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.

Question

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 × [latex]10^6[/latex] 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.

Accepted Answer

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 [latex]\sigma _{bend}=\frac{3FL}{2wh^2} [/latex] (6-14)

[latex]45,000 \ psi=\frac{3FL}{2wh^2}=\frac{(3)(F)(5 \ in.)}{(2)(0.5 \ in.)(0.375 \ in.)^2}=106.7F [/latex]
[latex]F=\frac{45,000}{106.7}=422 \ lb [/latex]

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

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

[latex]18 imes 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 } [/latex](6-15)
[latex]\delta =0.0278 \ in. [/latex]

In this calculation, we assumed a linear relationship between stress and strain and also
that there is no viscoelastic behavior.

Chapter 6

Q. 6.6

Q. 6.6

Step-by-Step

Verified Solution