Answer:
The maximum load is when the hole size is 12 mm so the largest bit such that the load is not exceeding the P value is 12 mm.
Explanation:
As the question is not complete, the complete question is found online and is attached herewith.
From the given data
Allowable stress is given as [tex]\sigma_{all}=145 MPa[/tex]
Diameters of the bits are 12 mm, 15 mm, 18 mm, 21 mm, 24 mm
For the fillet, the dimensions are given in the attached question which are as
radius rf=9 mm
Small diameter d=75 mm
Large Diameter D=112.5 mm
The ratios of the diameters are calculated as
D/d=1.5
rf/d=0.12
So for these values and from the table, the value of K is found as 2.1
The area is given as
A=d*t
where d=75 mm, t=12 mm
A=75*12=900x10^-6 m2
Now the value of the average stress is given as
[tex]\sigma_{av}=\dfrac{\sigma_{max}}{K}\\P=\sigma_{av}A\\P=\dfrac{\sigma_{max} A}{K}\\P=\dfrac{145 \times 10^6 *900\times 10^{-6}}{2.1}\\P=62.1 kN[/tex]
So for the various holes the value of r is the radius of the hole so
D=112.5 mm as given above whereas
d=D-2r
so the areas, value of k and respective values of P and σ are calculated for each radii and are attached with the solution.
As seen from the attached values, it is observed that the maximum load is when the hole size is 12 mm so the largest bit such that the load is not exceeding the P value is 12 mm.
