RCC member/structure design steps in detail w.r.t. IS code references

1. Unit 1- Design of Beam subjected to Torsion
Step
No.
Calculation / Procedure Reference
1 Find Me1 = Mu + Mt
Where 𝑀 = 𝑇
⁄
.
Clause no 41.4.2, IS 456: 2000
2 If Mt > Mu, Find Me2 = Mt – Mu…….. if required Clause no 41.4.2.1, IS 456:
2000
3 Find Ast for Me1 and Me2(if required)
𝐴 =
0.5 𝑓
𝑓
× 1 − 1 −
4.6 𝑀
𝑓 𝑏 𝑑
× 𝑏 𝑑
Me1 Ast1 (bottom steel) Me2  Ast2 (top steel) * if required
4 Find Equivalent Shear=Ve
𝑉 = 𝑉 + 1.6
𝑇
𝑏
Clause no 41.3.1, IS 456: 2000,
page no 75
5 Find 𝜏 = Check 𝜏 ≤ 𝜏 Clause no 40.1, IS 456: 2000
Table no.20 , IS 456: 2000
6 Find 𝜏
If % of steel is not given find Pt% as under
𝑃 % = 100 ×
𝐴
𝑏𝑑
Where Ast tension steel or longitudinal steel calculated from
maximum BM subjected to torsion
Table no.19 , IS 456: 2000
7 If 𝜏 > 𝜏 go for minimum shear reinforcement as per clause no.
26.5.1.6
Clause no 40.4 & 26.5.1.6, IS
456: 2000
8 If 𝜏 < 𝜏 find Sv from (a) & (b) and provide minimum value
If Assumed stirrups2 legged #8mm Asv = 2 x Aᵩ = 2x 50 =100
If Assumed stirrups2 legged #10mm Asv = 2 x 78 = 156
a) 𝐴 =
.
+
. .
b)
( )
.
Clause no 41.4.3, IS 456: 2000
9 Summary and detailing:
Ast1, Ast2 , Stirrups Top steel
Bottom steel
Stirrups
(As per design)
(As per design)
(As per design)

2. Unit-2- Continuous Beams
Steps Calculations/ procedure Reference
1 Cross sectional Dimensions
Effective depth = 𝑑 . =
Adopt, effective cover= dc = 50mm
Thus, D= dc + deffe
width of beam = width of wall (generally 230, 300 etc.)
2 Loads
Dead Load
(a) Self weight of beam = b x D x 25
(b) DL on beam = (given in kN/m)
Total DL = g = (a) + (b)
Live Load = q = (given in kN/m)
***Ultimate loads
Dead load = gu = 1.5 x g
Live load = qu = 1.5 x q
3 Design Bending Moment and Shear Force
 Maximum -ve BM at interior supports
𝑀 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 × 𝑔 × 𝐿 + 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 × 𝑞 × 𝐿
 Maximum +ve BM at Center of span
𝑀 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 × 𝑔 × 𝐿 + 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 × 𝑞 × 𝐿
 Maximum Shear at support (next to end)
𝑉 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 × 𝑔 × 𝐿 + 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 × 𝑞 × 𝐿
Table no. 12 and 13 of IS 456:2000, pp36
 C1, C2, C3 column of respective table
4 Reinforcement
Find Steel for M-ve and M+ve 𝐴 =
.
1 − 1 −
.
𝑏𝑑
Select diameter (Ф), find area (A Ф) and number of bars
No. of bars=
.
Dia. (Ф) 8 10 12 16 20 25
Area (A Ф) 50 78 113 201 314 490
5 Shear reinforcement
Find 𝜏 = Check 𝜏 ≤ 𝜏
Clause no 40.1, IS 456: 2000
Table no.20 , IS 456: 2000
Find 𝜏
If % of steel is not given find Pt% as under
Table no.19 , IS 456: 2000
C 1 C 2 C 3 C 4 C 4 C 3 C 2 C 1
C 1 C 2 C 2 C 1
C 3 C 4 C 3
B M C O E F F IC IE N T (C O L U M N N O . IN T A B L E 1 2 )
S F C O E F F IC IE N T (C O L U M N N O . IN T A B L E 1 3 )

