Analysis & Design of Reinforced Concrete Structures (1)

1. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
73
Dr. Muthanna Adil Najm
When compression steel is used, the nominal resisting moment of the steel is
assumed to be consisted of two parts.
1- The part due to the compression concrete and the balancing area of tensile
reinforcement.







2
11
a
dfAM ysn
2- The part due to compression steel and the balancing area of tensile
reinforcement.
 ddfAM ysn 2 if compression steel is yield
 ddfAM ssn 2 if compression steel is not yield
 ddfA
a
dfAMMM ysysnnn 






2
121
The strain in the compression steel is checked to determine whether or not it has
yielded. With the obtained strain, the compression stress is determined.
ss E
c
dc
f 










 
 003.0
And the value of 2sA is computed with the following expression:
ssys fAfA 2
Also, the net tensile strain in the extreme tensile steel ( εt ) should be calculated to
check the ductility of the section.
005.0t Ø = 0.9
005.0004.0  t Ø < 0.9
To determine the strain and thus the stress in both tensile and compression steel,
the location of the N.A ( the value of 'c') should be determined using the following
quadratic equation:
Strength Design Method
Analysis & Design of Doubly Reinforced Beams.
SA'
SA
=
S1A
d +
SA'
S2A
d-d'
d'

2. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
74
Dr. Muthanna Adil Najm
sscys E
c
dc
AcbffA 003.085.0 1 




 
 
Balanced and maximum steel ratio requirements:
y
s
bb
f
f 
  ( Balanced steel ratio)
y
s
f
f 
  maxmax ( Maximum steel ratio)
Analysis Procedure:
1- Find the location of the N.A ( the value of 'c' ) using the quadratic equation:
sscys E
c
dc
AcbffA 003.085.0 1 




 
 
2- Solve the above equation for 'c' then find 'a':
ca 1
3- Compute the strains in compression steel and tensile steel.
a- Check the yielding of compression steel; 003.0




 

c
dc
s
if
s
y
ys
E
f
  Then compression steel yields and ss AA 2
if
s
y
ys
E
f
  Then compression steel is not yield and ss E
c
dc
f 




 
 003.0
and s
y
s
s A
f
f
A 

2
b- Calculate the strain of the extreme tensile steel t to check section ductility.
003.0




 

c
cd
t
4- 21 sss AAA 
5- Calculate the design moment strength:
 











 ddfA
a
dfAM ysysn
2
1 if
s
y
ys
E
f
  and;
 











 ddfA
a
dfAM ssysn
2
1 if
s
y
ys
E
f
 

3. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
75
Dr. Muthanna Adil Najm
Analysis Examples:
Ex.1)
Determine the design moment capacity of the beam section shown below. Use
MPafc 21 and MPafy 420 .
Sol.)
2
407210184 mmAs 
2
12326162 mmAs 
Find the location of the N.A ;
sscys E
c
dc
AcbffA 003.085.0 1 




 
 
    000,200003.0
65
123235085.02185.04204072 




 

c
c
c
4804800073920037.53101710240 2
 ccc
04804800097104037.5310 2
 cc
096.904786.1822
 cc
   
mmc 4.223
2
87.26386.182
2
96.9047486.18286.182 2





  mmca 9.1894.22385.01  
Check the yielding of compression steel;
0021.0
200000
420
0021.0003.0
4.223
654.223
003.0 




 





 
 ys
c
dc

Compression steel yields ( ys ff  ) and 2
2 1232mmAA ss 
2
21 284012324072 mmAAA sss 
Check section ductility;
005.000519.0003.0
4.223
4.223610
003.0 




 





 

c
cd
t
 Section is ductile and 9.0
65 mm
545 mm
680 mm
2 Ø 28
4 Ø 36
350 mm

4. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
76
Dr. Muthanna Adil Najm
 











 ddfA
a
dfAM ysysn
2
1
  mkNMn .72.80610656104201232
2
9.189
61042028409.0 6












 

Ex.2)
Determine the design moment capacity of the beam section shown below. Use
MPafc 28 and MPafy 420 .
Sol.)
2
28287074 mmAs 
2
7603802 mmAs 
Find the location of the N.A ;
sscys E
c
dc
AcbffA 003.085.0 1 




 
 
    000,200003.0
65
76035085.02885.04202828 




 

c
c
c
2964000045600057.70801187760 2
 ccc
0296400007317605.7080 2
 cc
015.418635.1032
 cc
   
mmc 48.134
2
61.16535.103
2
15.4186435.10335.103 2





  mmca 3.11448.13485.01  
Check the yielding of compression steel;
0021.0
200000
420
00155.0003.0
48.134
6548.134
003.0 




 





 
 ys
c
dc

Compression steel is not yield and ;
MPaE
c
dc
f ss 31020000000155.0003.0 




 

