Chapter 8, Problem 8.6.3P

To determine

(a)

The available shear strength of the connection based on Allowed Stress Design.

Answer to Problem 8.6.3P

The available shear strength of the connection based on Allowed Stress Design.

is $51.75\text{kips}$.

Explanation of Solution

Given:

A $\text{W}16\times 45$ beam is connected to a $\text{W}10\times 45$ column.

The $20$ number of Group $\text{A}$ bolts of diameter equal to $\frac{7}{8}\text{in}$ are used.

Calculation:

The available shear strength of the connections is least of the following,

1. Strength of the bolt.
2. Plates shear yielding strength.
3. Plate shear rupture strength.
4. Plates block shear strength.
5. Strength of the weld.

Write the expression to obtain the bolt shear strength.

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{F_{\text{nv}}\times A_{\text{b}}}{\Omega }\\ =\frac{F_{\text{nv}}\times \frac{\pi d^2}{4}}{\Omega }\end{array}$ ..... (I)

Here, the cross section area of the unthreaded part of the bolt is $A_{\text{b}}$, the nominal shear strength of the bolt is $F_{\text{nv}}$, safety factor is $\Omega$ and diameter of the bolt is $d$.

Substitute $48\text{ksi}$ for $F_{\text{nv}}$ from manual for $A325-N$ type of bolts, $\frac{7}{8}\text{in}$ for $d$ and $2$ for $\Omega$ in Equation (I).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{48\text{ksi}\times \frac{\pi {\left(7/8\text{in}\right)}^2}{4}}{2}\\ =14.4312\text{kips}/\text{bolt}\end{array}$

The shear strength of the $4$ bolts is given as,

$\begin{array}{c}4\times \left(\frac{R_{\text{n}}}{\Omega }\right)=14.4312\text{kips}/\text{bolt}\times 4\text{bolts}\\ =57.7248\text{kips}\end{array}$ ..... (II)

Write the expression to obtain the bearing strength of the bolts on the plate.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2L_{\text{c}}t{F}_{\text{u}}\right)}{\Omega }\le \frac{\left(2.4dt{F}_{\text{u}}\right)}{\Omega }$ ..... (III)

Here, the safety factor is $\Omega$, the ultimate tensile strength of the connected part is $F_{\text{u}}$, thickness of the connected part is $t$, diameter of the bolt is $d$ and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute $2$ for $\Omega$, $5/16\text{in}$ for $t$, $58\text{ksi}$ for $F_{\text{u}}$, $1.7812\text{in}$ for $L_{\text{c}}$ and $7/8\text{in}$ for $d$ in Equation (III).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2\times 1.7812\text{in}\times \frac{5}{16}\text{in}\times \text{58}\text{ksi}\right)}{2}\le \frac{\left(2.4\times \frac{7}{8}\text{in}\times \frac{5}{16}\text{in}\times \text{58}\text{ksi}\right)}{2}\\ =19.371\text{kips}/\text{bolt}\le 19.031\text{kips}/\text{bolt}\end{array}$

Thus,

${\left(\frac{R_{\text{n}}}{\Omega }\right)}_1=19.031\text{kips}/\text{bolt}$

The bearing strength for the two bolts adjacent to the plate.

$\begin{array}{c}2\times {\left(\frac{R_{\text{n}}}{\Omega }\right)}_1=2\times 19.031\text{kips}/\text{bolt}\\ =38.062\text{kips}\end{array}$

Write the expression to obtain the bearing strength of the bolts other than on the plate.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2L_{\text{c}}t{F}_{\text{u}}\right)}{\Omega }\le \frac{\left(2.4dt{F}_{\text{u}}\right)}{\Omega }$ ..... (IV)

Here, the safety factor is $\Omega$, the ultimate tensile strength of the connected part is $F_{\text{u}}$, thickness of the connected part is $t$, diameter of the bolt is $d$ and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute $2$ for $\Omega$, $5/16\text{in}$ for $t$, $58\text{ksi}$ for $F_{\text{u}}$, $2.0625\text{in}$ for $L_{\text{c}}$ and $7/8\text{in}$ for $d$ in Equation (IV).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2\times 2.0625\text{in}\times \frac{5}{16}\text{in}\times \text{58}\text{ksi}\right)}{2}\le \frac{\left(2.4\times \frac{7}{8}\text{in}\times \frac{5}{16}\text{in}\times \text{58}\text{ksi}\right)}{2}\\ =22.4297\text{kips}/\text{bolt}\le 19.031\text{kips}/\text{bolt}\end{array}$

