Analysis of Rectangular Beams With Tension Reinforcement

Analysis of Singly Reinforced Rectangular Beam 3rd Stage
Analysis of rectangular beams with tension reinforcement

 Generally , in the analysis problem the following information are known:
1. Beam dimensions and reinforcement (b,h,d and As).
2. Materials strength (fy and Fc`).
As (Reinforcement area)
And following information are required

 Check the adequacy of the section according to ACI requirement.
 Compute design moment Mn.
 Compute the maximum live or dead load.
Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 1

Procedure Analysis for Rectangular Beams with tension
Reinforcement (Singly reinforcement)
1. Calculate ρ=
Where As = n
N = number of bars
D = diameter of reinforcement bar
Check if the provided ρ is in agreement with ACI requirements.

ρ ≤ ρmax = 0.85β1
u= 0.003
√
As ≥ As minimum= bwd ≥ bwd
2. Calculate
a=
c=
 according to ACI code can be calculated from table below
t= u where: u=0.003
 If t ≥0 005, then =0.9
 If t<0.005 then
=0.483+83.3 t

3. Calculate Mn
Mn can be calculated from:
Mn= Asfy(d- )
Or
Mn= 0.85fc`ab(d- )
Or
Mn= ρfybd2(1-0.59
4. Find Mu and compare it with φMn
If φMn≥Mu the section is Ok
If φMn<Mu the section is not Ok
Concrete Cover
To provide the steel with adequate concrete protection against corrosion, the
designer must maintain a certain minimum thickness of concrete to cover
outside of the outermost steel.
 As a general case, requirement for beams (that not exposed to

weather)=40 mm

Example 1: Check the adequacy of the beam shown below according to ACI
requirement, use fc`=25 Mpa, fy=400 Mpa, neglect the self-weight
Pu=100 kN
3φ20mm
Solution:
As = n =3 0 = 942 mm2
ρ= = = 5.71 10-3
ρmax = 0.85β1 = 0.85 0.85 = 19. 4 10-3
ρ< ρmax O.k
Asminimum= bw d = 300 550 = 525 mm2
As > Asminimum o.k
A= = = 59.1 mm
c= = = 69.5 mm
550 5
t= u= 0.003=20.7 10-3
5
t > 0.005
=0.9
Mn= As fy(d ) = 0.9 942 400 (550 ) 10-6 = 176 kN.m
Mu = = =125 kN.m Mn > Mu the section is O.k

Example 2: Check the adequacy for the beam below according to ACI
requirement, if the beam is subjected to uniform dead load (3)
kN/m (without self-weight) and uniform live load (2) kN/m, use
fc`=28 Mpa and fy=420 Mpa
WD=3 kN/m (without self-weight), WL=2 kN/m
Solution: 3φ25 mm
As = n =3 5 =1472.6 mm 2
d=h cover stirrups = 600 40 10 = 537.5 mm

ρ= = = 6.85 10-3
ρmax =0.85β1 = 0.85 0.85 = 0.0206
ρ< ρmax O.k
As minimum= bw d = 400 537.5 = 716.67 mm2
As > As minimum
a= = = 64.96 mm
c= = =76.43 mm
5 5
t= u= 0.003 = 0.018
t = 0.018> 0.005
=0.9
Mn= Asfy(d ) = 0.9 1472.6 420 (5 5 ) 10-6 = 281.1 kN.m
Find Mu
Mu=
Wu=1.2WD+1.6WL
WD self-weight = γ b d=24 0.4 0.6 =5.76 kN/m
WD total=5.76 + 3=8.76 kN/m
Wu=1.2 8.76 +1.6 2=13.712 kN/m
Mu= = = 42.85 kN.m Mn> Mu section is O.K. ▀
Example: check the adequacy of the cantilever shown below; the cantilever is
subjected to uniform dead load (3) kN/m (include self-weight) and
uniform live load (4) kN/m, fc`=28 Mpa and fy=420 Mpa
3φ20

