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## Technical Manual for Design and Construction of Road Tunnels - Civil Elements

### Appendix C - Cut-and-Cover Tunnel Design Example

##### 9.2 Shear Design (S5.8.3.3)

The nominal shear resistance, V_{n} shall be determined as the lesser of

LRFD eq. 5.8.3.3-1: V_{n} = V_{c} + V_{s}

or

LRFD eq. 5.8.3.3-2: V_{n} = 0.25 × f'_{c} × b_{v} × d_{v}

Note V_{p} is not considered

Where:

LRFD eq. (5.14.5.3-1)

Where

For slab concrete shear (V_{c}), refer to LRFD Section 5.14.5.

LRFD eq. (5.8.3.3-4)

Where for α = 90° and θ = 45°

Where:

A_{s} = area of reinforcing steel in the design width (in^{2})

b = design width (in)

d_{e} = effective depth from extreme compression fiber to centroid of tensile force in tensile reinforcement (in)

d_{e}= 27.75

V_{u} = shear from factored loads (kip)

M_{u} = moment from factored loads (kip-in)

A_{v} = area of shear reinforcement within a distance s (in^{2}) = 0 in^{2}

s = spacing of stirrups (in) = 12 in

b_{v} = effective web width taken as the minimum web width within the depth d_{v} (in)

d_{v} = effective shear depth taken as the perpendicular distance to the neutral axis (in)

d_{v} = 0.9 × d_{e} or 0.72 × h (LRFD section 5.8.2.9)

d_{v} = 24.98 in

Maximum shear and associated moment from analysis output:

V_{u} = 28 kip M_{u} = 63.0 kip-ft

V_{c} = 63.42 kip value controls

or V_{c} = 83.92 kip

V_{s} = 0.00 kip

V_{n} = 63.42 kip

V_{n} = 299.70 kip therefore V_{n} = 63.42 kip

Φ = 0.90

ΦV_{n} = 57.08 kip > V_{u} OK

#### 10. Bottom Slab Design

##### 10.1 Slenderness Check (S5.7.4.3)

K = 0.65

lu = 37.25 ft = 447 in

d = 1.75 ft = 21.0 in

I = 9261 in^{4}

r = 6.06 in

β_{1} = 0.85

d_{s} = 18.75 in

d'_{s} = 3.25 in

#8 bar dia. = 1.00 in

k × (l_{u} / r) = 47.93

From analysis output

where M_{1} = 13 kip-ft P_{1} = 23.6 kip

M_{2} = 57.1 kip-ft P_{2} = 23.6 kip

34 - 12 (M_{1}/M_{2}) = 31.27

Consider slenderness since k × (lu / r) is greater than 34 - 12 (M_{1}/M_{2})

Calculate EI:

E_{c} = 3834.25 ksi

I_{g} = 9261 in^{4}

c = 8 in EI = 3427836.25 kip-in^{2}

I_{s} = 202.34 in^{4} EI = 6855672.51 kip-in^{2}

M_{no} = 61.20 kip-ft

M_{2} = 57.10 kip-ft

Note: M_{no} does not include effects of vertical live load surcharge

β_{d} = M_{no}/M_{2} = 1.07

__Approximate Method (LRFD 4.5.3.2.2)__

P_{e} =

P_{e} = 801.51 kip

__Moment Magnification__

(The components for sidesway will be neglected. Bracing moment will not include lateral force influence. Live load surcharge is excluded also.)

C_{m} = 0.6 + 0.4 (M_{1}/M_{2}) = 0.69

P_{u} = 23.6 kip

δ_{b} = 1.00

M_{c} = δ_{b} × M_{2b} + Δ_{s} × M_{2s}

M_{c} = 28.32 kip-ft

__Factored flexural resistance__

Do not consider compression steel for calculating M_{n}.

c = 2.73 in

a = 2.32 i

M_{n} = 1667.36 kip-in = 138.95 kip-ft

M_{r} = ΦM_{n} = 125.05 kip-ft OK (≥ Mc) M_{r}>M_{u}

__Create interaction diagram__

Assume ρ_{min} = 1.0%

A_{smin} = 2.52 in^{2}

A_{sprov} (total) = 3.16 in^{2} choose #8 at 6"

