UMN_Capstone_4F_BKBM_FinalReport

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BBYZ Engineer INC
123 Main St
Minneapolis, MN 55345
May 7, 2015
Quentin Odes
5930 Brooklyn Blvd
Minneapolis, MN 55429
Re: 15th Ave Student Housing, Team 4
Dear Mr. Odes:
This letter is written in response to your request on January 20th
, 2015 for the structural design of
the 15th Avenue Housing. It is a 250,000 square foot, 6-story housing project, located near the
University of Minnesota campus, consisting of 5 levels of wood framing supported by a single
story precast concrete podium which transfer loads to a spread footing foundation.
After 15 weeks, a group of four project engineers completed the structural design. 470 hours
were spent with a cost of $70/hour/member, the total cost is $105,770. Attachment of the
complete report is included below.
It was a pleasure to work with you and your team. We’re looking forward to doing business with
you again in the future.
Sincerely,
Seung Ho Baek
Project Engineer
beakx017@umn.edu
(952) 345 6789

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15th Ave Student Housing
701 15th Avenue, SE
Minneapolis, MN
Prepared for:
BKBM Engineers
5930 Brooklyn Blvd
Minneapolis, MN
Prepared by:
Seung Ho Baek
Evan Bui
Jin Ken Yap
Sichen Zhong
May 7, 2015

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Certification Page
By signing below, the team members submit that this report was prepared by them and is their
original work to the best of their ability.
______________________________________________
Seung Ho Baek
Project Engineer
______________________________________________
Evan Bui
Project Engineer
_______________________________________________
Jin Ken Yap
Project Engineer
______________________________________________
Sichen Zhong
Project Engineer

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Executive Summary
This report stated the analysis and design of structural system to support the loads. In order to
design entire structural frame, multiple tasks were completed: gravitational load, joist, beam,
header, walls, and precast level. The loads were determined following Chapter 6 and Chapter 7
in ASCE 7-05. Each section had different methods to design. Based on the analysis, each section
details were determined.
Forte, the computer program, was used for the preliminary design of the wood joists, studs and
headers. To solve for the stresses in the wood components and compare to the design criteria
specified in IBC 2000 and NDS 2005, the program inputs include the loads, dimensions
associated with horizontal components and dimensions for vertical components.
The dimensions were given by the floor plan with measurement notations marked. These
dimensions included the spans over each region of units.

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Table of Contents
Introduction..............................................................................................................1
Background .............................................................................................................1
Methodology............................................................................................................1
Load Combinations..................................................................................................2
Wood Frame.............................................................................................................5
Precast Concrete Podium.........................................................................................8
Spread Footing.......................................................................................................11
Result .....................................................................................................................12
Analysis and Commentary.....................................................................................21
Sustainability Consideration..................................................................................22
Fire Hazard.............................................................................................................22
Schedule and Budget..............................................................................................23
Summary................................................................................................................23
References..............................................................................................................24
Appendix A – Joist Design ................................................................................. A-1
Appendix B – Wind Load....................................................................................B-1
Appendix C – Header Design ..............................................................................C-1
Appendix D – Beam Design ............................................................................... D-1
Appendix E – Bearing Wall Design.....................................................................E-1
Appendix F – Shear Wall Design ........................................................................F-1
Appendix G – Diaphragm Design....................................................................... G-1
Appendix H – Precast Concrete Plank Design ................................................... H-1
Appendix I – Precast Concrete Beam Design.......................................................I-1
Appendix J – Precast Concrete Column Design...................................................J-1

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Appendix K – Spread Footing Design................................................................ K-1
Appendix L – Budget ..........................................................................................L-1
Tables
Table 1 Weight details of roof and floor................................................................12
Table 2 Loads summary.........................................................................................13
Table 3 Coefficients used to determine snow load ................................................13
Table 4 Wood joist design ....................................................................................14
Table 5 Reference design value for Douglas Fir Larch as sawn lumber stud........14
Table 6 Coefficients for 2x6 Douglas Fir Larch sawn lumber used as stud..........15
Table 7 Allowable quantities for design of 2x6 Douglas Fir Larch sawn lumber.16
Table 8 Bearing wall design with detailed configuration ......................................16
Table 9 Jack and King Studs design ......................................................................17
Table 10 Shear wall panel configuration for east-west direction ..........................18
Table 11 Shear wall panel configuration for north-south direction.......................18
Table 12 Diaphragm design for all wood frame floors..........................................19
Table 13 Precast plank design................................................................................19
Table 14 Precast beam design................................................................................20
Figures
Figure 1 Plan view of a region of the floor plan with slab direction .......................9
Figure 2 Plan view of a region of the floor plan with beam direction...................10

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1. Introduction
Residential buildings, particularly in the form of student housing, represent a significant portion
of construction in the United States. The demand for student dormitory and apartment housing
around the University of Minnesota-Twin Cities East Bank area requires an increasing amount of
modern, comfortable, convenient reliable and sustainable residential building construction. This
Capstone Design Project focused on the design of a new student housing complex -- Radius,
located on 15th Ave near the University of Minnesota. This residential building includes one
level of precast concrete and five levels of wood frame construction. It is a fairly typical
configuration of student housing and the use of timber construction for buildings in excess at
stories has become commonplace.
The project goal for the Capstone group was to analyze and design a structural system that
supports loads. The group focused on the design of the precast concrete at the first level, the
design of the wood frame for the upper five levels and the conventional concrete foundations.
The design was initiated with determination of loads based on the International Building Code
2000 edition along with ASCE 7-05. The wood frame was designed following the guidelines and
specifications in the National Design Specification for Wood Construction 2005 edition. The
information on wood frames and precast concrete designs were given by the mentors from
BKBM Engineers. A number of manufactured building materials were selected by the mentors as
well.
2. Background
The Housing is proposed to be 250,000 square feet and 64 feet in height with 189 feet wide and
308 feet long. The building includes six levels and a parking basement, the top five floors are
wood frames resting on the precast concrete podium 1st
floor. The parking basement is out of the
scope of the project. The 6-story housing is supported by conventional spread footings.
3. Methodology
The project started with the loads analysis. The loads were divided into two parts; they were
gravity loads and lateral load. Dead load, live load, and snow load were considered as gravity
loads while wind load was considered as the lateral load.
The wood frame was designed using the method given in the National Design Specification for
Wood Construction (NDS 2005). The wood frame includes the wood joist, flush beam, door
header, window header, bearing wall, and shear wall. The first floor, precast concrete podium
design included the plank, beams and columns. The project ended with the footing design.
The project started with the wood frame design at the top floor. So, the load path started with the
gravity loads flowed from the joist to the beam and then to the load bearing walls. Then, all the
loads from the wood frame were transferred into the precast plank at the first level. At the precast

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level, the loads first flowed from the precast plank to the precast beam and then to the precast
column. The load path analysis ended at the footings.
3.1 Load Combinations
Allowable stress design (ASD) methodology was used for the design of the wood frame. In such,
unfactored loads were used considering the following load combinations given in Section
1605.3.1 of the International Building Code (IBC 2000).
𝐿𝐶1 = 𝐷 (1)
𝐿𝐶2 = 𝐷 + 𝐿 (2)
𝐿𝐶3 = 𝐷 + (𝐿 𝑟 𝑜𝑟 𝑆 𝑜𝑟 𝑅) (3)
𝐿𝐶4 = 𝐷 + (𝑊 𝑜𝑟 0.7𝐸) + 𝐿 + (𝐿 𝑟 𝑜𝑟 𝑆 𝑜𝑟 𝑅) (4)
𝐿𝐶5 = 0.6𝐷 + 𝑊 (5)
Where D is the dead load, L is the live load, Lr is the roof live load, S is the snow load, R is the
rain load, W is the wind load, and E is the earthquake.
Earthquake (E) and rain (R) were considered as rare events in Minnesota and therefore they were
negligible. Besides, the roof live load (Lr) was compared to snow load (S) using the values given
in the IBC (see Table 2) and found out that S was the controlling case. As a result, the non-
controlling load combinations were discarded. Shown below are the load combinations analyzed
for the design are:
𝐿𝐶1 = 𝐷 (1)
𝐿𝐶2 = 𝐷 + 𝐿 (2)
𝐿𝐶3 = 𝐷 + 𝑆 (6)
𝐿𝐶4 = 𝐷 + 𝑊 + 𝐿 + 𝑆 (7)
𝐿𝐶5 = 0.6𝐷 + 𝑊 (5)
3.1.1 Dead Load
Dead load is the self-weight of the structure itself, which is carrying the gravity load and must be
estimated before design. Section details in plan were used to select materials in order to decide
the total roof and floor dead load. The weight of each material from the section details was
selected from ASCE-07.

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3.1.2 Live Load
Live load is the weight of everything superimposed on, or temporarily attached to, a structure.
Live loads were determined from the IBC 2000 depending on building usage (Table 1607.1 in
IBC 2000).
ASCE 7-05 Section 4.6 states, “The full intensity of the appropriately reduced live load applied
only to a portion of a structure or member shall be accounted for if it produces a more
unfavorable effect than the same intensity applied over the full structure or member.” This means
that it is necessary to pattern the live loads to consider the maximum effect on the members. In
this project, there were multiple loading cases considered for constructing the shear and bending
moment envelopes for the continuous members. According to the ASCE 7-05 manual, there were
five load cases studied for the three span members shown in figure A-2 in appendix A.
The uniformly distributed live loads may be reduced according to the ASCE 7-05. The members
for which a value of KTTAT is 400 [ft2
] or more are permitted to be designed for a reduced live
load in accordance with the following formula:
𝐿 = 𝐿 𝑜(0.25 +
15
√𝐾 𝑇𝑇 𝐴 𝑇
(8)
where L is the reduced design live load per ft2
of area supported by the member, Lo is the
unreduced design live load per ft2
of area supported by the member, KLL is the live load element
factor (see Table 4-2 in ASCE 7-05), and AT is the tributary area in ft2
.
It was noted that L shall not be less than 0.5Lo for members supporting one floor and L shall not
be less than 0.4Lo for members supporting two or more floors.
3.1.3 Snow Load
It was assumed that the snow loads acted on a flat roof even though the roof had a slope surface
in reality because the roof slope is negligible. The snow load, S in lb/ft2
is calculated using the
following formula:
𝑆 = 0.7 𝐶𝑒 𝐶𝑟 𝐼𝑝 𝑔 (9)
Where Ce is the exposure factor, Ct is the thermal factor, I is the importance factor and pg is the
ground snow load [lb/ft2
].
All the values of Ce, Ct, I and pg were taken directly from the Chapter 7 of ASCE 7-05 manual
and tabulated in table 3 in the report.
A continuous beam system was used to investigate the effects of the different loading cases (see
A-1 in the appendix A).The first case was full snow along the entire span. The second case was a
full balanced snow load on either exterior span and half the balanced snow load on other spans

