Chapter 5, Problem 5.5.16P

To determine

(a)

The adequacy of the W 6 X12 beam for use as purlin using LRFD method.

Answer to Problem 5.5.16P

The beam is adequate to be used for purlin.

Explanation of Solution

Given:

A truss with a roof system supporting a total gravity load of 40 psf of roof surface, half dead load and half snow. Spacing = 10 ft on centers.

Calculation:

Let’s calculate the nominal flexural strength about the X and Y axes.

Determine the strong axis bending strength. As neither the beam design charts nor the Z tables include shapes under W8 compute the flexural strength of W 6 X12. As there is no foot note to indicate otherwise the shape is compact.

We need to determine what controls the lateral torsional buckling.

Computing the values of $L_p$ and $L_r$ for W 16 X 26, we have the results as follows:

$L_p=1.76\times r_y\times \sqrt{\frac{E}{F_y}}$

Substitute the values from the AISC Manual as

$r_y=0.918\text{in}\text{.}$

$E=29000\text{Ksi}\text{.}$

$F_y=50\text{Ksi}\text{.}$

$\begin{array}{l}{L}_p=1.76\times 0.918\text{in}\times \sqrt{\frac{29000Ksi}{50Ksi}}\\ L_p=38.912\text{in}\text{.}\\ L_p=3.243\text{ft}\text{.}\end{array}$

$L_r=1.95\times r_{ts}\times \frac{E}{\left(0.7\times F_y\right)}\times \sqrt{\frac{J\times c}{S_x\times h_o}+\sqrt{{\left(\frac{J\times c}{S_x\times h_o}\right)}^2+6.67\times {\left(\frac{0.7\times F_y}{E}\right)}^2}}$

Substitute the values from the AISC Manual as

$J=0.0903{\text{in}}^4.$

$S_x=7.31{\text{in}}^3.$

$h_o=5.750\text{in }\text{.}$

$r_{ts}=1.08\text{in}\text{.}$

$E=29000\text{Ksi}\text{.}$

$F_y=50\text{Ksi}\text{.}$

$\begin{array}{l}{L}_r=1.95\times 1.08\text{in}\times \frac{29000\text{ksi}}{\left(0.7\times 50\right)\text{ksi}}\times \sqrt{\frac{0.0903{\text{in}}^4\times 1.00}{7.31{\text{in}}^3\times 5.75\text{in}}+\sqrt{{\left(\frac{0.0903{\text{in}}^4\times 1.00}{7.31{\text{in}}^3\times 5.75\text{in}}\right)}^2+6.67\times {\left(\frac{0.7\times 50\text{ksi}}{29000\text{ksi}}\right)}^2}}\\ L_r=1.95\times 1.08\text{in}\times \frac{29000\text{ksi}}{\left(0.7\times 50\right)\text{ksi}}\times \sqrt{\left(2.15\times {10}^{-3}\right)+\sqrt{\left(4.62\times {10}^{-6}\right)+6.67\times \left(1.456\times {10}^{-6}\right)}}\\ L_r=2.106\text{in}\times \text{828}\text{.57}\times \text{0}\text{.0770}\\ L_r=134.438\text{in}\\ L_r=11.2\text{ft}\text{.}\end{array}$

Now calculate the plastic moment for the section, we have

$M_p=F_y\times Z_X$

Where, $Z_X$ is the plastic section modulus along the major principal axis and $F_y$ is the yield strength.

Substitute the values from the ASIC manual, we have

$\begin{array}{l}{M}_p=50\text{Ksi}\times 8.3{\text{in}}^3.\\ M_p=415\text{Ksi}{\text{in}}^3.\\ M_p=34.58\text{Ksi}{\text{ft}}^3.\\ M_p=34.58\text{Kips}\text{ft}.\end{array}$

Let’s compare the values $L_p$, $L_b$ and $L_r$, we have

$L_p=3.243\text{ft}$, $L_r=11.2\text{ft}$ and $L_b=10.00\text{ft}\text{, }{L}_b\text{is the unbraced length of the beam}$

$\therefore L_r>L_b>L_p$

Which implies that the strength is governed by inelastic Lateral- Torsional Buckling.

