Question 10.14: As part of the product development process for a new pump, t...

Objective     Compute \sigma_{max}, \sigma_{min},  and  \tau_{max}.

Given           Strain readings from a 0° , 45° , 90° strain-gage rosette

ε_{1} = 950 × 10^{−6}  mm/mm

ε_{2} = −375 × 10^{−6}  mm/mm,

ε_{3} = 525 × 10^{−6}   mm/mm

Aluminum 2014-T6: E = 73.1 \times 10^{9} N/m²; s_{y} = 414 \times 10^{6} N/m²  (Appendix A–14)  v = 0.33   (Table 2–1)

Analysis     Use the Procedure for Analyzing the Data from a Strain-Gage Rosette

Results       Steps 1 though 4 have already been completed

Step 5. In using Equations (10–22) through (10–24), we will show only the whole number part of the strain values. The unit is then microstrains, με.

The maximum principal strain is

\epsilon _{max} = \frac{\left(\epsilon _{1} + \epsilon _{3}\right) }{2} + \frac{\sqrt{\left(\epsilon _{1} – \epsilon _{2}\right)^{2} + \left(\epsilon _{2} – \epsilon _{3}\right)^{2}} }{\sqrt{2} } 

\epsilon _{max} = \frac{(950 + 525)}{2} + \frac{\sqrt{\left[(950-(-375))\right]^{2} + \left(-375 – 525\right)^{2}} }{\sqrt{2} } = 1870  \mu \epsilon 

The minimum principal strain is

\epsilon _{min} = \frac{\left(\epsilon _{1} + \epsilon _{3}\right) }{2} – \frac{\sqrt{\left(\epsilon _{1} – \epsilon _{2}\right)^{2} – \left(\epsilon _{2} – \epsilon _{3}\right)^{2}} }{\sqrt{2} } 

\epsilon _{min} = \frac{(950 + 525)}{2} – \frac{\sqrt{\left[(950-(-375))\right]^{2} + \left(-375 – 525\right)^{2}} }{\sqrt{2} } = -395  \mu \epsilon 

The angle from gage number 1 to the nearest principal strain axis is

\beta = \frac{1}{2} \tan^{-1} \left[ \frac{(\epsilon _{2}-\epsilon _{3})-(\epsilon _{1}-\epsilon _{2})}{(\epsilon _{1}-\epsilon _{3})} \right] 

\beta = \frac{1}{2} \tan^{-1} \left[ \frac{(-375 -525)-[950 -(-375)]}{(950 – 525)} \right]  = -39.6°

Step 6. The maximum principal stress is found from Equation (10–28):

\sigma _{max} = \frac{E}{1-v^{2}}(\epsilon _{max}+v\epsilon _{min})

\sigma _{max} = \frac{73.1 \times 10^{9}  Pa}{1-(0.33)^{2}}[1870 + 0.33(-395)(10^{-6})] = 143 MPa

The minimum principal stress is found from Equation (10–29):

\sigma _{max} = \frac{E}{1-v^{2}}(\epsilon _{min}+v\epsilon _{max})

\sigma _{max} = \frac{73.1 \times 10^{9}  Pa}{1-(0.33)^{2}}[-395 + 0.33(1870)(10^{-6})] = 18.2 MPa

Step 7. The maximum shearing strain is found from Equation (10–30):

\gamma_{max} = |(\epsilon _{max}-\epsilon_{min}) = |1870 – (-395)| = 2265 \times 10^{-6}   rad

Step 8. The maximum shearing stress in the plane of the initial element is found from Equation (10–32):

\tau_{max} = \frac{E \gamma_{max}}{2(1+v)} = \frac{(73.1 \times 10^{9} pa)(2265 \times 10^{-6})}{2(1+0.33)} = 62.3 MPa

Step 9. We note that both the maximum principal stress and the minimum principal stress are positive or tensile. Therefore, we must draw a supplemental Mohr’s circle to determine the true maximum shearing strain. See Figure 10–50. The circle is simply drawn by plotting both principal stresses on the horizontal axis and drawing the circle to include both. The supplementary circle is drawn through \sigma_{max} and the origin of the axes because that represents a zero stress perpendicular to the plane of the initial stress element. The true maximum shearing stress is equal to the radius of this circle, found from

\tau_{max} = ( \sigma_{max} – 0)/2 = (143  MPa)/2 = 71.5 MPa

Summary   The final results are

\sigma_{max} = 143 MPa    \sigma_{min} = 18.2 MPa  in the plane of the initial element

True \sigma_{max} = 0 perpendicular to the plane of the initial element

\tau_{max} = 71.5 MPa