Welcom

ANSYS Lesson 01

2009-11-11T16:26:00.001+03:00

Lesson 01

ANSYS Basics,

Sample 01: simple steel bracket analysis:

The bracket is two parts, the base plate: 200*200mm, 20 mm thick.

The second welded part: 100*200, 20 mm thick.

We will fix this to the concrete column of wall using 4 anchor bolts at the 4 corners

Idea of solution:

Modeling the structure using ANSYS, applying support as point support, and in this case we can get easy the reactions later at each bolt.

Of course we will check the steel bracket against the stresses,

Using a steel: E= 205000 N/mm²

Poisson ratio = 0.3

1- Using ANSYS:

You must note that the power of using ANSYS that the program is empty… strange? No that’s true. That mean the program doesn’t know the units you are using like force, area units… that mean you just start with any units you want then you will get the results in the same units.

For example we will use mm, N, N/mm² for E, and we will get the stress in N/mm²

Before you start I like to mention that the program looks very difficult and most of us heard that ANSYS is very advance and very difficult, forget that blab la bla. When you start you’ll be amazed it’s easy YES. And after using ANSYS you will start to feel the material behavior.

2- Understanding ANSYS:

ANSYS uses finite elements method (FEM) and before you start with ANSYS its better to have a look and understand what is FEM, its enough to read a few pages about FEM to start with ANSYS.

· Phase 1: Modeling the structure using 1:1 scale and here you have 2 choices:

a- Using the program options and drawing utility to do that. It’s a little long but easy if you have no idea about CAD-3D, and I will link some documents about how to do that later.

b- The easy way is to use AutoCAD 3d then export the file to (SAT), I did prepare the SAT file for you so you can start with it , download the file from here.

· Phase 2: assign material and mesh the structure by using one of the program elements.

· Phase 3: assign the loads and the support.

· Phase 4: analysis.

· Phase 5: reading result.

START NOW,

Open AutoCAD then

File > Export > [ choose ACIS(.sat) ] > select the bracket then ok

Open ANSYS:

- To choose the working directory:

File > change Directory

- Importing the SAT file:

File > import > SAT

- To see the solid shape of or structure:

plotCtrls > style > solid model facets > [choose normal faceting] OK

Plot > replot

Please try the command plot> . . . to see the difference between plotting lines, points..

Your structure shall look as the picture below.

You can take a picture of your structure:

plotCtrls > hard copy > to file

the picture shall be saved in the folder you choose at the beginning, this is important when you want to prepare a report showing the shape or the stresses..

now we have to prepare the volume (solid)

the shape must be one solid, check first how many solids you have

plotCtrls > numbering > VOLU volume numbers

Volume = solid, you will see two colors that mean your structure is two solids.

To make them one:

Preprocessor>Modeling>Operate>Booleans>Add>Volumes. Click on Pick All. now we have added all the volumes into a single volume.

Now you have only one solid

Defining the material:

Preprocessor>Material Properties>Material Models. In the window that

comes up, select

Structural>Linear>Elastic>Isotropic.

Put the numbers in the following:

You finished selecting the material you want to use, STEEL.

Now we have to choose an element for the FEM meshing.

The program have a lot of elements type, don’t worry you will get familiar with It soon, for now just choose the following which can cover almost all steel design.

These elements are used for the small pieces of meshing

Preprocessor>Element Type>Add/Edit/Delete... In the 'Element Types' window that opens click on Add... The following window opens:

After finishing the FEM element type we will do the meshing.

The program has a very useful tool:

Preprocessor>meshing>mesh tool

In the size controls > click SET and enter 10 in the box which is the FEM meshing element size, this number depend on the shape of the structure and the accurate of result you want. To know more check the FEM.

Click MESH > choose the structure > OK

You have the following.

Applying loads and support:

30KN= 30000N

Preprocessor>loads>define loads>apply>structural>force/moment>on nodes

Pick the point where the point load is applied and enter the load value –y

Click OK to see the following.

To apply to support conditions rotate the bracket to see the back of it, you can do that easely by holding Ctrl +right mouse click

Apply the support:

Preprocessor>loads>define loads>apply>structural>force>displacement>on nodes

Choose the four points at the four corners.. Click OK

Hold the movement in the three directions, X Y Z

Your structure is ready for analyzing, save now,

Analyzing:

Solution>solve>current LS

Wait till finish, some time it take a lot of time depending on the structure size and mesh size, and analyzing type.

Reading result:

General postproc>plot results>contour plot>nodal solu

We will choose: stress > Von Mises

Note here the value of 4985 (red color) forget it now we will know what is this later,

Assume we are using a steel 275 grade.

