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Technical Manual for Design and Construction of Road Tunnels - Civil Elements
Appendix C - Cut-and-Cover Tunnel Design Example
The purpose of this design example is to provide guidance to the application of the AASHTO LRFD Bridge Design Specifications when designing concrete cut and cover box tunnel structures.
Reference is made to the AASHTO LRFD specifications throughout the design example. Specific references to sections are denoted by the letter "S" preceeding the specification article.
1. Tunnel Section Geometry and Materials
The tunnel is a reinforced concrete double-chamber box structure. It is located entirely below grade and is built using cut and cover construction. Because the water table is located above the tunnel, hydrostatic soil pressures surround the structure. Figure 1 shows the internal dimensions for one of the openings. These dimensions serve as the starting point for the structural dimensions shown in Figure 2.
1.1 Tunnel Section Dimensions
Figure 2 shows the geometry of the underground cut and cover box cross-section.
1.2 Material Properties
2. Computer Model of Tunnel
The analysis of the tunnel subjected to applied loads and the design of the structural components are performed using a model generated by general purpose structural analysis computer software. Concrete walls and slabs are modeled as a rigid frame, composed of groups of members that are interconnected by a series of joints (see Section 4.0 Analysis Model Input and Section 5.0 Analysis Model Diagram). All joints are located along the centroids of the structural components. Members are modeled as one foot wide segments in the longitudinal direction of the tunnel to represent a one-foot-wide "slice" of the structure. AASHTO LRFD factored loads and load combinations are applied to the members and joints as required. The structure is analyzed to determine member forces and reactions, which will be used to design individual structural components of the tunnel.
2.1 Model Supports
Universal restraints are applied in the Y-translation and X-rotation degrees of freedom to all members. Spring supports located at joints spaced at 1'-0" on center are used to model soil conditions below the bottom slab of the tunnel. Springs with a K constant equal to 2600 k/ft are used, applied only in the downward Z direction. The spring support reaction will account for the earth reaction load.
3. Load Determination
The tunnel is located completely below grade and is subjected to loading on all sides. The self weight load of the concrete structure is applied vertically downward as component dead load. Vehicular live loads and vertical earth pressure are applied in the vertical downward direction to the top slab. Buoyancy forces are applied vertically upward to the bottom slab. Lateral forces from live load, soil overburden, horizontal earth pressure, and hydrostatic pressure are applied to the exterior walls. Load designations are referenced from LRFD Section 3.3.2 (see Figure 3).
3.1 Total Dead Loads
Dead loads are represented by the weight of all components of the tunnel structure and the vertical earth pressure due to the dead load of earth fill.
Concrete dead load (per foot length) (DC)
Vertical earth pressure (EV)
3.2 Live Load
Live load represents wheel loading from an HS-20 design vehicle. It is assumed that the wheels act as point loads at the surface and are distributed downward in both directions through the soil to the top slab of the tunnel. The load distribution is referenced from LRFD Section 184.108.40.206.6. Figure 4 shows the distribution of the wheel loads to the top slab.
Wheel Loads (LL)
Live Load Surcharge (LS)
Lateral earth pressure is typically represented by the equation:
σ = k0γn
The following lateral pressures are applied to the exterior walls of the tunnel (see Figure 5):
Calculate the lateral earth pressures:
EH1 = ko(γs × nsurch) = 0.080 ksf
EH2 = ko(γs × ns + γsat × nsat) = 0.494 ksf
EH3 = ko(γs × ns + γsat × nsat) = 1.314 ksf
EH4 = kw(γw × nw) = 1.825 ksf
3.4 Buoyancy Load WA
Area of water displaced, A
A = B × H = 1855.125 sq. ft.
Buoyancy = A × γw = 115.76 klf (along tunnel) OK
WA = Buoyancy/B = 1.513 klf
3.5 Load Factors and Combinations
Loads are applied to a model using AASHTO LRFD load combinations, referenced from LRFD Table 3.4.1-1. The loads, factors, and combinations for the applicable design limit states are given in Table 1.
Table 1: Load Factors and Load Combinations
4. Analysis Model Input
4.1 Joint Coordinates
The cross section of the tunnel model lies in the X-Z global plane. Each joint is assigned X and Z coordinates to locate its position in the model. See Section 5.0 and Figure 6 for a diagram of the model.
