C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\1_torsion.doc 6 p. 1 of Torsion of circular bar The cross-sections rotate without deformation. The deformation that does occur results from the difference in rotation from one cross-section to the net. This is a shear deformation. (Note how the lines that were originally perpendicular are no longer perpendicular.) Assumptions Kinematics o Displacements o Strains Constitutive relations & Structural constitutive relations Equilibrium BC s Assumptions Kinematics The appendi in this file has a proof for these approimations of the displacements due to rotations.
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\1_torsion.doc 6 p. 2 of This results in the following strains ε = ε = ε = 0 ε yy zz y u v dφ = + = z y d u w dφ ε z = + = y z d v w εyz = + = φ+ φ = 0 z y If you transform the strain to an r-theta coordinate system, you will find that ε θ =constant. (Note: we don t have to assume the rotation is zero at =0. A more general form is (0) Constitutive dφ φ = φ +.) d dφ Structural constitutive relation: We want to relate M to. d
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\1_torsion.doc 6 p. 3 of If the bar consists of homogeneous layers (i.e. concentric cylinders of different materials), you just do the integration piecewise. We could also have transformed to a polar coordinate system. Then we would have dφ M = σ θr rdrdθ where σ θ = G r d Comments: You might to take a look at the A&H tet.
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\1_torsion.doc 6 p. 4 of Torsion of a Bar with a Square Cross-Section The response is similar to the circular rod, but there are some differences. Cross sections warp some, but not too badly Load was applied in a way that causes more than simple torsion where the load is applied. Away from this location, the response becomes simple torsion. Formulas: http://www.roymech.co.uk/useful_tables/torsion/torsion.html More: http://www.freestudy.co.uk/
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\1_torsion.doc 6 p. 5 of Appendi: Approimation of Rotation z P θ y Suppose point P is in the yz plane. The coordinates can be given in terms of r and θ y = r cosθ z = r sinθ θ = θ + φ where θ = angle before twisting begins 0 0 If there is a change in the angle theta, what are the changes in the y- and z-coordinates of point P? define the change in theta (i.e the amount of rotation) to be φ. The easy way to calculate the change is to simply take the Taylor series of the epressions for y and z in terms of theta. The change in the y- coordinate is the displacement in the y-direction. The change in the z-coordinate is the displacement in the z-direction. dy y( θ + φ) y( θ) + φ dθ φ = 0 dy v= y( θ + φ) y( θ) = φ = rsin( θ) φ dθ = zφ φ = 0 dz z( θ + φ) z( θ) + φ dθ α = 0 dz w= z( θ + φ) z( θ) = φ = rcos( θ) φ dθ = yφ φ = 0 If you are not a believer in calculus (Please do not be that way!), you can always derive the same thing as follows: Now suppose the point is rotated about the origin by an angle φ. The new coordinates are Therefore, the displacements are ' ' y rcos( θ φ) and z rsin( θ φ) = + = + ' ( θ φ θ) ( θ φ θ) ' v = y y = r cos( + ) cos and w = z z = r sin( + ) sin Using trigonometric identities, this can be epressed as ( θ φ θ φ θ) and ( θ φ θ φ θ) v = r cos cos sin sin cos w = r sin cos + cos sin sin If φ is very small, cosφ 1, sin φ φ,so for very small rotation:
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\1_torsion.doc 6 p. 6 of ( ) ( ) v = r sinθ φ = zφ and w = r cosθ φ = yφ
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 1 of 9 Torsion of Single-Cell, Thin Wall, Closed Section Beams Shear flow (definition) The result that q=constant in this single cell depen on the assumption that the only non-zero stress is σ. That is why there are not more tractions indicated on the differential volume.
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 2 of 9 Torque due to shear flow T torque = r q = q r but r = 2dA => T = 2q da n n => T = 2qA where A= enclosed area The torque is independent of the location of O in the figure (i.e the location about which moment is calculated=> it is a free vector. n Note: n is the outward normal vector. rn is not always positive as one moves around the perimeter. Below is such an eample.