Thus,

${\left(\frac{R_{\text{n}}}{\Omega }\right)}_2=19.031\text{kips}/\text{bolt}$

The bearing strength for the two such bolts is,

$\begin{array}{c}2\times {\left(\frac{R_{\text{n}}}{\Omega }\right)}_2=2\times 19.031\text{kips}/\text{bolt}\\ =38.062\text{kips}\end{array}$

Thus, the total bearing strength of the bolts is:

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=38.062\text{kips}+38.062\text{kips}\\ =76.\text{124}\text{kips}\end{array}$ ..... (V)

Thus, the bolt strength is least of the shear strength and bearing strength of the bolts.

From Equations (II) and (V),

$\varphi R_{\text{n}}=57.7248\text{kips}$ ..... (VI)

Write the expression to obtain the plate shear yielding strength.

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.60F_{\text{y}}{A}_{\text{g}}\right)}{\Omega }\\ =\frac{\left(0.60F_{\text{y}}\left(w_{\text{g}}\times t\right)\right)}{\Omega }\end{array}$ ..... (VII)

Here, the gross cross section area of the plate perpendicular to the applied load is $A_{\text{g}}$ safety factor is $\Omega$, web plate yield strength is $F_{\text{y}}$, thickness of the plate is $t$ and gross width of the plate is $w_{\text{g}}$.

Substitute $1.50$ for $\Omega$, $36\text{ksi}$ for $F_{\text{y}}$, $13.5\text{in}$ for $w_{\text{g}}$ and $5/16\text{in}$ for $t$ in Equation (VII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.60\times 36\text{ksi}\left(13.5\text{in}\times \frac{5}{16}\text{in}\right)\right)}{1.50}\\ =60.751\text{kips}\end{array}$ ..... (VIII)

Write the expression to obtain the shear rupture strength of the plate.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.60F_{\text{u}}{A}_{\text{nv}}\right)}{\Omega }$ ..... (IX)

Here, the net area along the shear surface is $A_{\text{nv}}$ safety factor is $\Omega$ and plate ultimate strength is $F_{\text{u}}$.

Substitute, $2.0$ for $\Omega$, $58\text{ksi}$ for $F_{\text{u}}$ and $2.9688{\text{in}}^2$ for $A_{\text{nv}}$ in Equation (IX).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.60\times 58\text{ksi}\times \text{2}\text{.9688}{\text{in}}^2\right)}{2.0}\\ =51.6571\text{kips}\end{array}$ ..... (X)

Write the expression to obtain the block shear strength of the plate.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.6F_{\text{u}}{A}_{\text{nv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}\right)}{\Omega }\le \frac{\left(0.6F_{\text{y}}{A}_{\text{gv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}\right)}{\Omega }$ ..... (XI)

Here, the net area along the shear surface is $A_{\text{gv}}$, ultimate strength of the plate is $F_{\text{u}}$, the net area of the tension surface is $A_{\text{nt}}$, the reduction factor for the tensile strength is $U_{\text{bs}}$ and the gross area along the shear surface is $A_{\text{gv}}$ and yield strength of plate is $F_{\text{y}}$.

Substitute $2.0$ for $\Omega$, $58\text{ksi}$ for $F_{\text{u}}$, $2.422{\text{in}}^2$ for $A_{\text{nv}}$, $1.0$ for $U_{\text{bs}}$, $0.547{\text{in}}^2$ for $A_{\text{nt}}$, $3.5156{\text{in}}^2$ for $A_{\text{gv}}$ and $36\text{ksi}$ for $F_{\text{y}}$ in Equation (XI).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(\begin{array}{l}0.6\times 58\text{ksi}\times \text{2}\text{.422}{\text{in}}^2\\ +1.0\times 58\text{ksi}\times \text{0}\text{.547}{\text{in}}^2\end{array}\right)}{2.0}\le \frac{\left(\begin{array}{l}0.6\times 36\text{ksi}\times \text{3}\text{.5156}{\text{in}}^2\times 1.0\\ \times 58\text{ksi}\times \text{0}\text{.547}{\text{in}}^2\end{array}\right)}{\Omega }\\ =58\text{kips}\le 53.8315\text{kips}\end{array}$

Thus,

$\frac{R_{\text{n}}}{\Omega }=53.8315\text{kips}$ ..... (XII)

Write the expression to obtain the design shear strength of weld.

