. Let us consider the point P with coordinates y = R, z =0,0 x L. Evaluate the principal stresses and the principal stress directions.

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1 14 3. The linear 3-D elasticity mathematical model and substituting in (3.94) gives (n 1 ) =1 (n ) ( ) t n =λ 1 ( 1 (n ) ( ) ) + λ (n ) + λ 3 ( ) t n =λ 1 +(λ λ 1 )(n ) +(λ 3 λ 1 )( ). (3.95) Since λ 1 λ λ 3 we can conclude that t n λ 1 and hence the maximum value of t n that can be reached by varying the plane at the point (varying n) isλ 1 and, of course, is reached f the plane with nmal n 1. Analogously, substituting ( ) =1 (n 1 ) (n ) into (3.94) leads to t n =λ 3 +(λ 1 λ 3 )(n 1 ) +(λ 1 λ )(n ) which implies that t n λ 3 and hence λ 3 is the minimum value 13 of t n and, of course, is reached f the plane with nmal. Since λ 1 and λ 3 are the maximum and minimum stress values that can be reached, we call λ 1,λ and λ 3 principal values and denote these principal stresses as τ 1,τ and τ 3 respectively, and the associated nmals n 1, n, the principal stress directions. Example 3.6 Consider the cylinder as shown in Figure 3.3. The solid cylinder is subjected to self-equilibrating tsional moments M t at the end sections, i.e., M t = M t e x at x = L and M t = M t e x at x =. The stresses at a generic cross-section are given in Figure 3.51 (see Sectio.6). Therefe, at a generic cross-section there is only a shear stress distribution and the maximum value τ is related to the twisting moment by τ = M t I t R where I t = πr4. Let us consider the point P with codinates y = R, z =, x L. Evaluate the principal stresses and the principal stress directions. 13 The expression t n (n)= n T Tn is actually a Rayleigh quotient, and we have in general f the Rayleigh quotient ρ (v) =v T Av, with v T v =1 and A a symmetric matrix of der n, that λ n ρ (v) λ 1, where λ n and λ 1 are the smallest and largest eigenvalues of A, see Bathe, 1996 f a proof

2 3.3 Stresses 143 Fig Shear stress f a cross-section of a cylinder in tsion Solution The stress tens at point P is given by τ T = and the principal stresses can be obtained solving λ τ det(t λi) = det λ τ λ which leads to λ ( λ τ ) =. The roots of this equation are λ 1 = τ, λ = and λ 3 = τ. Hence, the principal stresses are τ 1 = τ, τ =,τ 3 = τ. To obtain the principal stress directions, we need to solve (T τ i I)n i = with n i =1 f i = 1 to 3. F the first principal stress, τ 1 = τ we have τ τ n 1 τ n τ τ (3.96)

3 The linear 3-D elasticity mathematical model τn 1 + τ = τn = τn 1 τ =. Of course, the first and third equations in the above system are the same. Considering the first and the second equations, we obtain n 1 =, n = and imposing that n 1 + n + n 3 = 1 (3.97) we obtain n 1 =1 n 1 = ±. Choosing n 1 = we have n 1 = e x + e z. Note that if we had used n 1 =, then we would have simply obtained n 1 as the solution. Considering τ we have τ n 1 n τ = = τn 1 = which yields n 1 =, = and equation (3.97) gives n = ±1.

4 3.3 Stresses 145 Therefe we can select n = e y. Finally, f τ 3 we have τ τ τ τ τ n 1 n τn 1 + τ = τn = τn 1 + τ = which gives n 1 =, n = and substituting in (3.97) n 1 = ± Selecting n 1 = yields = and we have in summary the following thonmal vects defining the principal directions, i.e., n 1 = e x + e z n = e y = e x + e z. These can be used as new base vects of a new reference codinate system, and in this system, T =. τ Of course, since τ 1 and τ 3 are the maximum and minimum nmal stresses, we have

5 The linear 3-D elasticity mathematical model max(t n ) = τ min(t n ) = τ. Mohr s circles In der to obtain further insight into the state of stress at a point, we study below the stresses f planes which contain one of the principal directions. F example, let us consider a generic plane π which contains. This situation is summarized in Figure 3.5. Fig Definition of a generic plane π which contains ; the angle α is varying; the stress state in the plane π is analyzed Since n 1 and n are thogonal to, they lie in a plane thogonal to π which we call ϕ. Let α be the angle between π and n 1 measured in the clockwise sense from n 1, see Figure 3.5. The stress vect acting on π can be decomposed in the nmal and tangential directions, i.e., t = t n +t s = t n n + t s s where n is the nmal to π and we let s be a unit vect in the direction of t s. The sense of s is chosen such that when t s is positive it tends to rotate the prism of triangular base shown in Figure 3.5 in the clockwise direction. The unit vects n and s and hence t lie in plane ϕ and the stress on the plane π is