MDSOLIDS Software Learning

Assignment No: 02

Name: Adil Iqbal

NUST No: ME-1243

Assignment No: 02

Semester: ME - V (A)

Total No. of Words:

EULER’s BUCKLING

The question at hand is as follows: Determine the critical load of an aluminum tube that is 1.5 m long and has a 16 mm outer diameter and a wall thickness of 1.25 mm. Use E = 70 GPa.

This question has been taken from Mechanics of Materials (4th Edition) by Beer, Johnston and DeWolf. The question number is 10.10 on page number 620.

Data –

Length = 1.5 m Outer diameter = 16 mm Wall Thickness = 1.25 mm E = 70 GPa

Required – Critical Load on the aluminium tube is to be found out.

Practical Solution: The question in hand can be solved practically through the software MDSolids. The first step to be taken is to decide which section to be chosen in the software; as the question is of Euler’s buckling of columns, thus from the main screen of the software the columns section is selected. The columns tab is opened and a new window pops up which is our primary solution window. The interface of the software and the columns section is illustrated below:

Columns Section is selected

Interface of the Column Section

The first step is to define the shape and dimensions of the section/column under consideration. For that, the cross-section tab is selected from the primary window. A drop down shows you two options to choose from (select and design) and from them the select option is chosen. This allows you to choose from predefined shapes and to alter their dimensions according to your requirements. As we have to work on a hollow tube we choose cylindrical pipe/tube from simple configurations.

Cylinder (Pipe/Tube) is chosen

Once the proper shape is decided and selected from the section properties module, its dimensions are to be input into the software. The parameters to be input are the outer diameter, inner diameter and the elastic modulus E. The input values are shown on the next page.

Select this to define column geometry

Once the geometry of the section is finalized, we return to the column buckling module where the column length parameter is fed, the supports are decided and the results are computed. The results that are obtained by the software is shown below:

Results Obtained

Thus, the critical Euler load value is equal to 0.4872 kN, while the slenderness ratio is 286.609.

Analytical Solution

Data –

Length = 1.5 m Outer diameter = 16 mm Wall Thickness = 1.25 mm E = 70 GPa

Pcr = ? Pcr = π2∗E∗IL2

I = π4 * ( Co

4 – Ci4 ) where Co is outer radius = 8 mm (0.008 m)

Ci is inner radius = 6.75 mm (0.00675 m)

Therefore, I = π4 * ( 0.0084 – 0.006754 ) = 1.586 x 10-9 m4

So Pcr = π2∗70E9∗1.58E-9

1.52 = 487.16 N = 0.48716 kN

Discussion

The problem of finding the critical load Pcr was done both practically and analytically and the solution matched with the approximate answer being 0.4872 kN. This shows that the assumptions taken by us (perfectly straight column and purely axial load) during analytical solution were correct.

The software is very helpful for solving column buckling questions as a variety of geometry can be selected and be analyzed. In addition to the Euler formula, we can also analyze the column with other formulas (Perry Robertson and AISC ASD formula etc.). The critical stress against slenderness ratio graph can be plotted/inspected and the explanation is comprehensive enough to determine whether the Euler equation is suitable for the problem being solved. The software falls short when calculating the deflection of the buckled column.

Critical Stress vs Slenderness Ratio Graph

Pressure Vessels (Thin/Thick Cylinders)

The question that is being solved is as follows: Determine the hoop and longitudinal stresses set up in a thin boiler shell of circular cross section, 5 m long and of 1.3 m internal diameter when the internal pressure reaches a value of 2.4 bar (240 kN/m2). What will then be its change in diameter? The wall thickness of the boiler is 25 mm. E = 210 GN/m2, v = 0.3.

This question has been taken from Mechanics of Materials Volume 1 (2nd Edition) by author E.J Hearn. The question number is 9.1 on page number 213.

Data –

Length L = 5 m Internal diameter d = 1.3 m Internal Pressure P = 240 kPa Wall Thickness t = 25 mm (0.025 m) E = 210 GPa, v = 0.3

Required – Hoop and Longitudinal stresses set up in the thin boiler shell and the change in diameter.

Practical Solution: The practical solution to the given problem is found out by solving it through MDSolids software. Therefore, from the main screen of the software, the pressure vessels section is chosen. Once in the pressure vessel section, three types of vessels are provided (cylindrical, open ended and spherical). The provided problem is to be solved through open ended section type thus that type is chosen. The type is illustrated below:

The values of inside radius (=0.65 m), wall thickness (=0.025 m), internal pressure (=240 kPa), E (=210 GPa) and v (=0.3) are input into the software. Make sure that the units of lengths, stresses and pressures are matching with the data above. The data is input in the illustration below:

The data is computed by pressing the compute button. In the description dialogue, the required stress and strain values are shown. The solution is illustrated in the screenshots below:

Strain Values for Open Ends Condition

Strain Values for Closed Ends Condition

Thus, the solution given by the software for the hoop stress is 6.24 MPa while the longitudinal stress which in this case is equal to the shear stress value is 3.12 MPa. The diametrical strain is equal to the hoop strain for a thin cylinder. For open end condition, the diametrical strain is 29.71 x 10-6 m/m while for closed ends the strain is 25.26 x 10-6 m/m. Thus,

For open end condition, the change in diameter = εH x d = 29.71 x 10-6 x 1300 = 0.0386 mm.