E_{s} = 29000 ksi

β_{1} = 0.85

Y_{t} = 10.5 in

0.85 × f'_{c} = 3.4 ksi

A_{g'} in^{2} = 252 in^{2}

A_{s} = A'_{s} = 1.6 in^{2}

__At zero moment point__

P_{o} = 1036 kip

ΦP_{o} = 725 kip

__At balance point calculate P _{rb} and M_{rb}__

c_{b} = 11.25 in

a_{b} = 9.56 in

f'_{s} = 62 ksi

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

A_{comp} = 114.75 in^{2}

y' = 4.78125 in

ΦP_{b} = 271 kip

ΦM_{b} = 3303 kip-in = 275 kip-ft

__At zero 'axial load' point (conservatively ignore compressive reinforcing)__

a = 2.3 in

ΦM_{o} = 1500.6 kip-in = 125 kip-ft

__At intermediate points__

a, in | c = a/b_{1} | A_{comp}, in^{2} | f'_{s}, ksi | f _{s}, ksi | f_{y}, ksi | ΦM_{n}, k-ft | ΦP_{n}, kips |
---|---|---|---|---|---|---|---|

125 | 0 | ||||||

2.3 | 2.7 | 27.6 | 36 | 552 | 60 | 118 | 24 |

3 | 3.5 | 36 | 48 | 423 | 60 | 139 | 63 |

4 | 4.7 | 48 | 58 | 317 | 60 | 162 | 107 |

5 | 5.9 | 60 | 64 | 254 | 60 | 178 | 139 |

6 | 7.1 | 72 | 68 | 212 | 60 | 190 | 168 |

7 | 8.2 | 84 | 70 | 181 | 60 | 200 | 196 |

8 | 9.4 | 96 | 72 | 159 | 60 | 207 | 225 |

9 | 10.6 | 108 | 74 | 141 | 60 | 212 | 253 |

10 | 11.8 | 120 | 75 | 127 | 60 | 215 | 282 |

11 | 12.9 | 132 | 76 | 115 | 60 | 215 | 310 |

12 | 14.1 | 144 | 77 | 106 | 60 | 213 | 339 |

13 | 15.3 | 156 | 78 | 98 | 60 | 208 | 368 |

14 | 16.5 | 168 | 79 | 91 | 60 | 201 | 396 |

15 | 17.6 | 180 | 79 | 85 | 60 | 192 | 425 |

16 | 18.8 | 192 | 80 | 79 | 60 | 180 | 453 |

0 | 725 | ||||||

End 1 | 13 | 24 | |||||

End 2 | 57 | 24 |

Note Φ may decrease from 0.90 to 0.75 as a increases from 0.0 to ab. Use 0.75 to be conservative.

##### 10.2 Shear Design (S5.8.3.3)

V_{n} = V_{c} + V_{s} or V_{n} = 0.25 × f'_{c} × b_{v} × d_{v}

d_{v} = 16.88 in

For slab concrete shear (V_{c}), see LRFD Section 5.14.5

Maximum shear and associated moment from analysis output:

V_{u} = 19.4 kip M_{u} = 30.3 kip-ft

V_{c} = 44.96 kip value controls

or V_{c} = 56.70 kip

Where A_{v} = 0 in^{2} and s = 12 in

V_{s} = 0.00 kip

V_{n} = 44.96 kip

V_{n} = 202.50 kip therefore V_{n} = 44.96 kip

ΦV_{n} = 40.46 kip > V_{u} OK

#### 11. Exterior Wall Design

##### 11.1 Slenderness Check (LRFD 5.7.4.3)

K = 0.65

lu = 22.13 ft = 265.5 in

d = 2.00 ft = 24.0 in

I = 13824 in^{4}

r = 6.93 in

β_{1} = 0.85

d_{s} = 21.75 in

d'_{s} = 3.25 in

#8 bar dia. = 1.00 in

k × (l_{u} / r) = 24.91

From analysis output

where M_{1} = 171.4 kip-ft P_{1} = 34.4 kip

M_{2} = 137.2 kip-ft P_{2} = 34.4 kip

34-12 (M_{1}/M_{2}) = 19.01

Consider slenderness since k × (l_{u} / r) is greater than 34-12 (M_{1}/M_{2})

Calculate EI:

E_{c} = 3834.25 ksi

I_{g} = 13824 in^{4}

c = 9.5 in EI = 7330894.82 kip-in^{2}

I_{s} = 285.29 in^{4} EI = 14661789.6 kip-in^{2}

M_{no} = 61.20 kip-ft

M_{2} = 137.20 kip-ft

Note: M_{no} does not include effects of vertical live load surcharge

β_{d} = M_{no}/M_{2} = 0.45

__Approximate Method (LRFD 4.5.3.2.2)__

P_{e} =

P_{e} = 4858.82 kip

__Moment Magnification__

(The components for sidesway will be neglected. Bracing moment will not include lateral force influence. Live load surcharge is excluded also.)

