Chapter 5, Problem 5.15.6P

To determine

(a)

Whether a $W6\times 12$ is adequate or not by using LRFD

Answer to Problem 5.15.6P

Adequate

Explanation of Solution

Given:

Total gravity load = 40 psf of roof surface

$W6\times 12$ of A992 steel

Formula used:

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

L_pis unbraced length in an inelastic behavior

$c=\frac{h_0}{2}\sqrt{\frac{I_y}{C_w}}$

$L_r=1.95r_{ts}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x{h}_0}+\sqrt{{\left(\frac{Jc}{S_x{h}_0}\right)}^2+6.76{\left(\frac{0.7F_y}{E}\right)}^2}}$

L_ris unbraced length in an elastic behavior

$M_n=C_b\left[M_p-\left(M_p-0.7F_y{S}_x\right)\left(\frac{L_b-L_p}{L_r-L_p}\right)\right]\le M_p$

M_n is nominal moment strength

M_pis plastic moment capacity

Calculation:

Determine the nominal flexural strength about x and y axes:

Neither the beam design charts nor the Z tables include shapes smaller than W8, so the flexural strength of the $W6\times 12$ must be computed.

From the dimensions and properties tables, the shape is compact.

The following properties of a $W6\times 12$ are below:

$\begin{array}{l}A=3.55in{.}^2\\ r_{ts}=1.08in.\\ h_0=5.75in.\\ S_x=7.31in{.}^3\\ Z_x=8.30in{.}^3\\ I_y=2.99in{.}^4\\ r_y=0.918in.\\ J=0.0903in{.}^4\\ S_y=1.50in{.}^3\end{array}$

A is Cross-sectional area

S_xis Elastic section modulus about X -axis

Z_xis Plastic section modulus about X -axis

I_yis Moment of inertia about Y -axis

r_yis Radius of gyration about Y -axis

S_yis Elastic section modulus about Y -axis

$r_{ts}$ is $\sqrt{\frac{\sqrt{I_y{C}_w}}{S_x}}$

C_wis Warping constant

h₀is Distance between centroid of flanges

J is Torsional moment of inertia

$\begin{array}{l}{L}_p=1.76r_y\sqrt{\frac{E}{F_y}}=1.76\left(0.918\right)\sqrt{\frac{29000}{50}}=38.91in.=3.243ft\\ L_r=1.95r_{ts}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x{h}_0}+\sqrt{{\left(\frac{Jc}{S_x{h}_0}\right)}^2+6.76{\left(\frac{0.7F_y}{E}\right)}^2}}\\ =1.95\left(1.08\right)\frac{29000}{0.7\left(50\right)}\sqrt{\frac{0.0903\times 1}{7.31\times 5.75}+\sqrt{{\left(\frac{0.0903\times 1}{7.31\times 5.75}\right)}^2+6.76{\left(\frac{0.7\left(50\right)}{29000}\right)}^2}}\\ =134.6in.=11.22ft.\end{array}$

For $L_b=14ft$, $L_p<L_b<L_r$, so

$M_n=C_b\left[M_p-\left(M_p-0.7F_y{S}_x\right)\left(\frac{L_b-L_p}{L_r-L_p}\right)\right]\le M_p$ (Inelastic Lateral torsional buckling)

From the below given figure in the textbook, $C_b=1.14$

$\begin{array}{l}{M}_p=F_y{Z}_x=50\left(8.30\right)=415in-kips=34.58ft-kips\\ M_n=1.14\left[415-\left(415-0.7\times 50\times 7.31\right)\left(\frac{10-3.23}{11.22-3.243}\right)\right]\\ =319.4in-kips=26.62ft-kips<M_p\end{array}$

For the y axis, since the shape is compact, there is no flange local buckling

$M_{ny}=M_{py}=F_y{Z}_y=50\left(2.32\right)=116in-kips=9.667ft-kips$

Check the upper limit:

$\begin{array}{l}\frac{Z_y}{S_y}=\frac{2.32}{1.50}=1.55<1.6\\ \therefore M_{ny}=M_{py}=9.667ft-kips\\ {\varphi }_b{M}_{nx}=0.90\left(26.62\right)=23.96ft-kips\\ {\varphi }_b{M}_{ny}=0.90\left(9.667\right)=8.7ft-kips\end{array}$

Roof load: Combination 3 controls

$w_u=1.2D+1.6S=1.2\times \frac{40}{2}+1.6\times \frac{40}{2}=56psf$

where,D is dead load and S is snow load

Tributary width = $\frac{\sqrt{17}}{4}\left(6\right)=6.185ft$

Purlin load = $56\left(6.185\right)=346.4lb/ft$

Component normal to roof = $w_{ux}=\frac{4}{\sqrt{17}}\left(346.4\right)=336.1lb/ft$

Component parallel to roof = $w_{uy}=\frac{1}{\sqrt{17}}\left(346.4\right)=84.01lb/ft$

Calculate factored bending moment about x axis and y axis

$\begin{array}{l}{M}_{ux}=\frac{w_{ux}{L}^2}{8}=\frac{0.3361{\left(10\right)}^2}{8}=4.201ft-kips\\ M_{uy}=\frac{w_{uy}{L}^2}{8}=\frac{0.08401{\left(10\right)}^2}{8}=1.05ft-kips\end{array}$

Use ½ of weak-axis bending strength in the interaction equation:

$\frac{M_{ux}}{{\varphi }_b{M}_{nx}}+\frac{M_{uy}}{{\varphi }_b{M}_{ny}}=\frac{4.201}{23.96}+\frac{1.05}{8.7/2}=0.417<1.0$ (OK)

Conclusion:

$W6\times 12$ is adequate.

