MIDAS USER

INTEGRATED SOLUTION SYSTEM FOR BRIDGE AND CIVIL ENGINEERING

Project Application

All Frame Girder as Plate All Plate

Cross Frame Substructure Modeling with Soil Substructure

Staged Seuencing - Pour Sequence

midas Civil I Bridging your Innovations to Realities

!

I

Moving Load

Daniel MariscalSenior Structural Engineer

URS

Corporation

Tampa, FL

In 2006, I was tasked to evaluate several structural analysis software

packages, and to select the best alternative for our office. Following several

weeks of trials, I selected Midas Civil, and after several years of usage, I am

very pleased with my choice.

Coming from using non-graphical interfaced structural analysis programs, the

selection of Midas Civil was an easy choice. The graphical interface was

powerful, intuitive, and very easy to use. The menu structure and the

maneuverability of the program were simple to learn and fast to get use to.

The capabilities and scope of the program have proven thru the years to be

everything a structural engineer needs by its side to provide an efficient and

effective structural design. The output results are handled via the graphical

interface or spreadsheets. Either way, output is detailed and easy to obtain.

Output and results are handled with a powerful graphical interface where

deformations, forces and stresses are viewed on any direction or thru a cross

section cut. The program can be described with one word: amazing.

Excellent customer technical support is as important as the software itself,

and Midas has provided thru the years excellent support. The support

personal are knowledgeable not only of the software itself, but of

engineering in general. Support answers to my technical questions have

been prompt, honest and always courteous.

Frank BlakemoreDesign Manager

US 69 Bridge Replacement D/B

Project Garver USA

Kansas City, MO

Garver first started using MIDAS a couple of years ago and we have been

impressed with its capabilities. So when our team was shortlisted for the US

69 Bridge project, we naturally chose MIDAS to develop a global model of

the bridge that would provide us reliable results. Now that we are in the final

design phase, we have further developed the MIDAS model and are using it

extensively in our design process.

Our engineers that have been developing the model have been very

pleased with its functionality and ability to create large models efficiently.

The ease of inputting and updating our cross-sections, where properties are

reliably calculated, has been benefi- cial in our iterative process for final

design. Also, the ability to import input data and export results quickly

allowed us to easily adapt our pre- and post-analysis tools to MIDAS.

The technical support has been great throughout our learning curve. Whenever

we have modeling questions, the support team has been very reliable in

getting back to us and generating solutions to any of our modeling issues.

The on-line tutorials and webinars have been very helpful whenever we

introduce younger engineers to MIDAS, and have gotten them up to speed

much quicker than we expected.

Output and results are handled with

a powerful graph-ical interface

where deforma-tions, forces and

stresses are viewed on any

direction or thru a cross section cut.

The ease ofinputting and updating our

cross-sections, where proper-

ties are reliably calculated, has been beneficialin our iterative

process for finaldesign.

Travis ButzBridge Engineer

Burgess and Niple, Inc.

Columbus, OH

I have been using Midas Civil for over six years, and I have found it to be a

powerful and flexible tool for bridge analysis and design. The program

has a user-friendly interface that allows new users to become

productive very quickly. It also has a wide range of analysis capabilities,

allowing users to perform complex tasks such as construction stage

analysis, buckling analysis, and analysis of time-dependent concrete

behavior.

My experience with Midas technical support has been outstanding. My

questions are always answered promptly, and numerous online tutorials

and demonstrations are available to help users take full advantage of the

software. Midas Civil is a great option for engineers looking for user-

friendly, highly functional structural analysis software.

Past Projects Using midas Civil:

- Rich Street Bridge, Columbus, Ohio

- Market Street Bridge, Steubenville, Ohio

- New River Gorge Bridge, Fayetteville, West Virginia

- LUC-75-0577, Toledo, Ohio

- Post-Tensioned Concrete Arch Design

- Steel Suspension Bridge Rehabilitation

- Steel Arch Bridge Rehabilitation

- Curved Steel Girder Design

Ken LeeSenior Bridge Engineer

Jacobs

Engineering

Denver, CO

I first used Midas Civil while working on the Ohio River DB project last

year, and am pleased to address that it provided a set of advanced and

practical solutions that has unique features among any other bridge

software.

The first distinguishing features were its user-friendly interface and

text editor that can handle a complex structural modeling easily.

Especially text editor can be interacted intuitively to graphic interface.

Defining composition section is also my favorable aspect of the

Midas Civil. With the feature composition sections can be defined

in an elaborate way for steel and concrete girder composite

bridges. Another notable feature I recognized during a recent

project is Midas includes multiple solution techniques to solve a

nonlinear static or dynamic problem.

It is for these reasons that Midas Civil will benefit to bridge engineers

to meet their satisfaction.

Daniel MariscalSenior Structural Engineer

URS

Corporation

Tampa, FL

In 2006, I was tasked to evaluate several structural analysis software

packages, and to select the best alternative for our office. Following several

weeks of trials, I selected Midas Civil, and after several years of usage, I am

very pleased with my choice.

Coming from using non-graphical interfaced structural analysis programs, the

selection of Midas Civil was an easy choice. The graphical interface was

powerful, intuitive, and very easy to use. The menu structure and the

maneuverability of the program were simple to learn and fast to get use to.

The capabilities and scope of the program have proven thru the years to be

everything a structural engineer needs by its side to provide an efficient and

effective structural design. The output results are handled via the graphical

interface or spreadsheets. Either way, output is detailed and easy to obtain.

Output and results are handled with a powerful graphical interface where

deformations, forces and stresses are viewed on any direction or thru a cross

section cut. The program can be described with one word: amazing.

Excellent customer technical support is as important as the software itself,

and Midas has provided thru the years excellent support. The support

personal are knowledgeable not only of the software itself, but of

engineering in general. Support answers to my technical questions have

been prompt, honest and always courteous.

Frank BlakemoreDesign Manager

US 69 Bridge Replacement D/B

Project Garver USA

Kansas City, MO

Garver first started using MIDAS a couple of years ago and we have been

impressed with its capabilities. So when our team was shortlisted for the US

69 Bridge project, we naturally chose MIDAS to develop a global model of

the bridge that would provide us reliable results. Now that we are in the final

design phase, we have further developed the MIDAS model and are using it

extensively in our design process.

Our engineers that have been developing the model have been very

pleased with its functionality and ability to create large models efficiently.

The ease of inputting and updating our cross-sections, where properties are

reliably calculated, has been benefi- cial in our iterative process for final

design. Also, the ability to import input data and export results quickly

allowed us to easily adapt our pre- and post-analysis tools to MIDAS.

The technical support has been great throughout our learning curve. Whenever

we have modeling questions, the support team has been very reliable in

getting back to us and generating solutions to any of our modeling issues.

The on-line tutorials and webinars have been very helpful whenever we

introduce younger engineers to MIDAS, and have gotten them up to speed

much quicker than we expected.

Output and results are handled with

a powerful graph-ical interface

where deforma-tions, forces and

stresses are viewed on any

direction or thru a cross section cut.

The ease ofinputting and updating our

cross-sections, where proper-

ties are reliably calculated, has been beneficialin our iterative

process for finaldesign.

Travis ButzBridge Engineer

Burgess and Niple, Inc.

Columbus, OH

I have been using Midas Civil for over six years, and I have found it to be a

powerful and flexible tool for bridge analysis and design. The program

has a user-friendly interface that allows new users to become

productive very quickly. It also has a wide range of analysis capabilities,

allowing users to perform complex tasks such as construction stage

analysis, buckling analysis, and analysis of time-dependent concrete

behavior.

