## Chapter 6

## Q. 6.6

Flexural Strength of Composite Materials

The flexural strength of a composite material reinforced with glass fibers is 45,000 psi, and the flexural modulus is 18 × 10^6 psi as determined using a three-point bend test. A sample, which is 0.5 in. wide, 0.375 in. high, and 8 in. long, is supported between two rods 5 in. apart. Determine the force required to fracture the material and the deflection of the sample at fracture, assuming that no plastic deformation occurs.

## Step-by-Step

## Verified Solution

Based on the description of the sample, w = 0.5 in., h = 0.375 in., and L = 5 in. From Equation 6-14:

Flexural strength for three point bend test \sigma _{bend}=\frac{3FL}{2wh^2} (6-14)

45,000 \ psi=\frac{3FL}{2wh^2}=\frac{(3)(F)(5 \ in.)}{(2)(0.5 \ in.)(0.375 \ in.)^2}=106.7F

F=\frac{45,000}{106.7}=422 \ lb

Therefore, the deflection, from Equation 6-15, is

Flexural modulus E _{bend}=\frac{L^3 F}{4wh^3\delta } (6-15)

18\times 10^6 \ psi=\frac{L^3 F}{4wh^3\delta } =\frac{(5 \ in.)^3 (422 \ lb)}{(4)(0.5 \ in.)(0.375 \ in.)^3 \delta } (6-15)

\delta =0.0278 \ in.

In this calculation, we assumed a linear relationship between stress and strain and also

that there is no viscoelastic behavior.