Given: Large plate in pure shear. Stress state perturbed by a small hole. The BCs are at r = a {\displaystyle r=a} (103) t r = t θ = 0   ;     n ^ = − e ^   r ⇒ σ r r = σ r θ = 0 {\displaystyle {\text{(103)}}\qquad t_{r}=t_{\theta }=0~;~~{\widehat {\mathbf {n} }}=-{\widehat {\mathbf {e} }}~r\Rightarrow \sigma _{rr}=\sigma _{r\theta }=0} at r → ∞ {\displaystyle r\rightarrow \infty } (104) σ 12 → S   ;     σ 11 → 0   ;     σ 22 → 0 {\displaystyle {\text{(104)}}\qquad \sigma _{12}\rightarrow S~;~~\sigma _{11}\rightarrow 0~;~~\sigma _{22}\rightarrow 0} We will solve this problem by superposing a perturbation due to the hole on the unperturbed solution. The effect of the perturbation will decrease with increasing distance from the hole, i.e. the effect will be proportional to r − n {\displaystyle r^{-n}\,} . (105) σ 11 = σ 22 = 0   ;     σ 12 = S {\displaystyle {\text{(105)}}\qquad \sigma _{11}=\sigma _{22}=0~;~~\sigma _{12}=S} Therefore, (106) σ 12 = − φ , 12 = S {\displaystyle {\text{(106)}}\qquad \sigma _{12}=-\varphi _{,12}=S} Integrating, (107) φ , 1 = − S x 2 + f ( x 1 ) ⇒ φ = − S x 1 x 2 + ∫ f ( x 1 ) d x 1 {\displaystyle {\text{(107)}}\qquad \varphi _{,1}=-Sx_{2}+f(x_{1})\Rightarrow \varphi =-Sx_{1}x_{2}+\int f(x_{1})dx_{1}} Since φ {\displaystyle \varphi } is a potential, we can neglect the integration constants (these do not affect the stresses - which are what we are interested in). Hence, (108) φ = − S x 1 x 2 = − S ( r cos ⁡ θ ) ( r sin ⁡ θ ) = − S r 2 2 sin ⁡ ( 2 θ ) {\displaystyle {\text{(108)}}\qquad \varphi =-Sx_{1}x_{2}=-S(r\cos \theta )(r\sin \theta )=-{\cfrac {Sr^{2}}{2}}\sin(2\theta )} or, (109) φ = − S r 2 2 sin ⁡ ( 2 θ ) {\displaystyle {\text{(109)}}\qquad \varphi =-{\cfrac {Sr^{2}}{2}}\sin(2\theta )} Note that we have arranged the expression so that it has a form similar to the Fourier series of the previous section. For this we have to add terms to φ {\displaystyle \varphi } in such a way that The unperturbed solution continues to be true as r → ∞ {\displaystyle r\rightarrow \infty \,} . The terms have the same form as the unperturbed solution,i.e., s i n ( 2 θ ) {\displaystyle sin(2\theta )\,} terms. The new φ {\displaystyle \varphi } leads to stresses that are proportional to r − n {\displaystyle r^{-n}\,} . Recall, φ = ∑ n = 0 ∞ f n ( r ) cos ⁡ ( n θ ) + ∑ n = 0 ∞ g n ( r ) sin ⁡ ( n θ ) {\displaystyle \varphi =\sum _{n=0}^{\infty }f_{n}(r)\cos(n\theta )+\sum _{n=0}^{\infty }g_{n}(r)\sin(n\theta )} where, f 0 ( r ) = A 0 r 2 + B 0 r 2 ln ⁡ r + C 0 + D 0 ln ⁡ r f 1 ( r ) = A 1 r 3 + B 1 r + C 1 r ln ⁡ r + D 1 r − 1 f n ( r ) = A n r n + 2 + B n r n + C n r − n + 2 + D n r − n   ,     n > 1 {\displaystyle {\begin{aligned}f_{0}(r)&=A_{0}r^{2}+B_{0}r^{2}\ln r+C_{0}+D_{0}\ln r\\f_{1}(r)&=A_{1}r^{3}+B_{1}r+C_{1}r\ln