A room tunnel form reinforced concrete
alternative to a steel framed residence
hall.
Drexel University Race Street Dormitory
Doug Tower
Structural Thesis
Presentation Content
1. Drexel’s Initial Requirements
2. Drexel’s Chosen Architectural Elements
3. Existing Structure
4. Existing Structural Plans
5. Redesign Goals
6. Slab System
7. Lateral Analysis
8. New Architecture
9. Construction Comparison
10. Advantages & Disadvantages of Proposal
Drexel’s Initial Requirements
• 10 stories of suites
• 11 to 13 suites per floor, 4 students per
suite
• Other typical residence hall elements
• Quickly assembled Design-Build project
Drexel’s chosen architectural
elements.
• Partial Ground Floor - electrical, bike
storage, trash compactor, etc.
• Elevated grade and extended First
Floor -major spaces including lobby,
security, mail, multipurpose room, laundry,
etc.
• 10 stories of suites above –
12 suites per floor
Existing Structure
• Steel Frame
• Brace and Moment Frames
• 2” topped 8” deep hollow core plank
flooring
• 2 individual lateral resisting structures
• First Floor height 10ft from lower grade
• Second Floor height 14ft from First
• Consecutive Floor to Floor heights, 9’4”
Existing Structure: The first floor is abutted
against a higher grade slab.
Slab
Existing Structure: Typical floor framing plan.
Each wing has an independent lateral system.
Is a tunnel form design a competitive
alternative to the existing design?
Primary Design Goals
• Low Cost
• Fast Construction
• Maintain architectural elements
• Low interference in dorm room space
Secondary Goals
• Effectively utilize the benefits of concrete
• Explore the tunnel form system through its
application
With tunnel forms, concrete is poured
everyday and forms are reused the next.
• Pre built, angled forms two per bay
• Wheels and braces
• High early strength
concrete is used
(>1000 psi after 24hr)
• Pole braces are
installed after form
is removed.
Tunnel form construction has other
advantages.
• Dimensional accuracy
• Improved fire rating for lower insurance
• Strong insulating
properties
• Acoustical isolation
• Durability
Preliminary Design
Preliminary Design
Preliminary Design
Preliminary Design
Problem! Little Longitudinal
Stiffness in wings, immense
transverse stiffness!
Preliminary Design
Problem! Little Longitudinal
Stiffness in wings, immense
transverse stiffness!
Fracture!
Typical floor structural redesign with an isolation joint.
Note the longitudinal walls. Walls are based around 5’
nominal form
width.
•All walls are 7 in thick to
carry required
reinforcement and fire
rating.
The first floor structural redesign. Note the large
number of beams and 5’ wall sections to support
those beams in order to open up spaces.
Ground floor structural redesign
The slab system
• Slab thickness controlled by deflection of
cantilever to isolation joint.
• Reinforcement patterns are for both one
way and two way action.
• ADOSS used for preliminary calculations.
• RAM concept confirm slab thickness and
reinforcement
Lateral Analysis
Essentially all walls are designed
as shear walls.
Sample Shear Wall Calculation
Sample Shear Wall Calculation
Load Combination= 0.9D+1.6W
Tributary
Width (in) =
Wall Line 1a, 1b
Height (in)
Thickness
1248
Length
7.00
60
A
526.08
No. Stories
Slab=
11
I
3682.56
Pu (lb)
84932159.86
H/L
Assume
d=0.8L
346.95
2.37
>?
0.2f'c (psi)
800.00
Compression
Mu/z+Pu (lb)
768076.165
537802.07
alpha
Assume z=0.6L
c=
350.72
Pu/A+Muy/I
>?
371.15
Tension Mu/z
(lb)
230274.0925
Pu/A (psi)
for Design
Displacement
H/400
Pu/A-Muy/I
-75.39
MuY/I (psi)
146.04
Vu=
58.24
2
No
Boundary
Elements
225.11
0.75 Vc/2 (k)
Vu<
Min Shear
Rein
Vu<
139.74
Acr*Sqrt(f'c)
232.92
Sample Shear Wall Calculation
Load Combination= 1.2D+1.6W+L+0.5S
Tributary Width
(in) =
Wall Line 1a, 1b
Height (in)
Thickness
1248
Length
7.00
60
A
526.08
No. Stories Slab=
I
3682.56
84932159.86
410.87
2.37
>?
0.2f'c (psi)
350.72
Pu/A+Muy/I
800.00
Compression
Mu/z+Pu (lb)
1213376.975
789625.79
alpha
Assume z=0.6L
c=
>?
439.53
Tension Mu/z (lb)
407861.7971
Pu/A
(psi)
Pu (lb)
H/L
Assume
d=0.8L
11
for Design
Displacement
H/400
MuY/I (psi)
214.42
Vu=
98.87
2
Boundary
Elements
(c>)
0.75 Vc/2 (k)
Vu<
Min
Shear
Rein
Pu/A-Muy/I
-10.69
225.11
Vu<
139.74
Acr*Sqrt(f'c)
232.92
Sample Shear Wall Reinforcement
calculations.
Horizontal Steel
min 0.0025
Vertical Steel
rho
horizontal
0.0025
rho vertical
0.0025
vertical
section
8736
horizontal
section
3682.56
As
Req.
Max
Spacing
21.84
As
Req.
18
Max
Spacing
9.206
4
Tensi
on
Steel
Spacing
16
3.83790
1542
79
18
# perp
rows
70
# long
rows
4
0.28
Area
Req./Bar
Spacing
18
Area
Area
Req./Bar
No. Bars
0.13
Bar
2
#9
Bar
Area/Bar
#5
0.31
Bar
Area/Bar
#4
0.2
Area
(in^2)
4
As
24.49
Actual rho
0.002803
As
14.07
0.003820
Shear Wall
Extra shear reinforcement
Questions?