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2P
2P
4P
A
B
C D E
G
H
F
3m
4m 4m
3m
4m
Calcular el valor máximo de P (en kg) en la armadura que se muestra en la figura, sabiendo
que en las barras de esfuerzo de tensión es de 300kgf/𝑐𝑚2, el esfuerzo de compresión es
de 150kgf/𝑐𝑚2
y la sección transversal de todas las barras son iguales de 10𝑐𝑚2
𝑭𝒙 = 𝟎
4P+ 𝐴𝑋 = 0
𝐴𝑋 = 4𝑃
𝑭𝒚 = 𝟎
𝐴𝑦 + 𝐹𝑌-2P-2P = 0
𝐴𝑦 + 𝐹𝑌 = 4𝑃
𝑴𝑨 = 𝟎
𝐴𝑋 = 4P
𝐴𝑌=0
𝐹𝑌 = 4P
-4P(6)-2P(4)-2P(8)+ 𝐹𝑌(12) = 0
𝐹𝑌 = 48𝑃/12
𝐹𝑌 = 4𝑃
𝐴𝑦 = 0
RESOLUCIÓN DEL PROBLEMA
𝐹𝐴𝐻
𝐹𝐴𝐵
4𝑃
53°
𝐹𝐴𝐻 𝑠𝑒𝑛53°
𝐹𝐴𝐻𝑐𝑜𝑠53°
𝑭𝒙 = 𝟎
-4P+ 𝐹𝐴𝐻𝑠𝑒𝑛53° = 0
𝐹𝐴𝐻 = −4P/sen53°
𝐹𝐴𝐻 = 5.00P 𝑡𝑒𝑛𝑠𝑖ó𝑛
𝑭𝒚 = 𝟎
𝐹𝐴𝐵 + 𝐹𝐴𝐻𝑐𝑜𝑠53° = 0
𝐹𝐴𝐵 = − 5.00𝑃 𝑐𝑜𝑠53° − 0.67𝑃
𝐹𝐴𝐵 = −3.00𝑃 Compresión
𝐹𝐵𝐶
𝐹𝐵𝐻
𝐹𝐴𝐵
𝑭𝒙 = 𝟎
𝐹𝐵𝐻 = 0
𝑭𝒚 = 𝟎
𝐹𝐵𝐶 = -3.00P
𝐹𝐹𝐸
𝐹𝐹𝐺
4P
37°
𝑭𝒙 = 𝟎
- 𝐹𝐹𝐺 − 𝐹𝐹𝐸𝑐𝑜𝑠37° = 0
−𝐹𝐹𝐸𝑐𝑜𝑠37° = 𝐹𝐹𝐺
𝐹𝐹𝐺 = 6.64𝑐𝑜𝑠37°
𝐹𝐹𝐺 = 5.30P
𝑭𝒚 = 𝟎
- 𝐹𝐹𝐸𝑠𝑒𝑛37° + 4𝑃 = 0
- 𝐹𝐹𝐸 = 4𝑃/ 𝑠𝑒𝑛37°
- 𝐹𝐹𝐸 = −6.64𝑃
𝐹𝐺𝐸
𝐹𝐺𝐻 𝐹𝐺𝐹
𝐹𝐺𝐷𝑠𝑒𝑛37°
𝐹𝐺𝐷𝑐𝑜𝑠37°
37°
𝐹𝐺𝐷
𝑭𝒙 = 𝟎
- 𝐹𝐺𝐻 − 𝐹𝐺𝐷𝑐𝑜𝑠37° + 𝐹𝐺𝐹 = 0
𝐹𝐺𝐻 = 3.32𝑐𝑜𝑠37° + 5.30
𝐹𝐺𝐻 = 7.95𝑃
𝑭𝒚 = 𝟎
𝐹𝐺𝐸 + 𝐹𝐺𝐷𝑠𝑒𝑛37 = 0
𝐹𝐺𝐷 = −3.32𝑃
𝐹𝐶𝐻𝑠𝑒𝑛53
37°
𝐹𝐶𝐻𝑠𝑒𝑛53
𝐹𝐶𝐷
𝐹𝐶𝐻
𝐹𝐶𝐵
4P
𝑭𝒙 = 𝟎
4P+𝐹𝐶𝐷 + 𝐹𝐶𝐻𝑠𝑒𝑛53° = 0
𝐹𝐶𝐷 = −𝐹𝐶𝐻𝑠𝑒𝑛53° − 4P
