clase-07-ejemplos de diseño - corte

8/12/2019 Clase-07-Ejemplos de Diseo - Corte

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G-13

Example G.8a Built-up Plate Girder with Transverse Stiffeners

Given:

A built up ASTM A36 I-shaped girder spanning 56 ft. has a uniformly distributed dead load of

0.92 klf and a live load of 2.74 klf in the strong direction. The girder is 36 in. tall with 12 in.

1 in. flanges and a c in. web. Determine if the member has sufficient available shearstrength to support the end shear, without and with tension field action. Use transverse

stiffeners, as required.

User note: This built-up girder was purposely selected with a thin web in order to illustrate the

design of transverse stiffeners. A more conventionally proportioned plate girder would have atleast a in. web and slightly smaller flanges.

Solution:

Material Properties:

Built-up girder ASTM A36 Fy= 36 ksi Fu = 58 ksi Manual

Table 2-3

Geometric Properties:Built-up girder tw= 0.313 in. d= 36.0 in. bft=bfc= 12.0 in.

tf= 1.50 in. h = 33.0 in.

Calculate the required shear strength at the support

LRFD ASD

Ru =[1.2(0.92 klf) + 1.6(2.74 klf)](28.0 ft)

= 154 kips

Ra =(0.92 klf + 2.74 klf)(28.0 ft)=102 kips

Determine if stiffeners are required

Aw= dtw = (36.0 in.)(0.313 in.) = 11.3 in.2

h/tw= 33.0 in. / 0.313 in. = 105

105 < 260 Therefore, kv= 5

1.37 /vk E Fy = 1.37 5(29,000 ksi) /(36 ksi) = 86.9

105 > 86.9 therefore, use Specification Eqn. G2-5 to calculate Cv

Section G2.1b


2/6

G-14

Cv= 21.51

( / )v

w y

Ek

h t F=

2

1.51(29,000 ksi)(5)

(105) (36 ksi)= 0.547

Calculate Vn

Vn= 0.6FyAwCv= 0.6(36 ksi)(11.3 in.2)(0.547) = 134 kips

Check the available shear strength without stiffeners

Eqn. G2-5

Eqn. G2-1

LRFD ASD

v= 0.90vVn= 0.90(134 kips) = 120 kips120 kips < 154 kips not o.k.

Therefore, stiffeners are required.

v= 1.67Vn/v= 134 kips / 1.67 = 80.0 kips80.0 kips < 102 kips not o.k.

Therefore, stiffeners are required.

Section G1

Manual Tables 3-16a and 3-16b can be used to select stiffener spacings needed to develop the

required stress in the web.

Limits on the Use of Tension Field Action:

Consideration of tension field action is not permitted if any of the following are true:

a) end panels in all members with transverse stiffeners

b) members when a/hexceeds 3.0 or [260/(h/tw)]2

c) 2Aw/(Afc+Aft)> 2.5

d) h/bfc orh/bft> 6.0

Select stiffener spacing for end panel

Tension field action is not permitted for end panels, therefore use Table 3-16a.

Section G3.1

LRFD ASD

Use Vu=vVn to determine the requiredstress in the web by dividing by the web area

v n

w

V

A

= u

w

V

A=

2

154 kips

11.3in.= 13.6 ksi

Use Va= Vn/v to determine the requiredstress in the web by dividing by the web area

n

v w

V

A= a

w

V

A=

2

102 kips

11.3 in.= 9.03 ksi

Use Table 3-16a from the Manual to select the required stiffener ratio a/hbased on the h/tratio

of the girder and the required stress. Interpolate and follow an available stress curve, v n

w

V

A

=

13.6 ksi for LRFD,

n

v w

V

A = 9.03 ksi for ASD, until it intersects the horizontal line for a h/tw

value of 105. Project down from this intersection and take the maximum a/hvalue of 1.80

from the axis across the bottom. Since h= 33.0 in., stiffeners are required at (1.80)(33.0 in.) =59.4 in. maximum. Therefore, use 59.0 in.