$\frac{R_{\text{n}}}{\Omega }=\left(0.9279\text{kips}/\text{in}\right)w$ ..... (XIII)

Here, the size of the fillet weld is $w$ in terms of sixteenths.

Substitute $0.25\text{in}$ for $w$ in Equation (XIII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\left(0.9279\text{kips}/\text{in}\right)0.25\text{in}\\ \text{=}\left(0.9279\text{kips}/\text{in}\right)\times \frac{1}{4}\text{in}\\ \text{=}\left(0.9279\text{kips}/\text{in}\right)\times 4\times \left(\frac{1}{16}\right)\text{in}\\ =\left(0.9279\text{kips}/\text{in}\right)\times 4\text{sixteenths}\end{array}$

Further solve the above equation.

$\frac{R_{\text{n}}}{\Omega }=3.711\text{6}\text{kips}/\text{in}$

For $2$ weld on both sides of the plate is given as,

$\begin{array}{c}2\times \frac{R_{\text{n}}}{\Omega }=2\times 3.7116\text{kips}/\text{in}\\ =7.4232\text{kips}/\text{in}\end{array}$ ..... (XV)

Write the expression to obtain the shear yielding strength for the base metal for $1\text{in}$ length.

$\frac{R_{\text{n}}}{\Omega }=0.40\times F_{\text{y}}\times t$ ..... (XVI)

Here, the thickness of the base metal is $t$ and the yield strength of the base metal is $F_{\text{y}}$.

Substitute $36\text{ksi}$ for $F_{\text{y}}$ and $5/16\text{in}$ for $t$ in Equation (XVI).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=0.40\times 36\text{ksi}\times \frac{5}{16}\text{in}\\ =4.5\text{kips}/\text{in}\end{array}$ ..... (XVII)

Write the expression to obtain the shear rupture strength of the base metal.

$\frac{R_{\text{n}}}{\Omega }=0.3\times F_{\text{u}}\times t$ ..... (XVIII)

Here, the thickness of the base metal is $t$ and the ultimate strength of the base metal is $F_{\text{y}}$

Substitute $58\text{ksi}$ for $F_{\text{u}}$ and $5/16\text{in}$ for $t$ in Equation (XVIII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=0.3\times 58\text{ksi}\times \frac{5}{16}\text{in}\\ =5.4\text{375}\text{kips}/\text{in}\end{array}$ ..... (XIX)

Write the expression to obtain the yielding strength for the base metal.

$\frac{R_{\text{n}}}{\Omega }=0.40\times F_{\text{y}}\times t\times L$ ..... (XX)

Here, the thickness of the base metal is $t$, the yield strength of the base metal is $F_{\text{y}}$ and the length of the plate is $L$.

Substitute $36\text{ksi}$ for $F_{\text{y}}$, $5/16\text{in}$ for $t$ and $13.5\text{in}$ for $L$ in Equation (XX).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=0.40\times 36\text{ksi}\times \frac{5}{16}\text{in}\times 13.5\text{in}\\ =60.7\text{5}\text{kips}\end{array}$ ..... (XXI)

From Equation (VI), (VIII), (X) and (XXI), the least value will be the available shear strength of the connections.

Thus,

$\frac{R_{\text{n}}}{\Omega }=51.75\text{kips}$

Conclusion:

Thus, the available shear strength of the connection is $51.75\text{kips}$.

To determine

(b)

The flexural strength of the member based on Allowed Stress Design.

Answer to Problem 8.6.3P

The available flexural strength of the member based on Allowed Stress Design.

is $121.59\text{ft-kips}$.