3. 𝑃 % = 100 ×
𝐴
𝑏𝑑
Where Ast tension steel or longitudinal steel calculated
from maximum BM subjected to torsion
If 𝜏 > 𝜏 go for minimum shear reinforcement as per
clause no. 26.5.1.6
Clause no 40.4 & 26.5.1.6, IS 456:
2000
If 𝜏 < 𝜏 find Sv from
𝑉 = 𝑉 − 𝜏 𝑏 d
If Assumed stirrups2 legged #8mm Asv = 2 x Aᵩ = 2x 50
=100
If Assumed stirrups2 legged #10mm Asv = 2 x 78 = 156
𝑉 =
0.87 𝑓 𝐴 𝑑
𝑆
Clause no 41.4.3, IS 456: 2000
6 Deflection Control
Find &
𝐿
𝑑
=
𝐿
𝑑
× 𝑘 × 𝑘 × 𝑘
<  then safe
𝐿
𝑑
= 26
𝑘 = 1
𝑘 = 1
7 Detailing
Curtail bars Curtail bars Curtail bars
Extra Top Extra Top Top Straight Bars
Bottom Straight
Bars
D
L1 L2 L3

4. For redistribution of Moments
Step no. 3 Changes as below
Find Fixed end moment for each beam AB, BC, CD,…. 𝐹𝐸𝑀 =
Find Span moments for each beam AB, BC, CD,…. 𝑆𝑝𝑎𝑛 𝑀𝑜𝑚𝑒𝑛𝑡𝑠 =
Create table Find Final Span moments and End Moments:
Support B C (if required)
process
Span Moment
FEM
(average of AB
& BC)
Span Moment
FEM
(average of CB
& CD)
Span Moment
X1 X2 X3 X4 X5
R %
redistribution
-X1 * R/100
-X3 * R/100
-X5 * R/100
-X3 * R/100
Carry over
-X3 * R/100*(1/2)
-X1 * R/100*(1/2)
-X5 * R/100*(1/2)
-X3 * R/100*(1/2)
Final Moment
Sum sum sum sum Sum
Span Moment
Support
Moment
Span Moment
Support
Moment
Span Moment

5. Unit No.3 - Design of Circular Water Tank
Flexible Base / Not casted monolithically /Not restrained at base
Step No. Calculation /Procedure Referance
1. Thk of Wall
a) 150 mm
b) 30mm per m depth + 50 mm
Provide greater of a) & b)
2. Hoop Tension 𝑇 =
. .
where 𝛾 = 9.81𝑘𝑁/𝑚
3. Find Ast required for 1 m ht.
𝐴 =
Minimum steel required = 0.3 % of gross area.
4. Select diameter (Ф) of bar and find spacing
Dia. (Ф) 8 10 12 16 20 25
Area (A Ф) 50 78 113 201 314 490
Spacing = S =
×
.
……. Provide spacing less than actual required.
Provide this steel near both the face.
5. Actual Ast Provided
𝐴 . =
𝐴 × 1000
𝑆
6. Tensile Stress in Concrete
𝐶 = ( × ) (( ) )
𝐹𝑜𝑟 𝑀20, 𝐶 = 1.2 𝑁/𝑚𝑚 , b = 1000,
𝐴 = 𝐴 .
Table 1, IS 3370 Part 2 , PP.7
7. Vertical Distribution Steel
At least 0.3% of gross area.
=
0.3
100
× 1000 × 𝑡
Select diameter (Ф) of bar and find spacing
8. Base Slab
i) 150 mm thick slab
Top mesh and bottom mesh of steel 10 𝑚𝑚 ∅ @ 300 𝑚𝑚 .
9 Detailing
See attached

6. Rigid Base / casted monolithically / restrained at base
Approximate Method
Step No. Calculation /Procedure Referance
1. Height of cantilever effect: 𝑜𝑟 1 𝑚 𝑤ℎ𝑖𝑐ℎ𝑒𝑣𝑒𝑟 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟
2. Hoop Tension 𝑇 =
.( ).
where, 𝛾 = 9.81𝑘𝑁/𝑚
3. Find Ast required for 1 m ht.
𝐴 =
Minimum steel required = 0.3 % of gross area.
4. Select diameter (Ф) of bar and find spacing
Dia. (Ф) 8 10 12 16 20 25
Area (A Ф) 50 78 113 201 314 490
Spacing = S =
×
.
……. Provide spacing less than actual required.
Provide this steel near both the face.
5. Actual Ast Provided, 𝐴 . =
×
6. Find thickness of Wall(t)
Tensile Stress in Concrete
𝐶 = ( × ) (( ) )
𝐹𝑜𝑟 𝑀20, 𝐶 = 1.2 𝑁/𝑚𝑚 , b = 1000,
𝐴 = 𝐴 .
Table 1, IS 3370 Part 2 , PP.7
7. Vertical Distribution Steel
At least 0.3% of gross area.
=
0.3
100
× 1000 × 𝑡
Select diameter (Ф) of bar and find spacing
8. Design of Bottom cantilever part ( For h )
Max. moment due to cantilever effect 𝑀 = ( × 𝑏 × ℎ) ×
Where, b = (𝛾 × 𝐻)
9. Area of Steel
𝐴
𝑀
𝜎 × 𝑗 × 𝑑
𝜎 − 𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 2 𝑃𝑃. 8
𝜎 = 115 𝐹𝑜𝑟 𝐹𝑒 − 250
𝜎 = 150 𝐹𝑜𝑟 𝐹𝑒 − 415
𝜎 = 205 𝐹𝑜𝑟 𝐹𝑒 − 500
10. Haunch Reinforcement
Haunch Reinforcement= Min. steel= (0.3% × 𝑏 × 𝑑)
11. Base Slab
i) 150 mm thick slab
Top mesh and bottom mesh of steel 10 𝑚𝑚 ∅ @ 300 𝑚𝑚 .
12 Detailing
See attached

