Solved example 4.2 The items 1 to 21 below constitutes the Navigation chart The details about how to use the Navigation chart is given 1 Preliminary width b 2 Preliminary Total depth D
#2
3 Concrete cover 4 Grade of concrete
#4
5 Preliminary effective depth 6 Preliminary effective span
#6
7 Checks on preliminary dimensions 8 Total factored load
#8
#3 #5
#7 #9
9 Maximum factored moment 10 Effective depth required
#10
11 Depth of NA, xu 12 Ast required
#12
13 Ast provided 14 Final effective depth
#14
15 Final effective span
#16
Final checks: 16 Spacing of bars
#17
#11 #13
#15
17 Minimum area of steel 18 Maximum area of steel
#18
19 Check for pt
#20
21 MuR of final section
#22
#19
#21
A rectangular reinforced concrete beam, is simply supported on two walls. The thickness of both the walls is 250 mm and the centre to centre distance between the walls is 4.25m. The beam has to carry a uniformly distributed live load of 7kN/m and dead load of 30kN/m. Design the beam section for maximum moment at midspan. Assume Fe415 steel and moderate exposure conditions Solution: Given:
here
Grade of steel
415 N/mm2
fy
#1
1
#2
2
#3
3
#4
4
#5
5
#6
6
#7
7
#8
8
Fixing up preliminary dimensions: Let width of the beam
b
=
250 mm
=
423
=
450 mm
Let D be rounded off from 'span/12' = 4230/10 = Rounding off to the next higher multiple of 50, we get
D
Concrete cover: The nominal cover Cc required to the main reinforcing bars can be obtained from clauses 26.4.1 and 26.4.2 of the code Table.4.1 This is given in The exposure condition is given as 'Moderate' So the nominal cover to be provided can be taken as Cc =
30 mm
This will also satisfy the requirements for normal fire resistance. (Cl 26.4.3 and Table 16 of the code) The grade of concrete to be used for 'Moderate' exposure condition is fck = 25 N/mm2 Preliminary value of effective depth: Assume
Ф Фl
Effective cover
Table.4.2
=
20 mm
=
10 mm
=
50 mm
=
400 mm
=
4250 mm
=
4400 mm
=
4250 mm
So effective depth = d= D – effective cover
Effective span ● c/c distance between supports ● Clear span + effective depth =4250 -250 +400 lesser of the above = l Checks on preliminary dimensions: (4.2) Check D/b ratio
=
1.8
Falls between 1.5 and 2. Hence OK
4.3
Check for deep beam l/D = 4250/450
=
9.44
● 60b
=
15000
● 250b2/d
=
39062.5
lesser of the above
=
15000
=
4000
=
2.81 kN/m
Not less than 2. Hence OK
4.4
Check for slender beam
Clear distance between lateral restraints =4250 -250 4000
<
15000 Hence OK
Bending moment calculation Self wt of the beam per metre = 25 bD
where b and D are in metres and Unit wt.of reinforced concrete is 25kN/cu.mt 30.00 kN/m
Additional Dead load = Total Dead load =
wDL
=
32.81 kN/m
Total Live load =
wLL
=
7.00 kN/m
Total load =
w
=
39.81 kN/m
=
59.72 kN/m
wu =1.5w
Total factored load =
#9
9
Maximum factored moment at midspan Mu
=
wu l 2
=
8
134833740.2 Nmm 134.83 kNm
Design steps: Step 1
d= Where
k =0.362f ck
(
√
Mu kb
x u , max d
)(
1−0.416
x u , max d
)
#10
10
k =0.362f ck
(
x u , max d
)(
1−0.416
Table 3.4
x u , max
xu,max /d So 'k'
d
)
=
0.4791
=
3.47
So required effective depth 'd'
=
Calculated 'd' is less than that in
