Exam paper: Biomechanics

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1 Exam paper: Biomechanics Tuesday August 10th 2010; AM Code: 8W020 Biomedical Engineering Department, Eindhoven University of Technology The exam comprises 10 problems. Every problem has a maximum score of 10 points. It is a closed book exam. problem 1: 10 points. Consider a Cartesian vector basis { e x, e y, e z }. With respect to this basis the following vectors are given: a = 7 e x + 3 e y + 4 e z b = 3 ey + 4 e z Determine the result of the following tensor-vector products: ( e x e z + e z e x ) a ( e x e x + e y e y ) b problem 2: 10 points. A rigid bar with length l is supported by a joint in point A and by a roll bearing in point B, see figure. A moment M 1 = 3 e z is acting on the bar at a distance 1 l from point A. A 3 second moment M 2 = 5 e z is acting on the bar at a distance 2 l from point A. 3 e y e x A M 1 M2 B 1 l 3 1 l 3 1 l 3 Calculate the reaction forces at the points A and B. 1

2 problem 3: 10 points. Consider the Cartesian basis { e x, e y, e z }. The following vectors a and b are given: a = 3 e x + 4 e y b = ex + 2 e y + 2 e z Determine the angle θ between the two vectors a and b. problem 4: 10 points. A man is preventing an oil drum with weight F G and diameter D from running down a slope (angle α) by pulling a rope that is rolled up around the drum. The man is pulling with a force F T. Assume, that there is no slip between the man and the slope and between the drum and the slope. In addition, we assume that the rope is parallel to the slope. D α Sketch a free body diagram of the drum with all the forces that act upon it. Determine F T as a function of the angle α and the weight F G. 2

3 problem 5: 10 points. To test the behaviour of visco-elastic biological material a stepwise strain test is performed that can be specified by means of: ε(t) = ε 0 [H(t) + H(t t 1 ) 2H(t t 2 )], with H(t) the Heaviside function and with ε 0 a constant strain measure. The load history F(t) associated with this strain history can be calculated by means of the relaxation function G(t). Assume, that: G(t) = c 2 + c 1 e ( t τ ), with: c 2, c 1 and τ positive material constants. Give an expression for the force F(t) for an arbitrary time t > 0. problem 6: 10 points. The Kelvin-Voigt model in figure (a) comprises a spring with constant c and a dashpot with constant c η. ε c η ε 0 0 t t c (a) (b) Give a sketch of the force F(t) after applying a strain rate history ε as given in figure (b), assuming that ε(t = 0) = 0. Determine the values of F(t) just before and after t = t. 3

4 problem 7: 10 points. A cylindrical tube with length l, Young s modulus E, inner radius r, outer radius R and density ρ (mass per unit volume) is standing in up-right position. The gravitational constant is g. The positioning results in a distributed load q in the tube, because of gravity. An external load F is applied in negative x-direction on the top of the bar at x = l. The weight of the top plate can be neglected. F q q l x r R Determine the displacement field u(x) of the tube expressed in F, ρ, r, R, l and E. problem 8: 10 points. A person exerts a horizontal force F = 150 N in the test apparatus shown in the drawing. Find the horizontal force M (magnitude and direction) that the flexor muscle exerts on his forearm (L = 0.30 m, h = m), assuming the elbow can be modelled as a perfect joint. 4

5 problem 9: 10 points. A man with weight G M is walking on a board between the points A and B (see figure). The angle between the board and the horizontal plane is α. The weight of the board is G B, applied in the middle of the board. The distance between point A and the point upon which the weight force G M acts is x. In point B the board rests on a roll bearing that runs on a support surface, making an angle of α with respect to the horizontal. B α A α G M x l Determine the reaction forces in points A and B as a function of G M, G B, l, α and x. problem 10: 10 points. A bar AB with length l 1, Young s modulus E 1 and cross section A 1 is in point B connected to a bar BC with length l 2, Young s modulus E 2 and cross section A 2. The first bar is clamped in point A. A force F 1 is acting on point B and a force F 2 is acting on point C (see figure). A l 1, E 1, A 1 l 2, E 2, A 2 B F 1 C F 2 Determine the normal force and the stress in a cross section between A and B and in a cross section between B and C. 5

6 Formulas and equations from Biomechanics Inner product: a b = a b cos(φ) Cross product or vector product: c = a b ; c = a b sin(φ) Triple product: a b c = ( a b) c Tensor-vector product: a b p = a( b p) Operations that are used once a Cartesian vector basis { e x, e y, e z } is defined: inner product a b = a x b x + a y b y + a z b z cross product a b = (a y b z a z b y ) e x + (a z b x a x b z ) e y + (a x b y a y b x ) e z The component F t of a force F in the direction of a unit vector e is given by: ( ) F t = F e e The moment of a force F acting on point Q with position vector x Q with respect to point P with position vector x P is given by: M = ( x Q x P ) F The force extension relation for a fibre in 1-D: ( ) l F = c 1 = c(λ 1) l 0 The force extension relation for an activated muscle fibre in 1-D: ( ) ( ) l λ F = c 1 = c 1 l c λ c The force extension relation of a fibre in 3-D for small displacements and rotations: F = c a ( u B u A ) a l 0 The force strain relation for an elastic spring: F = cε The force strain relation for a dashpot: F = c η ε Differential equation for a one-dimensional continuous elastic medium: ( d EA du ) + q = 0 dx dx 6

7 Exam Biomechanics, Tuesday August 10 th, 2010 Code: 8W020 Biomedical Engineering Department Eindhoven University of Technology Answers Problem 1: ( e x e z + e z e x ) a = e x ( e z a) + e z ( e x a) = 4 e x + 7 e z ( e x e x + e y e y ) b = e x ( e x b) + e y ( e y b) = 3 e y Problem 2: R A = 2 l e y, RB = + 2 l e y Problem 3: cos (θ) = a b a b = 5 15 = 1 3 ( 1 θ = arccos 3) Problem 4: F T = 1 2 F G sin (α) Problem 5: For t t 1 F(t) = ε 0 ( c 2 + c 1 e t/τ) For t 1 < t t 2 F(t) = ε 0 ( 2c 2 + c 1 e t/τ + c 1 e (t t 1)/τ ) For t > t 2 F(t) = ε 0 ( c 1 e t/τ + c 1 e (t t 1)/τ 2c 1 e (t t 2)/τ ) Problem 6: For 0 < t t F(t) = c η ε 0 + c ε 0 t and F(t ) = c η ε 0 + c ε 0 t For t > t F(t) = c ε 0 t and F(t + ) = c ε 0 t

8 Problem 7: u(x) = ρg E ( 1 F 2 x2 lx) Eπ(R 2 r 2 ) x Problem 8: M = L h F = 1125 [N] directed to the left Problem 9: V A = G M (1 x ) ( l cos2 (α) + G B 1 1 ) 2 cos2 (α) H A = H B = G M ( x l sin(α) cos (α) ) + G B ( 1 2 sin(α) cos (α) ) ( ) ( ) x 1 V B = G M l cos2 (α) + G B 2 cos2 (α) Resultant reaction force in B (perpendicular to the support surface of the roll bearing): R B = G M ( x l cos (α) ) + G B ( 1 2 cos (α) ) Problem 10: Between A and B: N = F 1 F 2, σ = F 1 F 2 A 1 Between B and C: N = F 2, σ = F 2 A 2