Thesis Paper on Longer Span Floor Beams System of Edge Supported Structures (Part 2)

lV) Reinforcement calculation:

A. Short direction steel

Steel for positive moment at mid span

Use # 3 bar which area = 0.11

Spacing

Use # 3 @  alternate cranked bars.

Alternative bars should be cranked both at continuous and discontinuous edges.

Steel for negative moment at both supports

So provided 1 # 3 bar extra top in between two ckd. bars.

B. Long direction steel

Steel for positive moment at mid span

(+)ve M_B = 105.02 lb – ft

Spacing

Use # 3 @  alternate cranked bars.

Steel for negative moment at both supports

-M_B = 214.37 lb – ft

Crank to crank spacing

So provide 1-# 3 extra top between two ckd. bars.

Slab Panel S -3          

I) Check for one-way slab        

Panel size                                            = 7′-0″×15′-6″

Beam width                                        = 12″

Length, L=15′-0″ and width, w          =7′-0″

Panel ratio,

So the slab will be designed as one-way slab.

II) Calculation of slab thickness:

Minimum slab thickness

Thickness correction for

Corrected thickness, h = 3″×0.80 =2.40″

Slab thickness let, h =4″

Effective depth, d = 4 -1=3″

III) Moment calculation:

L L= 40 psf

Factored load =1.2×DL+1.6×LL = 1.2×80+1.6×40 =160 psf = 0.16 ksf

(+)ve moment at mid span   

(-)ve  moment at interior support

lV) Check for d:

Maximum design moment= 0.784 k-ft

V) Reinforcement calculation

A. Temperature & shrinkage steel.

Use # 3 bar as Temperature steel.

Use # 3 bar @ 13.50” c/c.

B. Main bar

(-)ve Moment at Interior support = 0.784 k-ft

(-)ve A_s  for Int. support = ρbd = 0.0023×12×3=0.083 in²/ft

(+)ve Moment at mid span = 0.56 k-ft.

We provided # 3 as main steel

Spacing at mid span =

So spacing at mid span =

A detail of slab reinforcement arrangement of all panels (S-1 ~ S-3) is given in Table 6.1 and Figure 6.1.

Bar arrangement, cut-off and bent up, bar spacing etc. are done as per discussions and ACI/BNBC Codes provisions presented in Chapter III.

Design of floor beam:

All floor beams are analyzed and designed by following the ‘ACI Moment Coefficient procedure’ (Appendices I & XVI). Typical lay-out plan of all floor beams is given in Appendix-XI. Analysis and design of all beam groups (F. B – 1~F.B-6) are presented below.

Design data

Design procedure = ACI moment Co-efficient

Materials:

                                         =  60 ksi

= 3 ksi

= 150 pcf

= 120 pcf

Main wall thickness                = 5″

Floor beam F.B-1

Let depth of beam        =25″

Beam size                     =12″×25″

l) Load calculation:

Self weight of beam                 =

Main wall / Partition wall weight

Load coming from slabs

S-1                      DL      =111.25× 19′×0.5×0.81×2= 1712.13   lb / ft

Total dead load, DL      = 312.50+396+1712.13=2420.63 lb/ft

Live load, LL                = 40×19′-0″×0.5×0.81×2 = 615.60   lb / ft 

Factored load                 = 1.2DL+1.6LL = 1.2×2420.63+1.6×615.60 =3889.72 lb/ft

=3.89   k / ft

ll) Moment calculation & d check: 

(-) ve moment at Ext. support =

(+) ve moment at mid span =

(-) ve moment at Int. support =

_{T-beam check}

_{For interior beam}

center to center distance of the beam =20′-0″ =240″

will be the smallest of the above, = 81.5″

Let, a = slab thickness = 6.5″

And assume it is a singly beam.

So it is not a T-beam.

Considering double layers of steel, effective depth of the beam, d= 25″-4″= 21″

Check for‘d’

Maximum Moment = 287.16 k-ft

lll) Reinforcement calculation:

(+)ve steel at mid span

(- )ve Steel at exterior support

(- )ve Steel at Interior support

IV) Stirrup design:
Shear at 1st Interior support,

Critical shear at d distance,

Concrete shear strength,

Allowable strength,

Since, <  , So stirrup is required.

