Chapter 8, Problem 8.2.7P

To determine

(a)

The required diameter of group $A$ bearing-type bolt by elastic analysis using LRFD.

Answer to Problem 8.2.7P

The required diameter of the bolt by elastic analysis using LRFD is $\frac{7}{8}\text{in}$.

Explanation of Solution

Given data:

The live load is $18\text{kips}$.

The dead load is $40\text{kips}$.

Concept Used:

Write the equation of nominal bolt area.

$A_{\text{b}}=\frac{\text{π}}{4}{d}^2$ ...... (I)

Here, the diameter of bolt is $d$, the nominal bolt area is $A_{\text{b}}$.

Calculation:

Write the equation to calculate the factored load for LRFD.

$P_{\text{u}}=1.2\left(DL\right)+1.6\left(LL\right)$ ...... (II)

Here, the dead load is $DL$ and the live load is $LL$.

Substitute $18\text{kips}$ for $DL$ and $40\text{kips}$ for $LL$ in Equation (II).

$\begin{array}{c}{P}_{\text{u}}=1.2\left(18\text{kips}\right)+1.6\left(40\text{kips}\right)\\ =21.6\text{kips}+64\text{kips}\\ =85.6\text{kips}\end{array}$

Determine the centroid with respect to the lower left bolt.

Write the equation to calculate the distance of centroid in $x$ direction.

$\overline{x}=\frac{\sum \left(n\times d\right)}{\sum T}$ ..... (III)

Here, the number of bolts is $n$, the distance of the bolts is $d$ and the total number of bolts is $T$.

Substitute the values in Equation (III).

$\begin{array}{c}\overline{x}=\frac{\left(2\times 4\text{in}\right)+\left(3\times 6\text{in}\right)}{6}\\ =\frac{8\text{in}+18\text{in}}{6}\\ =\frac{26\text{in}}{6}\\ =4.33\text{in}\end{array}$

Write the equation to calculate the distance of the centroid in $y$ direction.

$\overline{y}=\frac{\sum \left(n\times d\right)}{\sum T}$ ..... (IV)

Substitute the values in Equation (IV).

$\begin{array}{c}\overline{y}=\frac{\left(\left(2\times 4\text{in}\right)+\left(1\times 8\text{in}\right)\right)}{6}\\ =\frac{16\text{in}}{6}\\ =2.67\text{in}\end{array}$

Write the expression to calculate the load applied in the $x$ direction.

$P_{\text{x}}=P\cos \theta$ ....... (V)

Here, the total load is $P$ and load component in $x$ direction is $P_{\text{x}}$.

Substitute $\frac{4}{5}$ for $\cos \theta$ and $85.6\text{kips}$ for $P$ in Equation (V).

$\begin{array}{c}{P}_{\text{x}}=85.6\text{kips}\times \left(\frac{4}{5}\right)\\ =85.6\text{kips}\times 0.8\\ =68.48\text{kips}\end{array}$

Write the expression to calculate the load applied in the $y$ direction.

$P_{\text{y}}=P\sin \theta$     ...... (VI)

Here, the and load component in $y$ direction is $P_{\text{y}}$.

Substitute $\frac{3}{5}$ for $\sin \theta$ and $85.6\text{kips}$ for $P$ in Equation (VI).

$\begin{array}{c}{P}_{\text{y}}=85.6\text{kips}\times \left(\frac{3}{5}\right)\\ =85.6\text{kips}\times 0.6\\ =51.36\text{kips}\end{array}$

Write the equation to calculate the horizontal direct shear component.

$P_{\text{cx}}=\frac{P_{\text{x}}}{n}$ ...... (VII)

Here, the load in $x$ direction is $P_{\text{x}}$ and the total number of bolts is $n$.

Substitute $68.48\text{kips}$ for $P_{\text{x}}$ and $6$ for $n$ in Equation (VII).

$\begin{array}{c}{P}_{\text{cx}}=\frac{68.48\text{kips}}{6}\\ =11.41\text{kips}\end{array}$

Write the equation to calculate the vertical direct shear component.

