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7/30/2019 design of beam reinforced concrete
1/27
Design Criteria:
fc= 21 MPa
fy = 275 MPa
d = 40 mm
L = 7 000 mm
= 0.85
= 0.9
=0.85 600 600
=0.85 0.85600
75 600 75
= 0.078
= 0.75
= 0.75 0.07
= 0.084
Assume =
=
0.084
= 0.04
=
= 0.0475
= 0.86
Mu= 396.50 kNm
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7/30/2019 design of beam reinforced concrete
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Vu= 427.24 kN
Try 450 mm x 900 mm
d = h(Cc + 12 +steel bar)
d = 900(40 + 12 + 18 )
d = 830 mm
= 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 96.50kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4
6 = 07.88
Determine the number of steel bars:
=
= 5. 6
Allowable Shear:
=6
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7/30/2019 design of beam reinforced concrete
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=6
45080
= 285.27 kN
=
0.85 85.7
= 121.24 kN < 427.24 kN
=
=47.4
0.85 .4
= 381.40 kN
=
=
4
= 6.9
=6.9 7580
8.40= 6 0
Use maximum spacing:
. =
=80
Smax= 415 mm say 410 mm
=0.0
=
46 = 07.88
=0.0 07.8875
= 1221.66 mm say 1300 mm
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7/30/2019 design of beam reinforced concrete
4/27
Min. Ld =0.06
=0.06 675
= 594 mm
Use 594 mm development length
0.003
c
d-c
a
d-0.5a
b
d
d'
As
T
C
Es
T = C
Asfy = 0.85
5303.7 (275) = 0.85 (21) (450) a
a = 181.58 mm
=
= 213.62 mm
From strain diagram:
=0.00
66.8
=0.00
.6
= 0.0087
=
=75
00 000
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7/30/2019 design of beam reinforced concrete
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= 0.001375
Mu= 450.68 kNm
Vu= 442.72 kN
Try 450 mm x 900 mm
d = h(Cc + 12 +steel bar)
d = 900(40 + 12 + 18 )
d = 830 mm
= 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 450.68kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4
6 = 07.88
Determine the number of steel bars:
=
= 5. 6
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7/30/2019 design of beam reinforced concrete
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Allowable Shear:
=6
=6
45080
= 285.27 kN
=0.85 85.7
= 121.24 kN < 442.72 kN
=
=44.7
0.85 .4
= 399.61 kN
=
= 4
= 6.9
=6.9 7580
99.6= 0
Use maximum spacing:
. =
=80
Smax= 415 mm say 410 mm
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7/30/2019 design of beam reinforced concrete
7/27
=0.0
=
46
= 07.88
=0.0 07.8875
= 1221.66 mm say 1300 mm
Min. Ld =0.06
=0.06 675= 594 mm
Use 594 mm development length
0.003
c
d-c
a
d-0.5a
b
d
d'
As
T
C
Es
T = C
Asfy = 0.85
5303.7 (275) = 0.85 (21) (450) a
a = 181.58 mm
=
= 213.62 mm
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7/30/2019 design of beam reinforced concrete
8/27
From strain diagram:
=0.00
66.8 =
0.00.6
= 0.0087
=
=75
00 000
= 0.001375
Mu= 211.80 kNm
Try 450 mm x 900 mm
d = h(Cc + 10 +steel bar)
= 900(40 + 12 + 18)
= 830 mm
Mc = 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 211.80 kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
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7/30/2019 design of beam reinforced concrete
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4
6 = 07.88
Determine the number of steel bars:
=
= 5. 6
=0.0
Ab =
6 = 07.88
=0.0 07.8875
= 1221 mm say 1300 mm
Min. Ld =0.06
=0.06 675
= 594mm
Use 594 mm development length
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7/30/2019 design of beam reinforced concrete
10/27
Design Criteria:
fc= 21 MPa
fy = 275 MPa
d = 40 mm
L = 7 000 mm
= 0.85
= 0.9
=0.85 600 600
=0.85 0.85600
75 600 75
= 0.078
= 0.75
= 0.75 0.07
= 0.084
Assume =
=
0.084
= 0.04
=
=0.0475
= 0.86
Mu= 390.23 kNm
Vu= 386.98 kN
Try 450 mm x 900 mm
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7/30/2019 design of beam reinforced concrete
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d = h(Cc + 12 +steel bar)
d = 900(40 + 12 + 18 )
d = 830 mm
= 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 90.23kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4
6 = 07.88
Determine the number of steel bars:
=
= 5. 6
Allowable Shear:
=6
=6
45080
= 285.27 kN
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7/30/2019 design of beam reinforced concrete
12/27
=0.85 85.7
= 121.24 kN < 386.98 kN
=
=86.98
0.85 .4
= 333.98 kN
=
=4
= 6.9
=6.9 7580
.98
= 55 50
Use maximum spacing:
. =
=80
Smax= 415 mm say 410 mm
=0.0
=4
6 = 07.88
=0.0 07.8875
= 1221.66 mm say 1300 mm
Min. Ld =0.06
=0.06 675
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7/30/2019 design of beam reinforced concrete
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= 594 mm
Use 594 mm development length
0.003
c
d-c
a
d-0.5a
b
d
d'
As
T
C
Es
T = C
Asfy = 0.85
5303.7 (275) = 0.85 (21) (450) a
a = 181.58 mm
=
= 213.62 mm
From strain diagram:
=0.00
66.8
=0.00
.6
= 0.0087
=
= 7500 000
= 0.001375
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7/30/2019 design of beam reinforced concrete
