Chapter 21, Problem 21.1P

To determine

Lightest steel structure section for the cantilever beam.

Answer to Problem 21.1P

   $W12\times 22$

Explanation of Solution

Given:

Tensile yield strength of beam $s_y=50ksi$

Point load on the beam $P=15kips$

Calculation:

The deflection of beam due to point load $P$ and reaction $R_A$ is given by,

  

Length of beam is given by,

   $\begin{array}{l}l=a+b\\ l=10+15\\ l=25ft\end{array}$

From the appendix, beam diagrams and formulas, than the deflection of the cantilever beam at $A$ due to the point load $P$ is given by,

   $\begin{array}{l}{\Delta }_1=\frac{P{b}^2}{6EI}\left(3l-b\right)\\ {\Delta }_1=\frac{15{\left(15\right)}^2}{6EI}\left(3\times 25-15\right)\\ {\Delta }_1=\frac{33750}{EI}\end{array}$

Than the deflection beam with load removed and due to reaction $R_A$ is given by,

   $\begin{array}{l}{\Delta }_2=\frac{R_A{l}^3}{3EI}\\ {\Delta }_2=\frac{R_A{25}^3}{3EI}\\ {\Delta }_2=\frac{5208.33}{3EI}{R}_A\end{array}$

Deflection will be zero at support $A$,

   $\begin{array}{l}{\Delta }_1={\Delta }_2\\ \frac{33750}{EI}=\frac{5208.33}{EI}{R}_A\\ R_A=\frac{33750}{5208.33}\\ R_A=6.48kips\uparrow \end{array}$

Sum of the forces acting along $Y$ direction,

   $\begin{array}{l}\sum F_y=0\\ R_A-P+V_B=0\\ 6.48-15+V_B=0\\ V_B=8.52kips\uparrow \end{array}$

Taking form left of the beam,

Cutting plane $1:$

   $0\le x\le 10ft$

Cutting plane $2:$

   $10ft\le x\le 25ft$

Consider free body diagram at $0\le x\le 10ft$,

  

Calculate the shear force (upward forces are taken as positive),

   $\begin{array}{l}{V}_X=R_A\\ V_X=6.48kips\end{array}$

To find bending moment we have,

   $\begin{array}{l}{M}_X=R_A\times x\\ M_X=6.48xk-ft\end{array}$

Hence for $0\le x\le 10ft$,

Therefore shear force,

   $V_0=V_{10}=6.48kips$

Therefore bending moment,

   $\begin{array}{l}{M}_0=0k-ft\\ x=0ft\\ M_{10}=64.8k-ft@x=10ft\end{array}$

Consider free body diagram at $10ft\le x\le 25ft$,

  

From the free body diagram we have,

Shear force calculation,

   $\begin{array}{l}{V}_X=R_A-P\\ V_X=6.48kips-15\\ V_X=-8.52kips\end{array}$

Bending moment calculation,

   $\begin{array}{l}{M}_X=R_A\times x-P\left(x-10\right)\\ M_X=6.48x-15\left(x-10\right)k-ft\\ M_X=-8.52x+150k-ft\end{array}$

Therefore shear force,

   $V_{10}=V_{15}=-8.52kips$

Therefore bending moment,

   $\begin{array}{l}{M}_{10}=64.8k-ft\\ x=10ft\\ M_{25}=-63k-ft@x=25ft\end{array}$

Shear force and bending moment diagram is given by,

  

Therefore maximum shear force acting on the beam,

   $V=8.52kips$

Maximum bending moment acting on the beam,

   $\begin{array}{l}M=64.8k-ft\\ M=777.6k-in\end{array}$

Required plastic section modulus is given by formula,

   $\begin{array}{l}{Z}_X=\frac{1.67M}{s_y}\\ Z_X=\frac{1.67\times 777.6}{50}\\ Z_X=25.97i{n}^3\end{array}$

From the dimensions and properties of W shapes table, for $W12\times 22$ shape, $Z_X=29.3i{n}^3$ then the values of $d=12.31in$ and $t_w=0.260in$.

Shear capacity is calculated by,

   $\begin{array}{l}{V}_{Cal}=0.4s_y\times d\times t_w\\ V_{Cal}=0.4\times 50\times 12.31\times 0.26\\ V_{Cal}=64.012kips\end{array}$

Since the calculate shear capacity is greater than shear force acting on the beam.

Conclusion:

Therefore the lightest steel structure section for the cantilever beam is $W12\times 22$.