap. For this reason a 1 kip*ft moment was added to the OL1 load category. This is to force top steel, as this affects the pile punching shear checks. If there is no reinforcement in the top then the program considers the cap unreinforced for punching shear calculations.


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Comparison Comparison of Results (Units in kips) Value RISAFoundation Text Value % DifferenceOne-way Beam Shear Capacity, φVn (kips) 180.629*0.75 = 135.47* 135.4 0.05 Pedestal Punching Shear Capacity, φVn (kips) 320/1.004 = 318.73** 319 0.08 Corner Pile Punching Shear Capacity, φVn (kips) 141.913 217 NA*** Table 4.1 – Results Comparison *The program gives Vn explicitly, so the Phi was multiplied in here to get Phi*Vn. **The Phi*Vn is not given explicitly. The program gives the demand and the code check, so the calculation above shows what Phi*Vn is in RISAFoundation. ***A couple of things are occurring here. For one, we are transforming the round punching shear perimeter into an equivalent square perimeter. Thus, this would create a difference. Second, and more importantly, the punching shear capacity is based on the smallest possible shear perimeter, bo. The PCA notes example assumes that the punching shear perimeter would occur all the way around the pile, as shown in Figure 4.2 below.

Figure 4.2 In reality, however, the crack will perpetuate through a distance “d” from the edge of the pile. D/2 occurs at midway along the crack and is used for calculation purposes. However, the crack would look like this in an elevation view, as shown in Figure 4.3.


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Figure 4.3 Because of this the punching shear perimeter can not be taken as shown in the PCA notes. Instead you really only have a partial perimeter because you will break out the corner before you get all the way around. In RISA, including the square perimeter adjustment, it would look as shown in Figure 4.4.

Figure 4.4


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Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design examples.

Verification Problem 5: Eccentrically Loaded Footing

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Footing Under Biaxial Moment This problem represents the case where a footing may be subjected to an axial force and biaxial moments about its x- and y-axes. This example comes from the Design of Reinforced Concrete Structures, copyright 1985 Hassoun (Example 13.7, p.409-413). Description/Problem Statement A 12” by 24” column of an unsymmetrical shed is subjected to an axial load PD = 220 kips and a moment MD = 180 k-ft due to dead load, and an axial load PL = 165 kips and a moment ML = 140 k-ft due to live load. The base of the footing is 5 ft. below final grade and the allowable soil bearing pressure is 5 ksf. The footing has strength of 4 ksi and a steel yield of 40 ksi. Note that the text does not account for the self-weight of the footing. Therefore, the RISA model has the density of the concrete material set to zero.

Figure 5.1 – RISAFoundation Detail Report View


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Comparison Comparison of Results (Units Specified Individually) Value RISAFoundation Text Value % DifferenceMethod 1 Soil Pressure, qn (ksf) 4.283 (87.1/90)*4.42= 4.277* 0.07 Method 1 Mu-xx (k*ft) 687.2 687.4 0.03 Method 1 Mu-zz (k*ft) 523.11 523.2 0.02 Method 2 Soil Pressure Max, qn (ksf) 4.43 4.42** 0.23 Method 2 Soil Pressure Min, qn (ksf) 1.973 1.98 0.35 Method 2 Mu-xx (k*ft) 873.6 873 0.07 Table 5.1 – Results Comparison *The text book calculates a required area of 87.1 in^2. They then choose an area of 90 in^2. Thus, their value has been adjusted. **The text book example has an error. They state that 3.20 + 1.22 = 4.22 ksf when calculating qmax for method 2. This should be 4.42 ksf. Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design examples.

Verification Problem 6: Cantilever Retaining Wall #1

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Design of a Cantilever Retaining Wall This example comes from the Principles of Foundation Engineering, 3rd Edition by Das, copyright 1995. This is example A.8 on P798. In this problem we will compare the serviceability checks for a retaining wall example to the output from RISAFoundation. Description/Problem Statement The cross section of a cantilever retaining wall is shown below. For this case, fy = 413.7 MN/m2 and f’c = 20.68 MN/m2. Notes:

- RISAFoundation uses Rankine’s method to calculate lateral soil pressure coefficients. This example uses Coulombs method. Because of this the KLat Toe was set to 2.04. - The coefficient of friction in this example is calculated as: Tan (2/3*φ) = 0.237. This is the value entered in the program. - The ultimate bearing pressure is in this example is calculated as 574.07, so this is entered as the allowable bearing in the program.



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Comparison Comparison of Results (Units Specified Individually) Value RISAFoundation Text Value % Difference Mresist Against Overturning (kN-m/m) 1030.034 1044.3 (1128.98)* 1.37 Moverturning (kN-m/m) 379.047 379.25 0.05 Vresist Against Sliding (kN/m) 147.278 433.17 – 171.39 –106.67 = 155.1** 5.04 Vsliding (kN/m) 158.853 158.95 0.06 Max Bearing Pressure (kPa) 199.349 189.2*** 5.36 Bearing UC .347 189.2/574.07 = .329*** 5.47 Table 6.1 – Results Comparison *The text book accounts for the sloping outer face of the wall, which RISAFoundation does not. Also, the vertical portion of the soil force in the text is assumed to act at the edge of the heel. In RISAFoundation we assume this force to act at the inside face of the wall. These differences would equal 1128.98 – 11.79 – 2.6*28.03 = 1044.312 kN-m/m. **The text book assumes cohesion. RISAFoundation assumes cohesion-less soil. They give a Vresist = 111.5 + 106.7 + 215 = 433.17 kN/m. The 106.7 is a cohesion term that RISA doesn’t account for. The 215 comes passive pressure including cohesion. The cohesion term = 171.39 kN/m which RISA doesn’t account for. Accounting for these cohesion differences between RISAFoundation and the text gives a value = 433.17 – 171.39 - 106.67 = 155.1 kN/m. ***The text uses the Mresist to calculate the bearing pressure. Because this is different, the pressure calculation is different. Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design examples after accounting for differences in calculation procedures.


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Design of Reinforced Concrete Cantilever Retaining Walls In this problem we will compare the serviceability checks for a retaining wall example to the output from RISAFoundation. This example comes from Reinforced Concrete Design, Third Edition, copyright 1992 by Spiegel and Limbrunner. This is design example 8-1 on P214. Description/Problem Statement Design Data: unit weight of earth we = 100 lb/ft3, allowable soil pressure = 4,000 psf, equivalent fluid weight Kawe = 30 100 lb/ft3, and surcharge load ws = 400 psf. The desired factor of safety against overturning is 2.0 and against sliding is 1.5.



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Note: The shear key has been omitted from the RISAFoundation model, as this will affect the calculations for sliding and overturning. The text example did not assume a key when performing those calculations. Comparison

Value RISAFoundation Text Value % Difference M Resist (k*ft) 131.169 131.7 0 M Overturn (k*ft) 48.6 48.6 0 V Resist (kips) 10.008 9.855 1.55 V Slide (kips) 7.02 7.02 0 Max Soil Pressure (ksf) 3.101 3.043 1.9 Mu of Heel (k*ft) 46.69 67.65 NA* Vu Heel (k*ft) 11.22 20.82 NA* ϕVn of Heel (kips)

18.301* (0.85/0.75) = 20.74** 20.76 0.1 As Top (in2) #7 Bars @ 8" oc #7 Bars @ 8" oc 0 Mu of Toe (k*ft) 18.473 20.476 NA*** Vu of Toe (kips) 6.47 13.07 NA**** ϕVn of Toe (kips)

17.315* (0.85/0.75) = 19.62** 19.64 0.1 As Bot (in2) #7 Bars @ 16" oc #7 Bars @ 16" oc 0 Mu Stem Base (k*ft) 63.4 63.431 0.05 Vu Stem Base (kips) 10.023 (LC2) 10.049 0.26 ϕVn of Stem (kips) 15.281*(0.85/0.75) = 17.318 17.391 0.42 As Stem (in2) #8 Bars @ 9" oc #8 Bars @ 9" oc 0 Table 7.1 – Results Comparison *In the text example the "relieving" moment due to the upward soil pressure on the heel is not accounted for. This is accounted for in RISA. **This value is being adjusted for the change in ϕshear from 0.85 to 0.75. ***In the text example the "relieving" moment due to the downward soil pressure on the toe is not accounted for. This is accounted for in RISA. ****In the text example the shear location is taken as the face of wall. In RISA we are coming out a distance "d" from the wall and check the shear at that location.

Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design example.

Verification Problem 8: Rectangular Footing #2

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Rectangular Reinforced Concrete Footings This problem represents a typical design of a rectangular spread footing. This example comes from Reinforced Concrete Design, Third Edition, copyright 1992 by Spiegel and Limbrunner. This is design example 10-4 on P310. Description/Problem Statement A concrete footing 4 ft. below the finished ground line supports an 18-in. square tied interior concrete column. The total footing thickness is 24 in. One dimension of the footing is limited to a maximum of 7 ft. Service DL = 175 kips Service LL = 175 kips f’c (footing and column) = 3000 psi Steel Yield fy = 60 ksi Longitudinal column steel = No. 8 bars Soil Density = 100 lb/ft3 Allowable Soil Pressure = 5 ksf Effective Allowable Soil Pressure = 4.50 ksf

Figure 8.1 – RISAFoundation Detail Report View Note that the self-weight and overburden were input as zero and the allowable soil pressure was added directly as 4.50 ksf.


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Comparison Comparison of Results (Units Specified Individually) Value RISAFoundation Text Value % Difference Factored Soil Pressure, qu (ksf) 6.739* 6.74 0.01 Shear Demand, Vu two-way (k) 474.921 475 0.02 Shear Capacity, ϕVn two-way (k) ϕ*666.031 = 566.13 (ϕ=0.85)** 566 0.02 Shear Demand, ϕVu one-way (k) 157.246 157.1 0.09 Shear Strength, ϕVn one-way (k) ϕ*184.035 = 156.43 (ϕ=0.85)** 156.4 0.17 Bending Moment, Mu long direction (k*ft) 589.67 590 0.06 Bending Moment, Mu short direction (k*ft) 293.05 293 0.02 As required long direction (in2) 6.884 6.9 0.23 As required short direction (in2) 3.303 4.4/(4/3) = 3.3*** 0.09 As required T & S (in2) 5.962 5.96 0.03 Footing Bearing Strength (in2) ϕ*1652.4 = 1156.68 (ϕ=0.70)**** 1157 0.03 Factored Bearing Load, Pu (k) 542.5 542.5 0.00 Table 8.1 – Results Comparison *To actually see this value, check the "Service" checkbox for LC 2 and solve the model. Then look at the detail report in the Soil Bearing section. When viewing the rest of the results, uncheck this checkbox and re-solve. **In RISAFoundation the Vc value is reported without the φ value. If the Vc value is multiplied by the text φ then there is good agreement. ***In the text they are multiplying by 4/3*As required as their value. RISAFoundation will do this as well when actually reinforcing the footing, however, we also report the As required itself. ****In RISAFoundation the Bc value is reported without the φ value. If the Bc value is multiplied by the text φ then there is good agreement.

Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the textbook design example.

Verification Problem 9: Square Footing #2

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Design for Base Area, Depth, and Reinforcement of Footing This problem represents a typical design of a square spread footing The hand calculation comparison of this example comes from the PCA Notes for the ACI 318-05 Example 22.1, 22.2 and 22.3 (all in one problem) on page 22-7. Description/Problem Statement Service Dead Load = 350 kips Service Live Load = 275 kips Service Surcharge = 100 psf Weight of Soil and Concrete above Footing Base = 130 lb/ft3 Net Allowable Soil Pressure = 3.75 ksf

Figure 9.1 – RISAFoundation Detail Report View Notes: • Because the example does not use the self-weight of the footing in the calculation and instead just gives an average weight between the soil and concrete, the density of


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concrete has been set to 0. The Overburden has also been set to zero. Thus, the allowable soil pressure is simply added directly as 3.75 ksf. • The dfoot value for footings in RISAFoundation = footing thickness – bottom cover – 1*db. The examples use a d = 28”, thus the bottom cover is set to 4”.

Comparison Comparison of Results (Units Specified Individually) Value RISAFoundation Text Value % Difference Ex 22.1: qs (ksf) 5.089* 5.1 0.22 Ex 22.2 Shear Demand, Vu one way (k) 242.564 243 0.18 Ex 22.2 Shear Capacity, φVn one way (k) φ*478.5 = 358.868 (φ = 0.75)** 359 0.04 Ex 22.2 Shear Demand, Vu two way (k) 778.014 780 0.25 Shear Capacity, φVn two way (k) φ*1082 = 811.593 (φ = 0.75)** 812 0.05 Ex 22.2 Bending Moment, Mu (k*ft) 1190.77 1193 0.12 Ex 22.3 As required (in2) 9.704 9.6 1.08 Table 9.1 – Results Comparison *To actually see this value, check the "Service" checkbox for LC 2 and solve the model. Then look at the detail report in the Soil Bearing section. When viewing the rest of the results, uncheck this checkbox and re-solve. **RISAFoundation presents the Vc value without φ. When you multiply Vc by φ you get agreement.

Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the PCA Notes design examples.

Verification Problem10: Cantilever Retaining Wall #3

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Design of a Cantilever Retaining Wall In this example we have a non-sloping back-filled retaining wall with a load surcharge and a water table present. The wall and footing are not poured monolithically. Footing dowels occur at both faces of the wall and are of the same size and spacing as the wall reinforcement. A load combination of 1.0*DL+ 1.0*LL + 1.0*HL is used for the service LC and a load combination of 1.2*DL + 1.6*LL + 1.6*HL is used for the strength LC. In this example RISAFoundation’s values are compared to the values obtained from a hand calculation done for soil pressures, stability and all design aspects of the wall. This hand calculation is located in Appendix A10 Description/Problem Statement This problem comes from a hand calculation verification. It is testing all results for retaining wall stability, soil pressure calculations and reinforcement design.

Figure 10.1 – RISAFoundation Detail Report View Note: The retaining wall is cantilevered and the base is not restrained against sliding.

Verification Problem10: Cantilever Retaining Wall #3

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Comparison This section is the tabular comparison of the RISAFoundation answers and the summary from the detailed validation results. Comparison of Results (Units Specified Individually)

Value RISAFoundation Hand Calculation % Difference Lateral Earth Pressures NA

KLat Heel 0.307 0.307 0

KLat Heel Sat 0.333 0.333 0

KLat Toe 3.255 3.255 0

Stability Checks Overturning SF Min/SF 0.659 0.659 0 Sliding SF Min/SF 1.176 1.176 0

Wall Design UC Max Int 1.664 1.678 0.834 Shear UC Max 0.624 0.627 0.478 Dowel Shear UC Max 0.455 0.455 0

Footing Soil Pressures qmax (ft)* 5.6 5.603 0.054 Lsoil Length (ft)* 9.09 9.090 0

Footing Design Shear UC Heel 0.746 0.746 0 Moment UC Heel 0.967 0.967 0 Shear UC Toe 0.597 0.597 0 Moment UC Toe 0.63 0.630 0 Table 10.1 – Results Comparison *Note that the values shown here can be seen graphically by looking at the detail report for load combination 2. **See Appendix A10 for an in depth hand calculation.

Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the hand calculated design example.

Verification Problem11: Pile Cap Design Example

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Verification Problem 11: Pile Cap Design Example

Design of a Pile Cap In this example we have a pile cap with 12 HP14x102 piles providing support. The piles have an 85 kip compression capacity, a 12 kip tension capacity and a 14 kip shear capacity. The pile cap is 42" thick with a 6" pile embedment and made from 4 ksi lightweight concrete. A load combination of 1.0*DL+ 1.0*LL is used for the service LC and a load combination of 1.2*DL + 1.6*LL is used for the strength LC. Description/Problem Statement In this example RISAFoundation’s values are compared to the values obtained from a hand calculation done for all aspects of the pile cap. This hand calculation is located in Appendix A11.

Figure 11.1 – RISAFoundation Model View

Verification Problem11: Pile Cap Design Example

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Comparison This section is the tabular comparison of the RISAFoundation answers and the summary from the detailed validation results. Comparison of Results (Units Specified Individually)

Value RISAFoundation Hand Calculation % Difference Flexural Checks

Muxx (k-ft) 1432.03 1438 0.42

Muzz (k-ft) 937.13 932.8 0.46

Asminx (in^2) 13.835 13.835 0

Asminz (in^2) 10.13 10.13 0

Asflexx bot (in^2) 20.588 20.588 0

Asflexz bot (in^2) 15.075 15.075 0

UC Mx 0.755 0.753 0.27 UC Mz 0.445 0.488 8.81

Punching Shear Checks Pedestal Punching UC 0.719 0.719 0 Pile 4 Punching Capacity

(kips) 220.284 220.284 0

Pile 4 Punching UC 0.399 0.399 0 One Way Shear Checks

Shear Capacity Vcx (kips) 1186.972 1187 0

Shear Capacity Vcz (kips) 585.931 591.221 0.89

Pedestal Shear Capacities Vc (kips) 48.952 48.952 0

Vs (kips) 50.658 50.658 0 Table 11.1 – Results Comparison *Note that the values shown here can be seen graphically by looking at the detail report for the pile cap. **See Appendix A11 for an in depth hand calculation. Conclusion In this example it is shown that the RISAFoundation calculations reasonably match the hand calculated design example.

AppendixA10CantileverRetainingWall#3Calculations_________________________________________________________________________

Inthisexamplewehaveanon-slopingback- illedretainingwallwithaloadsurchargeandawatertablepresent.Herewewillcalculateallsoilpressures,designallaspectsoftheretainingwallandcheckforoverturningandsliding.InputParameters

Theretainingwalliscantileveredandthebaseisnotrestrainedagainstsliding.Thewallandfootingarenotpouredmonolithically.Footingdowelsoccuratbothfacesofthewallandareofthesamesizeandspacingasthewallreinforcement.Inthisexamplewewillusealoadcombinationof1.0*DL+1.0*LL+1.0*HLfortheserviceLCandaloadcombinationof1.2*DL+1.6*LL+1.6*HLforthestrengthLC.

≔DLFactor 1.2 ≔LLFactor 1.6 ≔HLFactor 1.6

Geometry

≔Hwall ⋅16 ft Hwall=Hsoil ≔Ltoe ⋅3.5 ft ≔Wkey ⋅18 in

≔Hwater ⋅6 ft ≔Lheel ⋅5.5 ft≔Dkey ⋅18 in

≔twall ⋅18 in ≔tfoot ⋅18 in≔Lkey ⋅4.5 ft

≔Lwall ⋅10 ft Totallengthofwall≔Lfoot =++Ltoe Lheel tfoot 10.5 ft Overalllengthofthefooting

≔Offsetkey =−−+Lkey Wkey Ltoe twall 1 ft Thekeyoffsetfromtheinteriorfaceofwallandtheinteriorfaceofkey.Materials

≔γconc ⋅.15 ――kip

ft3≔fc ⋅4 ksi ≔fy ⋅60 ksi

A10.1

Soil

≔μ 0.5 Coefoffrictionw/soil ≔β 0 back illangle ≔Htoesoil ⋅2 ft

≔Soilallow ⋅5 ksf ≔q ⋅500 psf surcharge≔γw ⋅62.4 pcf

≔γm ⋅115 pcf

≔ϕm ⋅32 deg

≔γs ⋅125 pcf

≔ϕs ⋅30 deg

≔SF 1.5 Thisisthesafetyfactorrequiredforbothslidingandoverturning.Note:Themoistsoilpropertiesarealsousedforthetoesoil.

A10.2

WallReinforcingProperties

≔dbinside ⋅0.75 in≔s ⋅8 in spacingofverticalbars

≔dbhoriz ⋅0.5 in≔swallhoriz ⋅10 in spacingofhorizontalbars

≔dboutside 0.5 in

≔Numfaces 2 Twofacesofreinforcement≔Asinside =――――

⋅dbinside2

40.442 in2 #6barsinterior.

≔Asoutside =――――⋅dboutside

2

40.196 in2 #4barsexterior

≔Ashoriz =――――⋅⋅2 dbhoriz

2

40.393 in2 #4barshorizontaleachface

≔coverinside ⋅2 in ≔coveroutside ⋅1 in

Theouterbarsareinthehorizontaldirection.

A10.3

FootingReinforcingProperties

≔dbtop ⋅0.75 in ≔stop 8 in

≔dbbot ⋅0.75 in≔sbot 8 in

≔dblong 0.5 in≔slong 16 in

≔Astop =―――⋅dbtop

2

40.442 in2 #6barsat8"spacingtop

≔Asbot =―――⋅dbbot

2

40.442 in2 #6barsat8"spacingbot

≔Aslong =――――⋅⋅2 dblong

2

40.393 in2 #4barsat16"spacinglongitudinaleachface

≔covertop 2 in ≔coverbot 3 in

A10.4

Calculations

ThissectionbreaksdownallofthecalculationsthatoccurwithinRISAFoundationforretainingwalldesign.ForceCalculationsForOverturning,SlidingandWallDesign

LateralEarthPressureCoef icients

=β 0 =ϕm 32 deg =ϕs 30 deg

≔Kam =⋅cos ((β))

⎛⎜⎜⎜⎝

―――――――――――−cos ((β))

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾−((cos ((β))))

2

⎛⎝cos ⎛⎝ϕm⎞⎠⎞⎠2

+cos ((β))‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

−((cos ((β))))2

⎛⎝cos ⎛⎝ϕm⎞⎠⎞⎠2

⎞⎟⎟⎟⎠

0.307

≔Kpm =⋅cos ((β))

⎛⎜⎜⎜⎝

―――――――――――+cos ((β))

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾−((cos ((β))))

2

⎛⎝cos ⎛⎝ϕm⎞⎠⎞⎠2

−cos ((β))‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

−((cos ((β))))2

⎛⎝cos ⎛⎝ϕm⎞⎠⎞⎠2

⎞⎟⎟⎟⎠

3.255

≔Kas =⋅cos ((β))

⎛⎜⎜⎜⎝

―――――――――――−cos ((β))

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾−((cos ((β))))

2

⎛⎝cos ⎛⎝ϕs⎞⎠⎞⎠2

+cos ((β))‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

−((cos ((β))))2

⎛⎝cos ⎛⎝ϕs⎞⎠⎞⎠2

⎞⎟⎟⎟⎠

0.333

LateralPressureCalculations(Service)