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Mu= 580.44 kNm
Vu= 441.33 kN
Try 450 mm x 900 mm
d = h(Cc + 12 +steel bar)
d = 900(40 + 12 + 18 )
d = 830 mm
= 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 580.44 kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4
6 = 07.88
Determine the number of
steel bars:
=
= 5. 6
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7/30/2019 design of beam reinforced concrete
15/27
Allowable Shear:
=6
= 6 45080
= 285.27 kN
=0.85 85.7
= 121.24 kN < 441.33 kN
=
= 44.0.85
.4
= 398.02 kN
=
=4
= 6.9
=6.9 7580
98.0= 0
Use maximum spacing:
. =
=80
Smax= 415 mm say 410 mm
=0.0
=4
6 = 07.88
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7/30/2019 design of beam reinforced concrete
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=0.0 07.8875
= 1221.66 mm say 1300 mm
Min. Ld =0.06
=0.06 675
= 594 mm
Use 594 mm development length
0.003
c
d-c
a
d-0.5a
b
d
d'
As
T
C
Es
T = C
Asfy = 0.85
5303.7 (275) = 0.85 (21) (450) a
a = 181.58 mm
=
= 213.62 mm
From strain diagram:
=0.00
66.8
=0.00
.6
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7/30/2019 design of beam reinforced concrete
17/27
= 0.0087
=
=75
00 000
= 0.001375
Mu= 242.68 kNm
Try 450 mm x 900 mm
d = h(Cc + 10 +
steel bar)
= 900(40 + 12 + 18)
= 830 mm
Mc = 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 242.68 kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4 6 = 07.88
Determine the number of steel bars:
=
= 5. 6
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7/30/2019 design of beam reinforced concrete
18/27
=0.0
Ab =
6 = 07.88
=0.0 07.8875
= 1221 mm say 1300 mm
Min. Ld =0.06
=0.06 675
= 594mm
Use 594 mm development length
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7/30/2019 design of beam reinforced concrete
19/27
Design Criteria:
fc= 21 MPa
fy = 275 MPa
d = 40 mm
L = 7 000 mm
= 0.85
= 0.9
=0.85 600 600
=0.85 0.85600
75 600 75
= 0.078
= 0.75
= 0.75 0.07
= 0.084
Assume =
=
0.084
= 0.04
=
=0.0475
= 0.86
Mu= 715.64 kNm
Vu= 627.92 kN
Try 450 mm x 900 mm
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7/30/2019 design of beam reinforced concrete
20/27
d = h(Cc + 12 +steel bar)
d = 900(40 + 12 + 18 )
d = 830 mm
= 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 715.64 kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4
6 = 07.88
Determine the number of steel bars:
=
= 5. 6
Allowable Shear:
=6
=6
45080
= 285.27 kN
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7/30/2019 design of beam reinforced concrete
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=0.85 85.7
= 121.24 kN < 627.92 kN
=
=67.9
0.85 .4
= 617.52 kN
=
=4
= 6.9
=6.9 7580
67.5
= 84 80
Use maximum spacing:
. =
=80
Smax= 415 mm say 410 mm
=0.0
=4
6 = 07.88
=0.0 07.8875
= 1221.66 mm say 1300 mm
Min. Ld =0.06
=0.06 675
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7/30/2019 design of beam reinforced concrete
22/27
= 594 mm
Use 594 mm development length
0.003
c
d-c
a
d-0.5a
b
d
d'
As
T
C
Es
T = C
Asfy = 0.85
5303.7 (275) = 0.85 (21) (450) a
a = 181.58 mm
=
= 213.62 mm
From strain diagram:
=0.00
66.8
=0.00
.6
= 0.0087
=
= 7500 000
= 0.001375
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7/30/2019 design of beam reinforced concrete
23/27
Mu= 764.45 kNm
Vu= 641.87 kN
Try 450 mm x 900 mm
d = h(Cc + 12 +steel bar)
d = 900(40 + 12 + 18 )
d = 830 mm
= 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 764.45 kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4
6 = 07.88
Determine the number of steel bars:
=
= 5. 6
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7/30/2019 design of beam reinforced concrete
24/27
Allowable Shear:
=6
= 6 45080
= 285.27 kN
=0.85 85.7
= 121.24 kN < 641.87 kN
=
= 64.870.85
.4
= 633.87 kN
=
=4
= 6.9
=6.9 7580
6.87= 8 80
Use maximum spacing:
. =
=80
Smax= 415 mm say 410 mm
=0.0
=4
6 = 07.88
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7/30/2019 design of beam reinforced concrete
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=0.0 07.8875
= 1221.66 mm say 1300 mm
Min. Ld =0.06
=0.06 675
= 594 mm
Use 594 mm development length
0.003
c
d-c
a
d-0.5a
b
d
d'
As
T
C
Es
T = C
Asfy = 0.85
5303.7 (275) = 0.85 (21) (450) a
a = 181.58 mm
=
= 213.62 mm
From strain diagram:
=0.00
66.8
=0.00
.6
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7/30/2019 design of beam reinforced concrete
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= 0.0087
=
=75
00 000
= 0.001375
Mu= 370.02 kNm
Try 450 mm x 900 mm
d = h(Cc + 10 +
steel bar)
= 900(40 + 12 + 18)
= 830 mm
Mc = 0.59
Mc =0.9(21) (450) (830)2(0.186) [1-0.59(0.186)]
= 970.20 kNm 370.02 kNm
Compute for steel area required
As = bd = 0.04 45080
= 5303.7 mm2
Use 36 mm
4 6 = 07.88
Determine the number of steel bars:
=
= 5. 6
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7/30/2019 design of beam reinforced concrete
27/27
=0.0
Ab =
6 = 07.88
=0.0 07.8875
= 1221 mm say 1300 mm
Min. Ld =0.06
=0.06 675
= 594mm
Use 594 mm development length
