The document analyzes and designs reinforced concrete beams using the strength design method. It provides examples of designing a simply supported rectangular beam, a cantilever beam, and an overhanging beam. The solutions include calculating loads, moments, required reinforcement, checking deflection requirements, and verifying the strength of the designed sections.

1. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
56
Dr. Muthanna Adil Najm
Ex.7)
Design a simply supported rectangular beam with a span of 10 m and carrying a
dead load of 20 kN/m (not including beam self weight) and service live load of 30
kN/m . Use 014.0 , b = 500 mm , MPafc 21 and MPafy 420 .
Sol.)
Assume that the beam self weight is 10 kN/m.
mkND /301020 
    mkNLDwu /84306.1302.16.12.1 
  mkN
lw
M u
u .1050
8
1084
8
22









c
y
yu
f
f
bdfM

 59.012





 

21
420014.0
59.01420014.09.0101050 26
bd
32
237676927mmbd 
 b = 500 mm  d = 690 mm
 h = 690 + 90 = 780mm.
Beam self weight = mkNmkNwhbws c /10/36.92478.05.0. 
 Assumed s.w is OK.
2
4830690500014.0 mmbdAs  
 Use 8 Ø 28.
  2
.,
2
., 483049286168 mmAmmA reqsprovs 
Check bar spacing for one raw bars;
mmdS b 286.24
7
288122402500



 Not Good, Use two rows.
h = 690 + 40 + 12 + 28 + 25/2 = 783 mm
 Use h = 800 mm
Strength Design Method
Analysis & Design of Rectangular Section Beams.
10 m
D = 30 kN/m & L =30 kN/m
500 mm
707
mm
800
mm
8Ø28

2. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
57
Dr. Muthanna Adil Najm
d = 800 - 40 - 12 - 28 - 25/2 = 707 mm
Check tensile strain:
mm
bf
fA
a
c
ys
9.231
5002185.0
4204928
85.0






mmdt 734
2
28
1240800 
mm
a
c 8.272
85.0
9.231
1


    005.00051.0003.0
8.272
8.272734
003.0 




c
cd
t
 9.0
Check Section:
  mkNMmkN
a
dfAM uysn 











 10501101
2
9.231
70742049289.0
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 625
16
10000
16

mmhmmh requiredavailable 625800 
 Section is OK.
Ex.8)
Design the cantilever beam shown below.
Use b
2
1
 , b = 300 mm ,
MPafc 28 and MPafy 420 .
Sol.)
    mkNLDwu /2.23106.162.16.12.1 
  kNLPu 80506.16.1 
    mkNP
lw
M u
u
u .6.657380
2
62.23
3
2
22

0283.0
420600
600
420
2885.085.0
600
60085.0 1











 
















 

yy
c
b
ff
f

6.0 m
L= 50 kN
3.0 m 3.0 m
D=6 kN/m
L=10 kN/m

3. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
58
Dr. Muthanna Adil Najm
0142.0
2
0283.0
2
1
 b








c
y
yu
f
f
bdfM

 59.012





 

28
4200142.0
59.014200142.09.0106.657 26
bd
32
140121985mmbd 
 b = 300 mm  d = 683 mm say d = 690mm.
 h = 690 + 90 = 780mm.
Check beam thickness for deflection requirements:
From ACI Table 9.5.a, the minimum permissible beam thickness for cantilever
beam is:
mm
l
h 750
8
6000
8
 < mmhavailable 780  Section is OK.
2
29396903000142.0 mmbdAs  
 Use 6 Ø 25.
  2
.,
2
., 293929464916 mmAmmA reqsprovs 
Check bar spacing for one raw bars;
mmmmS 252.9
5
256122402300


 Not Good
 Use two rows.
Check tensile strain:
mm
bf
fA
a
c
ys
3.173
3002885.0
4202946
85.0






mmdt 715
2
25
1240780 
mm
a
c 9.203
85.0
3.173
1


    005.00075.0003.0
9.203
9.203715
003.0 




c
cd
t
 9.0
Check Section:
300 mm
690
mm
780
mm
5Ø26

4. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
59
Dr. Muthanna Adil Najm
  mkN
a
dfAM ysn 











 9.671
2
3.173
69042029469.0
2

mkNMmkNM un  6.6579.671
 Section is OK.
Ex.9)
Design the overhanging beam shown below. Use
y
c
f
f 
 18.0 , b = 400 mm
, MPafc 28 and MPafy 420 .
Sol.)
    mkNLDwu /68206.1302.16.12.1 
  kNR 612
2
1868

1- Mid span section:
mkNM .918
012.0
420
28
18.018.0 


y
c
f
f









c
y
yu
f
f
bdfM

 59.012





 

