BGSCM14C10ExProb

A Beginners Guide to the Steel Construction Manual last modified: 28 July 2008Example Problem 10.1 by: TBQ

Given:

Stl Beams Conc. Slab Studs

Span 40 ft 3.5 ksi 65 ksiSpacing 4 ft Slab t 4 in diam. 0.5 in

Section W16X26 0.85 0.1963

Steel A992 3267 ksi

50 ksi n 9 LoadsD L

(psf) (psf)Floor 75 50

Wanted:

Solution:

Beam Section Properties

7.68d 15.7 in

301

Part a) Compute bELeft Side 2 ftRight Side 2 ftbE 4 ft

48 in

Locate PNA relative to top of slab

3.16 in Good! PNA in Slab

Compute Moment Capacity Compute Internal Moment

384 k 300.0 plf

-384 k 200.0 plf

arm 10.51 in 500.0 plf

336.2 ft-k 100.0 ft-kW 1.67

201.3 ft-k

0.497 Good!

Part b) Compute Shear Capacity of a Stud

f'c Fu

b1 Asc in2

Ec

Fy

Ag in2

Ix in4

yPNA =

Ts wD

Cc wL

wa

Mn Ma

Mn / W

Ma / (Mn / W)

An interior floor beam as shown in BGSCM Figure 10.8.1.1. The W16x26 beams are spaced 4'-0" O.C. and span 40 ft. The beams are simply supported. The 4" thick normal weight concrete slab has a 28-day compressive strength, f'c, of 3.5 ksi. The studs are 1/2" diameter and have Fu = 65 ksi. The floor live load is taken as 50 psf and the floor dead load (including beam, slab, and other items) is 75 psf.


For a typical interior beam:

a) compute Ma / (Mn / W ).b) Determine the number and layout of studs required for full composite action/c) Determine the Live Load only deflection.



Conc., Qn 10.50 k/studStud, Qn 12.76 k/studQn 10.50 k/stud

Compute the Maximum Shear Force, V'Conc. 571.2 kSteel 384 kV' 384 k

req'd # studs per half span: 36.58req'd # studs for whole span: 73.16

Use: 74 studs

Single row spacing 6.58 in. O.C.Double row spacing 13.33 in. O.C.

min spacing 3.00 inmax spacing 32.00 in

Use single line of studs, evenly spaced along entire span

Part c) Compute Location of ENA relative to top of slabArea y y*A

(in)Block 21.33 2.00 42.67W16 7.68 11.85 91.01Total 29.01 133.67

yena = 4.61

y relative to:

Area top of slab ENA

(in) (in)Block 21.33 28.44 2.00 -2.61 145.03 173.47W16 7.68 301 11.85 7.24 402.86 703.86

Total 877.34

Compute Deflection

200.0 plf

0.453 in = L / 1060.1

360Allow/Act 0.340 Good!

(in2) (in3)

Compute ITR:

Io y2*A ITR

(in2) (in4) (in3) (in4)

ITR =

wL

DLL

DLL,allow = L /

p







A Beginners Guide to the Steel Construction Manual last modified: 28 July 2008Example Problem 10.2 by: TBQ

Given:

Bay Concrete: Steel:Span 50 ft weight 150 pcf Stl Type A992

Beam Spacing 5 ft Slab t 4.0 in 50f'c 3000.0 psi

Added Loads 3155.9 ksi

Dead 25 psf 0.85 65Live 50 psf Modular Ratio 9.19

Cons. Live 20 psf Use: 9.00

Shored? Y at third points

Wanted:

Solution:

Limit State SummaryASD LRFD

Best Sections: W16X26 W14X26

Current Section Results:Flexural at Construction 0.222 0.232Composite Flexural 1.002 0.902Shear 0.231 0.208Construction Deflection 0.090 0.090LL Only Deflection 0.795 0.795Maximum 1.002 0.902

Left Right Total(in) (in) (in)

L/8 75 75(C-C)/2 30 30

30 30 60 in

Beam Size W16X26 <----- Change this value until the best solution is found!

Beam Section PropertiesAs 7.68 in^2 J 0.262 in^4Wt 26 plf Sx 38.4 in^3d 15.7 in Iy 9.59 in^4

Fy

Ec

b1 Stud Fu

Determine bE

bE:

A floor system similar to that shown in BGSCM Figure 10.1.1. The slab is 4" thick and is made from concrete with f'c = 3,000 psi. The floor beams span 50 ft and are spaced 5 ft apart. Assume that the beams are shored at third points during construction. The beams only have lateral support at shoring and the ends during the initial construction. The imposed floor dead load is 25 psf , the construction live load is 20 psf, and the occupancy floor live load is 50 psf. Consider studs of 1/2", 5/8", 3/4", and 7/8" diameter.


Select the lightest I-shaped section for the application considering both ASD and LRFD. Also determine the number and spacing of shear studs needed for full composite action.


