Estructuras de Acero :
“MIEMBROS A TENSIÓN »
EXAMPLES
Mgº Ingº CARMEN CHILÓN MUÑOZ
PIURA-PERÚ
What is the net area An for the tension member
as shown in the figure?
Solution:
T T
Ag = 4(0.25) = 1.0 sq in.
Width to be deducted for hole
An = [Wg – (width for hole)] (thickness of plate)
Standard Hole for a -in. diam bolt. 434
4
1Plate (inches)
Example (T1):
in. sq. 78.025.087
4
For a group of staggered holes along the tension direction, one must determine the line that produces smallest “Net Area”.
Paths of failureon net section
EFFECT OF STAGGERED HOLES ON NET AREA :-
T T
B
A
T T
s
g
A
Cp
p p
BIn the above diagram:
p = Pitch or spacing along bolt lines = Stagger Between two adjacent bolt lines (usually s = P/2)g = gage distance transverse to the loading.
In case (a) above : An = (Gross width – Σ hole dia.) . t
In case (b) above : An = (Gross width – Σ hole dia.+ Σ s2/4g) . t
Determine the minimum net area of the plate shown in fig. 3.4.2, assuming
in,-diam holes are located as shown:
Figure 3.4.2 Example 3.4.1
Example (T2):
Solution. According to LRFD and ASD-B2, the width used in deducing for
holes in the hole diameter plus 1/16 in., and the staggered length correction
Is (s2/4g).
..50.225.0161
1615
212 insq
1) Path AD (two holes) :
2) Path ABD (three holes; two staggers) :
....
)(
).(
.
.insq432250
44
1252
524
1252
16
1
16
15312
22
....
)(
).(
.
.insq422250
44
8751
524
1252
16
1
16
15312
22
3) Path ABC (three holes; two staggers) :
Angles:
When holes are staggered on two legs of an angle, the gage length (g) for use In the (s2/4g) expression is obtained by using length between the centers of the holes measured along the centerline of the angle thickness, i.e., the distance A-B in Fig: 3.4.3. Thus the gage distance g is
tggt
gt
gg baba 22
Gage dimension for an angle
Every rolled angle has a standard value for the location of holes (i.e. gage distance ga and gb), depending on the
length of the leg and the number of lines of holes. Table shows usual gages for angles as listed in the AISC Manual*.
Determine the net area (An ) for the angle given in figure below
if holes are used?
Angle with legs shown *flattened* into one plane41
421
41
221
2* 1 tgg
*legs and thickness in mm.
Example (T3):
.,16
15diain
Solutions. For net area calculation the angle may be visualized as beingflattened into a plate as shown in Figure above.
t4g
sDtAA
2
gn
where D is the width to be deducted for the hole.
1) Path AC:
2) Path ABC:
.. 75.35.016
1
16
15275.4 insq
.. 96.35.0)25.4(4
)3(
)5.2(4
)3(5.0
16
1
16
15375.4
22
insq
Since the smallest An is 3.75 sq in., that value governs.
An =
An =
9.5"
When some of the cross section (and not all the section) is
connected, we need to use effective net area concept :-
Ae = U An
where, U = Reduction Factor.
When all elements of the section are connected, U = 1.0.
When not all elements are connected.
i) Transverse Weld Connection:-
Ae = UAU = 1.0A = Area of connected part only
e.g. A = 6 x 1/2 = 3 in2
ii) Longitudinal Weld Connection :-
Ae = Ag U
U = 1.0 for L 2 w
U = 0.87 for 2w L 1.5 w
U = 0.75 for 1.5w L w
6”
Gussetplate
Angle6x4x1/2
T
Gussetplate
Angle6x4x1/2
T
L
Weld
w
In bolted connections, the reduction factor (U) is a function
of the eccentricity ( ) in the connection.
B3.2) - (LRFD 9.01 L
xU
Thus:-
Where:= distance between centroids of elements to the plane of load transfer
L = Length of the connection in the direction of load.
(See Commentary C – B 3.1 & C – B 3.2)
x
LFRD Specification for Structural Steel Buildings, December 27, 1999American Institute of Steel Construction
xDetermination of for U.