My experience with Midas technical support has been outstanding. My

questions are always answered promptly, and numerous online tutorials

and demonstrations are available to help users take full advantage of the

software. Midas Civil is a great option for engineers looking for user-

friendly, highly functional structural analysis software.

Past Projects Using midas Civil:

- Rich Street Bridge, Columbus, Ohio

- Market Street Bridge, Steubenville, Ohio

- New River Gorge Bridge, Fayetteville, West Virginia

- LUC-75-0577, Toledo, Ohio

- Post-Tensioned Concrete Arch Design

- Steel Suspension Bridge Rehabilitation

- Steel Arch Bridge Rehabilitation

- Curved Steel Girder Design

Ken LeeSenior Bridge Engineer

Jacobs

Engineering

Denver, CO

I first used Midas Civil while working on the Ohio River DB project last

year, and am pleased to address that it provided a set of advanced and

practical solutions that has unique features among any other bridge

software.

The first distinguishing features were its user-friendly interface and

text editor that can handle a complex structural modeling easily.

Especially text editor can be interacted intuitively to graphic interface.

Defining composition section is also my favorable aspect of the

Midas Civil. With the feature composition sections can be defined

in an elaborate way for steel and concrete girder composite

bridges. Another notable feature I recognized during a recent

project is Midas includes multiple solution techniques to solve a

nonlinear static or dynamic problem.

It is for these reasons that Midas Civil will benefit to bridge engineers

to meet their satisfaction.

Pankaj Shrestha

Structural Engineer

GM2 Associates, Inc. Glastonbury, CT

"I believe in MIDAS" In our firm we not only design advanced structures but also we do complex structural

analysis. In our previous experiences with other processes, we found modeling irregular structures to be very difficult and likewise, analyzing those structures was extremely time consuming; not at all fast and efficient. Upon finding out about MIDAS. We were able to model structures very easily and the powerful processer actually allowed us to run through our results far more and beyond our previous software.

Along with the software itself, I'm always impressed with the support team. They answered my queries on a timely manner and even sent me feedback and suggestions to improve my model. I believe in MIDAS and I know that they will continue to service even more to the engineering community.

I have already used MIDAS for two different projects already; one is for the 1-95 over

Housatonic River, a curved steel tip girder bridge for load rating.

Behrooz Adnani Senior Project Engineer

Close Jensen & Miller

Wethersfield, CT

"Recommed use of this

software for all types of bridge

and complex structures" We were investigating for a finite element analysis and rating software to utilize for analyzing steel box girder superstructure bridges and complex multi-span truss bridges. We found out CTDOT is now utilizing MIDAS for rating/design of bridge type structures that AASHTOWare is not able to analyze. We purchased MIDAS software for rating and analyzing an old multi-span complex truss bridge with piers supported on concrete cap and piles.

The ease of graphic interface to model the structure has been extremely helpful and time saver. Of course, we are at an initial learning curve phase of utilizing the software however, the online technical support and technical staff have been extremely helpful and punctual in making our experience easier. We would recommend use of this software for all types of bridge and complex structures.

We are currently utilizing MIDAS on CTDOT Project 158-212, bridge 1349, route 136 over Saugatuck river in Westport CT.

We were able to model structures

very easily and the powerful

processor actually allowed us to run

through our results

The ease of graphic

interface to model the

structure has been extremely

helpful and time saver

Concrete Girder Design

Report

Concrete Bridge Design Report(Sample)

1. Design Condition

Design Code Element AASHTO-LRFD2012 26

• Section Properties

- Gross section

H (in.) B (in.)

czp (in.) czm (in.) czps (in.) czms (in.) A

g (in.2)

ly (in.4)

s, (in. 3)

Sb (in. 3)

s,s (in. 3)

sbs (in.3)

• Materials

- Concrete

Girder Slab

Before

106.30 334.65

37.85 68.45

9.905.E+03 1.118.E+07 2.955.E+05 1.634.E+05

f C

(ksi) 5.000 5.000

Node(I/J) I

After

118.11 334.65

25.37 80.93 37.18 25.37

1.386.E+04 1.664.E+07 4.409.E+05 1.382.E+05 4.475.E+05 6.559.E+05

Ee

(ksi)

Section Composite

- Transformed sectionBefore

H (in.) 106.30 B (in.) 334.65

czp (in.) 37.77 czm (in.) 68.53 czps (in.) czms (in.) A

g (in.2) 1.049.E+04 ly (in.4) 1.266.E+07

s, (in.3) 3.350.E+05 Sb (in.3) 1.847.E+05 s,s (in.3)

sbs (in.3)

f,=0.20\ir'c

(ksi) 4074.3 0.447 4074.3 0.447

* 131 is 0.85 in lower fc 4ksi, the others are 0.85-0.05(fc-4.0) 2':0.65

- Prestressing steel Information

No. Tendon Name Bond dp

Aps Strength (ksi)

Type (in.) (in.2) f

py fpu

1 S_B2L Bond 96.457 8.169 232.060 275.572 2 S_B1L Bond 88.583 8.169 232.060 275.572 3 S_B4L Bond 112.205 8.169 232.060 275.572 4 S_B3L Bond 104.331 8.169 232.060 275.572 * d

P is Distance from extreme compression fiber to centroid of prestressing tendon.

- Longitudinal non prestressed steel reinforcement Information.Girder Slab Bottom for Flexure Top for Flexure

Es fy Es fy ds As ds As (ksi) (ksi) (ksi) (ksi) (in.) (in.2) (in.) (in.2)

29000.0 60.001 29000.0 1 60.00 114.17 10.16 3.94 53.34 * ds or d's is Distance from extreme compression fiber to centroid of non prestressing reinforcement.

- Tranverse non prestressed steel reinforcement Information.

After

118.11 334.65

25.82 80.48 37.63 25.82

1.445.E+04 1.818.E+07 4.901.E+05 1.573.E+05 4.421.E+05 6.443.E+05

131

0.800 0.800

EP

(ksi) 29007.539 29007.539 29007.539 29007.539

Torsion A1

(in.2)

18.00

Shear Torsion Es f

y a Av

I A,

I (ksi) (ksi) (deg.) s s, (in.2

) (in.) (in.2) (in.)

* α is angle between longitudinal and stirrup.

■ Sectional forces/stresses due to effective prestress

* ep is Distance from centroid of section to centroid of tendon

2.Flexure design for a section■ Moment Direction : Positive- Method of Calculation : Code- The maximum Strength Limit Load Combination :- The maximum factored moment (Mu) :

1) Depth of neutral axis to compression face (c). (see. 5.7.3.1)

4 S_B3L 0.396 268.668 2194.873 214499.1663 S_B4L 0.396 269.153 2198.830 232199.5542 S_B1L 0.396 267.441 2184.847 179112.326

Apsfps(dp-c)

1 S_B2L 0.396 268.105 2190.269 196803.044

609.600 65574.617 1873.795 4996.142

No.Axial force in tendons(Bond) by Code

Tendon Name k fps Tps=Apsfps

Axial force in reinforcement steels tensile zone compressive zone

Ts=Asfs ΣAsfs(ds-c) Cs=A'sf's ΣA'sf's(c-d's)

6.603 5.283 1767.821 2.641 7513.239 29767.442

586268.632 (kips-in.)

