Assignment on Reinforced Concrete I
1
CIVIL SERVICE COLLAGE
DEPARTMENT OF URBAN ENGINEEREING
1. Determine the ultimate moment capacities of singly reinforced rectangular beams with
dimensions and quantity of reinforcement bars.
a) Using the parabolic-rectangular stress block for the concrete and elastic – perfect
strain curve for reinforcement.
b) Using the expressions of αc and βc developed using the parabolic – rectangular stress
c) Using the simplified-rectangular stress block for the concrete and elastic – perfect
strain curve for reinforcement.
Data:
Beam size: b = 250mm
h = 600mm
Concrete grade: C20
d
Steel grade:
S460
Clear cover: C = 25mm
Reinforcement:
i) 2 ф 16 bars
ii) 6 ф 25 bars
the following
plastic stressblock.
plastic stress-
250mm
Comment on the type of failure in each case.
Solution:
a. Using the parabolic-rectangular stress block for the concrete and elastic – perfect plastic stress-strain curve
for reinforcement.
i) 2 ф 16 bars
1. Design Strength :
f cd 
0.68 f c y
s

0.68 * 20
 9.07 N m 2
1.5
fyk 460

 400
s
1.15
EffectiveDepth
fyd 


d  h   c    600  25  8  567 mm
2

Let the strain distribution lie in zone - 3
c  0.0035
zone  3  
yd  s  0.01
X’
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Assignment on Reinforced Concrete I
2
Computation of forces
i.
For Rectangular part
Ccr  bx  x' fcd
But, from similarity of triangle
0.002
x  0.571x
0.0035
 Ccr  250 x  0.571x 9.07  927.7575 x
x' 
ii.
For parabolic
dF  fc y dA
x'
0.571
0
0
Ccp   fc( y )dA 
 fc( y)bdy
But fc(y)=1000єc(y)(1-250 єc(y)) fcd
But from similarity of triangle
y
 c  y   * 0.0035
x
y
y
 fcy  1000 * 0.0035 (1  250 * 0.0035 ) * 9.07
x
x
2
y
y
 fcy  31.745  27.78 2
x
x
x'
0.571
y
y2
Ccp   fc( y )dA   (31.745  27.78 2 ) * 250dy
x
x
0
0
0.571x

y2
y3 
Ccp  31.745
 27.78 2 
2x
3x  0

Ccp  862.75 x
Total Cc = CcR + Ccp
= 927.75x + 862.75x
= 1790.50x
Horizontal force equation as per assumption fs= fyd
16 2
Ts  Asfyd  2( *
) * 400
4
Ts  160849.544
But
Cc = Ts
1790.50x = 160849.54
x = 89.83 mm
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Assignment on Reinforced Concrete I
3
Check the assumption
From similarity of triangle
0.0035  s 0.0035

567
89.83
s  0.01855  0.01
The assumption is failed!
Let say the assumption of strain is lie in zone-2
0  c  0.0035
zon  2  
s  0.01
Computation of forces
i. For Rectangular part
Ccr  bx  x' fcd
But,from similarity of triangle
0.002(567  x)
x  0.2(567  x)
0.01
 Ccr  250 x  0.2(567  x) 9.07  2721x  257134.5
x' 
iii.
For parabolic
dF  fc y dA
x'
0.571
0
0
Ccp   fc( y )dA 
 fc( y)bdy
But fc(y)=1000єc(y)(1-250 єc(y)) fcd
But from similarity of triangle
0.01 y
 c y  
567  x
0.01 y
 0.01 y 
 fcy  1000 * 
) * 9.07
(1  250 *
567  x
 567  x 
90.7 y
226.75 y 2
 fcy 

567  x 567  x 2
x'
Ccp   fc( y )dA 
0
0.2 ( 667 x )

0
(
90.7 y
226.67 y 2

) * 250dy
567  x 567  x 2
0.2 ( 667 x )
 90.7 y
226.67 y 2 
Ccp  250 

2 
 567  x 567  x   0
Ccp  171375.75  302.25 x
Total Cc = CcR + Ccp
= 2418.75x – 85758.75
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Assignment on Reinforced Concrete I
4
Horizontal force equation as per assumption
fs= fyd
16 2
Ts  Asfyd  2( *
) * 400
4
Ts  160849.544
But
Cc = Ts
2418.75x – 85758.75 = 160849.54
x = 102 mm
Check the assumption
From similarity of triangle
0.01  c c

567
102
c  0.0022
0  0.0022  0.0035,........
The assumption is ok !
Moment resistance,
Mrd = M1 + M2 + M3
M 1  f cd b( x  x' )( x'
x  x'
)
2
102  58
)
2
M 1  281170 Nmm  0.28KN .m
dM 2  dF
M 1  9.07 * 250(102  58)(58 
For the rectangular:
x'
0.2 ( 567 x
0.2 ( 567 x
0
0
0
M 2   dFy 
0.2 ( 567 x
For parabolic:
M2 

0
 fcydAy   fcybydy
 90.7 y
226.67 y 2 

y * 250dy

2 
 567  x 567  x  
0.2 ( 567 x )
 90.7 y 3
226.67 y 4 
M 2  250 

2 
 3567  x  4567  x   0
M 2  8178710.625 Nmm  8.18 KNm
M 3  Ts(d  x)
For tension:
M 3  160849.54(465  102)
M 3  74795037.91Nmm  74.79 KNm
Moment resistance,
Mrd = M1 + M2 + M3
= 0.28 + 8.18 + 74.79
= 83.25KNm
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Assignment on Reinforced Concrete I
b. Using
5
the expressions of αc and βc developed using the parabolic – rectangular stress block.
Solution:
Effective depth, d = 567mm
Design strength:
fcd = 9.07N/mm2
fyd = 400N/mm2
Assume the strain profile in the ULS to lie in zone 2 (guessing the write profile)
It corresponds to case (ii) i.e.
cm ≥o and N.A with in the section
 αc =
3 cm  2
kx
3 cm
 βc = [
 cm (3 cm  4)  2
kx ]
2 cm (3 cm  2)
From the above strain diagram:
 cm 
x
10
dx
x
10  2
x
 c  d  x
x
d
3
10
dx
3

3x(10)  2(d  x) x
3x(10)
d

303  2 x  2d
30d

32 x  2(567)
30(567)
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Assignment on Reinforced Concrete I
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32 x  1134

17010
Equilibrium equations:
From ∑Fx = 0:
Cc = Ts
 αc fcd b d = As1 fyd

(32 x  2(567))
*9.07*250*567 = 246559.5 (as calculated before)
30(567)
 x = 102mm
 βc =
13.2273 x
33.30616 d
= 0.07144
 βcd = 40.5086mm
 Z = d - βcd
= 526.49mm
Therefore; MRd = Ts*Z
= 246559.5 *526.49
= 84685095.05Nmm
MRd = 84.68KNm
c. Using the simplified-rectangular stress block for the concrete and elastic – perfect
plastic stress-strain curve for reinforcement.
Assume the strain profile in the ULS to lie in zone 2 (guessing the write profile)
As calculated and checked in the above solution.
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Assignment on Reinforced Concrete I
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Cc  Ts
0.8 xfcd b  As f yd
0.8 x(9.07 * 250)  400 * 402.12
x  88.67mm
Mrd  Ts (d  0.4 x)  400 * 402.12 * (567  (0.4 * 88.67))  85495507.6 N / mm 2
Mrd  85.49 KN / m 2
ii) 6 ф 25 bars
a. Using the parabolic-rectangular stress block for the concrete and elastic – perfect plastic stressstrain curve for reinforcement.
Let say the assumption of strain is lie in zone-2
c  0.0035
zon  4  
0  s  yd
Computation of forces
i.
For Rectangular part
Ccr  bx  x' fcd
But,from similarity of triangle
0.002( x)
x  0.571x
0.0035
 Ccr  250 x  0.571x 9.07  927.7575 x
x' 
ii.
For parabolic
iii.
For parabolic
dF  fc y dA
x'
0.571
0
0
Ccp   fc( y )dA 
 fc( y)bdy
But fc(y)=1000єc(y)(1-250 єc(y)) fcd
But from similarity of triangle
y
 c  y   * 0.0035
x
y
y
 fcy  1000 * 0.0035 (1  250 * 0.0035 ) * 9.07
x
x
2
y
y
 fcy  31.745  27.78 2
x
x
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Assignment on Reinforced Concrete I
x'
Ccp   fc( y )dA 
0
8
0.571