Explanation of Solution

Calculation:

The available flexural strength of the connections is least of the following,

1. Strength of the bolt.
2. Tension on gross section of the plate.
3. Tension on net section of the plate.
4. The block shear of the flange plate.
5. The block shear of the beam flange.
6. The compression of the bottom flange plate.

Write the expression to obtain the tension force due to yielding of the gross section area of the plate.

$\begin{array}{c}\frac{P_{\text{n}}}{{\Omega }_{\text{t}}}=\frac{F_{\text{y}}\times A_{\text{g}}}{{\Omega }_{\text{t}}}\\ =\frac{F_{\text{y}}\times \left(b\times t\right)}{{\Omega }_{\text{t}}}\end{array}$ ..... (XXII)

Here, the gross sectional area of the plate is $A_{\text{g}}$, the yield strength of the flange plate is $F_{\text{y}}$, safety factor in tension is ${\Omega }_{\text{t}}$, breadth of the plate is $b$ and the thickness of the plate is $t$.

Substitute, $36\text{ksi}$ for $F_{\text{y}}$ for $A36$ steel, $5/8\text{in}$ for $t$, $7\text{in}$ for $b$ and $1.67$ for ${\Omega }_{\text{t}}$ in Equation (XXII).

$\begin{array}{c}\frac{P_{\text{n}}}{{\Omega }_{\text{t}}}=\frac{36\text{ksi}\times \left(7\text{in}\times \frac{5}{8}\text{in}\right)}{1.67}\\ =94.3114\text{kips}\end{array}$ ..... (XXIII)

Write the expression to obtain the tension rupture strength of the plate.

$\frac{P_{\text{n}}}{\Omega }=\frac{\left(F_{\text{u}}\times A_{\text{n}}\right)}{\Omega }$ ..... (XXIV)

Here, the net area of the plate is $A_{\text{n}}$ safety factor in tension is $\Omega$ and plate ultimate strength is $F_{\text{u}}$.

Substitute, $2.0$ for $\Omega$, $58\text{ksi}$ for $F_{\text{u}}$ and $3.125{\text{in}}^2$ for $A_{\text{n}}$ in Equation (XXIV).

$\begin{array}{c}\frac{P_{\text{n}}}{\Omega }=\frac{\left(58\text{ksi}\times 3.125{\text{in}}^2\right)}{\Omega }\\ =90.625\text{kips}\end{array}$ ..... (XXV)

Write the expression to obtain the bolt shear strength.

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{F_{\text{nv}}\times A_{\text{b}}}{\Omega }\\ =\frac{F_{\text{nv}}\times \frac{\pi d^2}{4}}{\Omega }\end{array}$ ..... (XXVI)

Here, the cross section area of the unthreaded part of the bolt is $A_{\text{b}}$, the nominal shear strength of the bolt is $F_{\text{nv}}$, safety factor is $\Omega$ and diameter of the bolt is $d$.

Substitute $48\text{ksi}$ for $F_{\text{nv}}$ from manual for $A325-N$ type of bolts, $7/8\text{in}$ for $d$ and $2.0$ for $\Omega$ in Equation (I).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{48\text{ksi}\times \frac{\pi {\left(7/8\text{in}\right)}^2}{4}}{\Omega }\\ =14.4312\text{kips}/\text{bolt}\end{array}$

The shear strength of the $8$ bolts is given as,

$\begin{array}{c}8\times \frac{R_{\text{n}}}{\Omega }=14.4312\text{kips}/\text{bolt}\times 8\text{bolts}\\ =115.4496\text{kips}\end{array}$ ..... (XXVII)

Write the expression to obtain the bearing strength of the bolts on the plate.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2L_{\text{c}}t{F}_{\text{u}}\right)}{\Omega }\le \frac{\left(2.4dt{F}_{\text{u}}\right)}{\Omega }$ ..... (XXVIII)