7. Rigid Base / casted monolithically / restrained at base
I S code Method
Step No. Calculation /Procedure Referance
1. Thk. Of wall (i) 150mm (ii) 30mm per M depth +50mm
Find 𝐻
𝐷𝑡
2. Hoop Tension 𝑇 =
.( ).
where, 𝛾 = 9.81𝑘𝑁/𝑚
3. Find Ast required for 1 m ht.
𝐴 =
Minimum steel required = 0.3 % of gross area.
4. Select diameter (Ф) of bar and find spacing
Dia. (Ф) 8 10 12 16 20 25
Area (A Ф) 50 78 113 201 314 490
Spacing = S =
×
.
……. Provide spacing less than actual required.
Provide this steel near both the face.
5. Actual Ast Provided 𝐴 . =
×
6. Find thickness of Wall(t)
Tensile Stress in Concrete
𝐶 = ( × ) (( ) )
𝐹𝑜𝑟 𝑀20, 𝐶 = 1.2 𝑁/𝑚𝑚 , b = 1000,
𝐴 = 𝐴 .
Table 1, IS 3370 Part 2 , PP.7
7. Vertical Distribution Steel: At least 0.3% of gross area =
.
× 1000 × 𝑡
Select diameter (Ф) of bar and find spacing
8. Design of Bottom cantilever part ( For h )
Max. moment due to cantilever effect = 𝑀 = ( × 𝑏 × ℎ) ×
Where, b = (𝛾 × 𝐻)
9. Area of Steel, 𝐴
× ×
𝜎 = 115 𝐹𝑜𝑟 𝐹𝑒 − 250, 𝜎 = 150 𝐹𝑜𝑟 𝐹𝑒 − 415,
𝜎 = 205 𝐹𝑜𝑟 𝐹𝑒 − 500
𝜎 − 𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 2 𝑃𝑃. 8
10. Distribution Steel, = × (0.3% × 𝑏 × 𝑑)
b=1000 , d=effective depth
11. Base Slab, = 150 mm thk
Top mesh and bottom mesh of steel 10 𝑚𝑚 ∅ @ 300 𝑚𝑚 .
12 Detailing
See attached

8. t
H
h
# -(dia)- mm@ -(spacing)- c/c
Hoop Tension Steel
(refer step no. 4)
# -(dia)- mm@ -(spacing)- c/c
Distribution Steel
(refer step no. 7)
Horizontal Steel
Vertical Steel
# -(dia)- mm@ -(spacing)- c/c
(refer step no. 8)
Base slab steel
thk. of
base slab
Detailing of circular water tank with flexible base
t
H
h
# -(dia)- mm@ -(spacing)- c/c
Hoop Tension Steel
(refer step no. 4)
# -(dia)- mm@ -(spacing)- c/c
Distribution Steel
(refer step no. 7)
Horizontal Steel
Vertical Steel
# -(dia)- mm@ -(spacing)- c/c
(refer step no. 11)
Base slab steel
thk. of
base slab
Detailing of circular water tank with rigid base
# -(dia)- mm@ -(spacing)- c/c
Haunch Reinforcement
(refer step no. 10)
# -(dia)- mm@ -(spacing)- c/c
Cantilever BM Steel
(refer step no. 9)
Vertical Steel

