Chapter 7, Problem 7.6.6P

To determine

(a)

If the members and its connections are satisfactory using LRFD.

Answer to Problem 7.6.6P

The members and its connections are satisfactory.

Explanation of Solution

Given:

Tension member is $\text{L6}\times \text{3}\frac{1}{2}\times \frac{1}{2}$.

Steel Used is $\text{A36}$.

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

Diameter of the bolt is $1\frac{1}{8}\text{inch}$.

Dead load is $20\text{kips}$.

Live load is $60\text{kips}$.

Wind load is $20\text{kips}$.

Length of the member is $9\text{ft}$.

Calculation:

The properties for $\text{L6}\times 3\frac{1}{2}\times \frac{1}{2}$ section from the AISC steel table are as follows:

The gross area is $4.5{\text{inch}}^2$.

The distance from the plane of connection to the centroid is $0.829\text{inch}$.

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 $36\text{ksi}$.

Write the expression for slenderness ratio.

$\frac{L}{r_{\min }}=\frac{L}{r_{\text{z}}}$     ...... (I)

Here, radius of gyration is $r_{\min }$, length of the member is $L,$ and radius of gyration about z-axis is $r_{\text{z}}$.

Substitute $9\text{ft}$ for $L$ and $0.756\text{inch}$ for $r_{\text{z}}$ in Equation (I).

$\begin{array}{c}\frac{L}{r_{\min }}=\frac{\left(9\text{ft}\right)\left(\frac{12\text{inch}}{1\text{ft}}\right)}{0.756\text{inch}}\\ =142.857\\ \approx 143\end{array}$

The slenderness ratio is less than $300$, so it is acceptable.

Write the expression for cross-sectional area of the bolt.

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

Here, cross-sectional area of the bolt is $A_{\text{b}}$ and diameter of the bolt is $d$.

Substitute $1.125\text{inch}$ for $d$ in Equation (II).

$\begin{array}{c}{A}_{\text{b}}=\frac{\pi }{4}{\left(1.125\text{inch}\right)}^2\\ =0.994{\text{inch}}^2\end{array}$

Write the expression for nominal shear capacity of one bolt.

$R_{\text{nv}}=\left(F_{\text{nv}}{A}_{\text{b}}\right)$     ...... (III)

Here, nominal shear stress of one bolt is $R_{\text{nv}}$ and nominal shear stress of the bolt material is $F_{\text{nv}}$.

Substitute $0.994{\text{inch}}^2$ for $A_{\text{b}}$ and $54\text{ksi}$ for $F_{\text{nv}}$ in Equation (III).

$\begin{array}{c}{R}_{\text{nv}}=\left(54\text{ksi}\right)\left(0.994{\text{inch}}^2\right)\\ =53.68\text{kips}\end{array}$

Write the expression for slip critical shear strength per bolt for class A surfaces.

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

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 $D_{\text{u}}$, $1.0$ for $h_{\text{f}}$, $0.3$ for $\mu$, $1$ for $n_{\text{s}}$ and $56\text{kips}$ for $T_{\text{b}}$ in Equation (IV).

$\begin{array}{c}{R}_{\text{n}}=0.3\times 1.0\times 1\times 1.13\times 56\text{kips}\\ =18.98\text{kips}\end{array}$

Write the expression to calculate the nominal bearing strength of an edge bolt $R_{{\text{n}}_1}$.

$R_{{\text{n}}_1}=1.2l_{\text{c}}t{F}_{\text{u}}$     ...... (V)

Here, nominal bearing strength is $R_{{\text{n}}_1}$, thickness of the gusset plate is $t$, ultimate tensile stress is $F_{\text{u}}$ and the distance between the edge of the adjacent hole to the edge of the bolt hole is $l_{\text{c}}$.

With the upper limit of nominal bearing strength of the edge bolt,

$R_{{\text{n}}_1}=2.4dt{F}_{\text{u}}$

Write the expression to calculate the value of $l_{\text{c}}$ for edge bolts.

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

Here, height of the bolt is $h$.

