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# Bridges & Structures

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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
Box interior width, x = 35.75 ft 20.00 ft 1.00 ft 2.00 ft 1.75 ft 2.50 ft 10.00 ft 5.00 ft 34.25 ft 76.50 ft 74.50 ft 24.25 ft 22.13 ft

Figure 2 shows the geometry of the underground cut and cover box cross-section.

##### 1.2 Material Properties
Unit weight of concrete ,γc = 150 pcf 130 pcf 62.4 pcf 67.6 pcf 0.5 1

#### 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.

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).

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.

Top slab = Bottom slab = 0.15 ksf × (76.5 × 2.5) = 28.69 kip 0.15 ksf × (76.5 × 1.75) = 20.08 kip 0.15 ksf × (1 × 20) = 3.00 kip 0.15 ksf × 2 × (2 × 20) = 12.00 kip

Vertical earth pressure (EV)

EV= Soil wt = 1.30 ksf 1.30 ksf × 76.50 = 99.45 kip

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.

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

 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 = kos × nsurch) = 0.080 ksf

EH2 = kos × ns + γsat × nsat) = 0.494 ksf
ns = 5.00 ft
nsat = 5.00 ft

EH3 = kos × ns + γsat × nsat) = 1.314 ksf
ns = 5.00 ft
nsat = 29.25 ft

EH4 = kww × nw) = 1.825 ksf
nw = 29.25 ft

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.

• LS - Live load surcharge
• EH - Horizontal earth pressure load
• WA - Water load and stream pressure
EVDCLSEHWA
Limit State
Strength 1A1.31.251.751.351
B1.31.251.750.91
C0.91.251.751.351
D0.91.251.750.91
E1.30.91.751.351
F1.30.91.750.91
G0.90.91.750.91
H0.90.91.751.351
Strength 2A1.31.251.351.351
B1.31.251.350.91
C0.91.251.351.351
D0.91.251.350.91
E1.30.91.351.351
F1.30.91.350.91
G0.90.91.350.91
H0.90.91.351.351
Strength 3A1.31.25n1.351
B1.31.25n0.91
C0.91.25n1.351
D0.91.25n0.91
E1.30.9n1.351
F1.30.9n0.91
G0.90.9n0.91
H0.90.9n1.351
Service 1 1.01111
Service 4 1.01n11

#### 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:

kos×ns + γsat×nsat) = = 0.494 ksf

Pressure at base of wall = 1.314 ksf (see calculation in Sec. 3.3)

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:

MemberStartEnd
3011.311.15
3021.150.99
3030.990.82
3040.820.66
3050.660.49
MemberStartEnd
501-1.31-1.15
502-1.15-0.99
503-0.99-0.82
504-0.82-0.66
505-0.66-0.49
##### 6.2 Exterior Wall Loads Due to Hydrostatic Pressure EH4

Calculate pressure at top of wall:

kww×nw) = = 0.312 ksf

Pressure at base of wall = 1.825 ksf (see calcs. in Sec. 3.3)

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:

MemberStartEnd
3011.831.52
3021.521.22
3031.220.92
3040.920.61
3050.610.31
MemberStartEnd
501-1.83-1.52
502-1.52-1.22
503-1.22-0.92
504-0.92-0.61
505-0.61-0.31

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 ksi

Yield 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 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):

Pn = 0.80 [0.85×fc×(Ag - Ast) + fy×Ast]

Where:

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. (5.7.4.4-1):

Pr = ΦPn Φ = 0.9 for flexure

Where:

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 (S5.7.4.3)
 K= 0.65 β1= 0.85 lu= 37.25 ft = 447 in ds= 27.75 in d= 2.50 ft = 30.0 in ds= 3.25 in I= (12×30)/12 = 27000 in4 #9 bar dia.= 1.13 in r= = 8.66 in

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. (5.7.4.3-1 and 5.7.4.3-2):

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 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, Pe, shall be determined as specified in article 6.9.5.1 of LRFD. For all other cases, Pe 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)

k

= effective length factor as specified in LRFD 4.6.2.5

lu = unsupported length of a compression member (in)

Pe = 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):

Mc = δb × M2b + δs × M2s

Mu = 215.00 kip-ft

MuLAT = -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, Cm:

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

Cm = 0.72

Where:

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 5.7.3.2.1)

The factored resistance Mr shall be taken as:

Mr = ΦMn

Where:

Φ = resistance factor = 0.9

Mn = 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 Mn.

Where:

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

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:

As= 2.0 in2

fy= 60.0 ksi

fc= 4.0 ksi

β1= 0.85

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. (5.7.4.5-2)

Φ=0.7

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

a, inc = a/b1Acomp, in2f's,ksifs,ksify, ksiΦMn, k-ftΦPn, kips
2370
2.93.434.8366576029230
33.536386356029836
44.748504766035590
55.9605738160401133
67.1726231760435167
78.2846627260461195
89.4966923860484224
1011.81207219060521281
1214.11447515960546338
1517.61807712760561424
1821.22167910660548509
1922.42287910060537538
2124.7252809160507595
2327.1276818360465652
2529.4300817660410709
01015
End 17728
End 225528

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:

Acomp = a × 12 in2

ΦPn = Φ (Acomp - A's) × 0.85 × f'c + As' × f's - As × fy kips

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Updated: 06/19/2013
Federal Highway Administration | 1200 New Jersey Avenue, SE | Washington, DC 20590 | 202-366-4000