Ex 1.
The central part of a femur is modelled as a straight tube with outer diameter D, inner diameter d and length. Assume that the cortical bone behaves like a linear elastic material with Young’s modulus E. In addition, assume that the bone is loaded with an axial compressive force P.
1. Determine the stress field σ(x).
2. Determine the displacement field u(x). Assume that at x =l the displacement u(l ) satisfies u(l ) = 0.
1. σ(x) = σ = F/[π(D²-d²)/4]
2. ε(x)= σ(x)/E = F/[πE(D²-d²)/4]
ε(x)=du/dx
u(x) = F/[πE(D²-d²)/4]( x - l )
Ex 2.
Consider a muscle/tendon complex as shown in the figure below. To find out how much the tendon and the muscle are extended when the complex as a whole is loaded with a force F, a very crude two bar model can be used. The muscle is modelled as a bar with length l1, Young’s modulus E1 and cross section A1. At point B the muscle is attached to the tendon, which is modelled as a second bar with length l2, Young’s modulus E2 and cross section A2.
At point A the muscle is attached to the bone, which we consider as a rigid fixation. At point C a force F is applied in the direction of the bar (tendon).
1. Determine the internal force in a cross section between A and B and in a cross section between B and C.
F
2. Determine the stress σ in a cross section between A and B and a cross section between B and C.
F/A1 and F/A2
3. What happens with the calculated forces and stresses if the Young’s moduli of both muscle and tendon are reduced to half of their original value?
Nothing
4. Determine the displacements at point B and C as a result of the applied load F at point C.
ε1 = F/(E1A1) ε2 = F/(E2A2)
uA = 0
uB = uA + ε1l1 = F l1/(E1A1)
uC = uB + ε2l2 = F [ l1/(E1A1)+ l2/(E2A2) ]
Ex 3.
A test is performed to determine the Young’s modulus E of a blood vessel (see figure). The vessel has an inner radius r and an outer radius R. It is clamped on the left-hand side at x = 0, while the right-hand side at x = l is displaced over a distance ul. The displacement is small compared with the length of the vessel.
1. Determine the force F needed to displace the point B by ul as a function of E, ul, l, R and r.
F = [πE(R²-r²)] ul /l
Ex 4.
To determine the mechanical properties of a heart valve, a small rectangular piece of tissue is cut out of the valve and clamped in a uniaxial testing machine. After clamping, it can be seen that the test specimen is a little more slender in the middle (point M) than near the clamps (point K; see figure). It is decided to model the geometry of the sample as found after clamping. Because of symmetry only the right half of the sample is analysed.
The following equilibrium equation describes the deformation in the x-direction:
The width b(x) is assumed to be described by:
With b0 and α positive constants. The thickness is constant and given by t.
The length of the right half of the sample is denoted by l.
For the point x = 0 the displacement u(0) is assumed to be suppressed, u(0) = 0.
1. Determine the displacement u(l) of the point x = l when a force F is applied at that point.
A(x) = b0(1+αx/l )h
d/dx ( Eb0(1+αx/l )h du/dx ) = 0 (1+αx/l ) du/dx = K
du/dx = K / (1+αx/l )
x=0 => du/dx = K = F/(Ehb0) du/dx = F/(Ehb0) / (1+αx/l ) u(x) = F l /(αEhb0) ln (1+αx/l ) u(l) = F l /(αEhb0) ln (1+α)
Ex 5.
In many biomechanical applications, it is important to know the stress in the fibres that constitute the material.
The reasons for this are plenty. The fibres play a major role in the mechanical behaviour of the tissues as a whole, and many tissues will remodel their fibre architecture as a result of differences in loading of the cells and the fibres.
Consider the material that is shown in the following figure (a). It is the collagenous fibre structure that is found in human skin and visualized with a fluorescence microscope. Figure (b) shows a detail from the image
with one specific fibre bundle. We would like to determine the longitudinal stress in that particular fibre bundle.
Assume that we have done some (typically numerical) analysis and that we have found the stress matrix with respect to the 3-dimensional x, y, z-coordinate system in point A:
1. Express the stress in the fiber oriented at 𝒏=[4 3 0]/5
𝒏 is a unit vector 𝒏. 𝒏=1
traction vector in the plane perpendicular to the fiber:
𝑻 = 𝝈. 𝒏
Tensile stress of the fiber is the normal component of the traction vector:
𝜎𝑓= 𝑻. 𝒏 = (𝝈. 𝒏) . 𝒏 = 𝒏. (𝝈𝑻. 𝒏) = 𝒏. (𝝈. 𝒏) = 𝝈: (𝒏⨂𝒏) (𝒏⨂𝒏) is the fiber structural tensor
𝒏⨂𝒏 = [
𝑛1𝑛1 𝑛1𝑛2 𝑛1𝑛3
𝑛2𝑛1 𝑛2𝑛2 𝑛2𝑛3 𝑛3𝑛1 𝑛3𝑛2 𝑛3𝑛3] = 1
25[
16 12 0 12 9 0
0 0 0
]
𝝈. 𝒏 =1 5 [
36 0 48
0 100 0
48 0 64
] ( 4 3 0
) =1 5 (
4 × 36 300 48 × 4
)
(𝝈. 𝒏) . 𝒏 = 1 25 (
4 × 36 300 48 × 4
) . ( 4 3 0
) =36 × 16 + 900 25
𝝈: (𝒏⨂𝒏) = 1
25[
36 0 48
0 100 0
48 0 64
] : [
16 12 0 12 9 0
0 0 0
] =36×16+90025 kPa = 59.04 kPa