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while the third case was half the balanced snow load on either exterior span and full balanced
snow load on all other spans. Last but not least, all possible combinations of full balanced snow
load on any two adjacent spans and half balanced snow load on all other spans would be the last
case to be determined.
3.1.4 Wind Load
All design values were taken from Chapter 6 of ASCE 7-05. The sections, tables and figures
referenced in this section can be found in Chapter 6 of ASCE 7-05.
The wind loads were taken as lateral loads acting on the entire enclosed structure (section 6.2).
Because the structure is to be located in the urban area of Minneapolis, the basic wind speed was
taken as 90 mph according to ASCE 7-05 Figure 6-1. The importance factor accounting for the
use of the building was taken as 1.0 with occupancy category II. (ASCE 7-05 Table 6-1, Table 1-
1) The surface roughness category was identified as B (Section 6.5.6.2), and the exposure
category was defined as B (Section 6.5.6.3). The velocity pressure exposure coefficients were
taken and computed using Table 6-3 and Table 6-2.
Kh = 0.93 and
𝐾𝑧 = 2.01 ∗ (
𝑍
1200
)
2
7
(10)
Repeatedly, the topographic factor was taken as Kzt = 1.0 according to section 6.5.7.2. The gust
effect factor was taken as G = 0.85 according to Section 6.5.8.1.
The internal pressure coefficient was taken as either GCpi = 0.18 or -0.18 with respect to i)
positive internal pressure on all internal surfaces or ii) negative internal pressure on all internal
surfaces.
The external pressure coefficient, Cp, which in accordance with the different wind direction, was
taken from ASCE Figure 6-6.
The velocity pressure throughout the height of building, qz, was a function of elevation.
However, the velocity pressure, qh, was a constant. They were computed by using the following
equations:
𝑞 𝑧 = 0.00256𝐾𝑧 𝐾𝑧𝑡 𝐾𝑑 𝑉2
𝐼 (11)
𝑞 𝑧 = 0.00256𝐾𝑧 𝐾𝑧𝑡 𝐾𝑑 𝑉2
𝐼 = 18.85
The wind loads on leeward walls, windward walls and sidewalls are given by the following
expression (Section 6.5.12.2.1):
𝑝 = 𝑞𝐺𝐶 𝑝 − 𝑞𝑖(𝐺𝐶 𝑝𝑖) (12)

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Where q was taken as qz for windward walls, or qh for leeward and sidewalls; qi was taken as qh
as a conservative design.
The overall wind load acting on the entire enclosed structure was determined by combining
pressure on windward walls and pressure on leeward walls.
3.2 Wood Frame
The residential building has five levels. Level 2 to Level 6 consisted of wood frames primarily
including stud walls, I-joists, beams and headers. The design was initiated using the computer
program which is Forte specifically suited for the wood frame design. With the preliminary
results of design from Forte, the size of structural members was checked using allowable stress
design method according to the instructions in the NDS.
A full set of architectural drawings of all floor plans and profile views was provided. Noticing
that the layout of internal walls at each level was consistent throughout all five levels, it was
known that the loads that each floor carried were the same. This feature provided a convenient of
joist design and beam configuration. With the designated value of joist spacing, the bearing wall
and shear wall design were completed.
3.2.1 Wood Joist
An engineered wood joist, which is commonly known as an I-joist, is a product designed and
used extensively in floor and roof framing. In this project, the TJI®
joists were used as the floor
and roof joists to transfer the gravity loads (i.e., dead load, live load and snow load) to the walls
under it.
3.2.1.1 Joist Layout
The layout of each unit in the apartment was provided in the architectural floor plan. Typically, a
continuous span with multiple supports was used. The direction of the joist span was chosen span
from demising wall to demising wall as shown in Figure A-1 (Appendix A). It was noted that
only some of the walls were load bearing while the rest were just partition.
3.2.1.2 Joist Design
The joist design was divided into two parts; they were roof joists and floor joists. The only
difference between them was the applied loads. The controlling load combination for the roof
joist was D + S while for the floors was D + L. Then, patterned snow and live loads on the
continuous member were analyzed using SAP2000 to determine the flexural and shear demand.
According to the architectural drawings, the floor plan for each floor was the same. Similarly, the
bearing walls were stacked from the second floor to the top floor. So, the resulting design was
applied to each floor.

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The joists were designed based on the worst case. This allows for consistent construction. The
worst case scenario was identified as the largest wood joist size required to support the loads.
After studying 40 different types of unit plans, the worst case scenario was identified as a
continuous member with two short spans at the ends and a long span in the middle (see Figure A-
2 in Appendix A).
Lumber strength was affected by the cumulative duration of the maximum variable loads
experienced during the life of the structure. When considering other loads with different duration
characteristics, it was necessary to modify the tabulated stresses for the TJI®
joists provided by
the supplier (Table 3, ESR-1153) by a load duration factor as shown in Table 2.3.2 (NDS 2005).
The joist design was conducted using allowable stress design and deflection considerations. The
actual bending stress or moment demand shall not exceed the adjusted bending design values.
Other than flexure, shear design was also important and its design procedure is specified in the
NDS.
The deflection was a concern as well for the joist. Deflection is calculated as follows:
∆=
22.5𝑊𝐿4
𝐸𝐼
+
12 𝑊𝐿2
𝐾𝑑∗105
(13)
Where W is the uniform load [plf], L is the clear span [ft], E is the Young’s modulus [psi], I is
the moment of inertia [in4
], d is the out-to-out depth of joist [in], and K is a constant value
obtained from the ICC-ES Evaluation Report on TJI joists.
The example calculation of roof joist design was shown in Appendix A1 and the example
calculation of floor joist design was shown in Appendix A2.
3.2.1.4 Corridor
Commodity lumber was used for the corridor because it was more economical, the corridor had a
different elevation compared to the floor of the units and was designed as a single span between
the adjacent walls.
3.2.2 Wood Flush Beam
A wood flush beam is a structural member positioned at the same level as the joists to support
the weight of the frame. For those locations where a structural wall was absent, it was necessary
to add a flush beam perpendicular to the joist to provide an intermediate support. Wood flush
beam is designed for bending, shear and bearing length. In the calculation, there is an assumption
which is beam stability factor is equal to one. After design check, the beam stability factor
should be calculated and it should be less than one because beams may be unstable under the
applied loads. Instability is due to the tendency of the compression edge of the beam to buckle
causing the beam to deflect laterally.

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3.2.3 Bearing Wall
The bearing walls of the building serve as one of many load-carrying components such as joist
and beam except bearing walls resist lateral forces while supporting cumulative load from above.
3.2.3.1 Wall Layout
The floor plan illustrates that the internal walls are running perpendicular such that the joists that
have been designed must be installed parallel to some of the walls while perpendicular to the
others. Orienting joists in such a manner allows the design group to focus on those walls that are
perpendicular to the joist direction. In addition, in every apartment unit there are a number of
walls designed to separate rooms, which are partition walls. The partition walls are excluded
from the structural design because they do not provide support to the joists.
3.2.3.2 Selection of Wood Species for Wall
A set of wood species is commonly available in North America and can be used as vertical
members. The National Design Specification provides a list that specifies the material properties
of the different kinds of wood. These material properties included reference design values for a
range of member sizes for each species, in terms of Young’s Modulus, allowable compressive
stress, and allowable flexural stress.
3.2.3.3 Stud Wall
The stud walls are mainly subject to vertical compression and a relatively minor lateral uniform
load under daily service. The design was started using factors that are associated with the
reference design value of a certain wood species. These factors are specified in Chapter 3 of the
NDS, where many aspects of service conditions are discussed in depth.
3.2.4 Header
Headers are horizontal structural members that transfer the load from the floor around the
opening. Header has gravity load from the floor above, design for bending and shear. There were
two different cases to consider for the headers: exterior window and interior door openings. For
window and interior doors, either Laminated Veneer Lumber or commodity lumber with
multiple plies was used. In order to decide the header size, tabulated design values and lumber
property adjustments were selected from the IBC and Table 4A of the NDS. Header sizes could
be decided by checking bending for floor load, horizontal shear, bearing, and deflection. In the
checking calculation, there was an assumption which is the later stability factor, CL, is equal to
one. Also, the lateral stability was calculated separately and it was less than one.

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3.2.4.1 Jack and King Stud
The number of jack and king studs installed on each side of a header should comply with IBC
standards. Gravity load from floor above transfers to jack studs so jack studs need to be designed
for axial load. The number of jack stud could be decided by comparing actual compression to
compression perpendicular to the grain. King studs are designed for bending and axial which will
be similar to studs in bearing walls.
3.2.5 Shear Wall
Shear wall is a structural system exist to resist lateral loads that act on the structure. Seismic and
wind loads are shear wall most commonly design for. However, the location in which the
proposed project will be built will eliminate seismic load.
The shear walls were selected from demising walls such that all partition walls did not contribute
to the shear resistance. Demising walls were defined to be those walls dividing two adjacent
units. The shear walls were designed according to Chapter 23 in International Building Code
2000.
3.2.6 Diaphragm
Diaphragm works together with shear wall to form a lateral system to withstand lateral loads. It
transmits lateral load to vertical resisting elements such as shear wall or frames. Typically, it’s
horizontal and does double duty as floor system or roof system. It make out of plywood usually.
To determine linear distributed wind load:
𝑤 = 𝑊 ∗ 𝐻 (14)
where w is horizontal distributed wind load [lb/ft], W is wind load [lb/ft2], and H is height of
story being design.
Using w to determine section with maximum shear force V, then
𝑣 = 𝑉/𝐿 (15)
where v is shear [lb/ft], V is total shear force [lb], and L is the width of building [ft].
Once v is determine, using IBC 2006 chapter 16 table 2306.3.1 to pick panel grade.
3.3 Precast Concrete Podium
The podium of this complex was a precast level and consisted of the precast planks, precast
beams and the precast columns. The precast concrete design was conducted using Load and
Resistance Factor Design (LRFD) to determine the required strength of the member. The load

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from the entire wood frame was transferred into the plank, then beams and eventually to the
columns.
3.3.1 Precast Plank
An extruded hollow core was used as the concrete plank. First, the direction of the plank was
determined. The direction of the precast plank followed the columns lines specified in the
architectural drawings as shown in figure 1.The entire layout for the floor can also be seen in
figure F-2 in appendix F.
Figure 1. Plan view of a region of the floor plan with slab direction
When superimposed load consisted of both dead and live loads, the uniform distributed loading
was modified using the equation below in order to use the safe load tables in the appendix F.
𝑤𝑠 =
1.2
1.6
𝑤𝑠𝑑 + 𝑤𝑙 (16)
Another parameter required to read the safe load table was the length of the precast plank. With
the uniformly distributed superimposed service load [psf] and precast plank length [ft]
parameters, the size of the extruded hollow core was determined.
3.3.2 Precast Beam
In order to have a flat slab, an inverted T-beam was used for the precast beam design. The beam
was designed to support the precast plank on both sides and the gravity loads from the timbers
structure above. So, the direction of the precast beam spans was perpendicular to the precast
plank direction as shown in figure 2.

