reinforced concrete design of pedestrian bridge

Structural Concrete Design Pedestrian Bridge Crossing University San Diego California

Sycamore Canyon Pedestrian Bridge

Brad Wilton & Pedro Mercado III SE151A Term Project 03/21/2014


Structural Concrete Design

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Structural Concrete Design <("(5 $2.7:7J' O28',: ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( 5N <(5 6'.7&, *2+ 9G'%+ ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( 5N N( ?7+4'+ 6'.7&, ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( 5N N(" P='S3+' 6'.7&, 2* B+7:7-%= 9'-:72,. 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1. References 1.1 Background This report looks at a design of a two span pedestrian bridge which is to be constructed in East County San Diego. The geometry of the bridge calls for a span covering a 5 lane road way, which runs along a valley. The concrete structure of the bridge shall be designed in accordance to the American Concrete Institute Code (ACI 1318), along with special design criteria proposed by the owner. It is due to these constraints that the elevation view of the bridge, with proper dimensions, will be as shown in Figure 1.

The design envelope consists of two 50 ft. spans supported at the abutments by neoprene bearings, then two 2 ft. sections of overhang. The mid-span of the bridge is supported by a bent-cap which is connected to a cast-in-steel column. This column is 22 ft. vertically and sits at 4 ft. off-center with a 6 ft. deep pile. The diameter of the circular column for this particular design will be 4’-6” across. The cross section of the bridge is a total of 16’-8’’ wide, consisting of long box girders across the entire span of the structure which can be seen with full dimensions in Figure 2. The height of the box girder will be determined in the design. The bridge will be designed to support the self-weight, the weight of the 4 in asphalt overlay (101 pcf), handrails (68.5 lbs/ft), and a live load of 86 lbs/ft 2. Live loads will be applied to generate the worst case positive and negative forces. Due to the fact that this is a preliminary design, torsional effects will not be considered in this report. The materials for this project is normal weight concrete (150 pcf) with a specified compressive strength of 5.0 ksi. ASCE A706 Grade 60 reinforcing steel will be used for all reinforcement in this structure.

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Figure 2: 2: Girder Cross Section with Slab idealized as simply supported beam on center lines The analysis of the bridge will be done using various combinations of multiple loading conditions as follow: self-weight, asphalt weight, handrail weight, and multiple live loads at multiple locations throughout the structure and applied using SAP2000. **All dimensions in figures and tables are listed in inches unless otherwise stated.

2. Dimensions and Section Properties 2.1 Slab As seen in Figure Fig ure 2, the slab is idealized as a simply supported s upported beam with supports lying lyin g on the center lines of the girders. ACI 1318 Table 9.5(a) lists minimum mi nimum thickness thickne ss is as follows: Both ends continuous = L/28

Cantilevered = L/10

Section AB & CD are cantilevered and BC has both ends continuous Using the dimensions from Figure 2 the thickness for each section is calculated.

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The minimum requirement for section BC is 8 in., therefore BC shall be 8 in. rather than 3.214 in. Additionally, Additionally, AB & CD are tapered to 6 in., which is greater than than the minimum calculated calculated 5.5 in., so 6 in will be used.

Figure 3 Assumed 3 Assumed Dimensions of Slab For simplicity, and to be conservative, assume that the whole depth of the slab is 8 in. as seen in Figure 3. The sectional properties of the slab are tabulated next.

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2.2 Girder ACI 1318 Table 9.5(a) lists minimum mi nimum thickness thickne ss for one ends continuous contin uous as L/28. Using this calculation the height of the girder is calculated.