Here, the safety factor is $\Omega$, the ultimate tensile strength of the connected part is $F_{\text{u}}$, thickness of the connected part is $t$, diameter of the bolt is $d$ and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute $2.0$ for $\Omega$, $5/8\text{in}$ for $t$, $58\text{ksi}$ for $F_{\text{u}}$, $1.7812\text{in}$ for $L_{\text{c}}$ and $7/8\text{in}$ for $d$ in Equation (XXVIII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2\times 1.7812\text{in}\times \frac{5}{8}\text{in}\times \text{58}\text{ksi}\right)}{2.0}\le \frac{\left(2.4\times \frac{7}{8}\text{in}\times \frac{5}{8}\text{in}\times \text{58}\text{ksi}\right)}{2.0}\\ =38.7411\text{kips}\le 38.0625\text{kips}\end{array}$

Thus,

${\left(\frac{R_{\text{n}}}{\Omega }\right)}_1=38.0625\text{kips}/\text{bolt}$

The bearing strength for the $4$ bolts adjacent to the plate is,

$\begin{array}{c}4\times {\left(\frac{R_{\text{n}}}{\Omega }\right)}_1=4\times 38.0625\text{kips}/\text{bolt}\\ =152.25\text{kips}\end{array}$

Write the expression to obtain the bearing strength of the bolts other than on the plate.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2L_{\text{c}}t{F}_{\text{u}}\right)}{\Omega }\le \frac{\left(2.4dt{F}_{\text{u}}\right)}{\Omega }$ ..... (XXIX)

Here, the safety factor is $\Omega$, the ultimate tensile strength of the connected part is $F_{\text{u}}$, thickness of the connected part is $t$, diameter of the bolt is $d$ and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute $2.0$ for $\Omega$, $5/8\text{in}$ for $t$, $58\text{ksi}$ for $F_{\text{u}}$, $2.0625\text{in}$ for $L_{\text{c}}$ and $7/8\text{in}$ for $d$ in Equation (XXIX).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(1.2\times 2.0625\text{in}\times \frac{5}{8}\text{in}\times \text{58}\text{ksi}\right)}{2.0}\le \frac{\left(2.4\times \frac{7}{8}\text{in}\times \frac{5}{2}\text{in}\times \text{58}\text{ksi}\right)}{2.0}\\ =44.8594\text{kips}/\text{bolt}\le 38.0625\text{kips}/\text{bolt}\end{array}$

Thus,

${\left(\frac{R_{\text{n}}}{\Omega }\right)}_2=38.0625\text{kips}/\text{bolt}$

The bearing strength for $4$ such bolts is,

$\begin{array}{c}4\times {\left(\frac{R_{\text{n}}}{\Omega }\right)}_2=4\times 38.025\text{kips}/\text{bolt}\\ =152.25\text{kips}\end{array}$

Thus, the total bearing strength of the bolts is:

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=152.25\text{kips}+152.25\text{kips}\\ =304.\text{5}\text{kips}\end{array}$ ..... (XXX)

Thus, the bolt strength is least of the shear strength and bearing strength of the bolts.

From Equation (XXVII) and (XXX),

$\frac{R_{\text{n}}}{\Omega }=115.4496\text{kips}$ ..... (XXXI)

Write the expression to obtain the block shear strength of the plate.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.6F_{\text{u}}{A}_{\text{nv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}\right)}{\Omega }\le \frac{\left(0.6F_{\text{y}}{A}_{\text{gv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}\right)}{\Omega }$ ..... (XXXII)

Here, the net area along the shear surface is $A_{\text{gv}}$, ultimate strength of the plate is $F_{\text{u}}$, the net area of the tension surface is $A_{\text{nt}}$, the reduction factor for the tensile strength is $U_{\text{bs}}$ and the gross area along the shear surface is $A_{\text{gv}}$ and yield strength of plate is $F_{\text{y}}$.

Substitute $2.0$ for $\Omega$, $58\text{ksi}$ for $F_{\text{u}}$, $9.6875{\text{in}}^2$ for $A_{\text{nv}}$, $1.0$ for $U_{\text{bs}}$, $0.9375{\text{in}}^2$ for $A_{\text{nt}}$, $14.0625{\text{in}}^2$ for $A_{\text{gv}}$ and $36\text{ksi}$ for $F_{\text{y}}$ in Equation (XXXII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(\begin{array}{l}0.6\times 58\text{ksi}\times 9.6875{\text{in}}^2\\ +1.0\times 58\text{ksi}\times \text{0}\text{.9375}{\text{in}}^2\end{array}\right)}{2.0}\le \frac{\left(\begin{array}{l}0.6\times 36\text{ksi}\times 14.0625{\text{in}}^2\times 1.0\\ \times 58\text{ksi}\times \text{0}\text{.9375}{\text{in}}^2\end{array}\right)}{2.0}\\ =195.75\text{kips}\le 179.0625\text{kips}\end{array}$

