Page 213 - Marks Calculation for Machine Design

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PRINCIPAL STRESSES AND MOHR’S CIRCLE
The rotated stresses found in Example 1, and verified in Example 2, can now be used to
design the weld joint itself. The rotated stress (σ x x ) provides the normal stress requirement

alongtheweld,therotatedstress(σ y y )providesthenormalstressrequirementperpendicular

to the weld, and the rotated stress (τ x y ) provides the shear stress requirement for the weld.

Design of welds is not covered in this book, however Marks’ Standard Handbook for
Mechanical Engineers, as well as the Standard Handbook of Machine Design, are excellent
references for the required analysis.
Maximum Stress Elements. If only a set of rotated stresses are needed from a known set
of unrotated stresses for a specified angle (θ), like in Example 1, then Eqs. (5.1) to (5.3) are
sufficient to provide this information and this section would be complete. However, what
the machine designer really wants to know is what are the maximum stresses acting on the
element, and it is expected that these maximum stresses will not occur at an angle (θ) equal
to zero. Again, the angle (θ) represents rotation from a direction natural to the machine
element under investigation. In Example 1 it was the axis of the pressurized tank. For a
beam in bending it typically would be the neutral axis.
To find the maximum stresses, and the special angle of the stress element on which they
act, Eqs. (5.1) to (5.3), which are only functions of the angle (θ), are differentiated with
respect to the angle (θ), then these derivatives are set equal to zero to find the special angle,
denoted (φ p ), that the unrotated element must be rotated to provide the element with the
maximum values of the stresses. This special angle is then substituted in Eqs. (5.1) to (5.3)
to provide the relationships required. Remember, it is assumed that the unrotated stresses,
(σ xx ),(σ yy ), and (τ xy ) are known.
Leaving out the details of the differentiations, and the bzillion algebra and trig steps, the
maximum normal stress, called the principal stress (σ 1 ) is given by Eq. (5.7),

2

σ xx + σ yy σ xx − σ yy
2
σ 1 = + + τ xy (5.7)
2 2
and the minimum normal stress, called the principal stress (σ 2 ) is given by Eq. (5.8),

2

σ xx + σ yy σ xx − σ yy
2
σ 2 = − + τ xy (5.8)
2 2
where the special angle (φ p ) for the rotated element on which the principal stresses (σ 1 )
and (σ 2 ) act is given by Eq. (5.9).
2τ xy
tan 2φ p = (5.9)
σ xx − σ yy
For the rotated element defined by the angle (φ p ), the shear stresses are zero.
Without providing the proof, the maximum and minimum shear stresses are on an element
rotated 45 degrees from the angle (φ p ), denoted by (φ s ), and given by Eq. (5.10).
σ xx − σ yy
tan 2φ s =− (5.10)
2τ xy
Again, without providing the proof, the relationship between the special angle (φ p ) for the
principal stresses (σ 1 ) and (σ 2 ) and the special angle (φ s ) for the maximum and minimum
shear stresses is given by Eq. (5.11).
φ s = φ p ± 45 ◦ (5.11)