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Figure 2. Plan view of a region of the floor plan with beam direction
The precast beam design procedure was similar to the precast plank design. With the uniformly
distributed superimposed service load [klf] and precast beam length [ft] parameters, the size of
the inverted T-beam was determined. Appendix I showed the steps taken in beam design.
3.3.3 Precast Column
A rectangular reinforced concrete was used to support the beams as shown in Figure J-1. Based on the
information provided by the supplier, the concrete column compressive strength, f’c was 9000[psi]
and the yield strength of the steel reinforcement, fy was 60000[psi]. Since the column was
supporting the beam, so the beam reaction method was used. The beam reaction method enabled
us to collect the reaction forces from adjacent beams as the axial load, P as shown in Figure J-7.
A load, P, was eccentric when its line of action was offset a distance, e, from the column. Therefore, it
was necessary to consider also the bending moment (Mu = P*e) produced by the eccentric load.
In addition, wind load was another factor when designing the column.
There were two load combinations to be considered. The first one was gravity loads only, as
shown in equation 17 while the second one was gravity plus wind loads, as shown in equation
18.
𝑈 = 1.2𝐷 + 1.6𝐿 (17)
𝑈 = 1.2𝐷 + 1.6𝑊 + 1.0𝐿 (18)

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The column was considered safe if the results (Pu and Mu) from each load combination fell under
the interaction curve of that column size. An example was shown in Figure J-8. Appendix J
showed the steps taken in column design.
3.4 Spread Footings
Footing refers to the structural element that exists at the bottom of the building. As the soil does
not and is not able to support the walls and columns of the 6-story building directly, footing must
be designed properly to handle the transition of load from structure to soil. The immediately
contacting structural elements to the footings are exterior walls and columns at the basement
parking garage. Accordingly, two types of spread footing were designed, the wall footings and
the column footings.
3.4.1 Wall Footing
The wall footings were designed in such a way only a foot-long section of the footing and wall
was analysed. Therefore, finding the linear reactional force along each exterior wall was the first
essential task. To achieve that, a SAP model was built and solved with dead load, live load, snow
load and wind load applied to the model respectively. The linear-foot-long reaction force was
then used to obtain the factored load combinations. Allowable Stress Design was used to obtain
the base area of the footing, also the width of the wall footing. Allowable soil pressure was
needed at this stage. The depth of the wall footings were determined primarily to resist the shear
action due to the distributed soil pressure from the bottom.
Load and Resistance Factor Design was used to check and adjust capacities of the footing after
the size of footing was chosen. These capacities include bearing capacity, one-way shear
capacity, two way shear capacity (only for column footings) and flexural capacity.
3.4.2 Column Footing
The column footings were designed in a similar way except the reaction force at each column
was treated as point force. In addition to the one-way shear action, a column footing also has the
two-way shear action, also known as punching shear action.
3.4.3 Design Scope
Bearing capacity is dependent on the width of the walls or the size of the columns. The width of
the walls and the size of the columns, because of the lack of information about the basement
columns, were assumed reasonably.
One-way shear action and two-way shear action are associated with the depth of the footings.
The deeper the footing is, the more shear capacity it will have. Sometimes to avoid designing a
footing that is too deep, some stirrups might be added to the design to handle the exceeding
portion of required shear stress. The difference between the two types of shear actions is the
critical planes of them. One way shear action always occurs at the face of the wall or the face of
the column. Two way shear action has its critical plane surrounding the column with its
perimeter in a shape that is geometrically similar to the column’s perimeter.

12
Flexural capacity was designed such that the minimum flexural reinforcement requirement was
met according to section 10.5.1 in ACI 318-11. If the flexure stress exceeds the capacity
provided by the minimum reinforcement, more steel bars are to be placed to the tension side of
footing.
Reinforcement for shrinkage and temperature change were provided in the design.
4. Results
4.1 Loads
Table 1. Weight details of Roof and floor
Material Floor Load [psf] Roof Load [psf]
¾” gyp-crete 6.3 N/A
plywood wood deck (⅛”) 3.2 N/A
wood joist (TJI) 1.4 N/A
1 layer of ⅝” type x gypsum board 3 N/A
2 layers of all 2x6 joists 4.2 N/A
carpet and pad 2 N/A
sprinkling system 1 N/A
wood partition 5 N/A
miscellaneous mechanical 2 N/A
rigid insulation 1.5 N/A
Ballasted single-ply N/A 10
rigid insulation (1”) N/A 4.8
plywood roof deck (⅛”) N/A 3.2
sprinkling system N/A 1.5

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2 layers ⅝” GYP board N/A 5.6
Table 2. Loads summary
Loads [psf]
Dead load for floor 30
Dead load for roof 26
Roof live load 20
Live load for residential 40
Live load for common area 100
Snow load 35
Table 3.Coefficients used to determine snow load
Value Reasons Reference
Exposure factor, Ce 1.0 Urban area, partially exposed to
snow
ASCE 7-05 Table 7-2
Thermal factor, Ct 1.0 Under normal thermal condition ASCE 7-05 Table 7-3
Importance factor, I 1.0 Category II ASCE 7-05 Table 7-4
Ground snow loads, pg
[psf]
50 Located in Minneapolis, MN ASCE 7-05 Figure 7-1
4.2 Wood Frame
4.2.1 Wood Joist
Table 4: Wood joist design
Wood Joist Result
Roof unit 16” TJI 110 with 16” spacing
Floor unit 16” TJI 110 with 16” spacing

14
Corridor 2 X 10 S-P.F No. 1
Common area 16” TJI 230 with 12” spacing
4.2.2 Wood Beam
The final design of the wood beam was shown in Table 5 after the design criteria stated in NDS
2005 were checked and satisfied. Lateral stability was calculated after design check as shown in
Appendix D. In result, a 2 X 6 D.F-L No.1 flush beam is used.
4.2.3 Bearing Wall
The species of wood chosen for the studs was the Douglas Fir Larch with a specific gravity of
0.5. All bearing walls fall into two categories. One was known as demising walls, which were
those walls dividing two adjacent units. The other category includes the walls run perpendicular
to the joists within the unit, and these walls were called in-unit walls in this project. For
convenience of workers and simplicity of construction, all in-unit bearing walls are designed
with 2 x 6 Douglas Fir Larch sawn lumber. The demising walls were designed to be double 2 x 4
Douglas Fir Larch sawn lumber. The stud spacing was identical to that of joists being 16 inches
on center.
The reference design values were found by the combination of the property of the selected wood
species, the use, and the size of the wood component. The reference design values for Douglas
Fir Larch were taken from NDS.
Table 5. Reference design value for Douglas Fir Larch as sawn lumber stud
Reference Design Value Value [psi]
Fb 700
Ft 450
FV 180
Fc perp 625
Fc 850

15
E 1,400,000
Emin 510,000
The coefficients that account for different aspects of service condition were calculated according
to NDS. These coefficient can affect the design in such a way each final allowable value for a
type of member is calculated by multiplying the reference design value by all applicable
coefficients. These coefficients include the load duration factor, CD, the wet service factor, CM,
the temperature factor, Ct, the beam stability factor, CL, the size factor, CF, the flat use factor,
Cfu, incising factor, Ci, the repetitive member factor, Cr, the column stability factor, CP, and
buckling area factor, CT. For instance, the allowable compressive stress was calculated by the
following expression:
𝐹𝑐
′
= 𝐹𝑐 ∗ 𝐶 𝑀 ∗ 𝐶𝑡 ∗ 𝐶 𝐹 ∗ 𝐶 𝑃 (17)
Table 6: Coefficients for 2 x 6 Douglas Fir Larch sawn lumber used as stud
reference
design
value
CD CM Ct CL CF Cfu Cr CP CT
load
duration
wet
service
temp
eratu
re
beam
stability
size flat use
repetitive
member
column
stability
buckling
area
Fb
Snow
load
1 1 1 1 1.15 1.15 - -
Ft 1.15
- - - -
- - -
FV
- - - - - - - - -
Fc perp
- - - - - - - - -
Fc
-
1 1
-
1
- -
0.7193 -

16
E
- - - - - - - - -
Emin
-
1 1
- - -
- 1.3047
The allowable values for ASD design were calculated as shown in table 7.
Table 7. Allowable quantities for design of 2 x 6 Douglas Fir Larch sawn lumber stud.
Allowable Quantities Value Unit
Fb' 1064.61 psi
Fc' 703.12 psi
Emin' 665403.80 psi
The final design of the bearing walls was determined after the design criteria stated in NDS 2005
were checked and satisfied for each level from level 2 to level 6. Table 8 shows the summary of
the bearing wall design including that of all wood frame levels.
Table 8: Bearing wall design with detailed configuration.
Bearing Wall Summary
Level 2nd 3rd 4th 5th 6th
Wood Species Douglas Fir Larch
Product Type Sawn Lumber
Nominal Size 2" x 6 "
Top Plate Double plates
Bottom Plate Double plates
Plies of Stud 3 3 2 2 1

17
4.2.4 Header
2 X 6 D.F-L No.1 was used for header. Bending for floor load, horizontal shear, bearing and
deflection were calculated in order to check (Appendix C).
4.2.4.1 Jack and King Stud
The final design of the Jack and King studs was shown in Table 9 after the design criteria stated
in NDS 2005 were checked and satisfied for each level from level 2 to level 6.
Table 9: Jack and King stud design
Wood Species Douglas Fir Larch
Product Type Sawn Lumber
Nominal Size 2" x 6 "
Plies of Jack stud 3 2 2 2 2
Plies of King stud 2 2 1 1 1
4.2.5 Shear Wall and Diaphragm
4.2.5.1 Shear Wall
In order to resist wind load in the east-west direction, four pairs of shear walls were selected for
each level. That is, for each level, four shear walls were located on the west wing of the building
are four identical shear walls located on the east wing. Each pair of these shear walls were
parallel and aligned such that they resist the wind load on the tributary area of the same width.
The shear resistance of a shear wall is usually given by a shear wall panel. Panel configurations
and the allowable shear of each panel configuration were provided in Table 2306.4.1 in IBC
2000. Every four pairs of shear walls was designed separately such that the allowable shear
exceeds the actual shear due to wind pressure.
Table 10: Shear wall panel (structural I grade) configuration for east-west direction.
Shear wall panel configuration - EW
Panel Thickness in 15/32 3/8 dbl 15/32 15/32 5/16

18
dbl
Fastener Penetration in 1 1/2 1 3/8 1 1/2 1 1/2 1 1/4
Nail Size 10d 8d 10d 10d 6d
Fastener Spacing at
Panel Edges
SW 1 in 3 2 2 4 4
SW 2 3 3 3 4 6
SW 3 3 3 2 4 4
SW 4 4 4 3 6 6
Shear wall design in the north-south direction was modified to fit the different features in the
floor plan. Demising walls, which were selected to be the shear walls, were not geometrically
symmetric in north-south direction. Therefore, the “worst scenario” approach was adopted.
The worst scenario was where the shear wall carries the wind load on the largest surface area,
which is equivalent to the widest surface for each level. With this method, the shear wall that
carries the widest surface of approximately 83 ft has a combined length of 53.8 ft. Based on this
case, the shear wall panel was selected again from Table 2306.4.1 in IBC.
Table 11: Shear wall panel (structural I grade) configuration for north-south direction.
Shear wall panel configuration - NS
Panel Thickness in 15/32 dbl 15/32 dbl 7/16 dbl 15/32 5/16
Fastener Penetration in 1 1/2 1 1/2 1 3/8 1 1/2 1 1/4
Nail Size 10d 10d 8d 10d 6d
Fastener Spacing at Panel Edges in 2 3 3 3 4
4.2.5.2 Diaphragm
Diaphragm is designed using north-south wind load at the sixth floor and maximum shear at the
worst case scenario span (conservative). The result is applied to all diaphragms. Calculations are
shown in appendix F.
Table 12: Diaphragm design for all wood frame floors.

