solucionario capitulo 9

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CHAPTER 9

x x x

P9.1. Determine the value of S , Z , " of the following shapes, using the dimensions given in the LRFDM.

(a) W36 × 230 (b)* W16 × 36

(c) WT18 × 115 (d) WT8 × 18

Solution

a. Section: W36×230

f f w

b = 16.5 in.; t = 1.26 in. d = 35.9 in.; t = 0.760 in.

1. Elastic section properties

Flange area = 16.5 × 1.26 = 20.8 in.2

w

Web depth, d = (35.9 - 2 × 1.26) = 33.4 in. Web area = 0.760 × 33.4 = 25.4 in.2

Total area, A = 20.8 × 2 + 25.4 = 67.0 in. 2

For the doubly symmetric section considered, the center of gravity, G, coincides with the pont of intersection of the two symmetry axes.

above the bottom fiber of the bottom flange

= 14,900 in. 4

(Ans.)

2. Plastic section properties

A y

To satisfy the condition I F dA = 0 for a homogeneous section, the PNA divides the section into two equal areas. For the doubly symmetric section considered, the PNA therefore coincides with the symmetry axis.

÷ = 18.0 in. above the bottom fiber of the bottom flange

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= 934 in. 3 (Ans.)

Shape factor, (Ans.)

x x x

The corresponding values given in LRFDM Table 1-1 are: I = 15,000in. ; S = 837 in. ; and Z =4 3

943 in. and take into account the contribution of the web-to-flange fillets. 3

b. Section: W16×36

f f w

b = 6.99 in.; t = 0.430 in.; d = 15.9 in.; t = 0.295

1. Elastic section properties

Flange area = 6.99 × 0.430 = 3.01 in.2

w

Web depth, d = (15.9 - 2 × 0.430) = 15.0 in. Web area = 15.0 × 0.295 = 4.43 in.2

Total area, A = 3.01 × 2 + 4.43 = 10.5 in. 2

For the doubly symmetric section considered, the center of gravity, G, coincides with the pont of intersection of the two symmetry axes.

above the bottom fiber of the bottom flange

= 443 in. 4

(Ans.)

2. Plastic section properties

A y

To satisfy the condition I F dA = 0 for a homogeneous section, the PNA divides the section into two equal areas. For the doubly symmetric section considered, the PNA therefore coincides with the symmetry axis.

÷ = 7.95 in. above the bottom fiber of the bottom flange

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= 63.2 in. 3 (Ans.)

Shape factor, (Ans.)

x x x

The corresponding values given in LRFDM Table 1-1 are: I = 448 in. ; S = 56.5 in. ; and Z =4 3

64.0 in. and take into account the contribution of the web-to-flange fillets. 3

x x x

P9.2. Determine the value of S , Z , " of the following built-up sections: (a) A W16 × 36 with one ½ × 12 in. plate welded to each flange. (b) A W16 × 36 with one ½ × 12 in. plate welded to the top of the flange (c) A W16 × 36 with a C12 × 20.7 with its web welded to the top flange.

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yx px

P9.3. Determine the values of M and M for the shapes given in Problem P9.1.

Solution

a. W36×230

y

Yield stress, F = 50 ksi

x

Elastic section modulus, S = 828 in. (from solution to P9.1a.)3

x

Plastic section modulus, Z = 934 in. (from solution to P9.1a.)3

y x y

Yield moment, M = S F = 828 × 50.0 ÷ 12 = 3450 ft-kips

p x y

Plastic moment, M = Z F = 934 × 50.0 ÷ 12 = 3892 ft-kips

b. W16×36

y

Yield stress, F = 50 ksi

x

Elastic section modulus, S = 55.7 in. (from solution to P9.1b.)3

x

Plastic section modulus, Z = 63.2 in. (from solution to P9.1b.)3

y x y

Yield moment, M = S F = 55.7 × 50.0 ÷ 12 = 232 ft-kips

p x y

Plastic moment, M = Z F = 63.2 × 50.0 ÷ 12 = 263 ft-kips

yx px

P9.4. Determine the values of M and M for the built-up shapes given in Problem P9.2. Assume A36 steel for all elements.

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P9.5. a) A W18x35 beam of A992 steel spans 30 ft and is connected to columns at either end by means of

standard web connections. Compute the uniformly distributed factored load that the member can resist. Assume continuous lateral bracing for the compression flange.

b) Find the maximum spacing of lateral supports for the design to still hold good.

Solution

Simply supported beam: Span, L = 30 ft u Loading: Uniformly distributed load, q

Compression flange continuously laterally braced. a.

Section: W21×62

f f w

From LRFDM Table 1-1 6 b / 2t = 7.06; h /t = 53.5 y

Material: A992 steel 6 F = 50 ksi

b px v n p

From LRFDM Table 5-3, for a W18×35, N M = 249 ft-kips; N V = 143 kips; L = 4.31 ft

Limiting b/t ratios for plate buckling:

Compression flange:

Web in flexure:

Web in shear:

Alternatively, these values can be read from Table 9.5.1.

f f pf w pw

As b / 2t < 8 and h /t < 8 , the section is compact. So, the design bending strength,

d b px

M = N M = 249 ft-kips

w pv

As h /t < 8 , the design shear strength is given by

d v n

V = N V = 143 kips

As per LRFD:

Maximum bending moment,

Maximum shear corresponding to this load is,

max u d

V = q L/ 2 = 2.21 × 30 ÷ 2 = 33.2 kips < V = 143 kips O.K.

So, the maximum uniformly distributed factored load the W18×35 of A992 steel can support is 2.21

klf. (Ans.)

b. The beam could be considered adequately braced for

b p

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P9.6. Determine the uniformly distributed factored load that can be carried by a W21×62 beam of A572 Grade 42 steel over a simply supported span of 24 ft with lateral supports at 6 ft.

