AISC-20

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 40 of 131 Design Columns for the lightest W10's and W12's section. Columns are to be sized for two options:Option I Continuous, Option II with Splices. Then prices are to be compared to select most economical option. Both explicit calculations or Column Design Tables may be used.

COLUMN B-2 LOAD TAKEOFF Dead load (psf) Live load (psf)

Penthouse Roof = 30 20 KLL= 4 Penthouse Floor = 100 80 Fy (ksi)= 50

Roof = 30 20

61 80 Bold Numbers are

61 80 Roof Live Loads

Tributary Areas (At):Width (ft) Length (ft)

At 1 = 36 30 1080At 2 = 18 15 270

At 3 = At 1 - At 2 = 810

Plan View At 1 (floors) : At 2 (penthouse floor) :

DL = 61 psf DL = 100 psf

LL = 80 psf LL = 80 psf

30 30

Penthouse At 3 (roof) :

DL = 30 psf

36 36 LL = 20 psf

Note: Live load reduction can be applied in accordance with ASCE 7-02 section 4.8.1

(see ASCE 7-02 Eq. 4-1)

Use the larger of:

L = reduced design live load value in psf L > 0.5*Lo (columns supporting one floor)

L > 0.4*Lo (columns supporting two or more floors)

if AI > 400 ft^2 live load reduction is allowed

AI (1) = 4320 >400 ft^2

AI (2) = 1080 >400 ft^2

L (2nd floor) = 38 psf > 32 psf, thus use = 38

L (3rd floor) = 38 psf > 32 psf, thus use = 38

L (penthouse floor) = 57 psf > 40 psf, thus use = 57

Red font indicates user input

3rd Floor =

2nd Floor =

L = Lo*(0.25+15/AI^0.5) ASCE 7-02 Eq. 4-1

Lo = unreduced live load in psf

AI = influence area

AI = Tributary area *KLL

Note: Roof Live loads are NOT allowed to be Reduced

At 1 At 3

At 2

PROJECT: STEEL BUILDING DESIGN CASE STUDY

SUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 41 of 131

COLUMN DESIGN

METHOD 1 - EXPLICIT CALCULATIONS

METHOD 2 - USE COLUMN DESIGN TABLES

CONTINUOUS - L =13 FEET

Notation:

Fcr - critical design force

Fy - specified yield strength, ksi

E - modulus of elasticity, ksi

K - effective length factor

l - Laterally unbraced length of member, in

r - governing radius of gyration about the axis of buckling, in

Begin Load Takeoff from roof and proceed downward:

From Ground floor to 2nd floor:

dead loads (psf) live loads (psf)

Pnth Roof 30 20

Pnth Floor 100 57

Roof 30 20

61 38

61 38

Pu (Pnth Roof) = 68 psf * A2 = 18 kipsPu (Pnth Floor) = 210 psf * A2 = 57 kips

Pu (Roof) = 68 psf * A3 = 55 kipsPu (3rd Floor) = 134 psf * A1 = 145 kipsPu (2nd Floor) = 134 psf * A1 = 145 kips

sum = 421 kips

PU = 421 kips

Note: Use column B2 for ALL Interior Columns.


fPn - design compressive strength, kips

lc - width-thickness ratio

Ag - gross area of member, in2

3rd Floor

2nd Floor

Pu = [1.2*dead load + 1.6*live load]

Pu = [1.2*dead load + 1.6* roof live load]

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 42 of 131 COLUMN DESIGN From Ground floor to 2nd floor:

may not be the lightest. If this is the case, do explicit calculations and then try a section from the Table 1-25 LRFD. Use W8x24 as the smallest size (do not use W10’s or W12’s with a nominal

Analysis(Equation E2-1 LRFD p.16.1-27)

(Equation E2-2 LRFD p.16.1-27)



Fy = 50 ksiK = 1

Pu = 421 kips L = 13.00 ftE = 29000 ksi

Assume KL/r = 60 to start0.7930

Fcr = 38.4 ksi

Ag = 12.88

LRFD p.4-25Check W 12:

Select W12x53 KL/rx = 29.8

Ag = 15.6 KL/ry = 62.9ry = 2.48 in 0.83

rx/ry = 2.11 in Fcr = 37.4rx = 5.23 in 496.4

Section OK



Ag = 14.4 KL/ry = 61.4ry = 2.54 in 0.81

rx/ry = 1.71 in Fcr = 37.9rx = 4.35 in 464.5

Section OK


Note: For small axial loads the smallest section from the column design Tables (Table 4-2 LRFD)

weight less than 24 lb/ft). If the column tables are used you only need to write down the section selected and its capacity.

fPn = 0.85 * Ag * Fcr

Fcr = (.658lc^2) * Fy when lc < 1.5

Fcr = (0.877 / lc2) * Fy when lc > 1.5

lc = (K * L/p * r) * (Fy / E)0.5

lc =

in2

in2

lc =

fPn = fPn > Pu ?

in2

lc =

fPn = fPn > Pu ?

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 43 of 131COLUMN DESIGN METHOD 1 - EXPLICIT CALCULATIONSMETHOD 2 - USE COLUMN DESIGN TABLESCONTINUOUS - L =13 FEETNotation:







From Ground 2nd floor to 3rd floor:


Pnth Roof 30 20

Pnth Floor 100 57

Roof 30 20

61 38

- -


Pu (Roof) = 68 psf * A3 = 55 kipsPu (3rd Floor) = 134 psf * A1 = 145 kips

sum = 275 kips

PU = 275 kips





3rd Floor

2nd Floor



PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 44 of 131COLUMN DESIGN From Ground 2nd floor to 3rd floor:






Fy = 50 ksiK = 1

Pu = 275 kips L = 13.00 ftE = 29000 ksi


Fcr = 38.4 ksi

Ag = 8.43



Ag = 11.7 KL/ry = 80.4ry = 1.94 in 1.06

rx/ry = 2.64 in Fcr = 31.2rx = 5.13 in 309.9

Section OK



Ag = 11.5 KL/ry = 78.8ry = 1.98 in 1.04

rx/ry = 2.16 in Fcr = 31.8rx = 4.27 in 310.4

Section OK



fPn = 0.85 * Ag * Fcr

Fcr = (.658lc^2) * Fy when lc < 1.5

Fcr = (0.877 / lc2) * Fy when lc > 1.5

lc = (K * L/p * r) * (Fy / E)0.5

lc =

in2

in2

lc =

fPn = fPn > Pu ?

in2

lc =

fPn = fPn > Pu ?

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 45 of 131COLUMN DESIGN METHOD 1 - EXPLICIT CALCULATIONSMETHOD 2 - USE COLUMN DESIGN TABLESCONTINUOUS - L =13 FEETNotation:







From 3rd floor to roof level:


Pnth Roof 30 20

Pnth Floor 100 57

Roof 30 20

- -

- -


Pu (Roof) = 68 psf * A3 = 55 kips

sum = 130 kips

PU = 130 kips





3rd Floor

2nd Floor



PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 46 of 131COLUMN DESIGN From 3rd floor to roof level:






Fy = 50 ksiK = 1

Pu = 130 kips L = 13.00 ftE = 29000 ksi


Fcr = 38.4 ksi

Ag = 3.99



Ag = 7.65 KL/ry = 103.3ry = 1.51 in 1.37

rx/ry = 3.42 in Fcr = 22.9rx = 5.17 in 149.0

Section OK



Ag = 8.84 KL/ry = 113.9ry = 1.37 in 1.51

rx/ry = 2.16 in Fcr = 19.4rx = 4.38 in 145.5

Section OK



fPn = 0.85 * Ag * Fcr

Fcr = (.658lc^2) * Fy when lc < 1.5

Fcr = (0.877 / lc2) * Fy when lc > 1.5

lc = (K * L/p * r) * (Fy / E)0.5

lc =

in2

in2

lc =

fPn = fPn > Pu ?

in2

lc =

fPn = fPn > Pu ?

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 47 of 131COLUMN DESIGN

From roof level to penthouse:


Pnth Roof 30 20

Pnth Floor - -

Roof - -

- -

- -

Pu (Pnth Roof) = 68 psf * A2 = 18 kips

sum = 18 kips

PU = 18 kips


3rd Floor

2nd Floor



PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 48 of 131COLUMN DESIGN

From roof level to penthouse:






Fy = 50 ksiK = 1

Pu = 18 kips L = 13.00 ftE = 29000 ksi


Fcr = 38.4 ksi

Ag = 0.56



Ag = 7.08 KL/ry = 96.9ry = 1.61 in 1.28

rx/ry = 2.12 in Fcr = 25.2rx = 3.42 in 151.5

Section OKLRFD p.4-26Check W 10:


Ag = 7.08 KL/ry = 96.9ry = 1.61 in 1.28

rx/ry = 2.12 in Fcr = 25.2rx = 3.42 in 151.5

Section OK



fPn = 0.85 * Ag * Fcr

Fcr = (.658lc^2) * Fy when lc < 1.5

Fcr = (0.877 / lc2) * Fy when lc > 1.5

lc = (K * L/p * r) * (Fy / E)0.5

lc =

in2

in2

lc =

fPn = fPn > Pu ?

in2

lc =

fPn = fPn > Pu ?

Lab Note: When beams and girders are framed to columns with simple shear connections, the columns are usually designed as concentrically loaded members.

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 49 of 131COLUMN DESIGN SPLICED vs CONTINOUS

Steel Prices per ton = $1,600

500

W10's

AnalysisContinous:Member Length (ft) Weight (tons) CostW10x49 52 1.274 $2,038

Wt. / ft (lb)= 49Total cost = $2,038

All columns spliced:Member Length (ft) Weight (tons) CostW10x49 13 0.3185 $510

Wt. / ft (lb)= 49W10x39 13 0.2535 $406

Wt. / ft (lb)= 39W10x30 13 0.195 $312

Wt. / ft (lb)= 30W8x24 13 0.156 $250

Wt. / ft (lb)= 243 splice = 0.75 $1,200

Total cost = $2,677

TOTAL COST = $2,038.40

Answer:Continuous columns are more economical

b -- indicates general value according to Modern Steel Construction. April 2000.

Wt of additional splice (lb)b =

PROJECT: STEEL BUILDING DESIGN CASE STUDYSUBJECT: INTERIOR COLUMN DESIGN (underneath penthouse) SHEET 50 of 131COLUMN DESIGN SPLICED vs CONTINOUS

Steel Prices per ton = $1,600

500

W12's

AnalysisContinous:Member Length (ft) Weight (tons) CostW12x53 52 1.378 $2,205

Wt. / ft (lb)= 53Total cost = $2,205

All columns spliced:Member Length (ft) Weight (tons) CostW12x53 13 0.3445 $551

Wt. / ft (lb)= 53W12x40 13 0.26 $416

Wt. / ft (lb)= 40W12x26 13 0.169 $270

Wt. / ft (lb)= 26W8x24 13 0.156 $250

Wt. / ft (lb)= 243 splice = 0.75 $1,200

Total cost = $2,687

TOTAL COST = $2,204.80

Answer:Continuous columns are more economical

Use : W10's sections Continuous columns are more economical

Total cost = $2,038.40

b - indicates general value according to Modern Steel Construction, April 2000

Wt of additional splice (lb)b =

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