Write the expression to calculate the value of $l_{\text{c}}$ for inner bolts.

$l_{\text{c}}=s-h$     ...... (VII)

Here, spacing between the bolts is $s$.

Write the expression to calculate the height of the bolt.

$h=d+\frac{1}{16}\text{inch}$     ...... (VIII)

Calculate the value of $h$.

Substitute $\frac{9}{8}\text{inch}$ for $d$ Equation (VIII).

$\begin{array}{c}h=\frac{9}{8}\text{inch}+\frac{1}{16}\text{inch}\\ =1.\text{188}\text{inch}\end{array}$

Calculate the value of $l_{\text{c}}$ for edge bolts.

Substitute $2\text{inch}$ for $l_{\text{e}}$ and $1.188\text{inch}$ for $h$ in Equation (VI).

$\begin{array}{c}{l}_{\text{c}}=2\text{inch}-\frac{1.188\text{inch}}{2}\\ =1.406\text{inch}\end{array}$

Calculate the nominal bearing strength for edge bolts.

Substitute $1.406\text{inch}$ for $l_{\text{c}}$, $\frac{3}{8}\text{inch}$ for $t$ and $58\text{ksi}$ for $F_{\text{u}}$ in Equation (V).

$\begin{array}{c}{R}_{{\text{n}}_1}=1.2\left(1.406\text{inch}\right)\left(\frac{3}{8}\text{inch}\right)\left(58\text{ksi}\right)\\ =36.70\text{kips}\end{array}$

Calculate the upper limit of nominal bearing strength for edge bolts.

Substitute $1.125\text{inch}$ for $d$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$.

$\begin{array}{c}{R}_{{\text{n}}_1}=2.4\left(1.125\text{inch}\right)\left(\frac{3}{8}\text{inch}\right)\left(58\text{ksi}\right)\\ =58.73\text{kips}\end{array}$

$$

Thus, the minimum value of $R_{{\text{n}}_1}$ is considered and is equal to $36.70\text{kips}$.

Write the expression to calculate the nominal bearing strength of inner bolt $R_{{\text{n}}_2}$.

$R_{{\text{n}}_2}=1.2l_{\text{c}}t{F}_{\text{u}}$     ...... (IX)

With the upper limit of nominal bearing strength of the inner bolt,

$R_{{\text{n}}_1}=2.4dt{F}_{\text{u}}$

Calculate the value of $l_{\text{c}}$ for inner bolts.

Substitute $3.5\text{inch}$ for $s$ and $1.188\text{inch}$ for $h$ in Equation (VII).

$\begin{array}{c}{l}_{\text{c}}=3.5\text{inch}-1.188\text{inch}\\ =2.312\text{inch}\end{array}$

Calculate the nominal bearing strength for inner bolts.

Substitute $2.312\text{inch}$ for $l_{\text{c}}$, $\frac{3}{8}\text{inch}$ for $t$, $1.125\text{inch}$ for $d,$ and $58\text{ksi}$ for $F_{\text{u}}$ in Equation (IX).

$\begin{array}{c}{R}_{{\text{n}}_2}=1.2\left(2.312\text{inch}\right)\left(\frac{3}{8}\text{inch}\right)\left(58\text{ksi}\right)\\ =60.34\text{kips}\end{array}$

Calculate the upper limit of nominal bearing strength for inner bolts.

Substitute $1.125\text{inch}$ for $d$, $58\text{ksi}$ for $F_{\text{u}}$ and $\frac{3}{8}\text{inch}$ for $t$.

$\begin{array}{c}{R}_{{\text{n}}_2}=2.4\left(1.125\text{inch}\right)\left(\frac{3}{8}\text{inch}\right)\left(58\text{ksi}\right)\\ =58.73\text{kips}\end{array}$

Thus, the minimum value of $R_{{\text{n}}_2}$ is considered and is equal to $58.73\text{kips}$.

Write the expression for total strength of connection.