2
(31.745
0
y
y
 27.78 2 ) * 250dy
x
x
0.571x

y2
y3 
Ccp  31.745
 27.78 2 
2x
3x  0

Ccp  862.75 x
Total Cc = CcR + Ccp
= 927.75x + 862.75x
= 1790.50x
Horizontal force equation
as per assumption fs is not yielded
fs  Es
from..similarity ..of ..triangle
s
c

dx x
c(d  x) 0.0035(562.5  x) 1.969  0.0035 x
s 


x
x
x
1.969  0.0035 x 393750  700 x
fs  Es  200000 *

x
x
393750  700 x   1159689431  2061670.1x
Ts  Asfs  2945.243 *
x
x
But
Cc = Ts
1159689431  2061670.1x
1790.51x 
x
x  413.79
fs  251.57
Ts  740933.71
Check the assumption
From similarity of triangle
0.0035(562.5  413.79
s 
413.79
s  0.001257  0.0035
The assumption is ok !
Moment resistance,
Mrd = M1 + M2 + M3
x  x'
)
2
M 1  9.07 * 250(0.429 * 413.79)(0.714 * 413.79)
M 1  f cd b( x  x' )( x'
For the rectangular:
M 1  118922158 Nmm  118.92 KN .m
For parabolic:
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Assignment on Reinforced Concrete I
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dM 2  dF
x'
0.571x
0.571x
0
0
0
M 2   dFy 
0.571x
M2 

0
 fcydAy   fcybydy
 31.745 y 27.78 y 2 


 y * 250dy
x
x2 

0.571x
 31.745 y 3 27.78 y 4 
M 2  250 


4x 2  0
 3x
M 2  52724762.66 Nmm  52.72 KNm
M 3  Ts(d  x)
For tension:
M 3  740933.71(562.5  413.79)
M 3  110184252.2 Nmm  110.18KNm
Moment resistance,
Mrd = M1 + M2 + M3
Mrd = 281.83KNm
b. Using
the expressions of αc and βc developed using the parabolic – rectangular stress block
Solution
Effective depth, d = 567mm
Design strength:
fcd = 9.07N/mm2
fyd = 400N/mm2
Assume the strain profile in the ULS to lie in zone 4 (guessing the write profile)
It corresponds to case (ii) i.e.
cm ≥o and N.A with in the section
 αc =
3 cm  2
kx
3 cm
 βc = [
 cm (3 cm  4)  2
kx ]
2 cm (3 cm  2)
For zone 3; εcm = εcu = 0.0035
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Assignment on Reinforced Concrete I
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Substituting this value in to the above expressions:
αc = 0.8095kx
βc = 0.416kx
Equilibrium equations:
From ∑Fx = 0:
Cc = Ts
 αc fcd b d = As1 fyd
 0.8095kx*9.07*250*562.5 = 2945.243(fs) (as calculated before)
But
393750  700 x
fs 
x
x
x
10324991.953k x x  1159689431  2061670.1x,......but.kx  
d 562.5
x  411.64mm
kx  0.732
kz  1  c
From ∑M = 0:
MRd = αc fcd b d2 kz
= αc fcd b d2 (1-βc )
= 0.8095kx*9.07*250*562.52(1-0.416*0.732)
= 295671813.1Nmm
= 295.67KNm
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Assignment on Reinforced Concrete I
d)
11
Using the simplified-rectangular stress block for the concrete and elastic – perfect plastic stressstrain curve for reinforcement.
Assume the strain profile in the ULS to lie in zone 2 (guessing the write profile)
As calculated and checked in the above solution
Cc  Ts
0.8 xfcd b  As f s
As  2945.243
fs 
393750  700 x
x
0.8 x(9.07 * 250)  2945.243
393750  700 x 
x
x  412.63mm
Ts  748493.663
Mrd  Ts (d  0.4 x)  748493.663 * (562.5  (0.4 * 412.63))  297487309.2 N / mm 2
Mrd  297.49 KN / m 2
2. Find the ultimate moment capacity for the RC slab shown in fig. below.
a.
Using the parabolic-rectangular stress block for the concrete and elastic – perfect plastic stressstrain curve for reinforcement.
b. Using the expressions of αc and βc developed using the parabolic – rectangular stress block.
c. Using
the
simplifiedrectangular stress block for
 12 c/c 180mm
the concrete and elastic –
perfect plastic stress-strain
h = 150mm
curve for reinforcement.
Data:
Thickness of slab = 150mm
Concrete grade: C25
Steel grade:
S300
Reinforcement: ф12 c/c 180mm
Clear cover: C = 20mm
Take 1m width for analysis purpose.
1m wide
Solution:
a. Using the parabolic-rectangular stress block
1. Design Strength :
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Assignment on Reinforced Concrete I
f cd 
0.68 f c y
s

12
0.68 * 25
 11.33 N m 2
1.5
fyk 300

 260.87 N / m 2
s
1.15
EffectivDepth
fyd 
d  124mm
Let say the assumption of strain is lie in zone-2
0  c  0.0035
zon  2  
s  0.01
Computation of forces
i.
For Rectangular part
Ccr  bx  x' fcd
But,from similarity of triangle
0.002(124  x)
x  0.2(124  x)
0.01
 Ccr  250 x  0.2(124  x) 11.33  13569 x  280984
x' 
ii.
For parabolic
dF  fc y dA
x'
0.2 (136 x )
0
0
Ccp   fc( y )dA 
 fc( y)bdy
But fc(y)=1000єc(y)(1-250 єc(y)) fcd
But from similarity of triangle
0.01 y
 c y  
124  x
0.01 y
 0.01 y 
 fcy  1000 * 
) * 11.33
(1  250 *
124  x
 124  x 
113.3 y 283.25 y 2
 fcy 

124  x 124  x 2
x'
0.2 (124 x )
0
0
Ccp   fc( y )dA 

113.3 y 283.25 y 2
(

) *1000dy
124  x 124  x 2
0.2 (124 x )
 113.3 y 283.25 y 2 
Ccp  1000 

2 
124  x 124  x   0
Ccp  187364  1511x
Total Cc = CcR + Ccp
= 12058x – 93620
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Horizontal force equation as per assumption fs= fyd
12 2
Ts  Asfyd  2( *
) * 260.87
4
Ts  163908.83
But
Cc = Ts
12058x – 93620= 163908.83
x = 21.36 mm
Check the assumption
From similarity of triangle
0.01  c
c

124
21.36
c  0.0021
0  0.0021  0.0035,........
The assumption is ok !
Moment resistance,
Mrd = M1 + M2 + M3……….about neutral axis
x  x'
M 1  f cd b( x  x' )( x'
)
2
21.36  12.63
For the rectangular: M 1  11.33 *1000(21.36  12.63)(12.63 
)
2
M 1  1741950.814 Nmm  1.742 KN .m
dM 2  dF
x'
0.2 (124 x )
0.2 (124 x )
0
0
0
M 2   dFy 
0.2 (124 x )
For parabolic:
M2 

0
 fcydAy   fcybydy
 113.3 y 283.25 y 2 

y * 1000dy

2 
124  x 124  x  
0.2 (124 x )
 113.3 y 3
283.25 y 4 
M 2  1000 

2 
 3124  x  4124  x   0
M 2  1987948.764 Nmm  1.99 KNm
M 3  Ts(d  x)
For tension:
M 3  160849.54(124  21.36)
M 3  16.82 Nmm
Moment resistance,
Mrd = M1 + M2 + M3
= 20.552KNm
Assume the strain profile in the ULS to lie in zone 2 (guessing the write profile)
It corresponds to case (ii) i.e.
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cm ≥o and N.A with in the section
 αc =
3 cm  2
kx
3 cm
 βc = [
 cm (3 cm  4)  2
kx ]
2 cm (3 cm  2)
From the above strain diagram:
 cm 
x
10
dx
x
10  2
x
 c  d  x

x
d
3
10
dx
3

3x(10)  2(d  x) x
3x(10)
d

303  2 x  2d
30d

32 x  2(124)
30(124)
 c 

32 x  248
3720
 Z = d - βcd = 115.54
13.2273 x
 0.06825
33.30616 d
Equilibrium equations:
From ∑Fx = 0:
Cc = Ts
 αc fcd b d = As1 fyd