(Commentary P16.1 – 177 AISC)
For bolted or riveted connections the following values for (U) may be used:-
a) W, M or S Shapes with flange width ≥ 2/3 depth, and structural tees cut from these shapes, provided connection to the flanges and has ≥ 3 fasteners per line in the direction of force, U = 0.90.
b) W,M or S Shapes where flanges width < 2/3 depth, and all other shapes, that has no fewer than 3 fasteners per line, U = 0.85
c) All members having only two fasteners in the line of stress U = 0.75
For short tension members such as Gusset plates the effective net
area equals (An), but must not exceed 0.85 of the gross area (Ag).
Example (T-4)Calculate the Ae values of the following section:-
7/8 bolts W 8 x 28→ flange width (6.54”) > 2/3 x depth (8.0”)→ Three bolts / lineU = 0.90Ag = 8.24 m2
An = gross area – hole area = 8.24 – (2 x 1.0 hole) x web tk 0.285
= 7.68 in2
Ae = U·An = 0.9 x 7.68 = 6.912 in2
hole dia = 7/8C 9 x 15
only 2 bolts / line, U = 0.75
Ag = 4.41 m2
An = 4.41 – (2 x 15/16) 0.285 = 3.875 in2
Ae = 0.75 x 3.875 = 2.907 in2
(i)
(ii)
(iii) x
3 3 ¾ dia bolt
x = 0.888L = 6 in (3+3)
xU = 1 - /L = 1 -0.888/6 = 0.852 < 0.9
Ag = 2.11 in2
An = 2.11 – 1 x (3/4 + 1/8) x 3/8 = 2.11 -0.328 = 1.782 in2
Ae = U·An = 0.852 x 1.782 = 1.518 in2
Alternative value of U = 0.85 (3 bolts / line)
(iv) w 10 x 33
7/8 dia. bolt
All sides connectedU = 1·0
Ag = 9.71 in2
An = 9.71 – 4 x 1.0 x 0.435 – 2 x 1.0 x 0.290
= 9.71 – 1.74 - 0.58 = 7.39 in2
Ae = U·An = 7.39 in2
Holesin flage
flage tk
holeHolesin web web tk.
The general philosophy of LRFD method:
iin QR For tension members: unt TT
where
t = resistance reduction factor for tensile members
Tn = Nominal strength of the tensile members
Tu = Factored load on the tensile members.The design strength tTn is the smaller of:
a) Yielding in the gross section;
t Tn = t Fy Ag = 0.9 Fy Ag
b) Fracture of the net section;
t Tn = t Fu Ae = 0.75 Fu Ae
This is to be followed by check of rupture strength (block shear failure),
and limitation of slenderness ratio ≤ 300.
Example (T-5):-
Find the maximum tensile capacity of a member
consisting of 2Ls (6 x 4 x ½) can carry for two cases:
(a) welded connection,
(b) bolted connection
1" dia bolts
Fy = 60 ksi
Fu = 75 ksi.
½”
5½
2½”
2”
1¾” 1¾”
Net area = gross area (all sides connected)
= 9.50 in2
Yielding Ft = 0.9 Fy Ag = 0.9 x 60 x 9.50 = 513 k
Fracture Ft = 0.75 Fu Ae = 0.75 x 75 x 9.5 = 534 k
Thus tension capacity, t Tn = 513 k (yielding controls)
(a) welded Connection
(b) Bolted Connection
Consider one L
‘An’ Calculation: Wg = gross width = 6 + 4 – ½ = 9.5 in.
Straight section : wn = 9.5 – 2 x = 7.25 in. 811
= 6.62 in. (Controls)
An = 6.62 x ½ = 3.31 in2 for one L
For 2Ls, An = 3.31 x 2 = 6.62 in2
All sides connected, U = 1.0, Ae = U.An = 6.62 in2
Calculation of t Tn :-
(i) Yielding: 0.9 Fy Ag = 0.9 x 60 x 9.50 = 513 k
(ii) Fracture:0.75 Fu Ae = 0.75 x 75 x 6.62 = 372 k.
(thickness)
Zig-Zag = 44
(1.75)
2.54
(1.75)139.5w
22
81
n
(2.5+2–0.5)2½”
4”
1.75” 1.75”
9½”
Design is an interactive procedure (trial & error), as we do not have the final connection detail, so the selection is made, connection is detailed, and the member is checked again.
