Chapter 7, Problem 7.7.2P

To determine

(a)

The design of a single angle tension member and a bolted connection using Load and Resistance Factor Design (LRFD) method.

Answer to Problem 7.7.2P

An angle $L8\times 6\times \frac{5}{8}$ and $10$ bolts of $1\frac{1}{4}\text{inch}$ diameter each of Group A in two lines with a spacing of $4\text{inch}$ and an edge distance of $2.5\text{inch}$ is used.

Explanation of Solution

Given:

Steel used is $\text{A36}$.

Thickness of gusset plate is $\frac{3}{8}\text{inch}$.

Dead load is $50\text{kips}$.

Live load is $100\text{kips}$.

Wind load is $45\text{kips}$.

Group A bolts are used.

Calculation:

The properties for $\text{A}36$ steel from the AISC steel table are as follows:

• The ultimate tensile stress is $58\text{ksi}$.
• The yield strength is $36\text{ksi}$.

From the AISC manual Table J3-1, the minimum tension force per bolt is $71\text{kips}$.

From the AISC manual Table D3-1, the reduction factor is equal to $0.8$.

Write the expression for total factored load by load combination 2.

$P_{\text{u}}=1.2D+1.6L$     ...... (I)

Here, total factored load is $P_{\text{u}}$, live load is $L,$ and dead load is $D$.

Substitute $100\text{kips}$ for $L$ and $50\text{kips}$ for $D$ in Equation (I).

$\begin{array}{c}{P}_{\text{u}}=\left(1.2\times 50+1.6\times 100\right)\text{kips}\\ =220\text{kips}\end{array}$

Write the expression for total factored load by load combination 4.

$P_{\text{u}}=1.2D+1.0W+0.5L$     ...... (II)

Here, the wind load is $W$.

Substitute $100\text{kips}$ for $L$, $50\text{kips}$ for $D,$ and $45\text{kips}$ for $W$ in Equation (II).

$\begin{array}{c}{P}_{\text{u}}=\left(1.2\times 50+1.0\times 45+0.5\times 100\right)\text{kips}\\ =155\text{kips}\end{array}$

Thus, the value of total factored load by load combination 2 is greater, so the total factored load is equal to $220\text{kips}$.

Write the expression for slip critical shear strength per bolt.

$\varphi R_{\text{n}}=\varphi \left(\mu D_{\text{u}}{h}_{\text{f}}{n}_{\text{s}}{T}_{\text{b}}\right)$     ...... (III)

Here, the ratio of mean actual bolt pre-tensioned to specified bolt pre-tensioned is $D_{\text{u}}$, slip shear strength per bolt is $R_{\text{n}}$, filler factor is $h_{\text{f}}$, number of slip planes is $n_{\text{s}}$, mean slip coefficient is $\mu ,$ and minimum tension force per bolt is $T_{\text{b}}$.

Substitute $1.13$ for $D_{\text{u}}$, $0.3$ for $\mu$, $1.0$ for $h_{\text{f}}$, $1.0$ for $\varphi$, $1$ for $n_{\text{s}}$ and $71\text{kips}$ for $T_{\text{b}}$ in Equation (III).

$\begin{array}{c}\varphi R_{\text{n}}=1.0\left(0.3\times 1.13\times 1.0\times 71\text{kips}\times 1\right)\\ =24.07\text{kips}\end{array}$

Write the expression for the design strength.

$B=N\times \varphi R_{\text{n}}$     ...... (IV)

Here, number of bolts are $N$ and design strength is $B$.

Substitute $24.07\text{kips}$ for $\varphi R_{\text{n}}$ and $10$ for $N$ in Equation (IV).

$\begin{array}{c}B=\left(10\times 24.07\right)\text{kips}\\ =240.7\text{kips}\end{array}$

Write the expression for required gross area of the member.

$A_{\text{g}}=\frac{P_{\text{u}}}{0.9F_{\text{y}}}$     ...... (V)

Here, the yield strength is $F_{\text{y}}$ and the gross area is $A_{\text{g}}$.

Substitute $36\text{ksi}$ for $F_{\text{y}}$ and $220\text{kips}$ for $P_{\text{u}}$ in Equation (V).