Solution:
As = n =3 0 = 942 mm2
d=600 40 10 =540 mm
ρ= = = 6.98 10-3
ρmax = 0.85β1 =0.85 0.85 = 0.0206
ρ < ρmax O.k
Asminimum= bw d= 250 540 = 450 mm2
As > Asminimum o.k
a= = = 66.5 mm
c= = =78.23 mm
5 0
t= u= 0.003 = 0.0177
t > 0.005
=0.9
Mn= Asfy(d ) = 0.9 942 420 (5 0 ) 10-6 = 180.44 kN.m
Calculate Mu
Mu =
Wu=1.2WD+1.6WL=1.2 3+1.6 4=10 kN/m
Mu= = =115.2 kN.m
Mn> Mu section is O.K. ▀
Tension side of beam where the

reinforcement shall be put

Example: check the adequacy of a simply supported beam shown in figure
below when subjected to a factored load of Wu=70 kN/m
(including self-weight), use fc`=28 Mpa and fy=420 Mpa
Wu=70 kN/m
5φ25 mm
Solution:
As= n* =5* 5 =2454 mm2
d=600-40-10-25- = 512 mm
ρ= = =0.0159
ρmax = 0.85β1 = 0.85*0.85* = 0.0206
ρ< ρmax O.k
As minimum= bw*d= *300*512 = 512 mm2
As > As minimum o.k
a= = =144 mm
c= = =169.83 mm
5 5
t= u= *0.003 = 6.5*10-3 > 0.005
Then =0.9
Mn=φAs*fy(d- ) = 0.9*2454*420*(512- )*10-6 = 408 kN.m
Mu= = =218.75 kN.m φMn> Mu section is O.K. ▀
dt

Example: for the precast beam shown in Figure below, the designer intended
to use 4φ20 ,check the adequacy of the beam according to ACI
requirement , the beam is subjected to uniform dead load (15) kN/m
(with self-weight) and uniform live load (20) kN/m
Assume in your solution:
 The beam is simply supported
 Fc`=28 Mpa and fy=420 Mpa
 Single layer of reinforcements
 Beam with 250 mm and effective depth 500 mm
Solution:
As = n* = 4* 0 =1256.63 mm2
ρ= = = 0.01
ρmax = 0.85β1 = 0.85*0.85* = 0.0206
ρ< ρmax O.k

As minimum= bw*d= *250*500=416.7 mm2
As > As minimum o.k
a= = =88.7 mm
c= = =104.35 mm
500 0 5
t= u= *0.003= 0.0113 > 0.005
0 5
Then =0.9
Mn= As*fy(d- ) = 0.9* 5 *420*(500- )*10-6 = 216.43 kN.m
Calculate Mu
WU=1.2WD+1.6WL=1.2*15+1.6*20=50 kN/m
Mu= = =225 kN.m
φMn<Mu the section is not O.K ▀
 beam dimensions must be increasing or using more reinforcement area .

Example: a rectangular beam with a width of 305 mm and an effective depth
of 444 m. it is reinforced with 4φ29mm (assume Abar=645 mm2) if
fc`=27.5 Mpa and fy=414 Mpa. Check the beam adequacy and
compute its design strength according to ACI requirements.
Solution:
As=n Abar=4 645=2580 mm2
ρ= = =0.019
ρmax = 0.85β1 =0.85 0.85 = 0.0205
ρ< ρmax O.k
As minimum= bw d = 305 444 = 458 mm2
As > As minimum o.k
a= = =150 mm
c= = =176 mm
t= u= 0.003= 0.00457 <0.005
Then:
=0.483+83.3 t
=0.483+83.3 0.00457
=0.86
Mn= As fy(d )=0.86 2580 414 (444 ) 10-6 = 339 kN.m

Example: Check the adequacy of the beam shown below according to ACI
requirement, fc`=28 Mpa ,fy=420 Mpa
3φ30 mm
Solution:
As = n =3 0 = 2120.57 mm2
d = 400 40 10 15 = 335 mm
ρ= = = 0.03165
ρ> ρmax
The section is not O.K and may not be used according to ACI requirements

Analysis of Rectangular Beams With Tension Reinforcement

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Analysis of Singly Reinforced Rectangular Beam 3rd Stage

Analysis of rectangular beams with tension reinforcement

And following information are required

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 1

Check if the provided ρ is in agreement with ACI requirements.

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 2

 As a general case, requirement for beams (that not exposed to

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 3

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 4

d=h cover stirrups = 600 40 10 = 537.5 mm

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 6

Tension side of beam where the

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Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 9

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 10

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 11

Lec. Hasanain M. Al-Musawi Reinforced Concrete Design I Page 12

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