Extended Calculation Question (Total: 20 marks) Determine the shear flow distribution for a torque of 56712.0Nm...

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Extended Calculation Question (Total: 20 marks) Determine the shear flow distribution for a torque of 56712.0Nm applied to the three cell section shown in the figure below. Note that the section has a constant shear modulus throughout. All answers should be given to 1 decimal place. Information about the section can be found in the table below: 2 3 4 1 III Wall Length (mm) Thickness (mm) Cell Area (mm2) 12 1084.0 1.215 I 105188.0 12 2104.0 1.733 II 201309.0 14,23 114.0 0.906 III 520351.0 34 752.0 0.906 34 752.0 0.906 Panel Inputs t=[1.215,1.733,0.906,0.906,0.906] ; %% vector of thickness in mm s=[1084,2104,114,752, 752]; %% length vector in mm delta=s./t Cell inputs % setup panelinputs and equation A=[103107,202303,525760]; Panels={[1,2,3,4]; % cell 1 [-3,5]; % cell 2 [6,-4,-5,-2];} C=table (A, Panels) cell 3 delta = 15 103 x 0.8922 1.2141 0.1258 0.8300 0.8300 Panels 3x1 cell array {14 double} {12 double} {14 double} Error using table (line 231) All table variables must have the same number of rows. b) (9 marks) Fill in the following equations for the rate of twist for cells: I, II and III respectively. You must use the shear flow sign convention shown in the figure above. Note: You should pay attention to the equations, particularly the constants at the front and the signs for each term (which are already correct). All the answers in part (b) should be greater than or equal to zero. Cell I: de _1 = dz 2105188.0G xqi- xqII- QIII) Cell II: de xqi+ xqII- XQIII) dz 2201309.0G Cell III: de 1 xqi- xqII+ xqIII) dz 2520351.0G Extended Calculation Question (Total: 20 marks) Determine the shear flow distribution for a torque of 56712.0Nm applied to the three cell section shown in the figure below. Note that the section has a constant shear modulus throughout. All answers should be given to 1 decimal place. Information about the section can be found in the table below: 2 3 4 1 III Wall Length (mm) Thickness (mm) Cell Area (mm2) 12 1084.0 1.215 I 105188.0 12 2104.0 1.733 II 201309.0 14,23 114.0 0.906 III 520351.0 34 752.0 0.906 34 752.0 0.906 Panel Inputs t=[1.215,1.733,0.906,0.906,0.906] ; %% vector of thickness in mm s=[1084,2104,114,752, 752]; %% length vector in mm delta=s./t Cell inputs % setup panelinputs and equation A=[103107,202303,525760]; Panels={[1,2,3,4]; % cell 1 [-3,5]; % cell 2 [6,-4,-5,-2];} C=table (A, Panels) cell 3 delta = 15 103 x 0.8922 1.2141 0.1258 0.8300 0.8300 Panels 3x1 cell array {14 double} {12 double} {14 double} Error using table (line 231) All table variables must have the same number of rows. b) (9 marks) Fill in the following equations for the rate of twist for cells: I, II and III respectively. You must use the shear flow sign convention shown in the figure above. Note: You should pay attention to the equations, particularly the constants at the front and the signs for each term (which are already correct). All the answers in part (b) should be greater than or equal to zero. Cell I: de _1 = dz 2105188.0G xqi- xqII- QIII) Cell II: de xqi+ xqII- XQIII) dz 2201309.0G Cell III: de 1 xqi- xqII+ xqIII) dz 2520351.0G