$\begin{array}{c}{A}_{\text{g}}=\frac{220\text{kips}}{0.9\times 36\text{ksi}}\\ =6.79{\text{inch}}^2\end{array}$

Write the expression for required effective area of the member.

$A_{\text{e}}=\frac{P_{\text{u}}}{0.75F_{\text{u}}}$     ...... (VI)

Here, ultimate strength is $F_{\text{u}}$ and the effective area is $A_{\text{e}}$.

Substitute $58\text{ksi}$ for $F_{\text{u}}$ and $220\text{kips}$ for $P_{\text{u}}$ in Equation (VI).

$\begin{array}{c}{A}_{\text{e}}=\frac{220\text{kips}}{0.75\times 58\text{ksi}}\\ =5.06{\text{inch}}^2\end{array}$

Write the expression for the minimum value of radius of gyration.

$r_{\text{min}}=\frac{L}{300}$     ...... (VII)

Here, the minimum radius of gyration is $r_{\text{min}}$ and the length of the member is $L$.

Substitute $240\text{inch}$ for $L$ in Equation (VII).

$\begin{array}{c}{r}_{\text{min}}=\frac{240\text{inch}}{300}\\ =0.8\text{inch}\end{array}$

Thus, try the section $L8\times 6\times \frac{5}{8}$, as per the above calculated values.

Write the expression for net area of $L8\times 6\times \frac{5}{8}$ section.

$A_{\text{n}}=A_{\text{g}}-A_{\text{holes}}$     ...... (VIII)

Here, net area of the section is $A_{\text{n}}$ and the area of the holes is $A_{\text{holes}}$.

Substitute $t{d}_{\text{h}}$ for $A_{\text{holes}}$ in Equation (VIII).

$A_{\text{n}}=A_{\text{g}}-2\left(t{d}_{\text{h}}\right)$     ...... (IX)

Here, diameter of the holes is $d_{\text{h}}$.

Substitute $\frac{5}{8}\text{inch}$ for $t$ and $8.14{\text{inch}}^2$ for $A_{\text{g}}$ in Equation (IX).

$\begin{array}{c}{A}_{\text{n}}=8.41{\text{inch}}^2-2\left(1\frac{1}{4}+\frac{1}{8}\right)\text{inch}\times \frac{5}{8}\text{inch}\\ =6.691{\text{inch}}^2\end{array}$

Write the expression for effective net area.

$A_{\text{e}}=A_{\text{n}}U$     ...... (X)

Here, the reduction factor is $U$ and the effective net area is $A_{\text{e}}$.

Substitute $0.8$ for $U$ and $6.691{\text{inch}}^2$ for $A_{\text{n}}$ in Equation (X).

$\begin{array}{c}{A}_{\text{e}}=\left(6.691\times 0.8\right){\text{inch}}^2\\ =5.35{\text{inch}}^2\end{array}$

Write the expression for minimum spacing.

$m_{\text{s}}=\left(2\frac{2}{3}\right)d$     ...... (XI)

Here, the minimum spacing is $m_{\text{s}}$ and the bolt diameter is $d$.

Substitute $1\frac{1}{4}\text{inch}$ for $d$ in Equation (XI).

$\begin{array}{c}{m}_{\text{s}}=\left(2.667\times 1\frac{1}{4}\right)\text{inch}\\ =3.334\text{inch}\end{array}$

Write the expression for design strength of edge bolts for an edge distance of $2.5\text{inch}$.

From the manual, Table 7-6,

$\varphi r_{\text{n}}=t\times 70.1\text{kips}/\text{inch}$     ...... (XII)

Here, load resistance factor is $\varphi$, the strength is $r_{\text{n}}$ and the thickness is $t$.

Substitute $0.375\text{inch}$ for $t$ in Equation (XII).

$\begin{array}{c}\varphi r_{\text{n}}=\left(0.375\text{inch}\times 70.1\text{kips}/\text{inch}\right)\\ =26.3\text{kips}\end{array}$

Write the expression for design strength of inner bolts for a spacing of $2\frac{2}{3}d$.

From the manual, Table 7-5,

$\varphi r_{\text{n}}=t\times 105\text{kips}/\text{inch}$     ...... (XIII)

Substitute $0.375\text{inch}$ for $t$ in Equation (XIII).