Figure 4 Sections 4 Sections of Girder For Centroid Calculations

Figure 5 Dimensions 5 Dimensions of Girder Table 1 Centroid 1 Centroid Calculations

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The moment of inertia is calculated by dividing the girder into cross sections that can be seen in Figure 4. Using the parallel axis theorem and dimensions from Figure 5, each section is calculated and then added up to find the total moment of inertia. The centroid is calculated by dividing A*y dividing A*y by total area. All dimensions dimensi ons are referenced from the bottom of the girder. All of these calculations can be seen below:

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2.3 Column For the column supporting the bridge at the center, there were two options for the diameter of the column. The diameters of 3’-6” or 4’-6” are available. For this bridge, we will be using 4’-6” columns. The sectional properties of this column can be seen below:

3. Design Loads 3.1 Slab

Figure 6 Conservative 6 Conservative Dimensions Used for Slab Design 3.1.1 Dead Load (concrete)

Figure 6 Idealized 6 Idealized Slab for Dead Load Calculations

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To be conservative, assume that the slab does not taper to 6 in. at the ends, but rather is a constant 8 in. across the entire span as seen in Figure 6. We also assume a 1 ft. cross-section on all sections for the slab as seen in Figure 7. Multiplying the depth and thickness of the slab by the density of concrete gives the distributed load for the slab.

3.1.2 Superimposed Dead Load (asphalt & handrails) The asphalt is a constant 4 in. thick as seen in Figure 6. Multiplying the depth and thickness of the asphalt by the density gives the distributed load for asphalt.

The distributed load for the asphalt spans the slab except the last 12 in. on either side. There are two handrails across the span of the bridge that can be seen in Figure 7 that can be treated as point loads. Once again, assume a 1 ft. cross-section.

3.1.3 Live Load ACI 1318-11 lists lis ts the live load (LL) for pedestrian pedestri an traffic as 86 lbs lb s per square foot. Assume a 1ft cross-section and multiply the live load by the depth of the slab.

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All the following figures are for the th e load cases of the slab and are ar e all listed in lbs or lbs/ft.

Figure 8 Concrete 8 Concrete Self Weight

Figure 9 Slab 9 Slab Asphalt Self Weight

Figure 10 Handrail 10 Handrail Weight

Figure 11 Live 11 Live Load Applied Everywhere

Figure 12 Live 12 Live Load Applied in Middle

Figure 13 Live 13 Live Load Applied to Outer Sections

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3.2 Girder In all cases for the girder assume 1 ft. cross sections as seen in Figure 14.

Figure 14 Girder 14 Girder Cross Section for Load Calculations 3.2.1 Dead Load (Concrete & Plug) Dead Load (Concrete) The area of the girder was previously calculated in Table 1. Multiplying the cross-sectional area of the girder by the density of concrete gives the distributed load for the concrete.

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Dead Load (Plug/Bent Cap) Calculating the volume of the plug, and multiplying by the density of the concrete, gives the weight of the plug that is treated as a point load in the center of the bridge.

3.2.2 Superimposed Dead Load (Asphalt & Handrails) Asphalt The asphalt is a constant 4 in. thick across the entire span of the bridge as previously stated. Figure 3 shows the asphalt dimensions on top of the girder. Using these distances, the distributed load for asphalt is calculated below:

Handrails There are two handrails across the entire span of the bridge. The load must be multiplied by two and is show below:

3.2.3 Live Load The live load can only occur inside the handrails. Figure 3 shows where on the slab this can occur. The distributed load due to live load is calculated below:

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All the following figures for the load loa d cases of the slab sla b are listed in lbs or lbs/ft.

Figure 15 Concrete 15 Concrete Self Weight

Figure 16 Weight From Bent Cap

Figure 17 Asphalt 17 Asphalt Self Weight W eight

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Figure 18 Handrail Self Weight

Figure 19 Live Load Everywhere

Figure 20 Live Load On Left Side & Alternative Span

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Figure 21 Live Load On Right Side & Alternative Span

4. Structural Analysis 4.1 Slab 4.1.1 Model Figure 22 references the dimensions of how SAP2000 was modeled with the assumptions that were used. The red dotted line represents the elements in SAP2000 and the blue dots indicate a node that connects each element. The slab was treated as a simply supported beam across the top of the girder.