To tell ANSYS: print the stress value between 0 and 275

Plotctrs>style>contours>uniform contours

Enter the numbers in the next box

Check the result you get:

Rotate the bracket and enjoy the stresses values, at each point.

Regarding the BIG Value of stress 4985, because we applied the load at one point only and that is impossible in the real world. So in this point I well get a really very big stress but you can just ignore it.

The gray color in the stress picture showing us where is the area with a stress more than the value we entered 275 (Stress failure).

Regards to all

Nov. 11^th 2009

ANSYS Tutorial تعليم أنسيس

2009-11-01T16:05:00.004+03:00

الزملاء الأكارم

قمت بتجهيز ملف تعليم مبسط للمبتدأين في برنامج

ANSYS

لأنه و بعد البحث الكثير في الانترنت و لقلة المراجع عن البرنامج و بالأخص العربية بدأت و أعددت الدرس الأول و قد أرفقت الدرس بمجموعة من الأمثلة بمجلد أخر و هو بالانكليزية و لكن لا أعتقد أن أحد سيجد صعوبة في فهم المطلوب إذا ما بدأ بالدرس الأول ثم انتقل إلى الأمثلة الأخرى ليزيد المهارة و توسيع المعرفة و يمكن اعتبار الدرس الأول هو كسر الجليد مع البرنامج الجميل
لتحميل الدرس
http://rapidshare.com/files/282805536/ANSYS01.rar.html

كلمة السر

wisam

و لتحميل الأمثلة
http://rapidshare.com/files/282558144/ANSYS_START.rar.html

و تلاحظون الصورة التالية لوصلة معدنية صممت باستخدام البرنامج

شكراً و حظ سعيد

Newton's Law

2009-08-05T19:14:00.002+03:00

Newton's Second Law

"Change of motion is proportional to the force applied, and take place along the straight line the force acts."

Newton's second law for the gravity force - weight - can be expressed as

F = m g (1)

where

F = force (weight)

m = mass

g = acceleration of gravity

The force caused by gravity - g - is called weight. Note! Mass - m - is a property.

The acceleration of gravity can be observed by measuring the change of velocity of a free falling object:

g = dv / dt (2)

where

dv = change in velocity

dt = change in time

A dropped object accelerate to a speed of 9.81 m/s or 32.174 ft/s in one second.

Acceleration of Gravity in SI Units

g = 9.81 m/s²

Acceleration of Gravity in Imperial Units

g = 32.174 ft/s²

Velocity and Distance Traveled of a Free Falling Object

The velocity of a free falling object can be expressed as:

v = g t (3)

where

v = velocity

The distance traveled by a free falling object can be expressed as:

s = 1/2 g t² (4)

where

s = distance traveled by the object

The velocity and distance traveled by a free falling object:

Note! The velocity is achieved without any aero-dynamical resistance (vacuum). The air resistance will be significant for higher velocities or for object with large surface area to mass ratio - feathers or similar.

Newton's First Law

"Every body continues in a state of rest or in a uniform motion in a straight line, until it is compelled by a force to change its state of rest or motion."

Newton's Third Law

"To every action there is always an equal reaction - if a force acts to change the state of motion of a body, the body offers a resistance equal and directly opposite to the force."

Concrete Properties

Properties of normal strength Portland cement concrete

· Density : 2240 - 2400 kg/m³ (140 - 150 lb/ft³)

· Compressive strength : 20 - 40 MPa (3000 - 6000 psi)

· Flexural strength : 3 - 5 MPa (400 - 700 psi)

· Tensile strength : 2 - 5 MPa (300 - 700 psi)

· Modulus of elasticity : 14000 - 41000 MPa (2 - 6 x 10⁶ psi)

· Permeability : 1 x 10^-10 cm/sec

· Coefficient of thermal expansion : 10^-5 ^oC^-1(5.5 x 10^-6 ^oF^-1)

· Drying shrinkage : 4 - 8 x 10^-4

· Drying shrinkage of reinforced concrete : 2 - 3 x 10^-4

· Poisson's ratio : 0.20 - 0.21

· Shear strain : 6000 - 17000 MPa (1 - 3 x 10⁶ psi)

· Specific heat capacity : 0.75 kJ/kg K (0.18 Btu/lb_m^oF (kcal/kg^oC))

Frictional force can be expressed as

F_f = μ N (1)

where

F_f = frictional force (N, lb)

μ = static (μ_s) or kinet

Basic Information you may need

2009-08-05T17:58:00.001+03:00

Original web site is engineering tool box.com

The content of the original site is very big so I pick up what structural engineer needs.

Copying the content for educational reasons. All rights belongs to the original owner.