Axial force in concrete (compressive zone)c a=β1c Ac ac Cc=0.85f'cAc Cc(c-ac)

81156.610Σ 0.541 4857.116 305294.184

cLCB1

4 S_B3L Bond -66.698 0.059 1216.774

61843.8583 S_B4L Bond -74.572 0.041 1241.750 92600.041

69693.676

No. Tendon Name

2 S_B1L Bond -50.950 0.241 1213.8111 S_B2L Bond -58.824 0.200 1184.780

BondType

ep

(in.)Apsfe(z-dir)

(kips)Apsfe(x-dir)

(kips)

5.906

Apsfe(x-dir)ep

(kips-in.)

29000.0 60.000 90.000 2.400 5.906 0.600

* fps = fpu(1-k c/dp) k = 2(1.04-fpy /fpu)

2) Flexural resistance▪ Nominal resistance(Mn). (see. 5.7.3.2)

Mn = ΣAsfs(ds-c)+ΣApsfps(dp-c)+Cc(c-ac)+ΣA'sf's(c-d's) =

▪ Resistance factor(Φ).dt = (in.)εt = 0.003(dt/c-1) = : Net tensile strain in the extreme tension steel

∴ Φ = : Tension Controlled Range, 0.005 ≤εt

▪ Factored resistance(M r = ΦMn)Mr = ≥ Mu =

3) Minimum reinforcement▪ Judgement (see. 5.7.3.3.2)

Mr = ≥ 1.33Mu =

∴ No check.

■ Moment Direction : Negative - Method of Calculation : Code - The maximum Strength Limit Load Combination : - The maximum factored moment (Mu) :

1) Depth of neutral axis to compression face (c). (see. 5.7.3.1)

* fps = fpu(1-k c/dp) k = 2(1.04-fpy /fpu)Σ 2366.747 22295.343

4 S_B3L 0.396 92.408 754.924 -7066.8223 S_B4L 0.396 -151.810 -1240.201 21374.8392 S_B1L 0.396 190.095 1552.974 9918.948

Apsfps(dp-c)

1 S_B2L 0.396 159.013 1299.049 -1931.623

3200.400 291341.162 609.600 11706.453

No.Axial force in tendons(Bond) by Code

Tendon Name k fps Tps=Apsfps

Axial force in reinforcement steels tensile zone compressive zone

Ts=Asfs ΣAsfs(ds-c) Cs=A'sf's ΣA'sf's(c-d's)

23.140 18.512 1166.135 9.256 4956.074 68811.606

cLCB60.000 (kips-in.)

Axial force in concrete (compressive zone)c a=β1c Ac ac Cc=0.85f'cAc Cc(c-ac)

922952.290 (kips-in.) 586268.63 (kips-in.) OK

922952.290 (kips-in.) 779737.28 (kips-in.)

1.00092

922952.29 (kips-in.)

114.1730.0489

1.00

Tps

6.603 7513.239 1873.795 609.600 8768.819

Σ 8768.819 822614.089

c (in.)Compression Force (kips.) Tensile Force (kips.) Tolerance

(C/T)Cc Cs Ts

2) Flexural resistance▪ Nominal resistance(Mn). (see. 5.7.3.2)

Mn = ΣAsfs(ds-c)+ΣApsfps(dp-c)+Cc(c-ac)+ΣA'sf's(c-d's) =

▪ Resistance factor(Φ).dt = (in.)εt = 0.003(dt/c-1) = : Net tensile strain in the extreme tension steel

∴ Φ = : Tension Controlled Range, 0.005 ≤εt

▪ Factored resistance(M r = ΦMn)Mr = ≥ Mu =

3) Minimum reinforcement▪ Judgement (see. 5.7.3.3.2)

Mr = ≥ 1.33Mu =

∴ No check.

3.Shear design for a section■ Maximum Shear- Section type : Segmental-Solid- The Strength Limit Load Combination : cLCB1- The factored Shear force : Vu =- Factored moment : Mu =- Factored axial force : Nu =- Resistance factor for shear : Φ =- Component of prestressing force

in direction of the shear force : Vp = ΣAps·fe(z-dir) = (kips.)1) Effective depth

bv = (in.) : Effective web width

0.9de , 0.72h ] : Effective depth for shearApsfps+Asfs

Apsfps+Asfs

dv = Max [Mn ,

0.90

0.14

62.99

de =Apsfpsdp+Asfsds = 101.309 (in.) : Effective depth for Bending

47.76 (kips.)578397.37 (kips-in.)

0.00 (kips.)

394154.564 (kips-in.) 0.00 (kips-in.) OK

394154.564 (kips-in.) 0.00 (kips-in.)

0.99974

394154.56 (kips-in.)

114.1730.0118

1.00

Tps

23.140 4956.074 609.600 3200.400 2366.747

c (in.)Compression Force (kips.) Tensile Force (kips.) Tolerance

(C/T)Cc Cs Ts

= , , ]=

2) Maximum spacing for transverse reinforcement (s max)▪ Shear stress on concrete (v u).

(Eq. 5.8.2.9-1)

▪ Maximum spacing for transverse reinforcement (s max)vu = < 0.125·f'c =

∴ = Min[0.8dv, 24.0(in.)] = (Eq. 5.8.2.7-1)

∴ = ≤

3) Minimum required transverse reinforcement (A v,min)

(Eq. 5.8.2.5-1)

∴ = (in.2) ≥ * Defining "OK" or "NG" will be checked inchecking Shear resistance by transverse reinforcement

4) Longitudinal Strain(εs)

∴ εs =where, = Max [ Mu , |Vu-Vp|dv ] = , ]

== =

5) Component of shear resisted by tensile stresses in concrete (V c)▪ Factor indicating diagonally cracked concrete to transmit tension (β).

(Eq. 5.8.3.4.2-1)

▪ Nominal shear resistance provide by concreteVc = 0.0316β √f'c bvdv = (Eq. 5.8.3.3-3)

6) Component of shear resisted by transverse reinforcement (V s)

Av ≥ Av,min(1+750εs)

2102.564 (kips.)

β =4.8

= 4.800 ∵

578397.373 (kips-in.)fpo 0.7fpu 192.900 (ksi.)

0.0 ≤ εs≤ 0.006 (Eq. 5.8.3.4.2-4)EsAs+EpAps

0.0000|Mu| Max [ 578397.373 4686.722

εs =(|Mu|/dv + 0.5Nu + |Vu-Vp|-Apsfpo) = -0.0003 ,

0.438 (in.2)fyAv 2.400 Av,min

s 5.906 (in.) smax OK

Av,min = 0.0316 √f'cbvs =

0.009 (ksi) 0.625 (ksi)smax 24.000 (in.)

vu =| Vu - ΦVp |

= 0.009 (ksi.)Φbvdv

Max [ 98.412 91.178 85.03998.412 (in.)

▪ Area of transverse reinforcement required (See 5.8.2.4)

Vu < 0.5Φ(Vc+Vp) ∴ No shear reinforcing

▪ Component of shear resisted by transverse reinforcement (V s)

(Eq. 5.8.3.3-4)where, θ = = (Eq. 5.8.3.4.2-3)

α = (deg.)

7) Shear resistance.▪ Nominal shear strength (Eq. 5.8.3.3.-1)

Vc +Vs +Vp = ≤

∴ Vn = =

▪ Factored shear resistance (V r).

∴ Vr = = ≥ Vu =

■ Minimum Shear- Section type : Segmental-Solid- The Strength Limit Load Combination : cLCB2- The factored Shear force : Vu =- Factored moment : Mu =- Factored axial force : Nu =- Resistance factor for shear : Φ =- Component of prestressing force

in direction of the shear force : Vp = ΣAps·fe(z-dir) = (kips.) 1) Effective depth

bv = (in.) : Effective web width

= , , ]=

Max [ 98.412 91.178 85.03998.412 (in.)

