Beam Design With Excel (Coefficient Method)

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Short Description

Beam Design...

Description

Prepared by,

Md. Humayoun Kabir

Insert Your Data

B.Sc. in Civil Engineering

Compressive Strength of : f'c

3000

psi

Yielding Strength of Steel: fy

50000

psi

IUBAT- International University of Business A

Negative Reinforcement in End Support 0 in2 (USD) 0 in2 (WSD)

1.36

Unit Weight on the Beam: ω

kip/f

4

Support Condition:

13.5 in (USD) Total Span Number:

1

Design the Span Number:

1

Length of the Span :

24

L

21.5 in (WSD)

f CC

Let The width of the Beam :

13.5

in

Clear Cover:

2.5

in

Reduction Factor for Beam φ

0.9

Modulus of Elasticity (Steel) Es

2.5

in

Beam Rein 29000000

Main Bar

2

Ext. Top

1

Beam Design with Fixed Dimention

Amount of St. in USD M

1.904 kip/f

Unit Weight on the Beam: Assume Dimen.

12

24

X

For USD

Your Assumption is OK

For WSD

Your Assumption is Wrong

Main Bar

3

Ext. Bottom

2

Amount of St. in USD M

Result for fixed dimension of Beam Finally Dimension:

12

X

Refresh-1

Stirrup

24 Beam Length:

Effective Depth:

21.5

in

WSD Posit. Rein.: Neg. Rein.:

--

in

--

in2

2

USD 2.01 in2 0

in2 Provide: Ø for:

10 5 6

Calculations

Refresh-1 Refresh-2

8

1. Smax

76.17001 in

2. Smax

10.25 in

3. Smax

24 in

Calculations Modulus of Elasticity (Conc.) Ec 3122018.6

10

fc

1350 psi

fs

20000 psi

n=

Es/Ec

9.2888621

10

r=

fs/fc

14.814815

15

k=

n/(n+r)

0.4

0.4

j=

1-(k/3)

0.8666667

0.866667

R

Positive Moment:

+M

234

97.92 kip/f*f 97.92 kip/f*f

Negative Moment:

-M

0 kip/f*f

d

19.286415

t

22.406415 in

So, Effective Depth will be: d

M+ 1175040 psi MCal d

22.41 in

24 in

20.88

d t Finaly d

Reinforcement: Positive As

Use: M/fsjd

3.24668435 in*in

Extra top:

Use:

Negative As:

M/fsjd

0 in*in As +

Final Dimension of the Beam is:

13.5

X

24

As -

Final Dim.

Stirrup Design:

Strr Wide of the beam:

10

in

Effective depth of the beam:

20.5

in

Uniformly distributed load:

1.89

kips/f

Span Length:

Reduction Factor:

21

f

f'c

3600

psi

λ

1

ф

0.75

fyt

72500 psi

72.5 ksi

The Maximum shear occurs in the end of the span:

Vu

19.845 kips

Vu

16.61625 kips

At the Critical Section,

Shear force varies linearly zero at mid span

12

Vc

Hence

фVc

24600 lb

18.45

kips

The point in which web reinforcement is no longer needed:

0.7380952 f

Zone: 1

The required spacing of web reinforcement is for vertical stirrups is:

-133.732

Reinforcement for zone two:

1.8 f

Vu= 19.845 kips

1. Smax

35.444444 in

2. Smax

10.25 in

3. Smax

24 in

Vu= 16.61625 kips

фVc=

18.45 kips

24.6 0.8 f

Method-02 1.07

Maximum Shear V

19.845 kips

10.5 f

9.7

Shear at d distance: x

17.8227 kips

Now Vc

66 psi

Vc

13530 lb

Here,

x > Vc

13.53 kips

So, Follow Next Step

Spacing:

USD Design 1 1

Technique for Coefficien

1 1

1

12

24

12

2

16

14

9

9

14

3

16

14

10

11

16

5 1

3 2

1 1 0 0

Unrestrained

1

Spandrel:

2

0

11

9

9

11

3

0

11

10

11

16

2

24

14

9

9

14

3

16

14

10

11

16

Positive Moment: Negative Moment:

97.92 k-f 0 k-f

Maximum Reinforcement Minimum Reinforcement

Factored Moment:

1175.04 k-in 0 k-in

0.016256 Max M 1175.04 k-in 0.004 Max Rein 0.0162563

Mu 141.6220194 11.90051 in

11.91 in

Total Depth:

15.03

Finally the dimension is

a

b

a

16

1175.04

0.088260911 0.013434004

b

X

c

991011.67 -100780.848 ρ1 ρ2

13.5

0.0883 0.01344

c

0.0883 0.01344

991011.67 -100780.848 0.101694915 0

0 0.1017 0

0.1017 0

al University of Business Agriculture and Technology

Email: [email protected]