$P_{\text{cy}}=\frac{P_{\text{y}}}{n}$ ...... (VIII)

Here, the load in $y$ direction is $P_{\text{y}}$ and the total number of bolts is $n$.

Substitute $51.36\text{kips}$ for $P_{\text{y}}$ and $6$ for $n$ in Equation (VIII).

$\begin{array}{c}{P}_{\text{cy}}=\frac{51.36\text{kips}}{6}\\ =8.56\text{kips}\end{array}$

Calculate the horizontal and vertical components of loads due to eccentricity.

$\begin{array}{c}{\displaystyle \sum \left(x^2+y^2\right)}=\left(\begin{array}{l}3\times {\left(1.67\text{in}\right)}^2+2\times {\left(0.33\text{in}\right)}^2+1\times {\left(4.33\text{in}\right)}^2+3\times {\left(2.67\text{in}\right)}^2\\ +2\times {\left(1.33\text{in}\right)}^2+1\times {\left(5.33\text{in}\right)}^2\end{array}\right)\\ =80.67{\text{in}}^2\end{array}$

Calculate the distance of the load from centroid of connection.

$\begin{array}{c}{e}_{\text{x}}=8\text{in}+2\text{in}+\text{2}\text{in}+4\text{in}-4.33\text{in}\\ =\text{11}\text{.67}\text{in}\end{array}$

$\begin{array}{c}{e}_{\text{y}}=2\text{in}+4\text{in}+4\text{in}-2.67\text{in}\\ =7.33\text{in}\end{array}$

Here, the eccentricity in $x$ direction is $e_{\text{x}}$ and the eccentricity in $y$ direction is $e_{\text{y}}$.

Write the expression for the moment for $x$ axis about midpoint of the connection between the bolts.

$M_{\text{x}}=P_{\text{y}}\times e_{\text{x}}$ ....... (IX)

Here, moment generated is $x$ direction is $M_{\text{x}}$.

Substitute $51.36\text{kips}$ for $P_{\text{y}}$ and $\text{11}\text{.67}\text{in}$ for $e_{\text{x}}$ in Equation (IX).

$\begin{array}{c}{M}_{\text{x}}=51.36\text{kips}\times \text{11}\text{.67}\text{in}\\ =599\text{kips-in}\end{array}$

Write the expression for the moment for $y$ axis about midpoint of the connection between the bolts.

$M_{\text{y}}=P_{\text{x}}\times e_{\text{y}}$ ....... (X)

Here, moment generated is $y$ direction is $M_{\text{y}}$.

Substitute $68.48\text{kips}$ for $P_{\text{x}}$ and $7.33\text{in}$ for $e_{\text{y}}$ in Equation (X).

$\begin{array}{c}{M}_{\text{y}}=68.48\text{kips}\times \text{7}\text{.33}\text{in}\\ =502\text{kips-in}\end{array}$

Write the expression for the moment for $y$ axis and $x$ axis about midpoint of the connection between the bolts.

$M=\sum \left(M_{\text{x}}-M_{\text{y}}\right)$ ....... (XI)

Here, the total moment generated is $M$.

Substitute $599\text{kips-in}$ for $M_{\text{x}}$ and $502\text{kips-in}$ for $M_{\text{y}}$ in Equation (XI).

$\begin{array}{c}M=599\text{kips-in}-502\text{kips-in}\\ =97\text{kips-in}\end{array}$

Write the equation to calculate the total fastener force.

$P=\sqrt{{\left({\displaystyle \sum P_{\text{x}}}\right)}^2+{\left({\displaystyle \sum P_{\text{y}}}\right)}^2}$ ...... (XII)

Here, the total fastener force is $P$ and the $x$ and $y$ direction of the total connection load $P$ are $P_{\text{x}}$ and $P_{\text{y}}$.

Write the equation to calculate the $x$ component of total connection load.