≔P1 =⋅Kam q 153.629 psf

≔P2 =⋅Kam⎛⎝ +q ⋅⎛⎝ −Hwall Hwater

⎞⎠ γm⎞⎠ 506.977 psf

≔P3 =⋅Kas⎛⎝ +q ⋅⎛⎝ −Hwall Hwater

⎞⎠ γm⎞⎠ 550 psf

≔P4 =++P3 ⋅⋅Kas Hwater⎛⎝ −γs γw⎞⎠ ⋅Hwater γw

⎛⎝ ⋅1.05 103 ⎞⎠ psf

≔P5 =++P4 ⋅⋅Kas tfoot ⎛⎝ −γs γw⎞⎠ ⋅tfoot γw⎛⎝ ⋅1.175 103 ⎞⎠ psf

≔P6 =++P5 ⋅⋅Kas Dkey⎛⎝ −γs γw⎞⎠ ⋅Dkey γw

⎛⎝ ⋅1.299 103 ⎞⎠ psf

A10.5

≔P7 =⋅⋅Htoesoil γm Kpm 748.555 psf

≔P8 =⋅⋅⎛⎝ +Htoesoil tfoot⎞⎠ γm Kpm⎛⎝ ⋅1.31 103 ⎞⎠ psf

≔P9 =⋅⋅⎛⎝ ++Htoesoil tfoot Dkey⎞⎠ γm Kpm

⎛⎝ ⋅1.871 103 ⎞⎠ psf

A10.6

LateralResultantForceLocationsforOverturning

≔H1 =⋅⎛⎝ +Hwall tfoot⎞⎠ 0.5 8.75 ft

≔H2 =++⋅⎛⎝ −Hwall Hwater⎞⎠⎛⎜⎝―1

3

⎞⎟⎠

Hwater tfoot 10.833 ft

≔H3 =⋅⎛⎝ +Hwater tfoot⎞⎠ 0.5 3.75 ft

≔H4 =⋅⎛⎝ +Hwater tfoot⎞⎠⎛⎜⎝―1

3

⎞⎟⎠

2.5 ft

≔H5 =⋅―1

3⎛⎝ +Htoesoil tfoot⎞⎠ 1.167 ft

A10.7

LateralForceSummationsforOverturning,SlidingandWallDesign

≔LF1 =⋅P1 ⎛⎝ +Hwall tfoot⎞⎠ 2.689 ――kip

ftThisvaluechangesforall3calculations.

≔LF1slide =+LF1 ⋅P1 Dkey 2.919 ――kip

ft

≔LF1wall =⋅P1 ⎛⎝Hwall⎞⎠ 2.458 ――

kip

ft Thisisthesamevalueforall3calculations.≔LF2 =⋅⋅―1

2⎛⎝ −P2 P1⎞⎠ ⎛⎝ −Hwall Hwater

⎞⎠ 1.767 ――kip

ft

≔LF3 =⋅⎛⎝ −P3 P1⎞⎠ ⎛⎝ +Hwater tfoot⎞⎠ 2.973 ――kip


≔LF3Slide =+LF3 ⋅⎛⎝ −P3 P1⎞⎠ Dkey 3.567 ――kip

ft

≔LF3wall =⋅⎛⎝ −P3 P1⎞⎠ Hwater 2.378 ――kip

ft

≔LF4 =⋅⋅⎛⎝ −P5 P3⎞⎠⎛⎜⎝―1

2

⎞⎟⎠⎛⎝ +Hwater tfoot⎞⎠ 2.342 ――

kip


≔LF4slide =⋅⋅⎛⎝ −P6 P3⎞⎠⎛⎜⎝―1

2

⎞⎟⎠⎛⎝ ++Hwater tfoot Dkey

⎞⎠ 3.372 ――kip

ft

≔LF4wall =⋅⋅⎛⎝ −P4 P3⎞⎠⎛⎜⎝―1

2

⎞⎟⎠Hwater 1.499 ――

kip

ft

≔LF5 =⋅⋅―1

2P8 ⎛⎝ +Htoesoil tfoot⎞⎠ 2.292 ――

kip


≔LF5slide =⋅⋅―1

2P9 ⎛⎝ ++Htoesoil tfoot Dkey

⎞⎠ 4.678 ――kip

ft

≔LF5wall =⋅⋅―1

2P7 Htoesoil 0.749 ――

kip

ft

A10.8

VerticalForceCalculations(ServiceandStrength)

≔w1 =⋅⋅Hwall twall γconc 3.6 ――kip

ft ≔w1f =⋅DLFactor w1 4.32 ――kip

ft

≔w2 =⋅⋅tfoot ⎛⎝ ++twall Ltoe Lheel⎞⎠ γconc 2.363 ――kip

ft≔w2f =⋅DLFactor w2 2.835 ――

kip

ft≔w3 =⋅⋅Wkey Dkey γconc 0.338 ――

kip

ft ≔w3f =⋅DLFactor w3 0.405 ――kip

ft

≔w4 =⋅⋅Ltoe Htoesoil γm 0.805 ――kip


kip

ft

≔qtotal =⋅q Lheel 2.75 ――kip

ft≔qtotalf =⋅LLFactor qtotal 4.4 ――

kip

ft

≔w5 =⋅⋅Lheel ⎛⎝ −Hwall Hwater⎞⎠ γm 6.325 ――

kip


kip

ft

≔w6 =⋅⋅Lheel Hwater γs 4.125 ――kip


kip

ft

A10.9

VerticalForceCentroids

≔D1 =+Ltoe ――twall

24.25 ft

≔D2 =――Lfoot

25.25 ft

≔D3 =+Lkey ――Wkey

25.25 ft

≔D4 =――Ltoe

21.75 ft

≔D5 =++Ltoe twall ――Lheel

27.75 ft

≔D6 =D5 7.75 ft

StabilityChecks

OverturningThischeckistakenfromthebaseofthetoeofthefooting.≔MR1 =+++⋅w1 D1 ⋅w2 D2 ⋅w3 D3 ⋅w4 D4 30.884 ⋅kip ―

ft

ft

≔MR2 =++⋅w5 D5 ⋅⎛⎝ +w6 qtotal⎞⎠ D6 ⋅LF5 H5 104.975 ⋅kip ―ft

ft

≔MR =+MR1 MR2 135.858 ⋅kip ―ft

ft

≔MOT =+++⋅H1 LF1 ⋅H2 LF2 ⋅H3 LF3 ⋅H4 LF4 59.667 ⋅kip ―ft

ft

≔OSF =――MR

MOT

2.277

≔UCOT =――SF

OSF0.659

Thisretainingwallpassestheoverturningcheckbecauseithasgreaterthana1.5safetyfactor.A10.10

SlidingThischeckistakenfromthebottomofkeyelevation.≔FSlide =+++LF1slide LF2 LF3Slide LF4slide 11.625 ――

kip

ft

≔R =++++++w1 w2 w3 w4 w5 w6 qtotal 20.305 ――kip

ftTotalverticalforce

≔FFriction =⋅R μ 10.153 ――kip

ft

≔LF8 =⋅⋅―1

2P9 ⎛⎝ ++Htoesoil tfoot Dkey

⎞⎠ 4.678 ――kip

ftTheforcesresistingslidingareduetobothfrictionandpassivepressureonthetoesideofthefooting.≔FResist =+FFriction LF5slide 14.831 ――kip

ft

≔SafetyFactorSliding =―――FResist

FSlide1.276

≔UCSliding =――――――SF

SafetyFactorSliding1.176

Thisretainingwallfailstheslidingcheckbecauseithaslessthana1.5safetyfactor.

