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Question 1

What are the strong form equations in 2D?

Answer 1

dσxx/dx + dσxy/dy + bx = 0

dσyx/dx + dσyy/dy + by = 0

Question 2

How do you integrate by parts the strong form equation in 2D?

Answer 2

• multiply by an arbitrary displacement and integrate over the volume
• Integrate by parts the first term in the x direction
• integrate by parts the second term in the y direction

Question 3

When converting between string and weak form in 2D what does the left and right hand side of the equation represent?

Answer 3

LHS: Represents K * d

RHS: Represents external forces. The first two terns are boundary forces and the last term is the body forces

Question 4

What do we know at every point on the boundary and for every direction?

Answer 4

We eitehr know the force or the displacement.
Therefore if u is known the du = 0 and if the distributed force is known n * σ = t, where t is the reactions (force).

Question 5

What does Γ, Γe and Γn mean?

Answer 5

Γ = total boundary
Γe = Boundary where ui is known
Γn = Boundary where t is known

Question 6

What does σ =?

Answer 6

σ = D * L * u
D = Elastic tensor E/(1+v)(1-2v) […]
L = Deformation tensor [d/dx,0,0 ; 0 d/dy ….]
u = displacement vector

Question 7

What does B and N =?

Answer 7

B = L * N = [ dN1/dx, 0 , dN2/dx ….] 3x4 matirx for N =2
L = [ d/dx,0 ; 0,d/dy ; d/dy,d/dx] 3x2
N = [N1,0,N2,0 ; 0,N1,0,N1] 2x4

Question 8

How is the isoparametric formulation achieved?

Answer 8

By mapping the geometry from a global (cartesian) XY-coordinate system to a ξη-coordinate system (local).

The region in the xy-plabe is the global domain and the ξη region from +1 to -1 is called the parent domain.

Question 9

How do yoiu denote going from global to local and vise versa?

Answer 9

From global to local:
ξ = ξ(x,y), η = η(x,y)

From local to global:
x = x(ξ,η), y = y(ξ,η)

Question 10

What are the shape functions in terms ξ & η for a 4-noded element?

Answer 10

N1(ξ,η) = 1/4 * (1 - ξ)(1- η)
N2(ξ,η) = 1/4 * (1 + ξ)(1- η)
N3(ξ,η) = 1/4 * (1 + ξ)(1+ η)
N4(ξ,η) = 1/4 * (1 - ξ)(1+ η)

Question 11

What does the Jacobian matrix equal and how is it used?

Answer 11

J = [ dx/dξ , dy/dη ; dx/dη , dy/dξ]

converts the shape function from local to glable coordinates

{dN/dx ; dN/dy} = J^-1* {dN/dξ ; dN/dη}

Question 12

How is the stiffness matrix [K] found using parental domain

Answer 12

[k] = ∫(1 to -1) ∫(1 to -1) B^T * D * B * det( J ) * dξdη

Question 13

When doing Guass integration what do the number of points, integration points and associated weight factor refer to?

Answer 13

The number of integration points:
Is the number of points along the element that you integrate under. This equal 2n - 1 where n is the order of the integral.

Integration points is the distance from 0 to the integration point

The associated weight factor is the length of the base pf the integration out of 2.

Question 14

How would you use Guass integration in 2D?

Answer 14

[K] = ∫∫ f(ξ,η) * dξdη = Σi=1Σj=1 wi * wj * f(ξi,ηj)

Question 15

How do you convert inertia forces (-ρu..) into weak form?

Answer 15

Integrate it over the volume and times it by an arbitrary displacement.

V ∫ du * (ρ * u..) dV
u.. = acceleration
ρ = mass density

Question 16

What is the difference between a consistent and lumped mass matrix?

Answer 16

A consistent mass matrix is defined by using the shape functions {N}. If you sum up all of the rows and columns you get 1.

Lumped mass matrix is when you sum up all the particle masses into the ith column. creating a diagonal matrix which also equals 1 if all the masses on the diagonal are summed.

Question 17

Why do we use lumped mass matrices over consistent mass matrices?

Answer 17

• It is an easier form to use and it occupies less storage.
• They yield better results in wave propagation problems because fewer spurious oscillations are generated.
• Better for central difference method

Question 18

How does the central difference method (explicit) work to calculate displacement, velocity and acceleration?

Answer 18

1. Calculate d..(0) using the initial conditions d(0) and
   d.(0).
2. Use d..(0) to calculate d.(t+1/2Δt)
3. Use d.(t+1/2Δt) to find d(t+Δt)
4. Find d..(t)
   and so on.

Question 19

What are some advantages and disadvantages of explicit methods?

Answer 19

Advanatges:
- They are simple and fast

Disadvantages:
- The time step Δt should be smaller than the critical time step Δtcr. if Δt > Δtcr, the intergration will be unstable and errors in the results will be obtained.

Question 20

How is the critical time step obtained?

Answer 20

Δtcr = h / ce

h = characteristic length of the shortest element in the mesh
ce = √ E / ρ

E = elastic modulus
ρ = Desnity of material

Question 21

What are the factors affecting the selection of the time step?

Answer 21

1. The time step Δt should be smaller than the critical time step Δtcr
2. The chosen time step size should appropriately represent the rate of variation of the applied load
3. The chosen time step size should appropriately represent the variation of the nonlinear damping and stiffness properties of nonlinear problems
4. The period of vibration of the structure should be considered while time step size is determined.

Question 22

How are displacements, velocities and accelerations found using Newmark Time Integration - constant average acceleration methods?

Answer 22

1. Find displacement at time t + Δt using:
   - the inverse of ([M] * 4/(Δt)^2 + [K])
   - Initial conditions
2. Calculate accelerations at time t + Δt using:
   - d(t + Δt)
   - d(t)
   - d.(t) a
   - d..(t)
3. Calculate velocities at time t + Δt using
   - d.(t)
   - d..(t)
   - d..(t + Δt)

and so on

Question 23

What is the difference between explicit and implicit methods?

Answer 23

• Explicit methods are simple and non-iterative, but implicit methods are more complex and iterative.
• Explicit methods are more robust for non-linear problems
• Implicit methods become more complicated with an increase in the degree of nonlinearities.
• Explicit methods are more appropriate for slow-speed dynamic problems
• Implicit methods are more appropriate for high-speed dynamic problems.

Question 24

Whats the difference between full and reduced integration?

Answer 24

• Full integration refers to the minimum Guass integration points for exact integration of the polynomial function for an undistorted element with linear material properties.
• Reduced integration refers to the Gauss integration order that is one order less than the full integration.

Question 25

What is shear locking and how is it solved?

Answer 25

• Fully integrated four-noded quadrilateral elements are too stiff in modelling bending about an axis perpendicular to the plane of the element, which results in non-physical shear strains and is called shear locking.
• This problem can be solved by using higher-order elements or reduced integration quadrilateral elements, but the latter has the drawback of giving rise to zero-deformation (hourglass) modes.

Question 26

What is Gaussian quadrature?

Answer 26

Gaussian quadrature replaces the integral by a sum of the integrand found at Gauss points times weighting factors for each Gauss point.

Question 27

How is shear locking prevented and what is hourglassing?

Answer 27

To prevent shear locking and save computational time, lower orders of quadrature can be used, but it introduces the problem of zero-energy modes (hourglass, keystone or zero energy modes) that are non-physical response modes and can grow without bound and destroy the solution unless they are controlled.
These modes occur when only one integration point at the centre of the element is used, and there is no stiffness predicted to resist the shear mode causing no strain at the centre.
These modes are not resisted by the stresses within the element and can grow without bound if not controlled.