${\displaystyle \sum P_{\text{x}}}=P_{\text{cx}}+P_{\text{mx}}$     ...... (XIII)

Here, the horizontal direct shear component is $P_{\text{cx}}$, the load coming on critical bolt in $x$ direction due to eccentricity is $P_{\text{mx}}$.

Write the equation to calculate the $y$ component of total connection load.

${\displaystyle \sum P_{\text{y}}}=P_{\text{cy}}+P_{\text{my}}$     ...... (XIV)

Here, the vertical direct shear component is $P_{\text{cy}}$, the load coming on critical bolt in y direction due to eccentricity is $P_{\text{my}}$.

The top right bolt is critical, thus check for its safety.

Calculate the distance of the bolt from the centroid.

$\begin{array}{c}x=4\text{in}+2\text{in}-4.33\text{in}\\ =1.67\text{in}\end{array}$

$\begin{array}{c}y=4\text{in}+4\text{in}-2.67\text{in}\\ =5.33\text{in}\end{array}$

Here, the distance of the top right bolt from the centroid in the x and y directions are $x$ and $y$ respectively.

Write the equation to calculate the load coming on the critical bolt in x direction due to eccentricity.

$P_{\text{mx}}=\frac{M\times y}{{\displaystyle \sum \left(x^2+y^2\right)}}$     ...... (XV)

Substitute $97\text{kips-in}$ for $M$, $5.33\text{in}$ for $y$ and $80.67{\text{in}}^2$ for ${\displaystyle \sum \left(x^2+y^2\right)}$ in Equation (XV).

$\begin{array}{c}{P}_{\text{mx}}=\frac{97\text{kips-in}\times 5.33\text{in}}{80.67{\text{in}}^2}\\ =\frac{517.01\text{kips}\cdot {\text{in}}^{\text{2}}}{80.67{\text{in}}^2}\\ =6.41\text{kips}\end{array}$

Write the equation to calculate the load coming on critical bolt in y direction due to eccentricity.

$P_{\text{my}}=\frac{M\times x}{{\displaystyle \sum \left(x^2+y^2\right)}}$     ...... (XVI)

Substitute $97\text{kips-in}$ for $M$, $1.67\text{in}$ for $x$ and $80.67{\text{in}}^2$ for ${\displaystyle \sum \left(x^2+y^2\right)}$ in Equation (XVI).

$\begin{array}{c}{P}_{\text{my}}=\frac{97\text{kips-in}\times 1.67\text{in}}{80.67{\text{in}}^2}\\ =\frac{161.99\text{kips}\cdot {\text{in}}^2}{80.67{\text{in}}^2}\\ =2.01\text{kips}\end{array}$

Calculate the $x$ component of the total connection load.

Substitute $-11.41\text{kips}$ for $P_{\text{cx}}$ and $6.41\text{kips}$ for $P_{\text{mx}}$ in Equation (XIII).

$\begin{array}{c}{\displaystyle \sum P_{\text{x}}}=-11.41\text{kips}+6.41\text{kips}\\ =-5\text{kips}\end{array}$

Calculate the $y$ component of total connection load.

Substitute $8.56\text{kips}$ for $P_{\text{cy}}$ and $2.01\text{kips}$ for $P_{\text{my}}$ in Equation (XIV).

$\begin{array}{c}{\displaystyle \sum P_{\text{y}}}=8.56\text{kips}+2.01\text{kips}\\ =10.57\text{kips}\end{array}$

Calculate the total fastener force.

Substitute $-5\text{kips}$ for ${\displaystyle \sum P_{\text{x}}}$ and $10.57\text{kips}$ for ${\displaystyle \sum P_{\text{y}}}$ in Equation (XII).

$\begin{array}{c}P=\sqrt{{\left(-5\text{kips}\right)}^2+{\left(10.57\text{kips}\right)}^2}\\ =\sqrt{\left(25{\text{kips}}^2\right)+\left(111.73{\text{kips}}^2\right)}\\ =11.693\text{kips}\end{array}$

Check the safety for the bottom right bolt.

Calculate the distance of the bolt from the centroid.

