Beam Design- @ Design Office

8/10/2019 Beam Design- @ Design Office

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Design of BeamReference

bf

hf

bw

hf = Heigth of slab (mm)= 125

Span = 16 (input) (Continous, flange bea

Effe. Depth

Effe. Depth = Span / 16 Span ( mm 4000

(d) = 250 mm

say d = 275 Input

Cover to r/f

As Durability

8110-1 Exposure Condition = Mild

Concrete Grade = C25

8110-1

Table 3.3 Cover = 25 mm

As fire resistance

8110-1 Fire Duration (Hr) = 1.5 Input

Table-3.4 Beam = Continouse

Cover = 20 mm

Cover Provided = 25 mm

Overall Height of beam = d + Cover + Dia/2 (bottom r/f ) + Dia (link)

Overall Height of beam = 318 mm

h = 325 mm Input

As fire resistance

Fire Duration (Hr) = 1.5

"T" Beam

Height of Beam (h)

8110-I

Table-3.9

Bredth of Beam (bw)


2/59

8110-1 bw > 200 mm

Figure 3.3

Say bw = 225 mm Input

8110-1

3.4.1.5

bf = Web width + lz/5

= 737.4 mm

Self weight = bwx ( h-hf) x L x 24 kN

= 4.169 kN

= 1.042 kN/m

Load from Slab

Slab Weight (Dead) =

= 29.28 kN

= 7.32 kN/m

Impose load =

= 36.60 kN

= 9.15 kN/m

Wall weight = Factor x Wall thickness x Height x Length

= 29.524 kN

= 7.381 kN/m

Total dead load = 15.743 kN/m

Total impose load = 9.150 kN/m

DL = 15.743 kN/m DL = 15.743

LL = 9.15 kN/m LL = 9.15

Effective width of flange beamBeam (bf)

a. for "T" beam

Load Analysis - Both "T" Beams or "L" Beam

Computation of self weight and Other dead loads

Dead load of slab x ( Area of Slab or slabs /2 )

No. of panal

live load x ( Area of Slab or slabs /2 ) x Factor

panal


3/59

A B C

4.000 4.000

Moments

Case -1 Maximum loading for both beam

design load = 1.4 DL + 1.6 LL

36.68

A B

L = 4.0 L =

For A - B , X

Mx MB

RA WL2/8 RB

Moments 73.3604

Differents 0

0

Corre. Moments 73.3604

MB = 73.360 kNm

Then , Moment at B

RA = (WL*L/2-MB)/L

55.020 kN

Then ,0


4/59

= 1.500 m

Then, M max

M max = -41.265225 kNm

For B- C , X

Mx MB

RA WL2/8 RB

MB = 73.360 kNm

Then , Moment at B

0 = WBCLBC2/2+RALAB-WABLAB

2/2-RCLBC

Rc = 55.02 kN

Then ,0


5/59

Case -II AB - Maximum

BC - Minimum


36.68

A B

L = 4.000 L =

For A - B , X

Mx MB

RA WL2/8 RB

Moments 73.3604

Differents -41.8744

-20.9372


MB = 52.423 kNm

Then , Moment at B

RA = (WL*L/2-MB)/L

60.255 kN

Then ,0


6/59

X = RA/W

= 1.643 m

Then, M max

M max = -49.4901448 kNm

For B- C , X

Mx MB

RA WL2/8 RB

MB = 52.423 kNm

Then , Moment at B

0 = WBCLBC2

/2+RALAB-WABLAB2

/2-RCLBCRc = 18.38 kN

Then ,0


7/59

Case -III AB - Minimum

BC - Maximum


15.74

A BL = 4.000 L =

For A - B , X

Mx MB

RA WL2/8 RB

Moments 31.486

Differents 41.8744

20.9372


MB = 52.423 kNm

Then , Moment at B

RA = (WL*L/2-MB)/L

18.380 kN

Then ,0


8/59

M Max @ dMx/dx=0

0 = WX-RA

X = RA/W

= 1.168 m

Then, M max

M max = -10.7295862 kNm

For B- C , X

Mx MB

RA WL2/8 RB

MB = 52.423 kNmThen , Moment at B

0 = WBCLBC2/2+RALAB-WABLAB

2/2-RCLBC

Rc = 60.25 kN

Then ,0


9/59

Hogging @ B

M max,Hog = 73.36 kNm

Saging A-B Mmax,sag = -49.49 kNm

Saging B-C Mmax,sag = -49.49 kNm

-60

-40

-20

0

20

40

60

0.

