Finding Q in shearing stress formula for composite cross-section

[wmsaqqa]

In this question below in the 2 files, i have a problem with the ( Q ) in the shearing stress.. can someone help me and explain for me how they get [Q=(32)(30)(15)] for point (a) and how about point (b) also !? what i know that Q=A.y , how did they get (32)(30) for the area and (15) for the height !

http://www.mediafire.com/?1m94xb7ndh7a310

http://www.mediafire.com/?2tgguunddyn2k3a

 

[SteamKing]

In this question below in the 2 files, i have a problem with the ( Q ) in the shearing stress.. can someone help me and explain for me how they get [Q=(32)(30)(15)] for point (a) and how about point (b) also !? what i know that Q=A.y , how did they get (32)(30) for the area and (15) for the height !

http://www.mediafire.com/?1m94xb7ndh7a310

http://www.mediafire.com/?2tgguunddyn2k3a

In calculating shear stress, the area A is not the area of the whole cross section, but the area of the cross section between a line thru the point where stress is to be calculated, parallel to the neutral axis (NA), and the outer edge of the section above this axis.

Thus, for point A, the area is 32(30). The centroid of this area above the NA of the section is 15, giving a Q at point A of 32(30)(15) mm^3. Check the diagram shown in the solution.

For point B, the area is 32(15), since point B is located 15 from the NA. Q at point B is then 32(15)(15+7.5) mm^3.

Since point C is located on the outer fiber, the area and the Q are both 0.