0.9de , 0.72h ] : Effective depth for shearApsfps+Asfs

Apsfps+Asfs

dv = Max [Mn ,

0.90

0.14

62.99

de =Apsfpsdp+Asfsds = 101.309 (in.) : Effective depth for Bending

OK

-443.83 (kips.)398010.09 (kips-in.)

0.00 (kips.)

7749.144 (kips.)Vc+Vs+Vp 6431.854 (kips.)

ΦVn 5788.668 (kips.) 47.764 (kips.)

29+3500εs 29.000 (deg.)90.000

6431.854 (kips.) 0.25·f'cbvdv + Vp =

Vs =Av·fy·dv(cotθ+cotα)sinα

= 4329.150 (kips.)s

0.5Φ(Vc+Vp) Φ(Vc+Vp) Vu

946.22 kips 1892.43 kips 47.76 kips

2) Maximum spacing for transverse reinforcement (s max)▪ Shear stress on concrete (v u).

(Eq. 5.8.2.9-1)

▪ Maximum spacing for transverse reinforcement (s max)vu = < 0.125·f'c =

∴ = Min[0.8dv, 24.0(in.)] = (Eq. 5.8.2.7-1)

∴ = ≤

3) Minimum required transverse reinforcement (A v,min)

(Eq. 5.8.2.5-1)

∴ = (in.2) ≥ * Defining "OK" or "NG" will be checked inchecking Shear resistance by transverse reinforcement

4) Longitudinal Strain(εs)

∴ εs =where, = Max [ Mu , |Vu-Vp|dv ] = , ]

== =

5) Component of shear resisted by tensile stresses in concrete (V c)▪ Factor indicating diagonally cracked concrete to transmit tension (β).

(Eq. 5.8.3.4.2-1)

▪ Nominal shear resistance provide by concreteVc = 0.0316β √f'c bvdv = (Eq. 5.8.3.3-3)

6) Component of shear resisted by transverse reinforcement (V s)▪ Area of transverse reinforcement required (See 5.8.2.4)

0.5Φ(Vc+Vp) Φ(Vc+Vp) Vu

946.22 kips 1892.43 kips 443.83 kips

Av ≥ Av,min(1+750εs)

2102.564 (kips.)

β =4.8

= 4.800 ∵

398010.092 (kips-in.)fpo 0.7fpu 192.900 (ksi.)

0.0 ≤ εs≤ 0.006 (Eq. 5.8.3.4.2-4)EsAs+EpAps

0.0000|Mu| Max [ 398010.092 43692.231

εs =(|Mu|/dv + 0.5Nu + |Vu-Vp|-Apsfpo) = -0.0015 ,

0.438 (in.2)fyAv 2.400 Av,min

s 5.906 (in.) smax OK

Av,min = 0.0316 √f'cbvs =

0.080 (ksi) 0.625 (ksi)smax 24.000 (in.)

vu =| Vu - ΦVp |

= 0.080 (ksi.)Φbvdv

Vu < 0.5Φ(Vc+Vp) ∴ No shear reinforcing

▪ Component of shear resisted by transverse reinforcement (V s)

(Eq. 5.8.3.3-4)where, θ = = (Eq. 5.8.3.4.2-3)

α = (deg.)

7) Shear resistance.▪ Nominal shear strength (Eq. 5.8.3.3.-1)

Vc +Vs +Vp = ≤

∴ Vn = =

▪ Factored shear resistance (V r).

∴ Vr = = ≥ Vu =

4.Torsional design for a section■ Case of Vmax.

- Section type : Segmental-Solid- The Strength Limit Load Combination : cLCB1- The factored Torsional moment : Tu =- The factored Shear force : Vu =- Factored moment : Mu =- Factored axial force : Nu =- Resistance factor for shear : Φ =- Component of prestressing force

in direction of the shear force : Vp = ΣAps·fe(z-dir) = (kips.)

1) NotationAo = Area enclosed by the shear flow path,

including any area of holes therein=

ph = Perimeter of the centerline of theclosed tranverse torsion reinforcement.

=Acp = Total area enclosed by outside

Perimeter of the concrete section.=

pc = The length of the outside perimeter ofconcrete section.

=

2) Checking Torsional Effects▪ Torsional cracking moment (T cr).

(Eq. 5.8.2.1-4)

where,= Acp

2/pc 121930.93 (in.3)

fpc)(1/2) = 64974.02 (kips-in.)

pc 0.125√f'cTcr = 0.125·√f'c

Acp2

( 1+

299.213 (in.)

13857.028 (in.2)

1240.157 (in.)

0.00 (kips.)0.90

0.14

5642.011 (in.2)

OK

16674.88 (kips-in.)47.76 (kips.)

578397.37 (kips-in.)

7749.144 (kips.)Vc+Vs+Vp 6431.854 (kips.)

ΦVn 5788.668 (kips.) -443.831 (kips.)

29+3500εs 29.000 (deg.)90.000

6431.854 (kips.) 0.25·f'cbvdv + Vp =

Vs =Av·fy·dv(cotθ+cotα)sinα

= 4329.150 (kips.)s

fpc = Compressive stress in concrete due to effective prestress only.

Tu = > = (Eq. 5.8.2.1-3)∴ Tu > 0.25Tcr, Consider Torsional Effects.

▪ Check torsional moment (Eq. 5.8.3.6.2-1)

where,θ = = (Eq. 5.8.3.4.2-3)

=

▪ Required longitudinal reinforcementIn solid section, (Eq. 5.8.3.6.3-1)ΣAps·fps+ΣAs·fy =

■ Case of Vmin.




(Eq. 5.8.2.1-4)

where,=


=ΣAps·fe(x-dir) 0.35

= 64974.02 (kips-in.)

(ksi)Ag Iy

0.125√f'c

Acp2/pc 121930.93 (in.3)

0.90

+ΣAps·fe(x-dir)·ep·yjoint =

0.14

Tcr = 0.125·√f'cAcp

2

( 1+fpc

)(1/2)

pc

-443.83 (kips.)398010.09 (kips-in.)

0.00 (kips.)

∴ ΣAps·fps+ΣAs·fy < NG

-16674.88 (kips-in.)

Apsfps+Asfy

9378.419 (kips.)

= 10360.64 (kips.)

0.9de , = 98.412

, 0.0 ≤ εs≤ 0.006

(in.) (See. 5.8.2.9)

(Eq. 5.8.3.4.2-4)EsAs+EpAps

0.00000

dv = Max [Mn , 0.72h ]

29+3500εs 29.000 (deg.)

εs =(|Mu|/dv + 0.5Nu + |Vu-Vp|-Apsfpo) = -0.0003

=Φ·2·Ao·At·fy·cotθ

= 111686.18 (kips-in.) OKst

Tu = 16674.88 (kips-in.) ≤ ΦTn

= 0.35 (ksi)Ag Iy

16674.882 (kips-in.) 0.25ΦTcr 14619.15 (kips-in.)

=ΣAps·fe(x-dir) +

ΣAps·fe(x-dir)·ep·yjoint

2 20.5 0.45

cot 0.52

u u u h up s

v o

M N V p TV V

d A

2 20.5 0.45

cot 0.52

u u u h up s

v o

M N V p TV V

d A



where,θ = = (Eq. 5.8.3.4.2-3)

=


■ Case of Tmax.




(Eq. 5.8.2.1-4)

where,=




-16674.882 (kips-in.) 0.25ΦTcr 14619.15 (kips-in.)

=ΣAps·fe(x-dir) 0.35

= 64974.02 (kips-in.)

