Ejercicio15

Ejercicio 6.15
Determine la fuerza en cada elemento de la armadura Pratt para
puente mostrada en la figura. Para cada uno de los elementosestablecer si se encuentra en tensión o compresión.
∑ MA = 0 = (9i x – 4j) + (18i x – 4j) + (27 i x –4 j) + (36i + Hj)
0 = -36 k – 72 k –108k + 36 Hk
216 = 36 H
H = 6 kips
∑ Fy = 0 = Ay – 4 kips – 4 kips – 4 kips + 6 kips
0 = Ay – 12 + 6
H = 6 kips
tan α =

12 ft = 1.34
9

α =55.13
sen 53.13° = 6
FAB
tan 53.13° = 6
FAC
FAB = 7.5 kips C
FAC = 4.5 kips T
12 ⎞
⎛ 12 ⎞
⎟ − FBC − FBE ⎜ ⎟
⎝ 15 ⎠
⎝ 15 ⎠

∑ Fx = 0 = FAB⎛⎜0 = 7.5 (0.8) – 4 – 0.8 FBE
0 = 6 – 4 – 0.8 FBE

∑ Fx = 0 = FAB⎛⎜

9⎞
⎛9⎞
⎟ + FBE ⎜ ⎟ − FBD
⎝ 15 ⎠
⎝ 15 ⎠

0 = 7.5 (0.6) + 2.5 (0.6) – FBDFBD = 6 kips
∑ Fx = 0 = FFH ⎛⎜

12 ⎞
⎛ 12 ⎞
⎟ − FFG − FEF ⎜ ⎟
⎝ 15 ⎠
⎝ 15 ⎠

0 = 7.5 (0.3) – 4 – FEF (0.8)
F0 = 2 – 0.8 FEF
FEF = 2.5 T
∑ Fx= 0 = FDB − FFH ⎛⎜

9⎞
⎛9⎞
⎟ − FEF ⎜ ⎟
⎝ 15 ⎠
⎝ 15 ⎠

0 = FDE – 7.5 (0.6) – 2.5 (0.6)
0 = FDE – 4.5 – 1.5
FDE = 6 kips C
tan α =

12 =1.34
9

α = 53.13° = 6
FHZ
tan 53.13° =

6
FGH

FFE = 2.5 kips T
FGH = 4.5 kips T
FAC = FCE = 4.5 kips T
FBC = 4 kips T

FFG = 4 kips T
FGE =FGH = 4.5 T
∑ Fy = 0 = FDE + FEF ⎛⎜

12 ⎞
⎛ 12 ⎞
⎟ + FBE ⎜ ⎟ − 4kips
⎝ 15 ⎠
⎝ 15 ⎠

∑ Fy = 0 = FDE + 2.5 (0.8) + 2.5 (0.8) – 4 kips
0 = 2 + 2– 4 + FDE
FDE = 0
FAB = FFH = 7.5 kips C
FAC = FGH = 4.5 kips T
FBC = FFG = 4 kips T
FBE = FEF = 2.5 kips T
FCE = FEG = 4.5 kips T