32 x  2(124)
*11.33*1000*124 = 163908.837 (as calculated before)
30(124)
 x = 21.31
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Therefore; MRd = Ts*Z
= 163908.827*115.54
= 18937515.78Nmm
= 18.93KNm
d. Using the simplified-rectangular stress block for the concrete and elastic – perfect
plastic stress-strain curve for reinforcement.
Assume the strain profile in the ULS to lie in zone 2 (guessing the write profile)
As calculated and checked in the above solution.
Cc  Ts
0.8 xfcd b  As f yd
0.8 x(11.33 * 250)  163908.83
x  18.08mm
Mrd  Ts (d  0.4 x)  163908.83(124  (0.4 * 18.08))  19139076.91N / mm 2
Mrd  19.14 KN / m 2
2. Determine the moment capacity of a simply supported beam of trapezoidal cross-section for the
following data.
a. Using the parabolic-rectangular stress block for the concrete and elastic – perfect
plastic stress-strain curve for reinforcement.
b. Using the simplified-rectangular stress block for the concrete and elastic – perfect plastic
stress-strain
curve
for
b2
reinforcement.
Data:
Concrete: C30
Steel: S400
h
b1/b2/h = 300/500/600
As1 = 5ф25 bars in one layer
525 bars
Cover, c = 25mm
Solution:
b1
From similarity of triangle from the fig.
a)
Using the parabolic-rectangular stress block for the concrete and elastic – perfect plastic stress-strain
curve for reinforcement.
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Assignment on Reinforced Concrete I
16
 b2  b1 


b  b 
h
 2 
 m  2 1 h  x  y 
m
hx y
2h
b h  b2  b1 
 b  b 

h  x  y 
bx  b1  m  b1   2 1 h  x  y  * 2  1
h
 2h

Computation of forces
dF = fcddA
dF2  f cd dA......., dA  bx dy
x
x
b1 h  b2  b1 
h  x  y dy
h
x'
x
C R   f cd dA  f cd  bx dy  f cd 
x'
CR 
x'
f cd
h
x
 b h  b
1
2
 b1 h  x  y dy
0.571x
C R  2917.2 x  0.406 x 2
y
y2
 41.65 2
x
x
.0.571x
2

y
y  b h  b2  b1 
h  x  y dy 
f cy dA    47.6  41.65 2  1
x
h
x 

0 
dF2  f cy dA......., dA  bx dy,.... fcy  47.6
x'
Cp  
0
.0.571x
Cp 

0

y
y2
 47.6  41.65 2
x
x

 300 * 600  500  300 

600  x  y dy 
600


0.571
1  4228 * 10 4 y 2 12495 * 10 3 y 3 8330 * y 4 17850 * y 3 9520 * y 2 
Cp 






600 
2x
3x
2
3x 2
4x 2
0
C p  2587.562 x  1.11x 2
C c  C p  C R  5504.762 x  1.516 x 2
Ts  Asfyd  5( *
25 2
) * 347.83
4
Ts  853693.6559
Cc=Ts, 5504.762x – 1.516x2 = 854120.503
x = 162.34mm
Moment resistance,
Mrd = M1 + M2 + M3……….about neutral axis
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Assignment on Reinforced Concrete I
17
600 530.32

 m  88.387  bm  2m  b1  476.77 mm
100
m
100 * 437.576
n
 72.929  bn  2n  b1  445.859mm
600
From similarity of triangl
b  bm
br  2
 488.387 mm
2
b  bn
bn  2
 461.316mm
2
For the rectangular:
M 1  f cd br (0.429 x)(0.7855 x)
M 1  2238.238 x 2
M 1  59048192.4 Nmm  59.05 KN .m
dM 2  dF
x'
0.571x
0
0
M 2   dFy 
0.571x
For parabolic:
M2 

0
0.571x
 fcydAy   fcybydy
0
 47.6 y 41.65 y 


 y * 461.316dy
x2 
 x
2
0.571x
 47.6 y 3 41.65 y 4 
M 2  461.316 


4x 2  0
 3x
M 2  22479916.13 Nmm  22.48 KNm
M 3  Ts(d  x)
For tension:
M 3  853693.6559(562.5  162.424)
M 3  341542343.1Nmm  341.54 KNm
Moment resistance,
Mrd = M1 + M2 + M3
= 423.07 KNm
b)
Using the simplified-rectangular stress block for the concrete and elastic – perfect plastic stress-strain
curve for reinforcement.
From similarity of triangle.
600 600  0.8 x

 m  100  0.133x
100
m
bm  2(100  0.133x)  300  500  0.267 x
br 
500  0.267 x  500
 500  0.133x
2
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
18
Cc  Ts
0.8 xfcd br  As f yd
0.8 x *13.6 * (500  0.133x)  853693.6559
x  164.113mm
Mrd  Ts (d  0.4 x)  853693.6559(562.5  (0.4 *164.113))  424161871.1N / mm 2
Mrd  424.16 KN / m 2
3. Compare the moment capacities for singly reinforced rectangular beams having the following
properties.
Beam
no.
1
2
3
4
5
a.
b
(mm)
300
300
300
300
300
d
(mm)
550
550
550
550
830
Reinforcement
bars
3 ф 22
4 ф 24
3 ф 22
3 ф 22
3 ф 22
Grade of
concrete
C20
C20
C20
C30
C20
Grade
of steel
S460
S460
S300
S460
S460
Taking beam 1 as the reference point, discuss the effects of changing the area of
reinforcement, grade of concrete, grade of steel and effective depth on the ultimate moment
capacities.
Note: Each beam has the same properties as beam 1 except for italicized quantity.
b.
Beam no.
What is the most effective way of increasing the ultimate moment capacities? What is the
least effective way?
b
(mm)
300
300
300
300
300
1
2
3
4
5
d
(mm)
550
550
550
550
830
Beam
no.
b
(mm)
d
(mm)
1
2
3
4
5
300
300
300
300
300
550
550
550
550
830
Reinforcement
bars
3 ф 22
4 ф 24
3 ф 22
3 ф 22
3 ф 22
Reinforcement
bars
3
4
3
3
3
ф
ф
ф
ф
ф
22
24
22
22
22
fctd
9.07
9.07
9.07
13.6
9.07
Grade of
concrete
C20
C20
C20
C30
C20
fyd
400
400
260.87
400
400
x
Grade of
steel
S460
S460
S300
S460
S460
fctd
fyd
9.07
9.07
9.07
13.6
9.07
400
400
260.87
400
400
Asf yd
x  0.636d
0.8 f cd b
( x  0.74d )
209.78
332.88
136.82
139.86
209.78
349.94
349.93
400
349.94
528.08
M  Asf yd d  0.4 x 
212.61
301.72
147.34
225.37
340.33
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
19
5. A two span continuous rectangular beam has been supported by three girders spaced 6m on centers as
shown in fig below. Check whether the beam is safe or not for the following data.
Data:
Loading:
Live load, LL = 15KN/m
Dead load, DL = 25KN/m (excluding the self weight of the beam)
Dimensions:
b/h/d = 300/600/550
Material:
Concrete grade: C25
Steel grade:
S460
A
A
B
B
6m
6m
420bars
420bars
550mm
550mm
300mm
Solution:
section A-A
300mm
Section B-B
1. Design Strength :
f cd 
0.68 f c y
s

0.68 * 25
 11.33 N m 2
1.5
fyk 460

 400 N / m 2
s
1.15
EffectivDepth
fyd 
d  550mm
2. Analysis:
Pd  1.3DL  1.6 LL



Pd  1.3 25  25 KN / m 3 * 0.3 * 0.6  1.6 * 15 KN / m
Pd  62.35 KN / m
By using Cross method
Relative stiffness:
K
I
1
L
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
-
20
Distribution Factor:
DFBA, BC 
-
K
1
  0.5
K 2
DF AB, BC  1
Fixed End Moment
WL2 62.35 * 6 2

 187.05KNm
12
12
WL2
62.35 * 6 2


 187.05KNm
12
12
M F AB  M F BC 
M F BA  M F CB
Jont
member
DF
FEM
CO
Final Moment
M
A
A
AB
1
187.05
-187.5
0
B
BA
0.5
-187.05
-93.525
-280.575
C
CB
1
187.05
187.05
0
BC
0.5
187.05
93.525
280.575
0
 R2 * 6  280.575  62.35 *
6
2
 R2  233.81KN
F
H
 0 : R1  R2  62.35 * 6
 R1  140.29 KN
To find maximum Moment and Shear force
Vx  R1  62.35 x  140.29  62.35 x
Mx  R1 x  62.35
x2
x2
 140.29 x  62.35
2
2
To obtained, maximum moment shear forc is zero, Vx=0
 140.29  62.35 x  0
x
140.29
 2.25m
62.35
M max  140.29 * 2.25  62.35
2.25 2
2
M max  157.83KNm
BMD
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
21
For Section A-A
Let the strain distribution
lie in zone - 3
c  0.0035
zone  3  
yd  s  0.01
Force equation:
Cc  Ts
0.8 xfcd b  As f yd
0.8 x(11.33 * 300)  4( *
20 2
)400
4
x  184.85mm
Check assumption:
From similarity of triangle
0.0035  s 0.0035

 s  0.0069
550
184.85
fyd
400
yd 

 0.002
Es 200 * 10 3
 yd  s  0.01
The assumption is ok!
Moment equation
Mrd  0.8xfcdb(d  0.4 x)  0.8 *184.85 *11.33 * 300(550  0.4 *185)
Mrd  239.45KNm  157.83KNm
The beam is safe for coming moment
For section B-B
Let the strain distribution lie in zone - 3
c  0.0035
zone  3  
yd  s  0.01
Cc  Ts
0.8 xfcd b  As f yd
0.8 x(11.33 * 300)  4( *
20 2
)400
4
x  184.85mm
Check assumption:
From similarity of triangle
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
22
0.0035  s 0.0035

 s  0.0069
550
184.85
fyd
400
yd 

 0.002
Es 200 * 10 3
 yd  s  0.01
The assumption is ok!
Moment equation
Mrd  0.8xfcdb(d  0.4 x)  0.8 *184.85 *11.33 * 300(550  0.4 *185)
Mrd  239.45KNm  157.83KNm
The beam is not safe for coming moment.
6.
Determine the ultimate moment capacity of a doubly reinforced rectangular section with the following
data.
Given:
b/h = 250/450mm
cover, c = 25mm
Concrete: C30
Steel:
S460
As1 = 8ф24 bars in two layers
As2 = 4ф18 bars
d-x
1, Design Strength: From similarity of triangle
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
f cd 
0.68 f c y
s