(ksi)Ag Iy

0.125√f'c

Acp2/pc 121930.93 (in.3)

0.90

+ΣAps·fe(x-dir)·ep·yjoint =

0.14

Tcr = 0.125·√f'cAcp

2

( 1+fpc

)(1/2)

pc

-355.70 (kips.)406684.84 (kips-in.)

0.00 (kips.)

∴ ΣAps·fps+ΣAs·fy ≥ OK

-16674.88 (kips-in.)

Apsfps+Asfy

9378.419 (kips.)

= 7535.03 (kips.)

0.9de , = 98.412

, 0.0 ≤ εs≤ 0.006

(in.) (See. 5.8.2.9)


0.00000

dv = Max [Mn , 0.72h ]

29+3500εs 29.000 (deg.)



= 111686.18 (kips-in.) OKst

Tu = -16674.88 (kips-in.) ≤ ΦTn

-16674.882 (kips-in.) 0.25ΦTcr 14619.15 (kips-in.)

2 20.5 0.45

cot 0.52

u u u h up s

v o

M N V p TV V

d A

2 20.5 0.45

cot 0.52

u u u h up s

v o

M N V p TV V

d A

where,θ = = (Eq. 5.8.3.4.2-3)

=


5.Crack check■ Bottom- The maximum Service Limit Load Combination : cLCB23- The maximum spacing of steel reinforcement in layer closet to the tension face(s max) (Eq. 5.7.3.4-1)

where,

dc = (in.)γe = : Exposure factorfss = (ksi.) : Tensile stress in bar at the service limit state

1.003.165

(in.) OK

= 1.049

(in.) ≥

1 +dc

202.921

0.7(h-dc)

3.937

∴ ΣAps·fps+ΣAs·fy ≥

βs·fss

βs =

OK

smax =700·γe - 2dc = suse = 7.874

Apsfps+Asfy

9378.419 (kips.)

= 7808.19 (kips.)

0.9de , = 98.412

, 0.0 ≤ εs≤ 0.006

(in.) (See. 5.8.2.9)


0.00000

dv = Max [Mn , 0.72h ]

29+3500εs 29.000 (deg.)



= 111686.18 (kips-in.) OKst

Tu = -16674.88 (kips-in.) ≤ ΦTn

2 20.5 0.45

cot 0.52

u u u h up s

v o

M N V p TV V

d A

2 20.5 0.45

cot 0.52

u u u h up s

v o

M N V p TV V

d A

Steel Girder Design Report

I. Design Condition (Positive Flexure)

1. Section Properties

1) Slab Properties

Bs = in

ts = in

th = in

fc' = ksi

Ec = ksi

Ar = in2

Fyr = ksi

2) Girder Properties

[Section]

bfc = in bft = in

tfc = in tft = in

D = in tw = in

[Design Strength]

Fyc = ksi (Compression Flange Yield Strength)

Fyw = ksi (Web Yield Strength)

Fyt = ksi (Tension Flange Yield Strength)

Es = ksi (Elastic Modulus of Steel)

3) Transverse Stiffener Properties

2. Elastic Section Properties

1) Steel Section

2) Short-term Composite Section

STop(n) (in3) 16087.410 SBot(n) (in) 4689.372

dTop(n) (in) 16.702 dBot(n) (in) 57.298

A(n) (in2) 255.949 Iy(n) (in

4) 268691.653 Iz(n) (in4) 121599.508

STop (in3) 3563.908 SBot (in3) 3830.890

dTop (in) 38.336 dBot (in) 35.664

- 90.000

A (in2) 133.812 Iy (in4) 136625.408 Iz (in

4) 2882.695

Web 1Side 36.000 5.000 1.500 -

4.500

3865.202

15.800

60.000

18.000 20.000

Fy (ksi) H (in) B (in) tw(in) tf(in) d0 (in)

50.000

50.000

50.000

29000.000

Position Type

Web A709-50W 0.563 50.000 70.000 less than 2 in.

Element 541Position I

Moment Type Beam

Tension Flange A709-50W 2.500 50.000 70.000

fy (ksi) fu (ksi) Note

Compression Flange A709-50W 2.500 50.000 70.000

2.500 2.500

69.000 0.563

Position Material Thick (in)

108.000

9.000

5.000

Steel composite Bridge Design Report(Sample)

Seolhee

Highlight

3) Long-term Composite Section(Es/Ec=3n)

II. Strength Limit State - Flexural Resistance

1. Flexure

■ Positive moment

1) Design Forces and Stresses

Positive load combination does not exist. Skip this check.

III. Design Condition (Negative Flexure)

1. Section Properties

1) Slab Properties

Bs = in

ts = in

th = in

fck = ksi

Ec = ksi

Ar = in2

Fyr = ksi

2) Girder Properties

[Section]

bfc = in bft = in

tfc = in tft = in

D = in tw = in

[Design Strength]

Fyc = ksi (Compression Flange Yield Strength)

Fyw = ksi (Web Yield Strength)

Fyt = ksi (Tension Flange Yield Strength)

Es = ksi (Elastic Modulus of Steel)

3) Transverse Stiffener Properties

2. Elastic Section Properties

1) Steel Section

- 90.000

A (in2) 133.812 Iy (in4) 136625.408 Iz (in

4) 2882.695

Web 1Side 36.000 5.000 1.500 -

Fy (ksi) H (in) B (in) tw(in) tf(in) d0 (in)

50.000

50.000

50.000

29000.000

Position Type

Web A709-50W 0.563 50.000 70.000 less than 2 in.

Tension Flange A709-50W 2.500 50.000 70.000

fy (ksi) fu (ksi) Note

Compression Flange A709-50W 2.500 50.000 70.000

2.500 2.500

69.000 0.563

Position Material Thick (in)

108.000

9.000

5.000

4.500

3865.202

15.800

60.000

20.000 18.000

STop(3n) (in3) 7242.680 SBot(3n)(in

3) 4348.148

dTop(3n) (in) 27.760 dBot(3n) (in) 46.240

A(3n) (in2) 174.525 Iy(3n) (in

4) 201057.767 Iz(3n) (in4) 42454.966

2) Short-term Composite Section

3) Long-term Composite Section(Es/Ec=3n)

4) Short-term Composite Section(Long. Reinforcement)

5) Long-term Composite Section(Long. Reinforcement/3)

IV. Strength Limit State - Flexural Resistance

1. Flexure

■ Negative moment


Loadcombination Name :

Loadcombination Type :

2) Cross-section Proportions

① Web Proportions (AASHTO LRFD Bridge, 2012, 6.10.2.1)

D

tw

② Flange Proportions (AASHTO LRFD Bridge, 2012, 6.10.2.2)

bf

2tw

bf = ≥ =

= 4.000 ≤ 12 ...... OK

18.000 D/6 11.500 ...... OK

-27.218

= 122.667 ≤ 150 ...... OK

StressesTop 16.292 4.618 6.047 26.958

Bot -15.157 -4.725 -7.337

-332.958 -218.967

Component fc,t (ksi)

