Design of a Wooden Pile-and-Plank Retaining Wall
Team Design Project 1
Dr. William Slaughter
ENGR0145 Statics and Mechanics of Materials II
March 2, 2015
Jacob Feid
Derek Nichols
ABSTRACT
The goal of this project was to design a retaining wall that met the strength and design requirements given
to us while minimizing the cost. We were allowed to use any dimension of Standard Structural Timber in
Table A-15 of “Statics and Mechanics of Materials: An Integrated Approach” by Riley in the lengths of
8, 10, and 12 feet for the use of our planks while limited to 10 foot long square cross sections for the use
of piles. We initially constructed free body, distributed load, shear force, and bending moment diagrams
for the piles (which acted as cantilevered beams) and the planks (which acted as simply supported
beams), and found the maximum bending moment for both situations. With these, we were able to find
the minimum section modulus for the planks and piles in terms of pile spacing and plank size. We
constructed a Microsoft Excel spreadsheet listing all sizes of allowable wood and calculated the section
modulus for each type. If the calculated section modulus was greater than the minimum for that sized
plank, then the cost of the plank was found. The minimum cost was found per number of piles, and then
the minimum of those was calculated to find the minimum possible cost.
INTRODUCTION
Retaining walls are structures used to alter topography and mediate erosion due to storm water runoff.
These walls are generally used to create a terrace in a steeply sloped area, as shown in Figure 1. These
walls can be constructed out of an assortment of materials such as stone, concrete, and wood. Regardless
of the material used, these walls must account for a high lateral backfill load and must be able to
withstand huge amounts of pressure on all parts of the wall. For this project we were required to design a
straight pile-and-plank retaining wall that is 80 feet in length. This wall will be used to create a 5 foot
high terrace, meaning the wall must be at least 5 foot high. The wall we design will be constructed using
pressure-treated Standard Structural Timber with an allowable flexural stress of 1200 psi. The lateral
pressure exerted by the soil on the retaining wall at the base of the wall is 500 lb/ft2 and decreases linearly
to 100 lb/ft2 5 feet above grade at the top of the retaining wall. This can be seen in Figure 1. The general
design of a pile-and-plank retaining wall calls for horizontal planks supported by vertical piles shown in
Figure 2.
Figure 1
Figure 2
Some design requirements for the piles and planks for the retaining wall are as follows:
 Piles
o For simplicity and aesthetics, all the piles should be the same, they should be evenly
spaced, and there should be a pile at either end of the retaining wall.

o
o
Planks
o
o
The piles are to be square in cross section.
The piles must extend 5 ft. below grade.
For simplicity, all of the planks should be the same size.
The length of the planks should be the same as the spacing between piles, so that each
plank is supported by piles at either end.
Standard Structural Timbers are available in 8, 10, and 12 ft. lengths and cost $14 per cubic foot. There is
an additional cost of $40 per pile for the concrete footing needed to support each pile. We will design this
retaining wall with a primary goal of minimizing the overall cost while still having a wall that is able to
remain structurally sound while supporting the pressure exerted by the backfill. We hypothesize that the
strength of the wall will increase as you add more piles and increase the cross sectional area of the planks.
The cheapest option will be a mix of many piles and a large cross sectional area for the planks.
Figure 3
Figure 4
ANALYSIS & DESIGN
Initial Variables:




x – distance from ground
L – plank length (ft)
b – plank base length (in)
h – plank height length (in)
The retaining wall had to be broken up into two separate problems in order to analyze the problem as a
whole. The first problem was considering the forces acting on the piles. Using Figure 3, we constructed
free body, distributed load, shear force, and bending moment diagrams for the piles acting as if they were
weightless cantilever beams (weight of the beam can be neglected since the beams are in reality standing
up). The free body diagram is simply Figure 2 rotated clockwise by 90°.
We can see that the distributed load resembles that of a trapezoid. The equation of the above line can be
represented with 𝑤(𝑥) = (500 − 80𝑥)𝐿. In order to analyze the reaction forces, the distributed load
must be represented as a concentrated load. In order to do this, the trapezoid is broken up into a single
rectangle and triangle.