$\begin{array}{c}x=4\text{in}+2\text{in}-4.33\text{in}\\ =1.67\text{in}\end{array}$

$y=2.67\text{in}$

Calculate the load coming on the critical bolt in x direction due to eccentricity.

Substitute $97\text{kips-in}$ for $M$, $2.67\text{in}$ for $y$ and $80.67{\text{in}}^2$ for ${\displaystyle \sum \left(x^2+y^2\right)}$ in Equation (XV).

$\begin{array}{c}{P}_{\text{mx}}=\frac{97\text{kips-in}\times 2.67\text{in}}{80.67{\text{in}}^2}\\ =\frac{258.99\text{kips}\cdot {\text{in}}^{\text{2}}}{80.67{\text{in}}^2}\\ =3.21\text{kips}\end{array}$

Calculate the load coming on critical bolt in y direction due to eccentricity.

Substitute $97\text{kips-in}$ for $M$, $1.67\text{in}$ for $x$ and $80.67{\text{in}}^2$ for ${\displaystyle \sum \left(x^2+y^2\right)}$ in Equation (XVI).

$\begin{array}{c}{P}_{\text{my}}=\frac{97\text{kips-in}\times 1.67\text{in}}{80.67{\text{in}}^2}\\ =\frac{161.99\text{kips}\cdot {\text{in}}^2}{80.67{\text{in}}^2}\\ =2.01\text{kips}\end{array}$

Calculate the $x$ component of the total connection load.

Substitute $11.41\text{kips}$ for $P_{\text{cx}}$ and $3.21\text{kips}$ for $P_{\text{mx}}$ in Equation (XIII).

$\begin{array}{c}{\displaystyle \sum P_{\text{x}}}=11.41\text{kips}+3.21\text{kips}\\ =14.62\text{kips}\end{array}$

Calculate the $y$ component of total connection load.

Substitute $8.56\text{kips}$ for $P_{\text{cy}}$ and $2.01\text{kips}$ for $P_{\text{my}}$ in Equation (XIV).

$\begin{array}{c}{\displaystyle \sum P_{\text{y}}}=8.56\text{kips}+2.01\text{kips}\\ =10.57\text{kips}\end{array}$

Calculate the total fastener force.

Substitute $14.62\text{kips}$ for ${\displaystyle \sum P_{\text{x}}}$ and $10.57\text{kips}$ for ${\displaystyle \sum P_{\text{y}}}$ in Equation (XII).

$\begin{array}{c}P=\sqrt{{\left(14.62\text{kips}\right)}^2+{\left(10.57\text{kips}\right)}^2}\\ =\sqrt{\left(213.74{\text{kips}}^2\right)+\left(111.73{\text{kips}}^2\right)}\\ =18.041\text{kips}\end{array}$

Therefore, the total shear force in the fastener is $18.041\text{kips}$.

Write the equation to calculate the nominal shear capacity of bolt.

$\varphi R_{\text{nv}}=\varphi F_{\text{nv}}{A}_{\text{b}}$     ...... (XVII)

Here, the nominal shear capacity is $R_{\text{nv}}$, the nominal shear strength is $F_{\text{nv}}$, the load resistance factor is $\varphi$ and the nominal bolt area is $A_{\text{b}}$.

Substitute $18.041\text{kips}$ for $\varphi R_{\text{nv}}$, $54\text{ksi}$ for $F_{\text{nv}}$ and $0.75$ for $\varphi$ in Equation (XVII).

$\begin{array}{l}18.041\text{kips}=0.75\times 54\text{ksi}\times A_{\text{b}}\\ A_{\text{b}}=\frac{18.041\text{kips}}{0.75\times 54\text{ksi}}\\ A_{\text{b}}=0.445{\text{in}}^{\text{2}}\end{array}$

Calculate the diameter of bolt.

Substitute $1.28{\text{in}}^{\text{2}}$ for $A_{\text{b}}$ in Equation (I).

