Mbd
13
= 24 kN/m 3 Brick = 2.6 = 2.7 kN/m 2 = 0 75 mm = 1.8 kN/m 2 0 4.5 kN/m 2 = 0.25 kN/m 2 = 0.25 kN/m 2 = 1.2 kN/m 2 = 1.2 kN/m 2 2.9 kN/m 2 = 4 kN/m 2 = 4 kN/m 2 3. Calculate Simply supported = wl 2 / 8 = wl/2 Brick Wall Imposed Load L Loading Data : Screed 1. Dead load 2. Live load Topping Slab self weight Ρ conc 205 H Core Slab M & E Ceiling Blockwall Moment Reaction INVERTED T-BEAM & L-BEAM DESIGN Slab Service Dead Load Imposed Load R kN/m 2 Floor Level m kN/m
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Transcript of Mbd
= 24 kN/m3Brick = 2.6
= 2.7 kN/m2= 0
75 mm = 1.8 kN/m20
4.5 kN/m2
= 0.25 kN/m2
= 0.25 kN/m2
= 1.2 kN/m2
= 1.2 kN/m2
2.9 kN/m2
= 4 kN/m2
= 4 kN/m2
3. Calculate
Simply supported
= wl2 / 8
= wl/2
Brick Wall
Imposed Load L
Loading Data :
Screed
1. Dead load
2. Live load
Topping
Slab self weight
Ρconc
205 H Core Slab
M & E
Ceiling
Blockwall
Moment
Reaction
INVERTED T-BEAM & L-BEAM DESIGN
Slab Service Dead Load
Imposed Load R
kN/m2
Floor Level m
kN/m
Beam Details
h = 600 mm Ømain = 32 mm
b = 650 mm Ølink = 10 mm
h1 = 280 mm Øu-tie = 10 mm
h3 = 280 mm Cover, C = 40 mm
b1 = 125 mm d = 534 mm
b3 = 125 mm fy = 460
h2 = 320 mm fyv = 250
h4 = 320 mm fcu = 40
b2 = 400 mm Ρconc = 24
bv = 525 βb = 1
bvd = mm2
L = 9 m
N/mm2
kN/m3
280350
h
h2
Beam Dimensions :
SECTION 2
N/mm2
N/mm2
h1 h3
h4
b1
SECTION 1
b2 b3
b
= 6 m
= 6 m
= 4.5 kN/m2
= 2.9 kN/m2
= 4.99 kN/m = 0
= 2.69 kN/m
= 52.1 (inclusive self weight)
= 59.8 (lower section)
= 4 kN/m2
= 4 kN/m2
= 24 kN/m
= 111
= 122
= 0.5 m equal 500 mm
= 104 kN
Moments,M :
= kNm
= kNm
kN/m
kN/m Section 1
Section 2
Section 1
Section 2
sagging
Total DL
column width
70.4
0
hcs R
hcs L
Others load data
Reaction and Moments
Imposed Load L
Slab
hcs & top(selfweight)
Service DL
Beam
Total LL
Imposed Load R
thus, W
hogging
2. Live Load
Reaction,V
1. Dead Load
kN/m
kN/m
Brick on Beam kN/m
K = M/bd2fcu BS 8110 : Part 1: 1997
= k' = Cl. 3.4.4.4
z = d {0.5 + √ (0.25 - K/0.9)}
=
< 0.95
x = (d-z)/ 0.45
= 59.3 mm
thus, z = 0.95 d
= 507 mm
As = M/0.95fyZ
= 317.6 mm2
Therefore use;
3 16 = mm2
ok
not ok
Min Reinforcement
0.2%bh = 480 mm2
Therefore use;
3 16 = 603 mm2 at top beam.
ok
not ok
nos. of T
> 0.95 use z = 0.95
0.009
0.99
no compression reinf. Required
Design for sagging reinf.