The area of the triangle and its center of mass are given by:
1
1
𝑙𝑏
𝐴1 = 2 𝑏ℎ = 2 (5 𝑓𝑡) (400 𝑓𝑡 2 ) = 1000
𝑙𝑏
𝑓𝑡
1
1
5
1
1
5
𝑦𝑐1 = 3 𝑏 = 3 (5 𝑓𝑡) = 3 𝑓𝑡
̅̅̅̅
The area of the rectangle and its center of mass are given by:
𝑙𝑏
𝑙𝑏
𝐴2 = 𝑏ℎ = (5 𝑓𝑡) (100 𝑓𝑡 2 ) = 500 𝑓𝑡
𝑦𝑐2 = 2 𝑏 = 2 (5 𝑓𝑡) = 2 𝑓𝑡
̅̅̅̅
Using these, we can calculate the equivalent concentrated load:
𝑁
𝐹𝑒𝑞𝑖𝑣 = ∑ 𝐴𝑖 = (1000 + 500)
𝑖=1
𝑁
𝑙𝑏
𝑙𝑏
= 1500 2 = 1500𝐿 𝑙𝑏
2
𝑓𝑡
𝑓𝑡
1
1
𝑙𝑏 5
𝑙𝑏 5
𝑦̅𝑐 =
∑ 𝐴𝑖 ̅̅
𝑦̅̅
[(1000 ) ( 𝑓𝑡) + (500 ) ( 𝑓𝑡)] = 1.944𝑓𝑡
𝑐𝑖 =
𝑙𝑏
𝐴𝑇
𝑓𝑡 3
𝑓𝑡 2
1500
𝑖=1
𝑓𝑡
The free body diagram can therefore be represented as a beam with a single load acting on its length and a
reaction force and couple acting at its contact with the ground. By summing the moments about the
ground and summing the forces acting on the pile, we were able to solve for the reaction forces acting on
the beam.
∑ 𝑀 = 0 = 𝑀𝑜 − 𝐹𝑒𝑞𝑖𝑣 𝑦̅𝑐 = 𝑀𝑜 − (1500 ∗ 𝐿
∑ 𝐹𝑦 = 0 = 𝑉 − 𝐹𝑒𝑞𝑖𝑣 = 𝑉 − 1500 ∗ 𝐿
𝑙𝑏
) (1.944 𝑓𝑡) → 𝑀𝑜 = 2916.67𝐿 𝑓𝑡 ∙ 𝑙𝑏
𝑓𝑡
𝑙𝑏
→ 𝑉 = 1500𝐿 𝑙𝑏
𝑓𝑡
Using this free body diagram, we can construct our distributed load, shear force, and bending moment
diagrams. The distributed load diagram can tell us the equivalent force (the area under the curve) which
we already solved for above.
The shear force diagram tells us the maximum shear force that the piles experience from the load. This
also shows us that this maximum shear force is experienced at the base of the pile. We can realize that
the area under the curve is equal to 2916.67𝐿 𝑓𝑡 ∙ 𝑙𝑏 since it must be equal to the moment from the
support reaction.
The bending moment diagram is used to observe that the maximum bending moment occurs at the base of
the beam.
From these diagrams, we can see that the maximum shear force and maximum bending moment
experienced by the piles both occur at the base and equal:
𝑉max 𝑝𝑖𝑙𝑒 = 1500𝐿 𝑙𝑏
𝑀max 𝑝𝑖𝑙𝑒 = 2916.67𝐿 𝑓𝑡 ∙ 𝑙𝑏
The second part of the problem involves analyzing the forces acting on the planks. Looking at Figure 4,
we can imagine each plank acting as a simply supported beam with a fixed support on each end. The
planks experience less and less force acting upon them as you move up the beam; therefore, we can select
𝑙𝑏
a beam such that it can withstand a constant stress of 500 𝑓𝑡 2 since this is the maximum stress it will
experience. We can construct free body, shear force, and bending moment diagrams for the simply
supported beams.
The free body diagram can be seen below. The distributed load acts over the entire beam, and since the
beam is simply supported, the support reactions each take half of the distributed load. Because the
𝑙𝑏
pressure on each beam is equal to 500 2 , we can multiply this by the depth (the height of each beam) in
𝑓𝑡
order to find the distributed load. The total force on each support is equal to half of the distributed load
times the length, so each supporting force is equal to 250ℎ𝐿.
