Chapter 7, Problem 7.6.5P

To determine

(a)

The maximum service load using LRFD.

Answer to Problem 7.6.5P

The maximum service load is $50.6\text{kips}$.

Explanation of Solution

Given:

Tension member is $\text{PL}\frac{1}{2}\times 6\frac{1}{2}$

Steel Used is $\text{A36}$.

Thickness of gusset plate is $\frac{5}{8}\text{inch}$.

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

Ratio of live load to dead load is $3.0$.

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 $36\text{ksi}$.

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

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

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 (I).

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

Write the expression for nominal shear capacity of one bolt.

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

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 (II).

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

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

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

Here, the ratio of the mean actual bolt pre-tensioned to the 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 (III).

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

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

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

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.

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

Here, height of the bolt is $h$.

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

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

Here, spacing between the bolts is $s$.

Write the expression to calculate the height of the bolt.

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

Calculate the value of $h$.

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

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

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 (V).

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

Calculate the nominal bearing strength for edge bolts.

Substitute $1.406\text{inch}$ for $l_{\text{c}}$, $\frac{1}{2}\text{inch}$ for $t$ and $58\text{ksi}$ for $F_{\text{u}}$ in Equation (IV).

$\begin{array}{c}{R}_{{\text{n}}_1}=1.2\left(1.406\text{inch}\right)\left(\frac{1}{2}\text{inch}\right)\left(58\text{ksi}\right)\\ =48.93\text{kips}\end{array}$

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 $\frac{1}{2}\text{inch}$ for $t$.

$\begin{array}{c}{R}_{{\text{n}}_1}=2.4\left(1.125\text{inch}\right)\left(\frac{1}{2}\text{inch}\right)\left(58\text{ksi}\right)\\ =78.3\text{kips}\end{array}$

Thus, the minimum value of $R_{{\text{n}}_1}$ is considered and is equal to $48.93\text{kips}$.

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

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

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

$R_{{\text{n}}_2}=2.4dt{F}_{\text{u}}$

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

Substitute $3\text{inch}$ for $s$ and $1.188\text{inch}$ for $h$ in Equation (VI).

$\begin{array}{c}{l}_{\text{c}}=3\text{inch}-1.188\text{inch}\\ =1.812\text{inch}\end{array}$

Calculate the nominal bearing strength for inner bolts.

Substitute $1.812\text{inch}$ for $l_{\text{c}}$, $\frac{1}{2}\text{inch}$ for $t$, $1.125\text{inch}$ for $d,$ and $58\text{ksi}$ for $F_{\text{u}}$ in Equation (VIII).

$\begin{array}{c}{R}_{{\text{n}}_2}=1.2\left(1.812\text{inch}\right)\left(\frac{1}{2}\text{inch}\right)\left(58\text{ksi}\right)\\ =63.06\text{kips}\end{array}$

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 $\frac{1}{2}\text{inch}$ for $t$.

$\begin{array}{c}{R}_{{\text{n}}_2}=2.4\left(1.125\text{inch}\right)\left(\frac{1}{2}\text{inch}\right)\left(58\text{ksi}\right)\\ =78.3\text{kips}\end{array}$

Thus, the minimum value of $R_{{\text{n}}_2}$ is considered and is equal to $68.06\text{kips}$.

Write the expression to calculate the nominal strength in yielding.

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

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

Substitute $tw$ for $A_{\text{g}}$ in Equation (IX).

$P_{\text{n}}=F_{\text{y}}tw$ ..... (X)

Here, the width of the section is $w$ and thickness of the section is $t$.

Substitute $0.5\text{inch}$ for $t$, $6.5\text{inch}$ for $w$ and $36\text{ksi}$ for $F_{\text{y}}$ in Equation (X).

$\begin{array}{c}{P}_{\text{n}}=\left[36\text{ksi}\left(0.5\text{inch}\times 6.5\text{inch}\right)\right]\\ =117\text{kips}\end{array}$

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.

$A_{\text{e}}=A_{\text{g}}-2\left(A_{\text{holes}}\right)$ ..... (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).

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

Substitute $\frac{1}{2}\text{inch}$ for $t$, $1.25\text{inch}$ for $d_{\text{h}},$ and $3.25{\text{inch}}^2$ for $A_{\text{g}}$ in Equation (XIII).

$\begin{array}{c}{A}_{\text{n}}=\left(3.25{\text{inch}}^2-\left(2\times \frac{1}{2}\text{inch}\times 1.25\text{inch}\right)\right)\\ =3.25{\text{inch}}^2-1.25{\text{inch}}^2\\ =2{\text{inch}}^2\end{array}$

Write the expression for nominal strength for fracture.

