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Computer Aided Design Beam

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WELCOME TO THE COMPUTER AIDED DESIGN BEAM (CADB)
VERSION 1 (EVALUATION)

DESIGNED BY

KEVIN TANTSEVI

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING
CARNEGIE MELLON UNIVERSITY

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Start

COPYRIGHT 2000

Member :

Project's name

: CAE

Designer's name

: Kevin Tantisevi

b1

Types of Beam:

Date:

3 spans

DESIGN CRITERIA:
Concrete Strength ( f'c )

=

6,000

psi

Elastic Modulus of Concrete (Ec)

=

4.42E+06

psi

Reinforcing bar Yield Strength ( fyr )

=

40,000

psi

Stirrups Yield Strength (fys)

=

40,000

psi

PARAMETER FOR FLEXURAL DESIGN :

1

=

0.75

b

=

0.066

min.

=

0.005



=

0.0328

w

=

0.2184

kn

=

1141.4

PARAMETER FOR SHEAR DESIGN :
B*f'c(1/2)

=

1161.90

Maximum number of bars in a row:

6

Beam Width:

07/16/15

15 inches

Load Data

A Single span beam
P1DL
P1LL
P2DL
P2LL
DL
LL

Dead Weight of Beam =
total factored Point Load 1 =
total factored Point Load 2 =
total factored Uniform Load =

kips
kips
kips
kips
kips / ft.
kips / ft.
0.52
0.00
0.00
0.72

L
a
b

ft.
ft.
ft.

kips/ft
kips
kips
kips

VA =
VB =

#DIV/0! kips
#DIV/0! kips
Max. Shear =
#DIV/0! kips
Max. ultimate positive moment =
#DIV/0! kips.ft
Max. ultimate negative moment =
0 kips.ft
hmin =
0 inches
req' d =
#DIV/0! inches
Determine depth (d) and thickness (h)
Use
Beam Depth =
#DIV/0! inches
Check the beam weight by recalculating the load
Determine As
req' As =

=

#DIV/0! inches

……. Passed

#DIV/0! inches.2

Bar No.
5
6
0.31
0.44
Area / bar (in2)
No. of bars*
#DIV/0!
#DIV/0!
Min. number of 2 steel bars is required
Use the top bar =
2 No. 14 bar
Check Shear
Check if Bd(f'c)1/2 > Vmax /
The shear reinforcement is
#DIV/0!
Bar No.
3
4
0.22
0.4
Area (in2)
Spacing of bars
#DIV/0!
#DIV/0!
Use the spacing of

Thickness

#DIV/0!

#DIV/0!

7
0.6
#DIV/0!

8
0.79
#DIV/0!

9
1
#DIV/0!

Use the bottom bar size

5
0.62
#DIV/0!

10
1.27
#DIV/0!
=

11
1.56
#DIV/0!
4 No. 14 bar

14
2.25
#DIV/0!

Load Data
A two span continuous beam
PDLa1
PLLa1
PDLb1
PLLb1
DL1
LL1
PDLa2
PLLa2
PDLb2
PLLb2
DL2
LL2

kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips

Factored Load Case
factored Beam weight
Case
Description
1
Full DL
2
Half Left LL
3
Full LL
4
Half Right LL

L1
a1
b1

ft.
ft.
ft.

L2
a2
b2

ft.
ft.
ft.

/ ft.
/ ft.

/ ft.
/ ft.

Pa1
0
0
0
0

0.72 kips / ft.
Pb1
0
0
0
0

UL1
0
0
0
0

Pa2
0
0
0
0

Pb2
0
0
0
0

UL2
0
0
0
0

Result of end moment
case
1
2
3
4
max. absolute

A

BB

C

0.00

727.01 -727.01

0.00

0.00

283.33 -283.33

0.00

0.00

850.00 -850.00

0.00

0.00

566.67 -566.67

0.00

0.00

1578.00 1578.00

0.00

Result of midspan moment and shear
case
1
2
3
4
max. absolute

VAR

MAB

VBL VBR

MBC

#DIV/0!

#DIV/0!

#DIV/0! #DIV/0!

#DIV/0!

#DIV/0!

VCL

#DIV/0!

#DIV/0!

#DIV/0! #DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0! #DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0! #DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0! #DIV/0!

#DIV/0!

#DIV/0!

Max. Shear =
#DIV/0! kips
Max.positive moment =
#DIV/0! kips.ft
Max. negative moment =
1578.00 kips.ft
hmin =
0 inches
req' d =
#DIV/0! inches
Determine depth (d) and thickness (h)
Use
Beam Depth =
#DIV/0! inches

Thickness

Check the beam weight by recalculating the load
Determine As
req' As+ =

#DIV/0! inches.2

=

#DIV/0! inches

……. Passed

req' As- =

#DIV/0! inches.2

Bar No.