Figure 22 Idealizing the Slab for SAP2000

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4.1.2 Load Combinations The following load combinations were used in SAP2000 to get the design envelope. Combo1 = 1.4(!self weight + !asphalt + !handrail) Combo2= 1.2(!self weight + !asphalt + !handrail) + 1.6(!liveload whole span) Combo3= 1.2(!self weight + !asphalt + !handrail) + 1.6(!live load on left) Combo4 =1.2(!self weight + !asphalt + !handrail) +1.6(!live load on right) Envelope= (Combo1 + Combo2 + Combo3 + Combo4) 4.1.3 Bending Moment Diagrams and Load Combinations

Figure 23 Combo 1

Figure 24 Combo 2

Figure 25 Combo 3

Figure 26 Combo 4

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4.1.4. Shear Force Diagrams of Load Combinations

Figure 27 Combo 1

Figure 28 Combo 2

Figure 29 Combo 3

Figure 30 Combo 4

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4.1.5 Design Envelopes

Figure 31 Shear Envelope

Figure 32 Moment Envelope

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4.2 Girder 4.2.1 Model Figure 33 references the dimensions of how SAP2000 was modeled with the assumptions that were used. The red dotted line represents the elements in SAP2000 and the blue dots indicate a node that connects each element. The bridge bearings were treated as rollers, and the base of the column was fixed.

Figure 33 Idealizing the Bridge for SAP2000 4.2.2 Load Combinations The following load combinations were used in SAP2000 to get the design envelope: Combo1 = 1.4(!self weight + !asphalt + !handrail) Combo2= 1.2(!self weight + !asphalt + !handrail) + 1.6(!liveload whole span) Combo3= 1.2(!self weight + !asphalt + !handrail) + 1.6(!live load on left) Combo4 =1.2(!self weight + !asphalt + !handrail) +1.6(!live load on right) Envelope= (Combo1 + Combo2 + Combo3 + Combo4) 4.2.3 Bending Moment Diagrams and Load Combinations

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Figure 34 Combo 34 Combo 1




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4.2.4 Shear Force Diagrams of Load Combinations




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Figure41 Combo Figure41 Combo 4

4.2.5 Design Envelopes

Figure 42 Shear 42 Shear Envelope

Figure 43 Moment 43 Moment Envelope

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5. Slab Design 5.1 Flexure Design of Critical Sections 5.1.1 Negative Moment

The bending moment can be observed from the design envelope. The maximum moment is approximately 2900 lb-ft for negative bending. A MATLAB function functi on was written for fiber fibe r discretization and was run to determine de termine the design de sign moment envelope for the slab with differing bar numbers and spacing. A copy of this function can be seen in the appendix. This code also checks to make sure that each design passes 1.2*M cr. A summary of some of these outputs is shown below in table 2. Table 2

The most economical design that passes these internal checks was selected for the design. It was determined that #4 bars should be used every 8 in. center-center with the design capacity listed below:

Another design check is to make mak e sure the design desi gn capacity is i s greater than 1.2 1. 2 times the cracking cra cking moment. For this project, 5000 psi concrete will be used. The elastic section modulus was previous calculated. Below is the equation for cracking moment:

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Because the capacity is greater than both 1.2 times the cracking moment (M cr), and the maximum moment (Mu), this design will be sufficient. We can check the design capacity by hand to verify the MATLAB function works as seen below:

We can see that the capacity is very close to the beam discretization and passes all design checks. #4 bars every 8 in. center to center is sufficient for for this design.

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5.1.2 Positive Moment The maximum positive bending very small, only 254 lb-ft. We will mirror the longitudinal reinforcement for negative bending on either side of the slab to be conservative in the design, as well as make construction simpler.

5.2 Design for Shear From the design envelope, the maximum shear on the slab at the face of the web is 944 lbs.

Since ultimate shear along the slab is less than the one half the shear capacity of the concrete, no shear reinforcement is needed in the slab.

6. Girder Design 6.1 Flexure Design of Critical Sections 6.1.1 Negative Moment The maximum negative bending moment can be observed from the design envelope. The maximum negative moment is 1667 kip-ft at the face of the bent cap. 6.1.1.1 Effective Width Negative Bending The effective width is 100 in. The total length of the top flange is 200 in. No portion needs to be ignored for the fiber discretization.