The shear force diagram can be made from this free body diagram. The diagram has an initial and final
jump of 250ℎ𝐿 meaning that the distributed load is linear between these two points. The area of each
triangle equals the bending moment contribution (the triangles are equal but opposite to each other).
1
1 𝐿
𝐴 = 𝑏ℎ = ( 𝑓𝑡) (250ℎ𝐿 𝑙𝑏) = 62.5ℎ𝐿2 𝑓𝑡 ∙ 𝑙𝑏
2
2 2
Since there are no concentrated moments, the only contributions to the bending moment diagram come
from the shear force diagram.
From these diagrams, we can see that the maximum shear force that the plank experiences occurs at the
supports while the maximum bending moment experienced by the planks occurs at the middle of the
plank:
𝑉max 𝑝𝑙𝑎𝑛𝑘 = 250ℎ𝐿 𝑙𝑏
𝑀max 𝑝𝑙𝑎𝑛𝑘 = 62.5ℎ𝐿2 𝑓𝑡 ∙ 𝑙𝑏
Strength Analysis
In order to perform the strength analysis, we must find the minimum section modulus for both the pile
and planks. We were given that the maximum allowable flexural stress in pressure-treated Standard
Structural Timber is 1200 psi.
𝜎𝑎𝑙𝑙 =
𝑆min 𝑝𝑖𝑙𝑒 =
𝑀𝑚𝑎𝑥
𝑀
→ 𝑆𝑚𝑖𝑛 = 𝑚𝑎𝑥⁄𝜎𝑎𝑙𝑙
⁄𝑆
𝑚𝑖𝑛
𝑖𝑛
𝑓𝑡
(2916.67𝐿 𝑙𝑏∙𝑓𝑡)(12 )
1200 𝑝𝑠𝑖
= 29.1667𝐿 𝑖𝑛3 (1)
𝑖𝑛
𝑆min 𝑏𝑒𝑎𝑚 =
(62.5ℎ𝐿2 𝑙𝑏∙𝑓𝑡)(12𝑓𝑡)
1200 𝑝𝑠𝑖
= 0.625ℎ𝐿2 𝑖𝑛3 (2)
With this information, we made a Microsoft Excel spreadsheet which included every type of Standard
Structural Timber available ranging from 2x4 to 12x20. The possible lengths included 8, 10, and 12 feet.
We first made a list of the possibilities for the number of piles in order to find the minimum number of
piles required. This list started at a minimum of eight piles which was the lowest amount possible in
order to fit a 12 foot beam between. We then calculated the minimum section modulus required for the
given number of beams using equation one. We then took the candidate for the piles (which had to be
square) which were 4x4, 6x6, 8x8, 10x10, and 12x12. We compared the given section modulus for these
beams to the minimum section modulus required to see what size pile is required for the given number of
piles. We found that you need at least 11 piles and that you need the following sized piles for the given
configurations:
𝟏𝟐𝐱𝟏𝟐 cross section for 𝟏𝟏 − 𝟏𝟕 piles
𝐶𝑟𝑜𝑠𝑠 𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑖𝑙𝑒 = {𝟏𝟎𝐱𝟏𝟎 cross section for 𝟏𝟖 − 𝟑𝟒 piles
𝟖𝐱𝟖 cross section for 𝟑𝟓 − 𝟒𝟏 piles
(3)
With this information, we were able to move on with our spreadsheet. We compiled the information for
every sized beam such as nominal size, actual size, cross sectional area, the number of beams needed to
𝑖𝑛
cover the height ( ROUNDUP( (5 𝑓𝑡)(12 𝑓𝑡)/ℎ ) ), and section modulus. We then added the three
possibilities for the length of the beam and found the volume for each possibility. This means that, for the
30 different cross sections, we had 90 different beam possibilities. We then went through and found
information for each beam for each number of piles. We started with 11 piles and found the length from
80 𝑓𝑡
pile to pile (
⁄(# 𝑝𝑖𝑙𝑒𝑠 − 1) ). With this information, we were able to find the minimum section
modulus for each plank using equation 2. If the plank’s actual section modulus was greater than the
minimum section modulus, then the word “Yes” was displayed, and the word “No” was displayed if the
minimum section modulus was greater than the actual. This process was repeated for every combination
of planks from 11 piles to 41 piles. By doing this, we were able to find all planks that were strong enough
to withstand the force of the soil.
