Maths problems 4

The five tests furnish data that appears to show that there is plastic deformation in the concrete, i.e. after cracks have occurred, and hence the deformation factor is no longer elastic, or proportional to the force applied.

Despite the above considerations, the question has requested a linear function of regression by least squares, which is presented below. Use the reference article from Wikipedia for notations and formulas.
Basically, the results of the linear regression is expressed in
y=F(x) = A+Bx


IP=xdelta=yx^2xy(x-x')(y-y')(x-x')^2F(x)=A+Bx
12.12.94.416.0913.61421634.0822443.764564227
26.223.8338.688423.8226-1.7182.9515244.80016467
37.175.9851.408942.8766-0.7680.5898245.038956034
410.55.71110.2559.9552.5626.5638445.875982605
513.77.74187.69106.0385.76233.2006446.680332464
Sum39.6926.16392.4473238.782219.45221677.38808 26.16
Average7.9385.232
D386.9404
B0.251359331
A3.236709632

You will notice that the sum of observed values of displacements is 26.16, and the sum of the values of F(x) of the linear fitted function is also 26.16, which is one of the properties of the linear regression formula, and furnishes a check of the calculations.

If you prefer to have the excel spreadsheet with all the formulae, PM me your e-mail address and I will be glad to send you the Excel file.