9. Design of Rectangular or Square Water Tank
Step No. Calculation /Procedure
1 Design constants
𝜎 − 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒 𝑛𝑜. 21 𝐼𝑆 456: 2000
𝜎 − 𝐼𝑆 3370 𝑃𝑎𝑟𝑡 𝐼𝐼 for Fe250=115, Fe415=150, Fe500=205
𝑚 = 280/(3𝜎 ),
𝑛 = ,
𝑗 = 1 − (𝑛/3)
2. Thk. Of wall (i) 150mm (ii) 30mm per M depth +50mm (iii) 60mmper meter length
of side
Provide maximum of (i) to (iii)
3 Design of Short wall
(a) Mid span section:
 𝑇 =
.( ).
where L-length of longer wall
 𝑀 =
.( ).
𝟏𝟔
where l-length of short wall
 𝑁𝑒𝑡 𝐵𝑀 = 𝑀 − 𝑇𝑥 where x = (t/2)-cover
 Steel for net BM 𝐴
× ×
 Steel for pull 𝐴
 Total Steel = 𝐴 = 𝐴 + 𝐴
(b) Corner section:
 𝑇 =
.( ).
where L-length of longer wall
 𝑀 =
.( ).
𝟏𝟐
where l-length of short wall
 𝑁𝑒𝑡 𝐵𝑀 = 𝑀 − 𝑇𝑥 where x = (t/2)-cover
 Steel for net BM 𝐴
× ×
 Steel for pull 𝐴
 Total Steel = 𝐴 = 𝐴 + 𝐴
4 Design of Long wall
(c) Mid span section:
 𝑇 =
.( ).
where l-length of short wall
 𝑀 =
.( ).
𝟏𝟔
where L-length of long wall
 𝑁𝑒𝑡 𝐵𝑀 = 𝑀 − 𝑇𝑥 where x = (t/2)-cover
 Steel for net BM 𝐴
× ×
 Steel for pull 𝐴
 Total Steel = 𝐴 = 𝐴 + 𝐴
(d) Corner section:
 𝑇 =
.( ).
where l -length of short wall
 𝑀 =
.( ).
𝟏𝟐
where L-length of long wall
 𝑁𝑒𝑡 𝐵𝑀 = 𝑀 − 𝑇𝑥 where x = (t/2)-cover
 Steel for net BM 𝐴
× ×

10.  Steel for pull 𝐴
 Total Steel = 𝐴 = 𝐴 + 𝐴
5 Design of Vertical Steel in both wall
Max. moment (cantilever effect) = 𝑀 = ( × 𝛾 × 𝐻 × ℎ) ×
𝐴 =
𝑀
𝜎 × 𝑗 × 𝑑
Minimum steel required = 0.3 % of gross area.
Ast > Astmin
6 Base Slab, = 200 mm thk
Top mesh and bottom mesh of steel 12 𝑚𝑚 ∅ @ 300 𝑚𝑚 .
7 Detailing
See attached
l
L
L1 L1
L2 L2
S1S2
S1S2
PLAN

11. t
H
h
Distribution Steel
Horizontal Steel
Vertical Steel
Base slab steel
thk. of
base slab
SECTION "L1-L1"
Cantilever BM Steel
Vertical Steel
SECTION "S1-S1"
LONG WALL SHORT WALL
Inner Face of wall
* *
*
* change steel in above case as per wall
t
H
h
Distribution Steel
Horizontal Steel
Vertical Steel
Base slab steel
thk. of
base slab
Cantilever BM Steel
Vertical Steel
SECTION "L2-L2" SECTION "S2-S2"
LONG WALL SHORT WALL
Outer face of wall
* change steel in above case as per wall
*
* *

12. Unit No.4 - Analysis of Prestress.
Cases/ Cable Profiles Calculation / Procedure
1. Tendon placed
concentric
Stress at top edge = +
Stress at bottom edge = −
2. Tendon placed at an
eccentricity
Stress at top edge = + −
.
Stress at bottom edge = − +
.
Where 𝑀 = at mid span (where, w = DL udl + LL udl)
𝑀 = 0 at end section
3. Prestress with Bent
Tendon
Stress at top edge = +
Stress at bottom edge = −
Where 𝑀 =
( ).
+
W- Point load at center
w = DL udl + LL udl
4. Prestress with
Parabolic Tendon
Stress at top edge = +
Stress at bottom edge = −
Where 𝑀 = and 𝑤 =
.