A10.11

DesigningtheWallStemThewallstemwaspouredseparatelyfromthefooting.Wherethewallispouredthefootinghasnotbeenintentionallyroughened.Footingdowelsoccuratbothfacesofthewallandareofthesamesizeandspacingasthewallreinforcement.=Lwall 10 ft =Hwall 16 ft =twall 1.5 ft

=Asinside 0.442 in2 #6barsinterior. =coverinside 2 in

=Asoutside 0.196 in2 #4barsexterior =coveroutside 1 in

=Ashoriz 0.393 in2 #4barshorizontaleachface=s 8 in

=swallhoriz 10 in=Numfaces 2 Theouterbarsareinthehorizontaldirection.

AxialandBendingDesign(perfoot)Thesearethecentroidheightsofeachportionofload.≔H1wall =――Hwall

28 ft

≔H2wall =+Hwater ―――――⎛⎝ −Hwall Hwater

⎞⎠3

9.333 ft

≔H3wall =―――Hwater

23 ft

≔H4wall =―――Hwater

32 ft

≔H5wall =―――Htoesoil

30.667 ft

A10.12

≔Pu ⋅0 kip

≔Mwalls −+++⋅LF1wall H1wall ⋅LF2 H2wall ⋅LF3wall H3wall ⋅LF4wall H4wall ⋅LF5wall H5wall

=Mwalls 45.787 ⋅kip ―ft

ft

=HLFactor 1.6

≔Mwallf =⋅HLFactor Mwalls 73.26 ⋅kip ―ft

ft

≔dcant =−−−twall coverinside dbhoriz ―――dbinside

215.125 in

≔dprime =++coveroutside dbhoriz ―――dboutside

21.75 in

≔awall =――――⋅Asinside fy

⋅⋅0.85 fc s0.975 in ≔aprime =――――

⋅Asoutside fy

⋅⋅0.85 fc s0.433 in

≔Mnwall ⋅――――――――

⋅⋅Asinside fy⎛⎜⎝

−dcant ――awall

2

⎞⎟⎠

12―12

s

=Mnwall 48.501 ⋅kip ―ft

ftThisisthemomentcapacityinthewallnotconsideringcompressionreinforcement

A10.13

≔ϕwall 0.9

Note:Theprogramtakesintoaccountcompressionreinforcementaswell,sotheprogramreportedvalueisalittlelarger(44.029).≔PhiMnwall =⋅ϕwall Mnwall 43.651 ⋅kip ―

ft

ft

≔BendingInteraction =――――Mwallf

PhiMnwall1.678

ReinforcementProvidedChecks(forentirewall)

HorizontalReinforcement

≔BarsHoriz1 =⋅Numfaces ―――Hwall

swallhoriz38.4

≔BarsHoriz =round⎛⎝BarsHoriz1⎞⎠ 38 Thetotalnumberofhorizontalbarsinthewall.≔Asprovh =⋅BarsHoriz ―――

Ashoriz

27.461 in2 Asprovided(H)

≔rhoprovh =―――――Asprovh

⋅⋅12 Hwall twall⋅1.799 10−4 RhoProvided(H)

≔rhominh .002 Rhomin(H)≔Asminh =⋅⋅rhominh Hwall twall 6.912 in2 Asmin(H)

InsideFaceVerticalReinforcement

≔BarsVertInt1 =――Lwall

s15

≔BarsVertInt =round⎛⎝BarsVertInt1⎞⎠ 15 Thetotalnumberofinteriorverticalbarsinthewall.≔Asprovint =⋅BarsVertInt Asinside 6.627 in2 IntAsProvided(V)≔rhoprovint =―――――

Asprovint

⋅⋅Lwall twall 12⋅2.557 10−4 IntrhoProvided(V)

A10.14

OutsideFaceVerticalReinforcement

≔BarsVertExt1 =―――⋅Lwall 12

s180

≔BarsVertExt =round⎛⎝BarsVertExt1⎞⎠ 180 Thetotalnumberofexteriorverticalbarsinthewall.≔Asprovext =⋅BarsVertExt Asoutside 35.343 in2 ExtAsProvided(V)≔rhoprovext =―――――

Asprovext

⋅⋅Lwall twall 120.001 ExtrhoProvided(V)

TotalVerticalReinforcement

≔rhominv .0015 rhomin(V)≔Asminv =⋅⋅⋅rhominv Lwall 12 twall 38.88 in2 Asmin(V)

ShearDesign

Concretecheck:

≔Vwallds1 =++⋅LF1wall ――――−Hwall dcant

Hwall

⋅LF3wall⎛⎜⎝―――――

−Hwater dcant

Hwater

⎞⎟⎠

LF2 5.91 ――kip

ft

≔Vwallds2 =−⋅―――−P4 P3

2

⎛⎜⎜⎝――――――⎛⎝ −Hwater dcant⎞⎠

2

Hwater

⎞⎟⎟⎠

⋅―P7

2――――――⎛⎝ −Htoesoil dcant⎞⎠

2

Htoesoil

0.833 ――kip

ft

Fortheconcretecheckweareusingtheshearforceatadistancedfromthebase.≔Vwallds =+Vwallds1 Vwallds2 6.743 ――

kip

ft ≔fc ⋅4000 ――lbf2

in4

≔Vwalldf =⋅HLFactor Vwallds 10.788 ――kip

ft ≔ϕv 0.75

≔Vc =⋅⋅2 ‾‾fc dcant⎛⎝ ⋅2.2958 104 ⎞⎠ ――

lbf

ft

A10.15

≔PhiVcwall =⋅ϕv Vc⎛⎝ ⋅1.7219 104 ⎞⎠ ――

lbf

ft

≔ShearConcInteraction =――――Vwalldf

PhiVcwall0.627

SteelCheck(shearfriction)Inthisexamplethewallisnotpouredmonolithicallywiththefooting.AllcodereferencesarepertheACI318-11.≔Vwallbases =−+++LF1wall LF2 LF3wall LF4wall LF5wall 7.353 ――

kip

ft

≔Vwallbasef =⋅HLFactor Vwallbases 11.765 ――kip

ft HereweareusingtheAsofthewallreinforcing,asthedowelsfromthefoundationmatchthewallr/f.≔Avf =―――――――――

⋅⋅⎛⎝ +Asinside Asoutside⎞⎠ 12 ―in

ft

s0.957 ――

in2

ft

=fy 60 ksi

≔μconc 0.6 Thisassumesthatthesurfaceofthefootingwherethewallispouredisnotintentionallyroughened.≔Vn =⋅⋅Avf fy μconc 113.056 ⋅―

1

mkip Equation11-25

PerSection11.6.6fymustbetaken<=60ksi.=twall 18 in ≔lwall ⋅12 ―

in

ft perfootdistance≔Ac =⋅twall lwall 216 ――

in2

ft≔fc ⋅4 ksi

Theequationsbelowarebasedonsection11.6.5.NotethattheprovisionsaredifferentintheACI318-02andACI318-05andcomefromsection11.7.5.

