1) The document provides three examples of determining the moment capacity of reinforced concrete beams using the strength design method. The examples calculate steel ratios, check code requirements, and determine moment capacities. 2) Design examples are also provided, including calculating reinforcement needed to resist a given moment and designing a beam to support specific service loads. Optimal dimensions are selected to maximize steel ratio within code limits. 3) Analysis and design procedures for rectangular reinforced concrete beams are demonstrated, including calculation of steel area, reinforcement ratios, strain checks and moment capacities.

1. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.6 Strength Design Method
49
Dr. Muthanna Adil Najm
Analysis Examples:
Ex.1)
Determine the moment capacity of a beam with 375mm width, 675mm depth and
reinforced with 4 Ø 28 mm bars, if MPafc 28 and MPafy 420 .
Sol.)
mmd 60075675  and
    2
2
24646164
4
28
4 mmAs 










011.0
600375
2464



bd
As

Check steel percentage for ACI requirements.
0033.0
420
4.14.1
0031.0
420
2825.025.0
min 


yy
c
ff
f

0033.0min 
0181.0
8
3
420
2885.085.0
005.0003.0
003.085.0 1
max 


















 

y
c
f
f

0181.0
0033.0
011.0
max
min






 
 
mm
bf
fA
a
c
ys
116
3752885.0
4202464
85.0



mm
a
c 4.136
85.0
116
1


Check net tensile strain:
Strength Design Method
Analysis & Design of Rectangular Section Beams.
375mm
600mm
84Ø2
tε
= 0.003cε
c
d-c
d675mm

2. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.6 Strength Design Method
50
Dr. Muthanna Adil Najm
    005.001.0003.0
6.136
4.136600
003.0 




c
cd
t Tension controlled
section.
mkN
a
dfAM ysn 











 9.560
2
116
6004202464
2
  mkNMn .8.5049.5609.0 
Ex.2)
Determine the moment capacity of the beam section shown below,
MPafc 28 and MPafy 420 .
Sol.)
    2
2
305410183
4
36
3 mmAs 










027.0
375300
3054



bd
As

Check steel percentage for ACI requirements.
0033.0
420
4.14.1
min 
yf

0206.0
7
3
420
2885.085.0
004.0003.0
003.085.0 1
max 


















 

y
c
f
f

0206.0027.0 max  
 Section is not ductile, Check tensile strain.
 
 
mm
bf
fA
a
c
ys
6.179
3002885.0
4203054
85.0



mm
a
c 3.211
85.0
6.179
1


    004.00023.0003.0
3.211
3.211375
003.0 




c
cd
t
 Section is not tension controlled, and not in the permissible transition zone
between strain of 0.004 to strain of 0.005.
 The section is not ductile and may not be used per ACI section 10.3.5
Ex.3)
Determine the moment capacity of the beam section shown below,
MPafc 28 and MPafy 420 .
Sol.)
300 mm
375mm
450 mm
3Ø36
375mm
450 mm
3Ø28

3. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.6 Strength Design Method
51
Dr. Muthanna Adil Najm
    2
2
18486163
4
28
3 mmAs 










02.0
375250
1848



bd
As

Check steel percentage for ACI requirements.
0206.0
7
3
420
2885.085.0
004.0003.0
003.085.0 1
max 


















 

y
c
f
f

0206.002.0 max  
 
 
mm
bf
fA
a
c
ys
4.130
2502885.0
4201848
85.0



mm
a
c 4.153
85.0
4.130
1


Check tensile strain:
   
004.0
005.0
0043.0003.0
4.153
4.153375
003.0







c
cd
t
Beam is in the permissible transition zone and
  841.0
3
250
002.00043.065.0 






mkN
a
dfAM ysn 











 5.240
2
4.130
3754201848
2
  mkNMn .3.2025.240841.0 
Design Examples:
Ex.4)
Calculate the required amount of reinforcement to resist a moment of
mkNMu .150 for a beam with section of b = 250 mm and h = 500 mm, , if
MPafc 28 and MPafy 420 .
Sol.)
Assume using 25 bars  d = 500 – 40 – 12 -25/2 = 435 mm
65.17
2885.0
420
85.0





c
y
f
f

 
52.3
4352509.0
10150
2
6
2




bd
M
R u
n


4. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.6 Strength Design Method
52
Dr. Muthanna Adil Najm
009.0
420
65.1752.32
11
65.17
12
11
1





