Chapter 8, Problem 8.4.9P

A simple beam with a rectangular cross section (width, 3,5 in_L; height, 12 in,) carries a trapczoi-dally distributed load of 1400 lb/ft at A and 1000 lb/ft at B on a span of 14 ft (sec figure).

Find the principal stresses 2 and the maximum shear stress r__ at a cross section 2 ft from the left-hand support at each of the locations: (a) the neutral axis, (b) 2 in. above the neutral axis, and (c) the top of the beam. (Disregard the direct compressive stresses produced by the uniform load bearing against the top of the beam.)

  

To determine

To find: Values of principal stress and maximum shear stress at neutral axis.

Answer to Problem 8.4.9P

Principal stress:

${\sigma }_1=218.71\text{ psi}$ ,${\sigma }_2=-218.71\text{ psi}$

Maximum shear stress${\tau }_{\max }\text{=}218.71\text{ psi}$

Explanation of Solution

Given Information:

Beam length$L=14\text{ ft}=168\text{ in}$

Dimensions of beam,

$\begin{array}{l}h=12\text{ in}\\ b=3.5\text{ in}\end{array}$

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

Load at point$C$ :

$W_C=1000+\frac{400\times 12}{14}=1342.86\text{ lb}$

From equilibrium;

$\begin{array}{l}{R}_A\times 14-\frac{1}{2}(1000+1400)\times 14\times \left[\frac{14}{3}\times \frac{1000+2\times 1400}{1000+1400}\right]=0\\ R_A=8866.67\text{ lb}\end{array}$

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

  $\begin{array}{l}M=R_A\times \text{2}-\frac{1}{2}(1342.86+1400)\times 2\times \left[\frac{2}{3}\times \frac{1342.86+2\times 1400}{1342.86+1400}\right]\\ M\text{=14971}\text{.43 lb-ft}\\ M\text{=179657}\text{.2 lb-in}\end{array}$

Shear force at$C\text{ is}$

  $V=R_A-\frac{1}{2}(1342.86+1400)\times 2=6123.8\text{ lb}$

Moment of inertia:

  $\begin{array}{l}I=\frac{b{h}^3}{12}\\ I=\frac{3.5\times {12}^3}{12}\\ I=504{\text{ in}}^{\text{4}}\end{array}$

First moment of area above neutral axis.:

  $\text{Q}=A_1\times {\text{y}}_{\text{1}}=3.5\times 6\times (6/2)=63{\text{ in}}^{\text{3}}$

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:

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{6123.8\times 63}{504\times 3.5}\\ {\tau }_{xy}=218.71\text{ psi}\end{array}$

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

Values of principal and normal stress are given by following equation:

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{0+0}{2}\pm \sqrt{{\left(\frac{0-0}{2}\right)}^2+{218.71}^2}\\ {\sigma }_{1,2}=0\pm 218.71\\ \\ {\sigma }_1=0+218.71=218.71\text{ psi}\\ {\sigma }_2=0-218.71=-218.71\text{ psi}\end{array}$

Maximum shear stress:

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{0-0}{2}\right)}^2+{218.71}^2}\\ {\tau }_{\max }=218.71\text{ psi}\end{array}$

Conclusion:

Hence, we get:

Principal stresses

${\sigma }_1=218.71\text{ psi}$ ,${\sigma }_2=-218.71\text{ psi}$

Maximum shear stress${\tau }_{\max }\text{=}218.71\text{ psi}$

To determine

To find: Values of Principal stress and maximum shear stress at 2 in above neutral axis.

Answer to Problem 8.4.9P

Principal stress:

${\sigma }_1=762.47\text{ psi}$ ,${\sigma }_2=-49.53\text{ psi}$

Maximum shear stress:

  ${\tau }_{\max }=406.0\text{ psi}$

Explanation of Solution

Given Information:

Beam length$L=14\text{ ft}=168\text{ in}$

Dimensions of beam,

$\begin{array}{l}h=12\text{ in}\\ b=3.5\text{ in}\end{array}$

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

Load at point$C$ :

$W_C=1000+\frac{400\times 12}{14}=1342.86\text{ lb}$

From equilibrium:

$\begin{array}{l}{R}_A\times 14-\frac{1}{2}(1000+1400)\times 14\times \left[\frac{14}{3}\times \frac{1000+2\times 1400}{1000+1400}\right]=0\\ R_A=8866.67\text{ lb}\end{array}$