19
UNBLOC
KED
DIAPHRA
GMS
Fasteners
spaced 6"
max. at
supported
edges
PANEL GRADE
COMMON
NAIL SIZE OR
STAPLE
LENGTH AND
GRADE
MINIMUM
FASTERNER
PENETRATION
IN FRAMING
(inches)
MINIMUM
NOMINAL
PANEL
THICKNESS
(inches)
MINIMUM
NOMINAL
WIDTH OF
FRAMING
MEMBERS AT
ADJOININIG
PANEL EDGES
AND
BOUNDARIES
(inches)
All other
configurati
ons (Cases
2,3,4,5 and
6)
Sheathing, single
floor and other
grades covered in
DOC PS1 and
PS2
1 1/2 16 Gage 1 5/16 2 90
4.3 Precast Concrete Podium
4.3.1 Precast Plank
Table 13: Precast plank design
Region Mu [k-ft] Plank Size Strands no. and size Mn [k-ft] Mn > Mu
A 615 12” x 48” 6-1/2 692 OK
B 240.7 8” x 48” 4-1/2 272.4 OK
C 280.7 8” x 48” 5-1/2 335.2 OK
D 455 8” x 48” 5-6/10 and 2-1/2 569.2 OK
E 78.8 8” x 48” 3-1/2 208 OK
F 224 8” x 48” 4-1/2 272.4 OK
G 555.2 8” x 48” 5-6/10 and 2-1/2 569.2 OK
H 122.3 8” x 48” 3-1/2 208 OK
I 689.2 12” x 48” 6-1/2 692 OK
J 675.5 12” x 48” 6-1/2 692 OK

20
4.3.1 Precast Beam
Table 14: Precast beam design
Beam # Mu [k-ft] Beam Size Strands # Mn [k-ft] Mn > Mu
1 569 24IT24 12 623.45 OK
2 608 24IT24 12 623.45 OK
3 654 24IT24 14 697.76 OK
4 533 24IT24 10 543.56 OK
5 267 24IT24 6 353.48 OK
6 583 24IT24 12 623.45 OK
7 305 24IT24 6 353.48 OK
8 538 24IT24 10 543.56 OK
9 657 24IT24 14 697.76 OK
10 556 24IT24 12 623.45 OK
11 376 24IT24 8 457.79 OK
12 612 24IT24 12 623.45 OK
13 630 24IT24 14 697.76 OK
14 899 24IT24 22 909.62 OK
4.3.2 Precast Column
All the columns used a 12” by 12” column with 4 no. 6 steel bars. The concrete compressive
strength, f’c is 9,000 psi while the steel yield strength is 60,000 psi.
4.4 Spread Footings
4.4.1 Provisions
Due to the fact that the sublevel was not designed in this project, information about column sizes
and wall thickness were to be assumed. Due to the lack of geologic report of the soil, at the stage
of determining the base area, an allowable soil pressure of 3000 psf was assumed according to
Table 1804.2 in IBC 2000 and the suggestion from mentors. In addition, a minimum footing
depth is specified in ACI 318-11.
The normal weight concrete (NWC) was assumed and used in the entire footing design. The
compressive strength of the concrete was assumed and required to be 4000 psi. For
reinforcement, grade 60 steel reinforcement bars were selected for any potential needs. The size
of the bars were subject to change during the design.
All calculations were done and recorded in Excel spreadsheets. See Appendix K for calculations.
4.5.2 Wall footing
Wall width was assumed to be 12 inches.
Given the magnitude of loads, 16-inch deep wall footing was initially assumed for wall footing
design and performed well through the design process. The shear stress did not exceed the shear

21
capacity provided by the concrete, therefore stirrup was not required in the wall footings.
Bearing capacity was checked and the thickness of wall successfully provided the bearing
capacity to the footing. Minimum flexural reinforcement was provided such that it provided
enough flexural capacity against bending actions that result in tensile stress in the bottom side of
the footing.
To enable the function of the reinforcement bars, the development length was calculated and
minimized by using 90 degree hook. See Figure K-1 for section detail of the designed wall
footing.
4.5.3 Column Footing
Column size was assumed to be 20 in. by 20 in. As a reasonable design, square footing was
selected.
The depth of the column footing was chosen to be 24 inches. The 2-foot-deep concrete provided
sufficient shear capacity for one-way shear action but failed to provide sufficient two-way shear
capacity, therefore stirrups were added to the design.
The flexural reinforcement was designed as prescribed in ACI 318-11. The shrinkage and
temperature reinforcement was provided by the flexural reinforcement as it runs in the transverse
direction as well. See Figure K-2 for section detail of the designed column footing.
5. Analyses and Commentary for the Structural Design
A continuous span was preferable to a simply supported span because it had more vertical load
capacity, less deflection and was more economical.
Because of its natural composition, wood is better able to resist higher short-term loads (i.e.,
transient live loads or impact loads) than long-term loads (i,e., dead loads and sustained live
loads). Under impact loading, wood can resist about twice as much stress as the standard 10-year
load duration (i.e, “normal duration”) to which wood bending stress properties are normalized in
the NDS. Therefore, the bending stresses must be adjusted by the load duration factor, CD.
Values of the CD for various load types are based on the total accumulated time effects of a given
type of load during the useful life of a structure. As a result, CD increases with decreasing load
duration.
Hollow core slabs are known for providing economical, efficient floor and roof systems
compared to a solid slab. This is due to their lower self-weight, less of raw materials, and thye
are quick to assemble and build. Structurally, a hollow core slab provided the efficiency of a pre-
stressed member for load capacity, span range, and deflection control. In addition, when properly
coordinated for alignment, the voids in a hollow core slab may be used for electrical and
mechanical runs.

22
6. Sustainability Consideration
Wood is a renewable resource. It grows naturally taking the energy from the sun and yet it is
recyclable. On the other hand, in concrete and steel production, fossil fuel is most likely the
source of energy being used. According to the U.S. General Services Administration, a report in
2009 showed that using wood in buildings, the cost is far less to operate. It features great energy
performance because it is an effective insulator and requires much less energy to produce
comparing to concrete or steel. The embodied energy required to extract, process, transport and
install wood is also less than that in buildings fabricated with brick, concrete, and steel.
Forests play a crucial role in Earth’s carbon cycle, during its lifecycle, trees absorb a great
amount of carbon dioxide and release oxygen through photosynthesis process. By choosing
wood over concrete or steel can limit or prevent greenhouse gases emissions during concrete and
steel production. Last but not least, tree can help reduce runoff during big rain event since it
promote infiltration into the aquifer.
6.1 Fire Hazard
One of the biggest drawbacks of choosing wood over steel and/or concrete is a higher risk of fire.
Therefore, the building will equipped with a fire sprinkler system installed. IBC 2006-Chapter 7
identified requirements for different structural components being installed in the building. In
addition, Weyerhaeuser is the provider warehouse and their TJI are fire resistance coated with
Flak Jacket Coating.

23
7. Schedules and Budget of the Structural Design
The structural design is done by a group of four project engineers took over the course of 15
weeks, a total of 470 hours. With a cost of $70/hr/member, it added up to a total of $105,770.
8. Summary
The 15th Ave Housing has one precast podium level as its level 2, and five upper levels are made
out of wood frame. The wood frame system consisting of roof joists, floor joists, beams, headers,
bearing walls, shear walls, and diaphragm.
The bearing walls of the entire building were designed to be assembled with Douglas Fir Larch 2
x 6 inch sawn lumber. The stud spacing was designed to be 16 in. o.c. following the joist
spacing. Each bearing wall had its doubled top plates and doubled bottom plates being the same
material as the stud. The number of stud ply varied through different levels. Bearing walls at
Level 2 use 3 plies; bearing walls at Level 3 used 3 plies; bearing walls at Level 4 used 2 plies;
bearing wall at Level 5 used 2 plies; and bearing walls at Level 6 used 1 ply.
The shear walls resisting the east-west wind load were designed individually. There were four
pairs of shear walls selected running east-west. These shear walls were designed to have
different shear wall panels installed. The thickness of panels, the minimum required penetration
of fastener, the nail size and the spacing of nails varied from levels and wall locations. Double
panels were used for walls in levels lower than Level 4.
The shear wall resisting north-south wind load was designed with one identical shear panel
configuration for each level. The configuration also varied from levels as the wind effect
increased toward the base of the structure.

24
9. References
ACI (American Concrete Institute). (2013) Building Code Requirements for Structural Concrete
(ACI 381-11) and Commentary. Fourth printing. ACI, Farmington Hills, MI.
ASCE (American Society of Civil Engineers). (2006) Minimum Design Loads for Buildings and
Other Structures, Standard ASCE/SEI 7-05. Third printing. ASCE, Reston, VA.
AWC (American Wood Council). (2005) National Design Specification (NDS) for Wood
Construction. AWC, Washington, DC.
ICC (International Code Council). (2000). International Building Code. ICC, Falls Church, VA.
ICC-ES (ICC Evaluation Service Evaluation Report). (2013). ESR-1153. IIC-ES, Brea, CA.
Macgregor, J.G. and Wight J.K. (2011). Reinforced Concrete: Mechanics & Design, 6th
Edition.
Prentice Hall, Englewood Cliffs, N.J.
Robert H. Falk. (2010). Wood Handbook-Wood as an Engineering Material-Chapter 1: Wood as
a Sustainable Building Material, Centennial Edition. Forest Products Laboratory, Madison,

A-1
Appendix A – Joist Design
The direction of the joist was selected to span between the load bearing walls as shown in Figure A-1.
Figure A-1: Plan view of a unit and direction of the joist spans
As discussed earlier, the joists were designed based on the worst case. The worst case was a
scenario which the largest wood joist size is required to support the loads. Figure A-2 showed
the two cases to be analysed in a unit. By inspection, the bottom case controlled because it
had a long span in the middle that created a large flexure and shear force.
Figure A-2. An example of a continuous span with four wall supports
The worst case for roof joist was found in the unit Type 3J and it was analyzed using Forte as
shown in figure A-3. Forte also computed the roof joist size that worked under the input
loads.

