Seleccion ione e un perfl ST que va a usarse usarse como como miembr miembro o a Ejercicio 4-21 Selecc tensión de 20 pies de longitud que soporte con seguridad las cargas de servicio en tensión: PD= 35lb ! P"= ##5lb $ PS= %5 lb&nieve'( "a cone)ión es a trav*s del pat+n con dos l+neas de tres tornillos de , pulg de di-metro! entre centros ( .se acero / 52 1rado 50 ( Desprecie el bloque de cortante DATOS L := 20ft
φ := 0.9
PD := 35kip ki p
Fy := 50ks i
PL := 115 115ki p
Fu := 65ks i
PS := 65ki p
l := 4in
Φ :=
3 4
Ω := 1.67
n := 2
in
SOLUCION
LRFD
PU1 := 1.4 ⋅ PD
ki p = 49 kip
PU2 := 1.2 ⋅ PD
+
1.6 ⋅ PL
+
PU3 := 1.2 ⋅ PD
+
1.6 ⋅ PS
= 146kip ki p
PU4 := 1.2PD
+
PL
+
PU5 := 1.2PD
+
PL
ki p = 157 157 kip
0.5P S
0.5 ⋅ PS
ki p(mayor valor) = 258.5 kip
ki p = 189.5 kip
PU6 := 0.9PD = 31.5 kip ki p
Agmin
:=
ST6X20.4 5.96in AST62!"4 :=
2
PU2
φ ⋅ Fy
−3 2 = 3.70 .706 × 10 m
1.51in #m :=
tf := 0.659in U := 1 −
#m l
A$gu%&'(s := n ⋅ Agmin
:=
Φ +
PU2
φ ⋅ Fu ⋅ U
+
1 8
in
⋅ tf
A$gu%&'(s
= 5.32 .324 ×
−3
10
2 m
FLUENCIA LRFD
ASD Fy ⋅ A ST10X33
φ ⋅ Fy ⋅ A ST10X33 = 436.05 kip
Ω
= 290.12 kip
(mayor al valor maximo)
FRACTURA U := 1 −
xm l
= 0.373
A := A ST10X33
− ⋅ Φ +
−3
⋅ !" = 5.354 × 10− 3 m2
2
m
ASD
LRFD 0.75 ⋅ A$ ⋅ F%
8
i
(bf es menor que 2/3d por lo ual se usa ! " 0.#$) 2 ⋅ # = 6.667 3
U := 0.85 A $ := A ⋅ U = 4.551 × 10
1
= 343.879kip
A $ ⋅ F% 2
= 229.253kip
(mayor al valor maximo)
Determinar ΦcPn $ PnΩ para cada una de las columnas ! usando la especifcación /S4 $ $ = 50 lbplg2 Ejercicio 5-13
a) %a&os ' Fy := 50
kip
φ & := 0.9
2
Ω& := 1.67
i 'L := 12"!
+
L := 6i
8i
= 152 ⋅ i 3
$ :=
8
i
Soluion ' a se ion unida &omo la forma de un &ubo on para *allar los momen&os de ineria usaremos los si+uien&es da&os que se pueden ver en la ima+en' ( $x! := L = 6 ⋅ i
) $x! := 2 ⋅ $
( i! := L − 2 ⋅ $
) i! := L = 6 i
= 5.25 i
A := ($x! ⋅ ) $x! − ( i! ⋅ ) i! 3
Ix :=
( $x! ⋅ )$x!
* x := 'L * x
12 Ix A
−
(i! ⋅ )i!
+
L = 6.75i
= 9 i2
3
12
3
4
= 59.273 i
= 2.566i
* y := 'L
= 59.229
* y
Iy := Iy A
) $x! ⋅ ($x! 12
−
)i! ⋅ (i! 12
3
= 2.337i
= 65.044
( Tomamos el mayor )
on ,/ry &endremos de la &abla 4-22 del manual S' FASD := 22k+i
,ASD := FASD ⋅ A
F&* Ω& = 198000l("
FLRFD := 33 ⋅ k+i
( φ ⋅ F&* )
,LRFD := FLRFD ⋅ A
4
= 49.148 i
= 297000l("
b) %a&os ' 1 0 X 2#.$ kip Fy := 50
2
φ & := 0.9
Ω& := 1.67
i A := 8.37i
2
= 8.37i2
xm := 1.12i
# := 10i
Ix := 126i
4
Iy := 11.3i
4
'L := 18"! * x := 'L * x
Ix A
= 3.88i
* y := 'L
= 55.671
* y
Iy A
= 1.162i
= 185.899
( Tomamos el mayor )
on ,/ry &endremos de la &abla 4-22 del manual S' FASD := 4.345k+i
,ASD := FASD ⋅ A ,ASDT-TAL
F&* Ω& = 36367.65l("
:= 2 ⋅ ,ASD = 72735.3 l("
FLRFD := 6.537 ⋅ k+i
( φ ⋅ F&* )
,LRFD := FLRFD ⋅ A ,LRFDT-TAL
= 54714.69l("
:= 2 ⋅ ,LRFD = 109429.38 l("
Ejercicio 5-15
a) /2 X 2
Fy := 36kip
2 1 # X 2.4
$ := 0.5i ( / := 12i
2
A 2C := 2 ⋅ 6.28i
A ,L := $ ⋅ ( /
= 6 i
# := 8i Ix2C := 6.58i
4
Iy2C := 61.5i
y m :=
$ 2
#
+
A ,L
L := 8i
2
= 12.56 i
x&m := 1.02i
2
A ,L ⋅
'L := 20"!
