Design Problem for the Continuous Beam for Concrete Technology-2 Design data:N = Registration no x 1.5 (Registration no = 35 then N = 1.5 x 35 = 52.5)

Design problem on the continuous reinforced beams: (25-Marks)

Design the following beams using ACI moment Co-efficient method.

Data Calculation:

Registration no = 35N = 1.5 x Registra No 52.5Column size Short dimension

376.25short beam span, Lx = 4.2long beam span, Ly = 5.8Cantilever Length = 903fc' = 25.1Steel Grade, fy = 420R= Lx/Ly 0.724137931034483Calculation of loads:Check of type of slab 1.38095238095238thickness of slab ( one way ) from front table 0thickness of slab ( two way ) for 420 P/165 and P/180 for 280 steel 120self weight of Slab 288Brick Ballast 135Floor Finish 138Any Partition or immovable load 0Total Dead load of slab 561live load 300Factored load of slab 11.4

Beam Dimensions and Loadbw (L/20 to L/15), mm 350

No = -700

M = N - No = 52.5 – ( -700) = 752.5

A roof system consists of M/2 x M/2 mm columns at spacing of M/130 m and M/180 in two mutually perpendicular directions. Beams run in both directions over the columns having more than two spans in each direction. Beams also supports a cantilever portion in both directions having projection of approximately 1.2xM mm. Building is to be used for the residential purpose. fc’ = M/30 MPa, Grade 420 steel, and live load to dead load ratio is less than 3.

a)      Continuous Interior short beam ( for N = 9,12,18,24,26,28,30,33,31)b)      Continuous Exterior short beam ( for N = 2,4,6,14,21,23,27,35)c)      Continuous Interior long beam ( for N = 7,10,17,19,20,25,29,39)d)      Continuous Exterior long beam ( for N = 3,8,11,13,22,32,34,36)

h, Depth, (L/12) 490cover, mm 75d , mm 415 Self weight of beam 4.037796Factored Self weight of beam W1 4.8453552skip this step is slab is one wayEquivalent Width Of Slab Supported By Beam For Two Way Slabs in metersfor Exterior Short beam

Lx/3 + Cantilever Length,m 2.303

skip this step is slab is two wayEquivalent Width Of Slab Supported By Beam For One Way Slabsfor Exterior beamLx/2+ Cantilever Length

Factored slab load on beam , Eq. Width * Factored slab load W2 30.04968156956

34.8950367695601

5.42375

Our Case is Figure (d) because beams are resting on columns and Assume fixity is providedFactors

16

14

10

11

16

Governing MomentCheck for depth of Beam

one end cont-dmin from table = L/18.5Dimension in mm 310

dmin for singly reinforced beams

dmin for singly RC beam230

Total Load on Beam W1+W2, wu

Clear Span=L=Ln

Maximum Capacity As Singly Reinforced Rectangular Section At Support

=

Effective Flange Width For T-Beam Behavior

Our case is T-beam here because beam is interioir

Effective Flange Width For T-Beam Behavior is lesser of , bf

1 0.4833333333333332 1.073 0.3521

bf 0.3521

Minimum Area, Amin

0.00333333333333333

580

Design for Moments:

Design For Positive Moment In Mid Spans

For Exterior Mid-Spanlet a = hf (mm) 120Mu+ (kN-m) = 73.3221094366744As (mm2) = 546.405167573399

As1= ρmaxbd

ρmin= 1.4 / fy

As,min

For positive moment, the flange of the T-beam will be in compression. However, it is more likely that for this smaller moment the N.A. will lie within the flange. The beam will act like a rectangular section of dimensions 1450 x 490 mm

Area of steel can not be calculated by using tables or curves as the steel ratio for the full section is much less.

a (mm)= 30.5495921163005As (mm2) = 485.267960528141a (mm)= 27.1314020090271As (mm2) = 483.201946056776

For Interior Mid-Spanlet a = hf (mm) 120Mu+ (kN-m) = 64.1568457570901As (mm2) = 478.104521626724a (mm)= 26.7308931017629As (mm2) = 422.590894433561a (mm)= 23.6271181591153As (mm2) = 420.964318120455

