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