Answer :
The answer to the given question is 3.9mPa
How to solve
Moment of inertia (using the right figure);
[tex]I = \frac{300(440)^3}{12} -\frac{(300-20)(400)^3}{12}[/tex]
[tex]\Rightarrow I =636266667\;\;mm^4[/tex]
The first moment of area (using left figure);
[tex]Q = A_1y_1 +A_2y_2[/tex]
[tex]\Rightarrow Q = (200\times20)\left ( \frac{200}{2} \right ) +(20\times300)\\left ( 200+\frac{20}{2} \right )[/tex]
[tex]\Rightarrow Q =1660000\;\;mm^3[/tex]
Shear stress will be;
[tex]\tau = \frac{VQ}{It} = \frac{30(10^3)(1660000)}{(636266667)(20)}[/tex]
[tex]\Rightarrow \tau = 3.91\;\;MPa[/tex]
The portion of stress that is parallel to a material's cross-section is known as shear stress. It results from the shear force, which is part of the force vector that is parallel to the cross-section of the material.
Contrarily, normal stress results from the force vector component that is perpendicular to the material cross-section that it affects.
Read more about maximum shear stress here:
https://brainly.com/question/22435706
#SPJ1
