Chapter 3, Problem 3.2.1P

To determine

(a)

The design strength using Load and Resistance Factor Design(LRFD).

Answer to Problem 3.2.1P

$85.05\text{kips}$.

Explanation of Solution

Given:

A36 steel PL ${\frac{3}{8}}^{\prime \text{}\prime }\times 7^{″}$ tension member with three $1-\text{inch}$ diameter bolt.

The effective area, $A_{\text{e}}=A_{\text{n}}$.

Concept Used:

Write the expression for the factored strength in yielding.

$P={\Phi }_{\text{y}}\times P_{\text{ny}}$ ...... (I)

Write the expression for the factored strength in rupture.

$P={\Phi }_{\text{f}}\times P_{\text{nf}}$ ...... (II)

Herethe strength of the material is $P$, the resistance factor for yielding is ${\Phi }_{\text{y}}$, the resistance factor for rupture is ${\Phi }_{\text{f}}$, the nominal strength in yielding is $P_{\text{ny}},$ and the nominal strength in rupture is $P_{\text{nr}}$

The design strength of LRFD is the minimum of Equation (I) and (II).

Write the expression for the nominal strength in yielding.

$P_{\text{ny}}=F_{\text{y}}\times A_{\text{g}}$ ...... (III)

Here, the yield strength in yielding is $F_{\text{y}}$, the gross area of the member is $A_{\text{g}}$.

Write the expression for the gross area of the member.

$A_{\text{g}}=L\times t$ ...... (IV)

Here, the length of the member is $L$, the thickness of the member is $t$.

Write the expression for the nominal strength in rupture.

$P_{\text{nf}}=F_{\text{u}}\times A_{\text{e}}$ ...... (V)

Here, the yield strength in rupture is $F_{\text{u}}$, the effective area is $A_{\text{e}}$.

Given that, $A_{\text{e}}=A_{\text{n}}$, hence,

$\begin{array}{c}{A}_{\text{e}}=A_{\text{n}}\\ =A_{\text{g}}-A_{\text{holes}}\end{array}$ ...... (VI)

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

Write the expression for the area of the holes.

$A_{\text{holes}}=t\times d_{\text{holes}}$ ...... (VII)

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

Write the expression for the diameter of the holes.

$d_{\text{holes}}=d+\frac{1}{8}\text{in}$ ...... (VIII)

Here, the diameter of the bolts is $d$ and $\frac{1}{8}\text{in}$ is the allowance for side clearance for bolts less than $1\text{in}$.

For the A36 steel resistance factors and the yield strength values are,

$\begin{array}{l}{\Phi }_{\text{y}}=0.90\\ {\Phi }_{\text{f}}=0.75\\ F_{\text{y}}=36\text{ksi}\\ F_{\text{u}}=58\text{ksi}\end{array}$

Calculation:

Calculate the gross area.

Substitute $7\text{in}$ for $L$ and $\frac{3}{8}\text{in}$ for $t$ in Equation (IV).

$\begin{array}{c}{A}_{\text{g}}=7\text{in}\times \frac{3}{8}\text{in}\\ =2.625{\text{in}}^2\end{array}$

Calculate the diameter of holes.

Substitute $1\text{in}$ for $d$ in Equation (VIII).

$\begin{array}{c}{d}_{\text{holes}}=1\text{in}+\frac{1}{8}\text{in}\\ =\frac{9}{8}\text{in}\end{array}$

Calculate the area of holes.

Substitute $\frac{3}{8}\text{in}$ for $t$ and $\frac{9}{8}\text{in}$ for $d_{\text{holes}}$ in Equation (VII).

$\begin{array}{c}{A}_{\text{holes}}=\frac{3}{8}\text{in}\times \frac{9}{8}\text{in}\\ =0.422{\text{in}}^2\end{array}$

Calculate the effective area.

Substitute $2.625{\text{in}}^2$ for $A_{\text{g}}$ and $0.422{\text{in}}^2$ for $A_{\text{holes}}$ in Equation (VI).

$\begin{array}{c}{A}_{\text{e}}=2.625{\text{in}}^2-0.422{\text{in}}^2\\ =2.203{\text{in}}^2\end{array}$

Calculate the yielding strength.

Substitute $36\text{ksi}$ for $F_{\text{y}}$, $2.625{\text{in}}^2$ for $A_{\text{g}}$ in the Equation (III).

$\begin{array}{c}{P}_{\text{ny}}=36\times 2.625\\ =94.5\text{kips}\end{array}$

Calculate the strength at rupture.

Substitute $58\text{ksi}$ for $F_{\text{u}}$, $2.203{\text{in}}^2$ for $A_{\text{e}}$ in the Equation (IV).

$\begin{array}{c}{P}_{\text{nf}}=58\times 2.203\\ =127.77\text{kips}\end{array}$

Calculate the design strength in yielding.

Substitute $94.5\text{kips}$ for $P_{\text{ny}}$ and $0.9$ for ${\Phi }_{\text{y}}$ in the Equation (I).

$\begin{array}{c}P=0.9\times 94.5\text{kips}\\ =85.05\text{kips}\end{array}$

Calculate the design strength in rupture.

Substitute $127.77\text{kips}$ for $P_{\text{nf}}$ and $0.75$ for ${\Phi }_{\text{f}}$ in the Equation (II).

$\begin{array}{c}P=0.75\times 127.77\text{kips}\\ =95.83\text{kips}\end{array}$

Conclusion:

Compare the design strengthin yielding and rupture. The design strength using LRFD is the smaller of the two.

Thus, the design strength using LRFD is $85.05\text{kips}$.

To determine

(b)

The allowable strength usingAllowable Strength Design (ASD).

Answer to Problem 3.2.1P

The allowable strength using ASD is $56.59\text{kips}$.

Explanation of Solution

Write the expression for the allowable strength inyielding.

$P=\frac{P_{\text{ny}}}{{\Omega }_{\text{y}}}$ ...... (IX)

Here, the safety factor in yielding is ${\Omega }_{\text{y}}$.

Write the expression for the allowable strength for rupture.

$P=\frac{P_{\text{nf}}}{{\Omega }_{\text{f}}}$ ...... (X)

Here, the safety factor in rupture is ${\Omega }_{\text{f}}$.

The allowable strength is the minimum of Equation (IX) and (X).

Calculation:

Calculate the allowable strength in yielding.

Substitute $94.5\text{kips}$ for $P_{\text{ny}}$ and $1.67$

${\Omega }_{\text{y}}$ for in Equation(IX).

$\begin{array}{c}P=\frac{94.5}{1.67}\text{kips}\\ =56.59\text{kips}\end{array}$

Calculate the allowable strength in rupture.

Substitute $127.77\text{kips}$ for $P_{\text{nf}}$ and $2$ for ${\Omega }_{\text{f}}$ in Equation (X).

$\begin{array}{c}P=\frac{127.77}{2}\text{kips}\\ =63.89\text{kips}\end{array}$

Conclusion:

Compare the allowable strength in yielding and rupture. The smaller values are the allowable strength using ASD.

Thus, the allowable strength using ASD is $56.59\text{kips}$.