Compute the nominal strength of beam using the equation given as follows:

$M_n=C_b\times \left[M_p-\left[M_p-\left[0.7\times F_y\times S_x\right]\times \left[\frac{L_b-L_p}{L_r-L_p}\right]\right]\right]\le M_p$

Substitute the values from the ASIC manual, we have

$\begin{array}{l}{M}_n=1.0\times \left[415Kips.in-\left[415Kips.in-\left[0.7\times 50Ksi\times 7.31i{n}^3\right]\times \left[\frac{\left(10-3.243\right)in}{\left(11.22-3.243\right)in}\right]\right]\right]\le 415Kips.in\\ M_n=319.4Kips.in<415.00Kips.in\end{array}$

For the Y − axis, there is no flange buckling since the shape is compact.

Calculate the flexural strength about y- axis as:

$M_{ny}=M_{py}$

Where, $M_{py}$ is the plastic moment about Y -axis and $M_{ny}$ is the nominal strength of beam about y- axis.

Now, calculating the plastic moment of section about minor principal axis as:

$M_{py}=F_y{Z}_y$

Substitute the values, we have

$\begin{array}{l}{M}_{py}=50\text{Ksi}\times 2.32{\text{in}}^{\text{3}}\\ M_{py}=116.00\text{Ksi}{\text{in}}^{\text{3}}.\\ M_{py}=9.667\text{Ksi}{\text{ft}}^{\text{3}}.\\ M_{py}=9.667\text{Kips}\text{ft}.\end{array}$

Calculate the flexural strength about y axis, we have

$M_{ny}=M_{py}$

$\therefore M_{ny}=M_{py}=9.667\text{Kips}\text{ft}.$

Checking the upper limit using the following :

$\frac{Z_y}{S_y}\le 1.6$

Substitute the values, we have

$Z_y=2.32i{n}^3$

$S_y=1.50i{n}^3$

$\begin{array}{l}\frac{Z_y}{S_y}=\frac{2.32i{n}^3}{1.50i{n}^3}\le 1.6\\ \frac{Z_y}{S_y}=1.546\le 1.6\\ 1.55\le 1.6\end{array}$

As the inequality is satisfied then the its OK.

Now using the LRFD method.

Following equation must be satisfied in order to know adhere to AISC specifications.

$\left(\frac{M_{ux}}{{\varphi }_b{M}_{nx}}+\frac{M_{uy}}{{\varphi }_b{M}_{ny}}\right)\le 1.0$

Where, $M_{ux}$ is the flexural load about x-axis, $M_{uy}$ is the flexural load about y-axis, ${\varphi }_b{M}_{nx}$ is the nominal flexural strength about x-axis and ${\varphi }_b{M}_{ny}$ is the nominal flexural strength about Y-axis.

Now we need to find the values to substitute them

$M_{ux}=\frac{w_{ux}{L}^2}{8}$

Where, $w_{ux}$ is the uniformly distributed load and $L$ is the length of the beam

$w_{ux}=w_p\mathrm{Sin}\theta$

Where, $\theta$ is the angle subtended by the load with the horizontal.

$w_p=w_u{b}_t$

$w_u=\left(1.2\times w_D\right)+\left(1.6\times w_S\right)$

Where $w_D$ is the uniformly distributed dead load on the roof and $w_S$ is the snow load on the roof.

As it is been given that half of load is dead load and half is snow load.

Therefore, as per the given conditions $w_D=\frac{40}{2}\text{psf}$ and $w_S=\frac{40}{2}\text{psf}$

$\begin{array}{l}{w}_u=\left(1.2\times \frac{40}{2}\text{psf}\right)+\left(1.6\times \frac{40}{2}\text{psf}\right)\\ w_u=\left(24\text{psf}\right)+\left(\text{32psf}\right)\\ w_u=56\text{psf}\text{.}\end{array}$

$b_t$ is the tributary length and is shown as, $b_t=H\mathrm{Sec}\theta$

Where, $H$ is the rise of truss and $\theta$ is the angle of purlin with vertical.

Following is the diagram from which we can find the value of angles.

Substitute the value for H = 6 ft and $\mathrm{Sec}\theta =\frac{\sqrt{17}}{4}$

$\begin{array}{l}{b}_t=6\text{ft}\times \frac{\sqrt{17}}{4}\\ b_t=6.185\text{ft}.\end{array}$

Substitute the values

$\begin{array}{l}{w}_p=56\text{psf}\times 6.185\text{ft}\\ w_p=346.341{\text{lb ft}}^{-1}.\end{array}$

$w_{ux}=w_p\mathrm{Sin}\theta$.

