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EGM 4592 BioSolid Mechanics Problem Set #
Assigned: January 23rd, 2007 Due: February 6th, 2007
- The components of a stress tensor are given as:
τ (^) ij kPa.
What is the magnitude of the traction (stress vector) acting on a surface whose normal vector is i j k
r r r
ν = 0. 1 + 0. 3 + 0. 9? What are the three components (in the
x-, y- and z-directions) of the stress vector acting on the surface? What is the normal stress acting on the surface? What is the resultant shear stress acting on the surface?
- A square piece of material is deformed from its initial state as shown in the three drawings below:
For cases 1,2, and 3, express the displacements in x 1 and x 2 in terms of a 1 and a 2 , and then determine ε 11 , ε 22 , ε 12 using Cauchy’s infinitesimal strain tensor formulation.
Case 1 : x 1 =1.05a 1 δx 1 / δa 1 = 1.05 & x 2 =1.01a 2 δx 2 / δa 2 = 1. So ε 11 = δx 1 / δa 1 = 1.05, ε 22 = δx 2 / δa 2 = 1.01, ε 12 = ε 21 = ½(δx 1 /δa 2 +δx 2 /δa 1 )=
Case 2 : x 1 =1.05a 1 -0.1a 2 δx 1 / δa 1 = 1.05, δx 1 / δa 2 = -0. x 2 =1.1a 2 δx 2 / δa 2 = 1. So ε 11 = δx 1 / δa 1 = 1.05, ε 22 = δx 2 / δa 2 = 1.01, ε 12 = ε 21 = ½(δx 1 /δa 2 +δx 2 /δa 1 )= -0.
Case 3 : x 1 =1.03a 1 +0.01a 2 δx 1 / δa 1 = 1.03, δx 1 / δa 2 = 0. x 2 =0.01a 1 +1.02a 2 δx 2 / δa 1 = 0.01, δx 2 / δa 2 = 1. So ε 11 = δx 1 / δa 1 = 1.03, ε 22 = δx 2 / δa 2 = 1.02, ε 12 = ε 21 = ½(δx 1 /δa 2 +δx 2 /δa 1 )= 0.
- Use Simulink and/or MATLAB to simulate the response of a Kelvin model for a viscoelastic material as shown:
a.) Determine the response of the system for unit step inputs of force (F) and displacement (u). Draw a magnitude and phase plot for the model for a range of sinusoidal inputs using the bode function in MATLAB. b.) Connect two of the given models in series (fixed at the left and moving at the right) and determine the response of the system for unit step inputs of force (F) and displacement (u) applied to the right terminus. Draw a magnitude and phase plot for the model for a range of sinusoidal inputs using the bode function in MATLAB.
Solution: From equation 7 in section 2.11 we have:
F +τ (^) ε F & = Er ( u + τ σ u &)
You can follow the derivation in section 2.12 to determine the expression for the complex modulus and then use that function in MATLAB, or you can take the Laplace transform of both sides and simplify to get the complex modulus as I showed in class:
( ) [ ] [ ] [ ] [ ]
[ ] [ ]
if youwanttogetu fromaknown F
F s
E s
s
us
if youwanttogetF fromaknown u or
u s
s
E s
so F s
F s s u sE s
F s s F s E us s us
r
r
r
r
σ
ε
ε
σ
ε σ
ε σ
τ
τ
τ
τ
τ τ
τ τ
(This should look a lot like equation 10, page 50)
5
Er = μ 0 = (^510) 1
=^1 =
μ
η τε 1 30 1
0 0
μ
μ μ
η τσ
In Simulink, your models could look like this:
30s+ 10s+ Transfer Fcn disp_step_output To Workspace
Step (^) Scope
5 Gain
10s+ 30s+ Transfer Fcn force_step_output To Workspace
Step (^) Scope
Gain
The ‘To Workspace’ blocks transfer the model output to the MATLAB workspace where it is easier to generate plots, and you can get something like this:
using the following commands: plot(tout,disp_step_outp ut) hold plot(tout,force_step_out put*100)
You also could do this entire bit in MATLAB with the following commands: s=tf('s') g=5(30s+1)/(10s+1) h=1/g step(g) hold step(h100)
which is the same figure but on a much longer time scale.
(^00 1 2 3 4 5 6 7 8 9 )
5
10
15
Time (seconds)
Step Response Force response to a displacement step input 100*Displacement response to a force step input
Step Response
Time (sec)
Amplitude
(^50 20 40 60 80 100 120 140 160 )
10
15
20
This plots say that if your sinusoidal force varies slowly, you will get a pretty large displacement. When your sinusoidal force varies quickly, you get less displacement. Again, slow stuff allows the damper to move while fast stuff is opposed by the damper. This happens in pretty much all the tissues we’ll study.
b) To model the connection of two Kelvin blocks in a row, we have to think about what we have. If we have a spring with equation F=kx we are inputting a displacement and computing a force (the relaxation response). If we add two springs (k 1 and k 2 ) in series, the equation becomes F=(1/k 1 +1/k 2 )-1x. So if we wish to input a displacement and measure a force, we need to combine the relaxation responses accordingly.
If we start with the equation x=F/k (the creep response) we are measuring the displacement with a force input. The reciprocal of stiffness (k) is compliance. If we add two compliances in series, the resulting compliance is just the sum of the two individual compliances.
In part a) we have a complex stiffness (g) and a complex compliance (h), but the combination rules are the same as for simple springs, so: gnew=(1/g+1/g)-1=g/2 and hnew=h+h=2h which can be modeled exactly the same as in part a) with a scale factor added.
