$10$:$46$

$.0$

$-||$ CH $6,11$PP

$5$

Therefore, the maximum shearing stress in the section is $13\mathrm{.}142$ ksi $.$

Step $6/7$

$($b$)$

Consider the area above point $"$a$"$ as shown in the figure,

Calculate the first moment of area using the relation,

$Q=Q_2+2Q_3$

$=8\mathrm{.}203+22\mathrm{.}25$

$=12\mathrm{.}703i{n}^3$

Step $\frac{7}{7}$

Calculate the shear stress at point $"$a$"$ using the formula,

${}_a=\frac{VQ}{I_{x}t}$

Substitute $25$ kips for $V,12\mathrm{.}703i{n}^3$ for $Q,113\mathrm{.}842$ in ${?}^4$ for $I_x,$ and $0\mathrm{.}25$ in$.$ for $t.$

${}_a=\frac{2512\mathrm{.}703}{113\mathrm{.}8420\mathrm{.}25}$

$=11\mathrm{.}158$ksi

Therefore, the shearing stress at point $a$ is