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Estructuras de Acero :

MIEMBROS A TENSIN

EXAMPLE

Mg Ing CARMEN CHILN MUOZ

PIURA-PER

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.

(inches)

Example (T1):

Evaluation of Net Area

For a group of staggered holes along the tension direction, one

must determine the line that produces smallest Net Area.

Paths of failure

on net section

Effect of staggered holes on Net Area :

EFFECT OF STAGGERED HOLES ON NET AREA :-

T

T

B

A

T

T

s

g

A

C

p

p

p

B

In the above diagram:

p = Pitch or spacing along bolt line

s = Stagger Between two adjacent bolt lines

(usually s = P/2)

g = gage distance transverse to the loading.

In case (a) above : An = (Gross width hole dia.) . t

In case (b) above : An = (Gross width hole dia.+ 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 3.4.2Example 3.4.1

Example (T2):

Effect of staggered holes on Net Area :

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).

1)Path AD (two holes) :

2)Path ABD (three holes; two staggers) :

3)Path ABC (three holes; two staggers) :

Net Area for Angles :

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

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*.

USUAL GUGES* FOR ANGLES, INCHES

(FROM 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 plane

*legs and thickness in mm.

Example (T3):

Example on Net Area:

Solutions. For net area calculation the angle may be

visualized as being

flattened into a plate as shown in Figure above.

where D is the width to be deducted for the hole.

1) Path AC:

2) Path ABC:

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.

EFFECTIVE NET AREA:

When not all elements are connected.

i) Transverse Weld Connection:-

Ae = UA

U = 1.0

A = Area of connected part only

e.g. A = 6 x 1/2 = 3 in2

For Welded Connections:-

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

Gusset

plate

Angle

6x4x1/2

T

Gusset

plate

Angle

6x4x1/2

T

L

Weld

w

In bolted connections, the reduction factor (U) is a function

of the eccentricity ( ) in the connection.

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)

For Bolted Connections:-

x

Examples for x & L of connections:

LFRD Specification for Structural Steel Buildings, December 27, 1999

American Institute of Steel Construction

x

Determination of for U.

(Commentary P16.1 177 AISC)

For bolted or riveted connections the following values

for (U) may be used:-

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).

Alternate Values for (U) :-

Example on Effective Net Areas:

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 / line

U = 0.90

Ag = 8.24 m2

An = gross area hole area

= 8.24 (2 x 1.0 hole) x web tk 0.285

= 7.68 in2

Ae = UAn = 0.9 x 7.68 = 6.912 in2

hole dia = 7/8

C 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

dia bolt

x

= 0.888

L = 6 in (3+3)

x

U = 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 = UAn = 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 connected

U = 10

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 = UAn = 7.39 in2

Holes

in flage

flage tk

hole

Holes

in web

web tk.

Design of Tension Members:

The general philosophy of LRFD method:

For tension members:

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 of strength calculation (capacity)

Example (T-5):-

Find the maximum tensile capacity of a member consisting of 2Ls (6 x 4 x ) can carry for two cases:

welded connection,

bolted connection

1" dia bolts

Fy = 60 ksi

Fu = 75 ksi.

5

2

2

1

1

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 = 9.5 in.

Solution:-

Straight section : wn = 9.5 2 x = 7.25 in.

= 6.62 in. (Controls)

An = 6.62 x = 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 =

(2.5+20.5)

2

4

1.75

1.75

9

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:-

Find required (Ag) from factored load .

Find required (Ae) from factored load .

Convert (Ae) to (Ag) by assuming connection detail.

From (ii) & (iii) chose largest (Ag) value

Find required (rmin) to satisfy slenderness

Select a section to satisfy (iv) and (v) above.

Detail the connection for the selected member.

Re-check the member again.

Design Procedure For Tension Members:-

Example on Design of Tension Members:

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.

Solution:-

Pu = 1.2 D + 1.6L = 1.2(18) + 1.6(52) = 104.8 kips

Because Ae = An for this member, the gross area corresponding to

the required net area is

Try t = 1 in.

Ag = 2.409 + 1(1) = 3.409 in.2

Solution:- (Cont.)

Because 3.409 > 3.235, the required gross area is 3.409 in.2, and

Round to the nearest 1/8 inch and try a 1 3 cross section.

Check the slenderness ratio:

Answer:

Use a 3 1 bar.

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 LLTu = 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:

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).

For single line 7/8 bolts ; Ag = An + (1)t = 2.16 + t = (Ag)2

Step 4) Find required rmin.

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

Ag = 2.67 in2 > 2.53 in2 OK

rmin = 0.727 in > 0.6 OK

Step 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.

Tu = 490 k; per channel, Tu = 245 k

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 in2

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):-

Required. rmin = (as a buildup section)

Try MC 10 x 25 ; Ag = 7.35 in2 ; tw = 0.38 and tf = 0.575, rx = 3.87 in.

Check capacity

An = 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 encontr que ocurra en las llamadas vigas copadas y es ahora aparente que este estado lmite tambin controla en algunos casos el comportamiento en los extremos conectados de los miembros en traccin.