Substitute, $w_p=346.341{\text{lb ft}}^{-1}and\mathrm{Sin}\theta =\frac{4}{\sqrt{17}}$

$\begin{array}{l}{w}_{ux}=346.341{\text{lb ft}}^{-1}\times \frac{4}{\sqrt{17}}\\ w_{ux}=336{\text{lb ft}}^{-1}.\end{array}$

Find the flexural load about x-axis, $M_{ux}=\frac{w_{ux}{L}^2}{8}$

$\begin{array}{l}{M}_{ux}=\frac{0.336{\text{Kips ft}}^{\text{-1}}\times {\left(10\text{ft}\right)}^2}{8}\\ M_{ux}=4.20\text{Kips ft}\text{.}\end{array}$

Similarly, for $M_{uy}$, we have

$M_{uy}=\frac{w_{uy}{L}^2}{8}$

Where, $w_{uy}$ is the uniformly distributed load and $L$ is the length of the beam

$w_{uy}=w_p\mathrm{Cos}\theta$

Substitute, $w_p=346.341{\text{lb ft}}^{-1}\text{and}\text{Cos}\theta =\frac{1}{\sqrt{17}}$

$\begin{array}{l}{w}_{uy}=346.341{\text{lb ft}}^{-1}\times \frac{1}{\sqrt{17}}\\ w_{uy}=84.0{\text{lb ft}}^{-1}\end{array}$

Find the flexural load about x-axis, $M_{uy}=\frac{w_{uy}{L}^2}{8}$

$\begin{array}{l}{M}_{uy}=\frac{0.0840{\text{Kips ft}}^{\text{-1}}\times {\left(10\text{ft}\right)}^2}{8}\\ M_{uy}=1.05\text{Kips ft}\text{.}\end{array}$

Now, find the values of

$\begin{array}{l}{\varphi }_b{M}_{nx}=0.9\times 26.62\text{kips}\text{.ft}\\ {\varphi }_b{M}_{nx}=23.958\text{kips}\text{.ft}\\ {\varphi }_b{M}_{ny}=0.9\times 9.667\text{kips}\text{.ft}\\ {\varphi }_b{M}_{ny}=8.70\text{kips}\text{.ft}\end{array}$

As we have found every value, now we can substitute the values and check the adequacy

$\left(\frac{M_{ux}}{{\varphi }_b{M}_{nx}}+\frac{M_{uy}}{{\varphi }_b{M}_{ny}}\right)\le 1.0$

$\begin{array}{l}\left(\frac{4.20\text{Kips ft}}{23.958\text{kips}\text{.ft}}+\frac{1.05\text{Kips ft}}{8.70\text{kips}\text{.ft}}\right)\le 1.0\\ \left(0.1753+0.121\right)\le 1.0\\ 0.2959\le 1.0\end{array}$

The equation is hence satisfied.

Conclusion:

Therefore, the beam is adequate.

To determine

(b)

The adequacy of W 6 X12 beam for use as purlin using ASD method.

Answer to Problem 5.5.16P

The beam is adequate to be used for purlin.

Explanation of Solution

Given:

A truss with a roof system supporting a total gravity load of 40 psf of roof surface, half dead load and half snow. Spacing = 10 ft on centers.

Calculation:

Now from Allowable stress design

$\left(\frac{M_{ax}}{\frac{M_{nx}}{{\Omega }_c}}+\frac{M_{ay}}{\frac{M_{ny}}{{\Omega }_c}}\right)\le 1.0$

Where $\frac{M_{nx}}{{\Omega }_c}$ is the nominal flexural strength about X-axis, $\frac{M_{ny}}{{\Omega }_c}$ is the nominal flexural strength about Y-axis, $M_{ax}$ is the flexural load about X-axis and $M_{ay}$ is the flexural load about Y-axis.

Now find the value of $M_{ax}=\frac{w_{ax}{L}^2}{8}$

Where, $w_{ax}$ is the uniformly distributed load and $L$ is the length of the beam

$w_{ax}=w_p\mathrm{Sin}\theta$

Where, $\theta$ is the angle subtended by the load with the vertical.

$w_a=w_D+w_S$

Where $w_D$ is the uniformly distributed dead load on the roof and $w_S$ is the snow load on the roof.

