Determine the forces in all members and supports of the truss by METHOD OF JOINT

Question

Determine the forces in all members and supports of the truss by METHOD OF JOINTS. Also indicate results (compression and tension) using arrows.

I've put the truss in an MCSolids software and printed out the answer to what the components should be... (C) = compression and (T) = tension, if there's anyone who could help me work out these solutions on paper it would be a HUGE help.

If someone can't work out the whole problem, then I would request you show me the work involved solving for Ay. But if you can show me each equation for compression and tension forces, it'd be a life-saver. Thanks.

Explanation / Answer

Let the reaction forces be Ray, Rbx, Rby

Net vertical force is zero.

Rby + Ray = 500 + 600 = 1100 lb ---- (1)

Rbx + 100 = 0

Rbx = -100 lb

Moment about point B is zero

Ray * 25 + 500 * 15 + 600 * 55 - 100 * 20 = 0

Ray = - (500 * 15 + 600 * 55 - 100 * 20)/25 = - 1540 lb

Rby = 1100 - Ray = 1100 - (-1540) = 2640 lb

Consider joint A

Net vertical force is zero

Fab * sin(1) + Ray = 0 [here tan1 = 20/25 = 4/5 so sin1 = 4/sqrt(41)]

Fab = - (-1540) * sqrt(41)/4 = 2465.2 lb

Net horizontal force is zero

Fab * cos1 + Fac = 0

Fac = - 2465.2 * 5/sqrt(41) = -1925 lb

Note: Here positive sign indicates member is in tension and negative sign indicates member is in compression

Consider Joint E

net vertical force is zero

Fed *sin2 - 600 = 0 [ Here Tan2 = 20/25 = 4/5 so sin2 = 4/sqrt(41)]

Fed = 600 *sqrt(41)/4 = 960.4 lb

Net Horizontal force is zero

- Fed * cos2 - Fec + 100 = 0

Fec = 100 - 960.4 * 5/sqrt(41) = -650 lb

Consider Joint D

for vertical = zero :

Fdc * sin3 + Fed * sin2 = 0  [ here Tan3 = 20/15 = 4/3 so sin3 = 4/5]

Fdc = - 750 lb

for horizontal = zero we get,

- Fdc * cos3 + Fec * cos2 - Fdb = 0

This gives

Fdb = 1200 lb

By taking Joint C and vertical force is zero we get

Fbc = 1375 lb

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