Result

V:16 13.5 in (USD) 13.5 in (WSD)

16 in (USD) 24 in (WSD)

Positive Reinforcement in Mid Span 2.45 in2 (USD) 3.25 in2 (WSD)

Beam Reinforcement (Negative) No. of #

5

bar

0.31

0.62 in2

No. of #

5

bar

0.31

0.31 in2

Total Area of Steel :

0.93

Amount of St. in USD Method:

0 in2 Required

WSD Method:

0 in2 Required

in2

Beam Reinforcement (Positive) No. of #

7

bar

0.6

1.8 in2

No. of #

8

bar

0.79

1.58 in2

Total Area of Steel : Amount of St. in USD Method:

2.45 in2 Required

WSD Method:

3.25 in2 Required

3.38

in2

of

1

Stirrup Design (Typical) 24

f

Span:

1

.

mm stirrup @ in c/c f

Provide: Ø for:

10 10 12

mm stirrup @ in c/c f

27 72

72 12 6

10 in

Area of Used Bar

0.62

in2

Strain of Concrete 0.62

Positive Moment Co-efficient:

0.13

Neg. Moment Co-Efficient:

0.00

Єu

Strain of Concrete

0.003

Design with Dimension 2

21.5 in

Wrong

Own Weight of column: Total Unit weight:

158.688 158.688 1904256 0 24.55206 M+ 24 in

M97.92

1.36 2.204 27.67206 28

10

24.88

20.88

10 No. of #

9

bar

1 in*in

2 No. of #

5

bar

0.31 in*in

d t 0 19.28642 19.28642 24.55206

d 24 24 28

20.88 20.88 24.88

Total Area of Steel:

10.62 in2

5.261538 in2 0 in2 X

24

Vu= Unit Load*( Total Span length/2)

Vu=Vu+(Vu/(Total beam length/2))*(Effective Depth in f)

24.6 kips

3

0.11

3

0.11

10

in

-133.732 -134 -133

Minimum:

-132.5

-133

10 in

New Sys.

S. 5

6.75 16

128

5.12

10

240

9.6

5.12 5 10

300

f

12

Technique for Coefficient 12

24

9

14

10

14

16 11

10

14

16

3 2

12 0.083333 24 0.041667

1 1

0

12

0

0

0 0

11

10

16

0

false

false

0 24

0

0 0

11

10

14

0

0.125

24

false

M+ M1175.04 1.36 2.204

β1

false

HM Max Re Min Re Max of pr Check D. 0 1175.04 0.0162563 0.004 0.016256 11.91 1175.04 0.0162563 0.004 0.016256 11.91 1904.256 0.0162563 0.004 0.016256 15.15 18.27 20 16.88

td ed

0.85

As+ 16 in

12.88 in

As-

20.88

13.5

991011.7 -100781 12.88

0.01344

+As

2.44944 in

20.88

2

a

b c 2604402 -264854 1904.256

ρ1 ρ2 0

-As

0.093909 0.007786

0.094 0.00779

0 in2 2604402 -264854 ρ1 ρ2

0.101695 0

0 0.1017 0

Єu

0.003

Design with Dimension

ght of column:

3.246684 3.246684 4.415632

300 plf 0.3 K-f 2.204 k/f

0 0 0

24

0.083333 0.041667

0

0

#DIV/0!

0

#DIV/0!

#DIV/0!

0

0.125

Simply

1 1 2

Pst. Row Neg. Row 0.01344 0 0.01344 0 0.01255 0

OK in 20.88 3.537594 in2 0 in2

21.5

0.016256 0.00779

0.00779

+As

2.00982 in2

0.016256 0.004

0.004

-As

0 in2

0

Theories For USD Method

Theories For WSD Method

Important Noti

This Microsof Excel spreadsheet will be able to design rcc beam in Co-Efficient m This sheet is following the ACI and BNBC code. This sheet is prepared by If any one wants to implement the result of this sheet in the practical field, it is hi

Md. Humayoun Kabir Bachelor of Science in Civil Engineering E-mail: [email protected]

IUBAT-International University of Business Agric

Watch Tutorials in YouTu

01. Column Design with Excel: https://www.youtube.com/watch?v=Jh

02. Beam Design with Excel: https://www.youtube.com/watch?v=kpb 03. Website: http://civil-school.blogspot.com/

tant Notice

cc beam in Co-Efficient method. et is prepared by the recent experience of the owner. the practical field, it is his/her own responsibility.

niversity of Business Agriculture and Technology.

ch Tutorials in YouTube

youtube.com/watch?v=Jh9uz1F-u3Q

utube.com/watch?v=kpbbzji8ZKs

m/

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