$R_{\text{nt}}=6\times R_{\text{n}}$     ...... (X)

Substitute $18.98\text{kips}$ for $R_{\text{n}}$ in Equation (X).

$\begin{array}{c}{R}_{\text{nt}}=6\times 18.98\text{kips}\\ =\text{113}\text{.9}\text{kips}\end{array}$

Write the expression to calculate the nominal strength in yielding.

$P_{\text{n}}=F_{\text{y}}{A}_{\text{g}}$     ...... (XI)

Here, yield strength is $F_{\text{y}}$, nominal strength is $P_{\text{n}}$ and gross area of the section is $A_{\text{g}}$.

Substitute $4.5{\text{inch}}^2$ for $A_{\text{g}}$ and $36\text{ksi}$ for $F_{\text{y}}$ in Equation (XI).

$\begin{array}{c}{P}_{\text{n}}=36\text{ksi}\times \text{4}\text{.5}{\text{inch}}^2\\ =162\text{kips}\end{array}$

Write the expression to calculate the diameter of the hole.

$d_{\text{h}}=d_{\text{b}}+S_{\text{c}}$     ...... (XI)

Here, diameter of the hole is $d_{\text{h}}$, side clearance is $S_{\text{c}},$ and diameter of the bolt is $d_{\text{b}}$.

Substitute $\frac{9}{8}\text{inch}$ for $d_{\text{b}}$ and $\frac{1}{8}\text{inch}$ for $S_{\text{c}}$ in Equation (XI).

$\begin{array}{c}{d}_{\text{h}}=\left(\frac{9}{8}\text{inch}+\frac{1}{8}\text{inch}\right)\\ =1.25\text{inch}\end{array}$

Write the expression to calculate the effective area.

$A_{\text{e}}=\left(A_{\text{g}}-A_{\text{holes}}\right)$     ...... (XII)

Here, area of holes is $A_{\text{holes}}$ and effective area is $A_{\text{e}}$.

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

$A_{\text{e}}=A_{\text{g}}-t{d}_{\text{h}}$     ...... (XIII)

Substitute $\frac{1}{2}\text{inch}$ for $t$, $1.25\text{inch}$ for $d_{\text{h}},$ and $4.5{\text{inch}}^2$ for $A_{\text{g}}$ in Equation (XIII).

$\begin{array}{c}{A}_{\text{n}}=\left(4.5{\text{inch}}^2-\left(\frac{1}{2}\text{inch}\times 1.25\text{inch}\right)\right)\\ =4.5{\text{inch}}^2-0.625{\text{inch}}^2\\ =3.875{\text{inch}}^2\end{array}$

Write the expression for the reduction factor.

$U=1-\frac{\overline{x}}{l}$     ...... (XIV)

Here, length of the connection is $l$, reduction factor is $U$ and distance from the plane of connection to the centroid is $\overline{x}$.

Substitute $17.5\text{inch}$ for $l$ and $0.829\text{inch}$ for $\overline{x}$ in Equation (XIV).

$\begin{array}{c}U=1-\frac{0.829\text{inch}}{17.5\text{inch}}\\ =1-0.0474\\ =0.9526\end{array}$

Write the expression for effective net area.

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

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

Substitute $0.9526$ for $U$ and $3.875{\text{inch}}^2$ for $A_{\text{n}}$ in Equation (XV).

$\begin{array}{c}{A}_{\text{e}}=3.875{\text{inch}}^2\times 0.9526\\ =3.691{\text{inch}}^2\end{array}$

Write the expression for nominal strength for fracture.

$P_{\text{n}}=F_{\text{u}}{A}_{\text{e}}$     ...... (XVI)

Here, nominal strength for fracture is $P_{\text{n}}$ and ultimate strength is $F_{\text{u}}$.

Substitute $3.691{\text{inch}}^2$ for $A_{\text{e}}$ and $58\text{ksi}$ for $F_{\text{u}}$ in Equation (XVI).

$\begin{array}{c}{P}_{\text{n}}=58\text{ksi}\times 3.691{\text{inch}}^2\\ =241.1\text{kips}\end{array}$

Write the expression for gross shear area.