Solution

Simply supported beam: Span, L = 24 ft b

Unbraced length, L = 6 ft Section, W21×62

y

Material: A572 Gr 42 6 F = 42 ksi

x y

From LRFDM Table 1-1, for a W21×62, Z = 144 in. ; 3 r = 1.77 in.

From Eq. 9.7.4, limiting unbraced length is,

b p

As L < L , the beam is considered adequately laterally braced. From Eq. 9.7.2, the design bending

strength is,

So, the maximum uniformly distributed factored load the W21×62 of A572 Gr 42 can support is 6.3

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P9.7. A W24×176 of A572 Grade 60 steel is used for a simple span of 36 ft. If the only dead load present is the weight o the beam, what is the largest service concentrated live load that can be placed at mid-span? Assume continuous lateral bracing. Deflection need not be checked.

Solution

Simply supported beam: Span, L = 36 ft. Compression flange continuously laterally braced. Section: W24×176

y Material: A572 Gr 60 steel 6 F = 60 ksi Loading:

D

Dead load: Self weight of the beam 6 q = 176 plf = 0.176 klf

L

Live load: A central concentrated load 6 Q Factored loads:

u

Uniformly distributed load 6 q = 1.2× 0.176 = 0.211 klf

u L

Central concentrated load 6 Q = 1.6 Q

Maximum bending moment at the center under factored loads is,

x w

From LRFDM Table 1-1 6 Z = 511 in. ;3 d = 25.2 in.; t = 0.750 in.

f f w

b / 2t = 4.81; h /t = 28.7 Limiting b/t ratios for plate buckling:

Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

As b / 2t < 8 and h /t < 8 , the section is compact. So, the design bending strength is,

d b px b x y

M = N M = N Z F = 0.90 × 511 × 60.0 = 27,594 in.-kips = 2300 ft-kips

w pv

d v n v w y

V = N V = N d t (0.6 F ) = 0.90 × 25.2 × 0.750 × 0.60 × 60.0 = 612 kips

As per LRFD:

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max u u

V = ½ q L + ½ Q = (0.5×0.211×36) + ( 0.5×1.6×157) d

= 129 kips < V = 612 kips O.K.

So, the maximum concentrated service live load the W24×176 of A572 Grade 60 steel can support

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P9.8. A triangular opening in the floor of an industrial building results in the factored loads on a W18×35 simple beam to be as shown in Fig. P9.8. The beam weight is not included. Use A588 Grade 50 steel. Assume full lateral support for the compression flange. Check the adequacy of the beam.

See Fig. P9.8 of the text book. Solution

Simply supported beam, AB. Span, L = 32 ft

Factored loads:

Concentrated load at center, C = 20 kips

Two triangular loads, with maximum intensity (at supports) of 1.5 klf. Total factored load on the beam = 20.0 + 2×½×1.5×16 = 44.0 kips Reaction,

Due to symmetry of the structure and the loading the maximum bending moment occurs at the center, C.

max A

Maximum shear, V = V = 22 kips

Additional bending moment at midspan due to self weight of the beam,

req

Required bending strength, M = 224 + 5.38 = 229 ft-kips From LRFDM Table 5-3 for a W18×35,

As the beam is continuously laterally braced,

Also d

V = 143 kips > 22 kips O.K.

The W18×35 beam is adequate for the given loads. (Ans.)

b

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P9.9. A 34 ft long W27×94 shape of A992 steel is used as a simply supported beam. It is subjected to a factored concentrated load of 80 kips at 12 ft from each support. In addition, it is subjected to a factored moment of 340 ft-kips, one at its left-end (anti-clockwise) and one at its right-end (clockwise). Neglect self weight of the beam in the calculation and check if the beam is safe. Assume full lateral support for the compression flange.

Solution

a. Data

Simply supported beam, AB: Span, L = 34 ft Compression flange continuously laterally braced. Loading:

Factored concentrated load at D and E (with AD = BE = a = 12 ft) = 80 kips Factored end-moment at B = 340 ft-kips (anti-clockwise)

Factored end-moment at E = 340 ft-kips (clockwise) u

Uniformly distributed load, q due to self-weight ( = 1.2 × 0.094 = 0.113 klf ) b. Required strengths

Due to the symmetry of the structure and the loading, the maximum bending moment occurs at the center, C. Consider the influence of self weight separately.

From the applied loads:

A B R = R = 80.0 kips A M = - 340 ft-kips C M = 80.0×17.0 - 80.0×5.0 - 340 = + 620 ft-kips

Due to self weight:

A B R = R = 0.113× 34 ÷ 2 = 1.92 kips C M So: u

Required bending strength, M = 620 + 16.3 = 636 ft-kips u

Required shear strength, V = 80.0 + 1.92 = 81.9 kips c. Design strengths

Section: W27×94

f f w

b px v n p

Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

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d b px

So, the design bending strength is: M = N M = 1040 ft-kips

w pv d v n

Also, as h /t < 8 , the design shear strength is: V = N V = 356 kips

u d u d

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P9.10. Repeat Problem P9.9, with the end moment acting at the left end only.