Manual

Table 3-16a

Select stiffener spacing for the second panel

Tension field action is allowed, but not required, since the second panel is not an end panel. Section G3.1


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G-15

Calculate the required shear strength at the start of the second panel, 59 in. from end

LRFD ASD

59.0in.154k 1.2(0.92klf)+1.6(2.74klf)12 in./ft

uV

= 127 kips

59.0 in= 102 - (0.92klf + 2.74klf)12 in./ft

a.V k

= 84.0 kips


LRFD ASD

v= 0.90

From previous calculation

vVn= 0.90(134 kips) = 120 kips120 kips < 127 kips not o.k.

Therefore additional stiffeners arerequired.

Use Vu=vVn to determine the requiredstress in the web by dividing by the web area

v n

w

V

A

= u

w

V

A=

2

127 kips

11.3in.= 11.2 ksi

v= 1.67


Vn/v= 134 kips / 1.67 = 80.0 kips80.0 kips < 84.0 kips not o.k.

Therefore additional stiffeners arerequired.

Use Va= Vn/v to determine the requiredstress in the web by dividing by the web area

n

v w

V

A= a

w

V

A=

2

84.0 kips

11.3 in.= 7.43 ksi

Section G1

Use Table 3-16b from the Manual to select the required stiffener a/hratio based on the h/tratio

of the girder and the required stress. Interpolate and follow an available stress curve,

v n

w

V

A

=11.2 ksi for LRFD, n

v w

V

A= 7.43 ksi for ASD, until it intersects the horizontal line for

a h/twvalue of 105. Because the available stress does not intersect the h/twvalue of 105, the

maximum value of 3.0 for a/h may be used. Since h = 33.0 in., an additional stiffener is

required at (3.0)(33.0 in.) = 99.0 in. maximum from the previous one.

Manual

Table 3-16b

Select stiffener spacing for the second panel

Tension field action is allowed, but not required, since the next panel is not an end panel. Section G3.1

Calculate the required shear strength at the start of the third pane, 158 in. from end

LRFD ASD

Vu= 154 kips

- [1.2(0.920 klf) + 1.6(2.74 klf)]158.0 in.

12 in./ft

= 81.7 kips

Va= 102 kips

- (0.920 klf + 2.74 klf)158.0 in.

12 in./ft

= 53.8 kips


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G-16


LRFD ASD

v= 0.90


vVn= 0.90(134 kips) = 120 kips120 kips >81.7 kips o.k.

Therefore additional stiffeners are not

required.

v= 1.67


Vn/v= 134 kips / 1.67 = 80.0 kips80.0 kips >53.8 kips o.k.

Therefore additional stiffeners are not

required.

Section G1

The four Available Shear Stress tables, Manual Tables 3-16a, 3-16b, 3-17a and 3-17b, are

useful because they permit a direct solution for the required stiffener spacing. Alternatively,

you can select a stiffener spacing and check the resulting strength, although this process islikely to be iterative. In the proof below, the stiffener spacings that were selected from the

charts in the example above are used.

Manual

Table 3-16


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6/6

G-18

Since /w

h t > 1.37 / 1.37 5.56 29,000 ksi / 36 ksi 91.7v yk E F ,

Cv= 21.51

( / )

v

w y

Ek

h t F=

2

1.51(29,000 ksi)(5.56)

(105) (36 ksi)= 0.613

Since / wh t > 1.10 /v yk E F =1.10 5.56(29,000ksi) /(36ksi) = 73.6, use Eqn. G3-2

2

10.6

1.15 1 ( / )

v

n y w v

CV F A C

a h

=2

2

1 0.6130.6(36 ksi)(11.3 in. ) 0.613

1.15 1 (3.0)

Vn = 176 kips

Calculate the available shear strength

Eqn. G2-5

Section G3.2

Eqn. G3-2

LRFD ASD

v= 0.90vVn= 0.90(176 kips) =158 kips156 kips > 127 kips o.k.

v= 1.67Vn/v= (176 kips) / 1.67 = 105 kips105 kips > 84.0 kips o.k.

Section G1

clase-07-ejemplos de diseño - corte

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