Cost Analysis
In order to conserve resources, the wood planks could be cut to their needed lengths, and if there was
enough left over, then the remaining wood could possibly be used again. In order to see how many times
each plank could be used, the ROUNDDOWN function in Excel was used. Each plank would be used to
span ROUNDDOWN(length of beam/length between piles) gaps. The total length was then divided by
the number of times each plank could be used and by the length between piles; therefore, the number of
boards needed for a given number of piles was found next using the equation:
Number of planks needed for length = ROUNDUP( 80 ft / (ROUNDDOWN(length of beam/length
between piles) * length between piles ).
If a beam passed the strength test, then the cost was displayed. The cost was calculated per cubic foot.
The volume was found by multiplying the cross sectional area by the length of the beam and the cost was
obtained by multiplying the volume by the number of planks needed for length, the number of planks
needed for height, and the cost per cubic foot.
𝑐𝑜𝑠𝑡 𝑓𝑜𝑟 𝑝𝑙𝑎𝑛𝑘𝑠 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑒𝑟 𝑝𝑙𝑎𝑛𝑘 ∗ # 𝑝𝑙𝑎𝑛𝑘𝑠𝑤𝑖𝑑𝑡ℎ ∗ # 𝑝𝑙𝑎𝑛𝑘𝑠ℎ𝑒𝑖𝑔ℎ𝑡 ∗ 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑖𝑛3
The minimum costs for planks were then found for each number of piles. After this, the costs for the piles
were found. The size of the pile depended on the number of piles used using equation 3. The equation for
the cost of the piles was:
𝑐𝑜𝑠𝑡 𝑓𝑜𝑟 𝑝𝑖𝑙𝑒𝑠 = # 𝑝𝑖𝑙𝑒𝑠 (𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑒𝑟 𝑝𝑖𝑙𝑒 ∗ 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑖𝑛3 + 𝑐𝑜𝑠𝑡 𝑓𝑜𝑟 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑝𝑒𝑟 𝑝𝑖𝑙𝑒)
The minimum cost for planks per pile was added to the cost of piles per number of piles. This gave us the
minimum cost for the wood needed for each number of piles. The minimum of these numbers was found.
This number represented the cost for the cheapest configuration that would meet the required strength.
The minimum cost was the searched for in the table, and the cross section, length, and number of piles
were found. Given the number of piles, the cross section of the piles was also found.
A copy of the important portion of this spreadsheet is attached with 11 and 12 piles included as examples.
All other pile amounts have the same setup as these two columns.
RESULTS
The cost for the wood per number of piles was plotted as number of piles versus cost. This allowed us to
see if there was any type of relationship between the number of piles and cost. As we can see, the cost
decreases as you add more piles. This is because as you add more piles, the length between piles
decreases, and as we can see from equation 2, this means that the minimum section modulus decreases
which allows smaller boards to be used. Smaller boards mean cheaper costs. The cost for the planks
decreased faster than the cost for the piles increased which lead to a negative relationship between the
number of piles and the cost. From our graph, we can see that the cheapest configuration is the one that
uses 37 piles. If we go back into our Excel spreadsheet, we can search for the minimum cost associated
with 37 piles which was $4,484.49 total with $981.09 for the planks. We then went through the working
cross section areas of planks that could be used for a 37 pile configuration to find which one had a
$981.09 price tag which happened to be a 10 foot long 2x12 plank. We also know that, since 37 piles are
being used, the cross section of the piles should be 8x8. The cheapest configuration uses 37 10 foot long
8x8 piles and 10 foot long 2x12 planks.
Bill of Materials:



Item
37 – 8x8 10 foot long wooden beams for the piles
54 – 2x12 10 foot long wooden beams for the planks
37 – Concrete footings
Cost
$2,023.41
$981.09
$1,480.00___
$4,484.50
Design:
The retaining wall would be designed with 37 piles spaced 2.22 feet apart spanning a distance of 80 feet.
The piles would be ten feet long with an 8x8 cross section. The piles would be buried five feet
underground and supported with a cement footing. Each plank would be 10 foot long with a 2x12 cross
section. Each plank can be used to span four 2.22 foot gaps meaning 1.112 feet are wasted from each
plank. Nine planks are needed to span the length of the wall while six are needed to span the height
meaning that 54 planks are needed total.