Machinery's Handbook, 31st Edition
DEFLECTIONS 219 Normal Stresses at Right Angles: This is shown in Fig. 11 at (2). This combination of stresses occurs, for example, in tanks subjected to internal or external pressure. Accord- ing to the Maximum Shear Stress theory, failure will occur in ductile materials due to shear, so this stress is critical to evaluate. The principal normal stresses are σ x = F 1 / A 1 , σ y = F 2 / A 2 , and σ z = 0 in this plane stress problem. Determine the values of these three stresses with their signs, order them algebraically, and then calculate the maximum shear stress: (23) Normal and Shear Stresses: The example in Fig. 11 at (3) shows a member subjected to a torsional shear stress, τ = T / Z p , and a direct compressive stress, σ = − F / A . At some point a on the member the principal normal stresses are calculated using the equation, (24) The maximum shear stress is calculated by using the equation, (25) The point a should ordinarily be selected where stress is a maximum value. For the example shown in Fig. 11 at (3), the point a can be anywhere on the cylindrical surface because the combined stress has the same value anywhere on that surface. 2 τ = largest smallest σ −σ 2 2 2 2 ! σ′ = σ σ + τ ` j 2 2 2 τ′ = σ + τ ` j
F 2
a T
a
F 2
y x
F 1
F 1
F 1
F 1
F
b
F 2
(1)
(2)
(3)
Fig. 11. Types of Combined Loading Tables of Combined Stresses.— Beginning on page 220 , these tables list equations for maximum nominal tensile or compressive (normal) stresses, and maximum nominal shear stresses for common machine elements. These equations were derived using gen- eral Equations (21), (22), (24), and (25). The equations apply to the critical points indicated on the figures. Cases 1, 2, 3, and 4 are cantilever beams. These may be loaded with a com- bination of a vertical and horizontal force, or by a single oblique force. If the single oblique force F and the angle θ are given, then horizontal and vertical forces can be calculated using the equations F x = F cos θ and F y = F sin θ . In cases 9 and 10 of the table, the equa- tions for σ a ′ can give a tensile and a compressive stress because of the ± sign in front of the radical. Equations involving direct compression are valid only if machine elements have relatively short lengths with respect to their sections; otherwise column equations apply. Calculation of Worst Stress Condition: Stress failure can occur at any critical point if either the tensile, compressive, or shear stress properties of the material are exceeded by the corresponding working stress. It is necessary to evaluate the factor of safety for each possible failure condition. If working with the von Mises equivalent stress, only tensile yield need be considered. The following rules apply to calculations using equations in the Table of Simple Stresses on page 216 and to calculations based on Equations (21) and (22). Rule 1: For every cal culated normal stress there is a corresponding induced shear stress; the value of the shear stress is equal to half that of the normal stress. Rule 2: For every calculated shear stress there is a corresponding induced normal stress; the value of the normal stress is equal to that of the shear stress. The tables of combined stress formulas below include equations for calculating both maximum nominal tensile or compressive stresses, and maximum nominal shear stresses.
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