0.68 * 30
 13.6 N m 2
1.5
fyk 460

 400 N / m 2
s 1.15
EffectivDepth
23
s1 
fyd 
y
4( y1 As1 )  4( y 2 As1 )
 47 mm
4( As1  As1 )
dx
cc  x 
x
d
1
s1
cc
fyd
400

 0,002
E
200 * 10 3
 xl im  0.636d  256.31mm
s1 
d  450  (47)  403mm
d 2  25 
18
 34mm
2
Force equation:
Cc  Cs  Ts
0.8 xfcd b  Asfyd  As f yd
18 2
24 2
)400  8( *
)400
4
4
 382.54mm  xlim  256.31mm..........ok!
0.8 x(13.6 * 250)  4( *
x req
Compression reinforcement is required.
M 1  0.8 xfcdb(d  0.4 x)
M 1  0.8 * 262.67 *13.6 * 250(403  0.4 * 262.67)
M 1  212861211.8KNm  212.86 KNm
from...similarity ...of ...triangle
s 2 
xlim  d 2
228.67
cc 
0.0035  0.003  yd  0.002
xlim
262.67
ItisYielde d
 M 2  As12 fyd (d  d 2 )  1832.95 * 400(403  34)
 M 2  270544010.4 Nmm  270.54 KNm
Cc  Ts1
0.8 xfcd b  As11 fyd
0.8 * 262.67 *13.6 * 250
 1786.156mm2
400
24 2
As12  As1  As11  8( *
)  1786.15  1832.95mm2
4
As11 
Mrd  M1  M 2  212.86  270.54  483.40KNm
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
24
7. Determine the flexural reinforcement for the rectangular concrete beam with the following data.
Given:
Dimensions:
b/h = 600/400mm
Cover, c = 25mm
Loading: Msd = 500KNm
Material Strengths: Concrete: C30
Steel: S-460
Solution:
d-x
f cd 
0.68 f c y
s

0.68 * 30
 13.6 N m 2
1.5
fyk 460

 400 N / m 2
s
1.15
Effectivdepth
fyd 
Let..us..use.. 20.bars
20
d  400  (25 
)  365mm
2
20
d 2  25 
 35mm
2
Ductility ..requirment :
x
 0.8(  0.44)  0.448
d
 163.52mm  xlim
Kxmax 
x max
M sd  0.8 xfcd b(d  0.4 x)
500 * 10 6  0.8 x *13.6 * 600 * (365  0.4 x)
xreq  x lim ...comression.
 x  912.5 x  191482.84
.re inf orcement..is..required
2
x  327.1mm  x req
Part I
F
 0,....Cc  Ts,.  0.8 xlim f cd b  As11 fyd
As11 
0.8 * 163.52 *13.6 * 600
 2668.65mm 2
400
H
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
25
M :
M 1  Cc * z1  0.8 xlim f cd b(d  0.4 x)
M 1  0.8 *163.52 *13.6 * 600 * (365  0.4 *163.52)
M 1  319.8KNm
M 2  M sd  M 1  500  319.8  180.2 KNm
Part II
M 2  Ts 2(d  d 2 )  As12 * fyd (d  d 2 )
180.2 * 102  As12 * 400 * 330
As12  1365.15mm 2
As1  As11  As12  4033.80mm 2
N o.. 20bar 
4033.8
 12.84...... provide..13 20..bar.. for..tension..re inf orcement.
20 2
*
4
Let assume that compression reinforcement is yielded, fs2 = fyd
F
H
 0,....As2 f yd  As12 fyd  As2  As12  1365.15mm2
Check the assumption
From similarity of triangle
s 2 
xlim  d 2
163.52  35
cc 
0.0035  0.00275  yd  0.002
xlim
163.52
ItisYielde d ..........the...asumption...is...ok!
 As 2  1365.15mm 2
 N o.. 20 
1365.15
 4.34......... provide....5 20...bars
20 2
*
4
8. Determine the quantity of reinforcement for the isolated T-beam shown in fig. below
Given:
bf
Msd = 550KN/m
Concrete: C25
hf
Steel: S460
d
Effective depth, d = 400mm
Width of web, bw = 250mm
Effective flange width, bf = 1000mm
Thickness of flange, hf = 100mm
bw
Use ф20 bars
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
26
Solution:
Assume that the Neutral axis lie in Flange
c
beff
h
f cd
Cc = 0.8xf cdbeff
x 0.8x
hf
N.A
d
z = (d - 0.4x)
s
As
fs
T s =Asf y d
bw
strains
x-section
stresses
resultant internal
forces
Using rectangular method:
Moment Equation
M sd  0.8 xfcd b(d  0.4 x)
550 * 10 6  0.8 x * 11.33 * 1000 * (400  0.4 x)
 3616 x 2  3616000 x  550 *10 6
x  187.113mm  x req  h f  100mm
The assumption is failed! The neutral axis lies in the web
beff
c
fcd
hf
hf
Cc1= h f(beff - bw)fcd
fcd
Cc2 = 0.8xfcdbw
x 0.8x
N.A
h d

As1
z1 = (d - 0.5h f) +
s
z2 = (d - 0.4x)
T s1 =Asffy d
T s2 =(As1 - Asf)fy d
bw
x-section
strains
Part I of the solution
Overhanging portion
Part II of the solution
web portion
For
part I
M 1  h f (beff  bw) f cd (d  0.5h f )
M 1  100(1000  250)11.33(400  0.5 * 100)  297412500 Nmm
M 1  297.41KNm
 M 1  Asf f yd (d  0.5h f )
Asf 
M1
297412500

2124.375mm 2
f yd (d  0.5h f ) 400(400  0.5 * 100)
2124.375
 6.76......... provide....7 20...bars
20 2
*
4
 M 1  550  297.41  252.59 KNm
 N o.. 20 
M 2  M sd
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ID-UEAR-0018/02
Assignment on Reinforced Concrete I
27
Ductility ..requirment :
x
 0.8(  0.44)  0.448
.
d
 179.2mm  xlim  x req  187.113mm
Kxmax 
x max
Compression reinforcement is required for Part II
Part II
d-x
For Part II – I
M 21  0.8 xfcd b(d  0.4 x)
M 21  0.8 * 179.2 *11.33 * 250 * (400  0.4179.2)  133.32 KNm
 As11 
0.8 xfcd b 0.8 *179.2 * 11.33 * 250

 1012.48mm 2
f yd
400
M 2 2  M 2  M 21  252.59  133.32  119.27 KNm
M 2 2  As12 f yd (d  d 2 )
As12 
M 22
119.27 * 10 6

 816.92mm 2
f yd (d  d 2 ) 400(400  35)
As1  As11  As12  1829.40mm 2
N o.. 20 
1829.40mm 2
 5.8......... provide....6 20...bars
20 2
*
4
Assume that compression reinforcement yielded, fs=fyd
F
H
 0,....Cc  Ts  As2 fyd  As12 f yd  As2  As12
 As2  816.92mm 2
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
28
Check assumption
s 2 
x lim  d 2
179.2  35
cc 
0.0035  0.00282  yd  0.002
x lim
179.2
ItisYielde d ..........the...asumption...is...ok!
 As 2  816.92mm 2
 N o.. 20 
9.
816.92
 2.6......... provide....3 20...bars
20 2
*
4
Compute the negative and positive moment capacities of the T-beam shown in fig. below.
Given:
Concrete: C25
Steel: S460
225 bars
625 bars
Overall depth, h = 550mm
Width of web, bw = 300mm
Thickness of flange, hf = 150mm
25 bars
Cover, c = 25mm
Note:
Neglect the effect of bottom
6m
bars as compression reinforcement
longitudinal view
at the supports
Consider the effect of the top
bars as compression reinforcement
at mid span
At the supports the bars are in
3m
3m
one layer
section view
At mid span, the bottom bars are
in two layers
Take the spacing b/n the bars equal to the diameter of the bar
Solution:
a)
Design strength
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
f cd 
0.68 f c y
s