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

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

Substitute $2.0{\text{inch}}^2$ for $A_{\text{e}}$ and $58\text{ksi}$ for $F_{\text{u}}$ in Equation (XIV).

$\begin{array}{c}{P}_{\text{n}}=58\text{ksi}\times 2.0{\text{inch}}^2\\ =116\text{kips}\end{array}$

Write the expression for gross shear area.

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

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

Substitute $5\text{inch}$ for $s_{\text{s}}$ and $\frac{1}{2}\text{inch}$ for $t$ in Equation (XV).

$\begin{array}{c}{A}_{\text{gv}}=2\left(\frac{1}{2}\text{inch}\times 5\text{inch}\right)\\ =5.0{\text{inch}}^2\end{array}$

Write the expression for the net area in shear.

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

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

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

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

Substitute $5{\text{inch}}^2$ for $A_{\text{gv}}$, $1.25\text{inch}$ for $d_{\text{h}},$ and $\frac{1}{2}\text{inch}$ for $t$ in Equation (XVII).

$\begin{array}{c}{A}_{\text{nv}}=\left(5{\text{inch}}^2-2\left(1.5\times \frac{1}{2}\text{inch}\times 1.25\text{inch}\right)\right)\\ =5{\text{inch}}^2-1.875{\text{inch}}^2\\ =3.125{\text{inch}}^2\end{array}$

Write the expression for net area in tension.

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

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

Substitute $3\text{inch}$ for $t_{\text{L}}$, $\frac{1}{2}\text{inch}$ for $t$ and $0.625{\text{inch}}^2$ for $A_{\text{holes}}$ in Equation (XVIII).

$\begin{array}{c}{A}_{\text{nt}}=\left(\frac{1}{2}\text{inch}\times 3\text{inch}\right)-0.625{\text{inch}}^2\\ =1.5{\text{inch}}^2-0.625{\text{inch}}^2\\ =0.875{\text{inch}}^2\end{array}$

Write the expression for the nominal strength in block shear.

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

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

Substitute $3.125{\text{inch}}^2$ for $A_{\text{nv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.875{\text{inch}}^2$ for $A_{\text{nt}},$ and $1$ for $U_{\text{bs}}$ in Equation (XIX).

$\begin{array}{c}{R}_{\text{nb}}=\left[\left(0.6\times 58\text{ksi}\times 3.125{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.875{\text{inch}}^2\right)\right]\\ =108.75\text{kips}+50.75\text{kips}\\ =159.5\text{kips}\end{array}$

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 $5{\text{inch}}^2$ for $A_{\text{gv}}$, $58\text{ksi}$ for $F_{\text{u}}$, $0.875{\text{inch}}^2$ for $A_{\text{nt}}$, $36\text{ksi}$ for $F_{\text{y}}$ and $1$ for $U_{\text{bs}}$ in Equation (XX).

$\begin{array}{c}{R}_{\text{ng}}=\left[\left(0.6\times 36\text{ksi}\times 5.0{\text{inch}}^2\right)+\left(1\times 58\text{ksi}\times 0.875{\text{inch}}^2\right)\right]\\ =108\text{kips}+50.8\text{kips}\\ =158.8\text{kips}\end{array}$

Thus, the nominal shear strength is taken as $158.8\text{kips}$, 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 $53.68\text{kips}$ for $R_{\text{n}}$ and $0.75$ for $\varphi$ in Equation (XXI).

$\begin{array}{c}B=0.75\times 53.68\text{kips}\\ =40.26\text{kips}\end{array}$

Calculate the design strength for slip per bolt.

Substitute $18.98\text{kips}$ for $R_{\text{n}}$ and $1.0$ for $\varphi$ in Equation (XXI).

$\begin{array}{c}B=1.0\times 18.98\text{kips}\\ =18.98\text{kips}\end{array}$

Write the expression for total bolt strength.

$B_{\text{t}}=4B$ ..... (XXII)

Here, total strength is $B_{\text{t}}$.

Substitute $18.98\text{kips}$ for $B$ in Equation (XXII).

$\begin{array}{c}{B}_{\text{t}}=4\times 18.98\text{kips}\\ =75.92\text{kips}\end{array}$

Write the expression for design strength in yielding.

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

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

Substitute $117\text{kips}$ for $P_{\text{n}}$ and $0.9$ for ${\varphi }_{\text{t}}$ in Equation (XXIII).

$\begin{array}{c}{B}_{\text{y}}=0.9\times 117\text{kips}\\ =105.3\text{kips}\end{array}$

Write the expression for design strength for fracture.

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

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

Substitute $116\text{kips}$ for $P_{\text{n}}$ and $0.75$ for $\varphi$ in Equation (XXIV).