C_{m} = 0.6 + 0.4 (M_{1}/M_{2}) = 1.10

P_{u} = 34.4 kip

δ_{b} = 1.11

M_{c} = δ_{b} × M_{2b} + δ_{s} × M_{2s}

M_{c} = 38.46 kip-ft

__Factored flexural resistance__

Do not consider compression steel for calculating M_{n}.

c = 2.73 in

a = 2.32 in

M_{n} = 1951.76 kip-in = 162.65 kip-ft

M_{r} = ΦM_{n} = 146.38 kip-ft OK (≥ M_{c}) M_{r} > M_{u}

__Create interaction diagram__

Assume ρ_{min} = 1.0%

A_{smin} = 2.88 in^{2}

A_{sprov} (total) = 3.16 in^{2} choose #8 at 6"

E_{s} = 29000 ksi

β_{1} = 0.85

Y_{t} = 12 in

0.85 × f'_{c} = 3.4 ksi

A_{g'} in^{2} = 288 in^{2}

A_{s} = A'_{s} = 1.6 in^{2}

__At zero moment point__

P_{o} = 1158 kip

ΦP_{o} = 811 kip

__At balance point calculate P _{rb} and M_{rb}__

c_{b} = 13.05 in

a_{b} = 11.09 in

f'_{s} = 65 ksi

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

A_{comp} = 133.11 in^{2}

y' = 5.54625 in

ΦP_{b} = 313 kip

ΦM_{b} = 4176 kip-in = 348 kip-ft

__At zero 'axial load' point (conservatively ignore compressive reinforcing)__

a = 2.3 in

ΦM_{o} = 1756.6 kip-in = 146 kip-ft

__At intermediate points__

a, in | c = a/b_{1} | A_{comp}, in^{2} | f'_{s}, ksi | f_{s}, ksi | f_{y}, ksi | ΦM_{n}, k-ft | ΦP_{n}, kips |
---|---|---|---|---|---|---|---|

146 | 0 | ||||||

2.3 | 2.7 | 27.6 | 36 | 612 | 60 | 179 | 24 |

3 | 3.5 | 36 | 48 | 449 | 60 | 211 | 63 |

4 | 4.7 | 48 | 58 | 315 | 60 | 248 | 107 |

5 | 5.9 | 60 | 64 | 235 | 60 | 273 | 139 |

6 | 7.1 | 72 | 68 | 181 | 60 | 293 | 168 |

7 | 8.2 | 84 | 70 | 143 | 60 | 310 | 196 |

8 | 9.4 | 96 | 72 | 114 | 60 | 324 | 225 |

9 | 10.6 | 108 | 74 | 92 | 60 | 335 | 253 |

10 | 11.8 | 120 | 75 | 74 | 60 | 343 | 282 |

11 | 12.9 | 132 | 76 | 59 | 60 | 348 | 310 |

13 | 15.3 | 156 | 78 | 37 | 60 | 348 | 368 |

15 | 17.6 | 180 | 79 | 20 | 60 | 336 | 425 |

17 | 20.0 | 204 | 80 | 8 | 60 | 312 | 482 |

19 | 22.4 | 228 | 81 | -2 | 60 | 276 | 539 |

21 | 24.7 | 252 | 81 | -10 | 60 | 227 | 596 |

0 | 811 | ||||||

Top of Wall | 171 | 34 | |||||

Bottom of Wall | 137 | 34 |

Note Φ may decrease from 0.90 to 0.75 as a increases from 0.0 to ab. Use 0.75 to be conservative.

##### 11.2 Shear Design (S5.8.3.3)

Maximum shear from analysis output:

V_{u} = 20.76 kip

Where β = 2

b_{v} = 12 in

d_{v} = 19.58 in

V_{c} = 0.0316 × β × f'_{c}^{0.5} × b_{v} × d_{v} LRFD eq. (5.8.3.3-3)

V_{c} = 29.69 kip

Where A_{v} = 0 in^{2} and s = 12 in

V_{s} = 0.00 kip

V_{n} = 29.69 kip

V_{n} = 234.90 kip therefore V_{n} = 29.69 kip

ΦV_{n} = 26.72 kip > Vu OK

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