Caculate of Concrete mix design
Step1 : Choice of slump . The slump is given and consistent with Table 3-7 is 20-100.
Step2 : Maximum aggregate size . The maximum aggregate size is 20mm , which meets the limitation of 1/5 of the minimum dimension between forms and 3/4 of the minimum clear space .
Step3 : Estimation of mixing water and air content . The concrete will be exposed to freezing and thawing , therefore , it must be air entrained . From Table 3-8 , the recommended mixing water amount is 200 kg/m3 , and the air content recommended for moderate exposure is 2% .
Step4 : Water / cement ratio (W/C) . According to both Table 3-1and
Table 3-3, the estimate of the required w/c ratio to give a 28-day compressive strength 40MPa is 0.35
Step5 : Calculation of cement content . Based on the step 3 and 4 , the required cement content is 200/0.35 = 571.42kg/m3
Step6 : Estimation of coarse aggregate content . From Table 3-11 , for fineness modulus of the fine aggregate of 2.60 , the volume of dry-rodded coarse aggregate per unit volume of concrete is 0.64 . Therefore , there will be 0.64m3 coarse aggregate is per volume concrete .And , the OD weight of the coarse is 0.59×1750 = 1032.5 kg . The SSD weight is
1032.5 ×1.003 = 1035.59kg .
Step 7: Estimation of fine aggregate content. The fine aggregate content can estimated by either the weight method or the volume method.
Weight method. The estimated concrete weight is 2280 kg/m3. Although for a first trial it is not general necessary to use the more exact calculation based on Equation ,this value will be used here:
Um = (10)(2.66)(100-2)+(200/0.35)(1-2.66/31)-200(2.66-1)
Um = 2797.65 kg/m3
Based on the already determined weights of water , cement ,and coarse aggregate ,the SSD weight of the fine aggregate is :
2797.65-200-(200/0.35)-1027= 999kg
Step 8: Adjustment for moisture in the aggregate. Since the aggregates will be neither SSD nor OD in the field, it is necessary to adjust the aggregate weights for the amount of water contained in the aggregate. Since absorbed water does not become part of the mix water, only surface water needs to be considered. For the given moisture contents, the adjusted aggregate weights become:
Coarse aggregate (stock):
From W(stock) = W(OD)[1+MC(OD)]
Get W(stock) = 1032.5×1.002= 1035kg
The extra water needed for coarse aggregate absorption is :
W(SSD)- W(stock) = 1035.59-1035
=0.59kg
Fine aggregate (stock): 999×1.02= 1019kg/m³
Extra water provided by fine aggregate: 1019 -999= 20kg
The mixing water is then: 200+0.59-20= 180.59kg
Thus, the estimated batch weights per m³ are as follows:
Water: 180.59kg
Cement: 571kg
Coarse aggregate: 1035kg
Fine aggregate: 1019kg
Total: 2805.59kg
ដោយ មាឌរបស់ធ្នឹមមួយមាន V = 200mm × 350mm × 4000mm
= 0.2m × 0.35m × 4m
V =0.28 m3
នាំឲ បរិមាណដែលគេប្រើក្នុងធ្នឹមមួយគឺៈ
2805.59×0.28 = 10020 kg/m3
Consider 7% amount extra
តែ ដោយគេបានបន្ថែមបរិមាណចំនួន 7% លើបរិមាណដែលទទួលបាន
នោះ គេបាន 10020×1.07 = 10721.4kg/m3
ដោយតាមបម្រាប់ គេមានធ្នឹមចំនួន 7 ដើម
We have 10721.4×7 = 75049.8kg/m3
ដូចនេះជាលទ្ធផលយើងត្រូវការបរិមាណសរុបទាំងអស់គឺៈ 75049.8 Kg/m3
Step 9: Trial mixes. Trial mixes should be carried out using the proportions calculated. The properties of the concrete in the trial mix must be compared with the desired properties, and the mix design must be corrected as described.
Step2 : Maximum aggregate size . The maximum aggregate size is 20mm , which meets the limitation of 1/5 of the minimum dimension between forms and 3/4 of the minimum clear space .
Step3 : Estimation of mixing water and air content . The concrete will be exposed to freezing and thawing , therefore , it must be air entrained . From Table 3-8 , the recommended mixing water amount is 200 kg/m3 , and the air content recommended for moderate exposure is 2% .
Step4 : Water / cement ratio (W/C) . According to both Table 3-1and
Table 3-3, the estimate of the required w/c ratio to give a 28-day compressive strength 40MPa is 0.35
Step5 : Calculation of cement content . Based on the step 3 and 4 , the required cement content is 200/0.35 = 571.42kg/m3
Step6 : Estimation of coarse aggregate content . From Table 3-11 , for fineness modulus of the fine aggregate of 2.60 , the volume of dry-rodded coarse aggregate per unit volume of concrete is 0.64 . Therefore , there will be 0.64m3 coarse aggregate is per volume concrete .And , the OD weight of the coarse is 0.59×1750 = 1032.5 kg . The SSD weight is
1032.5 ×1.003 = 1035.59kg .
Step 7: Estimation of fine aggregate content. The fine aggregate content can estimated by either the weight method or the volume method.
Weight method. The estimated concrete weight is 2280 kg/m3. Although for a first trial it is not general necessary to use the more exact calculation based on Equation ,this value will be used here:
Um = (10)(2.66)(100-2)+(200/0.35)(1-2.66/31)-200(2.66-1)
Um = 2797.65 kg/m3
Based on the already determined weights of water , cement ,and coarse aggregate ,the SSD weight of the fine aggregate is :
2797.65-200-(200/0.35)-1027= 999kg
Step 8: Adjustment for moisture in the aggregate. Since the aggregates will be neither SSD nor OD in the field, it is necessary to adjust the aggregate weights for the amount of water contained in the aggregate. Since absorbed water does not become part of the mix water, only surface water needs to be considered. For the given moisture contents, the adjusted aggregate weights become:
Coarse aggregate (stock):
From W(stock) = W(OD)[1+MC(OD)]
Get W(stock) = 1032.5×1.002= 1035kg
The extra water needed for coarse aggregate absorption is :
W(SSD)- W(stock) = 1035.59-1035
=0.59kg
Fine aggregate (stock): 999×1.02= 1019kg/m³
Extra water provided by fine aggregate: 1019 -999= 20kg
The mixing water is then: 200+0.59-20= 180.59kg
Thus, the estimated batch weights per m³ are as follows:
Water: 180.59kg
Cement: 571kg
Coarse aggregate: 1035kg
Fine aggregate: 1019kg
Total: 2805.59kg
ដោយ មាឌរបស់ធ្នឹមមួយមាន V = 200mm × 350mm × 4000mm
= 0.2m × 0.35m × 4m
V =0.28 m3
នាំឲ បរិមាណដែលគេប្រើក្នុងធ្នឹមមួយគឺៈ
2805.59×0.28 = 10020 kg/m3
Consider 7% amount extra
តែ ដោយគេបានបន្ថែមបរិមាណចំនួន 7% លើបរិមាណដែលទទួលបាន
នោះ គេបាន 10020×1.07 = 10721.4kg/m3
ដោយតាមបម្រាប់ គេមានធ្នឹមចំនួន 7 ដើម
We have 10721.4×7 = 75049.8kg/m3
ដូចនេះជាលទ្ធផលយើងត្រូវការបរិមាណសរុបទាំងអស់គឺៈ 75049.8 Kg/m3
Step 9: Trial mixes. Trial mixes should be carried out using the proportions calculated. The properties of the concrete in the trial mix must be compared with the desired properties, and the mix design must be corrected as described.
Post a Comment