ENGR 0145

STATICS AND MECHANICS OF MATERIALS 2

DR. SLAUGHTER

DESIGN PROJECT 1

Design of a Wooden Pile-and-Plank Retaining Wall

Blake Bernard

Daniel Leon

Stephen Mingey

Andrew Ragonese

February 29, 2016

ABSTRACT

This report contains an analysis of the design of a wooden pile and plank retaining wall. The

design constraints are outlined along with any assumptions made to simplify the process towards

a final design. Included is a description of how the cheapest effective materials were determined

and the pertinent data obtained along with the specific equations used for each of the points of

data. The thought process of how the calculated data was interpreted to select the best option is

included. The report ends with the selection of the cheapest material options to design a wooden

pile and plank retaining wall based on the given design constraints.

INTRODUCTION

Retaining walls are used for altering topography and typically create a terrace in steeply sloped

terrain. Going from a slope or hill the land can be excavated and leveled off to maximize the

amount of usable land. This particular assignment was to design a wooden pile and plank

retaining wall meeting all design constraints and requirements in the cheapest way possible.

The general design of a pile and plank retaining wall uses vertical piles supporting horizontal

planks, as shown below. Each pile requires a concrete footing to secure it underground.

In this idealized scenario, the cost of the wall is strictly dependent on the amount of lumber and

concrete used to build it. The amount of wasted material is included in these costs, so minimizing

this wasted material was one of the major factors throughout the design.

This particular retaining wall was to be 80 feet long with piles extending 5 feet above the ground

and reaching 5 feet below the ground. It is required that one pile is at either end of the retaining

wall. Additional requests for the piles include that they all be the same size, evenly spaced, and

square in cross section. Similarly, the planks must all be the same size and their length should be

the same as the spacing between piles. Standard structural timbers are available in 8, 10, and 12

ft lengths at $14 per cubic foot, and the concrete footing to be used costs $40 per pile.

The retaining wall will be constructed with pressure-treated Standard Structural Timber with an

allowable flexural stress of 1200 psi. (For simplicity, shear stresses will not be considered.) The

lateral pressure exerted by the soil against the retaining wall is 500 lb/ft2 at the base of the wall

and decreases linearly to 100 lb/ft2 at the top of the retained soil 5 ft above the grade, as shown

here.

Intuitively, it is difficult to measure what design will result in the lowest overall cost. If few piles

are used, the cost of the piles and their concrete footings will be low, but the high stresses placed

on the planks will require thicker beams and hence, higher costs. If many piles are used, the

lower stresses on the planks will allow for less material and lower costs in this area, but the sheer

number of piles will place a heavy demand on the cost of timber and concrete. One would

therefore assume that there is probably a middle ground in the number of piles that balances the

costs of the piles and planks. It turns out that this is true, but it requires a detailed analysis to

prove it.

ANALYSIS & DESIGN

Strength Analysis: Piles

To begin the design of the retaining wall, it makes sense to break up the analysis into two parts,

treating the piles and planks as separate problems. With the piles, note if the diagram is rotated

90 degrees it resembles a cantilevered beam with a trapezoidal load.

The trapezoidal load can be broken up into two separate loads comprised of a triangular load and

a rectangular load.

The resultant of this distributed load can be calculated as follows:

𝐴1 =1

2(5 𝑓𝑡) (400

𝑙𝑏

𝑓𝑡2) 𝐿 = (1000

𝑙𝑏

𝑓𝑡) 𝐿

𝐴2 = 𝑏ℎ = (100 𝑙𝑏

𝑓𝑡2) (5 𝑓𝑡)𝐿 = (500

𝑙𝑏

𝑓𝑡) 𝐿

𝑦𝑐1=

1

3𝑏 =

1

3(5 𝑓𝑡) =

5

3𝑓𝑡

𝑦𝑐2=

𝑏

2=

5

2 𝑓𝑡

Where 𝑦𝑐1 and 𝑦𝑐2

are the centroids of the triangular and rectangular loads respectively. Next, the

equivalent load (force), the location of the force, and the resulting moment can be found by

summing up the areas and finding the centroid of the complete trapezoidal load.

