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Technical Manual for Design and Construction of Road Tunnels - Civil Elements
Appendix G - Precast Segmental Lining Example
The following design example is intended to illustrate the application of the AASHTO LRFD Specifications to the design of a precast segmental concrete tunnel lining. The design scenario involves a tunnel constructed insoft ground using a tunnel boring machine. The roadway typical section approaching the tunnel is a 4-lane highway with full shoulders a median. The four lanes will be accomodated in two openings, each carrying two lanes of traffic. The tunnel section therefore will be sized to carry two 12'-0" traffic lanes with reduced shoulders on both sides. A 3'-3" wide walkway for maintenance will be included in the typical section. Emergencyegress will be accomodated either at the roadway level using the shoulders provided or through the adjacent bore. Access to the adjacent bore will be gained through cross passages located every 500' along the tunnel alignment. The tunnel will utilize jet fans in a longitudinal ventilation system. The jet fans will be suspended from the tunnel liner.
The analysis of the liner structure will be performed using the beam-spring model described in paragraph 10.??.??.
Figure 10E-1 provides the details of the typical section used in the example.
Design Example Typical Section
A typical dimension along the longitudinal axis of the tunnel for the segments is 5'-0". The structural analysis and modeling shown in the following sections of this design example will be based on a five foot length of tunnel. As usch applied loads and spring constants will be multiplied by 5 to account for the fact that the design section is five feet long.
Determine Number of Segments
Each segment must be fabricated at a casting yard or precast plant. Once it is fabricated, it must be stripped from the forms and moved to a curing area then to a storage yard. It must be transported form the storage yard to the tunnel site where a stockpile of segments is usually kept. At the tunnel site, it is loaded onto a materials cart that will transport the segment through the tunnel to the tunnel face where it will be erected to form part of ring. The segment must pass through all of the trailing gear associated with a tunnel boring machine on its way to the face. Segments are typically manufactured in advance of the mining operation so that there are suffecient segments on hand to allow the mining operation to proceed without stopping. It is not usual for segments to be damaged during handling and installation, so the number of segments produced is usually more than the total number of segments used in the tunnel. Therefore, segments must be handled several times, stored in at least two seperate locations, transported between the two seperate locations and transported through the tight space found inside a tunnel under construction.
Understanding this process helps to understand how determining the number of segments is a judgmnet decision that should balance minimizing the number of pieces in ring, keeping the length of each segment short enough that it can be practically stored, shipped and handled and making the piece light enough to be handled by the type and size machinery availble inside the tunnel to erect the segments.
Note that it is not unusual for a contractor to suggest a different arrangement of segments than that shown in the contract documents. Most owners allow the contractor to submit changes that are more in line with the means and methods used by a contractor.
For this example, the Inside Diameter = 35.00 ft Segment Length = 5 ft
Assume 8 Segments and a key segment.
Key segment subtends: 22.50 degrees
Other segments subtend: 42.188 degrees
Length of non-key segment along inside face of tunnel = 12.885 ft. This seems to be a reasonable length.
Number of joints = 9
This example problem will assume that the segments extend along 5 feet of the tunnel length. If 16 in. is assumed to be the thickness of the segments, then the weight of each segment is calculated as follows:
Length of segment along the centroid of the segment = 13.131 ft
Weight = 13.1309 x 5 x 150 = 9848.2 lbs = 4.92 tons
For a tunnel of this diameter, it should be practical to have equipment large enough to handle these segments at the face of the tunnel.
The example will follow through using 5 feet as the length of the lining along the length of the tunnel. As such input paramters including section properties, spring constants and loads will be based on a 5 foot length of lining being designed.
Determine Model Input Data
This section illustrates the development of the data required by most general purpose structural analysis programs. This type of program is required for the beam spring analysis used in this design example. Note that paragraph 4.4 of the AASHTO LRFD specifications describes the acceptable methods of structural analysis. The computer model used in this example for the analysis utilizes a matrix method of analysis which falls into the classical force and displacment category listed in paragraph 4.4. Paragraph 4.5 of the AASHTO LRFD specification describes the mathematical model requirements for analysis. This paragraph states that the model shall include loads, geometry and material behavior of the structure. The input required for these lements will be described below and include the calculation of loads, joint coordinates, the magnitude of the load at each joint, the modulus of elasticity of the concrete and the cross sectional area and moment of inertia of the liner segments.