29
0.68 * 25
 11.33 N m 2
1. 5
fyk 460

 400 N / m 2
s
1.15
EffectivDepth
fyd 
y1  25  12.5  37.5mm
y 2  25  25  12.5  62.5mm
y
y1 A1  y 2 A2
 50mm
A1  A2
d  550  (50)  500mm
25
 37.5mm
2
lo  0.85l  0.85 * 3000  2250mm

1 
1

bw   l o 2  300   * 2250   810mm
beffec  
5 
5

b
 actul  3000mm
d 2  25 
beffec  810mm
hf = 150mm
Cc  Cs  Ts
0.8 xfcd b  Asfyd  As f yd
25 2
25 2
)400  6( *
)400
4
4
 107.26mm  hf  150mm..........ok!
0.8 x(11.33 * 810)  2( *
x req
M 1  0.8 xfcdb(d  0.4 x)
M 1  0.8 *107.26 *11.33 * 810(500  0.4 *107.26)
M 1  359956590 KNm  359.96 KNm
359956590
 1968.71
400500  0.4 *107.26 
As12  As1  As11  6490.874   1968.71  976.533
As11 
Positive moment
 M t  M1  M 2
 M t  360.414 KNm
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
30
For negative moment
Cc  Cs  Ts
0.8 xfcd b  As f yd
25 2
0.8 x(11.33 * 810)  6( *
)400
4
x req  433.25mm..
d  550  37.5  512.5mm
 M  As12 fyd (d  0.4 x)
 M  6 * 490.874 * 400(512.5  4 * 433.25)
 M  399609978 Nmm  399.61KNm
10. A concrete floor system consists of one way slabs framed by beams as shown in fig below. The slab is
poured monolithically with the beam.
i) Design the slab for flexure
ii) Design the beam on axis-1 and axis-2 for flexure.
Data:
Dimensions:
Slab thickness = 150mm
6m
Beam web width, bw = 300mm
Over all depth, h = 450mm
Loading:
LL = 3.5KN/m2
DL from partition and finishing =
2kN/m2
6m
Material Strengths:
Concrete: C20
Steel:
S300
Weight of concrete = 25KN/m3
6m
Use ф 10 for the slab and ф20
for
the
beams
Note: Consider the different loading
cases
and
3m
3m
3m
the moment envelop
1
2
3
4
Solution:
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
1)
f cd 
fyd 
31
Design Strength:
0.68 f c y

s
0.68 * 20
 9.07 N mm 2
1 .5
fyk 300

 260.87 N / mm 2
s
1.15
2
2
0.21 fck 3
0.21( 20) 3
fctd 

 1.345 N / mm 2
s
1.15
2) Analysis:
Design Load:
Pd  1.3DL  1.6 LL



Pd  1.3 (2 KN / m 2 *1m)  25KN / m 3 * 0.15 *1  1.6(1m * 3.5KN / m)
Pd  13.075KN / m
-
By using Cross method Fixed End Moment
WL2 13.075 * 6 2

 9.81KNm
12
12
WL2
13.075 * 6 2


 9.81KNm
12
12
M F AB  M F BC  M F CD 
M F BA  M F CB  M F DC
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
Joint
member
Relative stiffness,
(End pine support,
B
BA
BC
C
CB
CD
3 I
( )
4 3
I
3
I
3
3 I
( )
4 3
Jont
A
member
DF
FEM
CO
Initial Moment
Balance
CO
Balance
CO
Balance
CO
Balance
Final Moment at
support
AB
1
9.81
-9.81
0
0
32
Total relative stiffness
I
)
K
3 I
( ))
4 K
(

I
4
I
( )
K
Disribiution
factor
I
)
K
I
K
3
7
4
7
4
7
3
7
(
7I
4
7I
4
I
4
B
BA
3/7
-9.81
-4.905
-14.715
2.102
0.6006
0.1716
0.049
-11.7918
C
BC
4/7
9.81
0
9.81
2.803
-1.4015
0.8009
-0.4004
0.2288
-0.1144
0.0654
11.7918
CB
4/7
-9.81
0
-9.81
-2.803
1.4015
-0.8009
0.4004
-0.2288
0.1144
-0.0654
-11.7918
D
CD
3/7
9.81
4.905
14.715
-2.102
-0.6006
-0.1716
-0.049
11.7918
DC
1
-9.81
9.81
0
0
Reaction forces
Rxn due to load
Rxn due to Moment
Final Rxn
A
19.6125
-3.93
RA= 15.68
B
19.6125 19.6125
3.93
0
RB= 43.155
C
19.6125 19.6125
0
3.93
RC= 43.155
D
19.6125
-3.93
RD= 15.68
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
33
To fined maximum Moment and Shear force
For span AB & CD
Vx  R1  13.075 x  15.68  13.075 x
Mx  R1 x  13.075
x2
x2
 15.68 x  13.075
2
2
To obtain, maximum moment shear force is zero, Vx=0
 15.68  13.075 x  0
x
M max  15.68 *1.2  13.075
15.68
 1.2m
13.075
1.2 2
2
M max  9.402 KNm
For span BC
Vx  R1  13.075 x  58.835  13.075 x
x2
x2
Mx  R1 x  43.155( x  3)  13.075
 58.835 x  129.465  13.075
2
2
To obtained, maximum moment shear force is zero, Vx=0.
 58.835  13.075 x  0
x
a)
58.835
 4.5m
13.075
M max  58.835 * 4.5  13.075
4.5 2
 129.465
2
M max  2.908KNm
Design Slab for flexure
Reinforcement
Effective Depth determination
d  150  (20 
10
)  125mm
2
Maximum, spacing
S max
S max
i.
 ab f yk
78.5 * 260.87

 327.82mm

0.5 *125
 0.5d
 2h  2 * 150  300mm
350mm


 300mm
For span AB & CD, Msd = 10.424KNm
By using general design chart Method.
 sd 
M sd
10.424 *10 6

 0.0735  0.295........ok!
bd 2 f cd 1000 *125 2 * 9.07
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
34
Kz  0.95
M sd
10.424 * 10 6
As 

 336.50mm 2
k z df yd 0.95 * 125 * 260.87
From GDC for μsd = 0.066,
S ( specing ) 
as b

As
*
10 2
* 1000
4
 233.41mm  300mm............ok!
336.5
Pr ovide....10c / c 230mm
ii.
For span BC
Msd = 5.463KNm
By using General Design Chart Method.
M sd
5.463 *10 6
 sd  2

 0.0385  0.295........ok!
bd f cd 1000 *125 2 * 9.07
From GDC for μsd = 0.0205, Kz  0.97
As 
M sd
5.463 * 10 6

 172.71mm 2
k z df yd 0.97 * 125 * 260.87
S ( specing ) 
as b

As
*
10 2
*1000
4
 454.75mm  300mm............is..not..ok!
172.71
Pr ovide....10c / c300mm
iii. For support B and C, Msd = 12.256KNm
By using General Design Chart Method.
 sd 
M sd
12.256 *10 6

 0.0865  0.295........ok!
bd 2 f cd 1000 *125 2 * 9.07
Kz  0.94
12.256 * 10 6

 399.84mm 2
0.94 *125 * 260.87
From GDC for μsd = 0.0205,
As 
M sd
k z df yd
S ( specing ) 
as b

As
*
10 2
* 1000
4
 196.43mm  300mm..............ok!
399.84
Pr ovide....10c / c195mm
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
35
Design for Shear
-
For span AB and CD, section 1-1 and 6-6
By similarity of triangle:
Vsd1  Vsd 6 
-
section 2-2 and 5-5
By similarity of triangle
Vsd 2  Vsd 5 
-
16.51 *1.612
 14.876
1.263
23.695 * 1.612
 22.0
1.737
section 3-3 and 4-4
By similarity of triangle
19.615 *1.375
 17.98
1.5
Vrd  0.25 f ctd bd  0.25 * 9.07 *1000 *125  283437.5
Vsd 3  Vsd 4 
Vrd  283.44KN  Vsd (allo)
Capacity of concrete for shear:
vc  0.25 f ctd k1k 2 bd
  10
k 2  1.6  d  1.475  1
ok!
k1  1  50
As  399.84
At support

399.84
1000 *125
k1  1.16  2.....ok!
As
399.84

bd 1000 *125
2
For slab section 2,3,4,5
vc  0.25 f ctd k1 k 2 bd
vc  0.25 * 0.8889 *1.16 *1.475 *1000 *125  47528.372 N
vc  47.528KN  Vsd .....ok!
For section 1&2
vc  0.25 f ctd k1k 2 bd
  10
At support
As  336.49

