Chapter 5, Problem 5.5.14P

To determine

(a)

Whether a $MC18\times 58$ is adequate or not, by using LRFD

Answer to Problem 5.5.14P

Inadequate

Explanation of Solution

Given:

${\text{F}}_y\text{ = 36 ksi}$

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:

All channel shapes in the Manual are compact.(There are no footnotes to indicate otherwise)

For an $MC18\times 58$,

$\begin{array}{l}A=17.1in{.}^2\\ d=18in.\\ t_f=0.625in.\\ r_{ts}=1.35in.\\ h_0=17.4in.\\ S_x=75in{.}^3\\ Z_x=95.4in{.}^3\\ I_y=17.6in{.}^4\\ r_y=1.02in.\\ J=2.81in{.}^4\\ C_w=1070in{.}^6\end{array}$

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

$L_p=1.76r_y\sqrt{\frac{E}{F_y}}=1.76\left(1.02\right)\sqrt{\frac{29000}{36}}=50.95in.=4.25ft$

For channels,

$\begin{array}{l}c=\frac{h_0}{2}\sqrt{\frac{I_y}{C_w}}=\frac{17.4}{2}\sqrt{\frac{17.6}{1070}}=1.116\\ L_r=1.95r_{ts}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x{h}_0}+\sqrt{{\left(\frac{Jc}{S_x{h}_0}\right)}^2+6.76{\left(\frac{0.7F_y}{E}\right)}^2}}\\ =1.95\left(1.35\right)\frac{29000}{0.7\left(36\right)}\sqrt{\frac{2.81\times 1.116}{75\times 17.4}+\sqrt{{\left(\frac{2.81\times 1.116}{75\times 17.4}\right)}^2+6.76{\left(\frac{0.7\left(36\right)}{29000}\right)}^2}}\\ =228.747in.=19.06ft.\end{array}$

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, $C_b=1.32$

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

$\begin{array}{l}{M}_p=F_y{Z}_x=36\left(95.4\right)=3434.4in-kips=286.2ft-kips\\ M_n=1.32\left[3434.4-\left(3434.4-0.7\times 36\times 75\right)\left(\frac{14-4.25}{19.06-4.25}\right)\right]\\ =1642.424in-kips=136.87ft-kips<M_p\\ {\varphi }_b{M}_n=0.90\left(136.87\right)=123.182ft-kips\\ M_u=\frac{w_u{L}^2}{8}+\frac{P_{u}L}{4}=\frac{1.2\times 0.058{\left(14\right)}^2}{8}+\frac{1.6\times 30\left(14\right)}{4}\end{array}$

$=169.7ft-kips>123.182ft-kips$ (N.G.)

Conclusion:

$MC18\times 58$ is not adequate.

To determine

(b)

Whether a $MC18\times 58$ is adequate by usingASD

Answer to Problem 5.5.14P

Inadequate

Explanation of Solution

Given:

${\text{F}}_y\text{ = 36 ksi}$

Formula used:

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

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

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

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

M_n is nominal moment strength

M_pis plastic moment capacity

Calculation:

All channel shapes in the Manual are compact. (There are no footnotes to indicate otherwise)

For an $MC18\times 58$,

$\begin{array}{l}A=17.1in{.}^2\\ d=18in.\\ t_f=0.625in.\\ r_{ts}=1.35in.\\ h_0=17.4in.\\ S_x=75in{.}^3\\ Z_x=95.4in{.}^3\\ I_y=17.6in{.}^4\\ r_y=1.02in.\\ J=2.81in{.}^4\\ C_w=1070in{.}^6\end{array}$

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

$L_p=1.76r_y\sqrt{\frac{E}{F_y}}=1.76\left(1.02\right)\sqrt{\frac{29000}{36}}=50.95in.=4.25ft$

For channels,

$\begin{array}{l}c=\frac{h_0}{2}\sqrt{\frac{I_y}{C_w}}=\frac{17.4}{2}\sqrt{\frac{17.6}{1070}}=1.116\\ L_r=1.95r_{ts}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x{h}_0}+\sqrt{{\left(\frac{Jc}{S_x{h}_0}\right)}^2+6.76{\left(\frac{0.7F_y}{E}\right)}^2}}\\ =1.95\left(1.35\right)\frac{29000}{0.7\left(36\right)}\sqrt{\frac{2.81\times 1.116}{75\times 17.4}+\sqrt{{\left(\frac{2.81\times 1.116}{75\times 17.4}\right)}^2+6.76{\left(\frac{0.7\left(36\right)}{29000}\right)}^2}}\\ =228.747in.=19.06ft.\end{array}$

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, $C_b=1.32$

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

$\begin{array}{l}{M}_p=F_y{Z}_x=36\left(95.4\right)=3434.4in-kips=286.2ft-kips\\ M_n=1.32\left[3434.4-\left(3434.4-0.7\times 36\times 75\right)\left(\frac{14-4.25}{19.06-4.25}\right)\right]\\ =1642.424in-kips=136.87ft-kips<M_p\\ \frac{M_n}{{\Omega }_b}=\frac{136.87}{1.67}=81.96ft-kips\\ M_a=\frac{w_a{L}^2}{8}+\frac{P_{a}L}{4}=\frac{0.058{\left(14\right)}^2}{8}+\frac{30\left(14\right)}{4}\end{array}$

$=106.421ft-kips>81.96ft-kips$

Conclusion:

$MC18\times 58$ is not adequate.