Solution

a. Data

Simply supported beam, AB: Span, L = 34 ft Compression flange continuously laterally braced. Loading:

Factored concentrated load at D and E (with AD = BE = a = 12 ft) = 80 kips Factored end-moment at B = 340 ft-kips (anti-clockwise)

u

Uniformly distributed load, q due to self-weight ( = 1.2 × 0.094 = 0.113 klf ) b. Required strengths

Consider the influence of self weight separately. From the applied loads:

A B R = 90.0 kips; R = 70.0 kips A M = - 340 ft-kips D M = 90.0 × 12.0 - 340 = + 740 ft-kips E M = 90.0×22.0 - 80.0×10.0 - 340 = + 840 ft-kips B M = 0 ft-kips

So, the maximum bending moment under applied load occurs at E. Due to self weight:

A B

R = R = 0.113× 34 ÷ 2 = 1.92 kips C

M

The maximum bending moment now occurs at the center, C. So:

u

Required bending strength, M = 840 + 16.3 = 856 ft-kips (conservatively) u

Required shear strength, V = 90.0 + 1.92 = 91.9 kips c. Design strengths

Section: W27×94

f f w

b px v n p

Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

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d b px

So, the design bending strength is: M = N M = 1040 ft-kips

w pv d v n

Also, as h /t < 8 , the design shear strength is: V = N V = 356 kips

u d u d

As, M < M and V < V , the W27×94 of A992 steel can still support the given loads.

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P9.11. A wide flange beam AB frames across an open well in a building and is subjected to the factored loads y

shown in Fig. P9.11. Determine if a W16×50 of F = 50 ksi steel is sufficient under these conditions. Assume continuous lateral support to the compression flange.

y

Material: F = 50 ksi steel Factored loads:

Concentrated load at D, 24 ft from A: 10 kips Uniformly distributed load over AD: 1.6 klf

o

Location of point of zero shear ( z from A):

Maximum moment,

Additional moment due to self-weight, u

Required bending strength, M = 298 + 12.0 = 310 ft-kips

max

Maximum shear, V = 30.9 kips

y

From LRFDM Table 5-3, for a W16×50 of F = 50 ksi steel,

As the section is compact and the compression flange continuously laterally supported

y

So, the W16×50 of F = 50 ksi steel is adequate. (Ans.)

P9.12. A simple beam consists of a W18×55 with a PL1×12 cover plate welded to each flange. Determine the factored uniform load the beam can support in addition to its own weight for a 28-ft simple span. Assume A572 Grade 50 steel and full lateral support.

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P9.13. A 36-ft long simple beam consists of a built-up section obtained by welding three A36 steel plates as shown in Fig. P2.6. Determine the factored uniformly distributed load the beam can support, in addition to its self-weight. The beam is continuously laterally supported

Simply supported beam. Span, L = 36 ft y

Material: A36 steel. F = 36 ksi Section: Built-up by welding three plates. a. Elastic section properties

Total area, A = 8.00 × 2.00 + 1.00×16.0 + 12.0×2.0 = 56.0 in. 2 A

To satisfy the condition I f dA = 0, the elastic neutral axis must pass through the center of gravity of the cross section.

= 8.71 in. above the bottom fiber of the bottom flange

= 3,500 in. 4

;

b. Plastic section properties

A y

To satisfy the condition I F dA = 0, for the unequal flanged I-section, the PNA divides the section into two equal areas. As, A /2 = 56/2 = 28.0 in. is greater than the area of the bottom flange, it2

p

follows that the PNA will be in the web. To locate the PNA, find the distance y , as measured from the bottom fiber of the bottom flange, such that the area above the PNA is equal to the area below. (12.0× 2.00) + l.00×( - 2.00) = 1.00× (18.0 - ) + 8.00× 2.00 ÷ = 6.0

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= 408 in. 3 c. Plastic moment

From LRFDS Eq. F1-1:

p y x y x min

Plastic moment, M = min [ F Z ; 1.5 F S ]

= min [ (36.0×408) ÷ 12; (1.5×36×310)÷ 12 ] = min [ 1224 ft-kips; 1395 ft-kips ] = 1224 ft-kips d. Design strengths

Width-to-thickness ratios of the plates: f f

Compression flange: b / 2t = 4.00 / 2.00 = 2.00 c w

Web in flexure: h /t = (2×12.0) / 1.00 = 24.0

Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

As b / 2t < 8 and h /t < 8 , the section is compact. So, the design bending strength,

d b px

M = N M = 0.90 × 1224 = 1100 ft-kips

w pv

d v n

V = N V = 0.9 × 16.0 × 1.0 × 0.60 × 36.0 = 311 kips

As per LRFD:

Maximum shear corresponding to this load is,

max u d

V = q L/ 2 = 6.79 × 36 ÷ 2 = 122 kips < V = 311 kips O.K. Self-weight of the beam = 56 × 3.4 = 190 plf = 0.190 klf

Factored value of the self-weight = 1.2 × 0.190 = 0.228 klf

So, the maximum uniformly distributed factored load the built-up section can support, in addition to

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P9.14. A W16×57 of A992 steel is used for the over hanging beam ABCD (Fig. P9.14) under the service loads (50% dead and 50% live ) shown. Assume continuous lateral support, and check the adequacy of the beam for bending and shear.

Overhanging beam ACBD. Main span AB: L = 20 ft Overhang BD: a = 8 ft Service loads:

Uniformly distributed load over AD: q = 2 klf

D

Concentrated load at free end, D: Q = 16 kips

C

Concentrated load at center of main span, C: Q = 32 kips

u

Factored distributed load, q = 1.2(1.0) + 1.6(1.0) = 2.80 klf

uC

Factored concentrated load at C, Q = 1.2(16.0) + 1.6(16.0) = 44.8 kips

uD

Factored concentrated load at D, Q = 1.2(8.0) + 1.6(8.0) = 22.4 kips

Maximum positive moment, occurs at C.

Maximum negative moment occurs at support B.