As
336.49

bd 1000 *125
k 2  1.6  d  1.475  1
ok!
k1  1  50
336.49
1000 *125
k1  1.135  2.....ok!
2
For slab section 2,3,4,5
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
36
vc  0.25 f ctd k1 k 2 bd
vc  0.25 * 0.8889 *1.135 *1.475 *1000 *125  46487.6644 N
vc  46.488KN  Vsd ..........ok!
Design of beam on axis 1-1
3) Analysis:
Design Load:
Pd  1.3DL  1.6 LL




Pd  1.3 (2 KN / m 2 *1.65m)  25KN / m 3 * 0.15 *1.65  (0.3 * 0.3 * 25)  1.6(1.65m * 3.5KN / m)
Pd  24.50 KN / m
-
By
using
Cross
method
Fixed End Moment
WL2 24.50 * 6 2

 73.50 KNm
12
12
WL2
24.50 * 6 2


 73.50 KNm
12
12
M F AB  M F BC  M F CD 
M F BA  M F CB  M F DC
joint
member
Relative stiffness,
(End pine support,
B
BA
BC
C
CB
CD
3 I
I
( )
4 6 12
I
6
I
6
3 I
I
( )
4 6 12
I
)
K
3 I
( ))
4 K
(
Total relative stiffness

I
( )
K
7I
24
7I
24
Disribiution
factor
I
)
K
I
K
3
7
4
7
4
7
3
7
(
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
Joint
A
member
DF
FEM
CO
Initial Moment
Balance
CO
Balance
CO
Balance
CO
Balance
CO
Balance
CO
Balance
Final Moment at
support
AB
1
73.50
-73.50
0
0
37
B
BA
3/7
-73.50
-36.75
-110.25
15.75
4.50
1.286
0.367
0.105
0.03
-88.212
C
BC
4/7
73.50
0
73.50
21.00
-10.50
6.00
-3.00
1.714
-857
0.49
-0.2450
0.14
-0.07
0.04
88.212
CB
4/7
-73.50
0
-73.50
-21.00
10.50
-6.00
3.00
-1.714
0.857
-0.49
0.2450
-0.14
0.07
-0.04
-88.212
D
CD
3/7
73.50
36.75
110.25
-15.75
-4.50
-1.286
-0.367
-0.105
-0.03
88.212
DC
1
-73.50
73.50
0
0
Reaction forces:
Rxn due to load
Rxn due to Moment
Final Rxn
A
67.62
-14.702
RA= 54.094
B
67.62
67.62
14.702
0
RB= 148.766
C
67.62
67.62
0
14.702
RC= 148.766
D
67.62
-14.702
RD= 54.094
To fined maximum Moment and Shear force
For span AB & CD
Vx  R1  24.50 x  15.68  24.50 x
Mx  R1 x  24.50
x2
x2
 54.09 x  24.50
2
2
To obtain, maximum moment shear force is zero, Vx=0
 54.09  24.50 x  0
x
M max  54.09 * 2.21  24.50
54.09
 2.21m
24.50
2.212
2
M max  59.71KNm
For span BC
Vx  R1  24.50 x  202.856  24.50 x
Mx  R1 x  148.766( x  6)  24.50
x2
x2
 54.09 x  892.596  24.50
2
2
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
38
To obtain, maximum moment shear forc is zero, Vx=0
 202.856  24.50 x  0
x
M max  54.09 * 9  892.596  24.50
202.856
 9.0m
24.50
92
2
M max  12.53KNm
For span CD
Vx  R1  24.50x  351.622  24.50x
a) Design Slab for flexure
Reinforcement
Effective Depth, use F6 for shear
d  450  (25 
20
 6)  409mm
2
Effective Width for L – Beam
For End span lo = 0.85l = 5100mm,
beffec
beffec
For middle span lo= 0.7l = 4200mm
1
1

1
1

bw  lo  300  5100  810mm

10
10
bw  l o  300  4200  720mm
beffec  
10
10

bw  bi  300  (150  1500  300)  4650mm
bw  bi  300  (150  1500  300)  4650mm
 810mm
beffec  720mm
i.
for span AB and CD,
Assume that the Neutral axis lie in Flange
c
beff
f cd
Cc = 0.8xf cdbeff
x 0.8x
hf
h
Msd = 59.71KNm
N.A
d
z = (d - 0.4x)
As
s
fs
T s =Asf y d
bw
x-section
strains
stresses
resultant internal
forces
By using GDC methode
M sd
59.71 *10 6
 sd  2

 0.0513  0.295........ok!
bd f cd 810 * 409 2 * 9.07
 Kz  0.95
From GDC for μsd = 0.066, 
 Kx  0.11
Check the assumption
x  K x * d  0.11* 409  44.99mm  hf  150mm
The assumption is ok !
As 
M sd
9.402 *10 6

 624.45mm 2
k z df yd 0.96 * 409 * 260.87
 N o.. 20 
624..45
 1.988......... provide....2 20...bars
20 2
*
4
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
ii.
39
For span BC
Msd = 12.53KNm
By using General Design Chart Methode
M sd
12.53 *10 6

 0.011  0.295........ok!
bd 2 f cd 720 * 409 2 * 9.07
 Kz  0.98
GDC for μsd = 0.0205, 
 Kx  0.05
 sd 
From
Check the assumption
x  K x * d  0.05 * 409  20.75mm  h f  150mm
The assumption is ok !
M sd
12.53 * 10 6
As 

 118.10mm 2
k z df yd 0.98 * 409 * 260.87
 N o.. 20 
iii.
118.10
 0.38......... for..stirrup.. provide....2 20...bars
20 2
*
4
For support B and C, Msd = 88.212
By using GDC method
M sd
88.212 *10 6

 0.188  0.295........ok!
bd 2 f cd 300 * 409 2 * 9.07
 Kz  0.89
GDC for μsd = 0.066, 
 Kx  0.26
 sd 
From
Check the assumption
x  K x * d  0.26 * 409  106.34mm  hf  150mm
M sd
88.212 * 10 6
As 

 915.51mm 2
k z df yd 0.89 * 409 * 260.87
 N o.. 20 
915.51
 2.91......... provide....3 20...bars
20 2
*
4
C) For axis 2 – 2, T- beam
4) Analysis:
Design Load:
Pd  1.3DL  1.6 LL




Pd  1.3 (2 KN / m 2 * 3m)  25KN / m 3 * 0.15 * 3  (0.3 * 0.3 * 25)  1.6(3m * 3.5KN / m)
Pd  42.15KN / m
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
-
40
By using Cross method Fixed End Moment
WL2 42.15 * 6 2

 126.45KNm
12
12
WL2
42.15 * 6 2


 126.45KNm
12
12
M F AB  M F BC  M F CD 
M F BA  M F CB  M F DC
joint
member
(
(End pine support,
B
BA
BC
C
CB
CD
3 I
( )
4 6
I
6
I
6
3 I
( )
4 6
Total relative stiffness
I
)
K
3 I
( ))
4 K
Relative stiffness,

I
8
(
I
)
K
I
)
K
I
K
3
7
4
7
4
7
3
7
(
7I
24
7I
24
I
8
Jont
A
member
DF
FEM
AB
1
126.45
BA
3/7
-126.45
BC
4/7
126.45
CB
4/7
-126.45
CD
3/7
126.45
CO
126.45
0
-
-63.225
0
0
63.225
DC
1
126.45
126.45
-189.675
27.096
-
-126.45
-36.129
18.0643
189.675
-27.096
-
0
-
0
7.743
2.213
0.632
0.180
0.052
-151.759
126.45
36.129
18.0643
10.3213
-5.161
2.948
-1.474
0.842
-0.421
0.241
-0.1205
0.0685
151.759
-10.3213
5.161
-2.948
1.474
-0.842
0.421
-0.241
0.1205
-0.0685
-151.759
-7.743
-2.213
-0.632
-0.180
-0.052
151.759
0
Initial Moment
Balance
CO
Balance
CO
Balance
CO
Balance
CO
Balance
CO
Balance
Final Moment at
support(KNm)
B
Disribiution
factor
C
D
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
41
Reaction forces:
A
126.45
-25.293
RA= 101.16
Rxn due to load
Rxn due to Moment
Final Rxn(KN)
B
C
126.45
126.45
25.293
0
RB= 278.19
126.45
126.45
0
25.293
RC= 278.19
D
126.45
-25.293
RD= 101.16
To fined maximum Moment and Shear force
For span AB & CD
Vx  R1  42.15 x  101.16  42.15 x
Mx  R1 x  42.15
x2
x2
 101.16 x  42.15
2
2
To obtain, maximum moment shear forc is zero, Vx=0
 101.16  42.15 x  0
M max
42.15
x
 2.4m
101.16
M max
2.4 2
 101.16 * 2.4  42.15
2
 121.39 KNm
For span BC
Vx  R1  42.15 x  379.35  42.15 x
Mx  R1 x  R2 ( x  6)  42.15
x2
x2
 379.35 x  1669.16  42.15
2
2
To obtain, maximum moment shear forc is zero, Vx=0
 379.35  42.15 x  0
x
M max  379.35 * 9  892.596  24.50
379.35
 9.0m
42.15
92
2
M max  12.53KNm
For span CD
Vx  R1  R2  R3  42.15x  657.55  42.15x
a) Design Slab for flexure
Reinforcement
Effective Depth, use F6 for shear
d  450  (25 
20
 8)  407 mm
2
Effective Width for L – Beam
For End span lo = 0.85l = 5100mm,
beffec
beffec
For middle span lo= 0.7l = 4200mm
1
1

