Sample Midterm Problem 2
edit
Solution
edit
Part (a)
edit
From the previous problem, for an angle of rotation of 30o {\displaystyle ^{o}} , the rotation matrix [ L ] {\displaystyle \left[L\right]} is
l i j = [ L ] = [ 3 / 2 1 / 2 0 − 1 / 2 3 / 2 0 0 0 1 ] {\displaystyle l_{ij}=\left[L\right]={\begin{bmatrix}{\sqrt {3}}/2&1/2&0\\-1/2&{\sqrt {3}}/2&0\\0&0&1\end{bmatrix}}} Therefore, the components of strain in the rotated co-ordinate system are given by
[ ε ] ′ = [ L ] [ ε ] [ L ] T or, ε i j ′ = l i p l j q ε p q {\displaystyle \left[{\boldsymbol {\varepsilon }}\right]^{'}=\left[L\right]\left[{\boldsymbol {\varepsilon }}\right]\left[L\right]^{T}~~{\text{or,}}~~\varepsilon _{ij}^{'}=l_{ip}l_{jq}\varepsilon _{pq}} Since we are given ε 30 o = ε 11 ′ {\displaystyle \varepsilon _{30^{o}}=\varepsilon _{11}^{'}} , we will calculate the value of this strain in terms of the original components of strain. Thus,
ε 11 ′ = l 1 p l 1 q ε p q = l 11 l 11 ε 11 + l 12 l 11 ε 21 + l 13 l 11 ε 31 + l 11 l 12 ε 12 + l 12 l 12 ε 22 + l 13 l 12 ε 32 + l 11 l 13 ε 13 + l 12 l 13 ε 23 + l 13 l 13 ε 33 = l 11 ( l 11 ε 11 + l 12 ε 12 + l 13 ε 13 ) + l 12 ( l 11 ε 21 + l 12 ε 22 + l 13 ε 23 ) + l 13 ( l 11 ε 31 + l 12 ε 32 + l 13 ε 33 ) = ( 3 2 ) [ ( 3 2 ) ( 0.01 ) + ( 1 2 ) ε 12 ] + ( 1 2 ) [ ( 3 2 ) ε 12 + ( 1 2 ) ( 0.02 ) ] = ( 3 / 4 ) ( 0.01 ) + ( 3 / 2 ) ε 12 + ( 1 / 4 ) ( 0.02 ) = ( 5 / 4 ) ( 0.01 ) + ( 3 / 2 ) ε 12 {\displaystyle {\begin{aligned}\varepsilon _{11}^{'}=&l_{1p}l_{1q}\varepsilon _{pq}\\=&l_{11}l_{11}\varepsilon _{11}+l_{12}l_{11}\varepsilon _{21}+l_{13}l_{11}\varepsilon _{31}+l_{11}l_{12}\varepsilon _{12}+l_{12}l_{12}\varepsilon _{22}+l_{13}l_{12}\varepsilon _{32}+\\&l_{11}l_{13}\varepsilon _{13}+l_{12}l_{13}\varepsilon _{23}+l_{13}l_{13}\varepsilon _{33}\\=&l_{11}(l_{11}\varepsilon _{11}+l_{12}\varepsilon _{12}+l_{13}\varepsilon _{13})+l_{12}(l_{11}\varepsilon _{21}+l_{12}\varepsilon _{22}+l_{13}\varepsilon _{23})+\\&l_{13}(l_{11}\varepsilon _{31}+l_{12}\varepsilon _{32}+l_{13}\varepsilon _{33})\\=&({\frac {\sqrt {3}}{2}})\left[({\frac {\sqrt {3}}{2}})(0.01)+({\frac {1}{2}})\varepsilon _{12}\right]+({\frac {1}{2}})\left[({\frac {\sqrt {3}}{2}})\varepsilon _{12}+({\frac {1}{2}})(0.02)\right]\\=&(3/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}+(1/4)(0.02)\\=&(5/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}\end{aligned}}} Therefore,
( 5 / 4 ) ( 0.01 ) + ( 3 / 2 ) ε 12 = ε 30 o = 0 {\displaystyle (5/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}=\varepsilon _{30^{o}}=0} Hence,
