Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

Q-1
Design and Calculate the following and confirm that either the failure mode will be ductile or brittle?

(i) Tensile force

(ii) Compressive force

(iii) Area of stress block/ Whitney stress block

(iv) Tensile steel is yielding or not

(v) Design moment

SOLUTION
(a)
GIVEN DATA
Effective Depth d = 17.5 in
Width Of Beam b = 12 in
Height Of Beam h = 20 in
(i)

CALCULATION OF TENSILE FORCE

Yield Stress of Steel fy = 60 ksi

8
Diameter Of Steel for #8 rebar Ds =
8
Ds = 1 in

π d2
Area Of Single Steel Rebar =
4

π (1)2
=
4
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

π 2
= ¿
4

π
Area Of four #8 rebars As = 4 ( )
4
As = 3.14 in2

Tensile Force T = As fy
T = (3.14)(60)
T = 188.4 k

(ii)

CALCULATION OF COMPRESSIVE FORCE

Compressive Strength Of Concrete fc’ = 4000 psi

According to equilibrium condition assumed

C=T
0.85 fc‘ Ac = T
0.85 fc’ (a)(b) = T
T
a =
0.85 f 'c b

188.4 ×10 3
a =
0.85(4000)(12)
a = 4.62 in
The Compression force will be equal to tension force as discussed above

Compressive Force C = 0.85 fc’ (a)(b)

C = 0.85 (4000) (4.62) (12)
C = 188.4 K

(iii)
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

AREA OF STRESS / WHITNEY’S BLOCK

Depth Of Stress Block a = 4.62 in

Width Of Stress Block b = 12 in
Area of Whitney’s Stress Block Ac = a×b
Ac = (4.62)(12)
Ac = 55.44 in2

(iv)

TENSILE STEEL IS YIELDING OR NOT

For fc ‘ = 4000 psi , β 1=0.85

a
c=
β1

4.62
c=
0.85
c = 5.44 in

d−c
Tensile Strain ε t = 0.003 ( ¿
c
17.5−5.44
ε t = 0.003 ( ¿
5.44
ε t = 0.0066

fy
Yield Strain Of Steel εy =
Es

60× 103
εy =
29000× 103
ε y = 0.002

As the Strain in Tension fiber is greater than the yielding Strain in the steel , Tensile Steel will yield .
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

As ε t = 0.0066

ε t ≥ ε y + 0.003

ε t ≥0.005

This contradicts that the beam is tension controlled , the mode of failure of beam will be ductile.

(v)

CALCULATION OF DESIGN MOMENT

Design Moment implies a safety factor to the calculated ultimate or nominal moment by multiplying
it by strength reduction factor () for the purpose of overcoming uncertainties due to applied
equation and material strength.

For tension controlled members  = 0.9

−a
Nominal Moment Mn = As fy ( d )
2
4.62
Mn = (3.14) (60 ×10 3) ( 17.5 – )
2
Mn = 2,862 k-in
Mn = 238.5 k-ft

Design Moment  Mn = 0.9 (238.5)

 Mn = 214.6 k-ft

(b)

GIVEN DATA
Effective Depth d = 500 mm
Width Of Beam b = 250 mm
Height Of Beam h = 565 mm

(i)

CALCULATION OF TENSILE FORCE

Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

Yield Stress Of Steel Rebar = 450 MPa

Diameter Of Steel for #25 rebar Ds = 25.4 mm

π d2
Area Of Single Steel Rebar =
4

π (25.4)2
=
4
= 510 mm2
Area Of four #8 rebars As = 3 (510 )

As = 15 30 mm2

Tensile Force T = As fy
T = (1530)(450)
T = 688.5 k
(ii)

CALCULATION OF COMPRESSION FORCE

Compressive Strength Of Concrete fc’ = 21.5 MPa

According to equilibrium condition assumed

C=T
0.85 fc‘ Ac = T
0.85 fc’ (a)(b) = T

The Compression force will be equal to tension force as discussed above

Compression Force C = 0.85 fc’ (a)(b)

C = 0.85 (21.5) (151) (250)
C = 688.5 K

(iii)
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

AREA OF WHITNEY’S STRESS BLOCK

Depth Of Stress Block a = 151 mm

Width Of Stress Block b = 250 mm
Area of Whitney’s Stress Block Ac = a×b

Ac = (151)(250)

Ac = 37 , 750 mm2

(iv)

TENSILE STEEL IS YIELDING OR NOT

For fc‘ = 21.5 MPa , β 1=0.85

a
c=
β1

151
c=
0.85
c = 177.64 mm

d−c
Tensile Strain ε t = 0.003 ( ¿
c
500−177.64
ε t = 0.003 ( ¿
177.64
ε t = 0.0054

fy
Yield Strain Of Steel εy =
Es

450 ×106
εy =
200 ×109
ε y = 0.0022

As the Strain in Tension Steel is greater than the yielding Strain of Steel , tensile steel will yield .

As ε t = 0.0054

ε t ≥ ε y + 0.003
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

ε t ≥0.005 2

This contradicts that the beam is tension controlled , the mode of failure of beam will be ductile.

(v)

CALCULATION OF DESIGN MOMENT

Design Moment implies a safety factor to the calculated ultimate or nominal moment by multiplying
it by strength reduction factor () for the purpose of overcoming uncertainties due to applied
equation and material strength.