32

En las conexiones de extremos, la senda de la menor resistencia

no siempre ser controlada por An o Ae, ms bien existe una senda

de falla que envuelve dos planos, Traccin en uno y Corte en el otro

plano perpendicular, puede ser ms crtica como se muestra en las figuras

La falla que involucra traccin en un plano y corte

simultneo 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 traccin (Fu)

Rn = 0.75 { 0.60 Fy. Agv + Fu . Ant}

2.- Fractura de corte (0.6Fu) + fluencia de traccin (Fy).

Rn = 0.75 { 0.60 Fu. Anv + Fy . Agt}

Nota: Como el estado lmite es la FRACTURA, la ecuacin gobernante

ser la que contenga el MAYOR TRMINO DE FRACTURA

Para determinar el Bloque de Corte de la Conexin en la Fig. 3.11:

Avg= rea total en corte = b.t

Ans = rea neta en corte = t [b (Naguj -1/2) (d + h)]

Atg = rea total en traccin = s.t

Ant= rea neta en traccin = 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= Dimetro del conector

t= Espesor

EJEMPLO

Determinar la Resistencia de Diseo del Bloque de Corte. Compare con la Resistencia de Diseo del perfil.

Ver Tablas de Propiedades de Perfiles Soldados.

Perfil Soldado CS300x74; Acero Fy = 2.53 t/cm2; Fu = 4.08 t/cm2

A = 94.5 cm2; tf = 0.95 cm; Pernos = 3/4" (1.90 cm); Huelgo = 1/16"

SOLUCION

-Bloque de Corte:

Fractura de Traccin + Fluencia de Corte:

t Pbc = 4*0.75*0.95 [ { 7.5 - 1/2*(1.90 + .16) }*4.08 + 27.5*0.6*2.53 ]

t Pbc = 194.2 t

Fractura de Corte + Fluencia de Traccin:

t Pbc = 4*0.75*0.95 [ { 27.5 - 3.5*(1.90 + .16) }*0.6*4.08 + 7.5*2.53 ]

t Pbc = 195.6 t ...... controla

Fluencia en la seccin del perfil:

t Pnf = 0.9*2.53*94.5 = 215.2 t

Fractura en el rea efectiva:

t Pnr = 0.75*0.9*[94.5 - 4*(1.90 + 0.32)*0.95]*4.08 = 237.0 t

Controla el bloque de corte :

t Pbc = 195.6 t

Ejemplo:

Determine la Resistencia de Diseo de la cartela sobre la cual se encuentra

soldado el ngulo de la Figura.

Acero A36. Fy = 36 ksi, Fu = 58 ksi

39

SOLUCION

Fluencia en la seccin total de los Ls:

t Pnf = 0.9*4.22 in2 * 36 = 137 kips

Fractura en los Ls:

U = 1 - 0.888/5 = 0.82

t Pnr = 0.75*0.82*4.22*58 = 150 kips

Bloque de corte en la cartela:

Fractura de Traccin + Fluencia de corte:

t Pbc = 0.75*(3/8) [58*3 + 0.6*36 (5 + 2)]

t Pbc = 91.5 kips

Fractura de corte + Fluencia de Traccin:

t Pbc = 0.75*(3/8) [0.6*58*7 + 3*36]

t Pbc = 98.9 kips

Controla: Bloque de corte : t Pbc = 98.9 kips

Mejorara si se aumentara el contacto entre el ngulo y la cartela

4

3

4

4

1

Plate

(

)

in.

sq.

78

.

0

25

.

0

8

7

4

=

-

=

.

.

50

.

2

25

.

0

16

1

16

15

2

12

in

sq

=

+

-

(

)

(

)

.

.

.

.

)

(

)

.

(

.

.

in

sq

43

2

25

0

4

4

125

2

5

2

4

125

2

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1

16

15

3

12

2

2

=

+

+

+

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(

)

(

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.

.

.

.

)

(

)

.

(

.

.

in

sq

42

2

25

0

4

4

875

1

5

2

4

125

2

16

1

16

15

3

12

2

2

=

+

+

+

-

t

g

g

t

g

t

g

g

b

a

b

a

-

+

=

-

+

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=

2

2

4

1

4

2

1

4

1

2

2

1

2

*

1

=

-

+

=

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+

t

g

g

.

,

16

15

dia

in

t

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s

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.

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.

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sq

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.

.

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.

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5

.

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.

4

(

4

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(

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.

2

(

4

)

3

(

5

.

0

16

1

16

15

3

75

.

4

2

2

in

sq

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+

-

B3.2)

-

(LRFD

9

.

0

1

-

=

L

x

U

i

i

n

Q

R

g

f

S

u

n

t

T

T

f

(

)

8

1

1

+

(

)

4

4

(1.75)

2.5

4

(1.75)

1

3

9.5

w

2

2

8

1

n

+

+

-

=

2

u

u

e

2

y

u

g

in.

2.409

0.75(58)

104.8

0.75F

P

A

Required

in.

3.235

0.90(36)

104.8

0.90F

P

A

Required

=

=

=

=

=

=

t

2.409

t

8

1

8

7

2.409

A

A

A

hole

n

g

+

=

+

+

=

+

=

in.

3.409

1

3.409

t

A

w

g

g

=

=

=

(OK)

300

239

0.2887

5.75(12)

r

L

Maximum

in.

0.2887

3.5

0.2917

A

I

r

obtain

we

,

Ar

I

From

in.

3.5

1(3.5)

A

in.

0.2917

12

3.5(1)

I

min

min

2

2

4

3

min