000

1.

000

2.

000

3.

000

4.

000

5.

000

6.

000

Moments

Length of Beam


10/59

=

h

(input)

m)

Input)

Dia. Bottom r/f (mm) = 16 Input

Dia of links (mm) = 10 Input


11/59

lz = 0.7 x L Input

Ls = 3660 Input (length of Slab)

Lb = 4000 Input (length of Beam)

Dead load

of slab= 4 Input (From salb design)

Factor = 1 Input (From salb design)

No.of Panal = 1 Input (From salb design)

live load of

slab= 5 Input (From salb design)

Factor = 1

No.of Panal = 1

thickness = 200 Input

height = 3355 Input

length = 4000 Input

Factor = 11 Input

kN/m Input

kN/m Input

Factor x

x No. of


12/59

36.6802

C

4.00 Input

WL2/8 RC

73.3604

0

73.3604


13/59

X

WL2/8 RC

WAB = 36.68 KN/m

WBC = 36.68 KN/m

LAB = 4.0 m

LBC = 4.0 m

3.000 4.000 5.000 6.500 8.000

0.00 73.36

73.36 0.00 -41.27 0.00


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15.743

C

4.000 Input

WL2/8 RC

31.486

20.9372

52.4232

9.

000

10.

000


15/59

WL2/8 RC

356.736 45.32

107.564 13.665

7.8715 45.32

3.285 4.000 5.665 6.832 8.000

0.00 52.42

52.42 0.00 -10.7 0.00


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36.6802

C4.000 Input

WL2/8 RC

73.3604

-20.9372

52.4232

7.

000

8.

000

9.

000

10.

000


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X

WL2/8 RC

233.19 12.7

691.76 37.7

37.68

2.335 4.000 4.710 6.357 8.000

0.00 52.42

52.42 0.00 -49.5 0.00


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7.

000

8.

000

9.

000

10.

000


19/59

hf = 125

d = 275

Cover = 25

h = 325

Results


20/59

bw = 225

bf = 737.4

Gk = 15.74

Qk = 9.15


21/59

Reference

bf

hf

bw

hf = Heigth of slab (mm)= 125

Span = 20.8 (input) (Continous, flange be

Effe. Depth

Effe. Depth = Span / 20.8 Span ( m 4000

(d) = 192 mm

say d = 300 Input

Cover to r/f

As Durability


Concrete Grade = 25

8110-1

Table 3.3 Cover = 25 mm

As fire resistance



Cover = 20 mm


Overall Height of beam = d + Cover + Dia/2 (bottom r/f ) + Dia (link)


h = 350 mm Input

As fire resistance


8110-1 bw > 200 mm

Figure 3.3 Say bw = 225 mm Input

Desig

"L" Beam

Height of Beam (h)

8110-I

Table-3.9

Bredth of Beam (bw)


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8110-1 a. for "t" beam

3.4.1.5 bf = Web width + lz/10

= 505 mm

Self weight = bwx ( h-hf) x L x 24 kN

= 4.238 kN

= 1.177 kN/m

Load from Slab


= 28.80 kN

= 8.00 kN/m

Impose load =

= 25.20 kN

= 7.00 kN/m

Wall weight = Factor x Wall thickness x Height x Length

= 26.05554 kN

= 7.24 kN/m



Max Shear = 61.53 kN Support at

IstructE.Manual Required width of beam = 1000 v

1985 2d

3.7.3 = 159.97 mm

Provided > Required

Satisfied

Grade of Concrete C = 25

= 35.00 kNm

Consider full width of Beam,

Then, M/ bd2 = 1.73 N/mm

2

8110-1

3.4.4.3 As = k' fcubd2

+ As'