For tension controlled members  = 0.9

−a
Nominal Moment Mn = As fy ( d )
2
151
Mn = (1530) (450) ( 500 – )
2
Mn = 2.922 ×108 N -mm
Mn = 292.2 kN-m

Design Moment  Mn = 0.9 (292.2)

 Mn = 263 kN-m

Q-2
Determine the cracking moments for the section shown if fc‘ =4000 psi and fr = 7.5 √ f c' .

SOLUTION
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

GIVEN DATA
Height Of Beam h = 21 in
Breath Of Beam b = 12 in
Effective Depth d = 18 in

GROSS MOMENT OF INERTIA

1 3
Ig = bh
12

Ig = Gross Inertia

fr = Modulus of rupture

1
Ig = (12)(21)3
12

Ig = 9,261 in4

MODULUS OF RUPTURE
Compressive Strength Of Concrete fc ‘ = 4000 psi

fr =7.5√ fc '
fr = 7.5√ 4000
fr = 474.34 psi

CRACKING MOMENT
f r Ig
Mcr =
yt

yt = Distance from Neutral Axis to extreme tension fiber

(474.34)(9,261)
Mcr =
10.5
Mcr = 4,18,367.88 lb-in
Dividing by 12 for conversion into lb-ft
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

Mcr = 34,864 lb-ft

Mcr = 34.86 k-ft

Q-3
Assume the sections have cracked and use the transformed-area method to compute their flexural
stresses for the loads or moments given.

SOLUTION
GIVEN DATA
Height Of Beam h = 20 in
Breath Of Beam b = 12 in
Effective Depth d = 17 in

8
Diameter Of Steel for #8 rebar Ds =
8
Ds = 1 in

π d2
Area Of Single Steel Rebar =
4

π (1)2
=
4
π 2
= in
4

π
Area Of four #8 rebars As = 4 ( )
4
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

As = 3.14 in2

TRANSFORMED AREA METHOD

This method is used to calculate the distance of Neutral Axis , from extreme compression fiber, but
first we will assume that NA is present at an arbitrary distance “x” from top of beam.

x
(14) (x) ( ) = (8) (3.14) (17-x)
2
7 x 2+ 25.12x– 427.04 = 0
x=6.22 in

TRANSFORMED MOMENT OF INERTIA

1
It = bh3 + ad2 + (nAs)d2
12
1 x
It = (14) (x)3 + (14)(x) ( )2 + (8) (3.14) (17-x)2
12 2
1 6.22 2
It = (14) (6.22)3 + (14)(6.22) ( ) + (8) (3.14) (17-6.22) 2
12 2
It = 280.74 + 842.24 + 2919.15
It = 4042.13 in4

FLEXURAL STRESSES
Applied Moment M = 60 k-ft = 7200 k-in

Modular Ratio n =8

COMPRESSIVE STRESS
M yc
fc =
It

yc = Distance from Neutral Axis to extreme compression fiber

(720,000)(6.22)
fc =
(4042.13)

fc = 1108 psi
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

TENSILE STRESS
M yt
fs = n
It

yt = Distance from Neutral Axis to centroid of tension steel

(720,000)(6.22)
fs = (8)
(4042.13)

fs = 15,361 psi

Q-4
Determine the flexural stresses in these members using the transformed-area method.

SOLUTION
GIVEN DATA
Height Of Flange Hf = 4 in
Breath Of Web bW = 12
Effective Depth d = 18 in
Effective Width b = 48 in

9
Diameter Of Steel for #8 rebar Ds =
8
Ds = 1.125 in

π d2
Area Of Single Steel Rebar As =
4
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

π (1.125)2
As =
4
As = 1 in2
Area Of three #9 rebars As = 3 (1 )
As = 3 in2

TRANSFORMED AREA METHOD

Let us suppose that the Neutral Axis lies in the web

x−4
(48) (4) (x−2) +(x−4) (12) ( ) = (10) (3) (18-x)
2
192x−384 + 6 x 2−48 x +96 = 540−30 x

6 x 2+ 174 x−828=0
x=4.16∈¿

TRANSFORMED MOMENT OF INERTIA

1 1
It = [ bh3 + Ad2 ]Flange + [ bh3 + Ad2 ]Between Neutral Axis And Flange + (nAs)d2
12 12

1 1 x−4 2
It = (48) (4 )3 + (48)(4 ) ( x−2)2 + (12) ( x−4)3 + (12)( x−4) ( ) + (10) (3) (18- x )2
12 12 2

1 1 4.16−4 2
It = (48) (4 )3 + (48)(4 ) ( 4.16 −2 )2 + (12) (4.16 - 4)3 + (12)(4.16−4) ( ) +
12 12 2
(10) (3) (18−4.16) 2

It = 1151.79 + 0.016 + 5746.37

It = 6898.17 in4

FLEXURAL STRESSES
Applied Moment M = 100 k-ft = 1200 k-in

Modular Ratio n = 10
Salman Ahmad Reinforced Concrete Design - I FA17-CVE-082

COMPRESSIVE STRESS
M yc
fc = I
t

yc = Distance from Neutral Axis to extreme compression fiber

(12,00,000)(4.16)
fc =
(6898.17)

fc = 723.67 psi
TENSILE STRESS
M yt
fS = n
It

yt = Distance from Neutral Axis to Centroid of Tension Steel

( 12,00,000)(13.84)
fs = (10)
(6898.17)

fs = 24,076 psi

THE END