To determine

(b)

Whether a $W6\times 12$ is adequate or not by using ASD

Answer to Problem 5.15.6P

Adequate

Explanation of Solution

Given:

Total gravity load = 40 psf of roof surface

$W6\times 12$ of A992 steel

Formula used:

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

L_pis unbraced length in an inelastic behavior

$c=\frac{h_0}{2}\sqrt{\frac{I_y}{C_w}}$

$L_r=1.95r_{ts}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x{h}_0}+\sqrt{{\left(\frac{Jc}{S_x{h}_0}\right)}^2+6.76{\left(\frac{0.7F_y}{E}\right)}^2}}$

L_ris unbraced length in an elastic behavior

$M_n=C_b\left[M_p-\left(M_p-0.7F_y{S}_x\right)\left(\frac{L_b-L_p}{L_r-L_p}\right)\right]\le M_p$

M_n is nominal moment strength

M_pis plastic moment capacity

Calculation:

Determine the nominal flexural strength about x and y axes:

Neither the beam design charts nor the Z tables include shapes smaller than W8, so the flexural strength of the $W6\times 12$ must be computed.

From the dimensions and properties tables, the shape is compact.

The following properties of a $W6\times 12$ are below:

$\begin{array}{l}A=3.55in{.}^2\\ r_{ts}=1.08in.\\ h_0=5.75in.\\ S_x=7.31in{.}^3\\ Z_x=8.30in{.}^3\\ I_y=2.99in{.}^4\\ r_y=0.918in.\\ J=0.0903in{.}^4\\ S_y=1.50in{.}^3\end{array}$

A is Cross-sectional area

S_xis Elastic section modulus about X -axis

Z_xis Plastic section modulus about X -axis

I_yis Moment of inertia about Y -axis

r_yis Radius of gyration about Y -axis

S_yis Elastic section modulus about Y -axis

$r_{ts}$ is $\sqrt{\frac{\sqrt{I_y{C}_w}}{S_x}}$

C_wis Warping constant

h₀is Distance between centroid of flanges

J is Torsional moment of inertia

$\begin{array}{l}{L}_p=1.76r_y\sqrt{\frac{E}{F_y}}=1.76\left(0.918\right)\sqrt{\frac{29000}{50}}=38.91in.=3.243ft\\ L_r=1.95r_{ts}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x{h}_0}+\sqrt{{\left(\frac{Jc}{S_x{h}_0}\right)}^2+6.76{\left(\frac{0.7F_y}{E}\right)}^2}}\\ =1.95\left(1.08\right)\frac{29000}{0.7\left(50\right)}\sqrt{\frac{0.0903\times 1}{7.31\times 5.75}+\sqrt{{\left(\frac{0.0903\times 1}{7.31\times 5.75}\right)}^2+6.76{\left(\frac{0.7\left(50\right)}{29000}\right)}^2}}\\ =134.6in.=11.22ft.\end{array}$

For $L_b=14ft$, $L_p<L_b<L_r$, so

$M_n=C_b\left[M_p-\left(M_p-0.7F_y{S}_x\right)\left(\frac{L_b-L_p}{L_r-L_p}\right)\right]\le M_p$ (Inelastic Lateral torsional buckling)

From the below given figure in the textbook, $C_b=1.14$

$\begin{array}{l}{M}_p=F_y{Z}_x=50\left(8.30\right)=415in-kips=34.58ft-kips\\ M_n=1.14\left[415-\left(415-0.7\times 50\times 7.31\right)\left(\frac{10-3.23}{11.22-3.243}\right)\right]\\ =319.4in-kips=26.62ft-kips<M_p\end{array}$

For the y axis, since the shape is compact, there is no flange local buckling

$M_{ny}=M_{py}=F_y{Z}_y=50\left(2.32\right)=116in-kips=9.667ft-kips$

Check the upper limit:

$\begin{array}{l}\frac{Z_y}{S_y}=\frac{2.32}{1.50}=1.55<1.6\\ \therefore M_{ny}=M_{py}=9.667ft-kips\\ \frac{M_{nx}}{{\Omega }_b}=\frac{26.62}{1.67}=15.94ft-kips\end{array}$

Roof load: Combination 3 controls

$w_a=D+S=\frac{40}{2}+\frac{40}{2}=40psf$

where, D is dead load and S is snow load

Tributary width = $\frac{\sqrt{17}}{4}\left(6\right)=6.185ft$

Purlin load = $40\left(6.185\right)=247.4lb/ft$

Component normal to roof = $w_{ax}=\frac{4}{\sqrt{17}}\left(247.4\right)=240lb/ft$

Component parallel to roof = $w_{ay}=\frac{1}{\sqrt{17}}\left(247.4\right)=60lb/ft$

Calculate factored bending moment about x axis and y axis

$\begin{array}{l}{M}_{ax}=\frac{w_{ax}{L}^2}{8}=\frac{0.24{\left(10\right)}^2}{8}=3ft-kips\\ M_{ay}=\frac{w_{ay}{L}^2}{8}=\frac{0.06{\left(10\right)}^2}{8}=0.75ft-kips\end{array}$

Use ½ of weak - axis bending strength in the interaction equation:

$\frac{M_{ax}}{M_{nx}/{\Omega }_b}+\frac{M_{ay}}{M_{ny}/{\Omega }_b}=\frac{3}{15.94}+\frac{0.75}{5.789/2}=0.447<1.0$ (OK)

Conclusion:

$W6\times 12$ is not adequate.