This presentation summarizes the design and analysis of reinforced concrete beams according to the ACI code. It discusses the objectives, which are to understand the USD method and gain concepts about analyzing and designing singly, doubly, and T-beams. It then covers design parameters, load calculations, analysis procedures for under- and over-reinforced beams, and design steps for doubly reinforced beams. Finally, it addresses when to consider a beam as a T-beam versus rectangular beam based on its dimensions.

1. Design of Concrete Structures-1
CE-2105

2. Presentation of design and analysis of RC beam(as per
ACI code)
Dr.Md. Rezaul Karim
Associate Professor
Department of Civil Engineering (CE)
Dhaka University of Engineering & Technology
Advisor

3. Our Group Member’s
Kazi Abu Monjur
Std Id: 121042
Md. Rabiul Islam
Std Id: 121044
Md. Irafanul Kabir
Std Id: 121045
Md. Shorab Hossain Pavel
Std Id. 121058
Mohammad Anis
Std Id. 121050

4. Contents of Presentation
 Objectives
 Singly Beam Design & Analysis
 Doubly Beam Design & Analysis
 T-Beam Design & Analysis

5. Objectives
• Gain the concept about USD method
• Gain the analysis concept of Singly, Doubly and T-Beam
• Gain the Design procedure of Singly, Doubly and T-Beam
• “In what sitution what Types of beam we should consider

6. As a Civil Engineer
What should I have
to consider ???
For Structures
Design?
 Design
 Strength
 Serviceability
 Durability
 Implacability
 Economy

7. Design Parameter
ß1=0.85-0.05(f’c-4000)/1000 and 0.65 ≤ß1 ≤0.85
Strength factor Φ should be adjust by ACI code provision

8. Analysis Over Reinforced as
Doubly
Analysis of Beam
ρ < ρ max
Analysis Under Reinforced as Singly
Mn=Asfy(d-a/2)
a=Asfy/.85f’cb
Mn=ΦMn
Adjust Φ by ϵt or c/dt
Assume b
Generally b=d/2
No
Yes

9. Design
Calculating the factored loads
Mu=It’s depends on beam support
Wu=1.2DL+1.6LL
Adjust Φ by ϵt or c/dtMu=Φ ρfybd2(1-0.59ρfy/f’c)
Determine
b and d
As= ρb𝑑2
Design complete

10. For Economy Design should
be revised
As=Mu/ Φfy(d-a/2)Assume a Check a
Alternate solution use lower ρ as
ρ=0.6 ρmax or ρ=0.75 ρmax etc

11. When a beam design as a Doubly reinforced beam ?

12. Step :1 Checking the Beam (Singly nor Doubly)
ρ> ρ max
ρ
N×As
ρmax= .85 βḟc/fy×
Ԑ𝑐𝑢
Ԑ𝑐𝑢+0.004
0.003
This condition is satisfy thus the
beam is doubly reinforced beam
Step :2 Check the Limits on reinforced ratio
If ρ>ῤ cy Compression bar yield
But ρ<ῤ cy Compression steel below yield stress
ῤcy= .85 βḟc/fy×
ď
𝑑
× Ԑ𝑐𝑢
Ԑ𝑐𝑢+0.004
+ ῤ
𝐴ś
𝑏𝑑

13. Step : 3 Determination the moment capacity
ρ>ῤcy
Where
   n s y s s s s
2
a
M A f A f d A f d d
          
 
OR ρ<ῤcy
   n s y s s s s
2
a
M A f A f d A f d d
          
 

14. Design step of Doubly reinforced beam
Step 1: Load Calculation
This condition is satisfy compression steel’s will be needed i.e doubly reinforced beam
Wu=1.2 DL+1.6LL
Mu= Its defends on beam supports characteristics
Step 2 : Check the Capacity of the Section
Mn>Mu
Wu=1.2 DL+1.6LL
ΦMn=Asfy(d-a/2)
ρ max × bd ρmax= .85 βḟc/fy×
Ԑ𝑐𝑢
Ԑ𝑐𝑢+0.05
b=width of beam
d= Effective depth

15. Step 3: Determine the area of steel
Excess Moment M1=Mu-Mmax
Aś=
𝑴𝟏
Φḟ 𝒔(𝒅−ď) ḟ𝒔=ἑsEs
Ԑu
(𝒄−ď)
𝒄
𝒄 =
𝒂
β𝟏
ď=2.5 “ (assume)
If ď < 21.5 then Compression reinforced yield stress
,so using lower compression stress ḟ𝒔
As=Aś+As(max)

16. Checking Design
Design of beam is OK
Φ must be adjust
Φ= 0.65+(Ԑt-0.002)
250
3
Ԑt=Ԑu(𝒅𝒕−𝒄)
𝒄
Now Mu=ΦMn=Φasfy(d-
𝑎
2
) >Design Moment
Hence Ok
If (As-Aś
ḟ 𝒔
𝒇𝒚
)<As
Wrong
True

17. - Beam
When and why A T-section consider as a T-beam or a rectangular Beam ??
hf<a consider as T-Beam hf>a consider as Rectangular-Beam

18. • b≤
• b ≤
•
𝐿
4
where L= Span of beam
𝐶𝑙𝑒𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑏𝑒𝑎𝑚
2
𝑏−𝑏𝑤
2
≤ 8hf
•
• b ≤
•
𝑏 − 𝑏𝑤≤
𝐿
2
where L= Span of beam
𝐶𝑙𝑒𝑎𝑟 𝑆𝑝𝑒𝑎𝑐𝑖𝑛𝑔 𝑜𝑓 𝑏𝑒𝑎𝑚
2
b-bw≤ 6hf
•
b≤ 4bw
bw≤ 2hf
For Symmetrical
For having a Slab on
side
For Isolated

19. Analysis of - Beam
Check is it
symmetric or Isolated
or hanging one
slab??
hf>a
???
No
Yes
Analysis consider as
rectangular beam
Analysis consider as T- beam

20. Mn1=Asf fy(d-hf/2)
Asw=As-AsfAsf=.85f’c/fy(b-bw)hf
Mn2=Aswfy(d-a/2)
Mn=Mn1+Mn2
Mu=ΦMn
Adjust Φ
by Ԑt or c/dt

21. Design of - Beam
Established the hf
generally slab
thickness
Determine b
according to ACI code
Choose bw & d

22. Calculate a trial As but
not exceed hf
For check the a trial As
Check Ԑ or c/d
Ensure that ρw≥ ρw min
hf<a
No
Yes

23. Thanks to All

24. Any Query ??