Answer:
point B where [tex]Q_B = 101.25 \ in^3[/tex] has the largest Q value at section a–a
Explanation:
The missing diagram that is suppose to be attached to this question can be found in the attached file below.
So from the given information ;we are to determine the point that has the largest Q value at section a–a
In order to do that; we will work hand in hand with the image attached below.
From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.
We also have block partitioned into different point segments . i,e A,B,C, D
For point A ;
Let Q be the moment of the Area A;
SO ; [tex]Q_A = Area \times y_1[/tex]
where ;
[tex]y_1 = (6 - \dfrac{1.5}{2})[/tex]
[tex]y_1 = (6- 0.75)[/tex]
[tex]y_1 = 5.25 \ in[/tex]
[tex]Q_A =(L \times B) \times y_1[/tex]
[tex]Q_A =(6 \times 1.5) \times 5.25[/tex]
[tex]Q_A =47.25 \ in^3[/tex]
For point B ;
Let Q be the moment of the Area B;
SO ; [tex]Q_B = Area \times y_2[/tex]
where ;
[tex]y_2 = (6 - \dfrac{1.5 \times 3}{2})[/tex]
[tex]y_2= (6 - \dfrac{4.5}{2}})[/tex]
[tex]y_2 = (6 -2.25})[/tex]
[tex]y_2 = 3.75 \ in[/tex]
[tex]Q_B =(L \times B) \times y_1[/tex]
[tex]Q_B=(6 \times 4.5) \times 3.75[/tex]
[tex]Q_B = 101.25 \ in^3[/tex]
For point C ;
Let Q be the moment of the Area C;
SO ; [tex]Q_C = Area \times y_3[/tex]
where ;
[tex]y_3 = (6 - \dfrac{1.5 \times 2}{2})[/tex]
[tex]y_3 = (6 - 1.5})[/tex]
[tex]y_3= 4.5 \ in[/tex]
[tex]Q_C =(L \times B) \times y_1[/tex]
[tex]Q_C =(6 \times 3) \times 4.5[/tex]
[tex]Q_C=81 \ in^3[/tex]
For point D ;
Let Q be the moment of the Area D;
SO ; [tex]Q_D = Area \times y_4[/tex]
since there is no area about point D
Area = 0
[tex]Q_D =0 \times y_4[/tex]
[tex]Q_D = 0[/tex]
Thus; from the foregoing ; point B where [tex]Q_B = 101.25 \ in^3[/tex] has the largest Q value at section a–a
