The image above represents flexural strength for circular cross-section in defects.

To compute for flexural strength for circular cross-section in defects, three essential parameters are needed and these parameters are **Load at fracture (F _{f}), Specimen radius (R)** and

**Distance between support Points (L).**

The formula for calculating flexural strength for circular cross-section in defects:

σ_{fs} = ^{FfL} / _{πR³}

Where:

σ_{fs} = Flexural Strength

L = Distance between Support Points

F_{f} = Load at Fracture

R = Specimen Radius

Lets’s solve an example;

Find the flexural strength when the distance between support points is 30, load at fracture is 21 and the specimen radius is 11.

This implies that;

L = Distance between Support Points = 30

F_{f} = Load at Fracture = 21

R = Specimen Radius = 11

σ_{fs} = ^{FfL} / _{πR³}

σ_{fs} = ^{(21)(30)} / _{π(11)³}

σ_{fs} = ^{(630)} / _{π(1331)}

σ_{fs} = ^{(630)} / _{(4181.4)}

σ_{fs} = 0.150

Therefore, the **flexural strength for circular cross-section **is **0.150 Pa.**

**Calculating the Distance Between Support Points when the Flexural Strength for Circular Cross-section, the Specimen Radius and the Load at Fracture is Given.**

L = ^{σfs x πR³} / _{F}_{f}

Where;

L = Distance between Support Points

σ_{fs} = Flexural Strength

F_{f} = Load at Fracture

R = Specimen Radius

Let’s solve an example;

Find the distance between support points when the flexural strength is 44, load at fracture is 3 and specimen radius is 20.

This implies that;

σ_{fs} = Flexural Strength = 44

F_{f} = Load at Fracture = 3

R = Specimen Radius = 20

L = ^{σfs x πR³} / _{F}_{f}

L = ^{44 x π x 3³} / _{20}

L = ^{44 x π x 27} / _{20}

L = ^{3732.2} / _{20}

L = 186.61

Therefore, the **distance between support point **is **186.61.**