Proposed Design Procedure:-
i) Find required (Ag) from factored load .
ii) Find required (Ae) from factored load .
iii) Convert (Ae) to (Ag) by assuming connection detail.
iv) From (ii) & (iii) chose largest (Ag) value
v) Find required (rmin) to satisfy slenderness
vi) Select a section to satisfy (iv) and (v) above.
vii) Detail the connection for the selected member.
viii) Re-check the member again.
Example (T-6):-
A tension member with a length of 5 feet 9 inches
must resist a service dead load of 18 kips and a service
live load of 52 kips. Select a member with a rectangular
cross section. Use A36 steel and assume a connection
with one line of 7/8-inch-diameter bolts.
Member length = 5.75 ft.
Pu = 1.2 D + 1.6L = 1.2(18) + 1.6(52) = 104.8 kips
2
u
ue
2
y
ug
in. 2.4090.75(58)
104.8
0.75F
P ARequired
in. 3.2350.90(36)
104.8
0.90F
P ARequired
Because Ae = An for this member, the gross area corresponding tothe required net area is
t2.409t8
1
8
72.409
AAA holeng
Try t = 1 in.
Ag = 2.409 + 1(1) = 3.409 in.2
Because 3.409 > 3.235, the required gross area is 3.409 in.2, and
in. 3.4091
3.409t
Aw g
g
Round to the nearest 1/8 inch and try a 1 3 ½ cross section.Check the slenderness ratio:
Use a 3 ½ 1 bar.
(OK) 3002390.2887
5.75(12)
r
L Maximum
in. 0.28873.5
0.2917
A
Ir
obtain we, Ar I From
in. 3.51(3.5) A
in. 0.291712
3.5(1)I
minmin
2
2
43
min
Select a single angle tension member to carry (40 kips DL) and (20 kips LL), member is (15)ft long and will be connected to any one leg by single line of 7/8” diameter bolts. Use A-36 steel.
Solution:
Step 1) Find Required (Tu):-
Tu = 1.2 DL + 1.6 LL Tu = 1.4 DL
= 1.2 x 40 + 1.6 x 20 or = 1.4 x 40
= 48 + 32 = 80k = 56k
Tu = 80k (Controls)
Example (T-7):-
Step 2) Find required Ag & Ae:
1g2
y
ureq.g )(A in 2.47
360.9
80
0.9F
T)(A
2ureq.e in 1.84
580.75
80
0.75Fu
T)(A
Step 3) Convert (Ae) to (Ag):
Since connection to single leg, then use alternative
(U) value = 0.85 (more then 3 bolt in a line).
2in 16.285.0
84.1)( U
AA en
For single line 7/8” bolts ; Ag = An + (1)t = 2.16 + t = (Ag)2
Step 4) Find required rmin.
in. 0 .6300
1215
300
Lrm in
Step 5) Select angle:
By selecting (t) we get Ag & rmin
t (Ag)1 (Ag)2
1/4 2.47 2.41
3/8 2.47 2.53
1/2 2.47 2.66
select t = 3/8” (Ag)2 = 2.53 in2
(Controls)
Selection
83
213L4
Ag = 2.67 in2 > 2.53 in2 OK
rmin = 0.727 in > 0.6 OKStep 6) Design the bolted connection: (chap. 4).Step 7) Re-check the section.
Select a pair of MC as shown to carry a factored ultimate load of 490 kips in tension. Assume connection as shown. Steel Fy = 50 ksi, Fu = 65 ksi (A572, grade 50) length = 30 ft.
1. Tu = 490 k; per channel, Tu = 245 k
2. Required, (Ag)1 = 245 / 0.9 x 50 = 5.44 in2
Required, (Ae) = 245 / 0.75 x 65 = 5.03 in2
Required, (An) = = 5.03 in2UAe
3. Assume that flange thickness ~ 0.5 in and web tk. ~ 0.3 in. (experience !)
An = (Ag)2 – 2 x 1.0 x 0.5 – 2 x 1.0 x 0.3
= (Ag)2 – 1.60
(Ag)2 = An + 1.60 = 5.03 + 1.60 = 6.63 in.(controls)
10” 2MC
7/8” bolt U = 1.0 (Well connected)
Example (T-8):-
4. Required. rmin = (as a buildup section)
5. Try MC 10 x 25 ; Ag = 7.35 in2 ; tw = 0.38 and tf = 0.575, rx = 3.87 in.
6. Check capacityAn = 7.35 – 2 x 1.0 x 0.575 – 2 x 1.0 x 0.38
= 7.35 – 1.910 = 5.44 in2. Ae = 5.44 in2.