Paragraph 4.5.1 of the AASHTO LRFD specifications also says that the model shall include the response characteristics of the foundation where appropriate. Since the surrounding ground is an integral part of the structural lining, the response characteristic of the ground is modeled by the springs installed in the model.
Calculate Joint Coordinates:
Joint coordinates are calculated along the centroid of the lininig segments. In order to calculate the joint coordinates for the initial analysis runs, a lining thickness must be assumed. If the lining thickness changes as a result of the design process, the analysis should be re-run using the parameters associated with the revised lining thickness. This process continues until the lining thickness will support the loads effects from the analysis.
Assume a lining thickness = 16" Radius to centroid of the lining (ro) = 18.17 ft
Joint coordinates are claculated as:
Y coordinate = ro x sinα X coordinate = ro x cosα See Figure 2
In order to keep the model mathematically stable, use a chord length between joint coordinates approximately equal to 1.5 times the thickness of the liner. See paragraph 10.? of the manual.
For a radius ro = 18.17 ft the angle subtended by chord length of c = 2sin-1(c/2ro)
For chord length = 2.00 ft subtended angle = 6.31 degrees
Number of joints = 360 / 6.31 = 57 say 72 joints at 5 degrees between joints.
72 joints was selected to provide analysis results at the invert, crown and springlines.
Tabulation of Joint Coordinates at the centroid of the lining:
|Joint||α (deg)||x (ft)||y (ft)||Joint||α (deg)||x (ft)||y (ft)|
Figure 10E-2 shows the arrangement of joints and members for the computer model.
Joints & Members - Computer Model
Calculate Spring Constants
The subsurface investigation revealed that the tunnel alignment traverses a very stiff clay.
The modulus of subgrade reaction of the clay supplied by the subsurface investigation program is 22 kcf.
Spring constants can be determined based on tributory projections on the x and y axis of each joint or alternately, if the analysis software being used supports the use of radial springs, then all spring constants will be the same. The following formulas can be used to to determine spring constants.
For orthoganal springs:
Spring constant in the Y direction = Ks(Xn + Xn+1)/2
Xn = |(xn - xn+1)|
Xn+1 = |(xn+1 - xn+2)|
Spring constant in the X direction = Ks(Yn + Yn+1)/2
Yn = |(yn - yn+1)|
Yn+1 = |(yn+1 - yn+2)|
In the above equations:
The coordinates for joint N = (xn, yn)
The coordinates for joint N+1 = (xn+1, yn+1)
The coordinates for joint N+2 = (xn+2, yn+2)
Figure E10-3 is a graphic representation of the above calculations of orthogonal spring constants.
Spring Constant Computation
The above computation for orthogonal spring constants uses the coordinates of the joints calculated as input for the model. Since these joints lie along the centroid of the lining and not the outside face where the contact with the surrounding ground occurs, the spring constants calculated using this method should be modified to more closely approximate the resistance provided by the surrounding ground. The modification factor would be the ratio of the outside radius to the radius at the centroid. For this example, the modification factor would be calculated as follows:
Radius at centroid = rc = 18.17 ft
Radius at outside face = ro = 18.83 ft
Modification factor = ro / rc = 18.83 / 18.17 = 1.04
For radial springs, since a one foot length of tunnel is being modeled, the computation of the tributary area for each joint is the same and is the length of the arc between joints.
This tributary area can be calculated as πroα/180
ro = Radius to the outside face of the lining = 18.83 ft
α = Angle subtended between joints
It is important to use the outside radius of the tunnel when calculating spring constants since this is the face that is in contact with the surrounding ground.
For this example, the tributary area = 3.14159 × 18.83 × 4 / 180 = 1.31481 ft2
|Clay||Es (kcf)||Radial Spring Constant (k/ft)|
|Gneiss||4000||5259.3||Run analysis using the values shown for Gneiss and Marble to bracket the actual ground conditions.|
When running the computer model, only springs that are in compression are considered active. A spring is in compression if the joint displacements at the location of that spring indocate movement away from the center of the tunnel. Joint displacements toward the center of tunnel indicatemovementaway from the ground and the spring at that location should not be active in the model. The analysis is performed with an initial assumption of active and non active springs. The results of the analysis, specifically the joint displacements are examined to determine if the spring assumptions correspond with the output values. If the correspondence does not match, then the asumptions for the springs is adjusted and the analysis re-run. This procedures continues until a solution is obtained where the input values for the springs matches the output values for the joint displacements.