$\begin{array}{l}0.445{\text{in}}^{\text{2}}=\frac{\text{π}}{4}{d}^2\\ d^2=\frac{0.445{\text{in}}^{\text{2}}\times 4}{\text{π}}\\ d=\sqrt{0.567{\text{in}}^{\text{2}}}\\ d=0.753\text{in}\end{array}$

Therefore, adopt the bolt diameter of $\frac{7}{8}\text{in}$.

Check for safety:

Write the equation to calculate the bearing deformation strength of the bolt.

$R_{\text{bd}}=2.4dt{F}_{\text{u}}$ ...... (XVIII)

Here, the bearing deformation strength is $R_{\text{bd}}$, the thickness of the connected part is $t$ and the ultimate tensile stress of gusset plate is $F_{\text{u}}$.

Assume A36 steel and the value of $F_{\text{u}}$ is $58\text{ksi}$.

Assume the thickness of the plate is $t$ and its value is $\frac{3}{8}\text{inch}$.

Calculate the bearing deformation strength of bolt.

Substitute $0.875\text{in}$ for $d$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$ in Equation (XVIII).

$\begin{array}{c}{R}_{\text{bd}}=2.4\times \left(0.875\text{in}\right)\times \left(\frac{3}{8}\text{in}\right)\times 58\text{ksi}\\ =0.7875{\text{in}}^2\times 58\text{ksi}\\ =45.675\text{kips}\end{array}$

Write the equation to calculate the tear-out strength of inner bolt.

$R_{\text{ti}}=1.2l_{\text{c}}t{F}_{\text{u}}$ ...... (XIX)

Here, the tear-out strength of inner bolt is $R_{\text{ti}}$ and the clear distance, in the direction parallel to the applied load from the edge of bolt hole to the edge of the adjacent hole or to the edge of the material is $l_{\text{c}}$ and the thickness of the plate is $t$.

Write the equation to calculate the clear distance for edge bolt.

$l_{\text{c}}=l_{\text{e}}-\frac{h}{2}$ ...... (XX)

Here, the bolt edge distance is $l_{\text{e}}$ and the bolt hole diameter is $h$.

Write the equation to calculate the hole diameter.

$h=\frac{15}{16}\text{in}$ ...... (XXI)

Calculate the clear distance for edge bolt.

Substitute $\frac{15}{16}\text{in}$ for $h$ and $2\text{in}$ for $l_{\text{e}}$ in Equation (X).

$\begin{array}{c}{l}_{\text{c}}=2\text{in}-\frac{\left(\frac{15}{16}\text{in}\right)}{2}\\ =\frac{4\text{in}-0.9375\text{in}}{2}\\ =1.53\text{in}\end{array}$

Calculate the tear-out strength of inner bolt.

Substitute $1.53\text{in}$ for $l_{\text{c}}$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$ in Equation (XIX).

$\begin{array}{c}{R}_{\text{ti}}=1.2\times 1.53\text{in}\times \frac{3}{8}\text{in}\times 58\text{ksi}\\ =0.6885{\text{in}}^2\times 58\text{ksi}\\ =39.933\text{kips}\end{array}$

Write the equation to calculate the tear-out strength of edge bolt.

$R_{\text{te}}=1.2l_{\text{c}}t{F}_{\text{u}}$ ...... (XXII)

Here, the tear-out strength of edge bolt is $R_{\text{te}}$.

Write the equation to calculate the clear distance for edge bolt.

$l_{\text{c}}=s-h$ ...... (XXIII)

Here, the spacing between bolts is $s$.

Substitute $3\text{in}$ for $s$ and $\frac{15}{16}\text{in}$ for $h$ in Equation (XXIII).

$\begin{array}{c}{l}_{\text{c}}=3\text{in}-\frac{15}{16}\text{in}\\ =2.0625\text{in}\end{array}$

Calculate the tear-out strength of edge bolt.

Substitute $2.0625\text{in}$ for $l_{\text{c}}$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$ in Equation (XXII).

$\begin{array}{c}{R}_{\text{te}}=1.2\times 2.0625\text{in}\times \frac{3}{8}\text{in}\times 58\text{ksi}\\ =0.928{\text{in}}^2\times 58\text{ksi}\\ =53.83\text{kips}\end{array}$

The minimum of Equations (XXII), (XIX) and (XVIII) is the value of $R_{\text{n}}$.