A-2
Roof joist
Figure A-3. Roof joist design using Forte
In addition, the bending member design procedure (NDS) was used to prove that the result
from Forte was valid. First, it was known that the controlling load combination for the roof
joist was D + S. The value for dead and snow loads were shown in table 2. According to
ASCE7-05, the patterned snow load on the continuous member was analyzed using SAP2000
to determine the flexural and shear envelope. Figure A-4 showed the load cases to be
considered for patterned snow loads.
Figure A-4: Patterned snow loads

A-3
The load analysis on patterned snow loads using SAP2000 gave the flexural and shear
envelope shown in figure A-5 and A-6.
Figure A-5: Moment envelope for patterned snow loads
Figure A-6: Shear envelope for patterned snow loads

A-4
From the moment and shear envelope, the maximum value for each load case under patterned
snow load was determined as shown in Table A-1 and Table A-2.
Table A-1: Maximum moment of patterned snow load
[k-in] M-ext R M+mid M-int L M-int R M+mid M-int L M-int R M+mid M-ext L
LC1 0 - -35.2 -35.2 27.6 -34.8 -34.8 - 0
LC2 0 - -25.2 -25.2 19.0 -25.4 -25.4 - 0
LC3 0 - -34.8 -34.8 28.2 -34.0 -34.0 - 0
LC4 0 - -35.5 -35.5 27.9 -33.7 -33.7 - 0
LC5 0 - -35.5 -35.5 27.9 -33.7 -33.7 - 0
max 0 - -35.5 -35.5 28.2 -34.0 -34.0 - 0
Table A-2: Maximum shear of patterned snow load
[k] V-ext R V+mid V-int L V-int R V+mid V-int L V-int R V+mid V-ext L
LC1 0.03 - 0.69 -0.92 - 0.91 -0.69 - 0.08
LC2 -0.07 - 0.58 -0.65 - -0.65 -0.60 - 0.16
LC3 0.12 - 0.59 -0.92 - 0.91 -0.57 - -0.03
LC4 0.04 - 0.69 -0.92 - 0.91 -0.57 - -0.03
LC5 1.04 - 0.69 -0.92 - 0.91 -0.57 - -0.03
max 1.04 - 0.69 -0.92 - 0.91 -0.69 - 0.16
The data from Table A-1 and A-2 were compared to the reference design values for TJI joists
provided by the wood product company, Weyerhaeuser. The table of the reference design
values for TJI joists was shown in Figure A-7.
Figure A-7: Reference design values for TJI joists

A-5
Example calculation for roof joist design
According to Forte, a 16” depth TJI®
110 joist was used
1) Check flexure
Mmax = 35.5 [k-in] = 2958 [lb-ft] (from Table A-1)
Mallow,old = 4280 [lb-ft] (from Figure A-7)
Note: load duration factor, CD need to be applied according to NDS
Mallow,new = 4280 [lb-ft] * CD where CD = 1.15 for snow load
= 4280 [lb-ft] * 1.15
= 4922 [lb-ft]
∴ Mallow,new > Mmax [OK]
2) Check shear
Vmax = 0.92 [k] = 920 [lb] (from Table A-2)
Vallow,old = 2145 [lb] (from Figure A-7)
Note: load duration factor, CD need to be applied according to NDS
Vallow,new = 2145 [lb] * CD where CD = 1.15 for snow load
= 2145 [lb] * 1.15
= 2467 [lb]
Vallow,new > Vmax [OK]

A-6
Floor joist
The worst case for floor joist was found in the unit Type 3J and it was analysed using Forte
as shown in figure A-8. Forte computed the floor joist size that worked under the input loads.
Figure A-8: Floor joist design using Forte
In addition, the bending member design procedure (NDS) was used to prove that the result
from Forte was valid. First, it was known that the controlling load combination for the floor
joist was D + L. The value for dead and live loads were shown in Table 2. According to
ASCE7-05, the patterned live load on the continuous member was analyzed using SAP2000
to determine the flexural and shear envelope. Figure A2-2 showed the load cases to be
considered for patterned live loads.
Figure A-9: Patterned live loads

A-7
The load analysis on patterned live loads using SAP2000 gave the flexural and shear
envelope shown in two figures below.
Figure A-10: Moment envelope for patterned live loads
Figure A-11: Shear envelope for patterned live loads

A-8
From the moment and shear envelope, the maximum value for each load case under patterned
live load was determined as shown in table A-3 and A-4.
Table A-3: Maximum moment of patterned live load
[k-in] M-ext R M+mid M-int L M-int R M+mid M-int L M-int R M+mid M-ext L
LC1 0 - -41.6 -41.6 32.7 -41.1 -41.1 - 0
LC2 0 - -18.9 -18.9 12.9 -19.8 -19.8 - 0
LC3 0 - -40.9 -40.9 33.9 -39.3 -39.3 - 0
LC4 0 - -42.5 -42.5 33.4 -38.7 -38.7 - 0
LC5 0 - -40.0 -40.0 33.2 -41.7 -41.7 - 0
max 0 - -42.5 -42.5 33.9 -41.7 -41.7 - 0
Table A-4: Maximum shear of patterned live load
[k] V-ext R V+mid V-int L V-int R V+mid V-int L V-int R V+mid V-ext L
LC1 0.04 - 0.81 -1.08 - 1.08 -0.81 - 0.09
LC2 -0.19 - 0.58 -0.47 - 0.48 -0.62 - 0.28
LC3 0.25 - 0.59 -1.09 - 1.08 -0.54 - -0.15
LC4 0.05 - 0.82 -1.09 - 1.07 -0.54 - -0.14
LC5 0.02 - 0.58 -1.08 - 1.09 -0.82 - 0.08
max 0.25 - 0.82 -1.09 - 1.09 -0.82 - -0.15
The data from above tables were compared to the reference design values for TJI joists
provided by the wood product company, Weyerhaeuser. The table of the reference design
values for TJI joists was shown in Figure A-7.
Example calculation for floor joist design
According to Forte, a 16” depth TJI®
110 joist was used
1) Check flexure
Mmax = 42.5 [k-in] = 3541.67 [lb-ft] (from Table A-3)
Mallow,old = 4280 [lb-ft] (from Figure A-7)
Mallow,new = 4280 [lb-ft] * CD where CD = 1.0 for live load
= 4280 [lb-ft] * 1.0
= 4280 [lb-ft] > Mmax [OK]
2) Check shear
Vmax = 1.09 [k] = 1090 [lb] (from Table A-4)
Vallow,old = 2145 [lb] (from Figure A-7)
Vallow,new = 2145 [lb] * CD where CD = 1.0 for live load
= 2145 [lb] * 1.0
= 2145 [lb] > Vmax [OK]

B-1
Appendix B –Wind Load
Pressures produced by 90 mph wind.
Figure B-1. North-south wind, Leeward pressure as a function of height
Figure B-2. North-south wind, Windward pressure as a function of height

B-2
Figure B-3. Combined pressure of windward and leeward due to north-south wind with
different internal pressure.
Figure B-4. East-west wind, Leeward pressure as a function of height

B-3
Figure B-5. East-west wind, Windward pressure as a function of height
Figure B-6. Combined pressure of windward and leeward due to east-west wind with
different internal pressure

C-1
Appendix C –Header Design
Header Calculation of Design
Extreme fiber stress in bending, Fb
Shear parallel to grain, Fv
Compression perpendicular to grain, Fc⊥
Modulus of elasticity, E
1. Determine tabulated design values of Douglas Fir-Larch by using NDS-S (Table 4A)
Fb = 1200 psi
Fv = 180 psi
Fc⊥ = 625 psi
E = 1800000 psi
Where, Extreme fiber stress in bending, Fb
Shear parallel to grain, Fv
Compression perpendicular to grain, Fc⊥
Modulus of elasticity, E
2. Determine lumber property adjustments
CT = 1.2
CD = 1. 25
CF = 1.2
CH = 2.0
Cb = 1.0
CL = 1.0
F 𝑏’ = F 𝑏C 𝐷C 𝑟 𝐶 𝐹C 𝐿 = 1200 psi ∗ 1.25 ∗ 1.2 ∗ 1.2 ∗ 1.0 = 2160 psi
F 𝑣’ = F 𝑣C 𝐷C 𝐻 = 180 psi ∗ 1.0 ∗ 2.0 = 360 psi
F 𝑐⊥’ = F 𝑐⊥C 𝑏 = 625 psi ∗ 1.0 = 625 psi
𝐸’ = 𝐸 = 1800000 psi
With two plies Fb can be increased by 5 percent
F 𝑏’ = 1.05 ∗ F 𝑏′ = 2160 psi ∗ 1.05 = 2268 psi
Where, CT, Buckling stiffness factor = 1.2
CD, Load Duration Factor

C-2
CF, Size Factor
CH, Horizontal Shear Factor
Cb, Bearing Area Factor
CL, Beam Stability Factor
3. Check bending for floor load with 2 X 6 of 8 feet
𝑀 𝑚𝑎𝑥 =
wl2
8
=
240 plf ∗ (8 𝑓𝑡)2
8
= 1920 lb ft
Solve for S and compare calculated S from tabulated value
f 𝑏 =
𝑀 𝑚𝑎𝑥
𝑆
=
1920 lb ft
𝑆
≤ 2268 𝑝𝑠𝑖
𝑆 =
1920 𝑙𝑏 𝑓𝑡 ∗
12 inch
1 𝑓𝑡
2268 𝑝𝑠𝑖
= 10.16 𝑖𝑛𝑐ℎ ≤ 2 ∗ 7.563
4. Check horizontal shear
𝑉𝑚𝑎𝑥 =
wl
2
=
240 plf ∗ 8 𝑓𝑡
2
= 960 lb
f 𝑣 =
3V
2𝐴
=
3 ∗ 960 lb
2 ∗ 8.250 𝑖𝑛2
= 174.55 𝑝𝑠𝑖 ≤ F 𝑣’ = 360 𝑝𝑠𝑖
5. Check for bearing
R1 = R2 = 𝑉𝑚 𝑎𝑥 = 960 lb
f 𝑐⊥ =
R
𝐴 𝑏
=
960 𝑙𝑏
𝑙 𝑏 ∗ 2 ∗ 1.5 𝑖𝑛
=
320
𝑙 𝑏
psi
f 𝑐⊥ = F 𝑐⊥’
320 psi
𝑙 𝑏
= 625 𝑝𝑠𝑖
𝑙 𝑏 = 1.95 𝑖𝑛
3 in bearing length is fine
6. Check for deflection
ρ 𝑚𝑎𝑥 =
5wl4
384𝐸𝐼
=
5 ∗ 240 𝑝𝑙𝑓 ∗ (8 𝑓𝑡)4
∗
1728 𝑖𝑛3
1 𝑓𝑡3
384 ∗ 1800000 𝑝𝑠𝑖 ∗ 20.80 𝑖𝑛4 ∗ 2
= 0.0029 𝑖𝑛

