clase-10-ejemplos de diseño viga columna

8/12/2019 Clase-10-Ejemplos de Diseo Viga Columna

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2/8

H -2

Example H.1a W-shape Subjected to Combined Compression and BendingAbout Both Axes (braced frame).

Given:

Verify if an ASTM A992 W 14 99 has sufficient available strength to support the axial forces and momentslisted below, obtained from a second order analysis that includes P- effects. The unbraced length is 14 ft andthe member has pinned ends. KL x = KL y = Lb = 14.0 ft

LRFD ASD P u = 400 kipsM ux = 250 kip-ftM uy = 80.0 kip-ft

P a = 267 kipsM ax = 167 kip-ftM ay = 53.3 kip-ft

Solution:

Material Properties:ASTM A992 F y = 50 ksi F u = 65 ksi

ManualTable 2-3

Try a W 14 99

Take combined strength parameters from Manual Table 6-1LRFD ASD

p = 30.886

10 kips at 14.0 ft

b x = 31.38

10 kip-ftat 14.0 ft

b y = 32.85

10 kip-ft

p = 31.33

10 kips at 14.0 ft

b x = 32.08

10 kip-ftat 14.0 ft

b y = 34.29

10 kip-ft

ManualTable 6-1

Check limit for Equation H1-1a Check limit for Equation H1-1a

u

c n

P P

=400 kips

1,130 kips = 0.354

Since uc n

P P

> 0.2,

pP u + b xM ux + b yM uy M 1.0

3

3

3

0.866400kips

10 kips

1.38 250kip-ft10 kip-ft

2.8580.0kip-ft

10 kip-ft

= 0.346 + 0.345 + 0.228 = 0.927 M 1.0 o.k .

/a

n c

P P

=267 kips751 kips

= 0.356

Since/

a

n c

P P

> 0.2,

pP a + b xM ax + b yM ay M 1.0

3

3

3

1.33267kips

10 kips

2.08 167kip-ft10 kip-ft

4.2953.3kip-ft

10 kip-ft

= 0.356 + 0.347 + 0.229 = 0.931 M 1.0 o.k.

ManualTable 4-1

Section H1.1

Manual Table 6-1 simplifies the calculation of Specification Equations H1-1a and H1-1b. Adirect application of these equations is shown in Example H.2 .


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H -3

Example H.1b W-shape Column Subjected to Combined Compression andBending Moment About Both Axes (braced frame)

Verify if an ASTM A992 W 14 99 has sufficient available strength to support the axial forces and momentslisted below, obtained from a second order analysis that includes second-order effects. The unbraced length is14 ft and the member has pinned ends. KL x = KL y = Lb = 14.0 ft

LRFD ASD P u = 400 kipsM ux = 250 kip-ftM uy = 80.0 kip-ft

P a = 267 kipsM ax = 167 kip-ftM ay = 53.3 kip-ft

Solution:


ManualTable 2-3

Take the available axial and flexural strengths from the Manual Tables

LRFD ASDat KLy = 14.0 ft,

P c = c P n = 1130 kips

at Lb = 14.0 ft,M cx = M nx = 642 kip-ft

M cy = M ny = 311 kip-ft

u

c n

P P =

400 kips1,130 kips = 0.354

Since uc n

P P

> 0.2, use Eqn. H1.1a

8 + +

9ryrxr

c cx cy

M M P P M M

M 1.0

400 kips 8 250 kip-ft 80.0 kip-ft+ +

1130 kips 9 642 kip-ft 311 kip-ft

=

80.354 + 0.389 + 0.257

9 = 0.929 < 1.0

o.k.

at KLy = 14.0 ft, P c = P n/ c= 751 kips

at Lb = 14.0 ft,M cx = M nx / = 428 kip-ft

M cy = M ny / = 207 kip-ft

/a

n c

P P =

267 kips751 kips = 0.356

Since/

a

n c

P P

> 0.2, use Eqn. H1.1a

8 + +

9ryrxr

c cx cy

M M P P M M

M 1.0

267 kips 8 167 kip-ft 53.3 kip-ft+ +

751 kips 9 428 kip-ft 207 kip-ft

=

80.356 + 0.390 + 0.257

9 = 0.931 < 1.0

o.k.

ManualTable 4-1

ManualTable 3-10

ManualTable 3-2

Eq. H1.1a


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H -5

Calculate the combined stress ratio

LRFD ASD

+ +a bw bz a bw bz

f f f F F F

M 1.0

13.7 ksi 19.1 ksi 17.4 ksi+ +

38.8 ksi 49.3 ksi 67.6 ksi= 0.998 < 1.0

o.k.

+ +a bw bz a bw bz

f f f F F F

M 1.0

9.18 ksi 12.8 ksi 11.6 ksi+ +

25.8 ksi 32.9 ksi 45.0 ksi= 1.00 o.k.

Eqn. H2-1

A comparison of these results with those from Example H.1 shows that Equation H1-1a will produce less conservative results than Equation H2-1 when its use is permitted.

Note: this check is made at a point. The designer must therefore select which point along thelength is critical, or check multiple points if the critical point can not be readily determined.


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H -9

Example H.4 W-Shape Subject to Combined Axial Compression andFlexure

Given: Select an ASTM A992 W -shape with a 10 in. nominal depth to carry nominal axial compressionforces of 5 kips from dead load and 15 kips from live load. The unbraced length is 14 ft and the ends are

pinned. The member also has the following nominal required moment strengths, not including second-ordereffects:

M xD = 15 kip-ft M xL = 45 kip-ftM yD = 2 kip-ft M yL = 6 kip-ft

The member is not subject to sidesway.

