The answer has been mistakenly calculated using sin56.3 The correct equation considering ∠JBK should be: ΣFx = 0 +10 + F(BK) + F(BJ) cos56.3 = 0 +10 + F(BK) + 18.02 cos56.3 = 0 F(BK) = -20 kN. In the interest of correctness, you can refer the above mentioned equations. Video editing for the error will be effected before the next course run.
Your observation is correct. The answer has been mistakenly calculated using sin56.3 The correct equation considering ∠JBK should be: ΣFx = 0 +10 + F(BK) + F(BJ) cos56.3 = 0 +10 + F(BK) + 18.02 cos56.3 = 0 F(BK) = -20 kN. In the interest of correctness, you can refer the above mentioned equations. Video editing for the error will be effected before the next course run.
@Rita Singh @AKASH RAUT Your observations are correct. The answer has been mistakenly calculated using sin56.3 The correct equation considering ∠JBK should be: ΣFx = 0 +10 + F(BK) + F(BJ) cos56.3 = 0 +10 + F(BK) + 18.02 cos56.3 = 0 F(BK) = -20 kN. In the interest of correctness, you can refer the above mentioned equations. Video editing for the error will be effected before the next course run.
@@rishabhbiddappa7342 thank you for your reply but i still didn't understand it well.how does the cos component of Fbj acts at a distance 1.5m. can you please help me out ?
A section cutting CD and ED would expose forces F(DC), F(EC), F(FD) and F(EF) in the free body of truss. In case you take moment about C, this would give an equation involving F(FD) and F(EF). Hence, one cannot solve for the force F(FD) at once using this section. Take note that any section selected must cut the truss completely into separate portions. See 4.10