C-3
ρ 𝑎𝑙𝑙 =
𝐿
240
=
8 𝑓𝑡 ∗
12 𝑖𝑛
1 𝑓𝑡
240
= 0.325 𝑖𝑛
ρ 𝑚𝑎𝑥 < ρ 𝑎𝑙𝑙

D-1
Appendix D - Beam Design
Beam Design Lateral Stability
Lateral stability of 2 X 6 beam
The slenderness ratio
R 𝐵 = √
l 𝑒d
𝑏2
= √
132.1 𝑖𝑛 ∗ 6 in
22
= 14.07 ≤ 50
The slenderness ratio does not exceed 50.
Single span beam with uniformly distributed load
Effective span length on 5’ span
If,
𝑙 𝑢
𝑑
=
70"
6"
= 11.67 ≥ 7
So the effective span length should be calculated with following equation
𝑙 𝑒 = 1.63 ∗ 𝑙 𝑢 + 3d = 1.63 ∗ 70 + 3 * 6 = 132.1 inch
Where, le, the effective span length
lu, Laterally unsupported span length of bending member
d, depth of beam
𝐶𝐿 =
1 + (
𝐹𝑏𝐸
𝐹𝑏
" )
1.9
−
√[
1 + (
𝐹𝑏𝐸
𝐹𝑏
" )
1.9
]
2
−
𝐹𝑏𝐸
𝐹𝑏
"
0.95
≈ 0.95 ≤ 1
𝐹𝑏𝐸 =
𝐾𝑏𝐸E
𝑅 𝐵
2 =
0.438 ∗ 1800000
14.072
= 3978.8 𝑝𝑠𝑖
Where, F 𝑏
"
= F 𝑏C 𝐷C 𝑟 𝐶 𝐹C 𝐿 = 1200 psi ∗ 1.25 ∗ 1.2 ∗ 1.2 ∗ 1.0 = 2160 psi
FbE = KbEE / RB
2
KbE = 0.438

E-1
Appendix E – Bearing Wall Design
Bearing wall
Table E-1: Calculations and checks for bearing wall design except Level 6.
Check design
values
Level 5th 4th 3rd 2nd
b width(depth) inche
s
5.5 5.5 5.5 5.5
t thickness inche
s
1.5 1.5 1.5 1.5
Levels above 1 2 3 4
number of plies 2 2 3 3
s joist spacing in 16 16 16 16
s joist spacing ft 1.34 1.34 1.34 1.34
Dead Load line load on floor
joist
plf 41.34 41.34 41.34 41.34
Snow Load line load on floor
joist
plf 0 0 0 0
Live Load line load on floor
joist
plf 53.34 53.34 53.34 53.34
q_D+L plf 94.67 94.67 94.67 94.67
R_z lb 2450.68 2450.687 2450.68 2450.68
R_zmax max vert. reaction lb 4560.41 7011.09 9461.78 11912.46
A cross-sectional area in^2 16.5 16.5 24.75 24.75
f_c actual comp psi 276.39 424.91 382.29 481.31
F_c' psi 639.16 639.16 639.16 639.16
F_c' > f_c ? 1 = OK, 0 = NG 1 1 1 1
Bending and
comp.
3.9.2
l_e effective length in 120 120 120 120

E-2
ft 10 10 10 10
l_e1/d_1 slenderness ratio 21.82 21.82 21.82 21.82
F_cE1 pg. 19 psi 1175.54 1175.54 1175.54 1175.54
F_cE1 > f_c
?
1 = OK, 0 = NG 1 1 1 1
R_B Eq. 3.3-5 8.56 8.56 5.71 5.71
F_bE for biaxial bending psi 11139.9
3
11139.93 25064.84 25064.84
e eccentricity in 1.38 1.38 1.38 1.38
M M due to vert load lb-in 3369.69 3369.69 3369.69 3369.69
y in 2.75 2.75 2.75 2.75
I M of Inertia in 41.6 41.6 62.4 62.4
f_b1 =M*y/I psi 222.79 222.79 148.53 148.53
F_bE > f_b1 1 = OK, 0 = NG 1 1 1 1
p_lat lateral pressure psf 5 5 5 5
w_lat lateral line load plf 6.67 6.67 6.67 6.67
M_mid =wL^2/8 lb-ft 83.33 83.33 83.33 83.33
< end M won't control lb-in 1000 1000 1000 1000
C_L 0.98957
6
0.989576
317
0.9957981
22
0.995798122
F_b1' =F_b for strong axis psi 1053.52 1053.52 1060.14 1060.14
(f_c/F_c')^2 0.19 0.44 0.36 0.57
f_b1/(F_b1'*(1-(f_c/F_cE1))) 0.28 0.33 0.21 0.24
the third term
= 0
0 0 0 0
sum of terms 0.46 0.78 0.57 0.80
Pass? 1 = OK, 0 = NG 1 1 1 1

F-1
Appendix F – Shear Wall Design
Shear wall
Figure F-1: Shear wall selected for West Wing, Level 2.

G-1
Appendix G - Diaphragm Design
Diaphragm Calculation
Since, 2nd to 5th floor has height of 10.5 feet and is the tallest, 11 feet is used (conservative)
for calculation. Also wind load north-south is 20.8 [𝑙𝑏/𝑓𝑡2
] at the top most therefore, it is
used to to design diaphragms.
Equation 14:
𝑤 = 𝑊 ∗ 𝐻 = 28.0
𝑙𝑏
𝑓𝑡2
∗ 11 𝑓𝑡 = 228. 8𝑙𝑏/𝑓𝑡
Figure G-1. Six spans with uniformly distributed load of 228.8 lb/ft2
in SAP2000.
Figure G-2. Maximum shear determination using SAP2000.
Maximum shear force, V determined using Sap2000 to be 7992.66 lb.
Equation 15:
𝑣 =
𝑉
𝐿
= 7992.66 𝑙𝑏/ 189 𝑓𝑡 = 42.29 𝑙𝑏/𝑓𝑡
The design is picked using table 2306.3.1 in IBC-2006 chapter 16.

H-1
Appendix H – Precast Concrete Plank Design
An extruded hollow core plank was used for the slab construction .The concrete plank
compressive strength, f’c was 9000[psi].
Figure H-1: Cross section of an extruded hollow core plank
First, the plank direction was determined using the column line provided in the architectural
drawing.
A B
C
I J
D
E
Empty Space
F
G
H
Figure H-2: Plank layout
Given wall load, plank self-weight, wood partition weight and live load, the loads acting on
the plank can be determined. In addition to the pre-existed uniform dead load (i.e. wood
partition weight and plank self-weight) and live load, the wall load was acting as a uniform
load or point load depending on the direction of the wall. The wall load is determined when
analyzing the wood studs. The Table H-1 showed the numerical values of loads acting on a
plank.
Legend
Plank Direction

H-2
Table H-1: Loads acting on a plank
Type of Loads Value Unit
Wall load 7800 plf
Plank self-weight (8” x 48”) 248 plf
Wood partition weight 5 plf
Live Load 40 plf
An example calculation of loads acting on a plank in region A is shown below and the result
is shown in Figure F-3.
Assuming a 12” x 48” plank is used and load obtained from Table F-1,
Uniform Dead Load = Wood partition weight + Plank self-weight
= 5 [plf] + 320 [plf] = 325[plf]
Uniform Live Load = 40 [plf]
However, it was specified by the concrete products supplier that the uniform loading was
required to be modified when superimposed load consists of both dead and live loads. So,
Modified Uniform Load =
1.2
1.6
D + L =
1.2
1.6
*(325 [plf]) + 45 [plf] = 288.75 [plf] = 0.289 [klf]
Figure H-3: Loads acting on a plank in region A
Based on the load analysis, the moment diagram were determined and shown in Figure F-4.

H-3
Figure H-4: Bending moment diagram of a plank in region A
From Figure F-4, it is known that the maximum moment demand (Mu) was 615 [k-ft]. Based
on the LRFD design method, it was required for the nominal moment capacity (Mn) to
exceed the moment demand (Mu). Therefore, the size of the hollow core plank is determined
based on the safe load table provided by the supplier.
Figure H-5: Safe load table for a 12” x 48” hollow core plank
Since the given nominal moment capacity (Mn) in figure H-5 was per unit, it was required to
adjust the value of Mn to per foot because the load analysis was done based on per foot. An
example calculation of adjusted moment capacity assuming a 12” x 48” plank with strands
no. and size of 6-1/2 (third row in figure H-5) is used is shown below.
Mn = 173 [k-ft per unit]
= 692 [k-ft per foot] because there was 4 feet in a unit of plank

H-4
The LRFD requirement of Mn > Mu was satisfied since Mn = 692 [k-ft] > Mu = 615 [k-ft].
In results, a 12” x 48” plank with strands no. and size of 6-1/2 was used for the plank in
region A.
The same design procedure was done for the planks in other regions; they were region B, C,
D, E, F, G, H, I and J (see Figure H-2). Figure H-6 showed the free body and bending
moment diagrams of the planks for each region.
Region B Region C Region D
Region E Region F Region G
Region H Region I Region J

H-5
Figure H -6: Free body and bending moment diagrams of a plank in different region
Beside the 12” x 48” plank, there were two other sizes available too; they were 8” x 48” and
16” x 48” sizes. The safe load tables for both these sizes were shown in figure H-7 and figure
H-8.

I-1
Appendix I – Concrete Beam Design
Precast Beam Design
A double ledge beam (inverted T-beam) was used to support the plank (see Figure G-1). The
concrete beam compressive strength, f’c was 9000[psi].
Figure I-1: Cross section of a double ledge beam
First, the beam direction was determined to be the perpendicular to the plank direction (see
Figure I-2).
A B
C
I J
D
E
Empty Space
F
G
H
Figure I-2: Beam layout
Given wall load, plank self-weight, beam-self weight, wood partition weight, and live load,
the loads acting on a beam can be determined. In addition to the pre-existed uniform dead
load (i.e. wood partition weight, plank self-weight, and beam self-weight) and live load, the
wall load was acting as a uniform load or point load depending on the direction of the wall.
The wall load is determined when analyzing the wood studs. The Table I-1 showed the
numerical values of loads acting on a beam.
Legend
Plank Direction
Beam Direction

I-2
Table I-1: Loads acting on a beam
Type of Loads Value Unit
Wall load 7800 plf
Plank self-weight (8” x 48”) 62 psf
Beam self-weight (24IT24) 590.3 plf
Wood partition weight 5 plf
Live Load 40 plf
An example calculation of loads acting on a beam, for example, beam 1 is shown below and
the result is shown in Figure I-3.
Figure I-3: Dimension of region A
Assuming a 24IT24 beam is used and knowing that a 12” x 48” plank is used for region A,
Tributary width = l1/2 = 28’-3”/2 = 14.125 [ft]
Tributary area = l1/2 × L1 = 14.125 [ft] × 27.25 [ft] = 384.8 [ft2
]
Uniform Dead Load = [(Wood partition weight [psf] × Tributary width [ft]
+ (Plank self-weight [psf] × Tributary width [ft])
+ (Beam self-weight [plf])]
= (5 [psf] × 14.125 [ft]) + (80 [psf] × 14.125 [ft]) + 590.3 [plf]
= 1790.925 [plf] = 1.79 [klf]
Know that KLL is 2 for edge beams without cantilever slabs and AT is 384.8 [ft2
],
KLLAT = 2 × 384.8 [ft2
] = 769.6 [ft2
] > 400[ft2
]
So, the uniformly distributed live load was reduced using Equation ##:
A
l1
L1
Dimension
L1 = 27’-3”
l1 = 28’-3”