$\begin{array}{c}\varphi r_{\text{n}}=\left(0.375\text{inch}\times 105\text{kips}/\text{inch}\right)\\ =39.4\text{kips}\end{array}$

Thus, the slip critical strength is taken into consideration as it has less value.

Write the expression for gross shear area.

$A_{\text{gv}}=2\left(t\times s_{\text{s}}\right)$     ...... (XIV)

Here, shear length is $s_{\text{s}}$ and gross shear area is $A_{\text{gv}}$.

Substitute $0.375\text{inch}$ for $t$ and $18.5\text{inch}$ for $s_{\text{s}}$ in Equation (XIV).

$\begin{array}{c}{A}_{\text{gv}}=\left(2\times 0.375\times 18.5\right){\text{inch}}^2\\ =13.88{\text{inch}}^2\end{array}$

Write the expression for the net area in shear.

$A_{\text{nv}}=A_{\text{gv}}-A_{\text{holes}}$     ...... (XV)

Here, net area in shear is $A_{\text{nv}}$.

Substitute $t{d}_{\text{h}}$ for $A_{\text{holes}}$ in Equation (XV).

$A_{\text{nv}}=A_{\text{gv}}-2\left(t{d}_{\text{h}}\right)$     ...... (XVI)

Substitute $13.88{\text{inch}}^2$ for $A_{\text{gv}}$ and $0.375\text{inch}$ for $t$ in Equation (XVI).

$\begin{array}{c}{A}_{\text{nv}}=\left(0.375\times \left(18.5-4.5\times 1.375\right)\times 2\right){\text{inch}}^2\\ =9.234{\text{inch}}^2\end{array}$

Write the expression for net area in tension.

$A_{\text{nt}}=\left(t\times t_{\text{L}}\right)-A_{\text{holes}}$     ...... (XVII)

Here, tension length is $t_{\text{L}}$ and net area in tension is $A_{\text{nt}}$.

Substitute $3\text{inch}$ for $t_{\text{L}}$ and $0.375\text{inch}$ for $t$ Equation (XVII).

$\begin{array}{c}{A}_{\text{nt}}=\left(0.375\times \left(3-1\times 1.375\right)\right){\text{inch}}^2\\ =0.6094{\text{inch}}^2\end{array}$

Write the expression for the nominal strength in block shear.

$R_{\text{nb}}=0.6F_{\text{u}}{A}_{\text{nv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}$     ...... (XVIII)

Here, reduction factor is $U_{\text{bs}}$ and nominal strength in block shear is $R_{\text{nb}}$.

Substitute $9.234{\text{inch}}^2$ for $A_{\text{nv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.6094{\text{inch}}^2$ for $A_{\text{nt}}$ and $1$ for $U_{\text{bs}}$ in Equation (XVIII).

$\begin{array}{c}{R}_{\text{nb}}=\left[\left(0.6\times 58\text{ksi}\times 9.234{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.6094{\text{inch}}^2\right)\right]\\ =321.35\text{kips}+35.35\text{kips}\\ =356.7\text{kips}\end{array}$

Write the expression for nominal shear strength in gross shear.

$R_{\text{ng}}=0.6F_{\text{y}}{A}_{\text{gv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}$     ...... (XIX)

Here, nominal shear strength in gross shear is $R_{\text{ng}}$ and yield strength is $F_{\text{y}}$.

Substitute $13.88{\text{inch}}^2$ for $A_{\text{gv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.6094{\text{inch}}^2$ for $A_{\text{nt}}$, $36\text{ksi}$ for $F_{\text{y}}$ and $1$ for $U_{\text{bs}}$ in Equation (XIX).

$\begin{array}{c}{R}_{\text{ng}}=\left[\left(0.6\times 36\text{ksi}\times 13.88{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.6094{\text{inch}}^2\right)\right]\\ =299.8\text{kips}+35.35\text{kips}\\ =335.2\text{kips}\end{array}$

Thus, the nominal shear strength is taken as $335.2\text{kips}$, as it has less value.

Write the expression for design strength per bolt.