4.2 Member Definition
Members are defined by a beginning joint and an end joint, Ji and Jj , respectively, where i and j represent joint numbers.
All members are composed of concrete and represent a one foot wide "slice" of the tunnel section.
5. Analysis Model Diagram
The computer model represents a one foot wide slice of the cross-section of the tunnel. Members are connected by series of joints at their endpoints to form a frame, and are located along the centroids of the walls and roof and floor slabs. Joints in the 100 series and 200 series represent the floor and roof slabs respectively. Joints in the 300 and 500 series represent the exterior walls, while the 400 series represents the interior wall. The bottom diagram of Figure 5 shows all joints in the structure, while the top diagram shows only the joints at the intersections of slabs and walls.
Joints 302, 402, and 502 at the base of the exterior walls and joints 305, 405, and 505 at the top of the exterior walls are included to determine shear at the face of the top and bottom slabs.
6. Application of Lateral Loads (EH)
Lateral pressures EH1 through EH4 from Section 3.3 are applied to the members of the model as shown below. See Figure 7 for the horizontal earth pressure and hydrostatic pressure load distributions.
6.1 Exterior Wall Loads Due to Horizontal Earth Pressure EH3
Calculate pressure at top of wall:
ko(γs×ns + γsat×nsat) = = 0.494 ksf
Pressure at base of wall = 1.314 ksf (see calculation in Sec. 3.3)
Calculate interval increment for loading all exterior wall members:
The two tables below show the lateral earth pressure values (ksf) at the beginning and end of each member of the exterior walls:
6.2 Exterior Wall Loads Due to Hydrostatic Pressure EH4
Calculate pressure at top of wall:
kw(γw×nw) = = 0.312 ksf
Pressure at base of wall = 1.825 ksf (see calcs. in Sec. 3.3)
Calculate interval increment for loading all exterior wall members:
Δ = = 0.303 ksf
The two tables below show the lateral hydrostatic pressure values (ksf) at the beginning and end of each member of the exterior walls:
Figure 7 shows the load distribution along the exterior walls (members 301 to 305 and 501 to 505) for horizontal earth pressure (EH3) and hydrostatic pressure (EH4).
7. Structural Design Calculations - General Information
7.1 Concrete Design Properties
Modulus of elasticity of steel, Es = 29000 ksiYield strength of steel reinforcement, fy = 60 ksi Compressive strength of concrete, f'c = 4 ksi
7.2 Resistance Factors
Resistance factors for the strength limit state using conventional concrete construction are referenced from AASHTO LRFD Section 220.127.116.11.
Flexure Φ = 0.90 (Φ) varies from 0.75 to 0.9 (0.75 is conserverative)
Shear Φ = 0.90
Compression Φ = 0.7 since no spirals or ties
8. Interior Wall Design
8.1 Factored Axial Resistance (S18.104.22.168)
For members with tie reinforcement using LRFD eq. (22.214.171.124-3):
Pn = 0.80 [0.85×fc×(Ag - Ast) + fy×Ast]
Ast = 1.76 in2 (#6 at 6", ea. face)
Ag = 144.00 in2
Where Ag= 12×12 in2 (assuming wall thickness = 1 foot)
Pn = 471.37 kip
Factored axial resistance of reinforced concrete using LRFD eq. (126.96.36.199-1):
Pr = ΦPn Φ = 0.9 for flexure
Pr = factored axial resistance
Pn = nominal axial resistance
Pu = factored applied axial force
Pr = 424.24 kip
Check Pu < Pr
Pu = from computer model output = 78.00 kip < Pr OK
9. Top Slab Design
9.1 Slenderness Check (S188.8.131.52)
k × (lu/r) = 33.55
34 - 12 (M,1/M2) = 30.38
Where M1 and M2 are smaller and larger end moments
From analysis output
where M1 = 77 kip-ft P1 = 28.4 kip
M2 = 255 kip-ft P2 = 28.4 kip
Consider slenderness since k × (lu / r) is greater than 34 - 12 (M1/M2)
Calculate EI using LRFD eq. (184.108.40.206-1 and 220.127.116.11-2):
Mno = 215.00 kip-ft
M2 = 255.00 kip-ft
Note: Mno does not include effects of vertical live load surcharge
βd = Mno/M2 = 0.84
Approximate Method (LRFD 18.104.22.168.2)
The effects of deflection on force effects on beam-columns and arches which meet the provisions of the LRFD specifications may be approximated by the Moment Magnification method described below.