Poisson's ratio is the ratio of the relative contraction strain, or transverse strain normal to the applied load, to the relative extension strain, or axial strain in the direction of the applied load.

Can be expressed as υ = - εt / εl (m/m) (1)

Where: υ = Poisson's ratio

εt = transverse strain

εl = longitudinal or axial strain

Strain can be expressed as ε = dl/L (2)

Where: dl = change in length

L = initial length

For most common materials the Poisson's ratio is in the range 0 - 0.5.

Material	Poisson's Ratio
Upper limit	0.5
Aluminum	0.334
Aluminum, 6061-T6	0.35
Aluminum, 2024-T4	0.32
Glass	0.24
Stainless Steel 18-8	0.305
Steel, cast	0.265
Steel, Cold-rolled	0.287
Steel, high carbon	0.295
Steel, mild	0.303
Concrete	0.2

Stress can be expressed as stress = F / A (2)

Stress = (N/m²) (lb/in², psi)

F = force (N) (lb)

A = area of object (m²) (in²)

Young's Modulus (Tensile Modulus)

Young's modulus or Tensile modulus can be expressed as

E = stress / strain = (F / A) / (dL / L) (3)

Where

E = Young's modulus (N/m²) (lb/in², psi)

Hooke's Law

One of the properties of elasticity is that it takes about twice as much force to stretch a spring twice as far. That linear dependence of displacement upon stretching force is called Hooke's law which can be expressed as

F_s = -k dL (4)

Where

F_s = force in the spring (N)

k = spring constant (N/m)

dL = elongation of the spring (m)

Yield strength

Yield strength, or the yield point, is defined in engineering as the amount of strain that a material can undergo before moving from elastic deformation into plastic deformation.

Ultimate Tensile Strength

The Ultimate Tensile Strength - UTS - of a material is the limit stress at which the material actually breaks, with sudden release of the stored elastic energy.

1 N/m² = 1x10^-6 N/mm² = 1 Pa = 1.4504x10^-4 psi

1 psi (lb/in²) = 144 psf (lb_f/ft²) = 6,894.8 Pa (N/m²) = 6.895x10^-3 N/mm²

Modulus of Rigidity (or Shear Modulus) is the coefficient of elasticity for a shearing force. It is defined as "the ratio of shear stress to the displacement per unit sample length".

Modulus of Rigidity can be experimentally determined from the slope of a stress-strain curve created during tensile tests conducted on a sample of the material.

Definition of Modulus of Rigidity:

The ratio of shear stress to the displacement per unit sample length

1 GPa = 0.145x10⁶ psi (lb_f/in²)

1 x 10⁶ psi = 6.9 GPa

The Bulk Modulus Elasticity can be expressed as

E = - dp / (dV / V) (1)

Where E = bulk modulus elasticity

dp = differential change in pressure on the object

dV = differential change in volume of the object

V = initial volume of the object

The Bulk Modulus Elasticity can be alternatively expressed as E = dp / (dρ / ρ) (2)

Where: dρ = differential change in density of the object

ρ = initial density of the object

An increase in the pressure will decrease the volume (1). A decrease in the volume will increase the density (2).

The SI unit of the bulk modulus elasticity is N/m² (Pa)

The imperial (BG) unit is lb_f/in² (psi)

1 lb_f/in² (psi) = 6.894 10³ N/m² (Pa)

A large Bulk Modulus indicates a relative incompressible fluid.

Bulk Modulus for some common fluids can be found in the table below:

Material	Shear Modulus
Material	(10⁶ psi)	(GPa)
Aluminum Alloys	3.9	27
Aluminum, 6061-T6	3.8	24
Aluminum, 2024-T4	4	28
Carbon Steel	11.2	77
Concrete	3	21
Glass, 96% silica	2.8	19
Plywood	0.09	0.62
Structural Steel	11.5	79.3
Stainless Steel	11.2	77.2
Steel, Cast	11.3	78
Steel, Cold-rolled	10.9	75

Mass and Weight - the difference! - What is weight and what is mass? An explanation of the difference between weight and mass.

Specific Weight can be expressed as

γ = ρ g (2)

where

γ = specific weight (N/m³)

ρ = density (kg/m³)

g = acceleration of gravity (m/s²)

The SI-units of specific weight are N/m³. The imperial units are lb/ft³. The local acceleration g is under normal conditions 9.807 m/s² in SI-units and 32.174 ft/s² in imperial units.

Example - Specific Weight Water

Specific weight for water at 39 ^oF (4 ^oC) is 62.4 lb/ft³ (9.81 kN/m³) in imperial units. Specific weight in SI units can be calculated like

γ = (1000 kg/m³) (9.81 m/s²)

= 9810 N/m³

= 9.81 kN/m³