Thus,

$\frac{R_{\text{n}}}{\Omega }=179.0625\text{kips}$ ..... (XXXIII)

Write the expression to obtain the block shear strength of the beam flange.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.6F_{\text{u}}{A}_{\text{nv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}\right)}{\Omega }\le \frac{\left(0.6F_{\text{y}}{A}_{\text{gv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}\right)}{\Omega }$ ..... (XXXIV)

Here, the net area along the shear surface is $A_{\text{gv}}$, ultimate strength of the plate is $F_{\text{u}}$, the net area of the tension surface is $A_{\text{nt}}$, the reduction factor for the tensile strength is $U_{\text{bs}}$ and the gross area along the shear surface is $A_{\text{gv}}$ and yield strength of plate is $F_{\text{y}}$.

Substitute $2.0$ for $\Omega$, $65\text{ksi}$ for $F_{\text{u}}$, $8.7575{\text{in}}^2$ for $A_{\text{nv}}$, $1.0$ for $U_{\text{bs}}$, $1.9775{\text{in}}^2$ for $A_{\text{nt}}$, $12.7125{\text{in}}^2$ for $A_{\text{gv}}$ and $50\text{ksi}$ for $F_{\text{y}}$ in Equation (XXXIV).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(\begin{array}{l}0.6\times 65\text{ksi}\times 8.7575{\text{in}}^2\\ +1.0\times 65\text{ksi}\times 1.9775{\text{in}}^2\end{array}\right)}{2.0}\le \frac{\left(\begin{array}{l}0.6\times 50\text{ksi}\times 12.7125{\text{in}}^2\times 1.0\\ \times 65\text{ksi}\times 1.9775{\text{in}}^2\end{array}\right)}{2.0}\\ =235.04\text{kips}\le 254.956\text{kips}\end{array}$

Thus,

$\frac{R_{\text{n}}}{\Omega }=235.04\text{kips}$ ..... (XXXV)

Write the expression to obtain the compressive strength of the member.

$\frac{P_{\text{n}}}{{\Omega }_{\text{c}}}=\frac{\left(F_{\text{y}}\times A_{\text{g}}\right)}{{\Omega }_{\text{c}}}$ ..... (XXXVI)

Here, the safety factor for compression is ${\Omega }_{\text{c}}$, yield strength of the member is $F_{\text{y}}$ and the gross sectional area of the plate is $A_{\text{g}}$.

Substitute, $1.67$ for ${\Omega }_{\text{c}}$, $36\text{ksi}$ for $F_{\text{y}}$ and $4.375{\text{in}}^2$ for $A_{\text{g}}$ in Equation (XXXVI).

$\begin{array}{c}\frac{P_{\text{n}}}{{\Omega }_{\text{0}}}=\frac{\left(36\text{ksi}\times 4.375{\text{in}}^2\right)}{1.67}\\ =94.3114\text{kips}\end{array}$ ..... (XXXVII)

From Equations (XXIII), (XXV), (XXXI), (XXXV) and (XXXVII), the least value will be the tension at the net section as $H$.

Thus,

$H=90.625\text{kips}$

Write the expression to obtain the moment transferred by flange plate.

$M_{\text{n}}=H\times Z$ ..... (XXXVIII)

Here, lever arm distance is $Z$.

Substitute $90.625\text{kips}$ for $H$ and $16.1\text{in}$ for $Z$ in Equation (XXXVIII).

$\begin{array}{c}{M}_{\text{n}}=90.625\text{kips}\times 16.1\text{in}\\ =1459.0625\text{in-kips}\\ \text{=1459}\text{.0625}\text{in-kips}\times \left(\frac{1\text{ft}}{12\text{in}}\right)\\ =121.59\text{ft-kips}\end{array}$

Conclusion:

Thus, the available flexural strength of the member is $121.59\text{ft-kips}$.