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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, Vn shall be determined as the lesser of

LRFD eq. 5.8.3.3-1: Vn = Vc + Vs

or

LRFD eq. 5.8.3.3-2: Vn = 0.25 × f'c × bv × dv

Note Vp is not considered

Where:

equation LRFD eq. (5.14.5.3-1)

Where equation

For slab concrete shear (Vc), refer to LRFD Section 5.14.5.

equation LRFD eq. (5.8.3.3-4)

Where for α = 90° and θ = 45° equation

Where:

As = area of reinforcing steel in the design width (in2)

b = design width (in)

de = effective depth from extreme compression fiber to centroid of tensile force in tensile reinforcement (in)
de= 27.75

Vu = shear from factored loads (kip)

Mu = moment from factored loads (kip-in)

Av = area of shear reinforcement within a distance s (in2) = 0 in2

s = spacing of stirrups (in) = 12 in

bv = effective web width taken as the minimum web width within the depth dv (in)

dv = effective shear depth taken as the perpendicular distance to the neutral axis (in)

dv = 0.9 × de or 0.72 × h (LRFD section 5.8.2.9)

dv = 24.98 in

equation

Maximum shear and associated moment from analysis output:

Vu = 28 kip Mu = 63.0 kip-ft

Vc = 63.42 kip value controls

or Vc = 83.92 kip

Vs = 0.00 kip

Vn = 63.42 kip

Vn = 299.70 kip therefore Vn = 63.42 kip

Φ = 0.90

ΦVn = 57.08 kip > Vu 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 in4

r = 6.06 in

β1 = 0.85

ds = 18.75 in

d's = 3.25 in

#8 bar dia. = 1.00 in

k × (lu / r) = 47.93

From analysis output

where M1 = 13 kip-ft P1 = 23.6 kip

M2 = 57.1 kip-ft P2 = 23.6 kip

34 - 12 (M1/M2) = 31.27

Consider slenderness since k × (lu / r) is greater than 34 - 12 (M1/M2)

Calculate EI:

Ec = 3834.25 ksi

Ig = 9261 in4

c = 8 in EI = 3427836.25 kip-in2

Is = 202.34 in4 EI = 6855672.51 kip-in2

Mno = 61.20 kip-ft

M2 = 57.10 kip-ft

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

βd = Mno/M2 = 1.07

Section Dimensions

Approximate Method (LRFD 4.5.3.2.2)

Pe = equation

Pe = 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.)

Cm = 0.6 + 0.4 (M1/M2) = 0.69

Pu = 23.6 kip

δb = 1.00

Mc = δb × M2b + Δs × M2s

Mc = 28.32 kip-ft

equation

Factored flexural resistance

Do not consider compression steel for calculating Mn.

c = 2.73 in

a = 2.32 i

Mn = 1667.36 kip-in = 138.95 kip-ft

Mr = ΦMn = 125.05 kip-ft OK (≥ Mc) Mr>Mu

Create interaction diagram

Assume ρmin = 1.0%

Asmin = 2.52 in2

Asprov (total) = 3.16 in2 choose #8 at 6"

Es = 29000 ksi

β1 = 0.85

Yt = 10.5 in

0.85 × f'c = 3.4 ksi

Ag' in2 = 252 in2

As = A's = 1.6 in2

At zero moment point

Po = 1036 kip

ΦPo = 725 kip

At balance point calculate Prb and Mrb

cb = 11.25 in

ab = 9.56 in

f's = 62 ksi
f's > fy; set f's = fy

Acomp = 114.75 in2

y' = 4.78125 in

ΦPb = 271 kip

ΦMb = 3303 kip-in = 275 kip-ft

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

a = 2.3 in

ΦMo = 1500.6 kip-in = 125 kip-ft

At intermediate points

a, inc = a/b1Acomp, in2f's, ksif s, ksify, ksiΦMn, k-ftΦPn, kips
      1250
2.32.727.6365526011824
33.536484236013963
44.7485831760162107
55.9606425460178139
67.1726821260190168
78.2847018160200196
89.4967215960207225
910.61087414160212253
1011.81207512760215282
1112.91327611560215310
1214.11447710660213339
1315.3156789860208368
1416.5168799160201396
1517.6180798560192425
1618.8192807960180453
      0725
End 11324
End 25724