I-3
𝐿 = 𝐿 𝑜 (0.25 +
15
√𝐾 𝐿𝐿 𝐴 𝑇
) = 40[psf] * (0.25 +
15
√769.6
) = 31.6 [psf] > 0.4*40 [psf] = 16[psf]
Therefore, the uniform live load for beam 1 was calculated as follows:
Uniform Live Load = Reduced live load [psf] × Tributary width [ft]
= 31.6 [psf] × 14.125 [ft] = 446.35 [plf] = 0.45[klf]
However, it was specified by the concrete products supplier that the uniform loading was
required to be modified when superimposed load consists of both dead and live loads. So,
Modified Uniform Load =
1.2
1.6
D + L =
1.2
1.6
(1.79 [klf]) + 0.45 [klf] = 1.79 [klf]
In addition to the uniform load, there was an 8 [ft] long wall located perpendicularly to the
beam, thus acted as a point load at the distance of 11 [ft] from the left end of the beam (see
Figure I-4). The point load was determined as follows:
Wall load [k] = Wall load [plf] × Wall length [ft] = 7800 [plf] × 8 [ft] = 62.4 [k]
Figure I-4: Loads acting on beam 1

I-4
Based on the load analysis, the moment diagram were determined and shown in figure I-5.
Figure I-5: Bending moment diagram of beam 1
From figure I-5, it is known that the maximum moment demand (Mu) was 569 [k-ft]. Based
on the LRFD design method, it was required for the nominal moment capacity (Mn) to
exceed the moment demand (Mu). Therefore, the size of the inverted T-beam was determined
based on the safe load table provided by the supplier (see figure I-6).
Figure I-6: Safe load table for a 24IT24 beam

I-5
The nominal moment capacity (Mn) was provided in the table in figure I-6. To ensure that
the LRFD requirement of Mn > Mu is satisfied, a 24IT24 beam with 12 strands was chosen
because its Mn = 623.45 [k-ft] was greater than Mu = 569 [k-ft].
The same design procedure was done for other beams. Figure I-7 showed the calculated loads
acting on the beams and Figure I-8 showed the free body and bending moment diagrams of
different beams.
Table I-2. Calculated loads acting on different beams
Beam
#
Beam
length
[ft]
Wall
length
into
the
page
[ft]
Tributary
width [ft]
Tributary
area [ft2
]
KLLAT
[ft2
]
Reduced
live load
[psf]
Modified
uniform
load [klf]
Wall
point
load [k]
1 27.25 8 14.125 384.8 769.6 31.6 1.79 62.4
2 27.25 8 27.625 752.78 1505.56 25.46 2.72 62.4
3 26.33 10 13.5 355.46 710.91 32.50 1.65 78.0
4 27 9 9 243.00 486.00 37.22 1.29 70.2
5 26 4 22.65 588.90 1177.80 27.48 2.36 31.2
6 27 8 18.07 487.89 975.78 29.21 2.00 62.4
7 27 4.42 4.42 119.34 238.68 48.84 0.91 34.5
8 27 8.13 8.125 219.38 438.75 38.64 1.22 63.4
9 27 10 22.46 606.42 1212.84 27.23 2.33 78.0
10 27 7 19.83 535.41 1070.82 28.34 2.13 54.6
11 27 5.5 5.5 148.50 297.00 44.82 1.00 42.9
12 26 10 13.5 351.00 702.00 32.65 1.65 78.0
13 27.25 8 27.67 754.01 1508.02 25.45 2.72 62.4
14 30.17 14 14.17 427.51 855.02 30.52 1.68 109.2

I-6
Beam # Free Body Diagram Bending Moment Diagram
1
2
3
4
5

I-8
11
12
13
14
Figure I-8: Free body and bending moment diagrams of different beams
Beside the 24IT24 beam, there were three other sizes available too; they were 24IT28,
24IT32, and 24IT36 sizes. The safe load tables for these three sizes were shown in Figure I-9,
Figure I-10, and Figure I-11.

I-9

J-1
Appendix J
Precast column design
A rectangular reinforced concrete column was used to support the beam. The concrete plank
compressive strength, f’c was 9000[psi] and the yield strength of the rebars, fy was
60000[psi].
Figure J-1: Cross section of a rectangular reinforced concrete column
First, the location of each column was specified in the architectural drawing (see figure J-2).
The arrow pointed at the location of a column.

J-2
Figure J-2: Example of column location in the floor plan
To analyze and design the column, a simplified column layout was created. Since the column
was supporting the beam, so the beam reaction method was used. Therefore, the column
analysis was done using the beam direction.
A B
C
I J
D
E
Empty Space
F
G
H
Figure J-3: Column frame layout
An example calculation was done for the column design for frame no. 1. Figure J-4 showed
the chosen frame (part of the frame no. 1) for exterior column design while figure J-5 showed
the chosen frame (part of the frame no.1) for interior column design. The design for these two
parts of frame was applied to the other part of frame no. 1.
Figure J-4: Part of frame no.1 for exterior column design
Legend
Plank Direction
Beam Direction
Frame Number

J-3
Figure J-5: Part of frame no.1 for interior column design
The reactions at the end supports were known because the load analysis has been done when
designing the beam. Figure J-6 and figure J-7 showed the loads acting on each part of frame.
Figure J-6: Loads acting on the part of frame no.1 for exterior column design

J-4
Figure J-7: Loads acting on the part of frame no.1 for interior column design
The NS wind load [k] was determined as follows assuming the tributary width is 25[ft].
NS wind load [k] = NS wind pressure at 12ft height [psf] ×height [ft] × tributary width [ft]
= 15.4 [psf] × 12 [ft] × 25 [ft]
= 4.62 [k]
Exterior column
It is assumed that a 12” by 12” column and a 2” by 1/2” thick bearing pad were used. Thus,
the eccentricity length was estimated as follows:
e = bcol/2 – distance from pad to edge of column – ½*bpad = 12”/2 – ½” – ½*(2”) = 4.5”
Mu, gravity = Pgravity * e = 16.11[k] * 4.5[in] = 72.5[k-in] = 6 [k-ft] (anti-clockwise)
Mu, wind = NS wind load [k] * height [ft] = 4.62[k] * 12 [ft] = 55.44 [k-ft] (anti-clockwise)
Pwind = Mu, wind [k-ft]/ e [in] = 147.84 [k]
Two load combinations to be considered:
1. Gravity loads, U = 1.2D + 1.6L
=> Pu = 16.11[k] and Mu = 6 [k-ft]
2. Gravity plus wind loads, U = 1.2D + 1.6W + 1.0L
=> Pu = 249.2 [k] and Mu = 93.42[k-ft]
The results of factored load for exterior column were plotted on a 12” x 12” interaction
diagram provided by the supplier (see Figure H-8).

J-5
Figure J-8: Comparison of column strength and factored loads for exterior column
Interior column
It is assumed that a 12” by 12” column and a 2” by 1/2” thick bearing pad were used. Thus,
the eccentricity length was estimated as follows:
eleft = bcol/2 – distance from pad to edge of column – ½*bpad = 12”/2 – ½” – ½*(2”) = 4.5”
eright = bcol/2 – distance from pad to edge of column – ½*bpad = 12”/2 – ½” – ½*(2”) = 4.5”
Pgravity = Rleft + Rright = 23.27 [k] + 66.37 [k] = 89.64 [k]
Mleft = Rleft * eleft = 23.27 [k] * 4.5 [in] = 104.7 [k-in] = 8.7 [k-ft]
Mright = Rright * eright = 66.37 [k] * 4.5 [in] = 298.7 [k-in] = 24.9 [k-ft]
Mu, gravity = Mright – Mleft = 24.9 – 8.7 = 16.2 [k-ft] (clockwise)
Mu, wind = NS wind load [k] * height [ft] = 4.62[k] * 12 [ft] = 55.44 [k-ft] (anti-clockwise)
Pwind = Mu, wind [k-ft]/ e [in] = 147.84 [k]
Two load combinations to be considered:
1. Gravity loads, U = 1.2D + 1.6L
=> Pu = 89.64 [k] and Mu = 16.2 [k-ft]
2. Gravity plus wind loads, U = 1.2D + 1.6W + 1.0L
=> Pu = 259.83 [k] and Mu = 53.1 [k-ft]

J-6
The results of factored load for interior column were plotted on a 12” x 12” interaction
diagram provided by the supplier (see Figure J-9).
Figure J-9: Comparison of column strength and factored loads for interior column
Beside the 12 x 12 column, there were four other sizes available too; they were 12 x 16, 16
x16, 20 x 20, and 24 x 24 sizes. The interaction diagrams for these sizes were shown in figure
J-10.

J-7
Figure J-10: Interaction diagrams for different column size

K-1
Appendix K – Spread Footing Design
Design Summary
The section details of wall footing and column footing are shown here.
Figure K-1: detail of wall footing (section view).

K-2
Figure K-2: detail of column footing (section view).
Calculations
All calculations were done in Excel spreadsheets.
Table K-1: Calculations and checks for wall footing.
Wall Footing
Immediate Loads South North East West
D column weight kips 9.86 9.86 9.86 9.86
D due to service DL klf 4.38 4.41 6.12 6.12
L due to service LL klf 2.21 2.24 3.74 3.74
S due to service SL klf 0.25 0.25 0.43 0.43
W due to wind load klf 0.26 0.26 0.15 0.15
S or W max(S,W) klf 0.26 0.26 0.43 0.43
Sizing the footing
size 1 D/3ksf ft 1.46 1.47 2.04 2.04
size 2 (D+L+S or W)/4ksf ft 1.71 1.73 2.57 2.57
controlling size ft 1.71 1.73 2.57 2.57
b_min required footing width ft 1.71 1.73 2.57 2.57
B design footing width in 36 36 36 36
B design footing width ft 3 3 3 3
A_design
sq
ft 3 3 3 3
A_design >? A_req 1 = OK, 0 = NG 1 1 1 1