So,

BC

Maximum shear, V = 63.8 kips. From LRFDM Table 5-3, for a W16×57,

As the section is compact and continuously braced,

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P9.15. A W14×30 of A992 steel is used as a simply supported, uniformly loaded beam with a span length of 28 ft. The service dead load (not including the self weight) is 0.4 klf and the service live load is 0.78 klf. Assume that the beam is continuously laterally supported by the floor deck. Check the adequacy of the beam for bending and shear. Calculate the maximum dead load and live load deflections of the beam.

Solution

Simply supported beam. Span, L = 28 ft

Self weight of the beam = 30 plf

D

Service dead load, q = 0.40 + 0.03 = 0.43 klf

L

Service live load, q = 0.78 klf

Since the dead load is less than 8 times the live load, load combination LC-2 controls:

For a simply supported beam, under uniformly distributed load, the maximum bending moment occurs at midspan, and is equal to:

Maximum shear, which occurs at support, is equal to

From LRFDM Table 5-3 for a W14×30

As the beam is continuously braced,

We have

O.K.

So, the W14×30 is satisfactory for bending and shear. (Ans.)

For a simply supported beam, under uniformly distributed load, the maximum deflection occurs at midspan (Case 1, Table 5-17 of the LRFDM). The central deflection under service live load is,

(Ans.)

The central deflection under service dead load is

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y

P9.16. A W30×148 of F = 50 ksi steel has been selected for the simply supported beam shown in Fig. P9.16. Loads Q consist of 50 kips dead load and 80 kips live load. Check the adequacy of the beam for bending and shear. Calculate the maximum dead load and live load deflections. Assume continuous lateral support for the compression flange.

See Fig. P9.16 of the text book.

Solution

Simply supported beam AB. Span, L = 26 ft

Concentrated loads, at center C, and at points D and E (CD = CE = 10 ft) Service load, Q: 50 kips dead load, and 80 kips live load

u

Factored load, Q = 1.2 × 50.0 + 1.6 × 80.0 = 188 kips Reaction at A,

D

Bending moment at D, M = 282×3.0 = 846 ft-kips Bending moment at C,

Bending moment at C due to self weight, So, the required bending strength, Required shear strength,

From LRFDM Table 5-3, for a W30×148:

We have

From Beam Diagrams and Formulas (Case 7 and Case 9 of LRFDM Table 5-17), the maximum deflection which occurs at C:

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P9.17. a) A W12×30 of A992 steel is used as a simply supported beam with a span length of 10 ft. It is subjected to two concentrated loads of 80 kips each, located 1 ft from each support. The loads given are service live loads. Neglect self weight of the beam in the calculations. Check the adequacy of the beam for bending and shear, assuming that the beam is continuously laterally supported.

b) Redesign, if necessary (limit the selection to W12s only). Solution

Loading: Two concentrated loads at C and D (AC = BD = a = 1 ft) L

Service live load, Q = 80 kips

u L

Factored loads, Q = 1.6Q = 1.6×80 = 128 kips

From LRFDM Table 5-3, for a W12×30

Shape is compact and laterally supported, therefore

d v n max

V = N V = 86.3 < V = 128 kips N.G.

So, the W12×30 is not satisfactory. (Ans.)

req

Entering LRFDM Table 5-4 with V = 128 kips and limiting the selection to W12 shapes, observe that a W12×72 with

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P9.18. A large steel beam simply spans 60 feet and carries an applied load of 1.5 klf ( D = 0.5 klf and L = 1.0 klf). Select the lightest adequate W-shape of A992 steel. Include the effect of member self weight. Assume that the beam is continuously laterally supported.

Solution

Simply supported beam. Continuously laterally supported. y

Material: A992 steel. 6 F = 50 ksi Span, L = 60 ft

As the span is large, assume a self-weight of 100 plf or 0.100 klf. D Dead load, q = 0.500 + 0.100 = 0.600 klf L Live load, q = 1.00 klf u Factored load, q = 1.2(0.600) + 1.6(1.00) = 2.32 klf Maximum factored moment,

Maximum shear,

req

Entering LRFDM Table 5-3 with M = 1044 ft-kips, we observe that a W30×90 is the lightest section that provides adequate flexural strength. For this shape,

As the weight of the beam (90 plf) is less than the value assumed (100 plf) the design is O.K.

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P9.19. Select the lightest W section to carry a uniform dead load of 0.3 klf and a live load of 0.6 klf on a simply supported span of 34 ft. The beam is continuously laterally supported. Assume no deflection limitations. Use A572 Grade 42 steel.

Solution

Simply supported beam. Continuously laterally supported. y

Material: A572 Grade 42 steel. 6 F = 42 ksi Span, L = 34 ft Assume a self-weight of 40 plf or 0.040 klf. D Dead load, q = 0.300 + 0.040 = 0.340 klf L Live load, q = 0.600 klf u Factored load, q = 1.2(0.340) + 1.6(0.600) = 1.37 klf Maximum factored moment,

Maximum shear,

From Eq. 9.7.18:

req

Entering LRFDM Table 5-3 with Z = 62.9 in. , we observe that a W18×35 (in bold face) is the3 lightest section that provides adequate flexural strength. For this shape, we have:

x w f f w

Z = 66.5 in. ; d = 17.7 in.; t = 0.300 in.; b /2t = 7.06; h /t = 53.5 3

y

As the section is compact for F = 50 ksi steel (no indication to the contrary in this table), it is compact for Grade 42 steel considered. So, the design bending strength is,

d b px b x y

M = N M = N Z F = 0.90 × 66.5 × 42.0 = 2514 in.-kips = 210 ft-kips

Limiting b/t ratio for plate (web) buckling in shear:

w pv

d v n v w y

V = N V = N d t (0.6 F ) = 0.90 × 17.7 × 0.300 × 0.60 × 42.0 = 120 kips

As,

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P9.20. A simply supported beam AB with a span of 48 feet must carry three column loads of 100 kips each (Fig. P9.20). The loads are factored loads. Assume full lateral support for the compression flange. Select the most economical W-section assuming A992 steel, if the nominal depth is limited to 30 in. Include the effect of member self-weight.

