Answer
$$a=1.75 \space in$$
$$b=7.50 \space in$$
Work Step by Step
1) Allowable Stresses
$\sigma_{all} = \frac{\sigma_{U}}{F.S.} = 10\space ksi$
$\tau_{all} = \frac{\tau_{U}}{F.S.} = 83.33\space psi$
2) Plate width (a)
Since plate is under tensile stress the effective cross sectional area is reduced by the width of the opening.
$A_{sxn} = 0.25 * (a - 0.75)$
$\sigma_{all} = \frac{P}{A_{sxn}} = \frac{2.5\space kip}{ 0.25 * (a - 0.75)\space in^2} = 10\space ksi$
Solving for the unknown value, a:
$a=\frac{2.5}{10}*4 +3/4 = \boxed{1.75\space in}$ $\leftarrow\space\space ANS 1$
3)Plate bonding depth (b)
For calculating shear between the plate/concrete, four surfaces are taken into account - the two large flats of the plate, and the two edges. Therefore the effective area for shear distribution is:
$A_{shear} = 2(0.25*b) + 2(a*b) = 2(0.25b) + 2(1.75b) = 4b$
$\tau_{all} = \frac{P}{A_{shear}} = \frac{2500\space lb}{4b\space in^2} = 83.33\space psi$
Solving for the unknown value b:
$b = \frac{2500}{4*83.33} = \boxed{7.50\space in}$ $\leftarrow\space\space\space ANS2$
