Mechanics of Solids II

Week 8MECH3361Recap Contact ProblemsCasesTangential defNormal Displacementcontact stress

Stress Methods (Flowchart): Select stress function (combination of several) Calculate stresses , , Use B.C. to determine coefficients Fully determine stress functions Hookes law to determine strains Strain-Displacement to derive displacement function Disp B.C. to fully determine displacement functions. Solution to a circular hole in a plate

When , reaches the maximum

Strain energy density (SED)

Distortion energy density (DED)

Chapter 7 Plasticity and Failure

7.1 Introduction Plasticity should be avoided and most of mechanical system should work in elastic regime (e.g. gear and spring). Plastic permanent deformation should be made in metal forming.

Questions: Are the basic equations still valid? Criteria to enable us to present Relationship of stress strain Can we still use the methodology established in the sites

7.2 Octahedral Shear Stress

Definition:

Considering the principal directions as the coordinate axes, a special oriented plane which normal vector makes equal angles with each of the principal axes, i.e. having direction cosines equal to , is called an octahedral plane.

Octahedral normal stress From Chapter 1 stress transformation, we have

Since the octahedral plane is rotated from the principal stress direction whose shear components are zero, thus we can have

Octahedral normal stress:

is the mean principal stress or hydrostatic stress. This value is the same in all eight octahedral planes.

Octahedral shear stress

The octahedral shear stress = resultant shear stress on an octahedral plane

Total stress on the Octahedral plane:

The octahedral shear stress can be calculated from the total stress and normal stress as

Octahedral shear stress:

7.2 Distortion Energy

Work-energy principle: External work = Internal Strain energy (if no heat energy loss)

Strain energy density (strain energy per unit volume) (SED)

Apply the Hookes law to replace strain by stress:

Or apply the Hookes law to replace stress by strain:

When the x, y, z are principal directions, U can be express in terms of principal stresses

where , is the first strain invariant and K is bulk modulus.

Since =volume change rate, therefore represents SED due to volume change caused by hydrostatic stress.

corresponds the shape change due to shear stress (recall the octahedral shear stress), namely distortion energy density (DED)

7.3 Plasticity under Simple Tension

Typical tensile testing diagram:The yielding occurs after yielding limit in the test as shown. Unloading leads to plastic strain (permanent strain) and elastic strain (recoverable strain).

Beyond this yielding point, material may behave in different ways of the plastic deformation as shown below:

Mild steelStainless steelAl alloy

To simplify such curves, people introduced some simplified models:

Remarks: When the unloading takes place after plasticity has occurred, the deformation will be completely elastic. Thus we can use the method of elastic analysis to study the stress and deformation during unloading. After unloading, then load it again. The strain-stress behaviour will be first elastic. If the material has a hardening behaviour, it will not yield until the reloading stress reaches new yield stress, namely Y. Since Y is higher than Y, the material in reloading shows a hardening behaviour. Increase in yield stress and loading-reloading phenomena is called work-hardening

The bilinear model sometime is not accurate enough. For this reason some nonlinear models may be needed.

Power law: where material constants A= strength coefficient and n=strain-hardening exponent.

Rambery-Osgood model: K and n are material constants to be determined by experiments.

Curve fitting: Sometime fit the testing curves is needed asExample 7.1

A cylinder bar made of steel is subjected to a uniform tension P along its axis. The maximum axial strain was measured to be . Youngs modulus E=210GPa and its yield stress is Y=600MPa and the materials stress-strain curve shows almost no hardening. (a) Calculate the plastic strain in the bar at the above max strain. (b) If the bar is unloaded at this max strain, what is the residual strain remaining in the bar after complete unloading (short answer question, 2009 Final Exam).

Soln: (a) No hardening, the steel can be idealised as an elastic/perfectly plastic material as shown.

Hence at the maximum strain of , the plastic strain in the tested bar is

(b) When unloading, all the elastic strain will disappear and the plastic strain remains. Thus the residual strain after complete unloading is equal to .

Example 7.2 For the structure loaded at joint O shown below, determine (a) elastic limit Pe; (b) plastic limit load Pp.

SolnStep 1: determine internal forces. From the F.B.D. (right sub-figure)

We cannot solve for three unknowns using the two equations. Other compatibility condition is needed.

Step 2: Compatibility condition of strain (initial length of Bars 1 and 3 is ):

Elongation of bar 1 and 3 is approximately ,

Step 3: Stress inside the bars Step 4: Calculate the stresses

From Hookes law: and

Plug into stress eqn#2:

We obtain: Thus: Step 5: Elastic deformation and initial yielding:

Since , , which implies that yielding occurs in Bar 2 first.