$A_{\text{gv}}=t\times s_{\text{l}}$     ...... (XVII)

Here, thickness of member is $t$, shear length is $s_{\text{l}}$ and gross shear area is $A_{\text{gv}}$.

Substitute $19.5\text{inch}$ for $s_{\text{l}}$ and $\frac{1}{2}\text{inch}$ for $t$ in Equation (XVII).

$\begin{array}{c}{A}_{\text{gv}}=\frac{1}{2}\text{inch}\times 19.5\text{inch}\\ =9.75{\text{inch}}^2\end{array}$

Write the expression for the net area in shear.

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

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

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

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

Substitute $9.75{\text{inch}}^2$ for $A_{\text{gv}}$, $1.25\text{inch}$ for $d_{\text{h}},$ and $\frac{1}{2}\text{inch}$ for $t$ in Equation (XVIX).

$\begin{array}{c}{A}_{\text{nv}}=\left(9.75{\text{inch}}^2-\left(\frac{1}{2}\text{inch}\times 0.5\times 1.125\text{inch}\right)\right)\\ =9.438{\text{inch}}^2\end{array}$

Write the expression for net area in tension.

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

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

Substitute $2.5\text{inch}$ for $t_{\text{L}}$, $\frac{1}{2}\text{inch}$ for $t$ and $0.3125{\text{inch}}^2$ for $A_{\text{h}}$ in Equation (XX).

$\begin{array}{c}{A}_{\text{nt}}=\left(\frac{1}{2}\text{inch}\times 2.5\text{inch}\right)-0.3125{\text{inch}}^2\\ =0.9375{\text{inch}}^2\end{array}$

Write the expression for the nominal strength in block shear.

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

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

Substitute $9.438{\text{inch}}^2$ for $A_{\text{nv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.9375{\text{inch}}^2$ for $A_{\text{nt}},$ and $1$ for $U_{\text{bs}}$ in Equation (XXI).

$\begin{array}{c}{R}_{\text{nb}}=\left[\left(0.6\times 58\text{kips}\times 9.438{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.9375{\text{inch}}^2\right)\right]\\ =382.8\text{kips}\end{array}$

Write the expression for nominal shear strength in gross shear.

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

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

Substitute $9.75{\text{inch}}^2$ for $A_{\text{gv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.9375{\text{inch}}^2$ for $A_{\text{nt}}$, $36\text{ksi}$ for $F_{\text{y}},$ and $1$ for $U_{\text{bs}}$ in Equation (XX).

$\begin{array}{c}{R}_{\text{ng}}=\left[\left(0.6\times 36\text{ksi}\times 9.75{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.9375{\text{inch}}^2\right)\right]\\ =210.6\text{kips}+54.375\text{kips}\\ =265\text{kips}\end{array}$

Thus, the nominal shear strength is taken as $265\text{kips}$, as it has less value.

Write the expression for design strength per bolt.

$B=\varphi R_{\text{n}}$     ...... (XXI)

Here, load reduction factor is $\varphi$ and design strength of bolt is $B$.

Substitute $132.9\text{kips}$ for $R_{\text{n}}$ and $1$ for $\varphi$ in Equation (XXI).

$\begin{array}{c}B=1\times 132.9\text{kips}\\ \approx 133\text{kips}\end{array}$

Write the expression for design strength in yielding.

$B_{\text{y}}={\varphi }_{\text{t}}{P}_{\text{n}}$     ...... (XXII)

Here, load reduction factor in yielding is ${\varphi }_{\text{t}}$ and design strength in yielding is $B_{\text{y}}$.

Substitute $162\text{kips}$ for $P_{\text{n}}$ and $0.9$ for ${\varphi }_{\text{t}}$ in Equation (XXII).

$\begin{array}{c}{B}_{\text{y}}=0.9\times 162\text{kips}\\ =146\text{kips}\end{array}$

Write the expression for design strength for fracture.

