Answer:
F =  8849 N 
Explanation:
Given: 
Load at a given point = F =  4250 N
Support span = L = 44 mm
Radius = R = 5.6 mm
length thickness of tested material = 12 mm
First compute the flexural strength for circular cross section using the formula below:
σ
σ = FL / π R³
 Putting the given values in the above formula:
σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³
   = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³
   = 4250 (44 x 1
/1000
)) / 3.141593 ( 5.6 x 10⁻³ ) ³
   = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³
   = 187 / 3.141593 ( 5.6 x 1 / 1000 )
³
   = 187 / 3.141593 (0.0056)³
   = 338943767.745358
   = 338.943768 x 10⁶
σ = 338 x 10⁶ N/m²
Now we compute the load i.e. F from the following formula:
 = 2 σ
 = 2 σ d³/3 L
 d³/3 L
F = 2σd³/3L
   = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)
   = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
   = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
   = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
   = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)
   = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)
   = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)
   = 146016  / 125 / 3 ( 44 x 1 / 1000  )
   = ( 146016  / 125 ) /  (3 ( 11 /  250 ))
   =  97344  / 11
F =  8849 N