For steel/concrete composite columns, the Euler buckling load, Pe, shall be determined as specified in article 22.214.171.124 of LRFD. For all other cases, Pe shall be taken as:
(LRFD eq. 126.96.36.199.2b-5)
E = modulus of elasticity (ksi)
I = moment of inertia about axis under consideration (in4)k
= effective length factor as specified in LRFD 188.8.131.52
lu = unsupported length of a compression member (in)
Pe = 2626.67 kips
Moment Magnification (LRFD 184.108.40.206.2b)
(The components for sidesway will be neglected. Bracing moment will not include lateral force influence. Live load surcharge is excluded also.)
The factored moments may be increased to reflect effects of deformations as follows:
LRFD eq. (220.127.116.11.2b-1):
Mc = δb × M2b + δs × M2s
Mu = 215.00 kip-ft
MuLAT = -35.08 kip-ft
LRFD eq. (18.104.22.168.2b-3)
For members braced against sidesway and without transverse loads between supports, Cm:
Cm = 0.6 + 0.4 (M1/M2) LRFD eq. (22.214.171.124.2b-6)
Cm = 0.72
M1= smaller end moment
M2= larger end moment
Pu = factored axial load (kip) = 28.4 kips
Φ = resistance factor for axial compression
Pe = Euler buckling load (kip)
δb = 1
M2b = moment on compression member due to factored gravity loads that result in no appreciable sidesway calculated by conventional first-order elastic frame analysis; always positive (kip-ft)
M2b = 179.92 kip-ft
Mc = 179.92 kip-ft
Factored flexural resistance (LRFD 126.96.36.199.1)
The factored resistance Mr shall be taken as:
Mr = ΦMn
Φ = resistance factor = 0.9
Mn = nominal resistance (kip-in)
The nominal flexural resistance may be taken as:
(LRFD eq. 188.8.131.52.2-1)
Do not consider compression steel for calculating Mn.
As = area of nonprestressed tension reinforcement (in2)
fy = specified yield strength of reinforcing bars (ksi)
ds = distance from extreme compression fiber to centroid of nonprestressed tensile reinforcement (in2)
a = depth of equivalent stress block (in) = β1 × c
β1 = stress block factor specified in Section 184.108.40.206 of LRFD
c = distance from the extreme compression fiber to the neutral axis
LRFD eq. (220.127.116.11.2-4)
As= 2.0 in2
fy= 60.0 ksi
fc= 4.0 ksi
b= 12.0 in
c = 3.46 in
a = 2.94 in
Mn = 3153.53 kip-in = 262.79 kip-ft
ΦMn = 236.51 kip-ft OK (≥ Mc)
Mr = 236.51 kip-ft Mr > Mu
Create interaction diagram
Assume ρmin = 1.0%
Asmin = 3.6 in2
Asprov (total) = 4.00 in2 choose #9 at 6"
Es = 29000 ksi
β1 = 0.85
Yt = 15 in
0.85 × f'c = 3.4 ksi
Ag' in2 = 360 in2
As = A's = 2.0 in2
At zero moment point using LRFD eq. (18.104.22.168-2)
Po = 0.85 × f'c × (Ag - Ast) + Ast × fy = 1450 kip
ΦPo = 1015 kip
At balance point calculate Prb and Mrb
cb = 16.65 in
ab = β1× cb = 14.15 in
= 70 ksi
f's > fy; set f's = fy
Acomp = c × b = 199.8 in2
y' = a / 2 = 7.07625 in
ΦPb = Φ [0.85 × f'c × b × ab × As' × f's - As × fy] = 485 kip
ΦMb = 7442 kip-in = 620 kip-ft
At zero 'axial load' point (conservatively ignore compressive reinforcing)
= 2.9 in
ΦMo = 2838.2 kip-in = 237 kip-ft
At intermediate points
Note Φ may decrease from 0.90 to 0.75 as a increases from 0.0 to ab. Use 0.75 to be conservative.
Acomp = a × 12 in2
ΦPn = Φ (Acomp - A's) × 0.85 × f'c + As' × f's - As × fy kips