Many computer programs will perform this iterative process automatically. For programs that do not support an automatic adjustment, it is useful to model the springs as orthoganl springs. Modeling the sprngs this way makes it easier to determine if a joint is moving toward or away from the center of the tunnel since each component of the meovement (x and y) can be examined seperately and the direction of the movement ascertained by inspection. When using orthogonal springs, each spring component is adjusted separately.
Calculate Liner Section Properties
Segment Thickness = 16 in Segment Length = 5 ft = 60 in
As described in section 10.?? the joints in the liner segments will act to reduce the stiffness of the ring.
Formula for reducing stiffness is as follows: Ie = Ij + I*(4/n)^2 (Formula 10 - ??)
where Ie is modified I
n is number of joints (more than 4)
Ij is joint stiffness - conservatively taken as zero
Unmodified Moment of Inertia = 60 × 163 / 12 = 20480 in4
Number of Joints = 9
Reduced Moment of Inertia = 20480 × ( 4 / 9 )2 = 4045.4 in4
Segment Area = (16.0 / 12 ) × 5 = 6.67 ft2
Assume concrete strength = 5000 psi
AASHTO LRFD specification paragraph 126.96.36.199 provides the method for the calculation of the modulus of elasticity.
Ec = 33,000K1wc1.5√fc'
K1 = 1.0
wc = 145 pcf (AASHTO LRFD specification Table 3.5.1-1)
fc' = 5000 psi
Ec = 4074281 psi
Poisson's Ratio is given in AASHTO LRFD specification paragraph 188.8.131.52 as 0.2.
The soil load and the hydrostatic pressure are applied to the outside face of the tunnel lining. The structural model is built at the centroid of the lining. Therefore, the surface area to which the rock and hydrostatic loads are applied is larger than the surface area along the centroid the model. The surface area at the location of the centroid is directly proportional to the surface area at the outside face in the ratio of the radius of the outside face to the radius at the centroid. To account for this difference between the modeled area and the actual area and to include the full magnitiude of the applied loads, multiply the rock and hydrostatic loads by the ratio of outside radius to centroidal radius.
Radius to Centroid (rc) = 18.17 ft
Radius to Outside face (ro) = 18.83 ft
Multiply Loads Applied to Outside of Tunnel by ro/rc: 18.83 / 18.17 = 1.03
Calculate Hydrostatic Loads:
Hydrostatic head at the tunnel invert = 40 ft = 2.50 ksf
Hydrostatic Load from ground water is applied to the outside of the tunnel.
Value at the invert = 2.50 ksf
Applied amount = 2.50 × 1.037 × 5 = 12.94 ksf Where 5' is the length of the segment
The water pressure magnitude at each joint is calculated based on the distnace of the joint from the invert:
Magnitude of the hydrostatic pressure at joint j = [Value at invert - |(yinvert - yj)| x 62.4] x ro/rc x segment length
yinvert = the y coordinate of the joint at the tunnel invert
yj = the y coordinate of the joint at which the hydrostatic pressure is being calculated
Since the hydrostatic pressure is applied perpendicular to the face of the tunnel, it may be necessary or convenient, dependening on the software being used, to calculate the horizontal and vertical components of the hydrostatic pressure at each joint. This value can be calculated at joint j as follows.
X component of Hydrostatic Pressure at joint j = Magnitude at joint j times cos(αj)
Y component of Hydrostatic Pressure at joint j = Magnitude at joint j times sin(αj)
Figure 10E-4 is the hydrostatic pressure loading diagram and also includes a depiction of j and α.
Hydrostatic Pressure Loading Diagram
Tabulation of Hydrostatic Pressure Input Loads
|Joint||α (deg)||x (ft)||y (ft)||Magnitude (ksf)||X Component (ksf)||Y Component (ksf)|
Calculate Earth Loads
Roof Load = 4.55 ksf
Applied Load = 4.55 × 5 × 1.04 = 23.58 ksf
The horizontal load is given as 1.0 times the vertical load = 23.58 ksf
This load is applied vertically to the lining members. Care should be taken in the input of this load to be sure that it is modeled correctly. The total applied load should be equal to the Roof Load times the Outside Diameter of the Tunnel times the Length of the Segment. Figure 10E-5 shows the loading diagram for this load.