28
420012.0
59.01420012.09.010918 26
bd
32
226427559mmbd 
 b = 400 mm  d = 752 mm say d = 760mm.
 h = 760 + 90 = 850mm.
D = 30 kN/m & L =20 kN/m
12 m3 m 3 m
= 68 kN/muW
612 kN
612 kN
-204 kN
204 kN
408 kN
-408 kN
306 kN.m 306 kN.m
918 kN.m
+
_ _

5. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
60
Dr. Muthanna Adil Najm
Check beam thickness for deflection requirements:
From ACI Table 9.5.a, the minimum permissible beam thickness for both end
continuous beam is:
mm
l
h 571
21
12000
21
 < mmhavailable 850
And for cantilever portion is:
mm
l
h 375
8
3000
8
 < mmhavailable 850  Section is OK.
2
3648760400012.0 mmbdAs  
 Use 6 Ø 28 mm
  2
.,
2
., 364836966166 mmAmmA reqsprovs 
Check bar spacing for one raw bars;
mmdmmS b 286.25
5
286122402400


 Not Good
 Use two rows.
Check tensile strain:
mm
bf
fA
a
c
ys
163
4002885.0
4203696
85.0






mmdt 784
2
28
1240850 
mm
a
c 8.191
85.0
163
1


    005.00093.0003.0
8.191
8.191784
003.0 




c
cd
t
 9.0
Check Section:
  mkNMmkN
a
dfAM uysn 











 918948
2
163
76042036969.0
2
 
Section is OK.
2- Section at support:
mkNM .306
Assume using 25 bars  d = 850 – 40 – 12 -25/2 = 785 mm

6. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
61
Dr. Muthanna Adil Najm
65.17
2885.0
420
85.0





c
y
f
f

 
38.1
7854009.0
10306
2
6
2




bd
M
R u
n

0034.0
420
65.1738.12
11
65.17
12
11
1





 










y
n
f
R 


Check steel percentage for ACI requirements.
0033.0
420
4.14.1
min 
yf

2
10687854000034.0 mmbdAs  
 Use 2 Ø 25 + 1 Ø 16
  2
.,
2
., 106811832014912 mmAmmA reqsprovs 
Check Section:
mm
bf
fA
a
c
ys
2.52
4002885.0
4201183
85.0






  mkNMmkN
a
dfAM uysn 











 3063.339
2
2.52
78542011839.0
2

 Section is OK.
400 mm
785
mm
850
mm
2Ø25
+
1Ø16
400 mm
760
mm850
mm
6Ø28
Support SectionMid span Section

7. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
62
Dr. Muthanna Adil Najm
Compressive strength of concrete and yield strength of steel bars:
cf  yf
British units SI units British units SI units
2500 psi = 2.5 ksi 17 MPa Grade 40 ksi 300 MPa
3000 psi = 3 ksi 21 MPa Grade 50 ksi 350 MPa
4000 psi = 4 ksi 28 MPa Grade 60 ksi 420 MPa
5000 psi = 5 ksi 35 MPa Grade 75 ksi 520 MPa
ksiEs 29000 MPaEs 200000
ACI equations:
85.01  for Psifc 4000
65.005.0
1000
4000
85.01 




 
 cf
 for Psifc 4000
y
c
f
f 

3
min 
yf
200
min 











yy
c
b
ff
f
000,87
000,87
85.0 1 yf in ( psi)
Example:
Determine the moment capacity of the beam section shown below,
psifc 4000 and psify 000,60 .
Sol.)
  2
2
8
9
3
4
3 inAs 










0033.0
000,60
200200
min 
yf

10''
15''
18''
3 # 9
Strength Design Method in British Units

8. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.7 Strength Design Method
63
Dr. Muthanna Adil Najm
0181.0
8
3
60
485.085.0
005.0003.0
003.085.0 1
max 


















 

y
c
f
f

02.0
375250
1848



bd
As
 0033.0
000,60
200
min  
But also > 0181.0max  for 005.0t
Check maximum steel percentage for 004.0t .
0206.0
7
3
60
485.085.0
004.0003.0
003.085.0 1
max 


















 

y
c
f
f

0206.002.0 max  
 
 
.29.5
10400085.0
600003
85.0
in
bf
fA
a
c
ys



85.01  for psifc 4000
.22.6
85.0
29.5
1
in
a
c 

Check tensile strain:
   
004.0
005.0
00423.0003.0
22.6
22.615
003.0







c
cd
t Beam is in transition
zone and
  836.0
3
250
002.000423.065.0 






ftkink
a
dfAM ysn 











 3.1859.2223
2
29.5
15603
2
  ftkMn  9.1543.185836.0