Question

A C9 x 20 tension member is connected with 9/8 inch-diameter bolts as shown in figure FY

= 50 Ksi and Fu = 70 Ksi. The member is subjected to the following service deal load = 36

Kips and live load = 110 kips. Analyze the member for adequacy with respect to applied 

load using ASD and LRFD.

Answer 1

Calculate the nominal strength for the net section.

$$ P_{n}=F_{y} A_{g} $$

Here, \(F_{u}\) is the ultimate strength.

Substitute \(5.87\) for \(A_{g}, 50\) for \(F_{y}\) :

$$ \begin{aligned} P_{n} &=F_{y} A_{g} \\ &=50(5.87) \\ &=293.5 \mathrm{kips} \end{aligned} $$

Calculate the area of net section using the formula.

$$ \begin{aligned} A_{n} &=A_{g}-\text { area of holes } \\ &=A_{g}-\sum t_{w} d_{h} \end{aligned} $$

Calculate the diameter of hole.

\(d_{h}=\) diameter of bolt \(+\) side clearence

$$ \begin{aligned} &=1 \frac{1}{8}+\frac{1}{8} \\ &=1.25 \mathrm{in} \end{aligned} $$

Substitute \(0.448\) for \(t_{w}, 1.3125\) in for \(d_{h}\), and \(5.87\) for \(A_{g}\).

$$ \begin{aligned} A_{n} &=A_{g}-\text { area of holes } \\ &=A_{g}-\sum t_{w} d_{h} \\ &=5.87-(0.448)(1.25) \\ &=5.31 \mathrm{in}^{2} \end{aligned} $$

Calculate the net area using another formula.

$$ \begin{aligned} A_{n} &=A_{g}-\sum\left(d-\frac{s^{2}}{4 g}\right) \\ &=5.87-(0.448)(1.25)-(0.448)\left(1.25-\frac{1.5^{2}}{4(4)}\right) \\ &=4.813 \mathrm{in}^{2} \end{aligned} $$

Calculate the reduction factor using the formula.

$$ U=1-\frac{\bar{x}}{l} $$

Here, \(\bar{x}\) is the distance from centroid of connected area to plane of connection, \(/\) is the length of connection.

Substitute \(0.583\) for \(\bar{x}, 6(1.5)\) for \(/\) :

$$ \begin{aligned} U &=1-\frac{0.583}{6(1.5)} \\ &=0.9352 \end{aligned} $$

Calculate the effective net area of section.

$$ A_{e}=A_{g} U $$

Substitute \(4.757\) for \(A_{g}, 0.9352\) for \(U\) :

$$ \begin{aligned} A_{e} &=A_{g} U \\ &=4.757(0.9352) \\ &=4.45 \mathrm{in}^{2} \end{aligned} $$

Calculate the nominal strength in yielding.

$$ P_{n}=F_{u} A_{e} $$

Here, \(F_{u}\) is the ultimate strength.

Substitute \(4.5\) for \(A_{g}, 70\) for \(F_{y}\) :

$$ \begin{aligned} P_{n} &=F_{u} A_{e} \\ &=70(4.5) \\ &=315 \mathrm{kips} \end{aligned} $$

(a)

Calculate the design strength based on yielding.

design strength \(=\phi_{t} P_{n}\)

Here, \(\phi_{t}\) is the resistance factor.

Substitute \(293.5\) for \(P_{n}, 0.90\) for \(\phi_{t}\) :

design strength \(=\phi_{t} P_{n}\)

$$ \begin{aligned} &=0.90(293.5) \\ &=264.2 \mathrm{kips} \end{aligned} $$

Calculate the design strength based on fracture.

design strength \(=\phi_{t} P_{n}\)

Substitute 315 for \(P_{n}, 0.75\) for \(\phi_{t}\) :

design strength \(=\phi_{t} P_{n}\)

$$ \begin{aligned} &=0.75(315) \\ &=236.25 \mathrm{kips} \end{aligned} $$

Calculate the factored load.

$$ \begin{aligned} P_{u} &=1.2 D+1.6 L \\ &=1.2(36)+1.6(110) \\ &=219.2 \mathrm{kips} \end{aligned} $$

Since, 219 kips \(<236.25\)kips.

Therefore, the member has enough strength.

(b)

Calculate the allowable strength based on yielding.

allowable strength \(=\frac{P_{n}}{\Omega}\)

Substitute \(1.67\) for \(\Omega, 293.5\) for \(P_{n}\) :

allowable strength \(=\frac{P_{n}}{\Omega}\)

$$ \begin{aligned} &=\frac{293.5}{1.67} \\ &=176 \mathrm{kips} \end{aligned} $$

Calculate the allowable strength based on fracture.

allowable strength \(=\frac{P_{n}}{\Omega}\)

Substitute 2 for \(\Omega, 315\) for \(P_{n}\) :

allowable strength \(=\frac{P_{n}}{\Omega}\)

$$ \begin{aligned} &=\frac{315}{2} \\ &=157.5 \mathrm{kips} \end{aligned} $$

Calculate the factored load.

$$ \begin{aligned} P_{a} &=D+L \\ &=36+110 \\ &=146 \mathrm{kips} \end{aligned} $$

Since, 146 kips \(<157.5\)kips.

Therefore, the member has enough strength.