As it is been given that half of load is dead load and half is snow load.

Therefore, as per the given conditions $w_D=\frac{40}{2}\text{psf}$ and $w_S=\frac{40}{2}\text{psf}$

$\begin{array}{l}{w}_a=w_D+w_S\\ w_a=\left(\frac{40}{2}\text{psf}\right)+\left(\frac{40}{2}\text{psf}\right)\\ w_a=\left(20\text{psf}\right)+\left(\text{20}\text{psf}\right)\\ w_a=40\text{psf}\text{.}\end{array}$

Calculate the load on the purlin as follows:

$\begin{array}{l}{w}_p=w_a{b}_t\\ w_p=40\text{psf}\times 6.185\text{ft}\\ w_p=247.4\text{lb}\text{.}{\text{ft}}^{\text{-1}}\end{array}$

Following is the diagram from which we can find the value of angles.

$b_t$ is the tributary length and is shown as, $b_t=H\mathrm{Sec}\theta$

Where, $H$ is the rise of truss and $\theta$ is the angle of purlin with vertical.

Substitute the value for H = 6 ft and $\mathrm{Sec}\theta =\frac{\sqrt{17}}{4}$

$\begin{array}{l}{b}_t=6\text{ft}\times \frac{\sqrt{17}}{4}\\ b_t=6.185\text{ft}.\end{array}$

Load on the purlin is as follows:

$\begin{array}{l}{w}_{ax}=w_p\mathrm{Sin}\theta \\ w_{ax}=247.40\text{lb}{\text{.ft}}^{\text{-1}}\times \frac{4}{\sqrt{17}}\\ w_{ax}=240.01\text{lb}{\text{.ft}}^{\text{-1}}\end{array}$

$M_{ax}=\frac{w_{ax}{L}^2}{8}$

Substitute the values, we have

$\begin{array}{l}{M}_{ax}=\frac{0.240\text{kips}{\text{.ft}}^{\text{-1}}\times {\left(10\text{ft}\right)}^2}{8}\\ M_{ax}=3.00\text{kips}\text{.ft}\end{array}$

Now, $M_{ay}=\frac{w_{ay}{L}^2}{8}$

Where, L is the length of the beam and $w_{ay}$ is the uniformly distributed load about Y- axis.

$w_{ay}=w_p\mathrm{Cos}\theta$

Where, $\theta$ is the angle subtended by the load with the vertical.

$\begin{array}{l}{w}_{ay}=w_p\mathrm{Cos}\theta \\ w_{ay}=247.40\text{lb}{\text{.ft}}^{\text{-1}}\times \frac{1}{\sqrt{17}}\\ w_{ay}=60.01\text{lb}{\text{.ft}}^{\text{-1}}\end{array}$

$M_{ay}=\frac{w_{ay}{L}^2}{8}$

Substitute the values, we have

$\begin{array}{l}{M}_{ay}=\frac{0.06\text{kips}{\text{.ft}}^{\text{-1}}\times {\left(10\text{ft}\right)}^2}{8}\\ M_{ay}=0.750\text{kips}\text{.ft}\end{array}$

Now find the value of

$\begin{array}{l}\frac{M_{nx}}{{\Omega }_c}=\frac{26.62\text{Kips}\text{.ft}}{1.67}\\ \frac{M_{nx}}{{\Omega }_c}=15.940\text{Kips}\text{.ft}\\ \text{And}\frac{M_{ny}}{{\Omega }_c}=\frac{9.667\text{Kips}\text{.ft}}{1.67}\\ \frac{M_{ny}}{{\Omega }_c}=5.789\text{Kips}\text{.ft}\end{array}$

As we have found every value, now we can substitute the values and check the adequacy

$\left(\frac{M_{ax}}{\frac{M_{nx}}{{\Omega }_c}}+\frac{M_{ay}}{\frac{M_{ny}}{{\Omega }_c}}\right)\le 1.0$

$\begin{array}{l}\left(\frac{3.00\text{kips}\text{.ft}}{15.94\text{kips}\text{.ft}}+\frac{0.750\text{kips}\text{.ft}}{5.789\text{kips}\text{.ft}}\right)\le 1.0\\ \left(0.1882+0.1295\right)\le 1.0\\ 0.3176\le 1.0\end{array}$

Hence, the equation is satisfied.

Conclusion:

Therefore, the beam is adequate.