A10.16

≔Vn1 =⋅⋅0.2 fc Ac 172.8 ――kip

ft

≔Vn2 =⋅⎛⎝ +⋅480 psi ⋅0.08 fc⎞⎠ Ac 172.8 ――kip

ft≔Vn4 =⋅⋅0.2 fc Ac 172.8 ――

kip

ft

≔Vn3 =⋅⋅1600 psi Ac 345.6 ――kip

ft≔Vn5 =⋅⋅800 psi Ac 172.8 ――

kip

ft

≔Vnrough min ⎛⎝ ,,,Vn Vn1 Vn2 Vn3⎞⎠ ≔Vnsmooth =min ⎛⎝ ,,Vn Vn4 Vn5

⎞⎠ 34.459 ――kip

ft

≔SteelConcInteraction =――――Vwallbasef

⋅ϕv Vnsmooth

0.455

DesigningtheFooting

SoilPressureCalculation(forFootingDesign)

≔MOTS =⋅HLFactor ⎛⎝ +++⋅H1 LF1 ⋅H2 LF2 ⋅H3 LF3 ⋅H4 LF4⎞⎠ 95.467 ―――⋅kip ft

ft

≔MRS1 ⋅DLFactor ⎛⎝ +++++⋅w1 D1 ⋅w2 D2 ⋅w3 D3 ⋅w4 D4 ⋅w5 D5 ⋅w6 D6⎞⎠

≔MRS2 +⋅⋅LLFactor qtotal D6 ⋅⋅HLFactor LF5 H5

≔MRS =+MRS1 MRS2 172.625 ―――⋅kip ft

ft

≔RS =+⋅DLFactor ⎛⎝ +++++w1 w2 w3 w4 w5 w6⎞⎠ ⋅LLFactor qtotal 25.466 ――

kip

ft

≔xRS =――――−MRS MOTS

RS

3.03 ft

≔e1S =−――Lfoot

2xRS 2.22 ft =――

Lfoot

61.75 ft

≔LbasesoilS =⋅3 xRS 9.09 ft

A10.17

≔qmaxS =|||||||||||

if

else

<e1S ――Lfoot

6‖‖‖‖

+――RS

Lfoot――――⋅6 ⎛⎝ ⋅RS e1S⎞⎠

Lfoot2

‖‖‖‖――――――

⋅4 RS

⋅3 ⎛⎝ −Lfoot ⋅2 e1S⎞⎠

5.603 ksf ≔qminS =||||||||

|

if

else

<e1S ――Lfoot

6‖‖‖‖

−――RS

Lfoot――――⋅6 ⎛⎝ ⋅RS e1S⎞⎠

Lfoot2

‖‖ ⋅0 ksf

0 ksf

A10.18

DesignoftheHeel(Shear)

=covertop 2 in

=dbtop 0.75 in

=dblong 0.5 in

=stop 8 in

=slong 16 in

=Astop 0.442 in2

=Aslong 0.393 in2

≔dheel =−−tfoot covertop ――dbtop

215.625 in

Becausethefootingwilltendtoshearoffasshownhere,theshearcheckshouldoccuratthefaceofwall.

A10.19

=qtotalf 4.4 ――kip

ft

=w5f 7.59 ――kip

ft

=w6f 4.95 ――kip

ft

≔Vuheel1 =+++w5f w6f qtotalf ⋅⋅⋅DLFactor γconc tfoot Lheel 18.425 ――kip

ft

≔LsoilheelS =−−LbasesoilS Ltoe twall 4.09 ft

≔qmaxheelS =⋅qmaxS ―――――――⎛⎝ −−LbasesoilS Ltoe twall⎞⎠

LbasesoilS2.521 ksf

≔Vuheel2 =⋅⋅―1

2LsoilheelS qmaxheelS 5.155 ――

kip

ft

≔Vuheel =−Vuheel1 Vuheel2 13.27 ――kip

ftVuheel1isthetotaldownwardshearforceontheheel.Vuheel2isthetotalupwardshearforceontheheel.Becausethenetforceisdownward,thelocationoftheshearingiscon irmed.A10.20

≔fc ⋅4000 ――lbf2

in4

≔Vcheel =⋅⋅2 ‾‾fc dheel 23.717 ――kip

ft

≔PhiVcheel =⋅ϕv Vcheel 17.788 ――kip

ft

≔ShearheelInteraction =―――Vuheel

PhiVcheel

0.746

DesignoftheHeel(Moment)

≔Muheel =−⋅Vuheel1 ――Lheel

2⋅⋅Vuheel2 ―

1

3LsoilheelS 43.642 ―――

⋅kip ft

ft≔fc ⋅4 ksi

≔aheel =――――――

⋅⋅――⋅12 in

stopAstop fy

⋅⋅⋅0.85 12 in fc0.975 in ≔Astop1 =――

Astop

⋅1 ft0.442 ――

in2

ft

Thereinforcementspacingisat8"oc,sothemomentcapacityisnormalizedtobeperfoot.≔Mnheel =⋅⋅⋅――

⋅12 in

stopAstop1 fy

⎛⎜⎝

−dheel ――aheel

2

⎞⎟⎠

50.157 ―――⋅kip ft

ft

=ϕwall 0.9

≔PhiMnheel =⋅ϕwall Mnheel 45.142 ―――⋅kip ft

ft

≔BendheelInteraction =――――Muheel

PhiMnheel0.967

A10.21

DesignoftheToe(Shear)

=coverbot 3 in

=dbbot 0.75 in

=dblong 0.5 in

=sbot 8 in

=slong 16 in

=Asbot 0.442 in2

=Aslong 0.393 in2

≔dtoe =−−tfoot coverbot ――dbbot

214.625 in

Becausethefootingwilltendtoshearoffasshownabove,theshearcheckshouldoccuratadistancedfromthefaceofwall.

A10.22

≔qtoedS =―――――――――⋅⎛⎝ +−LbasesoilS Ltoe dtoe⎞⎠ qmaxS

LbasesoilS4.197 ksf

≔VutoeOT =⋅⎛⎝ −Ltoe dtoe⎞⎠⎛⎜⎝

+qtoedS ⋅―1

2⎛⎝ −qmaxS qtoedS⎞⎠

⎞⎟⎠

11.179 ――kip

ft

≔VutoeR =⋅⎛⎝ +w4f ⋅⋅⋅DLFactor γconc tfoot Ltoe⎞⎠⎛⎜⎝――――

−Ltoe dtoe

Ltoe

⎞⎟⎠

1.246 ――kip

ft

≔fc ⋅4000 ――lbf2

in4

≔Vutoe =−VutoeOT VutoeR 9.933 ――kip

ft≔Vctoe =⋅⋅2 ‾‾fc dtoe 22.199 ――

kip

ft

≔PhiVctoe =⋅ϕv Vctoe 16.649 ――kip

ft

≔SheartoeInteraction =―――Vutoe

PhiVctoe

0.597

A10.23

DesignoftheToe(Moment)

≔qtoefaceS =―――――――⋅⎛⎝ −LbasesoilS Ltoe⎞⎠ qmaxS

LbasesoilS3.446 ksf

≔MutoeOS +⋅⋅Ltoe qtoefaceS⎛⎜⎝――Ltoe

2

⎞⎟⎠

⋅⋅⋅―1

2Ltoe ⎛⎝ −qmaxS qtoefaceS⎞⎠

⎛⎜⎝

⋅―2

3Ltoe

⎞⎟⎠

=MutoeOS 29.916 ―――⋅kip ft

ft

≔VutoeRbend =⋅VutoeR ――――Ltoe

−Ltoe dtoe1.911 ――

kip

ft

≔MutoeR =⋅VutoeRbend

⎛⎜⎝――Ltoe

2

⎞⎟⎠

3.344 ―――⋅kip ft

ft

≔Mutoe =−MutoeOS MutoeR 26.571 ―――⋅kip ft

ft≔fc ⋅4 ksi

≔Asbot1 =――Asbot

⋅1 ft0.442 ――

in2

ft≔atoe =―――――

⋅⋅――⋅12 in

sbotAsbot fy

⋅⋅⋅0.85 12 in fc0.975 in

≔Mntoe =⋅⋅⋅――⋅12 in

sbotAsbot1 fy

⎛⎜⎝

−dtoe ――atoe

2

⎞⎟⎠

46.844 ―――⋅kip ft

ft=ϕwall 0.9

≔PhiMntoe =⋅ϕwall Mntoe 42.16 ―――⋅kip ft

ft

≔BendtoeInteraction =――――Mutoe

PhiMntoe0.63

A10.24

AppendixA11PileCapDesignCalculations_______________________________________________________________________________Inthisexamplewehaveapilecapwith12HP14x102pilesprovidingsupport.Thepileshavean85kipcompressioncapacity,a12kiptensioncapacityanda14kipshearcapacity.Thepilecapis42"thickwitha6"pileembedmentandmadefrom4ksilightweightconcrete.Aloadcombinationof1.0*DL+1.0*LLis usedfortheserviceLCandaloadcombinationof1.2*DL+1.6*LLisusedforthestrengthLC.InthisexampleRISAFoundation’svaluesarecomparedtothevaluesobtainedfromahandcalculation doneforallaspectsofthepilecap.ThishandcalculationislocatedinAppendixA11.Geometry,MaterialsandCriteria

