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 3.6.1.2.6. Figure 4 shows the distribution of the wheel loads to the top slab.

Wheel Loads (LL)

LL1		0.04 ksf	(S3.6.1.2.6)
LL2		0.16 ksf	controls

Live Load Surcharge (LS)

LS =	0.16 76.50	= 12.240 kip
Surch. Ht =		= 1.231 ft

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):

• EH₁ = LL surcharge
• EH₂ = Lateral earth pressure due to soil overburden
• EH₃ = Horizontal earth pressure
• EH₄ = Hydrostatic pressure

Calculate the lateral earth pressures:

EH₁ = k_o(γ_s × n_surch) = 0.080 ksf

EH₂ = k_o(γ_s × n_s + γ_sat × n_sat) = 0.494 ksf
n_s = 5.00 ft
n_sat = 5.00 ft

EH₃ = k_o(γ_s × n_s + γ_sat × n_sat) = 1.314 ksf
n_s = 5.00 ft
n_sat = 29.25 ft

EH₄ = k_w(γ_w × n_w) = 1.825 ksf
n_w = 29.25 ft

3.4 Buoyancy Load WA

Area of water displaced, A

A = B × H = 1855.125 sq. ft.

Buoyancy = A × γ_w = 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

• EV - Vertical pressue from dead load of earth fill
• DC - Dead load of structural components and nonstructural attachments
• LS - Live load surcharge
• EH - Horizontal earth pressure load
• WA - Water load and stream pressure

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, J_i and J_j , 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:

k_o(γ_s×n_s + γ_sat×n_sat) = = 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:

0.164 ksf

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:

k_w(γ_w×n_w) = = 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, E_s = 29000 ksi

7.2 Resistance Factors

Resistance factors for the strength limit state using conventional concrete construction are referenced from AASHTO LRFD Section 5.5.4.2.

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 (S5.7.4.4)

For members with tie reinforcement using LRFD eq. (5.7.4.4-3):

P_n = 0.80 [0.85×f_c×(A_g - A_st) + f_y×A_st]

Where:

A_st = 1.76 in² (#6 at 6", ea. face)

A_g = 144.00 in²

Where A_g= 12×12 in² (assuming wall thickness = 1 foot)

P_n = 471.37 kip

Factored axial resistance of reinforced concrete using LRFD eq. (5.7.4.4-1):

P_r = ΦP_n Φ = 0.9 for flexure

Where:

P_r = factored axial resistance

P_n = nominal axial resistance

P_u = factored applied axial force

P_r = 424.24 kip

Check P_u < P_r

P_u = from computer model output = 78.00 kip < P_r OK

9. Top Slab Design

9.1 Slenderness Check (S5.7.4.3)

k × (l_u/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 × (l_u / r) is greater than 34 - 12 (M₁/M₂)

Calculate EI using LRFD eq. (5.7.4.3-1 and 5.7.4.3-2):

M_no = 215.00 kip-ft

M₂ = 255.00 kip-ft

Note: M_no does not include effects of vertical live load surcharge

β_d = M_no/M₂ = 0.84

Approximate Method (LRFD 4.5.3.2.2)

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, P_e, shall be determined as specified in article 6.9.5.1 of LRFD. For all other cases, P_e shall be taken as:

(LRFD eq. 4.5.3.2.2b-5)

Where:

E = modulus of elasticity (ksi)

I = moment of inertia about axis under consideration (in⁴)

= effective length factor as specified in LRFD 4.6.2.5

l_u = unsupported length of a compression member (in)

P_e = 2626.67 kips

Moment Magnification (LRFD 4.5.3.2.2b)

(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. (4.5.3.2.2b-1):

M_c = δ_b × M_2b + δ_s × M_2s

M_u = 215.00 kip-ft

M_uLAT = -35.08 kip-ft

Where:

LRFD eq. (4.5.3.2.2b-3)

Where:

For members braced against sidesway and without transverse loads between supports, C_m:

C_m = 0.6 + 0.4 (M₁/M₂) LRFD eq. (4.5.3.2.2b-6)

C_m = 0.72

Where:

M₁= smaller end moment

M₂= larger end moment

P_u = factored axial load (kip) = 28.4 kips

Φ = resistance factor for axial compression

P_e = Euler buckling load (kip)

δ_b = 1

M_2b = 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)

M_2b = 179.92 kip-ft

M_c = 179.92 kip-ft

Factored flexural resistance (LRFD 5.7.3.2.1)

The factored resistance M_r shall be taken as:

M_r = ΦM_n

Where:

Φ = resistance factor = 0.9

M_n = nominal resistance (kip-in)

The nominal flexural resistance may be taken as:

(LRFD eq. 5.7.3.2.2-1)

Do not consider compression steel for calculating M_n.

Where:

A_s = area of nonprestressed tension reinforcement (in²)

f_y = specified yield strength of reinforcing bars (ksi)

d_s = distance from extreme compression fiber to centroid of nonprestressed tensile reinforcement (in²)

a = depth of equivalent stress block (in) = β₁ × c

Where:

β₁ = stress block factor specified in Section 5.7.2.2 of LRFD

c = distance from the extreme compression fiber to the neutral axis

LRFD eq. (5.7.3.1.2-4)

Where:

A_s= 2.0 in²

f_y= 60.0 ksi

f_c= 4.0 ksi

β₁= 0.85

b= 12.0 in

c = 3.46 in

a = 2.94 in

M_n = 3153.53 kip-in = 262.79 kip-ft

ΦM_n = 236.51 kip-ft OK (≥ M_c)

M_r = 236.51 kip-ft M_r > M_u

Create interaction diagram

Assume ρ_min = 1.0%

A_smin = 3.6 in²

A_sprov (total) = 4.00 in² choose #9 at 6"

E_s = 29000 ksi

β₁ = 0.85

Y_t = 15 in

0.85 × f'_c = 3.4 ksi

A_g' in² = 360 in²

A_s = A'_s = 2.0 in²

At zero moment point using LRFD eq. (5.7.4.5-2)

Φ=0.7

P_o = 0.85 × f'_c × (A_g - A_st) + A_st × f_y = 1450 kip

ΦP_o = 1015 kip

At balance point calculate P_rb and M_rb

c_b = 16.65 in

a_b = β₁× c_b = 14.15 in

= 70 ksi

f's > fy; set f's = fy

A_comp = c × b = 199.8 in²

y' = a / 2 = 7.07625 in

ΦP_b = Φ [0.85 × f'_c × b × a_b × A_s' × f'_s - A_s × f_y] = 485 kip

ΦM_b = 7442 kip-in = 620 kip-ft

At zero 'axial load' point (conservatively ignore compressive reinforcing)

= 2.9 in

ΦM_o = 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.

Where:

A_comp = a × 12 in²

ΦP_n = Φ (A_comp - A'_s) × 0.85 × f'_c + A_s' × f'_s - A_s × f_y kips