2
4
⋅ A 2C ⋅ 2
+ A2C
= 5.495 i
3 2 $ ⋅ (/ L Ix := 2 ⋅ Ix2C + A 2C ⋅ − x&m + = 308.236i4 12 2 3 2 2 $ ⋅ (/ # $ Iy := 2 ⋅ Iy2C + A 2C ⋅ + $ − ym + + (/ ⋅ $ ⋅ ym + = 417.853i4 12 2 2
A := A ,L
* x := 'L * x
Ix A
+ A2C = 18.56 i2
= 4.075 i
* y := 'L
= 58.892
* y
Iy A
= 4.745i
= 50.581
( Tomamos el mayor )
on ,/ry &endremos de la &abla 4-22 del manual S' FASD := 13.8k+i
F&* Ω&
,ASD := FASD ⋅ A
= 256128l("
FLRFD := 20.8 ⋅ k+ i
( φ ⋅ F&* )
,LRFD := FLRFD ⋅ A
= 386048 l("
b) #X3 A 8X31 := 9.12i
Ix8X31 := 110i
Fy := 50kip
$ := 0.5i
2
# := 8i
4
'L := 18"!
4
Iy8X31 := 37.1i (" := 8i
# := 8i
#3 ⋅ $
2 (" + $ −4 4 Ix := Ix8X31 + 2 + #⋅$⋅ = 1.237 × 10 m 12 2 # ⋅ $3 −5 4 Iy := Iy8X31 + 2 ⋅ = 1.551 × 10 m 12
A := A 8X31
* x := 'L * x
Ix A
+
2$ ⋅ #
= 0.106 m
* y := 'L
= 51.845
* y
Iy A
= 0.037 m
= 146.401
( Tomamos el mayor )
on ,/ry &endremos de la &abla 4-22 del manual S' FASD := 7.05k+i
F&* Ω&
,ASD := FASD ⋅ A = 536882.556N
FLRFD := 10.6 ⋅ k+i
( φ ⋅ F&* )
,LRFD := FLRFD ⋅ A = 807227.673 N
6-1 al 6-3 .se el siguiente procedimiento de tanteos ! estime un valor 6"r ! determine los es7uer8os Φccr $ cr Ωc de la tabla 922 del /S4 ! determine el -rea requerida ! seleccione una sección de prueba ! seleccione otra sección en caso de ser necesario Ejercicio 6-1 Seleccione la sección ;#0 mas ligera para soportar las cargas a)iales de compresión PD: #00 lb !P": #%0 lb si 6 "=#5 pies $ se usa acero /<<2 grado 50
DATOS )1! PD := 100kip PL := 160kip *L := 15ft
5rado $0 aero 2 f y := 50ks i SOLUCION LRFD Pu := 1.2 ⋅ PD
+
Suponemos '
1.6 ⋅ PL *L
= 1.673 ×
6
10 m ⋅ kg ⋅ s
−2
"4$
R
φ + ⋅ F+'
de la Tabla 4.22 del S " 3#.# 7si
Pn := 38.8ks i A'&,u&'i-$
:=
Pu Pn
= 6.252 ×
−3
10
2 m
!samos una 0 X 33 on las si+uien&es ara&eris&ias A.1!33 := 9.71in
2
' #.1!33 := 4.19in
'y.1!33 := 14in
verifiamos si la seion resis&e es&a ar+a de ompresion
*L = 42.959 '#.1!33 Pn := 39.3ksi ⋅ A.1!33
de la &abla 4-22 se ob&iene
= 1.697 ×
6
10 m ⋅ kg ⋅ s
φ + ⋅ F+' " 3.3 7si
−2
( omo es&e valor es mayor al u se aep&a la seion)
ASD
p% := ,D
,L
= 1.157 ×
Suponemos '
'L "4$ R
φ & ⋅ F&*
+
6
10 m ⋅ k ⋅ +
−2
de la Tabla 4.22 del S " 2$.# 7si
, := 25.8k+i A *$%$*i# :=
,% ,
= 9.402 ×
−3
10
2
m
!samos una 0 X $4 on las si+uien&es ara&eris&ias A 10X54 := 15.8i
2
* x10X54 := 4.37i
* y10X54 := 2.56i
verifiamos si la seion resis&e es&a ar+a de ompresion
'L = 41.19 * x10X54 , := 26.46k+i ⋅ A 10X54
de la &abla 4-22 se ob&iene
= 1.86 ×
6
10 m ⋅ k ⋅ +
φ & ⋅ F&* " 26.46 7si
−2
( omo es&e valor es mayor al u se aep&a la seion)
Ejercicio 6-3 Seleccione la sección ; mas ligera para soportar las cargas a)iales PD: 5 lb $ P": #25 lb si 6"=#3 pies $ $=3%si DATOS
" y := 36k+i /8 ,D := 75kip
,L := 125kip
'L := 13"!