Shear carried by stirrup =71.85-20.71=51.14 k

Floor beam F.B-2

Let, beam width          =12″

Depth of beam=

Take beam size                        = 12″×20″

l) Load calculation:

Self weight of beam F.B-2     =

All main wall weight

Load coming from slab, S-1 = 111.25×19′-0″×0.81/2 = 856.06   lb / ft

Total dead load, DL             = 250+416.5+856.06 = 1522.56 lb / ft

Live load from slab, S-1      = 40×19′-0″x0.81/2 = 307.80   lb / ft

Total factored load              = 1.2DL+1.6LL

= 1.2×1522.56+1.6×307.80 = 2319.55 lb-ft =2.32   k / ft

ll) Moment calculation: 

(-)ve moment at Ext. support =       =

(+)ve moment at mid span     =       =

(-)ve moment at Int. support  =       =

III) T-beam& d check:

_{T-beam check}

Center to center beam distance=12+0.5×(19′-0″)×12=126″

will be the smallest of the above, =39.17″

Let, a= slab thickness = 6.5″ and assume it is a singly beam

So it is not a T-beam, and it will be designed as a singly beam

Effective depth of the beam, d =20-4 =16″ (considering double layer of steel)

Check for d

Maximum moment, =171.33 k-ft

lV) Reinforcement calculation:

(+)ve steel at mid span

(-)veSteel at Interior support

(-)ve Steel at exterior support

V) Stirrup design:
Shear at 1st Interior support,

Critical shear at ′d′ distance,

Concrete shear stress,

Allowable stress,

Since, <  , so stirrup is required.

Shear carried by stirrup =28.32-17.25=11.07 k

Floor beam F.B-3

Let, beam width   = 12″

Depth of beam     =15″

Size of beam        =12″×15″

l) Load calculation:

Self weight of beam F.B-3     =

All main wall weight

Load coming from slab, S-1 = 111.25×27′-2″×0.19/2 = 287.15   lb / ft

Total dead load, DL             = 187.50+437.50+287.15 = 912.15 lb / ft

Live load from slab, S-1      = 40×27′-2″×0.19/2 = 103.25   lb / ft

Total factored load              = 1.2DL+1.6LL

= 1.2×912.15+1.6×103.25= 1259.76 lb-ft =1.26   k / ft

ll) Moment calculation: 

(-)ve moment at Ext. support =       =

(+)ve moment at mid span     =       =

(-)ve moment at Int. support  =       =

_{III) T-beam& d check:}

_{T-beam check}

(Center to center beam distance) =12+0.5 × (27′-2″) ×12=175″

will be the smallest of the above, = 31″

Let, a = slab thickness = 6.5″

And assume it is a singly beam

So it is not a T-beam, it will be designed as a singly beam.

Now effective depth of the beam, d =15-2.5 = 12.5″ (considering one layer of steel).

Check for d

Maximum moment, =45.48 k-ft

lV) Reinforcement calculation:

(+)ve steel at mid span

(-)ve Steel at Interior support

(-)ve Steel at exterior support

V) Stirrup design:
Shear at 1st Interior support,

Critical shear at ′d′ distance,

Concrete shear stress,

Allowable stress,

Since, <  , so stirrup is required.

Shear carried by stirrup = 16.60-12.32 = 4.28 k

Floor beam F.B-4

Let, beam width   = 12″

Depth of beam     = 15″

Size of beam        = 12″×15″

l) Load calculation:

Self weight of beam F.B-4     =

Self weight of 5″ wall

Load coming from slab, S-1   = 111.25×27′-2″×0.19/2 = 287.15   lb / ft

Load coming from slab, S-3   = 80×7/2 = 280 lb/ft

Total dead load, DL               = 187.50+437.50+287.15+280 = 1192.15 lb / ft

Live load from slab, S-1         = 40×27′-2″x0.19/2 = 103.25   lb / ft

Live load from slab, S-3         =40×7/2=140 lb/ft

Total live load, LL                 = (103.25+140.00) =243.24 lb/ft

Total factored load                 = 1.2DL+1.6LL

= 1.2×1192.15+1.6×243.24= 1819.76 lb-ft =1.82 k / ft

ll) Moment calculation: 

(-)ve moment at Ext. support =       =

(+)ve moment at mid span     =       =

(-)ve moment at Int. support  =       =

_{Maximum moment,} = 65.70 k-ft

III) T-beam& d check:

_{T-beam check}

(Center to center beam distance)=12+0.5×(27′-2″)×12=175″

will be the smallest of the above, =31″

Let, a=slab thickness=6.5″ and assume it is a singly beam

So it is not a T-beam, and it will be designed as a singly beam

Now effective depth of the beam, d=15-2.5=12.5″ (considering one layer of steel)

Check for d

Maximum moment, =65.70 k-ft

lll) Reinforcement calculation :

(+)ve steel at mid span

(-)veSteel at Interior support

(-)veSteel at exterior support

IV) Stirrup design:
Shear at 1st Interior support,

Critical shear at d distance,

Concrete shear stress,

Allowable stress,

Since, <  , So stirrup is required.