Factored, concentrated load at center C = 100 kips

Factored, concentrated loads at points D and E (AD = BE = a = 12 ft) = 100 kips

Neglect influence of self-weight to start with. Total factored load on the beam = 300 kips

A

Due to symmetry of the structure and the loading, reaction, R = 150 kips Maximum bending moment occurs at midspan.

max

Maximum shear, V = 150 kips

req

Entering the LRFDM Table 5-3 with M = 2400 ft-kips, we observe that a W30×191(in bold face) is the lightest W30 shape that provides the required flexure strength. For this section:

Additional bending moment at C due to factored self-weight is,

As the revised maximum moment, 2400 + 66.0 = 2470 ft-kips is less than 2530 ft-kips, the

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P9.21. The 32 foot long steel beams AB of Figure P9.21 are spaced 8 feet on center. They are simply supported at points D and E, and must carry a service floor load of 100 psf (50% dead load and 50% live load) and a wall load of 0.6 kips per linear foot of wall (100% dead load). Assume adequate lateral support, A992 steel, and select the lightest W-12 shape for flexure and shear. The maximum service load deflection of the beam should be within ± ½ in.

Beam with overhangs, ADCEB.

1 2

Main span (DE), L = 24 ft; overhangs (DA, EB), L = 4 ft b

Spacing of beams, S = 8 ft

D

Service distributed dead load, q = 50.0 × 8.0 = 400 plf L

Service distributed live load, q = 50.0 × 8.0 = 400 plf Factored distributed load,

Wall load = 0.6 klf

Factored concentrated load at A and B, Symmetric and symmetrically loaded structure.

D E

Reactions, R = R = 5.76 + 1.12×16.0 = 23.7 kips

The maximum positive bending moment occurs at midspan (point C). We have:

Maximum negative bending moment occurs at the supports B and D:

max

Maximum moment, M = 48.6 ft-kips px

From LRFDM Table 5-3, a W12×14 with NM = 65.2 ft-kips > 48.9 ft-kips is the lightest

section that has adequate bending strength.

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P9.22. A simply supported beam with a span of 24 ft has a uniform load of 0.6 klf over the entire span. In addition, it has a 40 kip load 8 ft from left end and a 30 kip load 8 ft from the other end of the beam. Select a W-shape of A 572 Grade 60 steel that can safely support the given factored loads. Assume full lateral support.

Solution

Simply supported beam. Span AB: L = 24 ft

Section: W-shape. Compression flange continuously laterally braced. y

Material: A572 Grade 60 steel. 6 F = 60 ksi Factored loads:

Uniformly distributed over AB: = 0.6 klf uD

Concentrated load at point D (with AD = 8 ft): Q = 40.0 kips uE

Concentrated load at point E (with BE = 8 ft): Q = 30.0 kips Assume self weight of the beam = 50 plf = 0.05 klf

u

Factored distributed load, q = 1.2( 0.05) + 0.6 = 0.660 klf

Maximum moment occurs at D.

max A

Maximum shear, V = R = 44.5 kips.

From Eq. 9.7.18:

req

Entering LRFDM Table 5-3 with Z = 74.4 in. , we observe that a W18×40 (in bold face) is the3

lightest section that provides adequate flexural strength. For this shape, we have:

x w f f w

Z = 78.4 in. ; d = 17.9 in.; t = 0.315 in.; b /2t = 5.73; h /t = 50.9 3

Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

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d b px b x y

M = N M = N Z F = 0.90 × 78.4 × 60.0 = 4234 in.-kips = 353 ft-kips

w pv

d v n v w y

V = N V = N d t (0.6 F ) = 0.90 × 17.9 × 0.315 × 0.60 × 60.0 = 183 kips

As,

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P9.23. A simply supported beam AB, with a span of 28 ft, has a 40 kip load at a point D, 16 ft from end A. In addition, the beam is subjected to a uniformly distributed load of 1.6 klf over the portion AD. Select a W-shape of A572 Grade 50 steel that can safely support the given factored loads. Assume full lateral support to the compression flange.

Solution

y

Material: F = 50 ksi steel Factored loads:

u D

Concentrated load at D, 16 ft from A: Q = 40 kips Uniformly distributed load over AD: 1.6 klf

As the shear changes sign at D, the maximum bending moment under these loads occurs at D.

Assume self-weight of the beam = 50 plf = 0.05 klf Additional moment due to self-weight,

Additional shear due to self weight,

u

Required bending strength, M = 362 + 5.88 = 368 ft-kips (conservatively)

u

Required shear strength, V = 35.4 + 0.84 = 36.2 kips

req

Entering LRFDM Table 5-3 with M = 368 ft-kips, we observe that a W21×50 is the lightest compact section that provides adequate flexural strength. For this shape,

As the weight of the beam (50 plf) is the same as the value assumed (50 plf) the design is O.K.

So, select a W21×50. (Ans.)

Note: From LRFDM Table 5-3, we observe that a W21×48 of Grade 50 steel has adequate bending and shear strengths; and could be selected. However, the section is compact. Design of non-compact beams is considered in Section 10.5.2.

