If you solve the deflection by Elv-q, what is the correct boundary condition for
ID: 2086008 • Letter: I
Question
If you solve the deflection by Elv-q, what is the correct boundary condition for the cantilever beam (Note: Mult Answers)? 8 RA Question 2 lf you solve the deflection by Elv =-q, what is the correct boundary condition for the beam? the beam is supported by a cable (length is a) at the left end, the axial rigidity of the cable is EA (Note: Multiple Answers) EA d wp-o-FalEA F s the force in the cable Question 3 If you solve the deflection by Elv"-q, what is the correct boundary condition for the beam, spring stiffness is k (Note: Multiple Answers)? x-L)-Re/k, Re is the reaction force at BExplanation / Answer
1. This is a cantilever beam fixed at the right end and free at left. The deflection of the cantilever beam at the fixed end is zero, i.e v(x=0) = 0. This is one of the boundary condition. Also the slope of the cantilever beam at the fixed end is also zero, i.e v'(x=0) =0. This also another boundary condition. So we can totally apply 2 boundary conditions to this problem.
2.This is like a simply supported beam, but it has a cable instead of a roller at the left end. At the right end, it has a pin support. So there will not be any deflection in this. Therefore, v(x=L) = 0 is one of the boundary condition. The left end is a cable and the deflection will be equal to Fa/EA (remember the deflection formula PL/AE). So, v(x=0) = Fa/EA will be the other possible boundary condition.
3. The left end is a pin support, so there will not be any deflection in it. Therefore, v(x=0) = 0 will be one of the boundary condition. At the right end there is a spring and the deflection of the spring is given by RB/K, where RB is the reaction at the right end. So, the deflection v(x=L) = RB/K is the other boundary condition that can be used. The other boundary conditions are no correct to apply for the problem.
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