Lets assume that the corresponding load in this bar reaches the yield limit . This is

Step 6: Elastic-plastic deformation and plastic limit load:

Bar 2 reaches elastic limit and undergoes plastic deformation if further increasing the load. Since the material is elastic/perfectly plastic, we always have . Thus from

, one can have

This reaches plastic limit where all the bars start undergoing different degrees of plastic deformation:

Step 7: Comparison of Elastic-plastic loading:

For example, , , which means that if we allow the frame to work in its elastic-plastic deformation regime, the load-carrying capacity is much increased.

7.4 Plasticity under Complex Stress StatesYield Criterion and initial yield conditionIt is assumed that when the principal stresses and some parameters satisfy a critical condition, plastic yielding occurs. Mathematically,

is called yield criterion or initial yield condition (Initial = before the critical state is reached, deformation is purely elastic). If plot it graphically, is a surface in the space of coordinates (as shown), namely initial yield surface.

Tresca criterionTresca observed that material flow seems to be along the direction of the maximum shear stress.

Margin

Or:is called Tresca criterion.

To determine , one can conduct an uniaxial tension test, in which metal will yield when . Thus: Finally, the Tresca criterion becomes:

von Mises criterionExperiment has shown that metals do not yield under high hydrostatic stress. Huber (1904) proposed that plastic yielding occurs when the distortion energy density equal or exceeds that of the same material under uniaxial tension. In 1913, von Mises suggested that metal will yield under a combination of the principal stresses

which is called von Mises criterion.

To determine , one can conduct an uniaxial tension test in the 1st principal stress direction, where and vanish, as.

Thus . The von Mises criterion becomes

Or in Cartesian coordinate system

Example 7.3A thin square plate is subjected to a set of uniform stresses on its edges as show, where =0.5. Y=600MPa. Find the maximum stress beyond which plastic deformation appears (Final exam 2012).Soln

Step 1: determine the principal stresses: Since plane stress problem ,

Since

Thus principal stresses

Step 2: According to the Tresca criterion :

Thus the maximum stress must be:

Step 3: according to the von Mises criterion :

Tresca criterion is lower in this case.

7.5 Failure TheoriesDuctile materials: The maximum Shear Stress Theory:

where is the safety factor

The maximum distortion energy theory:

Brittle materialsThere is no yield for brittle material in general. The maximum normal stress theory

is the failure normal stress

The maximum normal strain theory

is the failure normal strain.

Since from Hookes law:

Chapter 8 Finite Element Method

8.1 IntroductionConceptThe finite element method (FEM), or finite element analysis (FEA), is based on the idea of building a complicated object with simple blocks, or, dividing a complicated object into small and manageable pieces. Application of this simple idea can be found everywhere in daily life and engineering. Lego (kids toy) for complex shapes/patterns Bricks for Buildings Approximation of the area of a circle:

Triangular area of each element: .

Why Finite Element Method? Applications of analytical solutions are limited. Experimental solutions are expensive and time-consuming. Design analysis: hand calculations, experiments, and computer simulations FEM/FEA is the most widely applied computer simulation method in engineering Closely integrated with CAD/CAM applications

Applications of FEM in Engineering Mechanical/Aerospace/Civil/Automobile Engineering Structure analysis (static/dynamic, linear/nonlinear) Thermal/fluid flows Electromagnetics Geomechanics Biomechanics

A Brief History of the FEM 1943 Courant (Variational methods) 1956 Turner, Clough, Martin and Topp (Stiffness) 1960 Clough (Finite Element, plane problems) 1970s Applications on mainframe computers 1980s Microcomputers, pre- and postprocessors 1990s Analysis of large structural systems

Procedures Divide structure into pieces (elements with nodes) Describe the behaviours of the physical quantities on each element Connect (assemble) the elements at the nodes to form an approximate system of equations for the whole structure Solve the system of equations involving unknown quantities at the nodes (e.g., displacements) Calculate desired quantities (e.g., strains and stresses) at selected elements

Computer Implementations Preprocessing (build FE model, material properties, loads and boundary conditions) FEA solver (assemble and solve the system of equations) Postprocessing (sort and display the results)

Commercial FEM Packages ANSYS (General purpose, PC and workstations) ABAQUS (General purpose, Nonlinear and dynamic analyses) Strand7 (General purpose, fast solver and user-friendly) Cosmosworks

Type of elements

8.2 Spring Element

Step 1: Single Spring System Two nodes: i, j Nodal displacements: ui, uj (m) (x-displacement only) Nodal forces: (N) Spring constant (stiffness) k (N/m)

Spring force-displacement relation:

where elongation Consider the equilibrium of forces for the spring.

At node i, we have

At node j,

In a matrix form:

Or: where: k = elemental stiffness matrixu = (element nodal) displacement vectorf = (element nodal) force vectorNote that k is symmetric. Is k singular or nonsingular?

Multiple spring system

For element 1:

For element 2:

where and is the (internal) force acting on local node i of element m (i = 1, 2).

Consider equilibrium of the spring system at nodes: (F1, F2, F3 are total forces in the nodes) At node 1: At node 2: At node 3:

Thus:

Or in a total matrix for:

i.e.where K is named the global stiffness matrix (structure matrix) for the spring system.