$B_{\text{d}}=\varphi R_{\text{n}}$     ...... (XX)

Here, design strength of bolt is $B_{\text{d}}$.

Substitute $335.2\text{kips}$ for $R_{\text{n}}$ and $0.75$ for $\varphi$ in Equation (XX).

$\begin{array}{c}{B}_{\text{d}}=0.75\times 335.2\text{kips}\\ =251.4\text{kips}\end{array}$

Thus, the design strength is equal to $251.4\text{kips}$ as it is greater than the total factored load.

Conclusion:

Thus, an angle $L8\times 6\times \frac{5}{8}$ and $10$ bolts of $1\frac{1}{4}\text{inch}$ diameter each of Group A in two lines with a spacing of $4\text{inch}$ and an edge distance of $2.5\text{inch}$ is used.

To determine

(b)

The design of a single angle tension member and a bolted connection using Allowable Stress Design (ASD) method.

Answer to Problem 7.7.2P

An angle $L8\times 6\times \frac{5}{8}$ and $10$ bolts of $1\frac{1}{4}\text{inch}$ diameter each of Group A in two lines with a spacing of $4\text{inch}$ and an edge distance of $2.5\text{inch}$ is used.

Explanation of Solution

Calculation:

Write the expression for allowable factored load.

$P_{\text{a}}=D+L$     ...... (XXI)

Here, allowable factored load is $P_{\text{a}}$.

Substitute $100\text{kips}$ for $L$ and $50\text{kips}$ for $D$ in Equation (XXI).

$\begin{array}{c}{P}_{\text{a}}=\left(50+100\right)\text{kips}\\ =150\text{kips}\end{array}$

Write the expression for allowable factored load by load combination 6.

$P_{\text{a}}=D+0.75L+0.75\times 0.6W$     ...... (XXII)

Substitute $100\text{kips}$ for $L$, $45\text{kips}$ for $W$ and $50\text{kips}$ for $D$ in Equation (XXII).

$\begin{array}{c}{P}_{\text{a}}=\left(50+0.75\times 100+0.75\times 0.6\times 45\right)\text{kips}\\ =145.3\text{kips}\end{array}$

Thus, the value of allowable factored load by load combination 2 is greater, so the allowable factored load is equal to $150\text{kips}$.

Write the expression for slip critical shear strength per bolt.

$\frac{R_{\text{n}}}{\Omega }=\frac{\left(\mu D_{\text{u}}{h}_{\text{f}}{n}_{\text{s}}{T}_{\text{b}}\right)}{\Omega }$     ...... (XXIII)

Here, the load safety factor is $\Omega$.

Substitute $1.13$ for $D_{\text{u}}$, $0.3$ for $\mu$, $1.0$ for $h_{\text{f}}$, $1.5$ for $\Omega$, $1$ for $n_{\text{s}}$ and $71\text{kips}$ for $T_{\text{b}}$ in Equation (XXIII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\frac{\left(0.3\times 1.13\times 1.0\times 71\times 1\right)}{1.5}\\ =16.05\text{kips}\end{array}$

Write the expression for the design strength.

$B=N\times \frac{R_{\text{n}}}{\Omega }$     ...... (XXIV)

Here, the design strength is $B$.

Substitute $16.05\text{kips}$ for $\frac{R_{\text{n}}}{\Omega }$ and $10$ for $N$ in Equation (XXIV).

$\begin{array}{c}B=\left(10\times 16.05\right)\text{kips}\\ =160.5\text{kips}\end{array}$

Write the expression for required gross area of the member.

$A_{\text{g}}=\frac{P_{\text{a}}}{0.6F_{\text{y}}}$     ...... (XXV)

Here, the gross area is $A_{\text{g}}$.

Substitute $36\text{ksi}$ for $F_{\text{y}}$ and $150\text{kips}$ for $P_{\text{u}}$ in Equation (XXV).

$\begin{array}{c}{A}_{\text{g}}=\frac{150\text{kips}}{0.6\times 36\text{kis}}\\ =6.94{\text{inch}}^2\end{array}$

Write the expression for required effective area of the member.

$A_{\text{e}}=\frac{P_{\text{a}}}{0.5F_{\text{u}}}$     ...... (VI)

Here, the effective area is $A_{\text{e}}$.