For closed end condition, the change in diameter = εH x d = 25.26 x 10-6 x 1300 = 0.0328 mm.

Analytical Solution

The data provided is

Length L = 5 m Internal diameter d = 1.3 m Wall thickness t = 0.025 m E = 210 GPa, v = 0.3 Internal Pressure P = 240 kPa

Hoop and Longitudinal Stresses are to be found out as well as the change in diameter is to be calculated.

Thus

ϬH = Pd2t ϬH =

240,000∗1.32∗0.025

ϬH = 6.24 MPa

For longitudinal stress, ϬL = Pd4 t ϬL =

240,000∗1.34∗0.025

ϬL = 3.12 MPa

To find the change in diameter, we use the following equation,

Change in Diameter = Pd2

4 tE * (2-v)

Thus change in diameter = 240,000∗1.32

4∗0.025∗210E9 * (2-0.3)

Change in diameter = 3.28E-5 m = 0.0328 mm

Discussion

The problem at hand was solved both practically and analytically. The solutions that were calculated through the software matched those that were found out by analysis. The stress values of hoop and longitudinal stresses were 6.24 and 3.12 MPa respectively while the change in diameter calculated through the two methods match within the limits of analytical assumptions.

If open ends are assumed for the cylinder, the change in diameter is 0.0386 mm while if closed ends are assumed the change in diameter is 0.0328 mm. Analytically, the change in diameter is 0.0328 mm which shows that the assumption that the cylinder is closed end is correct. The software MDSolids provides us with an easy interface and simple solution. The solutions of

stresses and pressures are given in normal and shear terms, thus, the exact geometry and nature of the cylinder should be inspected first and the user should have precise knowledge of the geometry. The basic data that can be input in the software is limited to three parameters so for a cylinder with an external and an internal pressure is difficult to be solved by MDSolids. Also, the software is limited to calculate stress and strains and basic dimension only; thus, no change in volume or diameter can be calculated directly.

MDSolids however provides us with immediate graphical solutions (Mohr’s Circle) too which is helpful in polishing the knowledge of a student. The failure theories can be tested and verified through the software in an easy and quick manner. The Tresca failure theory solutions for the above question is illustrated below

Therefore the software was very successful in developing the understanding and application of the question and providing quick and easy solutions.

Flexure (Unsymmetrical Bending)

The question that is to be analyzed is: A C10 x 15.3 channel section is subjected to a bending couple M = 15 in. kips having its vector at an angle θ = 10o to the z axis. Calculate the stresses ϬA and ϬB at points A and B respectively.

This question is selected from Mechanics of Materials by Timoshenko. The question number is example 1 on page no. 478.

Data –

Channel Section C10 x 15.3 Couple Moment 15 kips. in @ θ = 10o

Required - Stress set up at A and B.

Solution – The section is illustrated in the diagram below

Practical Solution: Practical analysis is done through the software MDSolids. From the main screen, the flexure module is selected. A new window would be opened as illustrated below

Point A

Point B

First of all we need to define the section geometry. Thus, we open the cross section option and choose a standard AISC Shape C10 x 15.3 from the section properties module. Once the section is finalized, we choose the unsymmetrical bending tab from the flexure module. The value of moment and the angle that it makes with the z axis is input with the proper units. The results are then computed. The input data is shown below:

The next illustrations show the summarized results obtained by the software.

ϬA = Max. Tension Stress = 3254.6 psi ϬB = Max. Compression Stress = 1923.8 psi Angle β = 79.17o

Analytical Solution

The data provided is

Channel Section C10 x 15.3 Couple Moment 15 kips. in @ θ = 10o

The stress values at A and B are to be calculated.

Solution – From table E-3, the following data is found out

Centroid C = 0.634 in. IY = 2.28 in4 , IZ = 67.4 in4 (as IZ in this case = I1-1 and IY = I2-2) Flange Width = 2.60 in

For the coordinates of points A and B:

Point A – yA = -0.5 * Web Length = -0.5 * 10 = -5.0 in ZA = Flange Width – centroid length C = 2.60 – 0.634 = 1.966 in

Point B – yB = 0.5 * Web Length = 0.5 * 10 = 5.0 in ZA = -1 * centroid length C = –0.634 in

Moments about z and y axis are given by

MY = M sin θ = 15 sin 10o = 2.605 in. k MZ = M cos θ = 15 cos 10o = 14.77 in. k.

Thus, stress at A is ϬA = MyZaIy −MzYaIz = 2.605∗1.966

2.28−14.77∗−5

67.4 = 3340 psi

Thus, stress at A is ϬB = MyZbIy −MzYbIz = 2.605∗−0.634

2.28−14.77∗5.0

67.4 = - 1820 psi

Angle β is found out by β = tan-1 IzIy∗tan θ = tan-1 67.42.28

∗tan 10o = 79.1o

Note : Stress at A is tensile while that in B is compressive.