M3-01__RCC DESIGN BASIC310311 final2.ppt

M3-01__RCC DESIGN BASIC310311 final2.ppt

1. RC Design Basic
Aug 27, 2025
Md. Syied Mahbub Morshed
Executive Engineer
Design Division-1
Public Works Department

2. Data Given: Moment M, Beam Width b
Step 1:
Find d:
Aug 27, 2025
(Working Stress Design : WSD)
Design of Singly reinf. Beam for a given moment
2
2
1
dbjkfM
cc
rn
n
k


Ec
Es
n
c
s
f
f
r
3
1
k
j
Step 2:
Find steel As: djfAM
sss

3. Aug 27, 2025
(Working Stress Design : WSD)
Design of Singly reinf. Beam for a given moment…..
Step 4:
Revised steel As:
When the difference between revised steel and original is
negligible, the steel area As is final.
djfAM
sss

Step 3:
The values of k and j will be revised to be consistent with the
estimate of the steel area:
bd
A
s
 nnnk  
2
2
3
1
k
j
)(2 rnr
n
p
b



4. Data Given: Moment M, Beam Width b, depth d
Step 1: Find Maximum resisted moment by concrete Mcc:
Aug 27, 2025
(Working Stress Design : WSD)
Design of Doubley reinf. Beam for a given moment and
size
2
2
1
dbjkfM
cc

rn
n
k


Ec
Es
n
c
s
f
f
r
3
1
k
j
Step 2: If Mc is less than M, that indicates nessecity of
compression steel with denoting M
c
as M
1
Step 3: Find steel corresponding M
1
djfAM
ss11


5. Step 4: Find Moment resisted by Compression steel
M
2
= M – M
1
Aug 27, 2025
(Working Stress Design : WSD)
Design of Doubley reinf. Beam for a given moment and
size…
Step 5: Find steel corresponding M
2
Step 6: Total tensile steel area As = A
s1 + A
s2
)'(
22
ddfAM
ss

Step 7: Now the stress in the compressive steel

sss
f
k
ddk
ff 



1
)/'(
2'

6. Step 8: Find Compression steel A
s
’
Aug 27, 2025
(Working Stress Design : WSD)
Design of Doubley reinf. Beam for a given moment and
size…
)'('
'
2
ddf
M
A
s
s



7. USD: Design Procedure for Singly reinforced (Tension
pnly)Beam:
0285.0
87000
8700085.0
'
1


















yy
c
bal
ff
f

00333.0
min

Problem: Select a rectangular beam size and required reinforcement As to carry
service load moments Mu = 1670 kips-in .
fc′ = 4000 psi fy = 60,000 psi
Aug 27, 2025
STEP 1: Find maximum steel ratio for tension controlled section
For tension controlled section :
01806.063375.0
max 
bal

8. Step 2: Choose a steel ratio in between p
max
and p
min
to find effective
depth d:
an increase of conc. Section by 14% achieves the steel saving by
20%.
 Typically 0.50 p
max to 0.75 p
max is economic ratio.  
32'2
5.3733/59.01 inbdffyfbdM
cyu
 
Step 3: Find steel area
Aug 27, 2025
USD: Design Procedure for Singly reinforced Beam:
b=10” and d=19.3”
2
max 737.150.0 inbdbdAs  
Revise steel ratio rounding up d.

9. Aug 27, 2025
USD: Design Procedure for Singly reinforced Beam(size fixed)
Problem: Select required reinforcement As to carry service load moments
Mu = 1670 kips-in . Given beam size : b=10”, h=20”
fc′ = 4000 psi fy = 60,000 psi
in
bf
fA
a
c
ys
51.3
'85.0

Step 1: Assume a: Let a=4.0”
2
99.1
)2/(
in
adf
M
A
y
u
s



Step 2: Check a:
Step 3: Revise A
s
:
2
96.1
)2/(
in
adf
M
A
y
u
s



Step 4: Check a:
in
bf
fA
a
c
ys
46.3
'85.0
 As=1.96 in2

10. Aug 27, 2025
USD: Design Procedure for Singly reinforced Beam..
Step 5: Check minimum reinf.: 00333.00112.0/  bdAs
Step 6: Check the  value: inac 07.485.0/46.3/
1 
Variation of  with strain in steel
Variation of  with 60 grade steel
And c/d
t
=0.232=0.90 OK

11. Aug 27, 2025
Prob: Determine the required area of reinforcement for moments Mu = 211 ft-kips.
Check crack control requirements of 10.6. given b=10” and h =20”
USD: Design Procedure for doubly reinforced Beam..