Substitute $58\text{ksi}$ for $F_{\text{u}}$ and $150\text{kips}$ for $P_{\text{u}}$ in Equation (XXVI).

$\begin{array}{c}{A}_{\text{e}}=\frac{150\text{kips}}{0.5\times 58\text{ksi}}\\ =5.17{\text{inch}}^2\end{array}$

Write the expression for minimum value of radius of gyration.

$r_{\text{min}}=\frac{L}{300}$     ...... (XXVII)

Here, minimum radius of gyration is $r_{\text{min}}$.

Substitute $240\text{inch}$ for $L$ in Equation (XXVII).

$\begin{array}{c}{r}_{\text{min}}=\frac{240\text{inch}}{300}\\ =0.8\text{inch}\end{array}$

Thus, try the section $L8\times 6\times \frac{5}{8}$, as per the above calculated value.

Write the expression for net area of $L8\times 6\times \frac{5}{8}$ section.

$A_{\text{n}}=A_{\text{g}}-A_{\text{holes}}$     ...... (XXVIII)

Here, net area of the section is $A_{\text{n}}$.

Substitute $t{d}_{\text{h}}$ for $A_{\text{holes}}$ in Equation (XXVIII).

$A_{\text{n}}=A_{\text{g}}-2\left(t{d}_{\text{h}}\right)$     ...... (XXIX)

Substitute $\frac{5}{8}\text{inch}$ for $t$ and $8.14{\text{inch}}^2$ for $A_{\text{g}}$ in Equation (XXIX).

$\begin{array}{c}{A}_{\text{n}}=8.41{\text{inch}}^2-2\left(1\frac{1}{4}+\frac{1}{8}\right)\text{inch}\times \frac{5}{8}\text{inch}\\ =6.691{\text{inch}}^2\end{array}$

Write the expression for effective net area.

$A_{\text{e}}=A_{\text{n}}U$     ...... (XXX)

Here, the effective net area is $A_{\text{e}}$.

Substitute $0.8$ for $U$ and $6.691{\text{inch}}^2$ for $A_{\text{n}}$ in Equation (XXX).

$\begin{array}{c}{A}_{\text{e}}=\left(6.691\times 0.8\right){\text{inch}}^2\\ =5.35{\text{inch}}^2\end{array}$

Write the expression for minimum spacing.

$m_{\text{s}}=\left(2\frac{2}{3}\right)d$     ...... (XXXI)

Substitute $1\frac{1}{4}\text{inch}$ for $d$ in Equation (XXXI).

$\begin{array}{c}{m}_{\text{s}}=\left(2.667\times 1\frac{1}{4}\right)\text{inch}\\ =3.334\text{inch}\end{array}$

Write the expression for design strength of edge bolts for an edge distance of $2.5\text{inch}$.

From the manual, Table 7-6,

$\frac{R_{\text{n}}}{\Omega }=t\times 46.8\text{kips}/\text{inch}$     ...... (XXXII)

Substitute $0.375\text{inch}$ for $t$ in Equation (XXXII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\left(0.375\text{inch}\times 46.8\text{kips}/\text{inch}\right)\\ =17.55\text{kips}\end{array}$

Write the expression for design strength of inner bolts for a spacing of $2\frac{2}{3}d$.

From the manual, Table 7-5,

$\frac{R_{\text{n}}}{\Omega }=t\times 70.3\text{kips}/\text{inch}$     ...... (XXXIII)

Substitute $0.375\text{inch}$ for $t$ in Equation (XXXIII).

$\begin{array}{c}\frac{R_{\text{n}}}{\Omega }=\left(0.375\text{inch}\times 70.3\text{kips}/\text{inch}\right)\\ =26.36\text{kips}\end{array}$

Thus, the slip critical strength is taken into consideration as it has less value.

Write the expression for gross shear area.

$A_{\text{gv}}=2\left(t\times s_{\text{s}}\right)$     ...... (XXXIV)

Here, gross shear area is $A_{\text{gv}}$.

Substitute $0.375\text{inch}$ for $t$ and $18.5\text{inch}$ for $s_{\text{s}}$ in Equation (XXXIV).