r+D_{1}r^{-1}\\f_{n}(r)&=A_{n}r^{n+2}+B_{n}r^{n}+C_{n}r^{-n+2}+D_{n}r^{-n}~,~~n>1\end{aligned}}} So the appropriate stress function for the perturbation is (110) φ = g 2 ( r ) sin ⁡ ( 2 θ ) = ( C 2 r − 2 + 2 + D 2 r − 2 ) sin ⁡ ( 2 θ ) {\displaystyle {\text{(110)}}\qquad \varphi =g_{2}(r)\sin(2\theta )=\left(C_{2}r^{-2+2}+D_{2}r^{-2}\right)\sin(2\theta )} or, (111) φ = ( C 2 + D 2 r − 2 ) sin ⁡ ( 2 θ ) {\displaystyle {\text{(111)}}\qquad \varphi =\left(C_{2}+D_{2}r^{-2}\right)\sin(2\theta )} Hence, the stress function appropriate for the superposed solution is (112) φ = − S r 2 2 sin ⁡ ( 2 θ ) + ( C 2 + D 2 r − 2 ) sin ⁡ ( 2 θ ) {\displaystyle {\text{(112)}}\varphi =-{\cfrac {Sr^{2}}{2}}\sin(2\theta )+\left(C_{2}+D_{2}r^{-2}\right)\sin(2\theta )} We determine C 2 {\displaystyle C_{2}} and D 2 {\displaystyle D_{2}} using the boundary conditions at r = a {\displaystyle r=a} . The stresses are (113) σ r r = 1 r ∂ φ ∂ r + 1 r 2 ∂ 2 φ ∂ θ 2 = ( S − 4 C 2 r − 2 − 6 D 2 r − 4 ) sin ⁡ ( 2 θ ) (114) σ θ θ = ∂ 2 φ ∂ r 2 = ( − S + 6 D 2 r − 4 ) sin ⁡ ( 2 θ ) (115) σ r θ = − ∂ ∂ r ( 1 r ∂ φ ∂ θ ) = ( S + 6 D 2 r − 4 ) cos ⁡ ( 2 θ ) {\displaystyle {\begin{aligned}{\text{(113)}}\qquad \sigma _{rr}&={\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial r}}+{\cfrac {1}{r^{2}}}{\cfrac {\partial ^{2}\varphi }{\partial \theta ^{2}}}=\left(S-4C_{2}r^{-2}-6D_{2}r^{-4}\right)\sin(2\theta )\\{\text{(114)}}\qquad \sigma _{\theta \theta }&={\cfrac {\partial ^{2}\varphi }{\partial r^{2}}}=\left(-S+6D_{2}r^{-4}\right)\sin(2\theta )\\{\text{(115)}}\qquad \sigma _{r\theta }&=-{\cfrac {\partial }{\partial r}}\left({\cfrac {1}{r}}{\cfrac {\partial \varphi }{\partial \theta }}\right)=\left(S+6D_{2}r^{-4}\right)\cos(2\theta )\end{aligned}}} Hence, (116) σ r r | r = a = 0 = ( S − 4 C 2 a − 2 − 6 D 2 a − 4 ) sin ⁡ ( 2 θ ) (117) σ r θ | r = a = 0 = ( S + 2 C 2 a − 2 + 6 D 2 a − 4 ) cos ⁡ ( 2 θ ) {\displaystyle {\begin{aligned}{\text{(116)}}\qquad \left.\sigma _{rr}\right|_{r=a}&=0=\left(S-4C_{2}a^{-2}-6D_{2}a^{-4}\right)\sin(2\theta )\\{\text{(117)}}\qquad \left.\sigma _{r\theta }\right|_{r=a}&=0=\left(S+2C_{2}a^{-2}+6D_{2}a^{-4}\right)\cos(2\theta )\end{aligned}}} or, (118) 4 C 2 a − 2 + 6 D 2 a − 4 = S (119) 2 C 2 a − 2 + 6 D 2 a − 4 = − S {\displaystyle {\begin{aligned}{\text{(118)}}\qquad 4C_{2}a^{-2}+6D_{2}a^{-4}&=S\\{\text{(119)}}\qquad 2C_{2}a^{-2}+6D_{2}a^{-4}&=-S\end{aligned}}} Solving, (120) C 2 = S a 2   ;     D 2 = − S a 4 2 {\displaystyle {\text{(120)}}\qquad C_{2}=Sa^{2}~;~~D_{2}=-{\cfrac {Sa^{4}}{2}}} Back substituting, (121) σ r r = S ( 1 − 4 a 2 r 2 + 3 a 4 r 4 ) sin ⁡ ( 2 θ ) (122) σ θ θ = S ( − 1 − 3 a 4 r 4 ) sin ⁡ ( 2 θ ) (123) σ r θ = S ( 1 + 2 a 2 r 2 − 3 a 4 r 4 ) cos ⁡ ( 2 θ ) {\displaystyle {\begin{aligned}{\text{(121)}}\qquad \sigma _{rr}&=S\left(1-4{\cfrac {a^{2}}{r^{2}}}+3{\cfrac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(122)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3{\cfrac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(123)}}\qquad \sigma _{r\theta }&=S\left(1+2{\cfrac {a^{2}}{r^{2}}}-3{\cfrac {a^{4}}{r^{4}}}\right)\cos(2\theta )\end{aligned}}} Consider the elastic plate with a hole subject to pure shear. The stresses close to the hole are given by (29) σ r r = S ( 1 − 4 a 2 r 2 + 3 a 4 r 4 ) sin ⁡ ( 2 θ ) (30) σ θ θ = S ( − 1 − 3 a 4 r 4 ) sin ⁡ ( 2 θ ) (31) σ r θ = S ( 1 + 2 a 2 r 2 − 3 a 4 r 4 ) cos ⁡ ( 2 θ ) {\displaystyle {\begin{aligned}{\text{(29)}}\qquad \sigma _{rr}&=S\left(1-4{\frac {a^{2}}{r^{2}}}+3{\frac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(30)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3{\frac {a^{4}}{r^{4}}}\right)\sin(2\theta )\\{\text{(31)}}\qquad \sigma _{r\theta }&=S\left(1+2{\frac {a^{2}}{r^{2}}}-3{\frac {a^{4}}{r^{4}}}\right)\cos(2\theta )\end{aligned}}} Show that the normal and shear traction boundary conditions far from the hole are satisfied by these stresses. Calculate the stress concentration factors at the hole, i.e., ( τ max / S {\displaystyle \tau _{\text{max}}/S} ) (shear) and ( σ max / σ 0 {\displaystyle \sigma _{\text{max}}/\sigma _{0}} ) (normal). Calculate the displacement field corresponding to this stress field (for plane stress). Plot the deformed shape of the hole. Far from the hole, r = ∞ {\displaystyle r=\infty } . Therefore, (32) σ r r = S sin ⁡ ( 2 θ ) (33) σ θ θ = − S sin ⁡ ( 2 θ ) (34) σ r θ = S cos ⁡ ( 2 θ ) {\displaystyle {\begin{aligned}{\text{(32)}}\qquad \sigma _{rr}&=S\sin(2\theta )\\{\text{(33)}}\qquad \sigma _{\theta \theta }&=-S\sin(2\theta )\\{\text{(34)}}\qquad \sigma _{r\theta }&=S\cos(2\theta )\end{aligned}}} To rotate the stresses back to the ( x 1 , x 2 ) {\displaystyle (x_{1},x_{2})} coordinate system, we use the tensor transformation rule (35) [ σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33 ] = [ cos ⁡ θ − sin ⁡ θ 0 sin ⁡ θ cos ⁡ θ 0 0 0 1 ] [ σ r r σ r θ σ r z σ r θ σ θ θ σ θ z σ r z σ θ z σ z z ] [ cos ⁡ θ sin ⁡ θ 0 − sin ⁡ θ cos ⁡ θ 0 0 0 1 ] {\displaystyle {\text{(35)}}\qquad {\begin{bmatrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}{\begin{bmatrix}\sigma _{rr}&\sigma _{r\theta }&\sigma _{rz}\\\sigma _{r\theta }&\sigma _{\theta \theta }&\sigma _{\theta z}\\\sigma _{rz}&\sigma _{\theta z}&\sigma _{zz}\end{bmatrix}}{\begin{bmatrix}\cos \theta &\sin \theta &0\\-\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}} Setting σ r z = 0 {\displaystyle \sigma _{rz}=0} and σ θ z = 0 {\displaystyle \sigma _{\theta z}=0} , we get the simplified set of equations (36) σ 11 = σ r r cos 2 ⁡ θ + σ θ θ sin 2 ⁡ θ − σ r θ sin ⁡ ( 2 θ ) (37) σ 22 = σ r r sin 2 ⁡ θ + σ θ θ cos 2 ⁡ θ + σ r θ sin ⁡ ( 2 θ ) (38) σ 12 = σ r r − σ θ θ 2 sin ⁡ ( 2 θ ) + σ r θ cos ⁡ ( 2 θ ) {\displaystyle {\begin{aligned}{\text{(36)}}\qquad \sigma _{11}&=\sigma _{rr}\cos ^{2}\theta +\sigma _{\theta \theta }\sin ^{2}\theta -\sigma _{r\theta }\sin(2\theta )\\{\text{(37)}}\qquad \sigma _{22}&=\sigma _{rr}\sin ^{2}\theta +\sigma _{\theta \theta }\cos ^{2}\theta +\sigma _{r\theta }\sin(2\theta )\\{\text{(38)}}\qquad \sigma _{12}&={\frac {\sigma _{rr}-\sigma _{\theta \theta }}{2}}\sin(2\theta )+\sigma _{r\theta }\cos(2\theta )\end{aligned}}} Plugging in equations (32-34) in the above, we have (39) σ 11 = − S [ sin ⁡ ( 2 θ ) cos ⁡ ( 2 θ ) − sin ⁡ ( 2 θ ) cos ⁡ ( 2 θ ) ] = 0 (40) σ 22 = S [ sin ⁡ ( 2 θ ) cos ⁡ ( 2 θ ) − sin ⁡ ( 2 θ ) cos ⁡ ( 2 θ ) ] = 0 (41) σ 12 = S [ sin ⁡ ( 2 θ ) sin ⁡ ( 2 θ ) + cos ⁡ ( 2 θ ) cos ⁡ ( 2 θ ) ] = S {\displaystyle {\begin{aligned}{\text{(39)}}\qquad \sigma _{11}&=-S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(40)}}\qquad \sigma _{22}&=S\left[\sin(2\theta )\cos(2\theta )-\sin(2\theta )\cos(2\theta )\right]=0\\{\text{(41)}}\qquad \sigma _{12}&=S\left[\sin(2\theta )\sin(2\theta )+\cos(2\theta )\cos(2\theta )\right]=S\end{aligned}}} Hence, the far field stress BCs are satisfied. The stresses at the hole ( r = a {\displaystyle r=a} ) are (42) σ r r = S ( 1 − 4 + 3 ) sin ⁡ ( 2 θ ) = 0 (43) σ θ θ = S ( − 1 − 3 ) sin ⁡ ( 2 θ ) = − 4 S sin ⁡ ( 2 θ ) (44) σ r θ = S ( 1 + 2 − 3 ) cos ⁡ ( 2 θ ) = 0 {\displaystyle {\begin{aligned}{\text{(42)}}\qquad \sigma _{rr}&=S\left(1-4+3\right)\sin(2\theta )=0\\{\text{(43)}}\qquad \sigma _{\theta \theta }&=S\left(-1-3\right)\sin(2\theta )=-4S\sin(2\theta )\\{\text{(44)}}\qquad \sigma _{r\theta }&=S\left(1+2-3\right)\cos(2\theta )=0\end{aligned}}} The maximum (or minimum) hoop stress at the hole is at the locations where d σ θ θ / d θ = − 8 S cos ⁡ ( 2 θ ) = 0 {\displaystyle d\sigma _{\theta \theta }/d\theta =-8S\cos(2\theta )=0} . These locations are θ = π / 4 {\displaystyle \theta =\pi /4} and θ = 3 π / 4 {\displaystyle \theta =3\pi /4} . The value of the hoop stress is (45) at   θ = π 4 σ θ θ = − 4 S (46) at   θ = 3 π 4 σ θ θ = 4 S {\displaystyle {\begin{aligned}{\text{(45)}}\qquad {\text{at}}~\theta ={\frac {\pi }{4}}&&\sigma _{\theta \theta }=-4S\\{\text{(46)}}\qquad {\text{at}}~\theta ={\frac {3\pi }{4}}&&\sigma _{\theta \theta }=4S\end{aligned}}} The maximum shear stress is given by (47) τ max = 1 2 | σ r r − σ θ θ | = 2 S {\displaystyle {\text{(47)}}\qquad \tau _{\text{max}}={\frac {1}{2}}\left|\sigma _{rr}-\sigma _{\theta \theta }\right|=2S} Therefore, the stress concentration factors are (48) σ max S = 4   ;     τ max S = 2 {\displaystyle {\text{(48)}}\qquad {\frac {\sigma _{\text{max}}}{S}}=4~;~~{\frac {\tau _{\text{max}}}{S}}=2} The stress function used to derive the above results was (49) φ = − S 2 r 2 sin ⁡ ( 2 θ ) + S a 2 sin ⁡ ( 2 θ ) − S a 4 2 r − 2 sin ⁡ ( 2 θ ) {\displaystyle {\text{(49)}}\qquad \varphi =-{\frac {S}{2}}r^{2}\sin(2\theta )+Sa^{2}\sin(2\theta )-{\frac {Sa^{4}}{2}}r^{-2}\sin(2\theta )} From Michell's solution, the displacements corresponding to the above stress function are given by (50) 2 μ u r = − S 2 [ − 2 r sin ⁡ ( 2 θ ) ] + S a 2 [ ( κ + 1 ) r − 1 sin ⁡ ( 2 θ ) ] − S a 4 2 [ 2 r − 3 sin ⁡ ( 2 θ ) ] (51) 2 μ u θ = − S 2 [ − 2 r cos ⁡ ( 2 θ ) ] + S a 2 [ ( κ − 1 ) r − 1 cos ⁡ ( 2 θ ) ] − S a 4 2 [ − 2 r − 3 cos ⁡ ( 2 θ ) ] {\displaystyle {\begin{aligned}{\text{(50)}}\qquad 2\mu u_{r}&=-{\frac {S}{2}}\left[-2r\sin(2\theta )\right]+Sa^{2}\left[(\kappa +1)r^{-1}\sin(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[2r^{-3}\sin(2\theta )\right]\\{\text{(51)}}\qquad 2\mu u_{\theta }&=-{\frac {S}{2}}\left[-2r\cos(2\theta )\right]+Sa^{2}\left[(\kappa -1)r^{-1}\cos(2\theta )\right]-{\frac {Sa^{4}}{2}}\left[-2r^{-3}\cos(2\theta )\right]\end{aligned}}} or, (52) u r = S r sin ⁡ ( 2 θ ) 2 μ [ 1 + ( κ + 1 ) a 2 r 2 − a 4 r 4 ] (53) u θ = S r cos ⁡ ( 2 θ ) 2 μ [ 1 + ( κ − 1 ) a 2 r 2 + a 4 r 4 ] {\displaystyle {\begin{aligned}{\text{(52)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+(\kappa +1){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(53)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+(\kappa -1){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}} For plane stress, κ = ( 3 − ν ) / ( 1 + ν ) {\displaystyle \kappa =(3-\nu )/(1+\nu )} . Hence, (54) u r = S r sin ⁡ ( 2 θ ) 2 μ [ 1 + ( 4 1 + ν ) a 2 r 2 − a 4 r 4 ] (55) u θ = S r cos ⁡ ( 2 θ ) 2 μ [ 1 + 2 ( 1 − ν 1 + ν ) a 2 r 2 + a 4 r 4 ] {\displaystyle {\begin{aligned}{\text{(54)}}\qquad u_{r}&={\frac {Sr\sin(2\theta )}{2\mu }}\left[1+\left({\frac {4}{1+\nu }}\right){\frac {a^{2}}{r^{2}}}-{\frac {a^{4}}{r^{4}}}\right]\\{\text{(55)}}\qquad u_{\theta }&={\frac {Sr\cos(2\theta )}{2\mu }}\left[1+2\left({\frac {1-\nu }{1+\nu }}\right){\frac {a^{2}}{r^{2}}}+{\frac {a^{4}}{r^{4}}}\right]\end{aligned}}} At r = a {\displaystyle r=a} , (56) u r = S a sin ⁡ ( 2 θ ) μ ( 2 1 + ν ) (57) u θ = S a cos ⁡ ( 2 θ ) μ ( 2 1 + ν ) {\displaystyle {\begin{aligned}{\text{(56)}}\qquad u_{r}&={\frac {Sa\sin(2\theta )}{\mu }}\left({\frac {2}{1+\nu }}\right)\\{\text{(57)}}\qquad u_{\theta }&={\frac {Sa\cos(2\theta )}{\mu }}\left({\frac {2}{1+\nu }}\right)\end{aligned}}} Now μ = E / 2 ( 1 + ν ) {\displaystyle \mu =E/2(1+\nu )} . Hence, we have (58) u r = 4 S a sin ⁡ ( 2 θ ) E (59) u θ = 4 S a cos ⁡ ( 2 θ ) E {\displaystyle {\begin{aligned}{\text{(58)}}\qquad u_{r}&={\frac {4Sa\sin(2\theta )}{E}}\\{\text{(59)}}\qquad u_{\theta }&={\frac {4Sa\cos(2\theta )}{E}}\end{aligned}}} The deformed shape is shown below Elasticity
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