(i) Yielding Tn = 0.9 x 50 x (2 x 7.35) = 661.5 k
(ii) Fracture Tn = 0.75 x 65 x (2 x 5.44) = 530.4 k
Pn = 530.4 k > 490 k. OK
Use 2 MC 10 x 25
Bloque de Corte
Bloque de Corte Este tipo de falla se encontró que ocurría en las llamadas vigas copadas y es ahora aparente que este estado límite también controla en algunos casos el comportamiento en los extremos conectados de los miembros en tracción.
En las conexiones de extremos, la senda de la menor resistencia no siempre será controlada por An o Ae, más bien existe una senda de falla que envuelve dos planos, Tracción en uno y Corte en el otro plano perpendicular, puede ser más crítica como se muestra en las figuras
La falla que involucra tracción en un plano y corte simultáneo en otro perpendicular se llama bloque de corte.
Una vez que ocurre una fractura en un plano, la fuerza entera se transfiere al otro plano para completarse la falla.
Resistencia por Bloque de Cortantes 1.- Fluencia de corte (0.6Fy) + Fractura de tracción (Fu) ø Rn = 0.75 { 0.60 Fy. Agv + Fu . Ant} 2.- Fractura de corte (0.6Fu) + fluencia de tracción (Fy). ø Rn = 0.75 { 0.60 Fu. Anv + Fy . Agt}
Nota: Como el estado límite es la FRACTURA, la ecuación gobernante será la que contenga el MAYOR TÉRMINO DE FRACTURA
Para determinar el Bloque de Corte de la Conexión en la Fig. 3.11:
Avg = área total en corte = b.t Ans = área neta en corte = t [b – (Nºaguj -1/2) (d + h)] Atg = área total en tracción = s.t Ant = área neta en tracción = t [s - 1/2 (d + h)] t = 0.75, factor de resistencia. h = Huelgo = 1/16" (0.16 cm), en vez de 0.32 cm d = Diámetro del conector t = Espesor
EJEMPLO Determinar la Resistencia de Diseño del Bloque de Corte. Compare con la Resistencia de Diseño del perfil. Ver Tablas de Propiedades de Perfiles Soldados.Perfil Soldado CS300x74; Acero Fy = 2.53 t/cm2; Fu = 4.08 t/cm2A = 94.5 cm2; tf = 0.95 cm; Pernos = 3/4" (1.90 cm); Huelgo = 1/16"
SOLUCION - Bloque de Corte: Fractura de Tracción + Fluencia de Corte: øt Pbc = 4*0.75*0.95 [ { 7.5 - 1/2*(1.90 + .16) }*4.08 + 27.5*0.6*2.53 ] øt Pbc = 194.2 t Fractura de Corte + Fluencia de Tracción: øt Pbc = 4*0.75*0.95 [ { 27.5 - 3.5*(1.90 + .16) }*0.6*4.08 + 7.5*2.53 ] øt Pbc = 195.6 t ...... controla Fluencia en la sección del perfil: øt Pnf = 0.9*2.53*94.5 = 215.2 t Fractura en el área efectiva: øt Pnr = 0.75*0.9*[94.5 - 4*(1.90 + 0.32)*0.95]*4.08 = 237.0 t Controla el bloque de corte : øt Pbc = 195.6 t
Ejemplo:Determine la Resistencia de Diseño de la cartela sobre la cual se encuentrasoldado el ángulo de la Figura. Acero A36. Fy = 36 ksi, Fu = 58 ksi
SOLUCION Fluencia en la sección total de los Ls: øt Pnf = 0.9*4.22 in2 * 36 = 137 kips Fractura en los Ls: U = 1 - 0.888/5 = 0.82 øt Pnr = 0.75*0.82*4.22*58 = 150 kips Bloque de corte en la cartela:Fractura de Tracción + Fluencia de corte: øt Pbc = 0.75*(3/8) [58*3 + 0.6*36 (5 + 2)] øt Pbc = 91.5 kips Fractura de corte + Fluencia de Tracción: øt Pbc = 0.75*(3/8) [0.6*58*7 + 3*36] øt Pbc = 98.9 kips Controla: Bloque de corte : øt Pbc = 98.9 kips
Mejoraría si se aumentara el contacto entre el ángulo y la cartela
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