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
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
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