Chapter 6, Problem 6.8.3P

A beam of wide-flange shape, W 8 x 28, has the cross section shown in the figure. The dimensions are b = 6.54 in., h = 8.06 in., f_w = 0.285 in., and tf = 0.465 in.. The loads on the beam produce a shear force V = 7.5 kips at the cross section under consideration.

1. Use center line dimensions to calculate the maximum shear stress r_aiaxin the web of the beam.

Use the more exact analysis of Section 5,10 in Chapter 5 to calculate the maximum shear stress in the web of the beam and compare it with the stress obtained in part .

  

To determine

The maximum shear stress ${\tau }_{\max }$ in the web of the beam.

Answer to Problem 6.8.3P

The maximum shear stress ${\tau }_{\max }$ in the web of the beam is, ${\tau }_{\max }=3447psi$

Explanation of Solution

Figure :

Given:

The section $W8\times 28$ , width $b=6.54in$ , height $h=8.06in$ , thickness out hide $t_w=0.285in$ , thickness of flange $t_f=0.465in$ , shear force $V=7.5k$ ,

Concept Used:

Annuity problem requires the use of the moment of inertia equation as follows:

  $I_Z=\frac{t_w{h}^3}{12}+\frac{b{t}_f{h}^2}{2}$

Here,

  $Heightofbeam=h$ , web thickness $=t_w$ , $width=b$ , $flangewidth=b_f$ the moment of inertia about z −axis = $I_Z$

Annuity problem requires the use of the maximum shear stress equation as follows:

  ${\tau }_{\max }=\left(\frac{b{t}_f}{t_w}+\frac{h}{4}\right)\frac{Vh}{2I_Z}$

Here,

  $Maximumshearstress={\tau }_{\max }$ , $Shearforce=V$

Calculation:

As per the given problem

  $t_w=0.285in$ , $t_f=0.465in$ , $b=6.54in$ , $h=8.06in$

Annuity problem requires the use of this formula based on centerline dimensions

  $I_Z=\frac{t_w{h}^3}{12}+\frac{b{t}_f{h}^2}{2}$

Substitute these values in the formula

  $I_Z=\left[\frac{0.465{\left(8.06\right)}^3}{12}+\frac{6.54\times 0.465{\left(8.06\right)}^2}{2}\right]=111.216i{n}^4$

  $I_Z=111.216i{n}^4$

Annuity problem requires the use of this formula

  $t_w=0.285in$ , $t_f=0.465in$ , $b=6.54in$ , $h=8.06in$ , $I_Z=111.216i{n}^4$ , $V=7.5k$

  ${\tau }_{\max }=\left(\frac{b{t}_f}{t_w}+\frac{h}{4}\right)\frac{Vh}{2I_Z}$

Substitute these values in the formula

  ${\tau }_{\max }=\left[\left(\frac{6.54\times 0.465}{0.285}+\frac{8.06}{4}\right)\left(\frac{8.06\times 7.5}{2\times 111.216}\right)\right]=3447psi$

  ${\tau }_{\max }=3447psi$

Conclusion:

The maximum shear stress ${\tau }_{\max }$ in the web is calculated by the formula : ${\tau }_{\max }=\left(\frac{b{t}_f}{t_w}+\frac{h}{4}\right)\frac{Vh}{2I_Z}$

To determine

The maximum shear stress ${\tau }_{\max }$ in the web of the beam and compare it with the stress obtain in part (a)

Answer to Problem 6.8.3P

The maximum shear stress ${\tau }_{\max }$ in the web of the beam is, ${\tau }_{\max }=3446psi$

Explanation of Solution

Figure:

Given:

The section $W8\times 28$ , width $b=6.54in$ , height $h=8.06in$ ,thickness out hide $t_w=0.285in$ , thickness of flange $t_f=0.465in$ , shear force $V=7.5k$ ,

Concept Used:

Annuity problem requires the use of the equation as follows:

  $h_2=h+t_f$

  $h_1=h-t_f$

Here,

  $Heightofbeam=h$ , web thickness $=t_w$

Annuity problem requires the use of the moment of inertia equation as follows:

  $I=\frac{1}{12}\left(b{h}_2{}^3-b{h}_1{}^3+t_w{h}_1{}^3\right)$

Here,

  $Heightofbeam=h$ , web thickness $=t_w$ , $width=b$ , the moment of inertia = $I$

Annuity problem requires the use of the maximum shear stress equation in the web as follows:

  ${\tau }_{\max }=\frac{V}{8I{t}_w}\left(b{h}_2{}^2-b{h}_1{}^2+t_w{h}_1{}^2\right)$

Here,

  $Maximumshearstress={\tau }_{\max }$ , shear force = V

Calculation:

Based on more exact analysis As per the given problem

  $h=8.06in$ , $t_f=0.465in$

Annuity problem requires the use of this formula:

  $h_2=h+t_f$

Substitute these values in the formula

  $h_2=8.06+0.465=8.5in$

  $h_2=8.5in$

Annuity problem requires the use of this formula

  $h_1=h-t_f$

Substitute these values in the formula:

  $h_1=8.06-0.465=7.6in$

  $h_1=7.6in$

Annuity problem requires the use of this formula

  $b=6.54in$ , $h_1=7.6in$ , $h_2=8.5in$ , $t_w=0.285in$

  $I=\frac{1}{12}\left(b{h}_2{}^3-b{h}_1{}^3+t_w{h}_1{}^3\right)$

Substitute these values in the formula

  $I=\frac{1}{12}\left(6.54\times {\left(8.5\right)}^3-6.54{\left(7.6\right)}^3+0.285{\left(7.6\right)}^3\right)=109.295i{n}^4$

  $I=109.295i{n}^4$

Annuity problem requires the use of this formula

  $b=6.54in$ , $h_1=7.6in$ , $h_2=8.5in$ , $t_w=0.285in$ , $I=109.295i{n}^4$ , $V=7.5k$

  ${\tau }_{\max }=\frac{V}{8I{t}_w}\left(b{h}_2{}^2-b{h}_1{}^2+t_w{h}_1{}^2\right)$

Substitute these values in the formula

  ${\tau }_{\max }=\frac{7.5}{8\times 109.295\times 0.285}\left(6.54{\left(8.5\right)}^2-6.54{\left(7.6\right)}^2+0.285{\left(7.6\right)}^2\right)=3446psi$

  ${\tau }_{\max }=3446psi$

Conclusion:

The maximum shear stress ${\tau }_{\max }$ in the web of the beam is calculated by putting values in the formula: ${\tau }_{\max }=\frac{V}{8I{t}_w}\left(b{h}_2{}^2-b{h}_1{}^2+t_w{h}_1{}^2\right)$