Section I.2 Torsion of NonCircular Bars
The observations made for torsion of members with circular cross sections do not hold for those with noncircular cross sections. Consider the following facts for members with noncircular cross sections:
 The shear stress is not constant at a given distance from the axis of rotation. As a result sections perpendicular to the axis of member warp, indicating outofplane displacement.
 The theory of elasticity shows that the shear stress at the corners is zero.
 Maximum shear strain and stress are not at the farthest distance from the rotational axis of a homogeneous noncircular member.
Members with rectangular cross sections:
For a rectangular member under torsion the corners do not distort; the corner square angles remain square after torque is applied. This indicates that shear strain is zero at the corners since there is no distorsion. This fact is illustrated in this figure.
St. Venant was the first to accurately describe the shear stress distribution on the cross section of a noncircular member using the Theory of Elasticity. For more information click here.
Theory of Elasticity shows that:
 the maximum shear strain and stress occur at the centerline of the long sides of the rectangular cross section
 the shear strain and stress at the corners and center of the rectangular cross section are zero.
 the strain and stress variations on the cross section are primarily nonlinear.
 the preceding statements are demonstrated in the following figure.
The theory of Elasticity has been applied to find analytical solutions for the torsion of rectangular elastic members. The resulting equations for shear stress and angle of twist are as follows:
Here is Table A6.1
NOTE: The equations described in this section can only be used for members having OPEN crosssectional shapes. For open cross sections composed of multiple thin plates refer to Section I.5.
Section I.3 Elastic Membrane Analogy
Prandtl showed that the Laplace equation describing the torsion of an elastic member is identical to that used to describe the deflection of an elastic membrane subjected to a uniform pressure.
The elastic membrane analogy allows the solution of a torsion problem to be determined in a simpler way than that found by the theory of elasticity which requires the availability of the warping function.
Elastic Membrane and Twisted Bar Relationships
Consider a tube which has the same crosssectional boundary as the bar. For example, if the bar has a solid square cross section of side dimension b, then the tube will have a hollow square cross section of side dimension b as well. Next we stretch an elastic membrane over the tube's cross section and apply internal pressure. The deflected shape of the membrane helps us visualize the stress pattern in the bar under torsion.
The analogy can be viewed as follows:
 Lines of equal deflection on the membrane (contour lines) correspond to shearing stress lines of the twisted bar.
 The direction of a particular shear stress resultant at a point is at right angle to the maximum slope of the membrane at the same point.
 The slope of the deflected membrane at any point, with respect to the edge support plane is proportional in magnitude to the shear stress at the corresponding point on the bar's cross section.
 The applied torsion on the twisted bar is proportional to twice the volume included between the deflected membrane and plane through the supporting edges.
Examples
Keep in mind that the slope at a point on the deflected membrane, and not the displacement from the base, is the parameter that is related to the shear stress in the bar. In all of the following examples, the slope is zero at the very top of the membrane, therefore the stress is zero, not the maximum, at the same location on bar's cross section.
 Figure 1
 Figure 2
 Figure 3 Notice that the stress is maximum at the inside corner and zero at the outside corners.
Section I.4 Torsion of Thin  Wall Open Sections
The sections shown below are extruded sections
EXAMPLE PROBLEMS
 Example Problem 1 Maximum shear stress and angle of twist of a FORMED section
 Example Problem 2 Maximum shear stress and angle of twist of an EXTRUDED section
Angle of twist per unit length
T = Torque
G = Modulus of rigidity
K = Apparent polar moment of inertia (given in table below)
 
 
 
side = a 

^{*} This is Table A6.3 in Chapter A6 of "Analysis and Design of Flight Vehicle Structures", by Bruhn.
A more enxtensive list of formulae for various solid and hollow shapes can be found in "Formulas for Stress and Strain", by Roark, 1954 Ed.
1 komentar:
Thanks for the max shear stresses, my textbook's table is very limited.
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