beam design- @ design office
TRANSCRIPT
-
8/10/2019 Beam Design- @ Design Office
1/59
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)
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
5/59
Case -II AB - Maximum
BC - Minimum
design load = 1.4 DL + 1.6 LL
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
Corre. Moments 52.4232
MB = 52.423 kNm
Then , Moment at B
RA = (WL*L/2-MB)/L
60.255 kN
Then ,0
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
7/59
Case -III AB - Minimum
BC - Maximum
design load = 1.4 DL + 1.6 LL
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
Corre. Moments 52.4232
MB = 52.423 kNm
Then , Moment at B
RA = (WL*L/2-MB)/L
18.380 kN
Then ,0
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
10/59
=
h
(input)
m)
Input)
Dia. Bottom r/f (mm) = 16 Input
Dia of links (mm) = 10 Input
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
12/59
36.6802
C
4.00 Input
WL2/8 RC
73.3604
0
73.3604
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
14/59
15.743
C
4.000 Input
WL2/8 RC
31.486
20.9372
52.4232
9.
000
10.
000
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
16/59
36.6802
C4.000 Input
WL2/8 RC
73.3604
-20.9372
52.4232
7.
000
8.
000
9.
000
10.
000
-
8/10/2019 Beam Design- @ Design Office
17/59
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
-
8/10/2019 Beam Design- @ Design Office
18/59
7.
000
8.
000
9.
000
10.
000
-
8/10/2019 Beam Design- @ Design Office
19/59
hf = 125
d = 275
Cover = 25
h = 325
Results
-
8/10/2019 Beam Design- @ Design Office
20/59
bw = 225
bf = 737.4
Gk = 15.74
Qk = 9.15
-
8/10/2019 Beam Design- @ Design Office
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
8110-1 Exposure Condition = Mild
Concrete Grade = 25
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 = 343 mm
h = 350 mm Input
As fire resistance
Fire Duration (Hr) = 1.5
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)
-
8/10/2019 Beam Design- @ Design Office
22/59
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
Slab Weight (Dead) =
= 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
Total dead load = 16.41 kN/m
Total impose load = 7.00 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
Effective width of flange beamBeam (bf)
Computation of self weight and Other dead loads
Dead load of slab x ( Area of Slab or slabs /2 )
No. of panal
Check Bredth of Beam (bw) for Shear
-
8/10/2019 Beam Design- @ Design Office
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
Consider full width of Beam,
For Sagging Movements at Mid spans
Hoggings
ectionDesign
HoggingsectionDesign
Use this V
Dont Us
-
8/10/2019 Beam Design- @ Design Office
24/59
Then,
M/ bd2 = 0.59 N/mm
2
k = M/bd2fcu
0.024
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
= 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
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
-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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
27/59
Results
h
(input) hf =
m)
Input)
d =
Cover =
Dia. Bottom r/f (mm) = 16 Input
Dia of links (mm) = 10 Input
h =
bw =
n of Beam
-
8/10/2019 Beam Design- @ Design Office
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)
Factor = 1 Input (From salb design)
No.of Panal = 1 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
-
8/10/2019 Beam Design- @ Design Office
29/59
z < 0.95 d
1.425
z < 0.95 d
285
lue
e
-
8/10/2019 Beam Design- @ Design Office
30/59
z < 0.95 d
285
z < 0.95 d
285
lue
e
-
8/10/2019 Beam Design- @ Design Office
31/59
-
8/10/2019 Beam Design- @ Design Office
32/59
-
8/10/2019 Beam Design- @ Design Office
33/59
125
300
25
350
225
-
8/10/2019 Beam Design- @ Design Office
34/59
505
16.41
7.00
-
8/10/2019 Beam Design- @ Design Office
35/59
-
8/10/2019 Beam Design- @ Design Office
36/59
-
8/10/2019 Beam Design- @ Design Office
37/59
-
8/10/2019 Beam Design- @ Design Office
38/59
-
8/10/2019 Beam Design- @ Design Office
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
8110-1 Exposure Condition = Mild
Concrete Grade C = 25
8110-1Table 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
Overall Height of beam = 450 mm
h = 450 mm Input
As fire resistance
Fire Duration (Hr) = 1.5
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)
-
8/10/2019 Beam Design- @ Design Office
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
Slab Weight (Dead) =
= 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,
Total dead load = 23.002 kN/m
Total impose load = 9.000 kN/m
Max Shear = 90.00 kN
Max Shear = 128.16 kN
Max Shear = 128.16 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
Consider full width of Beam,
Effective width of flange beamBeam (bf)
a. for "T" beam
Computation of self weight and Other dead loads
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
-
8/10/2019 Beam Design- @ Design Office
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'
No need Compression R/F
1. If Compression R/F is not Req.
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
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'))
-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
-
8/10/2019 Beam Design- @ Design Office
42/59
= -175.00 kNm
= 175.00 kNm
Consider full width of Beam,
Then,
M/ bd2 = 1.61 N/mm
2
k =M/bd
2
fcu0.064
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
= 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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
45/59
Table 3.9 Permisible Span = 26
Effec.Depth
13.382 < 26
Check is Ok
-
8/10/2019 Beam Design- @ Design Office
46/59
hf
h
150 (input)
, flange beam)
5500 Input)
(link) Dia. Bottom r/f (mm) = 16 Input
Dia of links (mm) = 6 Input
Ls1
Ls2 Lb
Design of eam
-
8/10/2019 Beam Design- @ Design Office
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)
Factor = 1 Input (From salb design)
No.of Panal = 2 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)
Factor = 1 Input (From salb design)
No.of Panal = 2 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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
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
-
8/10/2019 Beam Design- @ Design Office
50/59
-
8/10/2019 Beam Design- @ Design Office
51/59
Input
Input
-
8/10/2019 Beam Design- @ Design Office
52/59
Input
-
8/10/2019 Beam Design- @ Design Office
53/59
hf = 150
d = 411
Cover = 25
h = 450
bw = 225
Results
-
8/10/2019 Beam Design- @ Design Office
54/59
bf = 645
Gk = 23.00
Q k = 9.00
-
8/10/2019 Beam Design- @ Design Office
55/59
-
8/10/2019 Beam Design- @ Design Office
56/59
-
8/10/2019 Beam Design- @ Design Office
57/59
-
8/10/2019 Beam Design- @ Design Office
58/59
-
8/10/2019 Beam Design- @ Design Office
59/59