Steel (MD1) Long-term (MD2) Short-term Sum

Forces ( - ) -58063.925 -18627.322 -30216.698 -106907.944

scLCB1

MY-MIN

ComponentMu (kips·in) Vu

(kips)T

(kips·in)Steel (MD1) Long-term (MD2) Short-term Sum

STop(R3) (in3) 4033.488 SBot(R3)(in

3) 3942.619

dTop(R3) (in) 36.578 dBot(R3) (in) 37.422

A(R3) (in2) 139.079 Iy(R3) (in

4) 147538.813 Iz(R3) (in4) 7285.160

STop(R) (in3) 4996.615 SBot(R) (in) 4118.349

dTop(R) (in) 33.435 dBot(R) (in) 40.565

A(R) (in2) 149.612 Iy(R) (in

4) 167061.266 Iz(R) (in4) 16090.704

STop(n) (in3) 7242.680 SBot(n) (in3) 4348.148

dTop(n) (in) 27.760 dBot(n) (in) 46.240

A(n) (in2) 174.525 Iy(n) (in

4) 201057.767 Iz(n) (in4) 42454.966

STop(n) (in3) 16087.410 SBot(n) (in3) 4689.372

dTop(n) (in) 16.702 dBot(n) (in) 57.298

A(n) (in2) 255.949 Iy(n) (in

4) 268691.653 Iz(n) (in4) 121599.508

STop (in3) 3563.908 SBot (in3) 3830.890

dTop (in) 38.336 dBot (in) 35.664

tf = ≥ =

Iyc

Iyt

③ Minimum Negative Flexure Concrete Deck Reinforcement (AASHTO LRFD Bridge, 2012, 6.10.1.7)

Ars = ≥ = in2

in which :

Adeck = in2

3) Flexural Strength Limit State in negative flexure

▪ Section Classification (AASHTO LRFD Bridge, 2012 6.10.6.3)

= ≤ ksi

Iyc

Iyt

Es

Fyc

in which :

Iyc = in4

Iyt = in4

Dc = in∴ Compact or Noncompact section.

▪ Hybrid Factor, Rh (AASHTO LRFD Bridge, 2012, 6.10.1.10.1)

in which :

ρ = =

β = =

Dn = in (larger of distance from elastic NA to inside flange face)

Fn = ksi (yield stress corresponding to Afn)

Afn = in2 (flange area on the side of NA corresponding to Dn)

▪ Plastic Moment(Mp) (AASHTO LRFD Bridge, 2012, D6.1)

① Plastic Forces

- Plastic Forces

Prt = = kips

Prb = = kips

Pt = = kips

Pw = = kips

Pc = = kips

- Distance from the plastic neutral axis

2 · H · tw · Fyw 1940.625

bfc · tfc · Fyc 2500.001

50.000

Fyr Art 474.000

Fyr Arb 474.000

bft · tft · Fyt 2249.996

min( Fyw /Fn , 1.0 ) 1.000

2Dn·tw / Afn 0.856

38.065

50.000

Rh =12 + β(3ρ-ρ3)

= 1.00012+2β

137.274 ...... OKtw

1666.674

1214.998

34.678

2 Dc= 123.300 ≤ 5.7 √ =

1062.000

min ( Fyc, Fyt ) 50.000 70.0 ...... OK

= 1.37 ≥ 0.3 ...... OK

10.0 ...... OK

15.800 0.01Adeck 10.620 ...... OK

in4

12

0.1 ≤ = 1.372 ≤

12

Iyt =tft · bft

3

= 1214.998

2.500 1.1tw 0.619 ...... OK

Iyc =tfc · bfc

3

= 1666.674 in4

drt = in (distance from the PNA to the centerline of the top layer of reinforcement)

drb = in (distance from the PNA to the centerline of the bottom layer of reinforcement)

dt = in (distance from the plastic neutral axis to midthickness of the tension flange)

dw = in (distance from the plastic neutral axis to middepth of the web)

dc = in (distance from the plastic neutral axis to midthickness of the compression flange)

② Plastic Moment

- Check the case of the plastic neutral axis

= kips ≥ Pt + Prb + Prt = kips∴

PNA In Web

- Distance of the plastic neutral axis

D

2

- Plastic Moment

Pw

2D

▪ Yield Moment(My) (AASHTO LRFD Bridge, 2012, D6.2.2)

① Yield Moment of Top Flange

MD1

STop

MAD = kips·in

MyTop = MD1 + MD2 + MAD = kips·in

② Yield Moment of Bottom Flange

MD1

SBot

MAD = kips·in

MyBot = MD1 + MD2 + MAD = kips·in∴My = min ( MyTop, MyBot ) = kips·in

in which :

S : noncomposite section modulus (in3)Sn : long-term composite section modulus with longitudinal reinforcements (in3)MD1 : moment of noncomposite section (kips·in)

MAD : additional yield moment of short-term composite section (kips·in)

##########

##########

##########

ksiSBot(R) SBot(R) 3.831E+03 3.943E+03 4.118E+03

+-1.863E+04

+MAD

= 50.000

##########

##########

Fy = +MD2

+MAD

=-5.806E+04

50.000 ksiSTop(R) STop(R) 3.564E+03 4.033E+03 4.997E+03

-5.806E+04+

-1.863E+04+

MAD=

= ########## kips·in

Fy = +MD2

+MAD

=

Mp = · [ Y2 + (D - Y)2 ] + [ Prt · drt+ Prb · drb+ Pt · dt +Pc · dc]

...... OK

Y = · (Pc - Pt - Prt -Prb

+ 1 ) = 22.091 inPw

23.341

12.409

48.159

Pc + Pw 4440.626 3197.996

32.755

32.571

▪ Web Load-Shedding Factor, Rb (AASHTO LRFD Bridge, 2012, 6.10.1.10.2)

2 Dc Es

tw fyc

in which :

fyc = ksi (specified minimum yield strength of a compression flange)

Rb =

▪ Limiting Unbraced Length, Lp (AASHTO LRFD Bridge, 2012, 6.10.8.2.3)

E

Fyc

in which :

rt = effective radius of gyration for lateral torsional buckling

1

3

▪ Moment Gradient Modifer, Cb (AASHTO LRFD Bridge, 2012, 6.10.8.2.3)

Calculation of Stress (C6.4.10)

f0 = ksi

f2 = ksi

fmid = ksi

f1 = f0 ) = ksi

For fmid/f2 > 1.0

Cb =

▪ Second-order elastic compression-flange Lateral bending stress (AASHTO LRFD Bridge, 2012, 6.10.1.6)

i. Because of discretely braced flange

Mu

Sl

ii. check Lb

Lb = in

Lb ≤ Lb'

fl = fl1 = ksi

fl = ≤ = ksi

▪ Web Plastification Factor, Rpc and Rpt (AASHTO LRFD Bridge, 2012, A6.2)

E

Fyc Dcp

Dc

in which :

)2

Rh · My

λpw(Dcp) = = 86.313 ≤ λrw (

( 0.54Mp

-

6.452

6.452 0.6Fyf 30.000 ...... OK

√

) = 185.690

0.09

, 220.563 ) = 212.736 infbu/Fyc Mu/Myc

, 1.2Lp√Cb·Rb

) = min( 212.736

120.000

Lb' = min( 1.2Lp√Cb·Rb

-1075.338= 6.452 ksi

tft(bft)2/6 166.667

1.000

fl = =Muz

=

bfc·tfc

-13.705

11.302

-13.705

max( 2fmid-f2 , -13.705

=

bfc

= 5.431 in√ 12( 1 +

Dc·tw)

= 137.274

50.000

1.000

Lp = 1.0 rt √ = 130.800 in

= 123.300 ≤ λrw = 5.7 √

Rh =

E

Fyc

Dcp = in

Dc = in∴=

Dc

Dcp

∴=

Mp Mp

Myc Myc

∴Rpc =

Mp Mp

Myt Myt

∴Rpt =

▪ Flexural Resistance in Continuously Braced Tension Flange (AASHTO LRFD Bridge, 2012, A6.1.4)