13. Unit no. 5.- Losses in Prestress.
Cause Calculation / Procedure Pre-tension Post-
tension
1. Loss of stress due to
elastic shortening of
concrete
∆𝑓 = 𝑚. 𝑓
Where –
∆𝑓 = loss of prestress
m = modular ratio =
𝑓 = stress in concrete  
2. Loss of stress due to
creep of concrete
∆𝑓 = ∅ . 𝑚. 𝑓 if ∅ is given in coefficient
Or
∆𝑓 = ∅ . 𝑓 . 𝐸 if ∅ is given per 𝑁
𝑚𝑚
 
3. Loss of stress due to
shrinkage of concrete
∆𝑓 = 𝑆ℎ𝑟𝑖𝑛𝑘𝑎𝑔𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 × 𝐸
 
4. Relaxation Loss 1% to 5 % of initial stress.
Or whichever value given in example  
5. Loss due to anchorage
slip
∆𝑓 =
∆𝛼 . 𝐸
𝐿
Where, ∆𝛼 = effective slip
𝐸 = Young’s modulus for tendon
L = Length of Tendon
 
6. Loss due Friction
( Wobble effect /
length effect)
𝑃 = 𝑃 (1 − 𝜇𝛼 − 𝑘𝑥 )
Where -
𝑃 = prestressing force in tendon at any distance x
from jack.
𝑃 = prestressing force at jacking end.
𝜇 = co-efficient of friction in curve 0.55-for
smooth conc.& 0.30- for steel fixed in duct
𝛼 = cumulative angle in radian
𝑘= friction coefficient (0 to 0.0066 )
∆𝑓 = 𝑓 . 𝑘. 𝑥
𝑓 = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒
 
Total loss of stress ∆𝑓 = ∑(1+2+3+4) ∑ (2+3+4
+5+6)
% loss of stress
=
∆𝑓
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠
× 100

14. Unit no. 6.
Design of Pre-stressed member (Beam)
Step Procedure (for Rectangular beam section)
1 Calculate live load moment - ML
𝑀 =
𝑤𝑙
8
2 Section Modulus and width of Section: 𝑍 =
Choose convenient depth, 𝑑 = × 𝑙
Now find ‘b’ for section using formula, 𝑍 =
3 Prestressing Force: 𝑃 = × 𝐴
4 Steel requirements 𝐴 =
select dia. of wire and number of wires
5 Dead load Moment = MD
DL= w = b x d x 25  𝑀 =
6 Eccentricity (e)
𝑒 =
2 𝑀 + 𝑀
2 𝑃
7 Drawing:
e
b
d
PP
L
Longitudinal Sectional Elevation
Section

15. Unit no. 6.
Design of Pre-stressed member (Beam)
Step Procedure (for Symmetrical I beam section)
1 Safe Stresses
Concrete:
At transfer, safe compressive stress = fr = 0.5fck
At service, safe compressive stress = fc = 0.4fck
Permissible tensile stress = ft = 0.129 𝑓
Steel: Safe stress in steel at service = 60% of ultimate stress
Ultimate stress = 1500N/sqmm
2
Calculate live load moment - 𝑀 =
Dead load Moment = MD = 20% of ML
Total BM =Mt = MD +ML
3 Overall depth “d” in mm  𝑑 = 𝑘 𝑀
Where k = coefficient = 35 and Mt is in kNm
4 Prestressing Force (P) 𝑃 =
.
5 Cross Sectional Properties:
 assume value of ‘x’ in range of 120 mm to 150mm
 Area of concrete required =𝐴 = 𝑃/(0.5𝑓 )
 Find bf using calculated Ac, using equation below
𝐴 = (𝑥 × 𝑏𝑓) × 2 + (𝑑 − 2𝑥) × 𝑥
 Provide bf greater than required
 Find Actual area provided = A
 Find Section modulus = Z = I/y
6 Steel requirements:
Area of tendons =As = (P/ safe stress in steel at service)
Choose suitable diameter of wire and find number of wires
Number of wires = As/ Area of single wire
𝑃 = 𝑆𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑡𝑒𝑒𝑙 × 𝑎𝑐𝑡𝑢𝑎𝑙 𝑡𝑒𝑛𝑑𝑜𝑛 𝑎𝑟𝑒𝑎 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
𝑃 =
𝑃
𝑙𝑜𝑠𝑠 𝑓𝑎𝑐𝑡𝑜𝑟
=
𝑃
(100 − % 𝑙𝑜𝑠𝑠)
100
7 Check for stresses
 Assume eccentricity (e) of cable at certain dist below NA
(a) At initial stresses , at transfer
𝑓 / = ± ∓
.
< fr…..OK
(b) Final stresses, at services
𝑓 / = ± ± ∓
.
< fc….OK
8 Design Summary and drawing showing bf,d,e,x,P
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bf