Question

Determine the forces acting in all the members of the truss shown in Fig. 6–9a

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3 kN 3 kN 15 kN 15 kN Fc 45 1.5 kN 0 15 kN FDB FDA 4.10 kN Fig. 6-9

I need help understanding the bold parts below of the solution. Please explain the methods/math involved with steps:

Joint C

.Fx=0;

- FCD cos 30 + FCB sin 45=0

Fy=0;

1.5 kN + FCD sin 30 - FCB cos 45=0

These two equations must be solved simultaneously for each of the two unknowns. Note, however, that a direct solution for one of the unknown forces may be obtained by applying a force summation along an axis that is perpendicular to the direction of the other unknown force.

Fy'=0;

1.5 cos 30kN - FCB sin15=0

FCB= 5.019 kN = 5.02 kN (C)

Then,

Fx'=0;

-FCD+5.019 cos 15-1.5 sin 30=0;

FCD= 4.10 kN (T)

Joint D

Fx=0;

-FDA cos 30 + 4.10 cos 30 kN=0

FDA=4.10 kN (T)

Fy=0;

FDB - 2(4.10 sin 30 kN)=0

FDB= 4.10 kN (T)

Answer 1

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