Mu = ≤ =

in which :

Фf =

▪ Local Buckling Resistance base on Discretely Braced Compression Flange (AASHTO LRFD Bridge, 2012, A6.3.2)

E

Fyc

λf ≤ λpf Therefore, compact flange

= =


E

Fyc

in which :


1

3

▪ Lateral-Torsional Buckling Resistance based on Discretely Braced Compression Flange (AASHTO LRFD Bridge, 2012, A6.3.3)

bfc·tfc

=

bfc

= 5.431 in√ 12( 1 +

Dc·tw)

kips·in

Lp = 1.0 rt √ = 130.800 in

λpf = 0.38 √ = 9.152

Mnc(FLB) Rpc · Myc ##########

...... OK

1.000

λf =bfc

= 4.0002 tfc

0.871Mp λrw - λpw(Dc)

0.871

########## Фf · Rpt · Myt ########## kips·in

λw - λpw(Dc)) ] = 0.871 ≤ =

1.145Mp λrw - λpw(Dc)

1.028

Rpt = [ 1- ( 1-Rh · Myt

) (

λw - λpw(Dc)) ] = 1.028 ≤ =Rpc = [ 1- ( 1-

Rh · Myc) (

≤ λrw

λpw(Dc) 63.808

λw =2 Dc

= 123.300tw

> λpw(Dcp) Therefore, Compact Web Sectiontw

λpw(Dc) = λpw(Dcp) ( ) = 63.808

46.909

34.678

λpw(Dcp) 86.313

2 Dcp= 166.787

1.000

λrw = 5.7 √ = 137.274

Lb ≤ Lp Therefore, compact unbraced length

= =

Mnc = = ksi

▪ Flexural Resistance based on Discretely Braced Compression Flange (AASHTO LRFD Bridge, 2012, A6.1.1)

1

3

in which :

Sxc = = in3

Фf =

V. Strength Limit State - Shear Resistance

1. Shear

■ Max




2) Shear Resistance (AASHTO LRFD Bridge, 2012, 6.10.9)

▪ Ratio of the shear-buckling resistance to the shear yield strength, C (AASHTO LRFD Bridge, 2012, 6.10.9.3.2)

▪ Web Classification

:

:

Transverse Spacing = in < D = in

So, this web is considered

shear-buckling coefficient of stiffened Webs

d0

D

D E·k

tw Fyw

therefore,

E·k

D Fyw

tw

▪ Nominal Resistance of Stiffened interior Webs (AASHTO LRFD Bridge, 2012, 6.10.9.3.2)

Vp = = kips

therefore,

0.58Fyw · D · tw 1125.563

2D · tw= 0.817 ≤ 2.500

( bfc · tfc + bft · tft )

C =1.57

· ( ) = 0.480

( )2

= 122.667 > 1.40 √ = 95.000

stiffened web

k = 5 +

5

= 7.939( )2

Transverse Stiffener Exist

90.0 3 207.0

Forces -169.047 -62.076 -143.335 -374.457

Longitudinal Stiffener Not Exist

1.000

scLCB1

FZ-MIN

ComponentVu (kips)

Steel Long-term Short-term Sum

= ########## kips·in...... OK

Myc / Fyc 4222.094

Mu + fl · Sxc = ########## ≤ Фf · Mnc

Mnc(LTB) Rpc · Myc ########## kips·in

min( Mnc(FLB) , Mnc(LTB) ) ##########

d0

D

Vu = ≤ = kips

in which :

Фv =

3) Transverse Stiffeners (AASHTO LRFD Bridge, 2012, 6.10.11)

① Projecting Width (AASHTO LRFD Bridge, 2012, 6.10.11.1)

bt = in ≥ = in

= in ≥ bt = in < = in

in which :

D = in (height of steel section)

tp = in (thickness of stiffener)

bf = in (width of flange)

② Moment of Inertia and Radius of Gyration (AASHTO LRFD Bridge, 2012, 6.10.11.1.3)

It = = in4

J = max (Jcal, 0.5) =

It1 = = in4 (yielding of stiffener)

Fyw

E

in which :

ρt = =

Vu ≤ Therefore,

It = in4 ≥ = in4

VI. Service Limit State

■ Positive/Negative moment


There is no service load combination. Skip this check.

■ Negative moment

▪ Concrete Deck Effectiveness Check

fr, modulus of rupture = ksi

Service II factored longitudinal tensile stress in deck = ksi∴

Concrete deck is effective in tension




scLCB5

MY-MIN

ComponentMs (kips·in) / fc,t (ksi)

62.500 min( It1, It2 ) 6.140 ...... OK

0.509

0.497

Fys ) = 36.000 ksi(bt/tp)2

ΦvVcr

62.182 in4

40

max(Fyw / Fcrs , 1.0) 1.389

Fcrs = min (0.31E

,

(d0/D)2

0.500

b · tw3 · J 6.140

It2 =D4 · ρt

1.3

( )1.5 =

69.000

1.500

20.000

tp · bt3 / 3 62.500

Jcal =2.5

- 2.0 = -0.531

16.0 tp 24.000 5.000 bf / 4 5.000 ...... NG

...... OK

1.000

5.000 2.0 + D / 30.0 4.300 ...... OK

850.316 kips√ 1+( )2

-374.457 Фv · Vn 850.316

Vn = Vp[ C +

0.87(1-C)

] =

▪ check stress of the concrete deck

Compact composite section in positive flexure utilized in shored construction

fdeck = ≤ fc' = ksi

in which :

( ) · ( )

( ) · ( )

n = =

fc' = ksi

2) Permanent deformation (AASHTO LRFD Bridge, 2012, 6.10.4.2)

▪ This section is not a compact section in positive flexure, skip top and bottom flange stress check.

▪ Nominal Bend-buckling Resistance for webs (AASHTO LRFD Bridge, 2012, 6.10.4.2.2)∴Fcrw = ksi

in which :

Dc = in

k = =

fc = ksi ≤ Fcrw = ksi

in which :

fc = compression-flange stress at the section under consideration due to the Service II loads.

VII. Constructibility

1. Flexure

■ Negative moment


Construction Stage :

Step :

2) Check slenderness of web (AASHTO LRFD Bridge, 2012, 6.10.6.2.3-1)

Es

Fyc

in which :

Dc = in

137.274 ...... Compact or noncompact Webtw

33.164

2 · Dc= 117.917 ≤ 5.7 √ =

Force (-) -69676.710

Stress(fbu)

Top 19.551

Bot -18.188

...... OK

Stage3-1

2

ComponentMu (kips·in) / fc,t (ksi)

Steel Section Only

50.000

38.270

9 / ( DC / D )2 29.257

-19.913 50.000

ksi ≤ min ( Rh·Fyc, Fyw/0.7 ) = 50.000 ksi

Es / Ec 7.503

4.500

Fcrw =0.9 E · k

= 50.748( D / tw )2

28.202= 0.497 ksi

I · n ########## 7.503

-19.913

0.497 0.24 1.080 ...... OK

fdeck =M · y

=-35494.159

StressesTop 13.034 1.801 1.395 16.231

Bot -12.125 -3.001 -4.787

Forces ( + ) -46451.140 -13047.469 -22446.690 -81945.298

ComponentSteel Long-term Short-term Sum

3) Discretely Braced Flanges in Compression (AASHTO LRFD Bridge, 2012, 6.10.3.2.1)


In constructibility (AASHTO LRFD Bridge, 2012, 6.10.3.2.1)

Rb =


E

Fyc

in which :


