4 Displacements of beam systems

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Aerospace Structure –Exercise #4 1 Exercise #4 Displacements in beam systems Aerospace Structure –Exercise #4 2 Displacements in beam systems – Exercise 1 Considering the Problem #1 of Statically Indetermin ate System (Ex 3), evaluate the vertical displaceme nt at the free end r = 2 N/mm L1 = 500 mm r Sy = ? L2 = 1500 mm L 2 L1 SOLUTION The problem is solved by applying PCVW. The real system is the following. r L 2 L1 The real system is statically indeterminate, but th e solution can be found by applying PCVW, as it has been seen in Ex. 3. The value of the undeterminate reaction is: N rL L LrL rL Y 2375 500 0 . 21500 500 4 3500 0 . 21500 0 . 28 3 4 3 8 3 1 2 1 12 −=⋅+⋅+⋅− =      ++−= A single axis z is considered. A technique based on the principle of superposition will be therefore applied. 0<z<L 1 T = -rz ; M z = - 1/2 r z2 ; L 1<z<L 1 +L 2 T = - rz - Y; M x = - 1/2 r z2 - Y (z-L1) ; The dummy system is obtained by applying a vertical unit force to the free end. z Aerospace Structure –Exercise #4 3 1 0<z<L 1 T’ = -1 ; M’ x = - z ; L1<z<L 2 T’ = - 1 ; M’ x = -z ; Basing on PCVW: δW e = S y 1 ∫ ∫ + += 1 0 2 1 1 '' L LL L xxxx d dx EJ MMdx EJ MMWδ ( )[ ] ( )[ ] ( )[ ] EJ LLLYL EJ LLLY EJ LLLr EJ rL dz EJ YzL EJ Yz EJ rz dz EJ rz Sy LL L L 2388 22 2 1 2 121 3 1 3 12 4 1 4 12 4 1 21 1 1 231 0 3 −+−−++−++= =      −++= ∫∫ + S y =-0.0781 mm Aerospace Structure –Exercise #4 4 Displacements in beam systems – Exercise 2 Evaluate the displacement at the free end of the beam system that is sketched in the following figure. L 1 = 750 mm L2 = 1500 mm r = 0.08 N/mm E = 200000 Mpa G = 77000 Mpa All the beams have the circular section indicated in the figure. SOLUTION The properties of the beam section are evaluated from the following expression: ( ) 444 39761 15 44 mm…