Design For Negative Moment at SupportsThe T-beam will act like a rectangular section of dimensions 350×490, because the flange comes under tensionDirect formula of steel ratio can be usedDesign For Negative Moment At Exterior Support

Mu- (kn-m) 64.1568457570901ρ 0.0028979584812094A- (mm2) = 420.928469395665

Direct formula of steel ratio can be usedDesign For Negative Moment At first Interior Support

Mu- (kn-m) 102.650953211344ρ 0.00472414390997181A- (mm2) = 686.181902923406

Direct formula of steel ratio can be usedDesign For Negative Moment At Interior Supports

Mu- (kn-m) 93.3190483739492ρ 0.00427484924273454A- (mm2) = 620.921852507192

Calculation of number of bars and bars dameter

As 483.201946056776Positive steel at exterior midspan bar # Area

16 199As 420.964318120455Positive steel at interior midspanbar# Area

16 199As 420.928469395665Negative steel at Exterior supportbar # Area

16 199As 686.181902923406Negative Steel At first Interior Support

bar # Area 22 387

As 620.921852507192Negative Steel At other Interior Supports

bar # Area 22 387

short direction xlong direction y

Brick Ballast Thickness 75 mmFloor Finish Thickness 60 mmDead Load Factor 1.2Live Load Factor 1.6Density of Concrete 2400 kg/m3

No -700M 752.5long Dimension

376.25 mmm (Lx)n (m) = 3.82375m (Ly)n (m) = 5.42375mm 0.903 mMpaMpa

Two way0

mm hfkg/m2kg/m2kg/m2kg/m2kg/m2kg/m2kN/m2

L = Span of Int or Ext Beam

Design of Continuous Beams Excel Sheet prepared by:Muhammad Riaz AhmadMNS UET, Multan

IF anyone found mistake in this design then please infrom at riazahmad140@gmail.com

Let we are desiging Interior long beam

kN/mkN/m

for Interior Short beam for Exterior long beam for Interior Long beam

2Lx/3 2.8 2.635937

our case

for Interior beamLx

ACI moment and shear coefficients. kN/m

kN/m

m

Moments kN-m64.1568457570901

73.3221094366744

102.650953211344

93.3190483739492

64.1568457570901

102.650953211344

both end cont-L/21

270 mm

mm

(1 –R2/3)Lx

0.0123365646258503

1791.88601190476 mm2100.785714285714 mm

Effective width (b) 246.960417989732 kN-m

Effective width (b) will be minimum of the following:

Our case is T-beam here because beam is interioir

x (clear spacing of beams (Si) on both sides) = c/c spacing for beams at regular interval

mm Effective width (b) will be minimum of the following: m

m

Note: Only above discussion is different for isolated (pre- cast) T or L beam. Other discussion is same (analysis and design formula).

mm2

    T-Beams

1.             L/4

2.             16hf+ bw

3.             bw+

  L-Beams

1.         L/12

2.         6hf+ bw

3.         bw+ Sc/2 on one side

Where Sc is the clear distance to the next web.

design as rectangular beam as b x h

The T-beam will act like a rectangular section of dimensions 350×490, because the flange comes under

Steel Placement at negative and moment

Total Ara of steel no of bars483.201946056776 3

Total Ara of steel no of bars420.964318120455 3

Total Ara of steel no of bars420.928469395665 3

Total Ara of steel no of bars686.181902923406 2

Total Ara of steel no of bars620.921852507192 2

Design of Continuous Beams Excel Sheet prepared by:Muhammad Riaz AhmadMNS UET, Multan

IF anyone found mistake in this design then please infrom at riazahmad140@gmail.com

3.46587395957

Effective width (b) will be minimum of the following:

x (clear spacing of beams (Si) on both sides) = c/c spacing for beams at regular interval

Effective width (b) will be minimum of the following:

Note: Only above discussion is different for isolated (pre- cast) T or L beam. Other discussion is same (analysis and design formula).

/2 on one side

is the clear distance to the next web.

Steel Placement at negative and moment

Note: Only above discussion is different for isolated (pre- cast) T or L beam. Other discussion is same (analysis and design formula).