An alternative way of assembling the whole stiffness matrix: Enlarging the stiffness matrices for elements 1 and 2 according to the nodal numbers, we have:

Element #1:

Element #2: Add these two expanded equations:

ThusBoundary and load conditions (for a problem as shown below)

Fix the left end and the spring system is loaded by external forces:

Because of zero displacement in u1=0, the above matrix Eq can be separated (equivalent) into

and

The unknowns are and reaction force F1

Solve the equation set, we obtain the displacements:

Thus the reaction force:

Remarks Both elemental stiffness and global stiffness matrices are symmetric and positive definite. Check the equilibrium of the external forces Suitable for other stiffness analysis (e.g. torsional spring etc)

Example 8.1

For a three-spring system shown, , , , and . Determine: (a) global stiffness matrix, (b) and , (c) reaction forces at nodes 1 and 4, (d) force in spring #2.

SolutionStep 1: Elemental stiffness matrices

, , Step 2: Expand the elemental stiffness matrices according to the node numbers

, , Step 3: Superposition (add) all elemental stiffness matrices for global stiffness matrix

Thus the global stiffness matrix: which is symmetric and banded.

Step 4: Global equilibrium equation system

Step 5: Apply boundary condition

Delete (or separate) the 1st and 4th rows and columns, we have non-vanishing equations

i.e.

Step 6: Solve for the linear equation set to obtain unknown displacements

Solve for the equations: Step 7: Reaction forcesFrom the dropped equations:

Example 8.2For the spring system with arbitrarily numbered nodes and elements, as shown above, find the global stiffness matrix.

SolutionStep 1: Construct the element connectivity (4 elements and 5 nodes)

which specifies the global node numbers corresponding to the local node numbers for each element.Step 2: Elemental stiffness matrices

Step 3: Expand the elemental stiffness matricesFor the un-ordered stiffness matrices (the stiffness matrix should be positive definite)

Step 4: Superposition (add) for global stiffness matrix

8.2 Bar ElementsBar Elements in 1-D

Known: L-length of bar, A-cross-sectional area, E-Youngs modulus, displacement , strain , stress Strain-displacement: Hookes law: Finite Element Method 1Step 1: Assuming that the displacement u is varying linearly along the axis of the bar, i.e.,

in which, when Step 2: Calculate strain and stress

(=elongation = relative displacement between node i and j)

For a bar, , where stiffness Step 3: Elemental stiffness matrix:The bar is acting like a spring in this case and we conclude that element stiffness matrix is

Step 4: Equilibrium equation: Remarks Degree of Freedom (dof) - Number of components of displacement vector at a node. For 1-D bar element: one dof at each node (i.e. the u displacement). Physical Meaning of the Coefficients in k The jth column of k (here j = 1 or 2) represents the forces applied to the bar to maintain a deformed shape with unit displacement at node j and zero displacement at the other node.

Finite Element Method 2 A Formal ApproachLets derive the same stiffness matrix for the bar using a formal approach which can be applied to many other more complicated situations.Step 1: Displacement: Define two linear shape functions (interpolation function) as follows

,

where natural coordinate: (). As such, we change the coordinate system from [x1,x2] to [0,1]

Express displacement as (interpolation of displacement)

where when ,

when ,

in a matrix form:, where is nodal displacement vector.Step 2: Strain-displacement

(note that vector u is not a function of x)where B is the elemental strain-displacement matrix.

Step 3: Stress (Hookes law)

Step 4: Apply Strain energy-work principle

The work done by the two nodal forces is:

Since we have

Thus:

This is

where is the stiffness matrix

which is the same as we derived using the direct method.This is a general result which can be used for the construction of other types of elements. This expression can also be derived using other more rigorous approaches, such as the Principle of Minimum Potential Energy, or the Galerkins Method.Note that the strain energy in the element can be written as

Example 8.3 (2011 Quiz question) The three bar structure is fully clumped at ends A and D as shown. Solve the following problems by using finite element method. (1) Write the elemental stiffness matrices. (2) Give the global equilibrium equation. (3) Calculate the displacement at nodes B and C. (4)Calculate the reaction forces at points A and D.

Solution Step 1: Connectivity: Elements: AB=(1), BC=(2), CD=(3). Nodes: A=1, B=2, C=3, D=4.Step 2: Elemental stiffness matrices of bars

Bar elements:

Thus: , , Step 3: Expand the elemental stiffness matrices according to unknown vector u:

, , Step 4: Superposition (add) all elemental stiffness matrices for global stiffness matrix

i.e. the global stiffness matrix: , which is symmetric and banded

Step 5: Equilibrium equation system:

Step 6: Apply boundary condition

Delete the 1st and 4th rows and columns, i.e.Step 7: Solve for the linear equation set to obtain displacements

Solve for the equations: Step 8: Reaction forces: From the dropped equations:

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