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A MATLAB function functi on was written for fiber fibe r discretization and was run to determine de termine the design de sign moment envelope for the slab with differing bar numbers and spacing. A copy of this function can be seen in the appendix. This code also checks to make sure that each design passes 1.2*M cr. A summary of some of these outputs is shown below in table 3. Table 3

The most economical design that passes these internal checks was selected for the design. It was determined that 14 #9 bars should be used with 10 of those bars in the top layer and 4 bars below the others. This will make it easier later to try and remove some of the bars. The design capacity for this case is listed below:

The output from the MATLAB beam discretization algorithm can be checked by the following equation which is a rough estimate for the total capacity:

This approximation is a conservative estimate for the capacity. As you can see the capacity is slightly lower than the beam discretization value, but is close enough to show the discretization program is accurate. Another design check is to make mak e sure the design desi gn capacity is i s greater than 1.2 1. 2 times the cracking cra cking moment.

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Because the capacity is greater than both 1.2 times the cracking moment (Mcr), and the maximum moment (Mu), this design will be sufficient. 6.1.2 Positive Moment The maximum positive bending moment can be observed from the design moment envelope at 1280 kip-ft in the middle of the span on the right. 6.1.2.1 Effective Width Positive Bending The effective width is 50 in. The total length of the bottom flange is 104 in., so 4 in. can be ignored for the fiber discretization.

A MATLAB function functi on was written for fiber fibe r discretization and was run to determine de termine the design d esign moment envelope for the slab with differing bar numbers and spacing. A copy of this function can be seen in the appendix. This code also checks to make sure that each design passes 1.2*M cr. A summary of some of these outputs is shown below in table 4. Table 4

The most economical design that passes these internal checks was selected for the design. It was determined that 10 #9 bars should be used with 6 of those bars in the bottom most layer, and 4 bars above the others. This will help later to give the option of removing some of the reinforcement. The design capacity for this case is listed below:

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The output from the MATLAB beam discretization algorithm can be checked by the following equation which is a rough estimate for the total capacity.

Another design check is to make mak e sure the design desi gn capacity is i s greater than 1.2 1. 2 times the cracking cra cking moment.

Because the capacity is greater than both 1.2 times the cracking moment (Mcr), and the maximum moment (Mu), this design will be sufficient.

6.2 Design for Shear From the shear design envelope the maximum shear value along the slab is 182 kips at the bent cap.

From the above relationships for shear, we can see that we have shear in all three zones. To be conservative, and for ease of construction, we will design for Type 3 across the entire length of the bridge.

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Bar spacing for type 3 is as follows:

We need to round down, for simplicity let’s use 10 inches for the stirrup spacing.

For type 3 shear, the calculation for required area of stirrups is used below:

We see that we require 0.556 square inches. We will use 4 #4 bars for our stirrups.

Now we need to check the design strength for 4 #4 bars 8" center to center.

The shear capacity is more than adequate for this section. The design capacity is about 215 kips while the maximimum shear felt on the girder is only 182 kips.

7. Development of Longitudinal Reinforcement and Miscellaneous Requirements 7.1 Slab Reinforcement 7.1.1 Longitudinal Bar Development For negative bending, all distances are from the right or left of web of the box girder. Because there is negative moment over nearly every section of the slab, there will be reinforcement along the top of the entire slab. These calculations are assuming we are looking at left side of the idealized slab in Figure 21. The clear span distance to the left is 48 inches and 76 inches to the right.

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Longitudinal bar cut-off can theoretically happen in the center of the span 34.2 inches from the center of the left web, and 20.1 inches from the center of the right web. The cut-off distance calculations can be seen below: Theoretical cut-off distance to the right of the web

Bar development length to right of the web

Theoretical cut-off distance to the right of the web

Bar development length to right of the web We will ignore cutoff length because the distance from the face of the webs to the center of the span is only 38 inches. Following the development lengths, left and right of the web would connect them in the center of the span anyway.