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

Interaction Diagram

10.2 Shear Design (S5.8.3.3)

Vn = Vc + Vs or Vn = 0.25 × f'c × bv × dv
dv = 16.88 in

For slab concrete shear (Vc), see LRFD Section 5.14.5

equation

Maximum shear and associated moment from analysis output:

Vu = 19.4 kip Mu = 30.3 kip-ft

Vc = 44.96 kip value controls
or Vc = 56.70 kip

Where Av = 0 in2 and s = 12 in

Vs = 0.00 kip

Vn = 44.96 kip

Vn = 202.50 kip therefore Vn = 44.96 kip

ΦVn = 40.46 kip > Vu 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 in4

r = 6.93 in

β1 = 0.85

ds = 21.75 in

d's = 3.25 in

#8 bar dia. = 1.00 in

k × (lu / r) = 24.91

From analysis output

where M1 = 171.4 kip-ft P1 = 34.4 kip

M2 = 137.2 kip-ft P2 = 34.4 kip

34-12 (M1/M2) = 19.01

Consider slenderness since k × (lu / r) is greater than 34-12 (M1/M2)

Calculate EI:

Ec = 3834.25 ksi

Ig = 13824 in4

c = 9.5 in       EI = 7330894.82 kip-in2

Is = 285.29 in4       EI = 14661789.6 kip-in2

Mno = 61.20 kip-ft

M2 = 137.20 kip-ft

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

βd = Mno/M2 = 0.45

Section Dimensions

Approximate Method (LRFD 4.5.3.2.2)

Pe = equation

Pe = 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.)

Cm = 0.6 + 0.4 (M1/M2) = 1.10

Pu = 34.4 kip

δb = 1.11

Mc = δb × M2b + δs × M2s

Mc = 38.46 kip-ft

equation

Factored flexural resistance

Do not consider compression steel for calculating Mn.

c = 2.73 in

a = 2.32 in

Mn = 1951.76 kip-in = 162.65 kip-ft

Mr = ΦMn = 146.38 kip-ft OK (≥ Mc) Mr > Mu

Create interaction diagram

Assume ρmin = 1.0%

Asmin = 2.88 in2

Asprov (total) = 3.16 in2 choose #8 at 6"

Es = 29000 ksi

β1 = 0.85

Yt = 12 in

0.85 × f'c = 3.4 ksi

Ag' in2 = 288 in2

As = A's = 1.6 in2

At zero moment point

Po = 1158 kip

ΦPo = 811 kip

At balance point calculate Prb and Mrb

cb = 13.05 in

ab = 11.09 in

f's = 65 ksi

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

Acomp = 133.11 in2

y' = 5.54625 in

ΦPb = 313 kip

ΦMb = 4176 kip-in = 348 kip-ft

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

a = 2.3 in

ΦMo = 1756.6 kip-in = 146 kip-ft

At intermediate points

a, inc = a/b1Acomp, in2f's, ksifs, ksify, ksiΦMn, k-ftΦPn, kips
      1460
2.32.727.6366126017924
33.536484496021163
44.7485831560248107
55.9606423560273139
67.1726818160293168
78.2847014360310196
89.4967211460324225
910.6108749260335253
1011.8120757460343282
1112.9132765960348310
1315.3156783760348368
1517.6180792060336425
1720.020480860312482
1922.422881-260276539
2124.725281-1060227596
      0811
 Top of Wall17134
 Bottom of Wall13734

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

Interaction Diagram

11.2 Shear Design (S5.8.3.3)

Maximum shear from analysis output:

Vu = 20.76 kip

Where β = 2

bv = 12 in

dv = 19.58 in

Vc = 0.0316 × β × f'c0.5 × bv × dv     LRFD eq. (5.8.3.3-3)

Vc = 29.69 kip

Where Av = 0 in2 and s = 12 in

Vs = 0.00 kip

Vn = 29.69 kip

Vn = 234.90 kip therefore Vn = 29.69 kip

ΦVn = 26.72 kip > Vu OK

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