Muestra de informa final de estadistica

...

[pic 170]

[pic 171]

[pic 173][pic 172]

[pic 174]

[pic 175]

[pic 177][pic 176]

NODO A:

[pic 179][pic 180][pic 181][pic 182][pic 183][pic 184][pic 185][pic 186][pic 187][pic 188][pic 189][pic 190][pic 178]

[pic 191]

[pic 193][pic 192]

[pic 195][pic 196][pic 194]

NODO C:

[pic 197][pic 198][pic 199][pic 200]

[pic 202][pic 203][pic 204][pic 205][pic 206][pic 207][pic 208][pic 201]

[pic 209]

[pic 210]

[pic 211]

[pic 213][pic 212]

NODO B:

[pic 214][pic 215][pic 216][pic 217][pic 218][pic 219][pic 220][pic 221][pic 222][pic 223][pic 224][pic 225][pic 226][pic 227][pic 228][pic 229][pic 230][pic 231][pic 232][pic 233][pic 234][pic 235][pic 236][pic 237][pic 238][pic 239][pic 240][pic 241][pic 242]

[pic 243]

[pic 244]

[pic 245]

[pic 246]

[pic 247]

[pic 248]

NODO D:

[pic 250][pic 251][pic 252][pic 253][pic 254][pic 255][pic 256][pic 249]

[pic 258][pic 259][pic 260][pic 261][pic 257]

[pic 262]

Para las fuerzas restantes:

[pic 263][pic 264]

[pic 265][pic 266]

[pic 267][pic 268]

[pic 269][pic 270]

[pic 271][pic 272]

Problema 6:

En la armadura mostrada, determinar el máximo valor de P, sabiendo que las barras pueden soportar una carga máxima de compresión igual a 500N y tracción igual a 800N.[pic 273]

[pic 274]

[pic 275]

[pic 276]

[pic 277]

[pic 278]

[pic 279]

DCL A:[pic 280]

[pic 281]

[pic 282][pic 283][pic 284][pic 285]

[pic 286][pic 287][pic 288][pic 289]

[pic 290][pic 291][pic 292][pic 293]

[pic 294]

[pic 295]

DCL B:[pic 296]

[pic 297]

[pic 298][pic 299][pic 300]

[pic 301][pic 302][pic 303]

[pic 304]

DCLC: [pic 305]

[pic 306]

[pic 307][pic 308][pic 309][pic 310][pic 312][pic 313][pic 314][pic 315][pic 316][pic 311]

[pic 317][pic 318]

[pic 319]

[pic 321][pic 320]

[pic 323][pic 322][pic 324]

[pic 325]

DCLD:

[pic 326]

[pic 327]

[pic 328][pic 330][pic 331][pic 333][pic 338][pic 329][pic 332][pic 334][pic 335][pic 336][pic 337]

[pic 339]

[pic 340][pic 341][pic 342]

[pic 343]

[pic 344][pic 345]

DCLE:[pic 347][pic 348][pic 351][pic 346][pic 349][pic 350][pic 352][pic 353][pic 354][pic 355][pic 356][pic 357][pic 358][pic 359]

DCL G:[pic 361][pic 367][pic 369][pic 360][pic 362][pic 363][pic 364][pic 365][pic 366][pic 368]

[pic 370]

La armadura es simétrica por lo tanto las fuerzas son iguales tanto en el lado izquierdo como en el derecho por tal motivo solo se analizó el lado izquierdo de la ecuación y se determinaron dichas fuerzas.

AB = 2.6P (T) ≤ 800N ⇒ P ≤ 307.69N

AC = 2.4P(C) ≤ 500N ⇒ P ≤ 208.33N

BD = 2.6P (T) ≤ 800N ⇒ P ≤ 307.69N

BC = 0

CD = 3.905P (C) ≤ 500N ⇒ P ≤ 128.04N

CE = 5.4P (C) ≤ 500N ⇒ P ≤ 92.59N

DF = 0.65P (C) ≤ 500N ⇒ P ≤ 769.23N

DE = 1.25P (T) ≤ 800N ⇒ P ≤ 640N

EF = 1.6P(C) ≤ 500N ⇒ P ≤ 312.5N

EG = 4.4P (C) ≤ 500N ⇒ P ≤ 113.63N

FG = 0