≔Lcap 183 in ≔Wcap 134 in ≔tcap 42 in ≔embed 6 in

≔fy 60 ksi ≔fc 4 ksi ≔λ 0.75 ≔ρconc ⋅.11 ――kip

ft3

≔Hped 24 in ≔N 12 NumberofPiles≔Lped 24 in ≔dpile 14 in SideDimensionofPile≔Wped 24 in ≔dbar 0.75 in Diameterofreinforcement

≔lx 49 in Distancefromc/lofpedestaltoc/lofpilesinthexdirection.≔l1z 24.5 in Distancefromc/lofpedestaltoc/lof1stpilesinthezdirection.≔l2z 73.5 in Distancefromc/lofpedestaltoc/lof2ndpilesinthezdirection.≔wx =−lx ――

Wped

237 in Distancefrompilescentroidtofaceofpedestalinxdirection.

≔w1z =−l1z ――Wped

212.5 in Distancefrom1stpilescentroidtofaceofpedestalinzdirection.

A11.1

Distancefrom2ndpilescentroidtofaceofpedestalinzdirection.≔w2z =−l2z ――Wped

261.5 in

EffectiveDepthCalculations(forbending)

≔c 1.5 in Cover(topandbottom)≔d =−−−tcap embed c dbar 33.75 in Distancefromthetopofcaptocentroidofbottomreinforcement≔dtop =−tcap embed 36 in Distancefromthetopofcaptothetopofthepiles

AppliedLoads

≔Pd 250 kip ≔Vx 20 kip

≔Pl 350 kip ≔Vz 40 kip

≔Mx =⋅Vz

⎛⎜⎝

+Hped ――tcap

2

⎞⎟⎠

150 ⋅kip ft ≔Mz =⋅Vx

⎛⎜⎝

+Hped ――tcap

2

⎞⎟⎠

75 ⋅kip ft

≔wped =⋅⋅⋅Hped Lped Wped ρconc 0.88 kip ≔wcap =⋅⋅⋅Lcap Wcap tcap ρconc 65.562 kip

≔Ptot =+++Pd Pl wped wcap 666.442 kip

PileForces(Service)WewillassumetheindividualpileforcesarecorrectandusetheRISAFoundationoutput.≔Ppile1 54.1593 kip ≔Ppile8 59.2103 kip

≔Pu1 76.1068 kip ≔Pu8 84.1884 kip

≔Ppile2 56.6083 kip ≔Ppile9 49.5675 kip≔Pu2 80.0252 kip ≔Pu9 68.7599 kip




≔Ppile6 54.3124 kip≔Pu6 76.3517 kip

≔Ppile7 56.7613 kip≔Pu7 80.2701 kip

A11.2

PileCapFlexuralDesignForthe lexuraldesignwearesimplytakingtheworstcasemomentateitherfaceofthepedestalandcheckingagainstthat.TodothisIsimplycomparethepileforcesforeachsideofthepedesalandtaketheworstcaseforces.≔wucapresistx =⋅⋅⋅⋅1.2

⎛⎜⎝――――

−Wcap Wped

2

⎞⎟⎠Lcap tcap ρconc 32.292 kip

≔wucapresistz =⋅⋅⋅⋅1.2⎛⎜⎝――――

−Lcap Lped

2

⎞⎟⎠Wcap tcap ρconc 34.178 kip

≔Mux −+⋅⎛⎝ ++Pu3 Pu7 Pu11⎞⎠ w1z ⋅⎛⎝ ++Pu4 Pu8 Pu12⎞⎠ w2z ⋅wucapresistx ――――−Lcap Lped

4

=Mux⎛⎝ ⋅1.438 103 ⎞⎠ ⋅kip ft

≔Muz =−⋅⎛⎝ +++Pu1 Pu2 Pu3 Pu4⎞⎠ wx ⋅wucapresistz ――――−Wcap Wped

4932.815 ⋅kip ft

Herearethecalculationsforminimumsteelforbothtemperatureandshrinkageand lexure.≔Asminx =⋅⋅.0018 Lcap tcap 13.835 in2 ≔Asminz =⋅⋅.0018 Wcap tcap 10.13 in2

≔Asflexxbot =――――――

⋅⋅⋅200 ――lbf

in2Lcap d

fy20.588 in2 ≔Asflexzbot =――――――

⋅⋅200 ――lbf

in2Wcap d

fy15.075 in2

≔Asreqdxbot ⋅6.226 in2 Valuesgivenintheprogram≔Asprovxbot ⋅12.812 in2

≔ax =―――――⋅Asprovxbot fy

⋅⋅0.85 Lcap fc1.235 in

≔PhiMnx =⋅⋅⋅0.9 Asprovxbot fy⎛⎜⎝−d ―ax

2

⎞⎟⎠

⎛⎝ ⋅1.91 103 ⎞⎠ ⋅kip ft

≔UCMx =―――Mux

PhiMnx0.753

≔Asreqdzbot 9.609 in2 Valuesgivenintheprogram≔Asprovzbot ⋅14.137 in2

A11.3

≔az =―――――⋅Asprovzbot fy

⋅⋅0.85 Wcap fc1.862 in

≔PhiMnz =⋅⋅⋅0.9 Asprovzbot fy⎛⎜⎝−d ―az

2

⎞⎟⎠

⎛⎝ ⋅2.088 103 ⎞⎠ ⋅kip ft

≔UCMz =―――Muz

PhiMnx0.488

InthexdirectiontheAsreqd(andeven4/3Asreqd)islessthantheminimumtemperatureandshrinkagesteel,theprogramusesthatminimum.Inthezdirectionthe4/3*Asreq'disgreaterthantheAsS&T,thusweuse9.609*4/3=12.812in^2.PedestalPunchingShearCheck

≔db =d 33.75 in Effectivedepthofslabforpedestalpunching.≔L1 =+Wped db 57.75 in Sidedimensionsfortheshearperimeter.≔L2 =+Lped db 57.75 in

≔Pupileped =+++++++++Pu1 Pu2 Pu3 Pu4 Pu5 Pu8 Pu9 Pu10 Pu11 Pu12 783.109 kip

Thisvaluerepresentsthesumofthefactoredaxialforcesinpilesoutsideofthepedestalpunchingshearperimeter.≔wucapped =⋅⋅⋅1.2 ⎛⎝ −⋅Wcap Lcap ⋅L1 L2⎞⎠ tcap ρconc 67.975 kip

Thisistheself-weightofthepilecapthatisoutsideofthepedestalpunchingshearperimeter.≔Pupunch =−Pupileped wucapped 715.134 kip

≔Muxped =⋅1.6 Mx 240 ⋅kip ft Forceinthepedestal.≔Muzped =⋅1.6 Mz 120 ⋅kip ft

≔bo =⋅2 ⎛⎝ +L1 L2⎞⎠ 231 in Punchingshearperimeter.≔c1 =―L1

228.875 in Thisisthedistancefromcentroidtoextreme iber.