1

3

▪ Moment Gradient Modifer, Cb (AASHTO LRFD Bridge, 2012, 6.10.8.2.3)

Calculation of Stress (C6.4.10)

f0 = ksi

f2 = ksi

fmid = ksi

f1 = f0 ) = ksi

For fmid/f2 > 1.0

Cb =

▪ Second-order elastic compression-flange Lateral bending stress (AASHTO LRFD Bridge, 2012, 6.10.1.6)

i. Because of discretely braced flange

Mu

Sl

ii. check Lb

Lb = in

Lb ≤ Lb'

fl = fl1 = ksi

fl = ≤ = ksi

① Check flange nominal yielding

= ≤ = kips

in which :

Φf =

Rh =

② Check flexural resistance

= ≤ = ksi

in which :

...... OK

1.000

1.000

fbu + fl/3 18.665 Φf · Fnc 50.000

1.430

1.430 0.6Fyf 30.000 ...... OK

fbu + fl 19.618 Φf · Rh · Fyc 50.000 ...... OK

, 273.923 ) = 260.899 infbu/Fyc Mu/Myc

, 1.2Lp√Cb·Rb

) = min( 260.899

120.000

Lb' = min( 1.2Lp√Cb·Rb

-238.252= 1.430 ksi

tft(bft)2/6 166.667

1.000

fl = =Muz

=

)bfc·tfc

-19.551

16.164

-19.551

max( 2fmid-f2 , -19.551

in

=

bfc

= 5.445 in√ 12( 1 +

Dc·tw

1.000

Lp = 1.0 rt √ = 131.130

Φf =

Fnc = ksi

▪ Local Buckling Resistance based on Discretely Braced Compression Flange (AASHTO LRFD Bridge, 2012, 6.10.8.2.2)

λf = =

λpf = =


In constructibility (AASHTO LRFD Bridge, 2012, 6.10.3.2.1)

Rb =

λf ≤ λpf Therefore,

= = ksi

in which :

Rb =

Rh =


E

Fyc

in which :


1

3

▪ Lateral-Torsional Buckling Resistance based on Discretely Braced Compression Flange (AASHTO LRFD Bridge, 2012, 6.10.8.2.2)

Lb ≤ Lp Therefore, compact unbraced length

= = ksi

in which :

Rb =

Rh =

Fnc = = ksi

③ Check web bend buckling

For sections with compact or noncompact webs, shall not be checked.

4) Discretely Braced Flanges in Tension (AASHTO LRFD Bridge, 2012, 6.10.3.2.2)

▪ Flange Lateral bending Stress (AASHTO LRFD Bridge, 2012, 6.10.1.6)

Because of discretely braced tension flange.

Mu

Sl

fl = ≤ = ksi ...... OK

= 1.765 ksitft(bft)

2/6 135.000

1.765 0.6Fyf 30.000

min( Fnc(FLB), Fnc(LTB) ) 50.000

fl = =Muz

=-238.252

bfc·tfc

Fnc(LTB) Rb · Rh · Fyc 50.000

1.000

1.000

=

bfc

= 5.445 in√ 12( 1 +

Dc·tw)

Lp = 1.0 rt √ = 131.130 in

1.000

Fnc(FLB) Rb · Rh · Fyc 50.000

1.000

1.000

1.000

50.000

bfc / 2tfc 4.000

0.38√(E/Fyc) 9.152

① Check flange nominal yielding

= ≤ = kips

in which :

Φf =

Rh =

2. Concrete Deck

1) Check longitudinal stresses in concrete deck (AASHTO LRFD Bridge, 2012, 6.10.1.7)

fdeck = > = ksi

in which :

( ) · ( )

( ) · ( )

n = =

fr = = ksi

fc' = ksi

3. Shear

■ Max

1) Design Forces

Construction Stage :

Step :

2) Shear requirement for webs (AASHTO LRFD Bridge, 2012, 6.10.3.3)



d0

D

D E·k

tw Fyw

therefore,

E·k

D Fyw

tw

▪ Nominal Resistance of Unstiffened interior Webs or End pannel

Vp = = kips

Vn = Vcr = = kips

in which :

C = ratio of the shear-buckling resistance to the shear yield strength

=

Vu = ≤ = = kipsФv · Vn 540.759 ...... OK

0.58Fyw · D · tw 1125.563

C · Vp 540.759

0.480

-202.856 φv · Vcr

C =1.57

· ( ) = 0.480

( )2

= 122.667 > 1.40 √ = 95.000

k = 5 +

5

= 7.939( )2

2

ComponentVu (kips)

Steel Section Only

Force -202.856

Es / Ec 7.503

0.24 √ fc' 0.509

4.500

Stage3-1

= 0.975 ksiI · n 2.687E+05 7.503

fdeck =M · y

=-6.968E+04 28.202

1.000

1.000

0.975 0.9 · fr 0.458 ...... NG

fbu + fl 21.315 Φf · Rh · Fyt 50.000 ...... OK

in which :

Фv =

VIII. Fatigue Limit State

■ Fatigue moment




2) Load-Induced Fatigue (AASHTO LRFD Bridge, 2012, 6.6.1.2)

■ Top Flange

The stress from unfactored DL = ksi ( - : Compression)

The stress from fatige LCB = ksi

Check Load-Induced Fatigue. [The stress from unfactored DL is the tensile stress.]

(ADTT)SL( = ) > Constant-Amplitude Fatigue Thresholds from Table 6.6.1.2.5-3 ( = )

=> Check for fatigue I

For Fatigue I, according to Table 6.6.1.2.5-3 Constant-Amplitude Fatigue Thresholds

ksi∴

ksi

= ksi ＜ ksi (warping stress = ksi)

■ Bottom Flange

The stress from unfactored DL = ksi ( - : Compression)

The stress from fatige LCB = ksi

-14.114

0.000

γ(Δf) 1.369 (ΔF)n = 12.000 0.000 ...... OK

10000.00 745.00

(ΔF)TH = 12.000

(ΔF)n = (ΔF)TH = 12.000

0.000

No Category (ADTT)SL Number of stress (n)

1 C' 10000.000 1.000

scLCB8

ComponentVu

(kips)

Shear Force -301.542

13.823

0.000 0.000

Bot(Comp.) - -11.275 -2.838 -5.282 -5.282

Bot(Tens.) - 0.000 0.000 0.122

1.369

Top(Comp.) - 0.000 0.000 0.000 0.000Stresses

Top(Tens.) - 12.184 1.639 1.369

Bot(Tens.) - 0.000 0.000

Short-term Sum

Forces

Top(Tens.) - -46451.140 -13047.469 -25900.027 -25900.027

Top(Comp.)

1.000

scLCB8

Component LCBMu (kips·in) / fc,t (ksi)

Steel Long-term

0.122

Bot(Comp.) - -46451.140 -13047.469 -25900.027 -25900.027

- 0.000 0.000 0.122 0.122

Skip this check. [ (The compressive stress from unfactored DL) > (The tensile stress from fatige LCB)]

3) Special Fatigue Requirement for Webs with transverse stiffeners (AASHTO LRFD Bridge, 2012, 6.10.5.3)



d0

D

D E·k

tw Fyw

therefore,

E·k

D Fyw

tw

Vu = ≤ Vcr = =

in which :

Vu = shear in the web due to the unfactored permanent load plus the factored fatigue load(Fatigue I)

C = 0.480

-301.542 kips 0.58 C · Fyw · D · tw 540.759 kips ...... OK

C =1.57

· ( ) = 0.480

( )2

= 122.667 > 1.40 √ = 95.000

k = 5 +

5

= 7.939( )2

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