Chapter 5, Problem 5.5.15P

To determine

(a)

Whether a $W24\times 104$ of A992 steel for the given beam by using LRFD

Answer to Problem 5.5.15P

Inadequate

Explanation of Solution

Given:

A992 steel

Formula used:

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

$M_n$ is nominal moment strength

$M_p$ is plastic moment capacity

$S_x$ is Elastic section modulus about X -axis

$L_p$ is unbraced length in an inelastic behavior

$L_r$ is unbraced length in an elastic behavior

$C_b=\frac{12.5M_{\max }}{2.5M_{\max }+3M_A+4M_B+3M_C}$

$C_b$ is factor to account for non- uniform bending within the unbraced length

$M_{\max }$ is absolute value of the maximum moment within the unbraced length

$M_A$ is absolute value of the moment at the quarter point of the unbraced length

$M_B$ is absolute value of the moment at the midpoint of the unbraced length

$M_C$ is absolute value of the moment at the three- quarter point of the unbraced length

Calculation:

Determine the nominal flexural strength:

From the $Z_x$ table,

$\begin{array}{l}{L}_p=10.3ft\\ L_r=29.2ft\end{array}$

For $L_b=20ft$, $L_p<L_b<L_r$

This shape is compact because there is no footnote in the dimensions and properties tables to indicate otherwise.

Since $L_p<L_b<L_r$,

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

$M_p=F_y{Z}_x=50\left(289\right)=14450in-kips=1204.167ft-kips$

$\begin{array}{l}{M}_n=C_b\left[14450-\left(14450-0.7\left(50\right)\left(258\right)\right)\left(\frac{20-10.3}{29.2-10.3}\right)\right]\\ =C_b\left(9815.56\right)\end{array}$

Determine the factored point loads:

$P_u=1.2P_D+1.6P_L$

D is dead point load and L is live point load

$\begin{array}{l}{P}_u=1.2\left(12\right)+1.6\left(36\right)\\ =72kips\end{array}$

Determine the factored distributed loads:

$w_u=1.2w_D+1.6w_L$

$w_D$ is uniformly distributed dead load and $w_L$ is uniformly distributed live load

$\begin{array}{l}{w}_u=1.2\left(1+.104\right)+1.6\left(3\right)\\ =6.125kips/ft\end{array}$

Determine the beam reactions:

$\begin{array}{l}{R}_A+R_B=72+6.125\times 30=255.75k\\ {\displaystyle \sum M_A}=0\\ R_B\times 30-72\times 10-6.125\times 30\times 15=0\\ R_B=115.875k\\ R_A=255.75k-115.875k=139.875k\end{array}$

$M_u=139.875\left(11.08\right)-72\left(11.08-10\right)-6.125\left(\frac{{11.08}^2}{2}\right)=1096.08ft-kips$

Compute C_b :

$\begin{array}{l}{M}_{\max }=M_u=1096.08ft-kips\\ M_A=139.875\left(15\right)-6.125\left(\frac{{15}^2}{2}\right)-72\left(15-10\right)=1049.06ft-kips\\ M_B=139.875\left(20\right)-6.125\left(\frac{{20}^2}{2}\right)-72\left(20-10\right)=852.5ft-kips\\ M_C=139.875\left(25\right)-6.125\left(\frac{{25}^2}{2}\right)-72\left(25-10\right)=502.81ft-kips\\ C_b=\frac{12.5M_{\max }}{2.5M_{\max }+3M_A+4M_B+3M_C}\\ =\frac{12.5\left(1096.08\right)}{2.5\left(1096.08\right)+3\left(1049.06\right)+4\left(852.5\right)+3\left(502.81\right)}\\ =1.268\\ M_n=C_b\left(9815.56\right)=1.268\left(9815.56\right)=12446.13in-kips<M_p=14450in-kips\\ {\varphi }_b{M}_n=0.9\left(12446.13\right)=11201.517in-kips=933.46ft-kips\end{array}$

Since $M_u=1096.08ft-kips>{\varphi }_b{M}_n=933.46ft-kips$, $W24\times 104$ is not adequate.

