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Download Biomechanics: Concepts and Computation by Cees Oomens, Marcel Brekelmans, Frank Baaijens PDF

By Cees Oomens, Marcel Brekelmans, Frank Baaijens

This can be the 1st textbook that integrates either basic and particular themes, theoretical heritage and biomedical engineering functions, in addition to analytical and numerical methods. This quantitative process integrates the classical strategies of mechanics and computational modelling strategies, in a logical development via quite a lot of basic biomechanics ideas. on-line MATLAB-based software program besides examples and difficulties utilizing biomedical purposes will inspire undergraduate biomedical engineering scholars to perform and try their talents. The ebook covers issues corresponding to kinematics, equilibrium, stresses and lines, and in addition makes a speciality of huge deformations and rotations and non-linear constitutive equations, together with visco-elastic behaviour and the behaviour of lengthy narrow fibre-like constructions. this can be the definitive textbook for college kids.

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Extra resources for Biomechanics: Concepts and Computation

Example text

The forces W = −Wey , due to the weight of the arm, and W 0 = −W0 ey due to the lifted weight are both supposed to be known. The reaction force in the joint F J and the force in the muscle F M are both unknown. However, the direction of the force in the muscle is known since this force is oriented with respect to the arm at an angle θ. 9 Lifting a weight. 47 Exercises while the force in the muscle is given by F M = −FM cos( θ) ex + FM sin( θ) ey . Notice that both the x- and y-component of the joint reaction force, FJx and FJy , respectively, are unknown (the joint is modelled by a hinge), while for the muscle only the magnitude of the muscle force FM is unknown.

4 is replaced by a hinge as in point B. In that case an additional reaction force in the vertical direction must be introduced at point A, see Fig. 6. With the same values of the parameters as before, force and moment equilibrium yields −20 − HA − HB = 0 10 + VA + VB = 0 −2HB − 20 − 20 = 0. 6 Free body diagram of the simple loaded body, statically indeterminate case. 44 Static equilibrium The horizontal reaction forces HA and HB can, incidentally, still be calculated, giving, as before: HA = 0 [N], HB = −20 [N].

In this case the lineof-action of force F is drawn through point P and the resulting moment is given by a curved arrow in counterclockwise direction. 20(b) is statically, completely equivalent. 20(d) for a clockwise direction of the moment. 20 Drawing convention of the moment vector for different force vector orientations. Figures (a) and (b) indicate a statically equivalent load for a counterclockwise orientation of the moment vector. Figures (c) and (d) are equivalent for a clockwise orientation of the moment vector.

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