𝐹𝐶𝐷 = −7.99𝑃
𝑭𝒚 = 𝟎
-4𝐹𝐶𝐻𝑐𝑜𝑠53° − 𝐹𝐶𝐵 = 0
-𝐹𝐶𝐻𝑐𝑜𝑠53° = −
3.00𝑃
𝑐𝑜𝑠53°
𝐹𝐶𝐻 = 4.99𝑃
𝐹𝐷𝐻
𝐹𝐷𝐺
𝐹𝐷𝐸
𝐹𝐷𝐶
2𝑃
𝐹𝐷𝐺𝑠𝑒𝑛53°
𝐹𝐷𝐺𝑐𝑜𝑠53°
𝑭𝒚 = 𝟎
-2P− 𝐹𝐷𝐻 + 𝐹𝐷𝐺 cos 53° = 0
3.32cos 53° − 2𝑃 = 𝐹𝐷𝐻
𝐹𝐷𝐻 = 0.001P
53°
53°
𝐹𝐸𝐹
𝐹𝐸𝐺
𝐹𝐸𝐺 𝐹𝐸𝐹𝑠𝑒𝑛53°
𝐹𝐸𝐹𝑐𝑜𝑠53°
2𝑃
𝑭𝒙 = 𝟎
𝐹𝐸𝐷 + 𝐹𝐸𝐹𝑠𝑒𝑛53° = 0
𝐹𝐸𝐷 = −6.64𝑠𝑒𝑛53°
𝐹𝐸𝐷 = −5.30𝑃
𝑭𝒚 = 𝟎
-2P+𝐹𝐸𝐺 − 𝐹𝐸𝐹𝑐𝑜𝑠53° = 0
−𝐹𝐸𝐺= 𝐹𝐸𝐹𝑐𝑜𝑠53° + 2𝑃
𝐹𝐸𝐺 = −6.64𝑃(𝑐𝑜𝑠53°) . + 2𝑃
𝐹𝐸𝐺 = −6.64𝑃(𝑐𝑜𝑠53°) .
𝐹𝐸𝐺 = 2.00𝑃
BARRA DGFUERZA TIPO
AH 5.00P T
AB -3.00P C
BC -3.00P C
EF -6.64P C
FG 5.30P T
DE -5.30P C
EG 2.00P T
DG -3.32P C
GH 7.99P T
CD -7.99P C
DH 0
CH 4.99P T
BH 0
DATOS
𝜎𝑇 ≤ 300𝑘𝑔/𝑐𝑚2
𝜎𝐶 ≤ 150𝑘𝑔/𝑐𝑚2
A=10 𝑐𝑚2
𝐴𝑚𝑎𝑥 = ?
𝜎=
𝑃
𝐴
SOLUCIÓN:
Barra FG
𝜎 = 300𝑘𝑔/𝑐𝑚2
A=10 𝑐𝑚2
Remplazando datos en el
formulaso
300𝑘𝑔/𝑐𝑚2
=
5.30𝑃
10 𝑐𝑚2
𝑃𝑀𝐼𝑁 =566.03kg
Barra DE
DE= 5.30𝑃
FG=5.30P
𝜎𝐶 = 150𝑘𝑔/𝑐𝑚2
A=10 𝑐𝑚2
Remplazando datos en el
formulaso
150𝑘𝑔/𝑐𝑚2
=
5.30𝑃
10 𝑐𝑚2
𝑃𝑀𝐴𝑋 =283.01kg
Barra GH Barra CD
GH=7. 99P CD=7.99P
𝜎 = 300𝑘𝑔/𝑐𝑚2 𝜎𝐶 = 150𝑘𝑔/𝑐𝑚2
A=10 𝑐𝑚2 A=10 𝑐𝑚2
Remplazando datos en el
formulaso
Remplazando datos en el
formulaso
300𝑘𝑔/𝑐𝑚2=
7.99𝑃
10 𝑐𝑚2 150𝑘𝑔/𝑐𝑚2=
7.99𝑃
10 𝑐𝑚2
𝑃𝑀𝐼𝑁 =375.4kg 𝑃𝑀𝐴𝑋 =187.73kg
𝜎=
𝑃
𝐴