Rock Loading Diagram
Appurtenance Dead Load
For this example, the appertenances consist of the jet fans, the drainage system and the roadway slab. The jet fans and the roadway slab are considered as DC loads or the drainage syatem is considered a DW load as given in paragraph 3.3.2 of the AASHTO LRFD specifications.
Jet fan load consists of dead load and a dynamic allowance for when the fan starts operation. The dynamic allowance does not need to be treated separately from the dead load. The total anticipated load from the jet fans is 2,000 pounds applied vertically.
Using figure 1, the jet fan load is applied at a location that is 6'-0" on either side of the center line of tunnel. Assume that the supports for the jet fan lie 1'-0" on either side of the centerline of the jet fan. Apply the load as a joint load to the joints that x coordinates are closest to ±2.00 and ±4.00. The load applied at each of these joints will be one half of the jet fan load shown above. For this example, the loads will be applied at joints 15, 16, 22 & 23.
The drainage system consists of a 6" diameter standard weight steel pipe. Conservatively assume that the pipe is full of water to calculate the dead load.
Pipe weight = 18.97 plf
Inside Diameter = 6.065 in
Inside Area = 6.065 × 3.14159 / 2.00 = 9.53 in2
Weight of water in pipe = 9.53 / 144 × 62.4 = 4.13 plf
Load Applied to Liner = (18.97 + 4.13 ) × 5 = 115.49 pounds
The pipe weight will be applied at the end of the roadway slab. Referring to Figure 1 shows that the intersection of the center of roadway slab and the tunnel wall is located approximately at approximately 9.2 feet below the center of the trunnel. (Assuming a 15" thickness for the roadway slab.) Therefore, in this model the drainage system load can be applied at joints 42 and 68 to approximatel the effect of this load.
The roadway slab consists of three components, the slab, the vertical center support and the barrier/walkway shapes.
Slab: Assume thickness of roadway slab and center support = 15 in
The intersection of the center roadway slab and the tunnel wall is located approximately 9.2 feet below the center of the tunnel. Therefore in this model, the slab load should be applied at joints 42 and 68 to approximate the effect of this load.
The approximate length of the roadway slab would be the distance between joints 42 and 68 = 32.93 ft
Weight of roadway slab = 1.25 × 150 × 32.93 × 5 = 30871 lbs
Since the roadway slab is continuous and supported in the center, assume that 40% of this load is applied at the side walls and 60% is applied at the center support.
Load applied to the side walls = 30871.1 × 0.20 = 6174 lbs
Load applied to center support = 30871.1 × 0.60 = 18523 lbs
Weight of center support = 1.25 × 150 × 7.50 × 5 = 7031 lbs
Total load from center support = 18523 + 7031 = 25554 lbs
Because fo the invert slab, the load from the center support will be distributed over several joints. Apply this load to joints 51 to 59.
Live load from the roadway slab will be the result of the application of the design truck or design tandem coincident with the lane load as per paragraph 184.108.40.206 of the AASHTO LRFD specifications. The minimum spacing of the truck load axles is 14'. The maximum truck axle load is 14'. This means with a 5' long segment, only one truck axial can be on a ring at any time. The maximum truck axle load is 32 kips. The tandem axles are spaced at 4'-0" and weigh 25 kips each. Using the 4-foot spacing, both tandem axles for a total of 50 kips can be on a single ring at a time. Therefore, use the tandem axle arrangement for this example.
The dynamic load allowance (IM) for the limit states used in the tunnel of tunnel linings (i.e., all limit states except fatigue and fracture) is given in AASHTO LRFD specifications in Table 3.6.2.-1 as 33%. The dynamic load allowanc is applied only to the design tandem and not to the lane load. The computation of the live load effect then is as follows:
Live Load Case 1 - One Traffic Lane:
50.000 kips × 1.33 × 1.20 = 79.8 kip
0.640 klf × 5.00 × 1.20 = 3.84 kip
Total: 83.64 kip
Where the value of 1.20 is the Multiple Presence Factor (m) given in the AASHTO LRFD specifications in Table 220.127.116.11.2-1
Where the value of 5.00 is the length of a single ring.
Assign 40% of this value to joint 42 and 60% of this value to joints 51 to 59.
Load applied at joint 42 = 33.5 kip
Load applied to each of joints 51 to 59 = 5.6 kip
Live Load Case 2 - Two Traffic Lanes:
50.000 kips × 1.33 × 1.00 = 66.5 kip
0.640 klf × 5.00 × 1.00 = 3.2 kip
Total: 69.7 kip
Where the value of 1.00 is the Multiple Presence Factor (m) given in the AASHTO LRFD specifications in Table 18.104.22.168.2-1
Where the value of 5.00 is the length of a single ring.