0.87fyz

Computation of reinforcem

For Hogging reinforcements at first interior support -

live load x ( Area of Slab or slabs /2 ) x Factor

panal



Dead load of slab x ( Area of Slab or slabs /2 )

No. of panal

Check Bredth of Beam (bw) for Shear


23/59

k = M/bd2fcu

= 0.069

k' = 0.156

Now, k < k'

No need Compression R/F

1. If Compression R/F is not Req.

k,

z = d 0.5 + (0.25-k/0.9) mm

= 274.85 mm But

0.95d =

Then ,z = 1.43

As = M/(0.87 fyz)

318.20 mm2

25 0 0.00

20 0 0.00

16 0 0.0012 2 226.29

10 1 78.57

304.86

2. If Compression R/F is Req.

k', z = d 0.5 + (0.25-k'/0.9) mm

= 233.07 mm But

0.95d =

Then ,z = 233.07

x = ( d - z ) /0.45 mm

148.74 mm

As' = (k-k') fcubd2/(0.87 fy(d-d'))

-439.53 mm2

As = k' fcubd2

+ As'

0.87 fyz

= 846.71 mm2

25 0 0.00

20 0 0.00

16 4 804.57

12 2 226.29

10 0 0.00

1030.86

= -27.00 kNm

= 27.00 kNm


For Sagging Movements at Mid spans

Hoggings

ectionDesign

HoggingsectionDesign

Use this V

Dont Us


24/59

Then,

M/ bd2 = 0.59 N/mm

2

k = M/bd2fcu

0.024

k' = 0.156

Now, k < k'



K,

z = d 0.5 + (0.25-k/0.9) mm

= 291.86 mm But

0.95d =

Then ,z = 285.00

As = M/(0.87 fyz)

236.72 mm2

25 0 0.00

20 0 0.0016 0 0.00

12 2 226.29

10 1 78.57

304.86


K', z = d 0.5 + (0.25-k'/0.9) mm

= 233.07 mm But

0.95d =

Then ,z = 233.07

x = ( d - z ) /0.45 mm

-514.59 mm

As' = (k-k') fcubd2/(0.95 fy(d-d'))

-1375.33 mm2

As = k' fcubd2

+ As'

0.87 fyz

= 0.00 mm2

25 0 0.00

20 0 0.00

16 0 0.00

12 0 0.00

10 0 0.00

0.00

for span

x = 148.74 mm

d = 1.5 mm

Saggingsection

Design

Saggingsection

Design

Use this V

Dont Us

Provide Shear reinforcements


25/59

Assume:- Reinforcement have been provided as bent up bars.

Shear resistance

vb = Asb( 0.95 fy)( cos + sin cot ) (( d-d')/sb)

Let,

Asb = 82

mm2 Input

= 45 Degree Input

= 60 Degree Input

sb = 200 mm Input

fy = 460 N/mm

2

Input

Then, vb = -135.74 kN

No.of bent bars = ((d-d')/Tan )/sb

= 3.91

Provide No.of bent bars = 4

Then provide shear links with shear resistance greater than Vb

8110-1 100 As = 0.298

Table3.8 bwd

Table3.8 Vc = 0.45 N/mm2 Input

and V

V = N/mm2 Input

0.5 Vc = 0.225 N/mm2

0.8 fcu = 4.00 N/mm2

Check V < 0.8 fcu

0 < 4.00

Check is Ok

Check


26/59

V < 0.5 Vc

0 > 0

Not Ok

8110-1

3.4.5.5 Spacing = h agg + 5mm

25 mm

Max. Span = 12.000

Effec.Depth

Table 3.9 Permisible Span = 26 Input

Effec.Depth

12.000 < 26

Check is Ok

Deflection check


27/59

Results

h

(input) hf =

m)