Shear carried by stirrup =23.86-12.32=11.54 k  

Floor beam F.B-5

Let, beam width   = 12″

Depth of beam     =12″

Size of beam        =12″×12″

Effective depth, d =12-2.5=9.5″

l) Load calculation:

Self weight of beam F.B-5     =

Self weight of  5″ wall

Load coming from slab, S-1 = 111.25×27′-2″×0.19/2 = 287.15   lb / ft

= (150+437.50+287.15) = 874.65 lb/ft

Live load from slab, S-1       = 40×27′-2″x0.19/2 = 103.25   lb / ft

Factored load (B to C)          = 1.2DL+1.6LL

=1.2×874.65+1.6×103.24=1214.76 lb/ft

Dead load slab, S-3               = 0.55/2×92.50×6′-0″ = 152.62 lb / ft

Live load from slab, S-3      = 0.55/2×40×6′-0″ = 66 lb / ft

Factored load                                   = 1.2DL+1.6LL

=1.2×152.62+1.6×66 = 288.74 lb/ft

Sub total load (A to B)        =1214.76+288.74=1503.50 lb/ft=1.50 k/ft

Concerted load at point “D”

Dead load from stair beam =

Dead load from stair          =

Dead load from slab, S-3   =118×3                                =354.00 lb/ft

= (375.00+279.30+354.00) = 3528.30 lb

Live load from stair           =

Live load from slab, S-3    =51.03×3′-0″           =153.09 lb

= (1999.50+153.09)=2152.59 lb

Factored load                                 = 1.2DL+1.6LL

= 1.2×3528.30+1.6×2152.59= 7678.10 lb =7.68   kip

ll) Moment calculation & d check: (moment calculation by GRAPS software}

(-)ve moment at Int. support = 49.63 k-ft

(-)ve moment at Ext. support  =39.32 k-ft

(+)ve moment at mid span     =37.35 k-ft

(-)ve moment at point   “B”    = 34.53 k-ft

Check for d

Maximum moment, =49.63 k-ft

Ill) Reinforcement calculation:

(+)ve steel at mid span

(-)veSteel at Interior support

(-)veSteel at exterior support

IV) Stirrup design:

:∑

and
Shear at 1st Interior support,

Critical shear at d distance,

Concrete shear stress,

Allowable shear stress,

Since, <  , So stirrup is required.

Shear carried by stirrup =27.05-9.38=11.54 k

Floor beam F.B-6

Let, beam width   = 12″

Depth of beam     = So taken beam depth=10″

Size of beam        =12″×10″

Effective depth, d =10-2.5=7.5″

l) Load calculation:

Self weight of beam    FB-6        =

Load coming from slab, S-2       =

All main wall weight                  =

Total dead load, DL                 

Live load from slab, S-2

Total factored load                      = 1.2DL+1.6LL

ll) Moment calculation & d check: 

(-)ve moment at both. Supports =       =

(+)ve moment at mid span     =       =

Check for d

Maximum moment, = 4.18 k-ft

lll) Reinforcement calculation:

(+)ve steel at mid span

(-)veSteel at both support

IV) Stirrup design:
Shear at 1st Interior support,

Critical shear at d distance,

Concrete shear stress,

Allowable shear stress,

Since, > , so stirrup is not required.

Use 2 legs # 3 @ 3.5″ c/c throughout the beam length.

Floor beam F.B -7(Stair beam)

Let, beam width   = 12″

Depth of beam     =10″

Size of beam        =12″×10″

Effective depth, d =10-2.5 = 7.5″

l) Load calculation:

Self weight of beam                    =

Load coming from slab, S-2       =

Load coming from stair,             =

Total dead load, DL                 

Live load from slab, S-2

Live load from stair

Total live load, LL

Total factored load                      = 1.2DL+1.6LL

ll) Moment calculation & d check: 

(-)ve moment at both. Supports =       =

(+)ve moment at mid span     =       =

Check for d

Maximum moment, =14 k-ft

lll) Reinforcement calculation:

(+)ve steel at mid span

(-)veSteel at both support

IV) Stirrup design:
Shear at 1st Interior support,

Critical shear at d distance,

Concrete shear stress,

Allowable shear stress,

Since, <  , So stirrup is required.