The bar development length is also a problem because it is 17 inches long. There is moment near the end of the flanges of the slab due to the loading conditions. For this reason, we will detail the bar development length for a hook as follows:

The bar development length for hooks will be 10 inches for ease of construction. Bar development length for the positive bending is the same for negative because it is the same exact design. Since the positive moment ends at the same locations as the theoretical cutoff distances for negative bending, there is more than enough distance to develop the bars for positive bending. For ease of construction, we will extend positive longitudinal reinforcement to the ends of the slab.

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7.1.2 Shrinkage and Temperature For temperature and shrinkage, we will use #4 bars. The equation below is used to determine the spacing:

We will round the spacing down and use spacing of 12 inches center to center.

7.2 Girder Reinforcement 7.2.1 Longitudinal Bar Development 7.2.1.1 Negative Bending For negative bending bar development length, all distances are from the right or left of the bent cap at mid-span of the bridge as seen in Figure 33. The clear spans from the left and right are equal from the column face. We will try to remove approximately 1/3 of the total bars for negative reinforcement. We will go from 14 #9 bars to only having the top 10 #9 bars. The remaining 10 bars will be referred to as "A" bars. The output from the MATLAB beam discretization can be seen below for the "A" bars:

We notice immediately by removing the 1/3 of the bars and only having these “A” bars the capacity is now less than 1.2 times the cracking moment. Because of this, we will not reduce any number of bars throughout the negative reinforcement.

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Theoretical cut-off distance to the right of the column

Bar development length to right of the column

Theoretical cut-off distance to the right of the column

Bar development length to right of the column To be conservative, assume the larger value of 206.5 in. from either side of the bent cap is where we can end our negative bending reinforcement. For constructability, round bar cutoff lengths to 208 in. from either face of the column. 7.2.1.2 Positive Bending Next we will try to remove approximately 1/3 of the total bars for positive reinforcement. We will go from 10#9 bars to only having the top 6 #9 bars. The remaining 6 bars will be referred to as "A" bars. The output from the MATLAB beam discretization can be seen below for the "A" bars:

We notice immediately that by removing the 1/3 of the bars and only having these “A” bars, the capacity is now less than 1.2 times the cracking moment. Because of this we will not remove any bars throughout the negative reinforcement.

Bar development length for positive bending:

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An additional requirement is having a minimum m inimum bearing bea ring clearance from the center of the bearing to the hook for positive bending. The equation is listed below:

The actual bearing clearance is 21.5 in., which is greater than the minimum bearing clearance, as well as the development length for the hook Since we are not reducing any bars in the positive reinforcement, we will continue the reinforcement along the entire bottom section of the girder.

7.2.2 Shrinkage and Temperature For temperature and shrinkage bars, we will use #4 bars. The equation below is used to determine the spacing:

We will round the spacing down and use spacing of 12 inches center to center.

8 Column Design 8.1 Interaction Diagrams A MATLAB code cod e was used to discretize di scretize a column. colum n. For this design we chose a 4’-6” column. col umn. A driver function ran the discretization for the column code for multiple p l ratios which are the

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area of the steel divided by the gross area of the column. This code assumed 8 bars were used for reinforcement to run. The interaction diagram can be seen below:

Figure 44

8.2 Pu ,Mu Combinations The table below lists all the Pu and Mu combinations that are taken from the four different load combinations. For each value of Pu, there is a moment at the top and bottom of the columns. The absolute values are taken for all of the M u values and these values are plotted onto the

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interaction diagrams generated in MATLAB. These values are annotated with black x’s and can be seen in Figure 45. Table 5

Figure 45 P 45 Pu Mu Pairs Plotted on Interaction Diagram

From this Interaction Diagram it is obvious that all P u, Mu pairs lie within the 0.01 range. Below we tabulate the required area and bars needed for the column:

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We will try #11 bars.

Round up to even number of bars . We will use 16 #11 bars in the column design.

8.3 Shear Design Now we need to check the design strength for #4 bars with an unknown spacing. First, we will find the shear strength provided by the concrete.

