Draw the first three mode shapes using a computer program. You are free to assign the numerical parameters related to the mone shapes ( Matlab, mathematica)

  

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The boundary conditions for this system are u (0,t) = 0, mat These conditions yield (0) = 0 and (k-w²m) o (1) = -EAd' (1). The general orthogon relationship of Eq. 4.104 can be written as Eq. 4.104 = Or -ku (1, t) - EA au(lt) dz EA; (a) o (x) - EA; (x) o (x) dx = −w SpA; (x) 0₁ (x) dx E A'; (1) 0% (1)-EA; (0) (0) - EA; (a) o (x) dx =-w² SpA6; (x) x (x) dx By using the boundary conditions of this example, this orthogonality relationship can be written as (kw/m) o; (1)ox (1)-f EA'; (x) o (x) dx = wf pA; (x) or (a) da ko; (1)øk (1) + S EAø'; (x) o (x) dx = w} [mo; (1)¢x (1) + SpA6; (x) x (x) dx] Similar relationship can be obtained for mode k as ko,(D)x(l) + Jo EA%; (a) ¢ (2) da =ủa mọ;(lợi(l) + Sổ A4, (2), (2) da Subtracting this equation from the one associated with mode j, one of orthogonality relationships for j #k: and for jk, one has mo (1) + f pA ko(1) + SEA (x) dx = m; (x) dx = kj where m; and k; are, respectively, the modal mass and stiffness coeffi frequency of the system can be defined as mo; (1)ox (1) + f pAo; (x) x (x) dx = 0 ko; (1)ox (1) + SEA; (x) o (x) dx = 0 Eq. 4.109 w² = = mj kj mj = If m and k approach zero, the natural frequency w; approaches the v for the simple end conditions. ko (1)+ EA(z)dz mo (1)+pA(z)dz EAo da SPAodz j=1,2,3.... The boundary conditions for this system are u (0,t) = 0, mat These conditions yield (0) = 0 and (k-w²m) o (1) = -EAd' (1). The general orthogon relationship of Eq. 4.104 can be written as Eq. 4.104 = Or -ku (1, t) - EA au(lt) dz EA; (a) o (x) - EA; (x) o (x) dx = −w SpA; (x) 0₁ (x) dx E A'; (1) 0% (1)-EA; (0) (0) - EA; (a) o (x) dx =-w² SpA6; (x) x (x) dx By using the boundary conditions of this example, this orthogonality relationship can be written as (kw/m) o; (1)ox (1)-f EA'; (x) o (x) dx = wf pA; (x) or (a) da ko; (1)øk (1) + S EAø'; (x) o (x) dx = w} [mo; (1)¢x (1) + SpA6; (x) x (x) dx] Similar relationship can be obtained for mode k as ko,(D)x(l) + Jo EA%; (a) ¢ (2) da =ủa mọ;(lợi(l) + Sổ A4, (2), (2) da Subtracting this equation from the one associated with mode j, one of orthogonality relationships for j #k: and for jk, one has mo (1) + f pA ko(1) + SEA (x) dx = m; (x) dx = kj where m; and k; are, respectively, the modal mass and stiffness coeffi frequency of the system can be defined as mo; (1)ox (1) + f pAo; (x) x (x) dx = 0 ko; (1)ox (1) + SEA; (x) o (x) dx = 0 Eq. 4.109 w² = = mj kj mj = If m and k approach zero, the natural frequency w; approaches the v for the simple end conditions. ko (1)+ EA(z)dz mo (1)+pA(z)dz EAo da SPAodz j=1,2,3....

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To draw the first three mode shapes using a computer program we need to solve the system of equation... View the full answer