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Resistencia de material

  • 1. 2P 2P 4P A B C D E G H F 3m 4m 4m 3m 4m Calcular el valor máximo de P (en kg) en la armadura que se muestra en la figura, sabiendo que en las barras de esfuerzo de tensión es de 300kgf/𝑐𝑚2, el esfuerzo de compresión es de 150kgf/𝑐𝑚2 y la sección transversal de todas las barras son iguales de 10𝑐𝑚2 𝑭𝒙 = 𝟎 4P+ 𝐴𝑋 = 0 𝐴𝑋 = 4𝑃 𝑭𝒚 = 𝟎 𝐴𝑦 + 𝐹𝑌-2P-2P = 0 𝐴𝑦 + 𝐹𝑌 = 4𝑃 𝑴𝑨 = 𝟎 𝐴𝑋 = 4P 𝐴𝑌=0 𝐹𝑌 = 4P -4P(6)-2P(4)-2P(8)+ 𝐹𝑌(12) = 0 𝐹𝑌 = 48𝑃/12 𝐹𝑌 = 4𝑃 𝐴𝑦 = 0
  • 2. RESOLUCIÓN DEL PROBLEMA 𝐹𝐴𝐻 𝐹𝐴𝐵 4𝑃 53° 𝐹𝐴𝐻 𝑠𝑒𝑛53° 𝐹𝐴𝐻𝑐𝑜𝑠53° 𝑭𝒙 = 𝟎 -4P+ 𝐹𝐴𝐻𝑠𝑒𝑛53° = 0 𝐹𝐴𝐻 = −4P/sen53° 𝐹𝐴𝐻 = 5.00P 𝑡𝑒𝑛𝑠𝑖ó𝑛 𝑭𝒚 = 𝟎 𝐹𝐴𝐵 + 𝐹𝐴𝐻𝑐𝑜𝑠53° = 0 𝐹𝐴𝐵 = − 5.00𝑃 𝑐𝑜𝑠53° − 0.67𝑃 𝐹𝐴𝐵 = −3.00𝑃 Compresión
  • 4. 𝐹𝐹𝐸 𝐹𝐹𝐺 4P 37° 𝑭𝒙 = 𝟎 - 𝐹𝐹𝐺 − 𝐹𝐹𝐸𝑐𝑜𝑠37° = 0 −𝐹𝐹𝐸𝑐𝑜𝑠37° = 𝐹𝐹𝐺 𝐹𝐹𝐺 = 6.64𝑐𝑜𝑠37° 𝐹𝐹𝐺 = 5.30P 𝑭𝒚 = 𝟎 - 𝐹𝐹𝐸𝑠𝑒𝑛37° + 4𝑃 = 0 - 𝐹𝐹𝐸 = 4𝑃/ 𝑠𝑒𝑛37° - 𝐹𝐹𝐸 = −6.64𝑃
  • 5. 𝐹𝐺𝐸 𝐹𝐺𝐻 𝐹𝐺𝐹 𝐹𝐺𝐷𝑠𝑒𝑛37° 𝐹𝐺𝐷𝑐𝑜𝑠37° 37° 𝐹𝐺𝐷 𝑭𝒙 = 𝟎 - 𝐹𝐺𝐻 − 𝐹𝐺𝐷𝑐𝑜𝑠37° + 𝐹𝐺𝐹 = 0 𝐹𝐺𝐻 = 3.32𝑐𝑜𝑠37° + 5.30 𝐹𝐺𝐻 = 7.95𝑃 𝑭𝒚 = 𝟎 𝐹𝐺𝐸 + 𝐹𝐺𝐷𝑠𝑒𝑛37 = 0 𝐹𝐺𝐷 = −3.32𝑃
  • 6. 𝐹𝐶𝐻𝑠𝑒𝑛53 37° 𝐹𝐶𝐻𝑠𝑒𝑛53 𝐹𝐶𝐷 𝐹𝐶𝐻 𝐹𝐶𝐵 4P 𝑭𝒙 = 𝟎 4P+𝐹𝐶𝐷 + 𝐹𝐶𝐻𝑠𝑒𝑛53° = 0 𝐹𝐶𝐷 = −𝐹𝐶𝐻𝑠𝑒𝑛53° − 4P 𝐹𝐶𝐷 = −7.99𝑃 𝑭𝒚 = 𝟎 -4𝐹𝐶𝐻𝑐𝑜𝑠53° − 𝐹𝐶𝐵 = 0 -𝐹𝐶𝐻𝑐𝑜𝑠53° = − 3.00𝑃 𝑐𝑜𝑠53° 𝐹𝐶𝐻 = 4.99𝑃