Assign 40% of this value to joint 42 and 68 and 60% of this value to joints 51 to 59.
Load applied at joints 42 and 68 = 27.9 kip
Load applied to each of joints 51 to 59 = 4.6 kip
The following table represents the load combinations associated with the limit states to be investigated and the associated load factors. These load cases were entered into the structural analysis software to obtain the results that are presented below.
The designation 1 & 2 in the above table indicates the number of live load lanes.
Design will be performed for the following load cases:
- Maximum moment (Mmax) and associate axial load (P).
- Maximum axial load (Pmax) and associated moment (M).
- Maximum shear (Vmax).
The following are the results:
Mmax = 367.1 ft-kip
Pmax = 1496.1 kip
Vmax = 93.5 kip
P = 524.1 kip
M = 173.2 ft-kip
Strength IIa1 Joint 19
Strength IIa1 Joint 38
Strength IIa1 Joint 15
Appendix G Segmental Concrete Design Example
Design Process Calculations
References: AASHTO LRFD Bridge Design Specifications, 3rd Ed., 2004
Data: Segmental lining dimensions:
Segment Length = 5.00 ft
Lining Thickness = 1.33 ft
1. Structure Design Calculations
1.1 Concrete Design Properties:
|AASHTO LRFD Reference|
|Es =||29000 ksi||22.214.171.124|
|fy =||60 ksi|
|f'c =||5 ksi|
|γc=||145 pcf||Table 3.5.1-1|
1.2 Resistance Factors
AASHTO LRFD Reference 126.96.36.199
Flexure = 0.75 (ø) varies to 0.9
Shear = 0.90
Compr. = 0.75
1.3 Limits for Reinforcement
AASHTO LRFD Reference 188.8.131.52
For non-prestressed compression members, the maximum area of reinforcement is given by AASHTO LRFD Specification Equation 184.108.40.206-1 as:
For non-prestressed compression members, the minimum area of reinforcement is given by AASHTO LRFD Specification Equation 220.127.116.11-3 as:
- As = Area of nonprestressed tension steel (in2)
- Ag = Gross area of the concrete section (in2)
- fy = Specified yiled strength of the reinforcing bars (ksi)
- f'c = Specified compressive strength of the concrete (ksi)
2. Check for One Lining Segment
2.1 Following a Design calculation check will be performed for one lining segment:
2.2 Slenderness Check (LRFD 18.104.22.168):
k = 0.65
lu = 5.00 ft = 60 in
d = 1.33 ft = 16.0 in
I = 4096 in4
r = 4.62 in
β1 = 0.85
ds = 13.75 in
d's = 2.25 in
#8 bar dia. = 1.00 in
From analysis output:
where M1 = 58.8 kip-ft
M2 = 67.5 kip-ft
P1 = 2864.9 kip
P2 = 2864.9 kip
Where M1 and M2 are smaller and larger end moments
2.3 Calculate EI (LRFD 22.214.171.124):
2.4 Approximate Method (LRFD 126.96.36.199.2)
The effects of deflection on force effects on beam-columns and arches which meet the provisions of the LRFD specifications and 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 188.8.131.52 of LRFD. For all other cases, Pe shall be taken as:
lu = unsupported length of a compression member (in)
k = effective length factor as specified in LRFD Article 184.108.40.206
E = modulus of elasticity (ksi)
I = moment of inertia about axis under consideration (in4)
Pe = 44992.35 kip
From LRFD section 220.127.116.11.2b:
(The components for sidesway will be neglected. Bracing moment will not include lateral force influence)
The factored moments may be increased to reflect effects of deformations as follows:
LRFD eq. (18.104.22.168.2b-1):
Mc = δb*M2b + δs*M2s
Mc = 68.89 kip-ft
Mu = 67.50 kip-ft
where M2b = 67.50 kip-ft
δb = 1.020656
Pu= factored axial load (kip)
Pe= Euler buckling load (kip)
