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Finally, we showed that torsion problems are also often statically indeterminate, and even though the loading and deformation is different, the technique we established in the last section for solving problems with axial loading is the same technique for solving problems with torque loading. a rod) that spins with a given frequency under an applied torque. We also used a method of dimensional analysis to determine the power generated by a transmission shaft (i.e. Using Hooke's law, we can relate this strain to the stress within the rod. From this analysis we can develop relations between the angle of twist at any a point along the rod and the shear strain within the entire rod. This different type of loading creates an uneven stress distribution over the cross section of the rod – ranging from zero at the center to its largest value at the edge. We learned about torque and torsion in this lesson. See if you can work the rest of this problem out on your own: What is the torque in each half of the rod? Most importantly, we need to ask ourselves "what do we know about the deformation?" Well, since the rod is stuck to the wall at edge, the twist at A and B must be equal to zero (just like the displacement in the last section). For this equation, we should note that half the rod is solid, the other half is hollow, which affects how we calculate J for each half. So, we need to consider our deformations – for torsion, that means let's turn to our equation that describes the superposition of twist angles.
And statically indeterminate means, draw a free body diagram, sum the forces in the x-direction, and you'll get one equations with two unknown reaction forces. More supports than is necessary: statically indeterminate. Immediately upon inspection you should note that the rod is stuck to two walls, when only one would be necessary for static equilibrium. The relationship between torque and shear stress is detailed in section 5.2 of your textbook, and it results in the following relation: Since twist applies a shear strain, we expect that torque will apply a shear stress. To discuss the stress within a twisted rod we need to know how torque and stress relate. So far, we've focused our attention on displacements and strain. Finally, the longer the rod, the smaller the shear strain. Second, and this is the big difference between axial-loaded structures and torque-loaded ones, the shear strain is not uniform along the cross section. It is zero at the center of the twisted rod, and is at a maximum value at the edge of the rod. The first thing might be obvious: the more angle of twist, the larger the shear strain (denoted by the Greek symbol gamma, as before). We can immediately learn a few things from this equation. That's where the Greek symbol rho came from – it denotes the distance along the cross section, with rho=0 at the center and rho=c at the outer edge of the rod. Before we get into the details of this equation, it's important to note that because we're only discussing circular cross sections, we've switched from Cartesian coordinates to cylindrical coordinates.