Conclusion:

$W24\times 104$ of A992 steel for the given beam is not adequate.

To determine

(b)

Whether a $W24\times 104$ of A992 steel for the given beam by using ASD

Answer to Problem 5.5.15P

Inadequate

Explanation of Solution

Given:

A992 steel

Formula used:

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

$M_n$ is nominal moment strength

$M_p$ is plastic moment capacity

$S_x$ is Elastic section modulus about X -axis

$L_p$ is unbraced length in an inelastic behavior

$L_r$ is unbraced length in an elastic behavior

$C_b=\frac{12.5M_{\max }}{2.5M_{\max }+3M_A+4M_B+3M_C}$

$C_b$ is factor to account for non- uniform bending within the unbraced length

$M_{\max }$ is absolute value of the maximum moment within the unbraced length

$M_A$ is absolute value of the moment at the quarter point of the unbraced length

$M_B$ is absolute value of the moment at the midpoint of the unbraced length

$M_C$ is absolute value of the moment at the three- quarter point of the unbraced length

Calculation:

Determine the nominal flexural strength:

From the Z_xtable,

$\begin{array}{l}{L}_p=10.3ft\\ L_r=29.2ft\end{array}$

For $L_b=20ft$, $L_p<L_b<L_r$

This shape is compact because there is no footnote in the dimensions and properties tables to indicate otherwise.

Since $L_p<L_b<L_r$,

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

$M_p=F_y{Z}_x=50\left(289\right)=14450in-kips=1204.167ft-kips$

$\begin{array}{l}{M}_n=C_b\left[14450-\left(14450-0.7\left(50\right)\left(258\right)\right)\left(\frac{20-10.3}{29.2-10.3}\right)\right]\\ =C_b\left(9815.56\right)\end{array}$

Determine the factored point loads:

$P_a=P_D+P_L$

$P_D$ is dead point load and $P_L$ is live point load

$\begin{array}{l}{P}_u=12+36\\ =48kips\end{array}$

Determine the factored distributed loads:

$w_u=w_D+w_L$

$w_D$ is uniformly distributed dead load and $w_L$ is uniformly distributed live load

$\begin{array}{l}{w}_a=\left(1+.104\right)+3\\ =4.104kips/ft\end{array}$

Determine the beam reactions:

$\begin{array}{l}{R}_A+R_B=48+4.104\times 30=171.12k\\ {\displaystyle \sum M_A}=0\\ R_B\times 30-48\times 10-4.104\times 30\times 15=0\\ R_B=77.56k\\ R_A=171.12k-77.56k=93.56k\end{array}$

$M_u=93.56\left(11.101\right)-48\left(11.101-10\right)-4.104\left(\frac{{11.101}^2}{2}\right)=732.89ft-kips$

Compute C_b :

$\begin{array}{l}{M}_{\max }=M_u=732.89ft-kips\\ M_A=93.56\left(15\right)-48\left(15-10\right)-4.104\left(\frac{{15}^2}{2}\right)=701.7ft-kips\\ M_B=93.56\left(20\right)-48\left(20-10\right)-4.104\left(\frac{{20}^2}{2}\right)=570.4ft-kips\\ M_C=93.56\left(25\right)-48\left(25-10\right)-4.104\left(\frac{{25}^2}{2}\right)=336.5ft-kips\\ C_b=\frac{12.5M_{\max }}{2.5M_{\max }+3M_A+4M_B+3M_C}\\ =\frac{12.5\left(732.89\right)}{2.5\left(732.89\right)+3\left(701.7\right)+4\left(570.4\right)+3\left(336.5\right)}\\ =1.267\\ M_n=C_b\left(9815.56\right)=1.267\left(9815.56\right)=12439.995in-kips<M_p=14450in-kips\\ \frac{M_n}{{\Omega }_b}=\frac{12439.995}{1.67}=7449.0988in-kips=620.76ft-kips\end{array}$

Since $M_u=732.89ft-kips>{\varphi }_b{M}_n=620.76ft-kips$, $W24\times 104$ is not adequate.

Conclusion:

$W24\times 104$ of A992 steel for the given beam is not adequate.