compression reinf. Required
0.156
nos. of T use min. steel
Shear BS 8110 : Part 1: 1997
Cl. 3.4.5.2
V = 104 kN
v = V/bvd
= 0.37 kN/m2
= 0.79 (100 As/bd)1/3 (400/d)1/4 / λm
= 0.11 Note:
= 0.75 1) - minimum links
= 1.6 2)
= 1.25 3)
= 0.46 N/mm2
Link, R sv Asv /sv Link, R sv Asv /sv Link, R sv Asv /sv
= 5.06 N/mm2 6 50 1.13 8 50 2.01 10 50 3.14
or 57 75 0.75 101 75 1.34 157 75 2.1
5 N/mm2 mm2 100 0.57 mm2 100 1.01 mm2 100 1.57
125 0.45 125 0.8 125 1.26
= 0.86 150 0.38 150 0.67 150 1.05
175 0.32 175 0.57 175 0.9
thus 200 0.28 200 0.5 200 0.79
= 0.48 225 0.25 225 0.45 225 0.7
250 0.23 250 0.4 250 0.63
= mm2
Link, R sv Asv /sv Link, R sv Asv /sv Link, R sv Asv /sv
Lower link 6 50 2.26 8 50 4.02 10 50 6.29
V = 61 kN 113 75 1.51 201 75 2.68 314 75 4.19
v = V/bd mm2 100 1.13 mm2 100 2.01 mm2 100 3.14
= 0.29 kN/m2 125 0.91 125 1.61 125 2.51
150 0.75 150 1.34 150 2.1
thus 175 0.65 175 1.15 175 1.8
= 0.88 200 0.57 200 1.01 200 1.57
225 0.5 225 0.89 225 1.4
= mm2 250 0.45 250 0.8 250 1.26thus use R10 @
150mm c/c
157.00
Asv /sv
USE equation no.2
- Asv ≥ 0.4 bvsv / 0.95fyv0.5 vc < v < vc + 0.4
v < 0.5 vc
- Asv ≥ bvsv (v-vc) / 0.95fyvvc + 0.4 < v < 0.8√fcu
100 As /bvd
400 / d
λm
fcu / 25
R-link
2R- link
157.00thus use R10@
250m c/c
USE equation no.2
shear capacity, vc
Asv /sv
vc
0.8√fcu
vc + 0.4
Deflection
bw = 400 BS 8110 : Part 1: 1997
b = 650 Cl. 3.4.6.3bw/b = 0.62 > 0.3 Table 3.9
=
= 20
= 9 m
= 0.5 m
= 0.5 m
=
120
= 0.38
= 244
= 2.07
=
= 0.00
= 1
= x MFt x MFc
= 41.4
= 16.9 < 41.4
(span/d)basic Table 3.14
Rectangular Section
Clear span
Width of support :
R
L
Modification Factor for
Tension Reinforcement,
MFt
0.55 +(477-f s )
(0.9+M/bd2)
M/bd2
f s N/mm2
MFt
Modification Factor for
Compression
Reinforcement, MFc
Refer Table 3.11
100As'prov/bd
MFc
Allowable Ratio (span/d)basic
(span/d)actual DEFLECTION PASS!
Equation 2 0.48
Equation 3 -0.19
Equation 2 0.88
Equation 3 -0.4
Copy and paste
Copy and paste
K =
K' =
K > K' compression reinforcement required
d' = 66d = 534
z == 0.78 d
thus z = 415
x = (d-z)/ 0.45 = 0.5 d=
d'/x = 0.25 < 0.45
ok
d'/x exceed 0.37, the compression stress less than0.95fy., to obtain the value in fig 2.2 of BS 8100
As' =
= -5311 mm2
2nos. of T 16 = mm2
oknot ok
As =
= 1069 mm2
7nos. of T 32 = 5630 mm2
oknot ok
Min Reinforcement
0.2%bh = 416 mm2
Therefore use; ok
2 16 = 402 mm2 at top beam. not ok
0.156
0.009
Design for sagging reinf.
nos. of T
0 DATE
Use min steel
267.00
April-14
PAGE
4 TEL : 03-9057 0566, FAX: 03-9057 8566
EMAIL: [email protected]
0 BEAM MARK 0
CLIENT 0 MADE BY 0
(K-K')fcubd2 /0.95 fy(d-d')
K'fcubd2/095fyz + As'
PROJECT
ZASYAZ ENGINEERS SDN BHD (No. 769568-X)
No. 7-2, JALAN 2/149B, GATEWAY AVENUE, 57000
KUALA LUMPUR CHECKED BY
d{0.5 + √ (0.25 - K'/0.9)}