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 3 of 9 Torsional Deformation (New derivation) us u ε = + s where u = displacement in the -direction u = displacment in the "s" direction (tangential direction) s We will approimate u s as us = n rφ, where r= perpendicular distance from center of twist (see sketch on previous page). For simplicity, we will assume that the bar is prismatic, which means that every crosssection is the same. This means that rn can vary with s, but not with. rnφ u φ u Hence, the shear strain is approimated by ε = + = rn + s s Where is the center of twist? Let s address this question when we need to do so. We will assume that φ is the same for all points in the cross-section, so it is not a function of s. We will also assume that u is not a function of. This is equivalent to assuming that if the cross-section warps, the warping is the same for all cross-sections. Hence, the equation for the shear strain becomes dφ du ε = rn + d If we integrate ε around the perimeter of the cross-section, we will obtain dφ du dφ ε = rn + = rn + du d d But the second integral is zero, u is the same at the start and finish of the contour integral. Also d φ d does not vary with s. Hence, we obtain dφ ε = rn d Recall that r = 2A, which means the average shear strain is n avg 1 2Adφ ε = where P=perimeter P ε = P d σ q The average shear strain can also be calculated using the constitutive relation ε = =. G Gt avg 1 q ε = P Gt Equate the two epressions for average shear strain and solve for the rate of rotation to obtain 2Adφ 1 q dφ 1 q = P d P ==> = Gt d 2A Gt
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 4 of 9 For a thin-walled, single cell under pure torsion, the shear flow is constant. Thus q can be pulled out of the integral and some nice simplifications are possible. However, if there are multiple cells or there is transverse load, q is not constant and the simplified relations are useless. If you want to learn the smallest number of formulas, do not simplify the formulas for special cases. The torsional rigidity or GJ is defined by the relationship M M dφ = GJ. Hence, GJ =. d dφ d For a single cell under pure torsion M = 2Aq so the epression for torsional rigidity is simply GJ = 2 4A. Warning: Do not use this for multi-cell beams.
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 5 of 9 Torsional Deformation (Original derivation optional reading) us u ε = + s where u = displacement in the -direction u = displacment in the "s" direction (tangential direction) s We will approimate u s as us = n rφ, where r= perpendicular distance from center of twist (see sketch on previous page). For simplicity, we will assume that the bar is prismatic, which means that every crosssection is the same. This means that rn can vary with s, but not with. rnφ u φ u Hence, the shear strain is approimated by ε = + = rn + s s Where is the center of twist? Let s address this question in more detail later. q Recall that σ = Gεand q = tσ => ε = Now let's equate this to the epression for the strain in terms of displacments q φ u ε = = rn + s u We are going to try to simplify this in the net few steps. First, solve for. s u q φ = rn s u du If the bar is sufficiently long, we can assume that u is independent of.=> =. Physically, s this means that the warping of the cross-section is independent of. This means that we can write du q φ q φ = rn => du = rn Integrate both sides completely around the boundary u ( s2) s2 q φ du ( 1) = u s rn s1 Since we have gone completely around the perimeter, u ( s ) = u ( s ), sand s correspond u ( s2 ) to the same point. Hence, du = 0. We are left with u ( s1 ) 1 2 1 2
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 6 of 9 q φ q φ rn = 0 or rn = 0 φ but = constant for the cross-section and we showed earlier that rn = 2A where A=the enclosed area, resulting in q φ q φ rn = 0=> 2A = 0 This gives us an epression for the rate of twist φ 1 q = 2A We had also observed for the thin-walled, single cell configuration, the shear flow is constant, giving us φ 1 = q 2A Rearranging this, we obtain an epression that is like a force vs deformation relationship. q = 2A φ φ The term, which is the rate of twist, is a measure of deformation " q" = shear flow around the cell What we want to have is an epression that relates the torque M to the deformation. Recall that M = 2Aq, which gives us M = 2 4A φ The torsional rigidity or stiffness is This is our structural constitutive relation. 2 4A also use the terminologygj to make it fit the pattern (e.g. recall. This term will sometimes be referred to as the TR. I might EA and EI ).