𝐹𝑒𝑞 = ∑ 𝐴𝑖 = (1000 𝑙𝑏

𝑓𝑡+ 500

𝑙𝑏

𝑓𝑡) 𝐿 = (1500

𝑙𝑏

𝑓𝑡) 𝐿

𝑦𝑐 =∑ 𝐴𝑖 𝑦𝑖

∑ 𝐴𝑖=

(1000𝑙𝑏𝑓𝑡

) 𝐿 (53) + (500

𝑙𝑏𝑓𝑡

) 𝐿(52)

(1500 𝑙𝑏𝑓𝑡

) 𝐿= 1.944 𝑓𝑡

∑ 𝑀𝐴 = 𝑀𝐴 − (𝐹𝑒𝑞)𝑦𝑐 = 0 → 𝑀𝐴 = (1500𝑙𝑏

𝑓𝑡) 𝐿(1.944 𝑓𝑡) = (2917 𝑙𝑏)𝐿

This transforms the free body diagram as the following:

Giving distributed load, shear force, and bending moment diagrams resembling:

From these diagrams, it is clear that the maximum bending moment occurs at the point of

support:

𝑀𝑚𝑎𝑥 = 𝑀𝐴 = (2917 𝑙𝑏)𝐿

2917 L

Area = M = (2917 ft-lb) L

Now taking the maximum flexural stress of 1200 psi into account, together with the

aforementioned maximum bending moment, a formula relating the number of piles to the

minimum allowable section modulus can be derived.

𝜎𝑚𝑎𝑥 = 1200 𝑝𝑠𝑖 = 𝑀𝑚𝑎𝑥

𝑆𝑚𝑖𝑛→ 𝑆𝑚𝑖𝑛 =

(2917 𝑙𝑏)

(1200 𝑝𝑠𝑖)𝐿 →𝑆𝑚𝑖𝑛 =

(2917 𝑙𝑏)(80 𝑓𝑡)

(1200 𝑝𝑠𝑖)(𝑛 − 1)

Strength Analysis: Planks

For the following plank calculations, the following variables are defined as: x = width of cross

section, t = thickness, L = length of plank, y = length of cut. These variables are illustrated in the

diagram below:

As described in the introduction, the maximum lateral pressure exerted on the retaining wall is

500 lb/ft2 at the base of the wall, but decreases linearly toward the top. To simplify calculations,

it will be assumed that this maximum pressure is uniformly distributed over the planks at the

bottom of the retaining wall. In designing for the strength of this “worst case” scenario, all

planks experiencing a lateral pressure less than 500 lb/ft2 will also meet the strength requirement.

Hence, the bottom plank can be modeled as a simply supported beam in the following way:

Calculating for the unknown RA and RB, it can be determined that the forces are equal due to

symmetry, each with magnitude half of the equivalent force:

∑ 𝐹𝑦 = 0 = 𝑅𝐴 + 𝑅𝐵 − (500𝑙𝑏

𝑓𝑡2) 𝑥𝐿

𝑅𝐴 = 𝑅𝐵 =1

2(500

𝑙𝑏

𝑓𝑡2) 𝑥𝐿 = (250

𝑙𝑏

𝑓𝑡2)𝑥𝐿

500 (lb/ft2) * x

500 (lb/ft2) * x * L

250 (lb/ft2) * x * L 250 (lb/ft2) * x * L

Next, a cut is made in the beam at point O, from which the equation for the bending moment can

be determined:

∑ 𝑀𝑂 = 0 = 𝑀 + (500𝑙𝑏

𝑓𝑡2) 𝑥𝑦

𝑦

2 − (250

𝑙𝑏

𝑓𝑡2) 𝑥𝐿𝑦

𝑀(𝑦) = (250𝑙𝑏

𝑓𝑡2) 𝑥𝐿𝑦 − (250

𝑙𝑏

𝑓𝑡2) 𝑥𝑦2

By symmetry, the maximum value of the bending moment occurs at y=L/2 (setting the derivative

of M(y) equal to zero also yields this result). By substituting L/2 for y in the moment equation, a

maximum value for the bending moment can be determined:

𝑀(𝑚𝑎𝑥) = 𝑀 ( 𝐿

2 ) = (125

𝑙𝑏

𝑓𝑡2) 𝑥𝐿2 − (62.5

𝑙𝑏

𝑓𝑡2) 𝑥𝐿2

𝑀(𝑚𝑎𝑥) = (62.5𝑙𝑏

𝑓𝑡2) 𝑥𝐿2 = 𝐵𝑥𝐿2

𝑤ℎ𝑒𝑟𝑒 𝐵 = (62.5𝑙𝑏

𝑓𝑡2) = 0.434028 𝑝𝑠𝑖

The second moment of a rectangular cross section and the identification of the neutral axis are

also vital to the final calculation:

500 (lb/ft2) * x * y

250 (lb/ft2) * x * L

From this,

𝐼 =1

12𝑥𝑡3 , 𝑐 =

𝑡

2

With these variables now known, the equation for allowable stress can be set up and an

expression relating the number of piles to the minimum thickness can be determined:

𝜎𝑎𝑙𝑙 ≥ 𝑀𝑐

𝐼=

𝑀 (𝑡2)

(1

12) 𝑥𝑡3=

6𝑀

𝑥𝑡2=

6𝐵𝑥𝐿2

𝑥𝑡2=

6𝐵𝐿2

𝑡2

𝜎𝑎𝑙𝑙 ≥6𝐵𝐿2

𝑡2 → 𝑡2 ≥

6𝐵𝐿2

𝜎𝑎𝑙𝑙

𝑡𝑚𝑖𝑛 = √6𝐵𝐿2

𝜎𝑎𝑙𝑙= 𝐿√

6𝐵

𝜎𝑎𝑙𝑙 → 𝑡𝑚𝑖𝑛 = 𝐿√

6(0.434028 𝑝𝑠𝑖)

1200 𝑝𝑠𝑖→𝑡𝑚𝑖𝑛 =

80 𝑓𝑡

𝑛 − 1√

6(0.434028 𝑝𝑠𝑖)

1200 𝑝𝑠𝑖

Cost Analysis

To complete this analysis, the cost of the least expensive retaining wall will be calculated for

every possible value of the number of piles, n. Note that as discussed earlier in this analysis, the

distance between piles, L, is related to n by:

𝐿 =80 𝑓𝑡

𝑛 − 1↔ 𝑛 =

80 𝑓𝑡

𝐿+ 1

Assuming the distance between piles cannot be less than 2 ft:

𝑛 ≤ ⌊80 𝑓𝑡

2 𝑓𝑡⌋ + 1 = 41

Since the longest beam available is 12 ft long:

𝑛 ≥ ⌈80 𝑓𝑡

12 𝑓𝑡⌉ + 1 = 8

Note: the notation ⌊𝑥⌋ and ⌈𝑥⌉ will be used to denote the floor of x (rounding down) and the

ceiling of x (rounding up), respectively.

Therefore, retaining walls with n piles such that 8 ≤ 𝑛 ≤ 41 will be studied.

As stated in the introduction, the standard structural timbers to be used for the piles and the

planks are available in lengths of 8 ft, 10 ft, and 12 ft. For every possible n, the timber length that

will minimize the volume of timber needed can be calculated in the following way:

Let D be the length of the standard timber used for the planks. For simplicity, only one type of

standard timber will be purchased for the planks. For a given length of standard timber, D, and a

given distance between piles (or equivalently, the length of each plank), L, there can fit a

maximum of D/L planks per timber, rounding down because the excess timber that is shorter

than L cannot be used. Hence, the number of planks one can get from one standard timber is:

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑙𝑎𝑛𝑘𝑠 𝑝𝑒𝑟 𝑡𝑖𝑚𝑏𝑒𝑟 = ⌊𝐷

𝐿⌋

Now, for a given n, there are n – 1 spaces between piles. Therefore, when considering one span

of planks across the retaining wall, the number of standard timbers needed is (n – 1)/[number of

planks per timber], rounding up because one cannot buy a fraction of a timber. Hence, the

number of standard timbers needed for one span of planks is:

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑖𝑚𝑏𝑒𝑟𝑠 𝑝𝑒𝑟 𝑠𝑝𝑎𝑛 = ⌈𝑛 − 1

⌊𝐷𝐿⌋

⌉

Finally, the number of standard timbers multiplied by the length of the standard timbers gives the

total length of timber needed for one span of planks:

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡𝑖𝑚𝑏𝑒𝑟 𝑝𝑒𝑟 𝑠𝑝𝑎𝑛 = 𝐷 ⌈𝑛 − 1

⌊𝐷𝐿⌋

⌉

Where L is related to n as above.

For a given n, finding the length of timber per span for each standard timber length (D = 8 ft, 10

ft, or 12 ft) and finding the timber that yields the minimum value will yield the optimum standard

timber length for the planks.

Now, according to the equation derived for the minimum thickness of the planks,

𝑡𝑚𝑖𝑛 = 𝐿√6𝐵

𝜎𝑎𝑙𝑙

The width of the planks, x, has no effect on the planks’ minimum thickness. Hence, the optimum

standard timber width for the planks will be any width that divides evenly into 60 inches (5 ft:

the height of the retaining wall). Since 7.5 in (nominal size 8 in) is the only available width that

divides evenly into 60 inches, a nominal width of 8 inches will be used for all planks. Therefore,

the number of spans of planks needed to cover the entire wall is (60 in)/(7.5 in) = 8. Hence, for

every retaining wall with n piles, distance L between piles, and standard timber of length D, the

number of required timbers for the planks is:

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑖𝑚𝑏𝑒𝑟𝑠 (𝑓𝑜𝑟 𝑝𝑙𝑎𝑛𝑘𝑠) = 8 ⌈𝑛 − 1

⌊𝐷𝐿⌋

⌉

For the piles, since they must extend 5 ft below ground and at least 5 ft above ground, the

optimum standard timber length for the piles in 10 ft.