M2b= moment on compression member due to factored gravity loads that result in no appreciable sideway calculated by conventional first-order elastic frame analysis; always positive (kip-ft)
Φ= resistance factor for axial compression
Pu = 2864.9 kips
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 = 0.95
M1= smaller end moment
M2= larger end moment
Factored flexural resistance:
(From LRFD section 126.96.36.199.1)
The factored resistance Mr shall be taken as:
Mn= nominal resistance (kip-in)
Φ = resistance factor
The nominal flexural resistance may be taken as:
Do not consider compression steel for calculating Mn
Mn = 3754.15 kip-in
Mn = 312.85 kip-ft
Φ = 0.9
ΦMn = 281.56 kip-ft => OK
Mr= 281.56 kip-ft Mr > Mc
As= area of nonprestresses tension reinforcement (in2)
fy= specified yield strenght of reinforcing bars (ksi)
ds= distance from extreme compression fiber to the centroid of nonprestressed tensile reinforcement (in)
a= cβ1; depth of equivalent stress block (in)
β1= stress block factor specified in Article 188.8.131.52 of LRFD
c= distance from the extreme compression fiber to the neutral axis
As= 6.0 in2
fy= 60.0 ksi
f'c= 5.0 ksi
b= 12.0 in
c = 8.30 in
a = β1×c
a = 6.64 in
Create interaction diagram
Asmin = 10.8 in2
Asprov (total) = 12.00 in2 Choose #7 at 6 both faces
Es = 29000 ksi
β1 = 0.85
Yt = 8 in
0.85×f'c = 4.25 ksi
Ag, in2 = 960 in2
As = A's = 6.0 in2
At zero moment point From LRFD eq. (184.108.40.206-2):
Po = 0.85 × f'c ×(Ag - Ast ) + Ast × fy
Po = 4415 kip
ΦPo = 3311 kip
Φ = 0.75
At balance point calculate Prb and Mrb
cb = 8.25 in
ab = 7.01 in ab = β1 × cb
f's = 63 ksi
f's Es [(0.003/c)×(c-d')]
f's > fy; set at fy
Acomp = 420.75 in2 Acomp = c × b
y' = a/2 = 3.50625 in
ΦPb = Φ[0.85 × f'c × b ×ab + A's × f's - As × fy]
ΦPb = 1341 kip
ΦMb = 9046 kip-in
ΦMb = 754 kip-ft
At zero 'axial load' point (conservatively ignore compressive reinforcing)
a = 0.3 in
ΦMo = 3674.4 kip-in
ΦMo = 306 kip-ft
At intermediate points
|a, in||c = a/β1||Acomp, in2||f's, ksi||fs, ksi||fy, ksi||ΦMn, k-ft||ΦPn, kips|
Φ may decrease from 0.90 to 0.75 as "a" increases
Note: from 0.0 to ab. Use 0.75 to be conservative.
Acomp = a ×60 in2
f's = Es × (0.003/c) × (c-A's) ksi
fs = Es × (0.003/c) × (c-As) ksi
ΦPn = Φ(Acomp - A's) × 0.85 × f'c + A's × f's - Asfy kips
3. Shear Design (LRFD section 220.127.116.11)
The nominal shear resistance, Vn shall be determined as the lesser of:
LRFD eq. 18.104.22.168-1:
Vn = Vc + Vs
LRFD eq. 22.214.171.124-2:
or Vn = 0.25 × f'c × bv × dv
NOTE: Vp is not considered
For slab concrete shear (Vc), refer to LRFD Section 5.14.5:
As= area of reinforcing steel in the design width (in2)
de= effective depth from extreme compression fiber to the centroid of the tensile force in the tensile reinforcement (in)
Vu= shear from factored loads (kip)
Mu= moment from factored loads (kip-in)
b= design width (in)
bv= effective web width taken as the minimum web width within the depth dv (in)
dv=effective shear depth taken as the distance, measured perpendicular to the neutral axis (in)
Av= area of shear reinforcement within a distance s (in2)
s= spacing of stirrups (in)
dv = 0.9×de or 0.72×h (LRFD section 126.96.36.199)
dv = 12.38 in
de= 27.75 Av = 0 in2
Vu×de/Mu = 6.68
Use Vu×de/Mu = 1.00
s = 12 in
Max. shear and associated moment from analysis output:
Vu = 32.8 kip
Mu = 67.5 kip-ft
Vc = 80.14 kip
or Vc = 46.49 kip Controls
Vs = 0.00 kip
Vn = 46.49 kip Vn = 185.63 kip
therefore Vn = 46.49 kip
ΦVn = 41.84 kip > Vu OK
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