A11.4

≔Ac =⋅bo db⎛⎝ ⋅7.796 103 ⎞⎠ in2 Acistheperimeterareaoftheshearcone.

≔Jc =++―――――⋅db ⎛⎝ +Wped db⎞⎠

3

6―――――

⋅⎛⎝ +Wped db⎞⎠ db3

6―――――――――

⋅⋅db ⎛⎝ +Lped db⎞⎠ ⎛⎝ +Lped db⎞⎠2

2⎛⎝ ⋅4.704 106 ⎞⎠ in4

Jcisthepolarmomentofinertiaandthisequationcanbefoundinthecommentarytosection11.11.7.2oftheACI318-11.≔γ 0.4

≔υumax =++―――Pupunch

Ac――――

⋅⋅γ Muxped c1

Jc――――

⋅⋅γ Muzped c1

Jc0.102 ksi

Thisisthecriticalpunchingshearstress,combiningtheaxialandmomentforcestransmittedthroughthepedestal.Punchingequationscanbefoundinthecommentarytosection11.11.7.2oftheACI318-11.Notethatherewearecombiningthestressesduetothemomentstogettheworstcasestressatacornerofthepedestalpunchingshearperimeter.≔fc ⋅4000 ――

lbf2

in4

≔PhiVcpunch =⋅⋅⋅⋅0.75 4 ‾‾fc bo db⎛⎝ ⋅1.479 103 ⎞⎠ kip

≔PhiVny =―――――⋅λ PhiVcpunch

⋅bo db0.142 ksi

≔Punchcodecheck =―――υumax

PhiVny

0.719

PilePunchingShearCheckHerewewilldoapunchingshearcheckforpile4,theworstcaseone.TheprogramlooksateachpileandcalculatesapunchingshearperimeterforInterior,EdgeandCornerscenariosandchoosesthesmallestvalueforthecheck.Forroundpiles,wecalculateanequivalentsquaredimensionsuchthattheperimeterofbothareequal.=dpile 14 in ≔dtoppunch =−tcap embed 36 in

≔Lpile =++⋅11 in dpile ―――dtoppunch

243 in

Becausethereisnotopreinforcementinthepilecap,theslabisconsideredunreinforcedforpilepunching.BecauseofthisourPhifactorisnow0.55andweessentiallytake2/3oftheoriginalstrength(thus4goesto8/3).Theratioof2/3*(0.55/0.75)is0.4888.Intheprogramweuseablanket50%reduction.A11.5

≔ϕ 0.55 ≔bo1 =⋅2 Lpile 86 in

≔PhiVcpunch =⋅⋅⋅⋅⋅ϕ λ ―8

3‾‾fc bo1 dtop 215.389 kip Ifweweretocalculateitexactly.

≔PhiVcpunch2 =――――――――⋅⋅⋅⋅0.75 λ ⋅4 ‾‾fc bo1 dtop

2220.284 kip Thisisthevaluetheprogramreports.

=Pu4 87.862 kip

≔Puratio =――――Pu4

PhiVcpunch20.399

OneWayShearCheck

=w1z 12.5 in =wx 37 in

=d 33.75 in =d 33.75 in

Becauseinthexdirectionw>d,thecriticallocationisatadistancedfromthepedestal.Thismeansthatweneedtocalculatetheweightofthepilecapresistingtheshearatthislocation.≔wucapresistxshear =⋅⋅⋅⎛⎜⎝

−――――−Wcap Wped

2d⎞⎟⎠Lcap tcap ρconc 10.397 kip

≔Vux =−+++Pu1 Pu2 Pu3 Pu4 wucapresistxshear 317.54 kip

Becauseinthezdirectionw<d,thecriticallocationisatthefaceofthepedestal.Becauseofthiswecanusethewucapresistz thatweusedforthemomentcalculation.≔Vuz =−+++++Pu3 Pu4 Pu7 Pu8 Pu11 Pu12 wucapresistz 459.197 kip

=Mux⎛⎝ ⋅1.438 103 ⎞⎠ ⋅kip ft =Muz 932.815 ⋅kip ft

dz >wz,thereforethecriticallocationforshearatthefaceofthepedestal.≔Asprovidedz ⋅12.8122 in2 ≔Asprovidedx =Asminx 13.835 in2

≔ρprovz =――――Asprovidedz

⋅Wcap d0.002833 ≔ρprovx =――――

Asprovidedx

⋅Lcap d0.00224

A11.6

Shearstrengthinthexdirectiondz >wz,thereforethecriticallocationforshearatthefaceofthepedestalandCRSIDesignHandbookequation13-2onP.13-26isused.=―――

Mux

⋅Vuz d1.114 Mu/Vu*dmustbelessthanorequalto1.0,souse1.0.≔MVratio 1

≔υcx =⋅⋅⎛⎜⎝――d

w1z

⎞⎟⎠(( −3.5 ⋅2.5 MVratio))

⎛⎜⎝

+⋅⋅1.9 λ ‾‾fc ⋅⋅2500 ――lbf

in2ρprovz MVratio

⎞⎟⎠

262.46 psi

≔υcmax =⋅10 ‾‾fc 632.456 psi

≔Vc_x =⋅⋅υcx Wcap d ⎛⎝ ⋅1.187 103 ⎞⎠ kip

Shearstrengthinthezdirectiondz <wz,thereforethecriticallocationforshearisatadistancedfromthefaceofthepedestalandACI318-11Equation11-5isused.=―――

Muz

⋅Vux d1.044 Mu/Vu*dmustbelessthanorequalto1.0,souse1.0.≔MVratio 1

≔υcz1 =+⋅⋅1.9 λ ‾‾fc ⋅⋅⋅2500 ――lbf

in2ρprovx MVratio 95.725 psi

≔Vc_z1 =⋅⋅υcz1 Lcap d 591.221 kip

A11.7

PedestalDesign

Inputs

≔dpedlongbar 1 in ≔dpedshearbar 0.5 in ≔coverped 1.5 in =Wped 24 in

≔dped =−−−Wped coverped dpedshearbar ――――dpedlongbar

221.5 in =λ 0.75

ConcreteShearCapacity

≔Vcped =⋅⋅⋅⋅2 λ ‾‾fc Wped dped 48.952 kip

SteelShearCapacity

≔Asv =⋅2⎛⎜⎝―――――

⋅dpedshearbar2

4

⎞⎟⎠

0.393 in2 ≔spedshear 10 in

≔Vs1 =――――⋅⋅Asv fy dped

spedshear50.658 kip ≔Vsmax =⋅⋅⋅8 ‾‾fc dped Wped 261.078 kip

≔Vs =min ⎛⎝ ,Vs1 Vsmax⎞⎠ 50.658 kip

CombinedBendingandAxialForces

≔Puped =++⋅1.2 Pd ⋅1.6 Pl ⋅⋅⋅⋅1.2 Hped Wped Lped ρconc 861.056 kip

≔Muxp =⋅⋅1.6 Vz Hped 128 ⋅kip ft ≔Muvp =⋅⋅1.6 Vx Hped 64 ⋅kip ft

Forthispedestaltheinteractiondiagramactuallyproducesaworstcasecodecheckatthetopofthepedestal.Thus,theaxialforceinthepedestalisnotincludingthepedestalself-weightandthemomentatthetopiszeroinbothdirections.

A11.8

RISA Foundation Verification Problems

Documents

verification package

test problems

rectangular footing

square footing

strip footing design

square spread footing

loaded footing

pile cap shear