Input)

d =

Cover =

Dia. Bottom r/f (mm) = 16 Input


h =

bw =

n of Beam


28/59

lz = 0.7 x L Input

L = 4000 Input (length of Slab) bf =

L = 3600 Input (length of Beam)

Dead load

of slab= 4 Input (From salb design)



live load of

slab= 3.5 Input (From salb design)

Factor = 1

No.of Panal = 1

thickness = 175 Input

height = 3660 Input

length = 3600 Input

Factor = 11.3 Input

Gk =

34.181 Q k =

both Ends

nt percentage of critical sections

Rectangular

x No. of

Factor x


29/59

z < 0.95 d

1.425

z < 0.95 d

285

lue

e


30/59

z < 0.95 d

285

z < 0.95 d

285

lue

e


31/59


32/59


33/59

125

300

25

350

225


34/59

505

16.41

7.00


35/59


36/59


37/59


38/59


39/59

Reference

bf

bw

hf = Heigth of slab (mm)=

Span = 16 (input) (Continous

Effe. Depth

Effe. Depth = Span / 16 Span ( m

(d) = 344 mm

say d = 411

Satisfied

Cover to r/f

As Durability


Concrete Grade C = 25

8110-1Table 3.3 Cover = 25 mm

As fire resistance



Cover = 20 mm


Overall Height of beam = d + Cover + Dia/2 (bottom r/f ) + Dia


h = 450 mm Input

As fire resistance


8110-1 bw > 200 mm

Figure 3.3

Say bw = 225 mm Input

Design of

"T" Beam

Height of Beam (h)

8110-I

Table-3.9

Bredth of Beam (bw)


40/59

8110-1

3.4.1.5 bf = Web width + lz/5

= 645 mm

Self weight = bwx ( h-hf) x L x 24 kN= 8.910 kN

= 1.620 kN/m

Load from Slab


= 75.900 kN

= 13.800 kN/m

Impose load =

= 49.500 kN= 9.000 kN/m

Wall weight = Factor x Wall thickness x Height x Le

= 41.703 kN

= 7.582 kN/m

Then,



Max Shear = 90.00 kN



IstructE.Manual Required width of beam = 1000 v

1985 2d

3.7.3 = 155.91 mm

Provided > Required

Satisfied

Grade of Concrete C = 25

= 70.00 kNm



a. for "T" beam


Dead load of slab x ( Area of Slab or

No. of panal

live load x ( Area of Slab or slabs /

panal

Check Bredth of Beam (bw) for Shear

Computation of reinf

For Hogging reinforcements at first interior su


41/59

Then, M/ bd2 = 1.84 N/mm

2

8110-1

3.4.4.3 As = k' fcubd2

+ As'

0.87 fyz

k = M/(bd2fcu)

= 0.074

k' = 0.156

Now, k < k'



k,

z = d 0.5 + (0.25-k/0.9) mm

= 374.03 mm

Then ,z = 374.03

As = M/(0.87 fyz)

467.64 mm2

25 0

20 0

16 3

12 0

10 2


k', z = d 0.5 + (0.25-k'/0.9) mm

= 319.30 mm

Then ,z = 319.30

x = ( d - z ) /0.45 mm

203.78 mm

As' = (k-k') fcubd2/(0.87 fy(d-d'))

-541.48 mm2

As = k' fcubd2

+ As'

0.87 fyz

= 1159.99 mm2

25 0

20 0

16 4

12 2

10 0

Use

D

Hoggin

gsectionDesign

HoggingsectionD

esign

HoggingsectionDesign


42/59

= -175.00 kNm

= 175.00 kNm


Then,

M/ bd2 = 1.61 N/mm

2

k =M/bd

2

fcu0.064

k' = 0.156

Now, k < k'



K,

z = d 0.5 + (0.25-k/0.9) mm

= 379.20 mm

Then ,z = 379.20As = M/(0.87 fyz)

