Chapter 8, Problem 8.4.15P

A beam with a wide-flange cross section (see figure) has the following dimensions: b = 5 in., t = 0.5 in,, ft = 12 in., and /?, = 10.5 in. The beam is simply supported with span length L = 10 ft and supports a uniform load q = 6 kips/fL

Calculate the principal stresses *r_l and and the maximum shear stress t__ at a cross section located

[|] JA

3 ft from the left-hand support at each of the following locations: (a) the bottom of the beam, (b) the bottom of the web, and (c) the neutral axis

To determine

Find principal stresses and maximum shear stress at top of beam.

Answer to Problem 8.4.15P

Principal stresses

${\sigma }_1=6.8\text{ ksi}$ ,${\sigma }_2=0\text{ ksi}$

Maximum shear stress${\tau }_{\max }=6.8\text{ ksi}$

Explanation of Solution

Given Information:

Beam length$L=10\text{ ft}=120\text{ in}$

Uniform load$q=6\text{ kips/ft}$

Concept Used:

Bending stress${\sigma }_x=\frac{My}{I}$

Shear stress${\tau }_{xy}=\frac{VQ}{It}$

Principal normal stresses${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

Maximum shear stress${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

So bending moment at point$C\text{ is}$

  $M=\frac{qL}{2}\times 1-\frac{q}{2}\text{=}\frac{6\times 10}{2}\times 1-\frac{6}{2}\text{=}27\text{ kip-ft=324 kip-in}$

Shear force at point$C\text{ is}$

  $\text{V}=\frac{qL}{2}-q=\frac{6\times 10}{2}-6=24\text{ kip}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{5\times {12}^3}{12}-\frac{5\times {10.5}^3}{12}+\frac{0.5\times {10.5}^3}{12}\\ I=285.9{\text{ in}}^{\text{4}}\end{array}$

First moment of area at the top of beam shall be zero,$\text{Q}=0{\text{ m}}^3$

So bending stress at top,

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{324\times 6}{285.9}\\ {\sigma }_x=6.8\text{ ksi}\end{array}$

And shear stress at that point,

$\begin{array}{l}\tau =\frac{VQ}{It}\\ \tau =0\text{ ksi}\end{array}$

For this situation no stress in$y$ direction so${\sigma }_y=0$

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{6.8+0}{2}\pm \sqrt{{\left(\frac{6.8-0}{2}\right)}^2+0^2}\\ {\sigma }_{1,2}=3.4\pm 3.4\\ \\ {\sigma }_1=3.4+3.4=6.8\text{ ksi}\\ {\sigma }_2=3.4-3.4=0\text{ ksi}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{6.8-0}{2}\right)}^2+0^2}\\ {\tau }_{\max }=6.8\text{ ksi}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=6.8\text{ ksi}$ ,${\sigma }_2=0\text{ ksi}$

Maximum shear stress${\tau }_{\max }=6.8\text{ ksi}$

To determine

Find principal stresses and maximum shear stress at top of web.

Answer to Problem 8.4.15P

Principal stresses

${\sigma }_1=7.6\text{ ksi}$ ,${\sigma }_2=-1.65\text{ ksi}$

Maximum shear stress${\tau }_{\max }=4.626\text{ ksi}$

Explanation of Solution

Given Information:

Beam length$L=10\text{ ft}=120\text{ in}$

Uniform load$q=6\text{ kips/ft}$

Concept Used:

Bending stress${\sigma }_x=\frac{My}{I}$

Shear stress${\tau }_{xy}=\frac{VQ}{It}$

Principal normal stresses${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

Maximum shear stress${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

So bending moment at point$C\text{ is}$

  $M=\frac{qL}{2}\times 1-\frac{q}{2}\text{=}\frac{6\times 10}{2}\times 1-\frac{6}{2}\text{=}27\text{ kip-ft=324 kip-in}$

Shear force at point$C\text{ is}$

  $\text{V}=\frac{qL}{2}-q=\frac{6\times 10}{2}-6=24\text{ kip}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{5\times {12}^3}{12}-\frac{5\times {10.5}^3}{12}+\frac{0.5\times {10.5}^3}{12}\\ I=285.9{\text{ in}}^{\text{4}}\end{array}$

