Question 7.S-P.7: Using a 60° rosette, the following strains have been determi...
Using a 60° rosette, the following strains have been determined at point Q on the surface of a steel machine base:
\epsilon_{1}= 40 m \epsilon_{2} = 980 m \epsilon_{3} = 330 m
Using the coordinate axes shown, determine at point Q, (a) the strain components \epsilon_{x}, \epsilon_{y} , and g_{xy}, (b) the principal strains, (c) the maximum shearing strain.
(Use n = 0.29.)
a. Strain Components \epsilon_{x}, \epsilon_{y} , g_{xy}. For the coordinate axes shown
u _{1}=0 u _{2}=60^{\circ} u _{3}=120^{\circ}
Substituting these values into Eqs. (7.60), we have
\epsilon_{1}=\epsilon_{x} \cos ^{2} u _{1}+\epsilon_{y} \sin ^{2} u _{1}+ g _{x y} \sin u _{1} \cos u _{1}\epsilon_{2}=\epsilon_{x} \cos ^{2} u _{2}+\epsilon_{y} \sin ^{2} u _{2}+ g _{x y} \sin u _{2} \cos u _{2}
\epsilon_{3}=\epsilon_{x} \cos ^{2} u _{3}+\epsilon_{y} \sin ^{2} u _{3}+ g _{x y} \sin u _{3} \cos u _{3} (7.60)
\epsilon_{1}=\epsilon_{x}(1) \quad+\epsilon_{y}(0) \quad+ g _{x y}(0)(1)
\epsilon _{2}= \epsilon _{x}(0.500)^{2}+\epsilon_{y}(0.866)^{2}+ g _{x y}(0.866)(0.500)
\epsilon _{3}= \epsilon _{x}(-0.500)^{2}+ \epsilon _{y}(0.866)^{2}+ g _{x y}(0.866)(-0.500)
Solving these equations for \epsilon_{x}, \epsilon_{y} , and g_{xy}, we obtain
\epsilon_{x}=\epsilon_{1} \epsilon_{y}=\frac{1}{3}\left(2 \epsilon_{2}+2 \epsilon_{3}-\epsilon_{1}\right) g_{xy}=\frac{\epsilon_{2}-\epsilon_{3}}{0.866}
Substituting the given values for \epsilon_{1}, \epsilon_{2}, and \epsilon_{3}, we have
\epsilon_{x}=40 m \epsilon _{y}=\frac{1}{3}[2(980)+2(330)-40] \epsilon_{y}=+860 m
g _{x y}=(980-330) / 0.866 g _{x y}=750 m
These strains are indicated on the element shown.
b. Principal Strains. We note that the side of the element associated with \epsilon_{x} rotates counterclockwise; thus, we plot point X below the horizontal axis, i.e., X(40, -375) . We then plot Y(860, +375) and draw Mohr’s circle.
\epsilon _{\text {ave }}=\frac{1}{2}(860 m +40 m )=450 mR=\sqrt{(375 m )^{2}+(410 m )^{2}}=556 m
\tan 2 u _{p}=\frac{375 m }{410 m } 2 u _{p}=42.4^{\circ}\curvearrowright u_{p}=21.2^{\circ}\curvearrowright
Points A and B correspond to the principal strains. We have
\epsilon_{a}=\epsilon_{\text {ave }}-R=450 m -556 m \epsilon_{a}=-106 m
\epsilon_{b}=\epsilon_{\text {ave }}+R=450 m +556 m \epsilon _{b}=+1006 m
Since s _{z}=0 on the surface, we use Eq.(7.59) to find the principal strain \epsilon_{c}:
\epsilon_{c}=-\frac{ n }{1- n }\left(\epsilon_{a}+\epsilon_{b}\right) (7.59)
\epsilon _{c}=-\frac{ n }{1- n }\left(\epsilon_{a}+\epsilon_{b}\right)=-\frac{0.29}{1-0.29}(-106 m +1006 m ) \epsilon_{c}=-368 m
c. Maximum Shearing Strain. Plotting point C and drawing Mohr’s circle through points B and C, we obtain point D^{\prime} and write
\frac{1}{2} g _{\max }=\frac{1}{2}(1006 m +368 m ) g _{\max }=1374 m