Ix 301.0 in^4 Cw 565bf 5.5 in ho 15.4 intf 0.3 in rts 1.385 intw 0.3 in bf/2tf 7.97ry 1.12 in h/tw 56.8Z 44.2 in^3

Compute the Beam Loads and Internal Forces

Slab Weight 50 psf

Loading during concrete pour Load after concrete pour DL of beam 26 plf DL (bm+slab+extra) 401 plf Slab weight (LL) 250 plf LL 250 plf CLL 100 plf

ASD LRFD ASD LRFDw 376.0 591.2 plf w 651 881.2 plfShear, V 9.40 14.78 k Shear, V 16.28 22.03 k

Unshored Moment, M 117.50 184.75 ft-k Moment, M 203.44 275.38 ft-kShored 8.70 13.69 ft-k

Steel Beam Capacity w/o Composite Action During Construction

Solve for the moment capacity, Mcx Shored Span 16.67 ft

Y & LTB Lp 47.5 in Cb: Lateral support at ends only3.96 ft Mmax 184.75 ft-k

Mp 184.1667 ft-k Ma 138.56 ft-kLr 134.47 in Mb 184.75 ft-k

11.21 ft Mc 138.56 ft-kMr 112.00 ft-k Cb 1.14Cb 1.14Lb 16.67 ftFcr 20.46 ksiMnx 65.5 ft-k Capacity Check

FLB l 7.97 ASD LRFD

9.15 Modifier 1.67 0.9

24.08 Mcx 39.21 58.93kc 0.53 Mrx 8.70 13.69Mnx 184.2 ft-k Mrx/Mcx 0.222 0.232

Controlling Mnx 65.5 ft-k

Composite Beam Capacity

Assume yna<tsT = C = 384 ka = 2.51 iny(pna) = a/beta1 = 2.95 in measured from the topyna/ts 0.74 ... Assumption O.K.arm 10.60 inMn 4068.52 in-k

lpf

lrf

Mn 339.04 ft-k ... USE THIS ONE

Assume ts<yna<ts+tf = 4.345 iny(pna) = 3.80 inCheck Assumption ts/y(pna) 1.05 y(pna)/(ts+tf) 0.87 ... Assumption N.G.a 3.23 in (note: not all concrete is considered in compression!)

Magnitude y dist from Moment(k) (in) yna (in) (ft-k)

Cc 494.15 1.61 2.18 89.97Cs -55.08 3.90 -0.10 0.46T -439.08 10.85 -7.05 258.07Total 0.00 0.00 ... DO NOT USE.

Assume ts+tf<yna = 4.35 in-12.63 in

1.91 iny(pna) = 1.91 inCheck Assumption (ts+tf)/y(pna) 2.27 ... Assumption N.G.a -10.74 in

Magnitude y dist from Moment(k) (in) yna (in) (ft-k)

Cc -1642.53 -5.37 7.28 -996.25Csf 94.88 4.17 -2.26 -17.88Csw -30.43 3.13 -1.22 3.09Tw 0.00 10.63 -8.72 0.00Tf -94.88 19.53 -17.62 139.28Total -1672.96 -871.77 ... DO NOT USE.

Solution y(pna) 2.95 inMn 339.04 ft-k

Capacity CheckASD LRFD

Modifier 1.67 0.9Mcx 203.02 305.14Mrx 203.44 275.38Mrx/Mcx 1.002 0.902

Shear CapacityCapacity Check

59.2 ASD LRFD

73.8 Modifier 1.67 0.9

117.75 Vc 70.51 105.98 Equation G2-5 Vr 16.28 22.03

Vr/Vc 0.231 0.208

lp

lr Vn

Shear Connectors

Design Load.85f'cAc 612 kips <- Max Concrete ForceAsFy 384 kips <- Max Steel Force

Design for 384 kips1 Row 2 Row

Connector diameter Capacity Req'd # Use spacing spacing Sum Qn(in) (k/stud) per ½ bm per 1 bm (in O.C.) (in O.C.) (k)

1/2" dia x 2" 0.5 9.36 41.02 158 3.80 7.59 739.585/8" dia x 2.1/2" 0.625 14.63 26.25 102 5.88 11.76 746.023/4" dia x 3" 0.75 21.06 18.23 70 8.57 17.14 737.247/8" dia x 3.1/2" 0.875 28.67 13.39 52 11.54 23.08 745.43

Deflection Calculation

Total Load Deflection during ConstructionDeflection is based on Steel Beam properties onlyL 16.67 ftw 376.0 plfTL defl 0.07 in <--- conservative approx for shored const.L/240 0.833333 inActual/allowable 0.090 ... O.K.

LL Only Deflection under Composite Action

Compute the Elastic Transformed Section Properties

Transformed bE 6.67 inArea y(top) Ay Io y(na) Ay^2 Ix(in^2) (in) (in^3) (in^4) (in) (in^4) (in^4)

Concrete 26.67 -2.00 -53.33 35.56 2.20 129.36 164.91Steel 7.68 -11.85 -91.01 301.00 -7.65 449.16 750.16CoverPL 0.00 -19.70 0.00 0.00 -15.50 0.00 0.00

34.35 -4.20 -144.34 915.07

-4.20 in from the beam C.A.

915.07 in^4

Note at the "y" terms are computed relative to the centroidal axis of the stl section

distributed LL 250 plfActual LL only defl 1.32 inAllowable defl 1.667 in = L/360Actual/allowable 0.795 ... O.K.

yna

ITR

ksi

ksi





Documents

BGSCM14C10ExProb