Question

A rectangular beam made of ABS plastic ( ) is b=20mm deep and t=10mm thick. Loads in the plane of the 20mm depth cause a bending moment of . What is the largest throughthickness edge crack that can be permitted if a safety factor of 2.5 against brittle fracture is required? (You may use .) (20points)

Answer 1

Here is the full question

A rectangular beam made of ABS plastic ( ) has dimensions b = 20 mm deep and t = 10 mm thick. Loads in the plane of the 20mm depth cause a bending moment of 10 N.m. What is the largest through thickness edge crack that can be permitted if a factor of safety of 2.5 against fracture is required?

Answer:

1.62 mm

Explanation:

The formula for the stress intensity factor (K) for various bending, curves and equation label is expressed as:

$K = 1.12 S_g \sqrt{\pi a}$

where $S_g$ = gross section nominal stress for center cracked plate of ABS plastic rectangular beam and it's given as:

$S_g =\frac{6 M}{b^2 \ t}$

$S_g = \frac{6(10 \ N.m)}{(20 mm ( \frac{1m}{1,000mm})^2(10mm)(\frac{1m}{1000mm}) }$

$S_g = 15*10^6N/m^2$

$S_g = 15MPa$

To determine the stress intensity factor K ; we have:

K = $\frac{K_{IC}}{X_K}$

From the table of fracture toughness of polymers and ceramics at room temperature, we selected the fracture toughness of ABS plastics material to be :

$3MPa \sqrt{m}$

So; K = $\frac{3MPa \sqrt{m}}{2.5}$

$= 1.2 MPa \sqrt{m}$

The largest through thickness edge crack that can be permitted due to bending moment of 10 N.m  is :

$1.12(15MPa)\sqrt{\pi a} = 1.2 MPa \sqrt{m}$

a = $\frac{0.0051}{3.14}$

a = 0.00162 m

a = 1.62 mm

Thus , the  largest through thickness edge crack that can be permitted due to bending moment of 10 N.m  is 1.62 mm