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 7 of 9 Summary of Key Equations for Single Cell ε u s = + u s but strain is typically calculated using stress + constitutive relation M = T = q r = 2qA n φ 1 q = For single cells, q can be pulled out of integral 2A φ 1 q s If integrand is piecewise constant, we obtain = 2A i q = σ t = M GJ = = 2 4A 2 4A 2A φ φ Solving many problems involves just the following (these are for multicell) M = 2 qa i i φ 1 q = = rotation for cell "i" i 2A i and all rotations are the same i
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 8 of 9 dφ In this eample, the variable φ = rotation angle and θ = d
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\2_torsion_thinWall_ver_2.doc p. 9 of 9 Note: there is a typo in the solution for part (c) in A&H. I have it corrected above. Summary of steps for analyzing a single thin-wall cell If the beam is statically determinate, use statics or integrate differential equation of equilibrium (in terms of stress resultants) to determine M. Determine the shear flow using the relationship that M = 2Aq Calculate shear stress σ from the shear flow. Calculate strain from the shear stress φ 1 s2 q Calculate the rotation per unit length from = 2A s1 Calculate the rotation by integrating the rotation per unit length. How you calculate torsional rigidity is determined by what you have already calculated. If you have already calculated the rotation per unit length, then TR = M/(rotation per unit length). If you have not, then you can directly calculate TR using TR = GJ = 4A s2 s1 2
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\3_torsion_multiCell_a.doc p. 1 of 6 Torsion of Multicell Thin-Walled Sections Ref. A&H p. 200 The theory discussed here is known as the Bredt theory of torsion. Consideration of equilibrium at each juncture lea to the conclusion that the sum of shear flow at a juncture = 0. Consider the following case that involves three shear flows. Summation of forces in the -direction yiel ( q1+ q2 + q3) d= 0 => q1+ q2 + q3 = 0. It is interesting to note that you do not reach this conclusion by just looking at the cross-section. You need a 3D plot. The shear flows acting on the cross-section are related to the shear flows in the -direction. You are enforcing equilibrium of forces (actually forces per unit length) acting on planes that are not part of the cross-section. There are some assumptions and observations that are essential to keep in mind when using this theory. For completeness, the list below includes the equilibrium argument given above. There is no default direction for shear flow. A sketch of the cross-section must be labeled. At each juncture ( node ): q = 0 i
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\3_torsion_multiCell_a.doc p. 2 of 6 Total = summation of contributions for each cell. When calculating the stiffness of a particular cell, you ignore the other cells and you use the entire skin or web thickness. Torque = 2 qi Ai i The rotation angle is the same for each cell. This is a compatibility requirement that will give us the equations needed to solve for the unknown shear flows. 1 q Recall that the rotation angle for a single cell is 2A. In most cases, the integrand is Gt piecewise constant, so we end up with a simple summation of contributions along the various walls, as follows. 1 q 1 q s 2A = Gt 2A Gt i There seems to be a contradiction when you calculate the torque and the rate of rotation o The shear flow is not considered constant in a single cell when the rate of rotation is being calculated. o The shear flow is considered to be constant when calculating the contribution to the torque. o This is just an apparent contradiction. That is, if you follow the procedure you will get the right answer and you will account for the shear flow being different in the shared web and the rest of the cell. Some sample calculations are given below to illustrate this point. o It is a good thing that this is just an apparent contradiction, since when we have combined transverse shear and torque loading, you cannot use the formula M = 2 qa i i. You must consider one wall at a time and add up the contributions. If there were a contradiction, this method would fail for torque only. That would be BAD! If there are n cells, we will have n-1 compatibility equations that enforce equality of all of the rotations and one equilibrium equation (for the cross-section) that states that M = 2 qa. i i
C:\Users\whit\Desktop\Active\304_2012_ver_2\_Notes\4_Torsion\3_torsion_multiCell_a.doc p. 3 of 6 We will calculate the torque due to shear flow three ways for the following configuration. Here are the details: The solutions are identical, as epected. Later in this course we will have combined torsion and transverse shear. For those problems the formula M = 2 qa i i is not useful unless you separate the problem into two problems: one with transverse force only and the other with torsion only. To do this you will need to obtain statically equivalent loa (i.e. shear force and moment) with the shear force being applied through the shear center. The shear center is a special location. If you apply a shear force through the shear center, there is twisting due to the shear force. We will derive a formula for calculating the shear center in a few more days.
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