FE Review

Structural Steel Design

Overview of Reference Handbook

• LRFD only

• Form of Equations is sometimes different from AISC.

• Some tables are not in same form as they are in the Manual.

• 14th Edition Steel Construction Manual

• Pages of formulas followed by tables

Example: Find Cb for 30-ft segment.

A B C

30 ft 10 ft

16 k

4 k 12 k

30 60

90 120

0.367.1)90(3)60(4)30(3)120(5.2

)120(5.12

bC

Example

Given:

W18 x 55, Fy = 50 ksi, Cb = 1.67

Find bMn for

a. Lb = 5 ft

Solution:

5 ft < Lp = 5.90 ft

bMn = b Mp = 420 kip-ft.

Example

b. Lb = 10 ft.

Solution

ft-kip 420 use

ft-kip 420 ft kip 606

ft-kip 60690.5109.1342067.1

)ft5.17ft10ft90.5(

ft 10

nMb

pMb

nMb

rLpL

Or, from Charts,

kips-ft 420 Use

kips-ft 606)363(67.1

,67.1For

1for kips-ft 363

n

n

b

bn

M

M

C

CM

Example:

Lb = 13 ft

Mu = 405 ft-kips

Cb = 1

Fy = 50 ksi

Select a W shape

Solution:

Use a W21 x 62

Example:

The beam has continuous lateral support. Fy = 50 ksi

Select a shape

30 ft

wD = 1.00 kips/ft wL = 2.00 kips/ft

Solution: Try a W24 x 55

kips-ft 495)30)(4.4(8

1

kips/ft 4.4)2(6.1)1(2.1

2

u

u

M

w

(OK) kips 66 kips 251

kips 662

)30(4.4

kips-ft 503

nv

u

pbnb

V

V

MM

Use a W24 x 55

13 ft

Columns

Example:

W12 x 50 Fy = 50 ksi

What is the design strength?

Solution:

kips 489)6.14(46.33

ksi 46.33

)50(658.09.0

658.09.0

11367.63

11350

1.8021.802

67.6396.1

)1213(8.0

220,286

)67.63(50

220,286

)/(

2

2

gcrcnc

y

rKLF

crc

y

y

AFP

FF

F

r

KL

r

KL

y

Alternate: From Table 4-22,

ksi 5.334.337.3367.07.33or

ksi 4.33

crcF

Alternate: From Table 4-1, with KL = 0.8(13) = 10.4 ft,

kips 488 471) - 0.4(499 - 499or

kips 471

ncP

Design:

KL = 14.5 ft Pu = 300 kips Fy = 50 ksi

Select a W10 shape

Solution:

OK kips 332 45 x W10a Use

(N.G.) kips 294 kips 300

kips 2942

282305

39 x W10aTry

c

c

n

n

P

P

2 3 3 2

x

+ 3.5 D = 35 k L = 105 k

L6 x 6 x ½

A36 steel

A = 5.77 in.2

x bar = 1.67 in.

¾ - in. A325 bearing-type bolts

Determine design strength for all limit states.

t = 3/8 in.

kips 187)7.207(90.0

kips 7.207)77.5(36

n

gyn

T

AFT

1. Tension on Gross Area

2. Tension on Net Area

kips 168)5.224(75.0

kips 224.53.871) (58

in.871.3 )364.5(7217.0

7217.0)3(2

67.111

in. 364.5)16/14/3)(5.0(77.5

2

2

n

eun

ne

holegn

T

AFT

UAA

L

xU

AAA

2 3 3 2

3.5

t = 3/8 in.

3. Block Shear

2

2

2

in. 047.1)16/13(5.05.22

1

in. 984.2)16/13(5.23322

1

in. 0.43322

1

in. 16

13

16

1

4

3

16

1

nt

nv

gv

bh

A

A

A

dd

kips 123

047.10.1984.26.0)58(75.0

6.075.0

ntbsnvun AUAFT

Upper Limit:

kips 110 Use

kips 110

)047.1)(58(0.1)0.4)(36(6.075.0

6.075.0

n

ntubsgvyn

T

AFUAFT

Example. Staggered Bolts

Find the smallest net area. Holes are for 1-inch diameter bolts.

Hole diameter = 1 + 1/16 = 1.063 in. (?)

tg

sdbA hgn

4

2

For line abde, 2in. 41.10)75.0(063.1(216 nA

For line abcde, 22

in. 27.10 )75.0()5(4

)3(2063.1(316

nA