Introduction to Torsion
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Torsion: Elastic Deformation
Torsion occurs when two equal and opposite torques, T, are applied to an object. The application of torque will cause the body in question to twist, what we call torsion. Devices containing shafts under load will necessarily experience some level of torsion due to some rotational resistances.
Torsion causes two types of measurable deformation. First is the angle of twist, \phi, which is a measure of how much the shaft face has rotated. The second deformation measure is the helix angle, \gamma, drawn on the axial face of the shaft between the shaft axis helix caused by deformation. There is a geometric relationship between these two measurements, detailed in the figure below.
Based on simple trigonometry, for sufficiently small angles of twist:
\phi r=\gamma L
In fact the helix angle is the exact same measurement as shearing strain. The relationship between shearing stress and strain makes it clear that the shear stress distribution is purely a function of shaft radius and can be expressed using the following formula:
\tau (r)=\tau_{max}\dfrac{r}{r_{max}}
\tau_{max}=G\dfrac{\phi r_{max}}{L}
It should be noted that this expression only holds in the elastic region of the material (\tau<\tau_y).
Net Torque
The equal and opposite reaction torque can be interpreted as the sum of all internal shearing stresses.
\text{Torque}=\text{Stress}\cdot\text{Area}\cdot\text{Distance}
\displaystyle T=\int r \tau dA
The following expression is obtained by integrating shear stress expressed with regards to shaft radius:
T=\tau_{max}\dfrac{\pi r^4}{2r_{max}}=\tau_{max}\dfrac{J}{r_{max}}
Where J is the polar moment of inertia, for circles: J=\frac{\pi}{2}r^4.
Torsion: Perfectly Plastic Behaviour
Imagine a material that, once undergoing plastic deformation, ceases to increase in stress past the yield stress. One outcome of such a material existing will be that it will cease to strain harden, the stress-strain curve will remain flat after the yield point. Such a material is referred to as perfectly plastic, due to the simple, known nature of perfectly plastic materials it is a great baseline for understanding plastic deformation in shafts.
The above image shows a stress-radius diagram for a shaft undergoing perfectly plastic behaviour. The following should be noted:
- Past a certain yield torque (T_y) the shaft will start yielding at a radius lower than r_{max}.
- This lower radius, the radius of plastic deformation, is symbolized by r_p.
- The additional torque applied past the yield torque must be redistributed to parts of the shaft that can absorb more load, this is symbolized by the redistribution of the green triangle to the red traingle in the above figure.
Torque causing yielding can be calculated using the procedure established for elastic deformation. At yield torque:
T_y=\tau_y\dfrac{J}{r_{max}}
Past yield torque:
\displaystyle T=\int_0^{r_p} r\tau dA+\int_{r_p}^{r_{max}}r\tau dA
\displaystyle T=2\pi\tau_y \left(\dfrac{c^3}{3}-\dfrac{r^3_p}{12}\right)
\displaystyle T=\int_0^{r_p} r\tau dA+\int_{r_p}^{r_{max}}r\tau dA
\displaystyle T=\int_0^{r_p} r\tau_y\dfrac{r}{r_p} 2\pi rdr+\int_{r_p}^{r_{max}}r\tau_y 2\pi rdr
\displaystyle T=2\pi\tau_y\dfrac{r^3_p}{4} +2\pi\tau_y\left(\dfrac{c^3}{3}-\dfrac{r^3_p}{3}\right)
\displaystyle T=2\pi\tau_y\left(\dfrac{c^3}{3}-\dfrac{r^3_p}{12}\right)
Torsion: Unloading
Unloading a shaft can be thought of as applying an equal and opposite unloading torque to the shaft. Application of an unloading torque will produce a net zero torque. Applying a torque in the opposite direction will necessarily apply a negative shear stress, this unloading shear stress can be calculated as follows:
\displaystyle T=\int_0^{r_p} r\tau_y \dfrac{r}{r_p}2\pi rdr + \int_{r_p}^{r_{max}}r\tau_y 2\pi rdr-\int_0^{r_{max}}r\tau_{ul}\dfrac{r}{r_{max}}2\pi rdr=0
\tau_{ul}=\tau_y\left(\dfrac{4}{3}-\dfrac{1}{3}\dfrac{r^3_p}{r^3_{max}}\right)
\displaystyle T=\int_0^{r_p} r\tau_y \dfrac{r}{r_p}2\pi rdr + \int_{r_p}^{r_{max}}r\tau_y 2\pi rdr-\int_0^{r_{max}}r\tau_{ul}\dfrac{r}{r_{max}}2\pi rdr=0
\displaystyle 0=\tau_y\left(\dfrac{c^3}{3}-\dfrac{r^3_p}{12}\right)-\tau_{ul}\dfrac{c^3}{4}
\displaystyle \tau_{ul}=\tau_y\left(\dfrac{4}{3}-\dfrac{1}{3}\dfrac{r^3_p}{r^3_{max}}\right)
When fully yielded:
\displaystyle \tau_{ul}=\dfrac{4}{3}\tau_y
Stresses remain in the shaft due to unloading a plastically deformed shaft, these are called residual stresses. There will also a resulting residual angle of twist after unloading. The below image shows the result of initial plastic loading combined with unloading.
Residual stresses can cause shafts to fail prematurely. However, compressive residual stresses can be used to strengthen particular materials. Regardless of what the reason is, residual stresses should always be accounted for in designs that will undergo plastic deformation.
Torsion of Non-Cicular Shafts
Torsion of non-circular shafts is usually approached empirically. Typically torsion of non-circular shafts are based on the analysis of an equivalent rectangular shaft. The width, a, and thickness, b, of a rectangular shaft are used to determine the maximum shear stress and angle of twist in a non-circular shaft.
\tau_{max}=\dfrac{T}{c_1ab^2}
\psi=\dfrac{TL}{c_2ab^3G}
The constants, c_1 and c_2, are dependent on geometric dimensions a and b and calculated empirically. An accurate formula exists for \dfrac{a}{b}\ge 5.
c_1=c_2=\dfrac{1}{3}\left(1-0.63\dfrac{b}{a}\right)
Further information on the stress distribution requires the use of the so-called membrane analogy if so desired.
