Consider a plane truss in figure 1.31. The horizontal and vertical members have length 1, while...

 

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Consider a plane truss in figure 1.31. The horizontal and vertical members have length 1, while inclined members have length 21. Assume Young's modulus E = 100 GPA, cross-sectional area A = 1.0 cm, and 1 = 0.3 m. floog 1. Use an FE program to determine the deflections and element forces for the following three load cases. Present you results in the form of a table. gm gritori Load Case A) Fx13 = Fx14 = 10,000 N sidswolle od nech da snT Yo Widgisw ar saim al sd bloods 15denom no song tomolooz 240 Load Case B) Fy13 = Fy14 = 10,000 N - 51g Load Case C) Fx13 = 10,000 N and Fx14 = -10,000 N migo-bir 2. Assuming that the truss behaves like a cantilever beam, one can determine the equivalent cross- sectional properties of the beam from the results for cases A through C above. The three beam properties are axial rigidity (EA)eq (this is different from the AE of the truss member), flexural rigidity (ED)eq, and shear rigidity (GA)eq. Let the beam length be equal to L (L = 6 0.3=1.8 m). The axial deflection of a beam due to an axial force F is given by: 19 FL (EA) eg The transverse deflection due to a transverse force F at the tip is: 13 7 20 26 1 (1.73) In eq. (1.73) the first term on the RHS represents the deflection due to flexure and the second term, due to shear deformation. In the elementary beam theory (Euler-Bernoulli beam theory) we neglect the shear deformation, as it is usually much smaller than the flexural deflection. The transverse deflection due to an end couple C is given by: 14 3 8 21 27 2 15 Vtip = 9 22 28 5 3 Utip TO FL3 FL + 3(EI) eq (GA) egleno Vtip 8 16 7 CL2 2(EI) eq 10 23 29 4 Figure 1.31 Plane truss and design domain for Project 1.3 (10) 17 24 30 5 (12) dans vibizale to zuluboM geoda oldswollA A soe Isitial muminiM 18 11 12 25 31 6 14) 19 (1.72) (13) (1.74) Consider a plane truss in figure 1.31. The horizontal and vertical members have length 1, while inclined members have length 21. Assume Young's modulus E = 100 GPA, cross-sectional area A = 1.0 cm, and 1 = 0.3 m. floog 1. Use an FE program to determine the deflections and element forces for the following three load cases. Present you results in the form of a table. gm gritori Load Case A) Fx13 = Fx14 = 10,000 N sidswolle od nech da snT Yo Widgisw ar saim al sd bloods 15denom no song tomolooz 240 Load Case B) Fy13 = Fy14 = 10,000 N - 51g Load Case C) Fx13 = 10,000 N and Fx14 = -10,000 N migo-bir 2. Assuming that the truss behaves like a cantilever beam, one can determine the equivalent cross- sectional properties of the beam from the results for cases A through C above. The three beam properties are axial rigidity (EA)eq (this is different from the AE of the truss member), flexural rigidity (ED)eq, and shear rigidity (GA)eq. Let the beam length be equal to L (L = 6 0.3=1.8 m). The axial deflection of a beam due to an axial force F is given by: 19 FL (EA) eg The transverse deflection due to a transverse force F at the tip is: 13 7 20 26 1 (1.73) In eq. (1.73) the first term on the RHS represents the deflection due to flexure and the second term, due to shear deformation. In the elementary beam theory (Euler-Bernoulli beam theory) we neglect the shear deformation, as it is usually much smaller than the flexural deflection. The transverse deflection due to an end couple C is given by: 14 3 8 21 27 2 15 Vtip = 9 22 28 5 3 Utip TO FL3 FL + 3(EI) eq (GA) egleno Vtip 8 16 7 CL2 2(EI) eq 10 23 29 4 Figure 1.31 Plane truss and design domain for Project 1.3 (10) 17 24 30 5 (12) dans vibizale to zuluboM geoda oldswollA A soe Isitial muminiM 18 11 12 25 31 6 14) 19 (1.72) (13) (1.74)