Question

1. Often, material variability in property such as strength is encountered. A sample data (max failure loads) for clear wood specimens tested parallel to the grain is given below. All loads in kips (15 samples): 16.6, 15.2, 15.4, 12.1, 14.9, 13.2, 17.1, 15.5, 15.0, 14.8, 15.2, 16.9, 12.9, 15.5, 16.0. The Normal Distribution function (Gaussian) is often found to best describe variability that is found in nature, including material properties. Using such method estimate the mean and the standard deviation. For parts (ii) and (iii) below, use fer' -fe + ts (NOT ACI) and the t-Table in Unit 8 - here, vou DO NOT have to use modification factors for number of samples<30. ii) What is the probability that a specimen will test below 13.7 kips? iii) Determine the 5% exclusion limit for the above wood. 5% exclusion limit means the strength value (in kips) such that only 5% of test results will fall below that strength

Answer 1

Sarn Samples 16 6 16-2 15-f /9 13.2 17.1 5 5 H-8 '5L 69 12.9 8442-/5 94 9 4-6 28:28) a 37 rm 37 noler Cusve f(X ) = /2 13-2- 1.37 12 이 549 1 5.49 y

tm L. 1 、GS 57,