To build a muscle, one must bring it to near exhaustion. What are the levels of force intensity (stress) in the muscle, when the muscle force peaks? To answer this question, one must evaluate (a) the force generated by a muscle and (b) the cross-sectional area of the muscle normal to the direction of the muscle force. Average axial force intensity or average normal stress at a cross section is defined as
where o-av denotes the average axial stress and F is the force acting on the cross-sectional area A. In parallel muscles, the cross-sectional area is not constant along the axis of the fiber. However, a single fiber can be represented in the form of a uniform circular cylinder (Fig. 6.9). Let A denote the cross-sectional area that is perpendicular to the axis of the cylinder. Then the axial stress acting on this cross-sectional area is uniform and is given by Eqn. 6.24. Let us now consider the stress that acts on a cross section whose normal vector makes an angle 6 with the vertical axis (Fig. 6.9b). In this case, the cross-sectional area is given by the equation:
Next, let us resolve the force F that acts on cross-sectional area A6 into two components (Fig. 6.9c), one normal (Fn) and the other tangential (Ft)
Figure 6.9a-c. A circular cylindrical specimen under the application of a tensile force of magnitude F (a). The free-body diagram of the part of the specimen is shown in (b). The stress distribution acting on an inclined cross section is shown in (c).
to the area A6. The equations for these two force components are as follows:
Let us now define normal (a) and shear (r) stress components as follows: a = Fn/Ae = (F/A) cos2 6 (6.27a)
It is clear that the concept of stress is a bit more complex than the entities that can be represented as vectors. Two distinct directions (not just one) are associated with stress, the direction of the force and the direction of the surface area on which the force acts. Stress is a measure of force intensity on a small area element. It is a tensor of a second order. Scalars such as temperature and mass are said to be tensors of the zero order; vectors are tensors of first order.
Equation 6.27 shows that both the normal and the shear stress vary considerably with the orientation of the cross-sectional area. If we take the derivatives of Eqns. 6.27a and 6.27b with respect to Q and set them equal to zero, we can determine those directions in which either a or t is maximum (minimum). When Q = 0, then the normal stress takes its largest value (a = F/A) whereas the shear stress becomes equal to zero. On the other hand, when Q = v/4, then the shear stress assumes its maximum value (t = 0.5 F/A). The evaluation of the maximal values of shear and normal stress is important in the consideration of failure of a material. Some materials fail easily under tension (compression) and others fail readily under shear stress. Concrete is not resilient to tension whereas steel can withstand both tension and compression. Ceramics are known to have low resistance to shear stress. If the circular cylinder under consideration had low resistance to shear but high resistance to tension, it would fail in the form of a tear that occurs at 45° relative to the axis of the cylinder.
As illustrated by the previous examples, in many cases Newton's second law allows us to compute contact forces and the resultant internal forces at a cross section. Once we know the resultant force carried by a structural member, we can compute the average stress by dividing the value of the force with the cross-sectional area of the planar cross section. The question that comes to mind is how close is the average stress to the actual stress. The large body of literature in solid mechanics allow us to provide some insights. First, in the presence of cracks or holes, the average stress is not an accurate indicator of the actual stress. Stress will intensify in the vicinity of a fracture or rupture. Also, if the cross section cuts across a number of different materials with different stiffness, stress in the material with higher stiffness may be greater than the one with lower stiffness. We can illustrate this by considering a cylindrical specimen consisting of two materials under the action of a tensile force (Fig. 6.10). Imagine this to represent a long limb of the human body, a long bone surrounded by soft tissue. The average stress normal to the axis of the cylinder is given by Eqn. 6.24. What are the average stresses carried by the two concentric cylinders? The total force acting on the cross section must be equal to F, and this leads to the following equation:
in which r denotes the radial distance and the index refers to concentric cylinders made of materials 1 and 2. We need additional information to determine the values of a1 and a2. We assume that planar cross sections that are normal to the axis of the specimen remain plane and normal to the axis. Thus, every line element parallel to the axis of the specimen undergoes the same extension A. Axial strain e is defined as follows e = A/L (6.29)
in which L denotes the length of the cylinder.
Figure 6.10a,b. A circular cylindrical specimen made of two different materials is under tensile force (a). Stress distribution on a cross section that is normal to the long axis of the specimen is shown in (b). The material occupying the core of the cylinder is stiffer than that of the outer shell.
If both materials are linearly elastic, then according to Hooke's law we have
in which E1 and E2 denote, respectively, Young's modulus of materials 1 and 2. Combining Eqns. 6.28, 6.29, and 6.30, we obtain:
Notice that (oi/o2) = Ei/E2. If the cross-sectional areas of the two materials are comparable and if Ei is much greater than E2, then material i carries much of the force applied on the specimen. This would be the case of a relaxed limb that is under tension; bone would carry much of the applied load. Consider a human thigh under traction. Approximately 30% of the cross-sectional area is bone and the rest is composed of muscle and fat tissue. Because the fat tissue is much more compliant than muscle, it would carry practically no force. The cross-sectional area that effectively carries the traction force must be that of the cross-sectional area of the muscle and the bone.
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The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.