  2
2 561760
420
310
mmA
f
f
A s
y
s
s 


2
21 22675612828 mmAAA sss 
Check section ductility;
65 mm
545 mm
680 mm
2 Ø 22
4 Ø 30
350 mm

5. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
77
Dr. Muthanna Adil Najm
005.00106.0003.0
48.134
48.134610
003.0 




 





 

c
cd
t
 Section is ductile and 9.0
 











 ddfA
a
dfAM ssysn
2
1
  mkNMn .3.5891065610310760
2
3.114
61042022679.0 6












 

Design of doubly reinforced concrete beams:
Sufficient tensile steel can be placed in most beams and bigger section can be used
so that compression steel is not needed. But, if section dimensions is restricted and
the required steel area exceeded the maximum steel ratio max , then compression
steel should be used.
Design Procedure:
1- Calculate the factored applied moment.
2- Calculate max and check if max  then design as a doubly reinforced beam.
Or calculate Mu max and compare with Mu
3- calculate 1sA where: bdAs max1 
4- Assume 9.0 and calculate 1nM :








c
y
ynu
f
f
bdfMM max
2
max1 59.01 
5- 12 uuu MMM 
6- Find the location of the neutral axis and check the yielding of compression steel
bf
fA
a
c
ys


85.0
1
and
1
a
c 
003.0




 

c
dc
s
a- If
s
y
ys
E
f
  Then compression steel yields and
 ddf
M
AA
y
u
ss



2
2
b- If
s
y
ys
E
f
  Then compression steel is not yield and ss E
c
dc
f 




 
 003.0
 ddf
M
A
s
u
s



2
and s
y
s
s A
f
f
A 

2

6. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
78
Dr. Muthanna Adil Najm
7- 21 sss AAA 
Design Examples:
Ex.3)
A beam is limited to the dimensions shown below. If mkNMu .1225 , determine
the required steel area. Use MPafc 21 and MPafy 420 .
Sol.)
0135.0
005.0003.0
003.085.0 1
max 












 

y
c
f
f

 
94.7
7003509.0
101225
2
6
2




bd
M
R u
u

 
53.23
2185.0
420
85.0



c
y
f
f










y
u
f
R 


2
11
1
0135.00284.0
420
53.2394.72
11
53.23
1
max 




 
 
 Design as a doubly reinforced concrete beam.
2
max1 33077003500135.0 mmbdAs  








c
y
yu
f
f
bdfM max
2
max1 59.01 
  mkNMu .7.73510
21
420
0135.059.017003504200135.09.0 62
1 





 
mkNMMM uuu .3.4897.735122512 
Check to see if compression steel yields;
32.222
3502185.0
4203307
85.0
1






bf
fA
a
c
ys
mm
a
c 55.261
85.0
32.222
1


70 mm
700 mm
350 mm

7. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
79
Dr. Muthanna Adil Najm
0021.0
200000
420
0022.0003.0
55.261
7055.261
003.0 




 





 

s
y
ys
E
f
c
dc

 compression steel yields.
   
2
6
2
2 2054
707004209.0
103.489
mm
ddf
M
AA
y
u
ss 






 Use 3Ø30    2
21217073 mmAs 
2
21 536120543307 mmAAA sss 
 Use 8Ø30    2
56567078 mmAs 
Ex.4)
Design a rectangular beam with maximum permissible dimensions shown in the
Figure below for mkNMD .230 and mkNML .305 . Use MPafc 28 and
MPafy 420 .
Sol.)
    mkNMu .7643056.12302.1 
0181.0
005.0003.0
003.085.0 1
max 












 

y
c
f
f

  mkNMu .7.53810
28
420
0181.059.015003754200181.09.0 62
.max) 





 
mkNMmkNM uu .7.538.764 max) 
 Design as a doubly reinforced concrete beam.
2
max1 33945003750181.0 mmbdAs  








c
y
yu
f
f
bdfM max
2
max1 59.01 
100 mm
500 mm
375 mm
70 mm
700 mm
350 mm
2 Ø 30
8Ø 30

8. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.9 Strength Design Method
80
Dr. Muthanna Adil Najm
mkNMMM uuu .3.2257.53876412 
Check to see if compression steel yields;
7.159
3752885.0
4203394
85.0
1






bf
fA
a
c
ys
mm
a
c 9.187
85.0
7.159
1


0021.0
200000
420
0014.0003.0
9.187
1009.187
003.0 




 





 

s
y
ys
E
f
c
dc

 compression steel is not yield.
MPaE
c
dc
f ss 68.2802000000014.0003.0 




 

   
2
6
2
2230
10050068.2809.0
103.225
mm
ddf
M
A
s
u
s 






 Use 6Ø22    2
22803806 mmAs 
2
2 14902230
420
68.280
mmA
f
f
A s
y
s
s 


2
21 488414903394 mmAAA sss 
 Use 8Ø28    2
49286168 mmAs 
100 mm
500 mm
375 mm
6 Ø 22
8Ø 28