ε 12 = − ( 2.5 ) ( 0.01 ) / 3 {\displaystyle \varepsilon _{12}=-(2.5)(0.01)/{\sqrt {3}}} Next, for an angle of rotation of 60o {\displaystyle ^{o}} , the matrix [ L ] {\displaystyle \left[L\right]} is
[ L ] = [ cos ( 60 o ) sin ( 60 o ) cos ( 90 o ) − sin ( 60 o ) cos ( 60 o ) cos ( 90 o ) cos ( 90 o ) cos ( 90 o ) cos ( 0 o ) ] = [ 1 / 2 3 / 2 0 − 3 / 2 1 / 2 0 0 0 1 ] {\displaystyle {\begin{aligned}\left[L\right]&={\begin{bmatrix}\cos(60^{o})&\sin(60^{o})&\cos(90^{o})\\-\sin(60^{o})&\cos(60^{o})&\cos(90^{o})\\\cos(90^{o})&\cos(90^{o})&\cos(0^{o})\end{bmatrix}}\\&={\begin{bmatrix}1/2&{\sqrt {3}}/2&0\\-{\sqrt {3}}/2&1/2&0\\0&0&1\end{bmatrix}}\end{aligned}}} Therefore, ε 60 o = ε 11 ′ {\displaystyle \varepsilon _{60^{o}}=\varepsilon _{11}^{'}} , is given by
ε 11 ′ = l 1 p l 1 q ε p q = l 11 l 11 ε 11 + l 12 l 11 ε 21 + l 13 l 11 ε 31 + l 11 l 12 ε 12 + l 12 l 12 ε 22 + l 13 l 12 ε 32 + l 11 l 13 ε 13 + l 12 l 13 ε 23 + l 13 l 13 ε 33 = l 11 ( l 11 ε 11 + l 12 ε 12 + l 13 ε 13 ) + l 12 ( l 11 ε 21 + l 12 ε 22 + l 13 ε 23 ) + l 13 ( l 11 ε 31 + l 12 ε 32 + l 13 ε 33 ) = ( 1 2 ) [ ( 1 2 ) ( 0.01 ) + ( 3 2 ) ε 12 ] + ( 3 2 ) [ ( 1 2 ) ε 12 + ( 3 2 ) ( 0.02 ) ] = ( 1 / 4 ) ( 0.01 ) + ( 3 / 2 ) ε 12 + ( 3 / 4 ) ( 0.02 ) = ( 7 / 4 ) ( 0.01 ) + ( 3 / 2 ) ( − ( 2.5 ) ( 0.01 ) / 3 ) = ( 7 / 4 ) ( 0.01 ) − ( 5 / 4 ) ( 0.01 ) = ( 1 / 2 ) ( 0.01 ) = 0.005 {\displaystyle {\begin{aligned}\varepsilon _{11}^{'}=&l_{1p}l_{1q}\varepsilon _{pq}\\=&l_{11}l_{11}\varepsilon _{11}+l_{12}l_{11}\varepsilon _{21}+l_{13}l_{11}\varepsilon _{31}+l_{11}l_{12}\varepsilon _{12}+l_{12}l_{12}\varepsilon _{22}+l_{13}l_{12}\varepsilon _{32}+\\&l_{11}l_{13}\varepsilon _{13}+l_{12}l_{13}\varepsilon _{23}+l_{13}l_{13}\varepsilon _{33}\\=&l_{11}(l_{11}\varepsilon _{11}+l_{12}\varepsilon _{12}+l_{13}\varepsilon _{13})+l_{12}(l_{11}\varepsilon _{21}+l_{12}\varepsilon _{22}+l_{13}\varepsilon _{23})+\\&l_{13}(l_{11}\varepsilon _{31}+l_{12}\varepsilon _{32}+l_{13}\varepsilon _{33})\\=&({\frac {1}{2}})\left[({\frac {1}{2}})(0.01)+({\frac {\sqrt {3}}{2}})\varepsilon _{12}\right]+({\frac {\sqrt {3}}{2}})\left[({\frac {1}{2}})\varepsilon _{12}+({\frac {\sqrt {3}}{2}})(0.02)\right]\\=&(1/4)(0.01)+({\sqrt {3}}/2)\varepsilon _{12}+(3/4)(0.02)\\=&(7/4)(0.01)+({\sqrt {3}}/2)(-(2.5)(0.01)/{\sqrt {3}})\\=&(7/4)(0.01)-(5/4)(0.01)=(1/2)(0.01)=0.005\\\end{aligned}}} Therefore,
ε 60 o = 0.005 {\displaystyle {\varepsilon _{60^{o}}=0.005}} Part (b)
edit
The result is valid for all materials. {\displaystyle {\text{The result is valid for all materials.}}}