$\begin{array}{c}{B}_{\text{f}}=0.75\times 116\text{kips}\\ =87\text{kips}\end{array}$

Write the expression for design strength for block shear.

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

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

Substitute $158.8\text{kips}$ for $P_{\text{n}}$ and $0.75$ for $\varphi$ in Equation (XXV).

$\begin{array}{c}{B}_{\text{b}}=0.75\times 158.8\text{kips}\\ =119.1\text{kips}\end{array}$

Thus, the bolt strength is $75.92\text{kips}$ as it is the least value.

Write the expression for total factored load.

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

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

Substitute $3D$ for $L$ and $75.92\text{kips}$ for $P_{\text{u}}$ in Equation (XXVI).

$\begin{array}{c}75.92\text{kips}=1.2D+1.6\left(3D\right)\\ D=12.65\text{kips}\end{array}$

Thus,

$\begin{array}{c}L=3D\\ =3\times 12.65\text{kips}\\ =37.95\text{kips}\end{array}$

Calculate the total load.

$\begin{array}{c}D+L=12.65\text{kips}+37.95\text{kips}\\ =50.6\text{kips}\end{array}$

Conclusion:

Thus, the maximum service load is $50.6\text{kips}$.

To determine

(b)

The maximum service load using ASD.

Answer to Problem 7.6.5P

The maximum service load is $50.6\text{kips}$.

Explanation of Solution

Calculation:

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

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

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

Substitute $53.68\text{kips}$ for $R_{\text{n}}$ and $2.0$ for $\Omega$ in Equation (XXVII).

$\begin{array}{c}{B}_{\text{A}}=\frac{53.68\text{kips}}{2.0}\\ =26.84\text{kips}\end{array}$ ..... (XXVII)

Calculate the allowable strength for slip per bolt.

Substitute $18.98\text{kips}$ for $R_{\text{n}}$ and $1.5$ for $\Omega$ in Equation (XXVII).

$\begin{array}{c}{B}_{\text{A}}=\frac{18.98\text{kips}}{1.5}\\ =12.65\text{kips}\end{array}$

Calculate the allowable strength for bearing per bolt.

Substitute $48.93\text{kips}$ for $R_{\text{n}}$ and $2.0$ for $\Omega$ in Equation (XXVII).

$\begin{array}{c}{B}_{\text{A}}=\frac{48.93\text{kips}}{2.0}\\ =24.46\text{kips}\end{array}$

Thus, the minimum value of allowable strength for shear is considered and is equal to $12.65\text{kips}$.

Write the expression for total bolt strength.

$B_{\text{T}}={\displaystyle \sum \text{Strength}\text{of}\text{each}\text{bolt}}$ ..... (XXVIII)

Substitute $12.65\text{kips}$ for $\text{Strength}\text{of}\text{each}\text{bolt}$ in Equation (XXVIII).

$\begin{array}{c}{B}_{\text{T}}=4\times 12.65\\ =50.6\text{kips}\end{array}$

Write the expression for allowable strength in yielding.

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

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

Substitute $117\text{kips}$ for $P_{\text{n}}$ and $1.67$ for ${\Omega }_{\text{t}}$ in Equation (XXIX).

$\begin{array}{c}{B}_{\text{Y}}=\frac{117\text{kips}}{1.67}\\ =70.06\text{kips}\end{array}$

Calculate the allowable strength in fracture.

Substitute $116\text{kips}$ for $P_{\text{n}}$ and $2.0$ for ${\Omega }_{\text{t}}$ in Equation (XXIX).

$\begin{array}{c}{B}_{\text{Y}}=\frac{116\text{kips}}{2.0}\\ =58\text{kips}\end{array}$

Calculate the allowable strength for block shear.

Substitute $158.8\text{kips}$ for $P_{\text{n}}$ and $2.0$ for ${\Omega }_{\text{t}}$ in Equation (XXIX).

$\begin{array}{c}{B}_{\text{Y}}=\frac{158.8\text{kips}}{2.0}\\ =79.4\text{kips}\end{array}$

Calculate the total bolt strength in yielding.

Substitute $58\text{kips}$ for $\text{Strength}\text{of}\text{each}\text{bolt}$ in Equation (XXVIII).

$\begin{array}{c}{B}_{\text{T}}=4\times 58\\ =232\text{kips}\end{array}$

Thus, the design strength is $50.6\text{kips}$ as it is the least value.

Write the expression for total factored load.

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

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

Substitute $50.6\text{kips}$ for $P_{\text{a}}$ in Equation (XXX).

$D+L=50.6\text{kips}$

Conclusion:

Thus, the maximum service load is $50.6\text{kips}$.