Give step-by-step solution with proper explanation and final answer:
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Design a mix proportioning for a concrete of M
4
0
grade by IS Code
(
1
0
2
6
2
:
2
0
0
9
)
as per following data: Type of Cement
=
OPC
4
3
of specific gravity of
3
.
1
5
Slump
=
1
5
0
mm Type of coarse aggregate
=
crushed angular aggregate
2
0
mm
(
specific gravity of
2
.
7
4
)
Type of fine aggregate
=
zone
1
–
sand with specific gravity of
2
.
7
4
.
Exposure
=
Severe
(
for RCC
)
Method of concrete placing
=
Pumping Degree of supervision
=
Good Chemical Admixture
=
Super
-
plasticizer
Question:
Give step-by-step solution with proper explanation and final answer:
VIPSearch
Google
Advanced AI Help
Design a mix proportioning for a concrete of M
4
0
grade by IS Code
(
1
0
2
6
2
:
2
0
0
9
)
as per following data: Type of Cement
=
OPC
4
3
of specific gravity of
3
.
1
5
Slump
=
1
5
0
mm Type of coarse aggregate
=
crushed angular aggregate
2
0
mm
(
specific gravity of
2
.
7
4
)
Type of fine aggregate
=
zone
1
–
sand with specific gravity of
2
.
7
4
.
Exposure
=
Severe
(
for RCC
)
Method of concrete placing
=
Pumping Degree of supervision
=
Good Chemical Admixture
=
Super
-
plasticizer
This is AI generated Response.
Asked by: naveen
Created at: 2025-10-10 08:01:26
Final Answer:
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# Step-by-Step Mix Design for M40 Grade Concrete as per IS 10262:2009
Given Data:
- **Grade of concrete:** M40 (Characteristic compressive strength = 40 MPa)
- **Cement type:** OPC 43, Specific gravity = 3.15
- **Slump required:** 150 mm
- **Coarse aggregate:** Crushed angular, 20 mm, Specific gravity = 2.74
- **Fine aggregate:** Zone I, Specific gravity = 2.74
- **Exposure:** Severe (for RCC)
- **Method of placing:** Pumping
- **Degree of supervision:** Good
- **Admixture:** Super-plasticizer
---
## 1. Target Mean Strength
\[
\text{Target Mean Strength (f_{ck'})} = f_{ck} + 1.65 \times S
\]
Where \( S \) = Standard deviation (from IS 10262:2009, Table 1, for M40: \( S = 5. \) MPa)
\[
f_{ck'} = 40 + 1.65 \times 5 = 40 + 8.25 = 48.25 \text{ MPa}
\]
---
## 2. Selection of Water-Cement Ratio
- For M40 grade, use Table 5 of IS 456:200 and IS 10262:2009.
- For severe exposure, **maximum w/c ratio = .45**.
- Use trial value: **w/c = .38** (with super-plasticizer).
---
## 3. Selection of Water Content
- For 20 mm aggregate and slump 25–50 mm, max water = 186 kg/m³ (Table 2, IS 10262:2009).
- For **150 mm slump** (with super-plasticizer), **reduce water** by 18% due to admixture effect.
- Increase by 3% for pumping = net 15% reduction.
\[
\text{Adjusted Water} = 186 \times (1 - .15) = 186 \times .85 = 158.1 \text{ kg/m}^3
\]
---
## 4. Calculation of Cement Content
\[
\text{Cement Content} = \frac{\text{Water Content}}{\text{w/c Ratio}} = \frac{158.1}{.38} = 416 \text{ kg/m}^3
\]
- IS 456:200 for severe exposure: **min cement = 320 kg/m³**, max = 450 kg/m³.
- **Use 416 kg/m³** (within limits).
---
## 5. Aggregate Proportion (Volume Method)
### 5.1 Volume of Concrete
Total volume = 1 m³
Volume of all ingredients except air = 1 - .02 (air) = **.98 m³**
### 5.2 Volume of Cement
\[
\text{Volume of Cement} = \frac{\text{Cement}}{\text{Specific Gravity} \times 100}
= \frac{416}{3.15 \times 100} = .132 \text{ m}^3
\]
### 5.3 Volume of Water
\[
\text{Volume of Water} = \frac{158.1}{1 \times 100} = .158 \text{ m}^3
\]
### 5.4 Volume of Superplasticizer
Assume dosage = 1% by weight of cement:
\[
\text{Weight of admixture} = .01 \times 416 = 4.16 \text{ kg}
\]
\[
\text{Volume of admixture} = \frac{4.16}{1.145 \times 100} = .0036 \text{ m}^3
\]
*(Assume specific gravity of admixture = 1.145)*
### 5.5 Volume of Aggregates
\[
\text{Total volume of aggregates} = .98 - (.132 + .158 + .0036) = .98 - .2936 = .6864 \text{ m}^3
\]
### 5.6 Proportion of Fine Aggregate (FA) and Coarse Aggregate (CA)
From Table 3 of IS 10262:2009:
- For 20mm aggregate, Zone I sand, **FA:CA = .35:.65** (by volume of aggregate).
\[
\text{Volume of FA} = .35 \times .6864 = .2402 \text{ m}^3
\]
\[
\text{Volume of CA} = .65 \times .6864 = .4462 \text{ m}^3
\]
---
### 5.7 Convert Volumes to Mass
#### Fine Aggregate (FA)
\[
\text{Mass of FA} = \text{Volume} \times \text{Specific Gravity} \times 100 = .2402 \times 2.74 \times 100 = 658 \text{ kg}
\]
#### Coarse Aggregate (CA)
\[
\text{Mass of CA} = .4462 \times 2.74 \times 100 = 1222 \text{ kg}
\]
---
## 6. Mix Proportion by Weight (per m³)
- **Cement:** 416 kg
- **Water:** 158 kg
- **Fine Aggregate:** 658 kg
- **Coarse Aggregate:** 1222 kg
- **Superplasticizer:** 4.2 kg
**Mix Ratio (by weight):**
Cement : FA : CA :: 416 : 658 : 1222
Or, divide by cement for ratio:
\[
1 : 1.58 : 2.94 \text{ (by weight)}
\]
---
## 7. Final Answer (Mix Design)
### **Mix Proportion for M40 Concrete (per m³):**
| Material | Amount |
|--------------------|---------|
| Cement (OPC 43) | 416 kg |
| Water | 158 kg |
| Fine aggregate | 658 kg |
| Coarse aggregate | 1222 kg |
| Superplasticizer | 4.2 kg |
**Mix Ratio by weight:**
**1 : 1.58 : 2.94** (Cement : Fine Aggregate : Coarse Aggregate)
---
## 8. Notes
- **Adjust for site moisture/water absorption.**
- **Trial mixes should be cast and tested for workability and strength.**
- **Admixture dosage may be fine-tuned as per manufacturer’s recommendations.**
---
**References:**
- IS 10262:2009 "Concrete Mix Proportioning – Guidelines"
- IS 456:200 "Plain and Reinforced Concrete – Code of Practice"
---
**Final Answer:**
> **The mix proportion (by weight) for M40 grade concrete as per IS 10262:2009 is:**
> **1 : 1.58 : 2.94 :: Cement : Fine Aggregate : Coarse Aggregate**
> with 158 kg water and 4.2 kg superplasticizer per m³.
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