$B_{\text{f}}=\varphi P_{\text{n}}$     ...... (XXIII)

Here, design strength for fracture is $B_{\text{f}}$.

Substitute $214.1\text{kips}$ for $P_{\text{n}}$ and $0.75$ for $\varphi$ in Equation (XXIII).

$\begin{array}{c}{B}_{\text{f}}=0.75\times 214.1\text{kips}\\ =161\text{kips}\end{array}$

Write the expression for design strength for block shear.

$B_{\text{b}}=\varphi P_{\text{n}}$     ...... (XXIV)

Here, design strength for block shear is $B_{\text{b}}$.

Substitute $256\text{kips}$ for $P_{\text{n}}$ and $0.75$ for $\varphi$ in Equation (XXIV).

$\begin{array}{c}{B}_{\text{b}}=0.75\times 256\text{kips}\\ =199\text{kips}\end{array}$

Thus, the bolt strength is $133\text{kips}$ as it is the least value.

Write the expression for total factored load.

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

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

Substitute $60\text{kips}$ for $L$ and $20\text{kips}$ for $D$ in Equation (XXV).

$\begin{array}{c}{P}_{\text{u}}=1.2\left(20\text{kips}\right)+1.6\left(60\text{kips}\right)\\ =120\text{kips}\end{array}$

The total factored load is $120\text{kips}$ which is less than the design load.

Conclusion:

Thus, the members and its connections are satisfactory.

To determine

(b)

If the members and its connections are satisfactory using ASD.

Answer to Problem 7.6.6P

The members and its connections are satisfactory.

Explanation of Solution

Calculation:

Write the expression to calculate the allowable strength for shear per bolt.

$B_{\text{A}}=\frac{R_{\text{n}}}{\Omega }$     ...... (XXVI)

Here, allowable strength is $B_{\text{A}}$ and the load safety factor is $\Omega$.

Substitute $132.9\text{kips}$ for $R_{\text{n}}$ and $1.5$ for $\Omega$ in Equation (XXVI).

$\begin{array}{c}{B}_{\text{A}}=\frac{132.9\text{kips}}{1.5}\\ =88.6\text{kips}\end{array}$     ...... (XXVI)

Write the expression for allowable strength in yielding.

$B_{\text{Y}}=\frac{P_{\text{n}}}{{\Omega }_{\text{t}}}$     ...... (XXVII)

Here, safety factor is ${\Omega }_{\text{t}}$ and allowable strength in yielding is $B_{\text{Y}}$.

Substitute $162\text{kips}$ for $P_{\text{n}}$ and $1.67$ for ${\Omega }_{\text{t}}$ in Equation (XXVII).

$\begin{array}{c}{B}_{\text{Y}}=\frac{162\text{kips}}{1.67}\\ =97\text{kips}\end{array}$

Calculate the allowable strength in fracture.

Substitute $214.1\text{kips}$ for $P_{\text{n}}$ and $2.0$ for ${\Omega }_{\text{t}}$ in Equation (XXVII).

$\begin{array}{c}{B}_{\text{Y}}=\frac{214.1\text{kips}}{2.0}\\ =107\text{kips}\end{array}$

Calculate the allowable strength for block shear.

Substitute $265\text{kips}$ for $P_{\text{n}}$ and $2.0$ for ${\Omega }_{\text{t}}$ in Equation (XXVII).

$\begin{array}{c}{B}_{\text{Y}}=\frac{265\text{kips}}{2.0}\\ =132.5\text{kips}\end{array}$

Thus, the design strength is $88.6\text{kips}$ as it is the least value.

Write the expression for total factored load.

$P_{\text{a}}=D+L$     ...... (XXVIII)

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

Substitute $20\text{kips}$ for $D$ and $60\text{kips}$ for $L$ in Equation (XXVIII).

$\begin{array}{c}{P}_{\text{a}}=20\text{kips}+60\text{kips}\\ =80\text{kips}\end{array}$

The allowable factored load is $80\text{kips}$ which is less than the design load.

Conclusion:

Thus, the members and its connections are satisfactory.