Shear carried by stirrup  =8.71-7.38=1.33 k

Details of sectional dimensions and reinforcement arrangement of all floor beams (FB -1 ~ FB-7) are given in Table 6.2 and Figure 6.2~6.6.

Bar arrangement, cut-off and bent up, bar spacing etc. are done as per discussions and ACI/BNBC Codes provisions presented in Chapter III.

 Details of sectional dimensions and reinforcement arrangement of all floor beams (FB -1 ~ FB-7)

CHAPTER VII 

ESTIMATION & COST ANALYSYS 

General:

This chapter gives detailed estimation of volume of concrete and steel which are done separately both for Type-I and Type-II Structures. Finally, cost analyses are completed according to “schedule of rate for civil works”, 12^th edition, Public Works Department (PWD).

 Estimate of volume of concrete for Type-I structure.

Volumes of concrete works for different structural elements of the structure are estimated as below (A~F):

A. Volume of concrete of footings

 Footing base:

F-1                               = 4 x 59 x 69 x 1799                           = 170.00 cft

F-2                               = 10 x 69 x 79 x 2099                         = 701.40 cft

F-3                               = 4 x 89 x 99 x 2299                           = 527.04 cft

Volume of concrete of all footing bases                                 = 1398.44 cft

 Pedestal columns:

C-1                              = 4 x 1399 x 1399 x 39-799                = 16.81 cft

C-2                              = 10 x 1399 x 1599 x 39-499              = 45.09 cft

C-3                              = 4 x 1399 x 1799 x 39-999                = 19.46 cft

Volume of concrete of all pedestal columns                           = 181.36 cft

Total volume of concrete of all footings                                 = 1579.80 cft

B. Volume of concrete of grade beams

GB-1                           = 6 x 409 x 1099 x 1299                     = 200.00 cft

GB-2                           = 3 x 649 x 1099 x 1099                     = 132.27 cft

Volume of concrete of all grade beams                                  = 332.27 cft

C. Volume of concrete of columns

C-1                              = 4 x 1099 x 1099 x 89-899                = 24.08 cft

C-2                              = 10 x 1099 x 1299 x 89-899              = 72.25 cft

C-3                              = 4 x 1299 x 1499 x 89-899                = 40.45 cft

Volume of concrete of all columns                                         = 136.78 cft

D. Volume of concrete of floor beams

F.B-1 & F.B-3             = 4 x 409 x 1299 x 1199                     = 146.66  cft

F.B-2                           = 2 x 409 x 1299 x 1399                     =    86.67 cft

F.B-4 & F.B-5             = 3 x 649 x 1299 x799                        = 111.36  cft

Volume of concrete of all floor beams                                   = 344.69 cft

E. Volume of concrete of slabs

S-1                               = 2 x 149 x 229 x 599                         = 252.56 cft

S-2                               = 2 x 149 x 229 x 599                         = 252.56 cft

S-3                               = 2 x 149 x 189 x 599                         = 202.64 cft

S-4                               = 2 x 149 x 189 x 599                         = 202.64 cft

S-5                               = 89 x 189 x 599                                 =   59.04 cft

+ 89 x 99-699 x 599                            =   31.16 cft

Volume of concrete of all slabs                                              = 1000.6 cft

F. Volume of concrete of stair

Waist slab                    = 2 x 129-699 x 39-999 x 699            = 46.88 cft

Tread & Rise               = 2 x 0.5 x 1099 x 6’ x 39-999           = 14.04 cft

Volume of concrete of stair                                                    = 60.92 cft

Total volume of concrete (A to F) for ground floor           = 3455.06 cft

Total volume of concrete for typical floor                          = 1542.99 cft

 Estimate of volume of steel for Type-I structure.

Volumes of steel for different structural elements of the structure are estimated as below (A~F):

1. A.    Volume of steel of footings

I. Footing base:

Long direction: 

Short direction:     

Main bars

Tie bars

Volume of steel of all pedestal columns (main bars + ties) = 332.33 kg

Total volume of steel of all footings                                    = 1877.61 kg

B. Volume of steel of grade beams

Main steel:

Extra top bars:

Stirrups:

Volume of steel of all grade beams = 1750.81 kg

C. Volume of steel of columns

Main bars:

Tie bars:

Volume of steel of all columns = 667.04 kg

D. Volume of steel of floor beams

Main bars:

Extra top bars:

Stirrups:

Volume of steel of all floor beams = 2086.22 kg

E. Volume of steel of slab

Long direction:

Short direction:

Volume of steel of all slabs = 1442.51 kg

F. Volume of steel of stair