$\begin{array}{c}{A}_{\text{gv}}=\left(2\times 0.375\times 18.5\right){\text{inch}}^2\\ =13.88{\text{inch}}^2\end{array}$

Write the expression for the net area in shear.

$A_{\text{nv}}=A_{\text{gv}}-A_{\text{holes}}$     ...... (XXXV)

Here, net area in shear is $A_{\text{nv}}$.

Substitute $t{d}_{\text{h}}$ for $A_{\text{holes}}$ in Equation (XXXV).

$A_{\text{nv}}=A_{\text{gv}}-2\left(t{d}_{\text{h}}\right)$     ...... (XXXVI)

Substitute $13.88{\text{inch}}^2$ for $A_{\text{gv}}$ and $0.375\text{inch}$ for $t$ in Equation (XXXVI).

$\begin{array}{c}{A}_{\text{nv}}=\left(0.375\times \left(18.5-4.5\times 1.375\right)\times 2\right){\text{inch}}^2\\ =9.234{\text{inch}}^2\end{array}$

Write the expression for net area in tension.

$A_{\text{nt}}=\left(t\times t_{\text{L}}\right)-A_{\text{holes}}$     ...... (XXXVII)

Here, net area in tension is $A_{\text{nt}}$.

Substitute $3\text{inch}$ for $t_{\text{L}}$ and $0.375\text{inch}$ for $t$ Equation (XXXVII).

$\begin{array}{c}{A}_{\text{nt}}=\left(0.375\times \left(3-1\times 1.375\right)\right){\text{inch}}^2\\ =0.6094{\text{inch}}^2\end{array}$

Write the expression for the nominal strength in block shear.

$R_{\text{nb}}=0.6F_{\text{u}}{A}_{\text{nv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}$     ...... (XXXVIII)

Here, nominal strength in block shear is $R_{\text{nb}}$.

Substitute $9.234{\text{inch}}^2$ for $A_{\text{nv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.6094{\text{inch}}^2$ for $A_{\text{nt}}$ and $1$ for $U_{\text{bs}}$ in Equation (XXXVIII).

$\begin{array}{c}{R}_{\text{nb}}=\left[\left(0.6\times 58\text{ksi}\times 9.234{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.6094{\text{inch}}^2\right)\right]\\ =321.35\text{kips}+35.35\text{kips}\\ =356.7\text{kips}\end{array}$

Write the expression for nominal shear strength in gross shear.

$R_{\text{ng}}=0.6F_{\text{y}}{A}_{\text{gv}}+U_{\text{bs}}{F}_{\text{u}}{A}_{\text{nt}}$     ...... (XXXIX)

Here, nominal shear strength in gross shear is $R_{\text{ng}}$.

Substitute $13.88{\text{inch}}^2$ for $A_{\text{gv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.6094{\text{inch}}^2$ for $A_{\text{nt}}$, $36\text{ksi}$ for $F_{\text{y}}$ and $1$ for $U_{\text{bs}}$ in Equation (XXXIX).

$\begin{array}{c}{R}_{\text{ng}}=\left[\left(0.6\times 36\text{ksi}\times 13.88{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.6094{\text{inch}}^2\right)\right]\\ =299.8\text{kips}+35.35\text{kips}\\ =335.2\text{kips}\end{array}$

Thus, the nominal shear strength is taken as $335.2\text{kips}$, as it has less value.

Write the expression for allowable strength.

$B_{\text{a}}=\frac{R_{\text{n}}}{\Omega }$     ...... (XL)

Here, allowable strength of bolt is $B_{\text{a}}$.

Substitute $335.2\text{kips}$ for $R_{\text{n}}$ and $2.0$ for $\Omega$ in Equation (XL).

$\begin{array}{c}{B}_{\text{a}}=\frac{335.2}{2.0}\text{kips}\\ =168\text{kips}\end{array}$

Thus, the allowable strength is equal to $168\text{kips}$ as it is greater than the total allowable load.

Conclusion:

Thus, an angle $L8\times 6\times \frac{5}{8}$ and $10$ bolts of $1\frac{1}{4}\text{inch}$ diameter each of Group A in two lines with a spacing of $4\text{inch}$ and an edge distance of $2.5\text{inch}$ is used.