Question

EXAMPLE-2 The structural engineer has specified a concrete strength of 4,500psi. Determine the required average strength (f cr) for each of the following scenarios: A new concrete plant for which SS is unknown. A plant for which SS=520psi based on 17 test results. A plant with extensive history of producing concrete with SS=350psi. A plant with extensive history of producing concrete with SS=550psi.. TABLE 5.3.2.1 - REQUIRED AVERAGE COMPRESSIVE STRENGTH WHEN DATA ARE AVAILABLE TO ESTABLISH A SAMPLE STANDARD DEVIATION Specified compressive Required average compressive strength, psi strength, psi Use the larger value computed from Eq. (5-1) and (5-2) fi 5000 = +1.34s (5-1) fort +2.335,- 500 (5-2) Use the larger value computed from Eq. (5-1) and (5-3) > 5000 for 1.34s, (5-1) f = 0.90f +2.33s (5-3) TABLE 5.3.1.2-MODIFICATION FACTOR FOR SAMPLE STANDARD DEVIATION WHEN LESS THAN 30 TESTS ARE AVAILABLE Modification factor for sample No. of tests standard deviation Less than 15 Use Table 5.3 2 2 15 1.16 20 1.08 25 1.03 30 or more 1.00 Interpolate for intermediate numbers of tests. Modified sample standard deviation, Se, to be used to determined required average strength, from 5.3 2.1 TABLE 5.3.2.2 - REQUIRED AVERAGE PRESSIVE STRENGTH WHEN DATA ARE NOT AVAILABLE TO ESTABLISH A SAMPLE STANDARD DEVIATION Specified compressive Required average compressive strength, psi strength, psi fc <3000 +1000 3000 se s 5000 fer = fe +1200 f> 5000 1 = 1.101 + 700

Answer 1

fc' = 4000psi

1. for new plant SS is un konwn

as per table 5.3.2.2

for fc'less than 5000psi and greater than 3000psi

fcr' = fc'+1200

fcr'= 4500+1200=5700 psi

2. a plant with SS= 520psi and 17 test results

as per table 5.3.12 determine modification factor

for no. tests 15, m=1.16

for no of tests 20, m=1.08

interpolate , for no. of tests =17

m=0.06+1.08=1.14

nodified SS= 1.14*520= 592.8psi

as per table 5.3.2.1, for fc'< 5000psi

fcr'= fc'+ 1.34ss

fcr'= 4500 +[ 1.34*592.8]= 5294.352 psi------------------[1]

fcr'= fc'+2.33ss-500

fcr'= 4500+[ 2.33*592.8] -500

fcr'= 5381.224--------------------------------------[2]

select larger of [1] or [2]

fcr' = 5381.224

3. a plant with history of producing ss= 350psi

for fc' < 5000 psi

fcr'= fc'+ 1.34ss

fcr'= 4500 +[ 1.34*350]= 4969 psi------------------[1]

fcr'= fc'+2.33ss-500

fcr'= 4500+[ 2.33*350] -500

fcr'= 4815.5 psi --------------------------------------[2]

select larger of [1] or [2]

fcr'= 4815.5 psi

4. plant with ss= 550psi

for fc' < 5000 psi

fcr'= fc'+ 1.34ss

fcr'= 4500 +[ 1.34*550]= 5237 psi------------------[1]

fcr'= fc'+2.33ss-500

fcr'= 4500+[ 2.33*550] -500

fcr'= 5281.5 psi --------------------------------------[2]

select larger of [1] or [2]

fcr'= 5281.5 psi