1153.17 mm2

32 -

25 -

20 4

16 -

12 -

10 -

0

2. If Compression R/F is Req.K', z = d 0.5 + (0.25-k'/0.9) mm

= 319.30 mm

Then ,z = 319.30

x = ( d - z ) /0.45 mm

203.78 mm

As' = (k-k') fcubd2/(0.87 fy(d-d'))

-603.45 mm2

As = k' fcu

bd2

+ As'

0.87 fyz

= 3,325.30 mm2

25 0

20 0

16 0

12 0

10 0

Use

D

Saggingsectio

nDesign

Saggingsection

Design

For Sagging Movements at Mid span


43/59

for span

x = 203.78 mm

d = 411 mm

Assume:- Reinforcement have been provided as bent up bars.

Shear resistance

vb = Asb( 0.95 fy)( cos + sin cot ) (( d-

Let,

Asb = 82

mm2

= 45 Degree

= 60 Degree

sb = 200 mm

fy = 460 N/mm2

Then, vb = 1010.36 kN

No.of bent bars = ((d-d')/Tan )/sb

= 5.64

Provide No.of bent bars = 6

Then provide shear links with shear resistance greater than Vb

Then provide shear links with shear resistance greate

100 As = 1.359

bwd

Table3.8 Vc = (0.79/rm)(100As/bwd)1/3

(400/d)1/4

Provide Shear reinforcements


44/59

100As = 1.359

8110-1 bwd

rm = 1.25

400/d = 0.97 but 400/d < 1

Hence = 0.97

fu/25 = 1.00

Then, Vc = 0.695 N/mm2

Vc+0.4 = 1.095 N/mm2

0.8 fcu = 4.000 N/mm2

Max shear

V = 1.266 N/mm2

0.5 Vc = #VALUE! N/mm2

0.8 fcu = 4.00 N/mm2

Check V < 0.8 fcu

1.2657665 < 4.00

Check is Ok

Check

V < 0.5 Vc

1.2657665 > 0

Not Ok

8110-1

3.4.5.5 Spacing = h agg + 5mm

25 mm

Max. Span = 13.382

Effec.Depth

Deflection check


45/59

Table 3.9 Permisible Span = 26

Effec.Depth

13.382 < 26

Check is Ok


46/59

hf

h

150 (input)

, flange beam)

5500 Input)

(link) Dia. Bottom r/f (mm) = 16 Input


Ls1

Ls2 Lb

Design of eam


47/59

lz = 0.7 x L Input

Ls (average) = 3000 Input (length of Slab)

Lb = 5500 Input (length of Beam)

Slab Weight = 3.6 Input (From salb design)

Finishing = 1 Input (From salb design)

Dead load

of slab= 4.6 Input (From salb design)



live load of

slab= 2 Input (From salb design)

Finishing = 1 Input (From salb design)Total Live = 3 Input (From salb design)



gth thickness = 200 Input

height = 3355 Input

length = 5500 Input

Factor = 11.3 Input

46.6032

Input

Support at both Ends

slabs /2 ) x Factor x

) x Factor x No. of

orcement percentage of critical sections

pport - Rectangular


48/59

But z < 0.95 d

0.95d = 390.45

0.00

0.00

603.43

0.00

157.14

760.57

But z < 0.95 d

0.95d = 390.45

0.00

0.00

804.57

226.29

0.00

1030.86

this Value

ont Use


49/59

167.063

But z < 0.95 d

0.95d = 390.45

0.00

0.00

1257.14

0.00

0.00

0.00

1257.14

But z < 0.95 d

0.95d = 390.45

0.00

0.00

0.00

0.00

0.00

0.00

this Value

ont Use

s


50/59


51/59

Input

Input


52/59

Input


53/59

hf = 150

d = 411

Cover = 25

h = 450

bw = 225

Results


54/59

bf = 645

Gk = 23.00

Q k = 9.00


55/59


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Beam Design- @ Design Office

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Transcript of Beam Design- @ Design Office