Calculate the design of strength of bolt by substituting $0.75$ for the resistance factor $\varphi$.

$\begin{array}{c}\varphi R_{\text{n}}=0.75\times 39.933\text{kips}\\ =29.95\text{kips}\end{array}$

The strength is greater than the design strength.

$29.95\text{kips}>18.041\text{kips}$

Therefore, adopt $\frac{7}{8}\text{in}$ diameter bolt.

Conclusion:

Thus, the size of bolt is $\frac{7}{8}\text{in}$.

To determine

(b)

The size of bolt by elastic analysis using ASD.

Answer to Problem 8.2.7P

The size of bolt by elastic analysis using ASD is $\frac{7}{8}\text{in}$.

Explanation of Solution

Calculation:

Write the equation to calculate the factored load.

$P_{\text{u}}=DL+LL$ ...... (XXIV)

Here, the dead load is $DL$ and the live load is $LL$.

Substitute $18\text{kips}$ for $DL$ and $40\text{kips}$ for $LL$ in Equation (XXIV).

$\begin{array}{c}{P}_{\text{u}}=\left(18\text{kips}\right)+\left(40\text{kips}\right)\\ =58\text{kips}\end{array}$

Write the expression to calculate the load applied in $\text{x}$ direction.

$P_{\text{x}}=P\cos \theta$ ....... (XXV)

Here, the total load is $P$ and load component in $\text{x}$ direction is $P_{\text{x}}$.

Substitute $\frac{4}{5}$ for $\cos \theta$ and $58\text{kips}$ for $P$ in Equation (XXV).

$\begin{array}{c}{P}_{\text{x}}=58\text{kips}\times \left(\frac{4}{5}\right)\\ =58\text{kips}\times 0.8\\ =46.4\text{kips}\end{array}$

Write the expression to calculate the load applied in $\text{y}$ direction.

$P_{\text{y}}=P\sin \theta$     ...... (XXVI)

Here, the and load component in $\text{y}$ direction is $P_{\text{y}}$.

Substitute $\frac{3}{5}$ for $\sin \theta$ and $58\text{kips}$ for $P$ in Equation (XXVI).

$\begin{array}{c}{P}_{\text{y}}=58\text{kips}\times \left(\frac{3}{5}\right)\\ =58\text{kips}\times 0.6\\ =34.8\text{kips}\end{array}$

Calculate the horizontal direct shear component.

Substitute $46.4\text{kips}$ for $P_{\text{x}}$ and $6$ for $n$ in Equation (VII).

$\begin{array}{c}{P}_{\text{cx}}=\frac{46.4\text{kips}}{6}\\ =7.73\text{kips}\end{array}$

Calculate the vertical direct shear component.

Substitute $0.890\text{kips}$ for $P_{\text{y}}$ and $6$ for $n$ in Equation (VIII).

$\begin{array}{c}{P}_{\text{cy}}=\frac{34.8\text{kips}}{6}\\ =5.8\text{kips}\end{array}$

Write the expression for the moment for $x$ axis about midpoint of the connection between the bolts.

$M_{\text{x}}=P_{\text{y}}\times e_{\text{x}}$ ....... (XXVII)

Here, moment generated is $x$ direction is $M_{\text{x}}$.

Substitute $34.8\text{kips}$ for $P_{\text{y}}$ and $\text{11}\text{.67}\text{in}$ for $e_{\text{x}}$ in Equation (XXVII).

$\begin{array}{c}{M}_{\text{x}}=34.8\text{kips}\times \text{11}\text{.67}\text{in}\\ =406.116\text{k-in}\end{array}$

Write the expression for the moment for $y$ axis about midpoint of the connection between the bolts.

$M_{\text{y}}=P_{\text{x}}\times e_{\text{y}}$ ....... (XXVIII)

Here, moment generated is $y$ direction is $M_{\text{y}}$.