  • 7. 𝐹𝐷𝐻 𝐹𝐷𝐺 𝐹𝐷𝐸 𝐹𝐷𝐶 2𝑃 𝐹𝐷𝐺𝑠𝑒𝑛53° 𝐹𝐷𝐺𝑐𝑜𝑠53° 𝑭𝒚 = 𝟎 -2P− 𝐹𝐷𝐻 + 𝐹𝐷𝐺 cos 53° = 0 3.32cos 53° − 2𝑃 = 𝐹𝐷𝐻 𝐹𝐷𝐻 = 0.001P 53°
  • 8. 53° 𝐹𝐸𝐹 𝐹𝐸𝐺 𝐹𝐸𝐺 𝐹𝐸𝐹𝑠𝑒𝑛53° 𝐹𝐸𝐹𝑐𝑜𝑠53° 2𝑃 𝑭𝒙 = 𝟎 𝐹𝐸𝐷 + 𝐹𝐸𝐹𝑠𝑒𝑛53° = 0 𝐹𝐸𝐷 = −6.64𝑠𝑒𝑛53° 𝐹𝐸𝐷 = −5.30𝑃 𝑭𝒚 = 𝟎 -2P+𝐹𝐸𝐺 − 𝐹𝐸𝐹𝑐𝑜𝑠53° = 0 −𝐹𝐸𝐺= 𝐹𝐸𝐹𝑐𝑜𝑠53° + 2𝑃 𝐹𝐸𝐺 = −6.64𝑃(𝑐𝑜𝑠53°) . + 2𝑃 𝐹𝐸𝐺 = −6.64𝑃(𝑐𝑜𝑠53°) . 𝐹𝐸𝐺 = 2.00𝑃
  • 9. BARRA DGFUERZA TIPO AH 5.00P T AB -3.00P C BC -3.00P C EF -6.64P C FG 5.30P T DE -5.30P C EG 2.00P T DG -3.32P C GH 7.99P T CD -7.99P C DH 0 CH 4.99P T BH 0 DATOS 𝜎𝑇 ≤ 300𝑘𝑔/𝑐𝑚2 𝜎𝐶 ≤ 150𝑘𝑔/𝑐𝑚2 A=10 𝑐𝑚2 𝐴𝑚𝑎𝑥 = ? 𝜎= 𝑃 𝐴 SOLUCIÓN: Barra FG 𝜎 = 300𝑘𝑔/𝑐𝑚2 A=10 𝑐𝑚2 Remplazando datos en el formulaso 300𝑘𝑔/𝑐𝑚2 = 5.30𝑃 10 𝑐𝑚2 𝑃𝑀𝐼𝑁 =566.03kg Barra DE DE= 5.30𝑃 FG=5.30P 𝜎𝐶 = 150𝑘𝑔/𝑐𝑚2 A=10 𝑐𝑚2 Remplazando datos en el formulaso 150𝑘𝑔/𝑐𝑚2 = 5.30𝑃 10 𝑐𝑚2 𝑃𝑀𝐴𝑋 =283.01kg
  • 10. Barra GH Barra CD GH=7. 99P CD=7.99P 𝜎 = 300𝑘𝑔/𝑐𝑚2 𝜎𝐶 = 150𝑘𝑔/𝑐𝑚2 A=10 𝑐𝑚2 A=10 𝑐𝑚2 Remplazando datos en el formulaso Remplazando datos en el formulaso 300𝑘𝑔/𝑐𝑚2= 7.99𝑃 10 𝑐𝑚2 150𝑘𝑔/𝑐𝑚2= 7.99𝑃 10 𝑐𝑚2 𝑃𝑀𝐼𝑁 =375.4kg 𝑃𝑀𝐴𝑋 =187.73kg 𝜎= 𝑃 𝐴