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P9.24. A 20 ft long cantilever beam AB is fixed at end A, and free at end B. It is subjected to a concentrated factored load of 24 kips at point C, where AC = 12 ft. In addition, the beam is subjected to a uniformly distributed factored load of 2.5 klf over its entire length. Select a W-shape of A572 Grade 60 steel that can support the loads. Assume adequate lateral support to the beam flanges.

Solution

Cantilever beam. Span AB: L = 20 ft

Section: W-shape. Flanges adequately laterally braced. y

Material: A572 Grade 60 steel. 6 F = 60 ksi Factored loads:

Uniformly distributed over AB: = 2.50 klf uC

Concentrated load at point C (with AC = 12 ft): Q = 24.0 kips Assume self weight of the beam = 100 plf = 0.10 klf

u

Factored distributed load, q = 1.2( 0.10) + 2.5 = 2.62 klf

From Eq. 9.7.18:

req

Entering LRFDM Table 5-3 with Z = 176 in. , we observe that a W24×68 (in bold face) is the3 lightest section that provides adequate flexural strength. For this shape, we have:

x w f f w

Z = 177 in. ; d = 20.1 in.; t = 0.415 in.; b /2t = 7.66; h /t = 52.0 3

Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

d b px b x y

M = N M = N Z F = 0.90 × 177 × 60.0 = 9558 in.-kips = 797 ft-kips

w pv

d v n v w y

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As,

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P9.25. A simply supported beam ABCD with an overhanging end is supported at points B and D such that AB = 8 ft and BC = CD = 12 ft. The beam is subjected to a uniformly distributed, factored load of 3 klf over the entire length. In addition, it is subjected to concentrated, factored loads of 10 kips and 40 kips, at points A and C, respectively. Select a W-shape of A572 Grade 50 steel that can support the factored loads. Assume adequate lateral support for the compression flange.

Solution

Section: W-shape. Compression flange adequately laterally braced. y

Material: A572 Grade 50 steel. 6 F = 50 ksi Overhanging beam ABCD.

Overhang AB: a = 8 ft Main span BD: L = 24 ft Factored loads:

Uniformly distributed over AD: = 3.0 klf uA

Concentrated load at free end, A: Q = 10 kips uC

Concentrated load at center of main span, C: Q = 40 kips Assume self weight of the beam = 50 plf = 0.05 klf

u

Factored distributed load, q = 1.2( 0.05) + 3.0 = 3.06 klf

So,

max BC

Maximum shear, V = V = 98.6 - 10.0 - 3.06 × 8.0 = 64.1 kips.

req

Entering LRFDM Table 5-3 with M = 371 ft-kips, we observe that a W21×50 is the lightest compact section that provides adequate flexural strength. For this shape,

As the weight of the beam (50 plf) is the same as the value assumed (50 plf) the design is O.K.

So, select a W21×50. (Ans.)

Note: From LRFDM Table 5-3, we observe that a W21×48 of Grade 50 steel has adequate bending and shear strengths; and could be selected. However, the section is compact. Design of non-compact beams is considered in Section 10.5.2.

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P9.26. Select the most economical wide flange section for the 30-ft-long overhang beam, with supports at the left end and 10 ft from the right end. It is subjected to the following service loads: 1.5 klf dead load and 2 klf live load uniformly distributed over the entire length of the beam. In addition, the beam is subjected to

concentrated live loads of 30 kips and 10 kips at 10 ft and 30 ft respectively from the left end. The beam is continuously laterally braced. Assume A572 Grade 42 steel.

Solution

Section: W-shape. Compression flange continuously laterally braced. y

Material: A572 Grade 42 steel. 6 F = 42 ksi Overhanging beam ACBD.

Main span AB: L = 20 ft Overhang BD: a = 10 ft Service loads:

Uniformly distributed over AD: Dead load = 1.5 klf L

Live load, q = 2.0 klf

LD

Concentrated live load at free end, D: Q = 10 kips

LC

Concentrated live load at center of main span, C: Q = 30 kips Assume self weight of the beam = 60 plf = 0.06 klf

u

Factored distributed load, q = 1.2(1.5 + 0.06) + 1.6(2.0) = 5.07 klf

uC

Factored concentrated load at C, Q = 1.6(30.0) = 48.0 kips

uD

Factored concentrated load at D, Q = 1.6(10.0) = 16.0 kips

So,

max BC

Maximum shear, V = V = 162 - 16.0 - 5.07 × 10.0 = 95.3 kips.

From Eq. 9.7.18:

req

Entering LRFDM Table 5-3 with Z = 131 in. , we observe that a W24×55 (in bold face) is the3 lightest section that provides adequate flexural strength. For this shape, we have:

x w f f w

Z = 135 in. ; d = 23.6 in.; t = 0.395 in.; b /2t = 6.94; h /t = 54.1 3

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Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

d b px b x y

M = N M = N Z F = 0.90 × 135 × 42.0 = 5103 in.-kips = 425 ft-kips

w pv

d v n v w y

V = N V = N d t (0.6 F ) = 0.90 × 23.6 × 0.395 × 0.60 × 42.0 = 211 kips

As,

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P9.27. Select the lightest W-shape for a 18-ft-long cantilever beam of A572 Grade 42 steel. It is subjected to concentrated, service live loads of 16 kips and 10 kips acting at 12 ft and 18 ft, respectively, from the fixed end. The beam is continuously laterally braced. The maximum service load deflection is limited to 1/600 of its span length. Neglect beam weight in all the calculations.

P9.28. In a building, roof beams are spaced at 10-ft intervals and are connected to girder webs by simple connections. The span of the beams is 24 ft. The beam is designed for the following service loads: dead load (not including the self weight), 65 psf; roof live load, 20 psf; snow load, 40 psf; and rain load, 25 psf. Deflection of the beam is limited to L/240. Select a suitable W-shape of A242 Grade 42 steel. Assume the beam is continuously laterally supported by the roof decking. Specify any camber provided.