Substitute $46.4\text{kips}$ for $P_{\text{x}}$ and $7.33\text{in}$ for $e_{\text{y}}$ in Equation (XXVIII).

$\begin{array}{c}{M}_{\text{y}}=46.4\text{kips}\times 7.\text{33}\text{in}\\ =340.112\text{k-in}\end{array}$

Write the expression for the moment for $\text{y-axis}$ and $\text{x-axis}$ about midpoint of the connection between the bolts.

$M=\sum \left(M_{\text{x}}-M_{\text{y}}\right)$ ....... (XXIX)

Here, the total moment generated is $M$.

Substitute $406.116\text{k-in}$ for $M_{\text{x}}$ and $340.112\text{k-in}$ for $M_{\text{y}}$ in Equation (XXIX).

$\begin{array}{c}M=\sum \left(406.116\text{k-in}-340.112\text{k-in}\right)\\ =66.004\text{k-in}\end{array}$

Check the safety for the bottom right bolt.

Calculate the distance of the bolt from the centroid.

$x=1.67\text{in}$

$y=2.67\text{in}$

Calculate the load coming on the critical bolt in x direction due to eccentricity.

Substitute $66.004\text{kips-in}$ for $M$, $2.67\text{in}$ for $y$ and $80.67{\text{in}}^2$ for ${\displaystyle \sum \left(x^2+y^2\right)}$ in Equation (XV).

$\begin{array}{c}{P}_{\text{mx}}=\frac{66.004\text{kips-in}\times 2.67\text{in}}{80.67{\text{in}}^2}\\ =\frac{176.23\text{kips}\cdot {\text{in}}^{\text{2}}}{80.67{\text{in}}^2}\\ =2.185\text{kips}\end{array}$

Calculate the load coming on critical bolt in y direction due to eccentricity.

Substitute $66.004\text{kips-in}$ for $M$, $1.67\text{in}$ for $x$ and $80.67{\text{in}}^2$ for ${\displaystyle \sum \left(x^2+y^2\right)}$ in Equation (XVI).

$\begin{array}{c}{P}_{\text{my}}=\frac{66.004\text{kips-in}\times 1.67\text{in}}{80.67{\text{in}}^2}\\ =\frac{110.23\text{kips}\cdot {\text{in}}^2}{80.67{\text{in}}^2}\\ =1.366\text{kips}\end{array}$

Calculate the $x$ component of the total connection load.

Substitute $7.73\text{kips}$ for $P_{\text{cx}}$ and $2.185\text{kips}$ for $P_{\text{mx}}$ in Equation (XIII).

$\begin{array}{c}{\displaystyle \sum P_{\text{x}}}=7.73\text{kips}+2.185\text{kips}\\ =9.915\text{kips}\end{array}$

Calculate the $y$ component of total connection load.

Substitute $5.8\text{kips}$ for $P_{\text{cy}}$ and $1.366\text{kips}$ for $P_{\text{my}}$ in Equation (XIV).

$\begin{array}{c}{\displaystyle \sum P_{\text{y}}}=5.8\text{kips}+1.366\text{kips}\\ =7.166\text{kips}\end{array}$

Calculate the total fastener force.

Substitute $9.915\text{kips}$ for ${\displaystyle \sum P_{\text{x}}}$ and $7.166\text{kips}$ for ${\displaystyle \sum P_{\text{y}}}$ in Equation (XII).

$\begin{array}{c}P=\sqrt{{\left(9.915\text{kips}\right)}^2+{\left(7.166\text{kips}\right)}^2}\\ =\sqrt{\left(98.307{\text{kips}}^2\right)+\left(51.35{\text{kips}}^2\right)}\\ =12.233\text{kips}\end{array}$

Write the equation to calculate the nominal shear capacity of bolt.

$\frac{R_{\text{nv}}}{\Omega }=\frac{F_{\text{nv}}{A}_{\text{b}}}{\Omega }$     ...... (XXX)

Here, the nominal shear capacity is $R_{\text{nv}}$, the nominal shear strength is $F_{\text{nv}}$, the safety factor is $\Omega$ and the nominal bolt area is $A_{\text{b}}$.