12. Aug 27, 2025
  
kftkft
ffyfbdM
cyu


211175
/59.01
'2


Step 1: Find max moment that can be resisted with tension reinf. only
01806.063375.0
max

bal
where

So compression reinf. required
Step 2:Moment to be resisted by comp. steel is: M
u1 =175 ft-k
Required steel A
s1 =p
maxbd =0.01806*10*(20-16)=2.88 in2
USD: Design Procedure for doubly reinforced Beam..

13. Design of Reinforced Concrete
Beams for Shear
Aug 27, 2025

14. Uncracked Elastic Beam BehaviorUncracked Elastic Beam Behavior
Look at the shear and bending
moment diagrams. The acting shear
stress distribution on the beam.
Aug 27, 2025

15. Strength of Concrete in Shear (No Strength of Concrete in Shear (No
Shear Reinforcement)Shear Reinforcement)
Shear span to depth ratio, a/d (M/(Vd))
5
l
effect little has Rato 2
d
a
requireddesign detail more spansshear deep 2
n



d
If
d
a
Aug 27, 2025
Design approach will be as deep beam

16. Location of Maximum Shear for Beam Design
Non-pre-stressed members:
Sections located less than a distance d from face of support may be
designed for same shear, V
u, as the computed at a distance d.
When
Compression fan carries
load directly into support.
Aug 27, 2025
1.The support reaction introduces compression into the end regions of the
member
2. No concentrated load occurs with in d from face of support

17. Location of Maximum Shear for Beam
Design
Compression from support
at bottom of beam tends to
close crack at support
Aug 27, 2025

18. Design Procedure for ShearDesign Procedure for Shear
(1)Calculate V
u
(2) Calculate V
c
:
(3) Check




done. no, If
4) to(goent reinforcem webadd yes, If
2
1
is
cu
VV
Aug 27, 2025
bdf
c
'2V
c


19. Design Procedure for ShearDesign Procedure for Shear
(4)
required reinf.shear
2
1
If
uc
VV
 








v
w
ysv
y
w
v
A
b
fA
s
f
sb
A min for
50
or 50
maxmin
Also:
(Done)
"24
2
max

d
s
Aug 27, 2025

20. Design Procedure for ShearDesign Procedure for Shear
(5)
cus
scnu
scu
VVV
VVVV
VVV





 d)(req' calulate , If
Check:
  section theincrease otherwise, 8 dbfV
wcs

Aug 27, 2025

21. Design Procedure for ShearDesign Procedure for Shear
(6) Solve for required stirrup spacing(strength) Assume # 3, #4, or #5
stirrups
(7) Check minimum steel requirement

s
ysv
V
dfA
s

50
max
w
ysv
b
fA
s
Aug 27, 2025

22. Design Procedure for ShearDesign Procedure for Shear
(8) Check maximum spacing requirement (ACI 11.5.4)
(9) Use smallest spacing from steps 6,7,8
"12
4
4 If
"24
2
4 If
maxc
maxc


d
sdbfV
d
sdbfV
ws
ws
Aug 27, 2025

23. Aug 27, 2025
Design for Torsion

24. Cracks
•generated due to pure torsion
follow the principal stress
trajectories
•The first cracks are observed at
the middle of the longer side.
•Next, cracks are observed at
the middle of the shorter side.
TORSIONAL Eff
ects
Aug 27, 2025

25. Design for Torsion and shear both:
Data available: V
u, T
u
i) Check if ΦTcr ≥ Tu
T
cr = 4√ (f
c′)A
2
cp/p
cp
No Torsional reinforcement is required if T
u
< ΦTcr/4
ii) Check if V
u ≥
(iii) Check for size of beam:

The size of the beam is O.K if:
√ [{V
u
/ (b
w
d)} 2 + {T
up
h/ (1.7A
oh
2)} 2] ≤ Φ {2√ (f
c
′) + 8√ (f
c
′)}
Where, V
u and T
u at critical section.
required reinf.shear
2
1
If
cuu
VV c
V
2
1

26. Aug 27, 2025
(a) Torsional reinforcement (at critical section,)
A
t
= Tus/(Φ2fyvAo cotθ)
(c) Toatal shear & Torsion Reinf.:
Atotal = 2At + Av
(d) Maximum spacing allowed
For torsion s
max
= p
h
/8 or 12”
(b) Shear reinforcement (at critical section) Av

yv
w
f
Sb50


27. Aug 27, 2025
(e) Longitudinal reinforcement.

2
cot

















yl
yv
h
t
l
f
f
p
s
A
A




























yl
yvl
h
yl
cpc
l
f
f
S
A
p
f
Af
A
'5
min
(f)
Distribute the torsional longitudinal reinf. At different layers(top, middle and
bottom but not more than 12”)

28. THANK YOU

Aug 27, 2025