Next, we need to determine the shear strength provided by the reinforcement. There are two regions along the column: the critical and non-critical regions. The critical region is defined as lo, which is the distance from the bottom and the top of the column and is calculated as follows:

There is a spacing for the critical section, S cr_min, and a different spacing for the non-critical section, Smin. Both of them are defined below: below:

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To be conservative, we will use the Scr_min value which is smaller than the S min distance. Shear spacing for the entire column will be 4” center to center.

8.4 Longitudinal Bar Development The bar development for longitudinal bars in compression is calculated below from ACI 12.3.2:

Therefore, a 22 in. minimum development length is needed from the column into the bottom of the bent cap and into the top of the pile.

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9 Drawings 9.1 Slab

%&'()* +, 9=%> 6':%7= %: P%-' 2* U'>

9.2 Girder

%&'()* +- B+2.. 9'-:72, 2* ?7+4'+ Z7:G )'7,*2+-'8',:

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%&'()* +. ?7+4'+ B+2.. 9'-:72, 6':%7= 2* )'7,*2+-'8',:

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9.3 Full Span

%&'()* +/ ?7+4'+ P3== 9;%, Z7:G )'7,*2+-'8',:

%&'()* 01 6':%7= 2* 0'%+7,& B='%+%,-' 67.:%,-'

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%&'()* 02 6':%7= *2+ T'&%:7J' )'7,*2+-'8',: B3:2**

9.4 Column

%&'()* 03 B2=38, )'7,*2+-'8',: D%M23:

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%&'()* 04 B+2.. 9'-:72, 2* B2=38,

%&'()* 0+ B2=38, 6':%7=

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Appendix A.1 Time Sheet

A.2 MATLAB Functions A.2.1 Discretize Disc retize Girder Function Fu nction function[n,phi_M_N] function [n,phi_M_N] = fiber_discretize_girder_negative(bar_number,N1,N2) f_c=5; %ksi h=33; %inches f_y = 60; %ksi d_b = bar_number/8; %inches A_s1 = N1*(pi*d_b ^2)/4; A_s2 = N2*(pi*d_b ^2)/4; for n for n = 2:100 Beta = 0.85 - (f_c-4)/20; a=n*h/100; y_c=linspace(h/200,h-h/100,100); Area(1:24)=200*h/100; %inches^2 %Effective flange width is only 50" not 52" per the code Area(25:82)=2*14*h/100; Area(83:101)=104*h/100; y=(0:h/100:h); a = n*h/100; %depth of compression center c = a/Beta; si = .003/c; for i=1:101 for i=1:101 strain_steel(i) = si*(h - c - y(i)); end location_bar_bottom = ceil((1.5+0.5+d_b/2)/(h/100)); %Include an extra 1/2 inch for the shear stirrups location_bar_top = floor((8-1.5-d_b/2)/(h/100)); if ((location_bar_top - location_bar_bottom)*.33 < d_b) && ((location_bar_top - location_bar_bottom)*.33 < 1) fprintf('***********************************not fprintf('***********************************not enough bar spacing ==>> go to jail*************************** **********\n' **********\n'); ); fprintf('***********************************not fprintf('***********************************not enough bar spacing ==>> go to jail************************** ***********\n' ***********\n'); ); fprintf('***********************************not fprintf('***********************************not enough bar spacing ==>> go to jail************************** ***********\n' ***********\n'); ); fprintf('***********************************not fprintf('***********************************not enough bar spacing ==>> go to jail************************** ***********\n' ***********\n'); ); 'top layer cannot be in the bottom must go in the web' location_bar_top = ceil(((1.5+0.5+d_b+max(1,d_b)+d_b/2)/(h/100))) %Reassigns top layer if it can't fit in the bottom flange end strain_steel(1:location_bar_bottom-1)=0; strain_steel(location_bar_bottom+1:location_bar_top-1)=0; strain_steel(location_bar_top+1:101)=0; E_steel = 29000; %ksi for i for i =1:101 f_steel(i)= sign(strain_steel(i))*min(abs(strain_steel(i))*E_steel,f_y); %stress of the steel end A_s(1:101)=0; A_s(location_bar_bottom)=A_s1; A_s(location_bar_top)=A_s2; for i=1:101 for i=1:101 F_steel(i)=A_s(i)*f_steel(i); %forces of steel end F_concrete(1:101)=0; for i=101:-1:103-n for i=101:-1:103-n F_concrete(i)=-.85*f_c*Area(i); %forces in concrete end Compression_concrete = sum(F_concrete); Compression_steel = F_steel*(F_steel(:)<0); %sums only the negative values of steel forces Compression = Compression_concrete + Compression_steel; %Total compressive forces Tension = F_steel*( F_steel(:)>0); %sums all the positive forces in the steel