P9.29. A simple beam in a research lab is to support a dead load of 1 klf and a live load of 1.5 klf. In addition, it is to carry a concentrated live load (movable) of 6 kip. The load consists of precision machinery which requires that the deflection be limited to a maximum of L/800. The beam has an effective span of 40 ft and is continuously laterally supported. Use A992 steel and select a suitable section.

P9.30. A W-shape is used to support a factored load of 12 klf on a 20 ft simple span. The architect specifies that the beam be no more than 18 in. in depth. Assume full lateral support to the compression flange and select a suitable A 992 steel beam. Is the deflection OK, if the beam carries a plastered ceiling? Live load is b rd the total load on the beam.

P9.31. A simply supported beam with continuous lateral support for the entire span of 30 ft is subjected to a factored, uniformly distributed load of 2.4 klf. The beam depth is limited to not more than 14 in. (nominal), and the deflection is limited to L/360. Select a suitable A992 shape.

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P9.32. To save construction depth with a precast plank system, a structural tee is used for the overhang beam ABC with supports at A and B, as shown in Fig. P9.32. The beam is subjected to a dead load of 0.6 klf and a live load of 0.3 klf. The dead load includes provision for self-weight of the beam. Assume continuous lateral support, and select a suitable WT6 of A992 steel.

y

Yield stress of material, F = 50 ksi Beam with overhang, ABC.

1

Main span (AB) = L = 20 ft

2 Overhang (BC) = L = 6 ft Loading: D Dead load, q = 0.6 klf L Live load, q = 0.3 klf u Factored load, q = 1.2 × 0.6 + 1.6 × 0.3 = 1.20 klf Total load = 1.20×26.0 = 31.2 kips

;

Location of point of zero shear, from support A 6 Maximum positive moment,

Maximum negative moment occurs at support B,

req

Required bending strength, M = 49.5 ft-kips As the beam is continuously laterally supported

x

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P9.33. The bay size for a shopping center is 32'×30'. The beams are spaced at 8 ft centers and have a span of 30 ft. All the beams and girders are simply supported. The floor deck consists of 5-in.-thick slab of light weight concrete (100 pcf) directly supported by the beams. Dead load from flooring and ceiling is 10 psf. The floor live load, including provision for partitions, is 120 psf. Design an interior beam without counting on composite action between the slab and the steel beam, but assuming that lateral buckling of the beam is prevented. Limit the live load deflection to L/360. Use A992 steel.

See Fig. 3.2.2 of the text book for guidance. Solution

Simply supported beam. b Span, L = 30 ft. Spacing, S = 8 ft. Weight of slab =

Weight of flooring and ceiling = 10 psf Assume the self-weight of the beam to be 40 plf. Distributed dead load,

Distributed live load,

Distributed factored load on the beam,

From LRFDM Table 5-3 select a W18×35 (in bold face):

Design bending strength,

Allowable deflection,

Maximum deflection, under nominal live load:

Required

req x

Enter LRFDM Table 5-2 with I = 602 in. and observe that a W18×40 with I = 612 in. is the4 4 lightest section that satisfies this condition.

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P9.34. Design a spandrel beam of the shopping center given in Problem P9.33. The spandrel beam supports, in addition to floor loads, a facia panel weighing 0.6 klf. There are no clearance limitations, but the total service load deflection is to be limited to ½ in. Assume continuous lateral support.

See Fig. 3.2.2 of the text book for guidance. Solution

Simply supported beam. Span, L = 30 ft

b Spacing of beams, S = 8 ft

b

Tributary width for the spandrel beam = ½ S = 4.00 ft

Weight of slab = (5/12) ×100 × 4.00 = 167 plf Weight of flooring and ceiling = 10.0×4.00 = 40.0 plf

Facia panel weight = 600 plf

Self-weight of spandrel beam (assumed) = 60.0 plf D

Total dead load, q = 0.867 klf

L

Total live load, q = 120 × 4= 480 = 0.480 klf

s

Total service load, q = 0.867 + 0.480 = 1.35 klf

u

Total factored load, q = 1.2 × 0.867 + 1.6 × 0.480 = 1.81 klf

all

Allowable deflection under total service load, * = 0.5 in. As this deflection limitation is quite stringent, we will select a section to satisfy the deflection criteria and check for strength.

x

So, I $ 1700 in.4

req

Enter LRFDM Table 5-2 with I = 1700 in. and select a W24×68 for which:4

x

I = 1830 in. > 1700 in. O.K.4 4

b px v n

N M = 664 ft-kips; N V = 266 kips

Maximum shear,

So, select a W24×68 of A992 steel. (Ans.)

P9.35. Design an interior girder of the shopping center described in Problem 9.33. Due to clearance limitations the nominal depth of the girder is limited to 24 inches. Limit the live load deflection to L/360. Determine the camber to be specified. Assume continuous lateral support.

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P9.36. The floor framing system of an office building has bay sizes of 27 ft by 34 ft. The floor beams are at 9 ft centers and have a span of 34 ft. Assume that the beams and girders are simply supported. The floor deck consists of 5-in.-thick reinforced concrete “one way” slab of ordinary concrete, directly supported by the beams, and carries a live load of 100 psf. Design the beam without counting on composite action between the slab and the steel beam, but assuming that lateral buckling of the beams is prevented. Check deflection, and specify camber. Use A992 steel.

Solution

Simply supported beam. b Span, L = 34 ft. Spacing, S = 9 ft. Slab thickness, t = 5 in.

