Программа "Booty Challenge" от Ashy Bines. Introduction.
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O scurta introducere, destinata a umple momente de asteptare, inainte de a incepe prelegerea in cadrul Zilelor Biz- 2016
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An introduction to learn quickly python.
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Intro to XTRACT program
1. Use Xtract on server (located in TRC folder) 2. Download trial version from http://www.imbsen.com/software‐XTRACT.html
SECTION ANALYSIS USING XTRACT (A QUICK INTRO) Need to input info in all 5 lines in order to proceed.
Enter the name of the section, usually start from a template or a previously defined section and select preferred units system.
SECTION TYPE AND TRANSVERSE REINFORCEMENT Select section type (e.g. circular or rectangular columns, rectangular beams, T‐beams, rectangular walls and enclosed composite)
Select transverse reinforcement layout.
Specify transverse reinforcement bar size and spacing (s). (Shown: hoops and crossties #4@3”) If allowed to, the program will use this info to calculate effective confinement of the core. Usually we input the confinement parameters (i.e. effective confining stresses) by hand using the procedures presented in the Course.
SECTION DIMENSIONS AND LONGITUDINAL REINF.
Specify dimensions and clear cover.
Specify amount of longitudinal reinforcement and bar size (dblong). (Shown: 8#8) Check the summary of the section properties and the longitudinal reinforcement ratio (total steel area divided by gross column area)
MATERIALS DEFINITION: UNCONFINED CONCRETE Add new unconfined cover concrete material. This is just a reference value used by the solver algorithm to define a “First Yield Criterion ”. Usually set to 70% o.
fr= 7.5(f’c)1/2 (or around 10%f’c) Note the sign convention! cu = 0.004 (ACI value typically taken as 0.003) sp = 0.004 (other value typically used is 0.006) Ec = 57(f’c)1/2 (f’c in psi) Alternative
When dealing with confined sections, set to a very large number (e.g. equal 1) to avoid the solver algorithm from reaching a “false limiting value”. This will not give unrealistic higher capacities because the strength is set to zero for such large strains (i.e. Post Crushing Stess=0)
Cf’c (large column nominal concrete compressive strength with reduction factor C=0.85. For beams use f’c)
cu = 0.004 sp = 0.006
MATERIALS DEFINITION: CONFINED CONCRETE Add new confined core concrete material.
MESH SIZE & SECTION CREATION Here the size of the discretization mesh can be adjusted (i.e. define the size of the triangles into which the section is divided). In general try to avoid making it much smaller than the default value because it usually shows an error that has to due with memory management Click Here to Begin!
EXERCISE 1: INCREMENTAL AXIAL LOADING The objective of this example is to show the response of a column with confined core under incremental axial loading (similar to Problem 5 – Homework 3) Go to Loading/Moment Curvature (Crtl+T) Give a name to the loading pattern Initial conditions prior to the incremental axial load analysis. For this example we will start from zero. (Usually, we set an axial load for columns or walls before doing incremental moment loading) Here we set which type of incremental demand we will impose to the section
(+) for compression and (–) for tension incremental loading
EXERCISE 1: INCREMENTAL AXIAL LOADING Hit run, then check the project manager and explore the outputs of the analysis
The failure mechanism is governed by the material that first reaches the ultimate strain. In this example it could be either the confined concrete reaching cu = 0.029 or the steel reaching su = 0.14. Here we can analyze what is the stress‐strain relation of each material at every stage of the response. Gives a summary of the response of the section (Py, Pu, ductility capacity u/y, etc.)
EXERCISE 2: MOMENT-CURVATURE ANALYSIS The objective of this example is to show the moment‐curvature (M‐) response of a column with an applied axial load Go to Loading/Moment Curvature (Crtl+T) Give a name to the loading pattern Apply a typical axial load of 10%Agf’c. (+) for compression Will apply moment increments until either the confined concrete or the longitudinal steel reach the limiting strain (cu = 0.029 or su = 0.14)
(–) for compression of the extreme top fiber
EXERCISE 2: MOMENT-CURVATURE ANALYSIS Hit run, then check the project manager and explore the outputs of the analysis
EXERCISE 2: MOMENT-CURVATURE ANALYSIS
Step 6 in the M‐ relation State of the section @Step6: bottom steel yielded, neutral axis above middepth steel
State of the concrete @Failure. This is why the analysis ended!!!
State of the bottom steel @Failure
State of the bottom steel @Step6
State of the extreme fiber of confined concrete in compression @Step6
EXERCISE 2: MOMENT-CURVATURE ANALYSIS
EXERCISE 3: PM-INTERACTION DIAGRAM The objective of this example is to show how to construction an axial‐load‐moment (PM) interaction diagram. Go to Loading/PM Interaction (Crtl+T) Give a name to the loading pattern Only half diagram is required because the section is symmetric about the horizontal and vertical axes. 0 deg will apply the moment increments around the X‐axis
Here the reduction factor given by the building codes can be applied.
Default values will give a smooth and accurate curve
EXERCISE 3: PM-INTERACTION DIAGRAM Hit run, then check the project manager and explore the outputs of the analysis