 










y
n
f
R 


Check steel percentage for ACI requirements.
0033.0
420
4.14.1
0031.0
420
2825.025.0
min 


yy
c
ff
f

0033.0min 
0181.0
8
3
420
2885.085.0
005.0003.0
003.085.0 1
max 


















 

y
c
f
f

0181.0
0033.0
009.0
max
min






2
979435250009.0 mmbdAs  
Use 2 Ø 25 bars;
    2
2
., 9824912
4
25
2 mmA provs 










2
.,
2
., 979982 mmAmmA reqsprovs 
Ex.5)
Design a simply supported rectangular beam with a span of 5 m and carrying a
total service dead load of 18 kN/m and service live load of 30 kN/m , if
MPafc 28 and MPafy 420 .
Sol.)
    mkNLDwu /70306.1182.16.12.1 
  mkN
lw
M u
u .8.218
8
570
8
22

The concrete dimensions will depend on the designer choice of reinforcement
ratio. To minimize the concrete section it is desirable to select the maximum
permissible steel ratio.
5 m
D = 18 kN/m & L =30 kN/m
250 mm
435
mm
500
mm
5Ø22

5. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.6 Strength Design Method
53
Dr. Muthanna Adil Najm
0181.0
8
3
420
2885.085.0
005.0003.0
003.085.0 1
max 


















 

y
c
f
f

nu MM 








c
y
yu
f
f
bdfM

 59.012





 

28
4200181.0
59.014200181.09.0108.218 26
bd
32
38079682mmbd 
b d
200 mm 436 mm d > 2b  try greater ' b ' value
250 mm 390 mm 3/2 b < d < 2b  OK.
300 mm 356 mm d < 3/2 b  Not Good
 Use section with b = 250 mm and d = 390 mm
2
17653902500181.0 mmbdAs  
 Use 3 Ø 28 mm
    2
.,
2
2
., 176518486163
4
28
3 mmAmmA reqsprovs 










h = 390 + 40 (cover) + 12 (Stirrups) + 28/2 (bar diameter /2) = 456 mm
 Use h = 460 mm
d = 460 - 40 (cover) - 12 (Stirrups) - 28/2 (bar diameter /2) = 394 mm
Check Section:
mm
bf
fA
a
c
ys
4.130
2502885.0
4201848
85.0






  mkNMmkN
a
dfAM uysn 











 8.2187.229
2
4.130
39442018489.0
2

 Section is OK.
Check deflection requirements:
From ACI Table 9.5.a, the minimum
permissible beam thickness for simply
supported beam is:
mm
l
h 5.312
16
5000
16

mmhmmh requiredavailable 5.312460 
250 mm
394
mm
460
mm
Ø283

6. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.6 Strength Design Method
54
Dr. Muthanna Adil Najm
 Section is OK.
Ex.6)
Resolve example (4) using steel percentage of max5.0  
Sol.)
    mkNLDwu /70306.1182.16.12.1 
  mkN
lw
M u
u .8.218
8
570
8
22

0181.0
8
3
420
2885.085.0
005.0003.0
003.085.0 1
max 


















 

y
c
f
f

009.0
2
0181.0
2
max


nu MM 








c
y
yu
f
f
bdfM

 59.012





 

28
420009.0
59.01420009.09.0108.218 26
bd
32
69881141mmbd 
b d
200 mm 591 mm d > 2b  try greater ' b ' value
300 mm 482 mm 3/2 b < d < 2b  OK.
350 mm 447 mm d < 3/2 b  Not Good
 Use section with b = 300 mm and d = 490 mm
2
1323490300009.0 mmbdAs  
 Use 3 Ø 25 mm
    2
.,
2
2
., 132314734913
4
25
3 mmAmmA reqsprovs 










h = 490 + 40 (cover) + 10 (Stirrups) + 25/2 (bar diameter /2) = 553 mm
 Use h = 560 mm
d = 560 - 40 (cover) - 10 (Stirrups) - 25/2 (bar diameter /2) = 498 mm
Check Section:
mm
bf
fA
a
c
ys
6.86
3002885.0
4201473
85.0







7. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.6 Strength Design Method
55
Dr. Muthanna Adil Najm
  mkNMmkN
a
dfAM uysn 











 8.2182.253
2
6.86
49842014739.0
2

 Section is OK.
Check deflection requirements:
From ACI Table 9.5.a, the minimum
permissible beam thickness for simply
supported beam is:
mm
l
h 5.312
16
5000
16

mmhmmh requiredavailable 5.312560 
 Section is OK.
300 mm
498
mm
560
mm
5Ø23