Weight of slab =

Assume the self-weight of the beam to be 50 plf. Distributed dead load,

Distributed live load,

Distributed factored load on the beam,

From LRFDM Table 5-3 select a W21×44 (in bold face):

Design bending strength,

Allowable deflection,

Maximum deflection, under nominal live load:

Deflection under nominal dead load, Provide a camber of ¾ in.

Use a W21×44 of A992 steel. As the self-weight assumed is close to the actual value, no revision is

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P9.37. Design an interior girder of the floor framing system of Problem 9.36. Due to clearance limitations, depth of girder cannot exceed 27 in. (nominal). Assume continuous lateral support.

P9.38. Design a spandrel girder of the floor framing system of Problem 9.36. The spandrel girder supports, in addition to beam reactions, a facia panel weighing 0.8 klf. There are no clearance limitations. Assume continuous lateral support.

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P9.39. Select the most economical W12 section that can carry a concentrated dead load of 40 kips and a live load of 30 kips at the third point of a 6 ft simple span. Check for bending, shear and local buckling. Neglect self weight of beam. Assume adequate lateral bracing to the compression flange.

Solution

Loading:

A concentrated load at D (with AD = a = 2 ft; BD = b = 4 ft).

D

Dead load, Q = 40 kips

L

Live load, Q = 30 kips u

Factored loads, Q = 1.2 × 40 + 1.6×30 = 96 kips

req

Entering LRFDM Table 5-3 with M = 128 ft-kips, we observe that a W12×26 is the lightest W12-shape that provides adequate flexural strength. For this shape,

f f w

From LRFDM Table 1-1, for a W12×26 6 b / 2t = 8.54; h /t = 47.2 Limiting b/t ratios for plate buckling:

Compression flange:

Web in flexure:

Web in shear:

f f pf w pw

As b / 2t < 8 and h /t < 8 , the section is compact. So, the design bending strength used above is O.K.

w pv

As h /t < 8 , the design shear strength used above is also O.K.

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P9.40. Select the lightest beam section to carry a uniformly distributed, factored load of 20 klf on a simple span of 8 ft. Compression flange is adequately supported. Use A992 steel.

P9.41. Select the lightest W section to support a factored uniformly distributed load of 2.6 klf over a simple span of 25 ft. Assume A572 Grade 60 steel and continuous lateral support. Make all checks. The given load does not include self weight of the beam.

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P9.42. Determine the size of the bearing plate required for an end reaction, under a factored load of 90 kips, for a W16×45 A992 steel beam. The beam rests on a concrete wall with a 28-day compressive strength of 3.0 ksi.

Solution

a. Data

From Table 1-1 of the LRFDM for a W16×45 : f d = 16.1 in.; b = 7.04 in. w f t = 0.345 in.; t = 0.565 in. 1 k = 0.967 in.; k = 13 ÷ 16 = 0.813 in.

b. Length of bearing plate, N

Assume , which requires N > 0.2 × 16.1 = 3.22 in.

From LRFDM Table 9-5: Beam End Bearing Constants, for a W16×45 of A992 steel: v n NV = 150 kips 1 2 NR = 41.7 kips; NR = 17.3 kli r 5 r 6 NR = 49.8 kips; NR = 6.52 kli

Minimum length of bearing plate to prevent local web yielding,

Minimum length of bearing plate to prevent web crippling,

Try N = 8 in.

Check: N /d = 8.00 /16.1 = 0.497 > 0.2, as assumed. O.K. c. Width of bearing plate, B

Assume conservatively that the bearing plate covers full area of the support. Hence, the confinement factor $ = 1 and the required plate width can be found from:

6 B $ 7.35 say 8 in. rounding to the nearest inch.

The flange width of a W24×62 is 7.04 in., making the bearing plate slightly wider than the beam flange.

d. Thickness of the bearing plate, t

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The required thickness is

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P9.43. A W14×426 column of A992 steel supports a factored axial load of 5310 kips. Design a base plate for the column if the supporting concrete has a cylinder strength = 5 ksi. Assume that (a) The full area of the concrete support is covered by the base plate, (b) The support will be a 4 ft by 4 ft concrete pier. Solution

a.

1. Data

u

Factored axial load, P = 5310 kips W14×426 column f b = 16.7 in.; d = 18.7 in. f f b d = 312 in. ;2 b + d = 35.4 in. Concrete: = 5.0 ksi y

Base plate: A36 steel F = 36.0 ksi

As the full area of the support is covered by the base plate, $ = 1.0

O.K.

2. Plan dimensions of base plate

Optimize plan dimensions.

f

> b = 16.7 in. O.K.

N $ > d = 18.7 in. O.K.

So, the plate dimensions are 44 × 48 in.

O.K.

(44)

n = * (conservatively)

Provide t = 2½ in. > 2.28 in.

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P9.44. A W10×45 column of A992 steel supports a factored axial load of 565 kips. Design a base plate for the column if the supporting concrete has a cylinder strength . Assume that (a) The full area of the concrete support is covered by the base plate, (b) The footing size is 20 in. by 20 in.

Solution a.

1. Data

u

Factored axial load, P = 565 kips W10×45 column f b = 8.02 in.; d = 10.1 in. f f b d = 81.0 in. ;2 b + d = 18.1 in. Concrete: = 3.0 ksi y

Base plate: A36 steel F = 36.0 ksi

As the full area of the support is covered by the base plate, $ = 1.0

O.K.

2. Plan dimensions of base plate

Optimize plan dimensions.

f

> b = 8.02 in. O.K.

N $ > d = 10.1 in. O.K.

So, the plate dimensions are 18 × 21 in.

O.K.

3. Plate thickness

dp dp 1

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n = *

Provide t = 1¾ in. . 1.76 in.