5

6

7

8

9

10

11

14

Area / bar (in2)

0.31

0.44

0.6

0.79

1

1.27

1.56

2.25

No. of bottom bars

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

No. of top bars

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

#DIV/0!

Min. number of 2 steel bars is required

Use top bar =
7
Check Shear
Check if Bd(f'c)1/2 > Vmax /
The shear reinforcement is

No. 14 bar

Use Bottom bar

#DIV/0!

Bar No.

3

4

5

Area (in2)

0.22

0.4

0.62

Spacing of bars

#DIV/0!

#DIV/0!

#DIV/0!

Use the spacing of

#DIV/0!

#DIV/0!

=

5 No. 11 bar

Load Data
A three span continuous beam
PDLa1
PLLa1
PDLb1
PLLb1
DL1
LL1
PDLa2
PLLa2
PDLb2
PLLb2
DL2
LL2
PDLa3
PLLa3
PDLb3
PLLb3
DL3
LL3

100
100
0
0
5
5
0
0
0
0
5
5
0
0
100
200
5
5

kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips
kips

Factored Load Case
factored Beam weight
Case
Description
1
Full DL
2
1 and 3 LL
3
1 and 2 LL
4
2 LL
5
2 and 3 LL

L1
a1
b1

15 ft.
10 ft.
0 ft.

L2
a2
b2

15 ft.
0 ft.
0 ft.

L3
a3
b3

12 ft.
0 ft.
4 ft.

/ ft.
/ ft.

/ ft.
/ ft.

/ ft.
/ ft.

Pa1
140
170
170
0
0

0.5 kips / ft.
Pb1
0
0
0
0
0

UL1
7
8.5
8.5
0
0

Pa2
0
0
0
0
0

Pb2
0
0
0
0
0

UL2
7
0
8.5
8.5
8.5

Pa3
0
0
0
0
0

Pb3
140
340
0
0
340

Result of end moment
case
1
2
3
4
5

A

max. absolute

B B

C C

0.00

357.35 -357.35

251.03 -251.03

0.00

D

0.00

278.82 -278.82

415.39 -415.39

0.00

0.00

489.70 -489.70

0.00

91.07 -91.07

-2.72 2.72

0.00

91.07 -91.07

0.00

0.00

-28.74 28.74

600.26 -600.26

0.00

0.00

848.00 848.00

852.00 852.00

0.00

Result of midspan moment and shear
VAR

case
1
2
3
4
5

VBL VBR

MBC

VCL VCR

MCD

-43.65

49.05 159.17

-186.45

70.66

9.10 312.28

-381.88

129.72

173.30 63.23

101.83

-133.70

195.67 -9.10

55.39

87.77

-119.64

209.73 96.58

-159.74

30.92 -0.23

-0.23

0.23

-6.07

6.07

6.07 63.75

-147.23

63.75 7.59

7.59

-7.59

181.00

-1.92

-1.92 21.82

-79.34

105.68 327.69

-366.48

114.31

241.00

384.00 160.00

204.00

155.00 487.00

569.00

201.00

384.00 kips
569.00 kips.ft
852.00 kips.ft
10 inches
27 inches

Determine depth (d) and thickness (h)
Use
Beam Depth =
27 inches

Thickness

Check the beam weight by recalculating the load
Determine As
req' As =

=

29 inches

……. Passed

8 inches.2

req' As- =

11 inches.2

Bar No.

5

6

7

8

9

10

11

14

Area / bar (in2)

0.31

0.44

0.6

0.79

1

1.27

1.56

2.25

No. of bottom bars

26

19

14

11

8

7

6

4

No. of top bars

36

25

19

14

11

9

8

5

Min. number of 2 steel bars is required

Use top bar =
5
Check Shear
1/2
Check if Bd(f'c) > Vmax /
The shear reinforcement is

No. 14 bar

Use Bottom bar

3

4

5

Area (in2)

0.22

0.4

0.62

Spacing of bars

11

13

13

11

=

4 No. 14 bar

not necessarily required but using the min. spacing stirrups instead

Bar No.

Use the spacing of

VDL

-107.06

1.92

max. absolute

Max. Shear =
Max.positive moment =
Max. negative moment =
hmin =
req' d =

MAB

78.98

inches. of No. 3 bar c/c

1

A

2

1
15 ft.

5 No. 14 bar

11 inches. of No. 3 bar c/c
4 No. 14 bar

15

C

15 ft.

Section 1-1, 2-2, 3-3

29

2

B

3

Drawing details for b1

3

D
12 ft.

1

A

B

1

1

A

B

1

1

A

1

2

B

2

C

A

1

B

2

C

B

2

2

2

2

C

3

C

3

D

2

C

3

D

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