#6
394.15 mm
#11
11
Step 2: So the preliminary dimensions are now the 'Finalized dimensions'. Step 3:
A = 0.1506 bd2 fck
=
150600000
B = - 0.362 bd2 fck
=
-362000000
C = Mu
=
134833740.2
Solution 1
=
1.9429
Solution 2
=
0.4608
So 'xu/d'
=
0.4608
xu
=
184.32 mm
#12
12
=
1155.05 mm2
#13
13
Solution 2 is the appropriate solution
Step 4:
Ast =
0.362 f ck b x u 0.87 f y
Alternate method for calculating steel using SP16: TABLE 3 FLEXURE−REINFORCEMENT PERCENTAGE pt FOR SINGLY REINFORCED SECTIONS
Mu bd 2 Mu/bd2
pt
3.36
1.153
=
3.37
3.37
1.158
3.38
1.162
By linear interpolation, required value of pt =
( )
Ast =
pt
100
This is close to the value in
bd
=
1.158 1157.88 mm2
#13
Arrangement of bars: No. of bars of type 1
= Ф1
Dia. Of bars of type 1
=
No. of bars of type 2
= Ф2
Dia. Of bars of type 2 So area of steel provided =
Note that new value of
25 mm 1 16 mm 1182.81 mm2
Ast,p
=
d
=
400 mm
=
8 mm
Final effective depth See calculation
=
2
#14
14
#15
15
#16
16
Here Фl
Final effective span: As d remains the same, l will also remain the same. Final value of l Designed section is shown in the fig below:
=
4250 mm
At this stage we can perform the various checks: ● The three checks on preliminary dimensions were done in
#8
● All the preliminary dimensions and the effective span remains the same ● So the results of those checks will not change. We will now do the other checks 1. Spacing of bars
#17
17
Minimum clear space required: For this, we refer cl.26.3 of the code Also see
Eq.4.13
Sh = [b-(2Cc +2Φl +2Φ1 +Φ2)] /2
=
54 mm
● Larger bar diameter
=
25 mm
● Max. size of aggregate+5 mm=20+5
=
25 mm
Larger of the above
=
25 mm
25
<
54
Hence OK
Maximum clear space allowable: ● For this, we refer cl.26.3.3 of the code ● From table 15 of the code, (row for Fe 415 steel and column for zero percent redistribution) we get maximum allowable clear distance as 180 mm 54
<
180
Hence OK
2. Check for the Minimum area of flexural reinforcement
#18
18
#19
19
For this we refer cl.26.5.1.1 of the code Here Also see
As = #14 204.82
<
1182.81
0.85bd fy
=
204.82 mm2
Ast,p
=
1182.81 mm2
Hence OK
3. Check for the Maximum area of flexural reinforcement For this we refer cl.26.5.1.b of the code Here Also see 0.04bD
=
4500 mm2
Ast,p
#14
<
1182.81
4500
=
1182.81 mm2
Hence OK
4. Check whether pt is less than ptlim
p t=
100 Ast , p bd pt,lim
Table 3.5 1.18
<
1.201
=
1.18
=
1.201
For this we refer cl 23.2.1 of the code Here Also see For singly reinforced rectangular beams with span less than 10m, (l/d)actual ≤ [(l/d)basic] kt (l/d)basic
=
20
To find kt:
1.1
#13
Ast,required
=
1155.05 mm2
#14
Ast,p
=
1182.81 mm2
#20
From Fig.4 of the code
fst
=
235.05
pt
=
1.18
kt
=
0.971
=
19.42
=
10.63
So l/d = 20 x kt (l/d)actual = Divide
4250
10.63
<
6. Check for MuR
by 19.4
20
#21
21
Hence OK
5. Check for deflection control
For simply supported spans,
#20
400 Hence OK
MuR of the final section: Here See calculation From 134.83
<
137.29 kNm
#10
Mu
137.29
Hence OK
=
134.83 kNm
This completes the design process and the checks. copyright©2015limitstatelessons.blogspot.in Home
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