Solution:


Calculate the required strength, not considering second-order effects

LRFD ASD P u = 1.2(5.00 kips) + 1.6(15.0 kips)

= 30.0 kipsM ux = 1.2(15.0 kip-ft) + 1.6(45.0 kip-ft)

= 90.0 kip-ftM uy = 1.2(2.00 kip-ft) + 1.6(6.00 kip-ft)

= 12.0 kip-ft

P a = 5.00 kips + 15.0 kips= 20.0 kips

M ax = 15.0 kip-ft + 45.0 kip-ft= 60.0 kip-ft

M ay = 2.00 kip-ft + 6.00 kip-ft= 8.00 kip-ft

Try a W 10 33

Geometric Properties:W 10 33 A = 9.71 in. 2 S x= 35.0 in.

3 Z x = 38.8 in.3 I x = 171 in.

4

S y= 9.20 in.3 Z y = 14.0 in.

3 I y = 36.6 in.4

L p = 6.85 ft Lr = 12.8 ft

ManualTable 1-1Table 3-1

Calculate the available axial strength

For a pinned-pinned condition, K = 1.0.

Since KL x = KL y = 14.0 ft and r x > r y, the y-y axis will govern.

CommentaryTableC-C2.2

LRFD ASD Manual

P c = c P n = 253 kips P c = P n/ c = 168 kips Table 4-1


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H -10

Calculate the required flexural strengths including second order amplification

Use Amplified First-Order Elastic Analysis procedure from Section C2.1b. Since themember is not subject to sidesway, only P - amplifiers need to be added.

111 /

m

r e

C B

P P

C m = 1.0

X-X axis flexural magnifier

2 42

1 2 2

1

29,000ksi 171in.

1.0 14.0 ft 12 in./ft x

e

x

EI P

K L = 1730 kips

Eqn. C2-2

Eqn. C2-5

LRFD ASD

= 1.0

11.0

1 1.0 30.0kips /1730 kips B = 1.02

M ux = 1.02(90.0 kip-ft) = 91.8 kip-ft

= 1.6

11.0

1 1.6 20.0kips/1730kips B = 1.02

M ax = 1.02(60.0 kip-ft) = 61.2 kip-ft

Eqn. C2-2

Y-Y axis flexural magnifier

2 42

1 2 2

1

29,000 ksi 36.6in.

1.0 14.0 ft 12 in./ft

ye

y

EI P

K L = 371 kips Eqn. C2-5

LRFD ASD

= 1.0

11.0

1 1.0 30.0kips/ 371kips B = 1.09

M uy = 1.09(12.0 kip-ft) = 13.1 kip-ft

= 1.6

11.0

1 1.6 20.0kips/ 371kips B = 1.09

M ay = 1.09 (8.00 kip-ft) = 8.76 kip-ft

Eqn. C2-2

Calculate the nominal bending strength about the x-x axis

Yielding limit state

M nx = M p = F y Z x = 50 ksi (38.8 in. 3) = 1940 kip-in or 162 kip-ft

Lateral-torsional buckling limit state

Since L p < Lb < Lr , Equation F2-2 applies

From Manual Table 3-1, C b = 1.14

Eqn. F2-1

ManualTable 3-1


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H -11

M nx = -

- -0.7 -

b pb p p y x

r p

L LC M M F S

L L

M M p

M nx = 3 14.0ft - 6.85ft1.14 1940kip-in. - 1940kip-in. - 0.7 50ksi 35.0 in. 21.8 ft - 6.85ft

= 1820 kip-in.M

1940 kip-in., therefore use:M nx = 1820 kip-in. or 152 kip-ft controls

Local buckling limit state

Per Manual Table 1-1, the member is compact for F y = 50 ksi, so the local buckling limit statedoes not apply

Calculate the nominal bending strength about the y-y axis

Since a W 10 33 has compact flanges, only the yielding limit state applies.

M ny = M p = F y Z y M 1.6 F yS y = 50 ksi (14.0 in. 3) M 1.6 (50 ksi )(9.20 in. 3) = 700 kip-in < 736 kip-in., therefore

Use M ny = 700 kip-in. or 58.3 kip-ft

Eqn. F2-2

ManualTable 1-1

Section F6.2

Eqn. F6-1

LRFD ASD

b = 0.90M cx = bM nx = 0.90(152 kip-ft) = 137 kip-ftM cy = bM ny = 0.90(58.3 kip-ft) = 52.5 kip-ft

b = 1.67M cx = M nx/ b = 152 kip-ft/1.67 = 91.0 kip-ftM cy = M ny/ b = 58.3 kip-ft/1.67 = 34.9 kip-ft

Section F2

Check limit for Equation H1-1a

LRFD ASD

ur

c c n

P P P P

=30.0 kips253 kips

= 0.119, therefore,

use Specification Equation H1.1b

+ +2

ryrxr

c cx cy

M M P P M M

M 1.0

30.0 kips 91.8 kip-ft 13.1 kip-ft + +

2(253 kips) 137 kip-ft 52.5 kip-ft

0.0593 + 0.920 = 0.979 M 1.0 o.k .

/ar

c n c

P P P P

=20.0 kips168 kips

= 0.119, therefore

use Specification Equation H1.1b

+ +2

ryrxr

c cx cy

M M P P M M

M 1.0

20.0 kips 61.2 kip-ft 8.76 kip-ft + +

2(168 kips) 91.0 kip-ft 34.9 kip-ft

0.0595 + 0.924 = 0.983 M 1.0 o.k.

Section H1.1

Eqn. H1.1b

clase-10-ejemplos de diseño viga columna

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