First moment of area of flange,

  $\text{Q}=A_1\times {\text{y}}_{\text{1}}=5\times \frac{12-10.5}{2}\times (\frac{10.5}{2}+\frac{12-10.5}{4})=21.1{\text{ in}}^{\text{3}}$

So bending stress at top of web,

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{324\times 5.25}{285.9}\\ {\sigma }_x=5.95\text{ ksi}\end{array}$

And shear stress at that point,

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{24\times 21.1}{285.9\times 0.5}\\ {\tau }_{xy}=3.5425\text{ ksi}\end{array}$

For this situation no stress in$y$ direction so${\sigma }_y=0$

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{5.95+0}{2}\pm \sqrt{{\left(\frac{5.95-0}{2}\right)}^2+{3.5425}^2}\\ {\sigma }_{1,2}=2.975\pm 4.626\\ \\ {\sigma }_1=2.975+4.626=7.6\text{ ksi}\\ {\sigma }_2=2.975-4.626=-1.65\text{ ksi}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{5.95-0}{2}\right)}^2+{3.5425}^2}\\ {\tau }_{\max }=4.626\text{ ksi}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=7.6\text{ ksi}$ ,${\sigma }_2=-1.65\text{ ksi}$

Maximum shear stress${\tau }_{\max }=4.626\text{ ksi}$

To determine

Find principal stresses and maximum shear stress at neutral axis.

Answer to Problem 8.4.15P

Principal stresses

${\sigma }_1=4.7\text{ ksi}$ ,${\sigma }_2=-4.7\text{ ksi}$

Maximum shear stress${\tau }_{\max }\text{=}4.7\text{ ksi}$

Explanation of Solution

Given Information:

Beam length$L=10\text{ ft}=120\text{ in}$

Uniform load$q=6\text{ kips/ft}$

Concept Used:

Bending stress${\sigma }_x=\frac{My}{I}$

Shear stress${\tau }_{xy}=\frac{VQ}{It}$

Principal normal stresses${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

Maximum shear stress${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

So bending moment at point$C\text{ is}$

  $M=\frac{qL}{2}\times 1-\frac{q}{2}\text{=}\frac{6\times 10}{2}\times 1-\frac{6}{2}\text{=}27\text{ kip-ft=324 kip-in}$

Shear force at point$C\text{ is}$

  $\text{V}=\frac{qL}{2}-q=\frac{6\times 10}{2}-6=24\text{ kip}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{5\times {12}^3}{12}-\frac{5\times {10.5}^3}{12}+\frac{0.5\times {10.5}^3}{12}\\ I=285.9{\text{ in}}^{\text{4}}\end{array}$

First moment of area for the section above the neutral axis,

  $\begin{array}{l}\text{Q}=A_1\times {\text{y}}_{\text{1}}+A_2\times {\text{y}}_2\\ Q=5\times \frac{12-10.5}{2}\times (\frac{10.5}{2}+\frac{12-10.5}{4})+0.5\times \frac{10.5}{2}\times (\frac{10.5}{4})\\ Q=28{\text{ in}}^{\text{3}}\end{array}$

So bending stress at neutral axis,

$\begin{array}{l}{\sigma }_x=\frac{M(y=0)}{I}\\ {\sigma }_x=0\text{ ksi}\end{array}$

And shear stress at that point,

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{24\times 28}{285.9\times 0.5}\\ {\tau }_{xy}=4.7\text{ ksi}\end{array}$

For this situation no stress in$y$ direction so${\sigma }_y=0$

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{0+0}{2}\pm \sqrt{{\left(\frac{0-0}{2}\right)}^2+{4.7}^2}\\ {\sigma }_{1,2}=0\pm 4.7\\ \\ {\sigma }_1=0+4.7=4.4\text{ ksi}\\ {\sigma }_2=0-4.7=-4.7\text{ ksi}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{0-0}{2}\right)}^2+{4.7}^2}\\ {\tau }_{\max }=4.7\text{ ksi}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=4.7\text{ ksi}$ ,${\sigma }_2=-4.7\text{ ksi}$

Maximum shear stress${\tau }_{\max }\text{=}4.7\text{ ksi}$