Substitute $12.233\text{kips}$ for $\frac{R_{\text{nv}}}{\Omega }$, $54\text{ksi}$ for $F_{\text{nv}}$ and $2$ for $\Omega$ in Equation (XXX).

$\begin{array}{l}12.233\text{kips}=\frac{54\text{ksi}\times A_{\text{b}}}{2}\\ A_{\text{b}}=\frac{12.233\text{kips}\times 2}{54\text{ksi}}\\ A_{\text{b}}=0.453{\text{in}}^{\text{2}}\end{array}$

Calculate the diameter of bolt.

Substitute $0.453{\text{in}}^{\text{2}}$ for $A_{\text{b}}$ in Equation (I).

$\begin{array}{l}0.453{\text{in}}^{\text{2}}=\frac{\text{π}}{4}{d}^2\\ d^2=\frac{0.453{\text{in}}^{\text{2}}\times 4}{\text{π}}\\ d=\sqrt{0.577{\text{in}}^{\text{2}}}\\ d=0.759\text{in}\end{array}$

Therefore, adopt the bolt diameter of $\frac{7}{8}\text{in}$.

Check for safety:

Calculate the bearing deformation strength of bolt.

Substitute $0.875\text{in}$ for $d$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$ in Equation (XVIII).

$\begin{array}{c}{R}_{\text{bd}}=2.4\times \left(0.875\text{in}\right)\times \left(\frac{3}{8}\text{in}\right)\times 58\text{ksi}\\ =0.7875{\text{in}}^2\times 58\text{ksi}\\ =45.675\text{kips}\end{array}$

Calculate the hole diameter.

$h=\frac{15}{16}\text{in}$ ...... (XXI)

Calculate the clear distance for edge bolt.

Substitute $\frac{15}{16}\text{in}$ for $h$ and $2\text{in}$ for $l_{\text{e}}$ in Equation (X).

$\begin{array}{c}{l}_{\text{c}}=2\text{in}-\frac{\left(\frac{15}{16}\text{in}\right)}{2}\\ =\frac{4\text{in}-0.9375\text{in}}{2}\\ =1.53\text{in}\end{array}$

Calculate the tear-out strength of inner bolt.

Substitute $1.53\text{in}$ for $l_{\text{c}}$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$ in Equation (XIX).

$\begin{array}{c}{R}_{\text{ti}}=1.2\times 1.53\text{in}\times \frac{3}{8}\text{in}\times 58\text{ksi}\\ =0.6885{\text{in}}^2\times 58\text{ksi}\\ =39.933\text{kips}\end{array}$

Calculate the clear distance for edge bolt.

Substitute $3\text{in}$ for $s$ and $\frac{15}{16}\text{in}$ for $h$ in Equation (XXIII).

$\begin{array}{c}{l}_{\text{c}}=3\text{in}-\frac{15}{16}\text{in}\\ =2.0625\text{in}\end{array}$

Calculate the tear-out strength of edge bolt.

Substitute $2.0625\text{in}$ for $l_{\text{c}}$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$ in Equation (XXII).

$\begin{array}{c}{R}_{\text{te}}=1.2\times 2.0625\text{in}\times \frac{3}{8}\text{in}\times 58\text{ksi}\\ =0.928{\text{in}}^2\times 58\text{ksi}\\ =53.83\text{kips}\end{array}$

The minimum of Equations (XXII), (XIX) and (XVIII) is the value of $R_{\text{n}}$.

Calculate the allowable strength of bolt by substituting $2$ for the safety factor $\Omega$.

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{39.933\text{kips}}{2}\\ =19.967\text{kips}\end{array}$

The strength is greater than the design strength.

$19.967\text{kips}>12.233\text{kips}$

Therefore, adopt $\frac{7}{8}\text{in}$ diameter bolt.

Conclusion:

Thus, the size of bolt is $\frac{7}{8}\text{in}$.