1
1

bw  lo  300  5100  810mm

10
10
bw  l o  300  4200  720mm
beffec  
10
10

bw  bi  300  (150  1500  300)  4650mm
bw  bi  300  (150  1500  300)  4650mm
 810mm
beffec  720mm
iv.
For span AB and CD,
Assume that the Neutral axis lie in Flange
Msd = 121.39KNm
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
42
c
beff
h
f cd
Cc = 0.8xf cdbeff
x 0.8x
hf
N.A
d
z = (d - 0.4x)
s
As
fs
T s =Asf y d
bw
strains
x-section
stresses
resultant internal
forces
By using GDC method
M sd
121.391 *10 6

 0.0612  0.295........ok!
bd 2 f cd 1320 * 407 2 * 9.07
 Kz  0.96
GDC for μsd = 0.066, 
 Kx  0.12
 sd 
From
Check the assumption
x  K x * d  0.12 * 407  48.84mm  hf  150mm
The assumption is………..ok!
As 
M sd
121.39 *10 6

 1190.94mm 2
k z df yd 0.96 * 407 * 260.87
 N o.. 20 
For
1190.94
 3.79......... provide...4 20...bars
20 2
*
4
span BC
Msd = 37.92KNm
By using General Design Chart Method
M sd
37.92 *10 6

 0.022  0.295........ok!
bd 2 f cd 1140 * 407 2 * 9.07
Kz  0.98
GDC for μsd = 0.0205, 
Kx  0.07
 sd 
From
Check the assumption
x  K x * d  0.07 * 407  28.49mm  h f  150mm
The assumption is ok !
As 
M sd
37.92 *10 6

 364.43mm 2
k z df yd 0.98 * 407 * 260.87
 N o.. 20 
c.)
364.43
 1.16........... provide....2 20...bars
20 2
*
4
For support B and C, Msd = 151.76KNm
By using GDC methode
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
 sd 
M sd
151.76 *10

 0.3367  0.295.......is..not.ok!
2
bd f cd 300 * 407 2 * 9.07
From GDC for μsd = 0.3367,
M
-
*
u
43
6
 Kz  0.89
 Kx  0.26


s 2  2.9%
d 2 d 1  0.106
  f c d bd 2  0.295 * 9.07 * 300 * 407 2 =132.97KNm
*
us
is allowable capacity of concrete without reinforcement
Compression reinforcement
s 2  2.9%  yd  2%  fs2  fyd   yielded
M  Mu * 151.76  132.9710 6
As 2  sd

 198.54mm2
d  d 2 fyd 407  43260.87
Since
 N o.. 20 
198.54
 0.632.........
20 2
*
4
provide....2 20...bars
For teing stirrup
-
tension reinforcement
As1 
M sd
M  Mu *
 sd
 1605.61  198.54  1804.15mm 2
k z df yd d  d 2 fyd
 N o.. 20 
1804.15
 5.74......... provide...6 20bar
20 2
*
4
Design for shear
For span AB and CD, section 1-1 and 6-6
By similarity of triangle
Vsd1  Vsd 6 
-
section 2-2 and 5-5
By similarity of triangle
Vsd 2  Vsd 5 
-
101.16 *1.993
 80.005
2.4
151.75 * 3.193
 134.594
3.6
section 3-3 and 4-4
By similarity of triangle
126.45 * 2.593
 109.295
3
Vrd  0.25 f ctd bd  0.25 * 9.07 * 300 * 407  276862 N  276.862 KN
Vsd 3  Vsd 4 
Vrd  276.862KN  Vsd (allo)
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
44
Capacity of concrete for shear at support B& D,
k 2  1.6  d  1.193  1
vc  0.25 f ctd k1k 2 bd
ok!
  20
1804.15
2
300 * 407
k1  1.7388  2.....ok!
k1  1  50
As  1804.15mm 2
At support

As 1804.15

bd 300 * 407
For slab section 2,3,4,5
vc  0.25 f ctd k1 k 2 bd
vc  0.25 * 0.8889 *1.7388 *1.193 * 300 * 407  56285.77 N
vc  56.286 KN
For slab section 1,6
k 2  1.6  d  1.193  1
vc  0.25 f ctd k1k 2 bd
ok!
  20
At support
As  2 * 314.16  628.32mm

628.32
2
300 * 407
k1  1.2573  2.....ok!
k1  1  50
2
As
628.32

bd 300 * 407
vc  0.25 f ctd k1 k 2 bd
vc  0.25 * 0.8889 *1.2573 *1.193 * 300 * 407  40699.39 N
vc  40.699 KN
S max = <,
 af yk 100 * 300

 200mm

 0.5b 0.5 * 300
0.5d  203.5mm

Design shear for span AB & CD
S max = 200mm
Vsd,max = 134.594
Vc,max = 56.286
Vs  Vsd  Vc  134.594  56.286  78.308KN
adf yd 100 * 407 * 260.87
S

 135mm  S max  200mm
Vs
78.308 *10 3
Use  8c / c...135mm
Design shear for span BC
Vsd,max = 109.295
Vc,max = 56.286
Vs  Vsd  Vc  109.295  56.286  53.009KN
adf yd 100 * 407 * 260.87
S

 200mm  S max  200mm
Vs
53.009 *10 3
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
Use
45
 8c / c...200mm
ANCHERAGE LENGTH
a)
Basic anchorage length, lb
b 
b)
f yd
4f bd

20 * 260.87
 733.60mm ,
4 *1.7778
fbd  2 fctd  2 * 0.8889  1.7778
Minimum anchorage length ,lb,min for tension
0.3b  0.3 * 733.6  220.08mm

b, min  10  10 * 20  200mm
200mm

b, min  220mm
Minimum anchorage length ,lb,min for comprasion
0.6b  0.6 * 733.6  440mm

b, min  10  10 * 20  200mm
200mm

c). Required anchorage length,
b, net 
b, min  440mm
b, net
abAs, cal
, where a= 1.0 for straight bar anchorage
As, req
 At support B & C
As, cal  1804.15mm 2
As, provided  6 * 314.16  1884.95mm 2
abAs, cal 1 * 733.6 *1804.15
b, net 

 702.15mm  b, min  220mm..... ...ok!
As, req
1884.96
 At span AB & CD
As, cal  1190.94mm 2
As, provided  4 * 314.16  1256.64mm 2
abAs, cal 1 * 733.6 *1190.94
b, net 

 695.24mm  b, min  220mm..... ...ok!
As, req
1256.64
 At span BC
As, cal  364.43mm2
As, provided  2 * 314.16  628.32mm2
abAs, cal 1 * 733.6 * 364.43
b, net 

 420.82mm  b, min  220mm..... ...ok!
As, req
628.32
As, cal  198.54mm 2
 At support B & C for compression
As, provided  2 * 314.16  628.32mm2
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
46
abAs, cal 1 * 733.6 *198.54
b, net 

 231.8mm  b, min  220mm..... ...ok!
As, req
628.32
BAR CURTAILMENT
A) For span BC no bar curtailment needed
B) At support B & C
6 20 , let curtail 50% of the bar, 3 20 , Kz = 0.78 , d = 407mm
Mrd   As f yd K z d  3 * 314.16 * 0.78 * 407 * 260.87  78.03KNm
No of bar =
Mx  101.16 x  42.15
For X1,
x2
2
 78.03  101.16 x  42.15
x2
 1669.16
2
x2
X2,  78.03  379.35 x  42.15
 1669.16
2
x 2  6.654m
l 2  x 2  L   b, net  al
Mx  379.35 x  42.15
x2
2
,for
x1  5.495m
l1  L  x1   b, net  al
l1  6 - 5.495  0.70215  0.30525
l1  6.654 - 6  0.70215  0.30525
l1  1.51m
l1  1.66m
Total curtailment
lt  l1  l 2  3.17m
C) At span AB & CD
No of bar =
4 20 ,
let curtail 50% of the bar,
2 20 ,
Kz = 0.96 , d = 407mm
Mrd  As f yd K z d  2 * 314.16 * 0.96 * 407 * 260.87  64.04 KNm
x2
Mx  101.16 x  42.15
2
For X1, and X2,
64.04  101.16 x  42.15
x2
2
,
x1  0.75m
x 2  4.05m
lt  x 2  x1   b, net  al 2
lt  4.05 - 0.75  0.70215  0.305252
lt  5.84m
 SURVICIBILITY CHECK
At span AB and CD
DL = 19.5 KN, LL = 10.5 KN b/h = 300/450 ,
a. Un factored load
Ps = DL + LL = 30 KN
b.
Design constant
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
47
Ecm  29Gpa
2
2
fctk  0.21 fck 3  0.21 *16 3  1.333KN / m 2
Es  210Gpa
n
c.
Es
210