RESULTS

Using the equation relating Smin to the optimum beam for the piles and the equation relating tmin

to the optimum beam for the planks, along with the previously described method for obtaining

the optimum standard timber length for the planks and calculating the number of timbers

required, the cheapest overall cost for a retaining wall with all possible numbers of piles was

calculated. The results of these calculations are displayed in the table below:

Number

of Piles

Piles Planks Total

Cost

($)

Optimum

Beam

Cost*

($)

Optimum

Beam

Optimum

Standard Timber

Length (ft)

Number of

Timbers

Required

Cost

($)

8 #N/A** #N/A 8 x 8 12 56 3675 #N/A

9 #N/A #N/A 8 x 8 10 64 3500 #N/A

10 #N/A #N/A 6 x 8 10 72 2888 #N/A

11 12 x 12 1854 6 x 8 8 80 2567 4421

12 12 x 12 2023 6 x 8 8 88 2823 4846

13 12 x 12 2191 6 x 8 8 96 3080 5271

14 12 x 12 2360 4 x 8 8 104 2199 4559

15 12 x 12 2529 4 x 8 12 56 1776 4305

16 12 x 12 2697 4 x 8 12 64 2030 4727

17 12 x 12 2866 4 x 8 10 64 1692 4557

18 10 x 10 2299 4 x 8 10 72 1903 4203

19 10 x 10 2427 4 x 8 10 72 1903 4330

20 10 x 10 2555 4 x 8 10 80 2115 4669

21 10 x 10 2683 4 x 8 8 80 1692 4374

22 10 x 10 2810 4 x 8 12 56 1776 4587

23 10 x 10 2938 4 x 8 8 88 1861 4799

24 10 x 10 3066 4 x 8 8 96 2030 5096

25 10 x 10 3194 4 x 8 10 64 1692 4885

26 10 x 10 3321 4 x 8 10 72 1903 5224

27 10 x 10 3449 4 x 8 10 72 1903 5352

28 10 x 10 3577 4 x 8 12 56 1776 5353

29 10 x 10 3705 2 x 8 12 56 796 4501

30 10 x 10 3832 2 x 8 12 64 910 4742

31 10 x 10 3960 2 x 8 8 80 758 4718

32 10 x 10 4088 2 x 8 8 88 834 4922

33 10 x 10 4216 2 x 8 10 64 758 4974

34 10 x 10 4343 2 x 8 8 88 834 5177

35 8 x 8 3314 2 x 8 12 56 796 4110

36 8 x 8 3409 2 x 8 12 56 796 4205

37 8 x 8 3503 2 x 8 10 72 853 4357

38 8 x 8 3598 2 x 8 12 64 910 4508

39 8 x 8 3693 2 x 8 12 64 910 4603

40 8 x 8 3788 2 x 8 12 64 910 4698

41 8 x 8 3882 2 x 8 8 80 758 4641

*The cost of the piles includes the cost of the concrete footing.

**For retaining walls with very few piles, the required strength of the piles was greater than the strength

of any available beam, so no retaining wall can be built for these numbers of piles.

For a more concise summary of this table, refer to the plot of the total cost of the retaining wall

versus the number of piles, shown below:

As shown in the table and the graph above, the cheapest overall retaining wall uses 35 piles.

Such a design will use 35 8”x8”x10’ standard structural timbers for the piles, 35 concrete

footings (one for each pile), and 56 2”x8”x12’ standard structural timbers for the planks. This

information, along with the cost per item, is summarized in the following table:

Item Quantity Volume per Item

(ft^3) Unit Cost ($) Total Cost ($)

8" x 8" x 10' Standard Structural Timber

35 3.91 14 1914

2" x 8" x 12' Standard Structural Timber

56 1.02 14 796

Concrete Footing 35 40 1400

The total cost for this design is $4110.

BestDesign

3600

4000

4400

4800

5200

5600

0 5 10 15 20 25 30 35 40 45

Tota

l Co

st (

$)

Number of Piles

Total Cost vs Number of Piles