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Structural Concrete Design Error(n) = Tension + Compression; M_n(n) = -(F_concrete(2:101)*y_c' + F_steel(2:101)*y_c')/12; %Moment in kip-ft M_N = (((M_n(n-1) - M_n(n))*(0 - Error(n)))/(Error(n-1) - Error(n)) + M_n(n)); %Interpolation for zero error d_t = h - 3; phi_calc = 0.65 + 0.25*(1/(c/d_t) - 5/3); if phi_calc phi_calc > 0.65 && phi_calc < 0.9 phi=phi_calc; end if phi_calc phi_calc < .65 %these statements say that 0.65 < phi < 0.9 phi=.65; end if phi_calc phi_calc > 0.9 phi= .9; end if Error(n)*Error(n-1) Error(n)*Error(n-1) < 0 %if there is a sign change in error leave the loop break end end phi_M_N= phi*M_N; if n n > 99 fprintf('***********************************not fprintf('***********************************not enough reinforcement ==>> go to jail************************** ***********\n' ***********\n'); ); fprintf('***********************************not fprintf('***********************************not enough reinforcement ==>> go to jail************************** ***********\n' ***********\n'); ); fprintf('***********************************not fprintf('***********************************not enough reinforcement ==>> go to jail************************** ***********\n' ***********\n'); ); fprintf('***********************************not fprintf('***********************************not enough reinforcement ==>> go to jail************************** ***********\n' ***********\n'); ); fprintf('***********************************not fprintf('***********************************not enough reinforcement ==>> go to jail************************** ***********\n' ***********\n'); ); fprintf('***********************************not fprintf('***********************************not enough reinforcement ==>> go to jail************************** ***********\n' ***********\n'); ); phi_M_N=0; end if phi_M_N phi_M_N < 1.2*1149 %M_cr in Kip-ft fprintf('***********************************does fprintf('***********************************does not pass M_cr check ==>> go to jail *************************************\n' ); end end

A.2.2 Discretize Disc retize Column Function function[phi_M_N,phi_P_n] function [phi_M_N,phi_P_n] = fiber_discretize_column(A_s,n) N=8; %# of bars f_c=5; %ksi f_y = 60; %ksi layers=N/2 +1; A_s1=A_s/8; d_b = sqrt(4*A_s1/pi); %inches d= 54; %diameter of colun in inches r=d/2; %radius of column in inches y_c=linspace(d/200,d-d/200,100); yy=linspace(r-r/100,r/100,50); %finds the first 51 centroid points of y for i for i = 1:50 x(i)=(r^2-yy(i)^2)^(1/2); end x_c(1:50)=x; x_c(51:100)=fliplr(x); for i for i = 1:100 Area(i)=x_c(i)*2*d/ 100; end Beta = 0.85 - (f_c-4)/20; y=(0:d/100:d); a = n*d/100; %depth of compression center c = a/Beta;

