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:
[tex]K = 1.12 S_g \sqrt{\pi a}[/tex]
where [tex]S_g[/tex] = gross section nominal stress for center cracked plate of ABS plastic rectangular beam and it's given as:
[tex]S_g =\frac{6 M}{b^2 \ t}[/tex]
[tex]S_g = \frac{6(10 \ N.m)}{(20 mm ( \frac{1m}{1,000mm})^2(10mm)(\frac{1m}{1000mm}) }[/tex]
[tex]S_g = 15*10^6N/m^2[/tex]
[tex]S_g = 15MPa[/tex]
To determine the stress intensity factor K ; we have:
K = [tex]\frac{K_{IC}}{X_K}[/tex]
From the table of fracture toughness of polymers and ceramics at room temperature, we selected the fracture toughness of ABS plastics material to be :
[tex]3MPa \sqrt{m}[/tex]
So; K = [tex]\frac{3MPa \sqrt{m}}{2.5}[/tex]
[tex]= 1.2 MPa \sqrt{m}[/tex]
The largest through thickness edge crack that can be permitted due to bending moment of 10 N.m is :
[tex]1.12(15MPa)\sqrt{\pi a} = 1.2 MPa \sqrt{m}[/tex]
a = [tex]\frac{0.0051}{3.14}[/tex]
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