K-3
Required strength
U1 1.4D klf 6.13 6.18 8.57 8.57
U2 1.2D + 1.6L + 0.5(S or W) klf 8.92 9.01 13.55 13.55
U controls klf 8.92 9.01 13.55 13.55
U ksf 2.97 3.00 4.52 4.52
Design for shear one-way action
h footing depth in 16 16 16 16
t_wall wall thickness (assumed) ft 1 1 1 1
b for foot-long wall section in 12 12 12 12
d shear depth in 12.19 12.19 12.06 12.06
V_c 2*sqrt(f'c)b_o*d klf 18.50 18.50 18.31 18.31
phi*V_c klf 13.87 13.87 13.73 13.73
V_u klf -0.05 -0.05 -0.02 -0.02
phi*V_c >? V_u 1 = OK, 0 = NG 1 1 1 1
Design for bearing
A1 wall area
sq
in 144 144 144 144
A2 developed area
sq
in 528 528 528 528
sqrt(A2/A1) 1.91 1.91 1.91 1.91
true sqrt(A2/A1) 1.91 1.91 1.91 1.91
B_r phi*.85*f'c*A1*sqrt(A2/A1) klf
609.3
8
609.3
8
609.3
8
609.3
8
B_u =U klf 8.92 9.01 13.55 13.55
B_r >? B_u 1 = OK, 0 = NG 1 1 1 1
bar size bottom layer #4 #4 #5 #5
d_b,bot rebar diameter bottom layer in 0.5 0.5 0.625 0.625
A_b bar area
sq
in 0.2 0.2 0.31 0.31
bar size upper layer #5 #5 #5 #5
d_b,up rebar diameter upper layer in 0.625 0.625 0.625 0.625
A_b bar area
sq
in 0.31 0.31 0.31 0.31
Design for flexure
b for foot-long wall section in 12 12 12 12
d depth of tension bars in 12.75 12.75
12.68
75
12.68
75
b_wing one side width of the footing ft 1 1 1 1
M_u moment at wall face
k-
ft/ft 1.49 1.50 2.26 2.26

K-4
M_n req =M_u/phi
k-
ft/ft 1.65 1.67 2.51 2.51
R_n req M_n req/(bd^2) ksi 0.01 0.01 0.02 0.02
rho 1/m*(1-sqrt(1-2mR_n/f_y))
0.000
170
0.000
171
0.000
260
0.000
260
A_s req R/F area = rho*b*d
sq
in 0.03 0.03 0.04 0.04
A_s min .0018*b*h
sq
in 0.35 0.35 0.35 0.35
A_s req
sq
in 0.35 0.35 0.35 0.35
s_max bar spacing in 6.94 6.94 10.76 10.76
s design tension bar spacing in 6.5 6 10 10
A_s for foot-long wall section
sq
in 0.37 0.40 0.37 0.37
A_s >? A_s min 1 = OK, 0 = NG 1 1 1 1
Check if tension
controlled
rho design
0.002
41
0.002
61
0.002
44
0.002
44
a A_s*f_y/(.85*f'c*b) in 0.54 0.59 0.55 0.55
x a/beta1 in 0.64 0.69 0.64 0.64
ep_t strain in bar 0.06 0.05 0.06 0.06
ep_t >? 0.005 1 = OK, 0 = NG 1 1 1 1
a/d_t 0.039 0.043 0.040 0.040
.375beta1 0.319 0.319 0.319 0.319
a/d_t <?
.375beta1 1 = OK, 0 = NG 1 1 1 1
Shrinkage &
Temperature
n S&T number of bars for S&T 4 4 4 4
A_s S&T reinforcement for S&T
sq
in 1.24 1.24 1.24 1.24
A_s/Bh
0.002
153
0.002
153
0.002
153
0.002
153
A_s/bh >? .0018 1 = OK, 0 = NG 1 1 1 1
s tension bar spacing in 6.5 6 10 10
s_max min(3h, 18") in 18 18 18 18
s <? s_max 1 = OK, 0 = NG 1 1 1 1
s_req S&T required bar spacing for S&T in 9.458 9.458 9.375 9.375
s S&T bar spacing for S&T in 9 9 9 9
s <? s_req S&T 1 = OK, 0 = NG 1 1 1 1
s_max S&T min(5h, 18") in 18 18 18 18
s <? s_max S&T 1 = OK, 0 = NG 1 1 1 1

K-5
Development
length
l_d
(fy*Phit*Phie/(25la*sqrt(f'c)))*d_
b in 18.97 18.97 23.72 23.72
l_d ft 1.58 1.58 1.98 1.98
l_provided provided length in 9 9 9 9
l_d <? L_prov 1 = OK, 0 = NG 0 0 0 0
hooked bars tension bars are hooked to minimize the development length
l_dh cal
(0.02Phi_e*f_y/(la*sqrt(f'c)))*d_
b in 9.49 9.49 11.86 11.86
multiplier
side cov. > 2.5 in., 90-deg hook,
hook cov > 2 in. 0.7 0.7 0.7 0.7
l_dh req l_dh cal*multiplier in 6.64 6.64 8.30 8.30
l_dh min max(8d_b, 6) in 6 6 6 6
l_dh max(l_dh req, l_dh min) in 6.64 6.64 8.30 8.30
l_d <? L_prov 1 = OK, 0 = NG 1 1 1 1
l_h hook length in 6 6 7.5 7.5
c_h
hook cover = d + d_b/2 - 4d_b -
l_h in 5 5 3 3
c_h >? 2 in 1 = OK, 0 = NG 1 1 1 1
Table K-2: Calculations and checks for column footing.
Column Footing
Immediate Loads Critical
D column weight kips 9.86
D due to service DL kips 368.52
L due to service LL kips 266.20
S due to service SL kips 29.32
Sizing the footing
size 1 D/3ksf ft 122.84
size 2 (D+L+S)/4ksf sq ft 166.01
controlling size sq ft 166.01
b_min required footing width ft 12.88
b design footing width in 156
b design footing width ft 13
A_design sq ft 169
A_design >? A_req 1 = OK, 0 = NG 1
Required strength
U1 1.4D kips 515.93
U2 1.2D + 1.6L + 0.5S kips 882.80

K-6
U controls kips 882.80
U ksf 5.22
Design for shear two-way action
h footing depth in 24
d_c square column width in 20
d shear depth in 19.125
b_o 4(d_c+d_c) in 160
V_c (2+4/beta)sqrt(f'c)b_o*d kips 1161.2
(alpha_s*d/b_o)sqrt(f'c)b_o*d kips 1354.7
4*sqrt(f'c)b_o*d kips 774.1
V_c control kips 774.1
phi*V_c kips 580.6
V_u kips 824.8
phi*V_c >? V_u 1 = OK, 0 = NG 0
bar size #8
d_b1 rebar diameter bottom layer in 1
rebar diameter upper layer in 1
A_b bar area
sq
in 0.79
one-way action
b_w footing width in 156
d shear depth in 19.125
V_c 2*sqrt(f'c)b_w*d kips 377.4
phi*V_c kips 283.0
V_u kips 276.6
phi*V_c >? V_u 1 = OK, 0 = NG 1
Design vertical stirrups for shear
reinforcement
size of stirrups #3
d_vb stirrup diamter 0.375
A_vb
sq
in 0.11
n_v number of stirrups crossing crack 8
A_v total stirrup cross area
sq
in 0.88
V_s req 2 V_u/phi - V_c kips 325.6
s_req 2 spacing of stirrups in 3.1
s_design 2 designed spacing in 3
V_s 2 designed kips 336.6
V_n 2 V_c + V_s kips 1110.7
phi*V_n >? V_u 1 = OK, 0 = NG 1
V_s req 1 V_u/phi - V_c kips -8.6
s_req 1 spacing of stirrups in -29.3

K-7
s_design 1 in 6
V_s 1 designed kips 42.1
V_n 1 V_c + V_s kips 419.5
phi*V_n >? V_u 1 = OK, 0 = NG 1
Check stirrup spacing
s_max min(d/2, 24") in 9.6
s <? s_max 1 = OK, 0 = NG 1
Design for bearing
A1 column area
sq
in 400
A2 developed area
sq
in 4624
sqrt(A2/A1) 3.4
true sqrt(A2/A1) 2
B_r phi*.85*f'c*A1*sqrt(A2/A1) kips 1768
B_u =U kips 882.8
B_r >? B_u 1 = OK, 0 = NG 1
Design for flexure
b in 156
d for upper layer (conserv) in 19.125
b_wing one side width of the footing ft 5.67
M_u moment at column face k-ft 1090.3
M_n req =M_u/phi k-ft 1211.4
R_n req M_n req/(bd^2) ksi 0.25478
rho 1/m*(1-sqrt(1-2mR_n/f_y)) 0.004419
A_s req R/F area = rho*b*d
sq
in 13.18
A_s min .0018*b*h
sq
in 6.74
A_s >? A_s min 1 = OK, 0 = NG 1
n required number of bars in one dir
16.68692
1
n actual number of bars in one dir 17
Check if tension controlled
A_s design
sq
in 13.43
rho design 0.004501
a A_s*f_y/(.85*f'c*b) in 1.519
x a/beta1 in 1.787
ep_t strain in bar 0.0291
ep_t >? 0.005 1 = OK, 0 = NG 1
a/d_t 0.07549
.375beta1 0.3188

K-8
a/d_t <? .375beta1 1 = OK, 0 = NG 1
Shrinkage & Temperature
A_s/bh 0.003587
A_s/bh >? .0018 1 = OK, 0 = NG 1
s_req bar spacing required by room in 9.75
s in 9
s_max min(3h, 18") in 18
s <? s_req 1 = OK, 0 = NG 1
s <? s_max 1 = OK, 0 = NG 1
s_max S&T min(5h, 18") in 18
s <? s_max S&T 1 = OK, 0 = NG 1
Development length
l_d (fy*Phit*Phie/(20la*sqrt(f'c)))*d_b in 47.43
l_d ft 3.95
l_provided provided length in 65
l_d <? L_prov 1 = OK, 0 = NG 1

L-1
Appendix L - Budget
Table L-1: Cost of structural design
Team
member
Hourly
billing
rate
[$]
Baek, Seung 70.00
Bui, Evan 70.00
Yap, Jin 70.00
Zhong, Sichen 70.00
Project Task
Projected
time
expenditure
[hr]
Projected
Cost [$]
Responsible
team
member(s)
Actual time
expenditure
[hr]
Actual
cost
PDW 6 1,680.00 All 4 4 1,120.00
Meet with Mentor(s) 44 12,320.00 All 4 64 17,920.00
Biweekly Project Reports 13 3,640.00 All 4 17 4,760.00
Report 1st Draft 40 11,200.00 All 4 24.5 6,860.00
Report 2nd Draft 28 7,840.00 All 4 27 7,560.00
Final Report 16 19,600.00 All 4 39 2,730.00
Midterm Presentation 12 3,360.00 All 4 27 1,890.00
Final Presentation 20 5,600.00 All 4 37 2,590.00
Task #1: Gravity Loads,
Lateral Loads, and Select
Wood Species and Grade
56 15,680.00 All 4 41.5 11,620.00
Task #2: 5 levels of wood
framing (floors, roofs,
and walls)
128 35,840.00 All 4 143.5 40,180.00
Task #3: 1st floor precast
concrete (beams, planks,
and columns)
64 17,920.00 Jin Yap 25.5 7,140.00
Task #4: Conventional
Foundation
64 17,920.00 Sichen Zhong 20 1,400.00
Total 491 152,600.00 470 105,770.00

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