SOLUCION
LRFD ,% := 1.2 ⋅ ,D
+ 1.6 ⋅ ,L = 1.12 ×
Suponemos '
φ & ⋅ F&*
8
10 i ⋅ l( ⋅ +
−2
'L "$0 R
de la Tabla 4.22 del S " 2#.4 7si
, := 28.4k+i A *$%$*i# :=
,% ,
= 10.211 i2
!samos una # X 3$ on las si+uien&es ara&eris&ias A 8X35 := 10.3i
2
* x8X35 := 3.51i
* y8X35 := 2.03i
verifiamos si la seion resis&e es&a ar+a de ompresion
'L = 44.444 * x8X35
de la &abla 4-22 se ob&iene
, := 29.21k+i ⋅ A8X35
= 1.162 ×
8 −2 10 i ⋅ l( ⋅ (+ omo es&e valor es mayor al u se aep&a la seion)
ASD p % := ,D
+ ,L = 7.722 ×
Suponemos '
φ & ⋅ F&*
7
10 i ⋅ l( ⋅ +
−2
'L "$0 R
de la Tabla 4.22 del S " #.6 7si
, := 18.9k+i
φ & ⋅ F&* "26.2 7si
A *$%$*i# :=
,% ,
= 9.899 ×
−3
10
2
m
!samos una 0 X 4# on las si+uien&es ara&eris&ias A 10X48 := 14.1i
2
* x10X48 := 3.61i
* y10X48 := 2.08i
verifiamos si la seion resis&e es&a ar+a de ompresion
'L = 43.213 * x10X48 , := 19.58k+i ⋅ A10X48
de la &abla 4-22 se ob&iene
= 1.228 ×
6
10 m ⋅ k ⋅ +
φ & ⋅ F&* " .$# 7si
−2
( omo es&e valor es mayor al 3u se aep&a la seion)
Ejercicio 6- 5 >epita el problema 2 usando las tablas disponibles en el ?anual del /S4 ! especialmente las de la parte 9
4on 6"=#3 ! 7$=50 si $ ubicando un perfl en la tabla 9 del manual del acero ! que soporte la carga de obtenemos un perfl ;@35 que satis7ace tanto la combinación por el m*todo ">D 2<0 ips como /SD 200ips ABercicio % Can a disearse varias columnas de edifcio ! usando acero /<<2 $ la especifcación /S4 ! seleccione las secciones ; mas ligeras disponibles $ estable8ca la resistencia de diseo ΦcPn $ la resistencia permisible /SD ! PnΩc! para esas columnas que se describen a continuación:
,D := 170kip
,L := 80kip
,% := 1.2 ⋅ , D
+
L := 16"!
&ipo 8#
1.6 ⋅ ,L
p %ASD := ,D + ,L
' := 1
= 332 ⋅ kip
= 250 ⋅ kip
' ⋅ L = 16 ⋅ "!
3or S%' #X$# 3or 9:% #X4#
( L := 25"!
,D 220kip 100kip L :=
' := 2
&ipo 84
,% := 1.2 ⋅ , D
+
1.6 ⋅ ,L
p %ASD := ,D + ,L
= 472 ⋅ kip
' ⋅ L = 50 ⋅ "!
3or S%' 4X;4 3or 9:% 4X;4
= 320 ⋅ kip
& ,D 120kip L := 100kip
L := 25"!
' := 0.5
&ipo 82 ,% := 1.2 ⋅ ,D + 1.6 ⋅ ,L
p %ASD := ,D
+
,L
= 304 ⋅ kip
' ⋅ L = 12.5 ⋅ "!
or S%' 2X$3 or 9:% 2X$0
= 220 ⋅ kip
# ,D 250kip L := 125kip
L := 18.5"!
,% := 1.2 ⋅ ,D + 1.6 ⋅ ,L
p %ASD := ,D
+
,L
= ⋅ kip
= ⋅ kip
' := 1
'⋅ L=
&ipo 84
⋅ "!
or S%' 4X#2 or 9:% 4X;4