 7.2  8
Ecm 29
Design load / moment
Psl 2 30 * 6 2
Mk 

 135KNm
8
8
d. Section property
Section 1 – 1
A1  n  1As  8  13 * 314.16  6598.24mm 2
A2  n  1As  8  17 * 314.16  8796.48mm 2
A3  bh  300 * 450  135000mm 2
y1  25  8  10  43mm
y 2  450  25  8  10  407mm
y3 
y
e.
450
 225mm
2
 Aiyi  6598.24 * 43  8796.48 * 407  135000 * 225  227.66mm
6598.24  8796.48  135000
 Ai
Moment of inertia
d 1  y  y1  184.66mm
d 2  y  y 2  179.34mm
d 3  y  y 3  2.66mm
 300  450 2 
  6598.24 *184.66  8796.48 *179.34  135000 * 2.66
Iut   Ii   Aidi 2  
12


Iut  5.0889 *10 8
f. Section modulus
Iut 5.0889 *10 8

 2.235 *10 6
227
.
66
yt
Mcr  1.7 f ctk Z  1.7 *1.333 * 2.235  5.065KNm
Z
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
48
g. Computation of Deflection
i)

Short term deflection
s  i  ii   max
5
48
Where δi- is the deflection due to the theoretical cracking moment Mcr acting on the
un-cracked transformed section.
5
* 5100 2 * 5.065 * 10 6
i 
 48
 0.89mm
E cm Iut
29000 * 5.0889 *10 8
le 2 Mcr
δii - is the deflection due to the balance of the applied moment over and above
the cracking value and acting on a section with an equivalent stiffness of
75% of the cracked value.


x   n  n   n  82.41mm
As  As2 3  4 314.16
 1

 0.00257
bd
300 * 407
x
Z  d   379.53mm
3
le 2 Mk  Mcr 
ii 
0,75 EsAsZ d  x 
5
* 5100 2 * 135  5.065 * 10 6
48
ii 
0.75 * 210000 * 4 * 314.16 * 379.53407  82.41
ii  13.06mm
2
δmax - is the deflection of fully cracked section
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
49
5
* 5100 2 *135 *10 6
48
 max 

 10.8
EsAsd  x  210000 * 4 * 314.16 * 379.53(407  82.41)
i  ii  13.95mm
shrt  
shrt  10.8mm
 max  10.8mm
le 2 Mk

Long term deflection




l   2  1.2

As2
As1


3314.16 
 s   2  1.2
10.8  11.88mm

4314.16 


Total deflection
total  s  l  22.68mm
Checking with the limit deflection
all 
le
5100

 25.5mm  total  22.68mm...............ok!
200 200
Section 2 – 2
A1  n  1As  8  13 * 314.16  6598.24mm 2
A2  n  1As  8  12 * 314.16  4398.24mm 2
A3  bh  300 * 450  135000mm 2
y1  25  8  10  43mm
y 2  450  25  8  10  407mm
450
 225mm
2
 Aiyi  6598.24 * 43  4398.24 * 407  135000 * 225  222.26mm
y
6598.24  4398.24  135000
 Ai
y3 

Moment of inertia.
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
50
d 1  y  y1  179.26mm
d 2  y  y 2  184.74mm
d 3  y  y 3  2.74mm
 300  450 2
Iut   Ii   Aidi 2  
12

Iut  3.632 *10 8


  6598.24 *179.26 2  4398.24 *184.74 2  135000 * 2.66 2



Section modulus.
Iut 3.632 *10 8

 1.634 *10 6
222.26
yt
Mcr  1.7 f ctk Z  1.7 *1.333 *1.634  3.70 KNm
Z

Computation of Deflection
i)

Short term deflection
s  i  ii   max
5
48
Where δi - is the deflection due to the theoretical cracking moment Mcr acting on the
un-cracked transformed section.
5
* 4200 2 * 3.70 *10 6
le Mcr 48
i 

 0.62mm
E cm Iut
29000 * 3.632 *10 8
2
δii - is the deflection due to the balance of the applied moment over and above
the cracking value and acting on a section with an equivalent stiffness of
75% of the cracked value.


x   n  n   n  181.69mm
As  As2 3  2314.16
 1

 0.0018
bd
300 * 407
x
Z  d   346.44mm
3
2
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
ii 
51
le Mk  Mcr 
2
0,75 EsAsZ d  x 
5
* 4200 2 * 108  3.70  *10 6
48
ii 
0.75 * 210000 * 2 * 314.16 * 346.44407  181.69 
ii  23.82mm
δmax - is the deflection of fully cracked section
5
* 4200 2 *108 *10 6
le Mk
48
 max 

 18.50mm
EsAsd  x  210000 * 2 * 314.16 * 346.44(407  181.69)
i  ii  24.44mm
shrt  
shrt  18.5mm
 max  18.50mm
2
ii)
Long term deflection


As  

3314.16  
18.5  3.70mm
l   2  1.2 2  s   2  1.2
As1  
2314.16 



iii)
total deflection
total  s  l  22.20mm
Checking with the limit deflection
all 
le
4200

 21mm  total  22.20mm...............ok!
200 200
11. A petroleum station consists of RC slab which cantilevers 3m on each side of central RC beam and is
monolithic with it. The RC beam is simply supported on columns over a clear span of 6m. Design the slab
for flexure.
Data:
Dimensions:
Material Strength
Slab thickness = 100/300m
Concrete: C25
Beam web width, bw = 400mm
Steel: S460
Over all depth of beam is limited to 500mm
Loading:
Unit weight of concrete = 25KN/m3
Use ф 10 bars for the slab and ф20 bars for the beam
LL = 2KN/m2
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
52
DL from finishing = 1KN/m2
Solution:
Slab design
A) Design load and Moment
f cd 
fyd 
0.68 f c y
s

0.68 * 25
 11.33 N m 2
1 .5
fyk
460

 400
s
1.15
To make a uniform load take 1m strip
0 .5  0 .1
* 1 * 25 ) + 1.6(2*1)
2
Pd * l 2 14.25 * 3.2 2
M 

 72.96 KNm
2
2
h f  100mm
Pd= 1.3 (1*1 +
= 14 95KN/m
300  100
 200mm
2
d  h  25  5)  170mm
h
By using General Design Chart Methode
M sd
72.96 *10 6
 sd  2

 0.223  0.295........ok!
bd f cd 1000 *11.33 *170 2
 Kz  0.87
From GDC for μsd = 0.0205, 
 Kx  0.31
Check the assumption
x  K x * d  0.31 *170  52.7mm  h f  100mm
The assumption is ok !
M sd
72.96 *10 6
As 

 1233.27mm 2
k z df yd 0.87 *170 * 400
S ( specing ) 
asb

As
10 2
* 1000
4
 63.68mm  300mm..............ok!
1233.27
*
Pr ovide....10c / c60mm
Beam design consider as a T - beam
lo  2l  2 * 3.2  6.4  6400mm

1 
1

bw   l o 2  400   * 6400   2960mm
beffec  
5 
5

b
 actul  6400 * 2  12800mm
beffec  2960mm
hf = 100mm
Endeshaw Tafesse
ID-UEAR-0018/02
Assignment on Reinforced Concrete I
53
Design load and moment
Pd  1.36.40.1 * 25  0.5 * 0.2 * 25  0.2 * 0.4 * 25  1 * 6.4  1.62 * 6.4  73KN / m
73 * 6 2
M 
 328.5KNm
8
d  500  25  10  465mm
By using General Design Chart Methode
M sd
328.5 *10 6

 0.0453  0.295........ok!
bd 2 f cd 2960 *11.33 * 465 2
 Kz  0.975
GDC for μsd = 0.0205, 
 Kx  0.105
 sd 
From
Check the assumption
x  K x * d  0.105 * 465  46.5mm  h f  100mm
The assumption is ok !
M sd
328.5 * 10 6
As 

 1811.414mm 2
k z df yd 0.975 * 465 * 400
 N o.. 20 
1811.414
 5.76......... provide....6 20...bars
20 2
*
4
24.50 KN/m
A
6m
RA
B
x
RB
6m
C
RC
D
6m
RD
x
x
Endeshaw Tafesse
ID-UEAR-0018/02