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

  $\begin{array}{l}M=R_A\times \text{2}-\frac{1}{2}(1342.86+1400)\times 2\times \left[\frac{2}{3}\times \frac{1342.86+2\times 1400}{1342.86+1400}\right]\\ M\text{=14971}\text{.43 lb-ft}\\ M\text{=179657}\text{.2 lb-in}\end{array}$

Shear force at$C\text{ is}$

  $V=R_A-\frac{1}{2}(1342.86+1400)\times 2=6123.8\text{ lb}$

Moment of inertia:

  $\begin{array}{l}I=\frac{b{h}^3}{12}\\ I=\frac{3.5\times {12}^3}{12}\\ I=504{\text{ in}}^{\text{4}}\end{array}$

First moment of area above given point:

  $\text{Q}=A_1\times {\text{y}}_{\text{1}}=3.5\times 4\times (2+2)=56{\text{ in}}^{\text{3}}$

So, bending stress at given point:

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{179657.2\times 2}{504}\\ {\sigma }_x=712.93\text{ psi}\end{array}$

And shear stress at that point:

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{6123.8\times 56}{504\times 3.5}\\ {\tau }_{xy}=194.41\text{ psi}\end{array}$

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

Values of principal and normal stress are given by following equation:

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{712.93+0}{2}\pm \sqrt{{\left(\frac{712.93-0}{2}\right)}^2+{194.41}^2}\\ {\sigma }_{1,2}=356.47\pm 406.0\\ \\ {\sigma }_1=356.47+406.0=762.47\text{ psi}\\ {\sigma }_2=356.47-406.0=-49.53\text{ psi}\end{array}$

Maximum shear stress ::

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{712.93-0}{2}\right)}^2+{194.41}^2}\\ {\tau }_{\max }=406\text{ psi}\end{array}$

Conclusion:

Hence, we get;

Principal stresses

${\sigma }_1=762.47\text{ psi}$ ,${\sigma }_2=-49.53\text{ psi}\text{.}$

Maximum shear stress${\tau }_{\max }=406.0\text{ psi}\text{.}$

To determine

To find: Values of principal stress and maximum shear stress at top of beam.

Answer to Problem 8.4.9P

Values of principal stress;

${\sigma }_1=2138.8\text{ psi}$ ,${\sigma }_2=0\text{ psi}$

Maximum shear stress${\tau }_{\max }=1069.4\text{ psi}$

Explanation of Solution

Given Information:

Beam length$L=14\text{ ft}=168\text{ in}$

Dimensions of beam,

$\begin{array}{l}h=12\text{ in}\\ b=3.5\text{ in}\end{array}$

Concept Used:

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

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

Values of principal stress are:${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

1. Maximum shear stress:

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

Load at point$C$ :

$W_C=1000+\frac{400\times 12}{14}=1342.86\text{ lb}$

From equilibrium:

$\begin{array}{l}{R}_A\times 14-\frac{1}{2}(1000+1400)\times 14\times \left[\frac{14}{3}\times \frac{1000+2\times 1400}{1000+1400}\right]=0\\ R_A=8866.67\text{ lb}\end{array}$

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

  $\begin{array}{l}M=R_A\times \text{2}-\frac{1}{2}(1342.86+1400)\times 2\times \left[\frac{2}{3}\times \frac{1342.86+2\times 1400}{1342.86+1400}\right]\\ M\text{=14971}\text{.43 lb-ft}\\ M\text{=179657}\text{.2 lb-in}\end{array}$

Moment of inertia:

  $\begin{array}{l}I=\frac{b{h}^3}{12}\\ I=\frac{3.5\times {12}^3}{12}\\ I=504{\text{ in}}^{\text{4}}\end{array}$

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

So, bending stress at top::

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{\text{179657}\text{.2}\times 6}{504}\\ {\sigma }_x=2138.8\text{ psi}\end{array}$

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${\sigma }_y=0$

Values of principal and normal stress are given by following equation::

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{2138.8+0}{2}\pm \sqrt{{\left(\frac{2138.8-0}{2}\right)}^2+0^2}\\ {\sigma }_{1,2}=1069.4\pm 1069.4\\ \\ {\sigma }_1=1069.4+1069.4=2138.8\text{ psi}\\ {\sigma }_2=1069.4-1069.4=0\text{ psi}\end{array}$

Maximum shear stress:

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{2138.8-0}{2}\right)}^2+0^2}\\ {\tau }_{\max }=1069.4\text{ psi}\end{array}$

Conclusion:

Hence, we get:

Values of principal stress:

${\sigma }_1=2138.8\text{ psi}$ ,${\sigma }_2=0\text{ psi}$

Maximum shear stress${\tau }_{\max }=1069.4\text{ psi}$ .