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Structural Concrete Design si = .003/c; R=r-1.5-.5; vert_dist1 = r-(r-1.5-0.5-d_b/2)*sind(45);%vert_space r-(r-1.5-0.5-d_b/2)*sind(45);%vert_space = (d-2*1.5-2*.5-2*d_b); vert_dist2 = r; vert_dist3 = r+(r-1.5-0.5-d_b/2)*sind(45); for i=1:100 for i=1:100 strain_steel(i) = si*(d - c - y_c(i)); end location_bar_bottom = ceil((1.5+0.5+d_b/2)/(d/100)); location_bar_middle1 = ceil((vert_dist1)/(d/100)); location_bar_middle2 = ceil((vert_dist2)/(d/100)); location_bar_middle3 = ceil((vert_dist3)/(d/100)); location_bar_top = floor((d - 1.5 - 0.5 - d_b/2)/(d/100)); strain_steel(1:location_bar_bottom-1)=0; strain_steel(location_bar_bottom+1:location_bar_middle1-1)=0; strain_steel(location_bar_middle1+1:location_bar_middle2-1)=0; strain_steel(location_bar_middle2+1:location_bar_middle3-1)=0; strain_steel(location_bar_middle3+1:location_bar_top-1)=0; strain_steel(location_bar_top+1:100)=0; E_steel = 29000; %ksi for i for i =1:100 f_steel(i)= sign(strain_steel(i))*min(abs(strain_steel(i))*E_steel,f_y); %stress of the steel end A_s(1:100)=0; A_s(location_bar_bottom)=A_s1; A_s(location_bar_top)=A_s1; A_s(location_bar_midd le1)=2*A_s1; A_s(location_bar _middle2)=2*A_s1; A_s(location_bar_midd le3)=2*A_s1; for i=1:100 for i=1:100 F_steel(i)=A_s(i)*f_steel(i); %forces of steel end F_concrete(1:100)=0; for i=100:-1:101-n for i=100:-1:101-n F_concrete(i)=-.85*f_c*Area(i); %forces in concrete end Compression_concrete = sum(F_concrete); Compression_steel = F_steel*(F_steel(:)<0); %sums only the negative values of steel forces Compression = Compression_concrete + Compression_steel; %Total compressive forces Tension = F_steel*(F_steel(:)>0); %sums all the positive forces in the steel P_n = Tension + Compression; M_n = -(F_concrete*y_c' + F_steel*y_c' + -P_n*d/2)/12; %Moment in kip-ft d_t = d - 3; phi_calc = 0.65 + 0.25*(1/(c/d_t) - 5/3); if phi_calc phi_calc > 0.65 && phi_calc < 0.9 phi=phi_calc; end if phi_calc phi_calc < .65 %these statements say that 0.65 < phi < 0.9 phi=.65; end if phi_calc phi_calc > 0.9 phi= .9; end phi_M_N= phi*M_n; phi_P_n= phi*P_n; end

A.2.3 Generate Genera te Interaction Plot Function Functi on function[phi_M_N,phi_P_n] function [phi_M_N,phi_P_n] = interaction_plots() close all f_c=5; f_y=60; %ksi d=54; %diameter in inches Ag= pi*d^2/4; %Gross Area in inches for i for i = 1:6 Pn_max = -0.8*0.65*(.85*f_c*(Ag-Ag*i/100) + Ag*(i/100)*f_y); %max value for compression for n for n = 1:100

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Structural Concrete Design [phi_M_N(n,i),phi_P_n(n,i)] = fiber_discretize_column(Ag*i/100,n) end end hold all plot(phi_M_N(:,1),phi_P_n(:,1)) plot(phi_M_N(:,2),phi_P_n(:,2)) plot(phi_M_N(:,3),phi_P_n(:,3)) plot(phi_M_N(:,4),phi_P_n(:,4)) plot(phi_M_N(:,5),phi_P_n(:,5)) plot(phi_M_N(:,6),phi_P_n(:,6)) xlabel('phi*M_n' xlabel('phi*M_n'),ylabel( ),ylabel('phi*P_n' 'phi*P_n'),title( ),title('Interaction 'Interaction Diagram Column Diameter 54"') 54"' ) hleg1 = legend('0.01' legend('0.01',,'0.02' '0.02',,'0.03' '0.03',,'0.04' '0.04',,'0.05' '0.05',,'0.06' '0.06'); ); Pu=-[302;302;388;388;323;323;323;323]; set(gca,'YDir' set(gca,'YDir',,'reverse' 'reverse'); ); Mu=[801;423;1028;550;687;683;1027;287]; plot(Mu,Pu,'x' plot(Mu,Pu, 'x')) end

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reinforced concrete design of pedestrian bridge

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