Chapter 8, Problem 8.5.34P

A compound beam ABCD has a cable with force P anchored at C The cable passes over a pulley at D, and force P acts in the —x direction, There is a moment release just left of B. Neglect the self-weight of the beam and cable. Cable force P = 450 N and dimension variable L = 0.25 m. The beam has a rectangular cross section (b = 20 mm, it = 50 mm).

(a) Calculate the maximum normal stresses and maximum in-plane shear stress on the bottom surface of the beam at support A.

(b) Repeat part (a) for a plane stress element located at mid-height of the beam at A.

(c) If the maximum tensile stress and maximum in-plane shear stress at point A are limited to 90 MPa and 42 MPa, respectively, what is the largest permissible value of the cable force P?

  

To determine

The maximum normal stresses and maximum in-plane shear stress on the bottom of the beam at fixed support A.

Answer to Problem 8.5.34P

Maximum normal stresses, ${\sigma }_1=0$ , ${\sigma }_2={\sigma }_x=-108.4{\rm M}{\rm P}a$

Maximum in-plane shear stress ${\tau }_{\max }=-54.2{\rm M}{\rm P}a$

Explanation of Solution

Given information: A compound beam ABCD has a cable having force P anchored at C as shown in the figure below:

Cable force $P=450N$

Dimension variable $L=0.25m$

The cross section of the beam $b=20mm$ and $h=50mm$

To calculate the maximum normal stress and in-plane shear stress first taken the cross-sectional properties of the beam.

Area of the rectangular beam $A=bh$

  $\begin{array}{l}A=20\times 50\\ A=1000m{m}^2\end{array}$

Moment of inertia of the rectangular beam is $I=\frac{b{h}^3}{12}$

  $\begin{array}{l}I=\frac{20\times {(50)}^3}{12}20\times 50\\ I=208333.333m{m}^4\end{array}$

Now reactions at fixed support A

Horizontal force $H_A=P$

  $H_A=450{\rm N}$

Shear force $V_A=\frac{4L}{3L}P$

  $\begin{array}{l}{V}_A=\frac{4}{3}\times 450\\ V_A=600{\rm N}\end{array}$

And moment at point A, $M_A=8PL$

  $\begin{array}{l}{M}_A=8\times 450\times 0.25\\ M_A=900{\rm N}-m\end{array}$

Now, maximum normal stresses on the bottom of the beam at A is,

  ${\sigma }_x=\frac{-H_A}{A}-\frac{-M_A\left(\frac{h}{2}\right)}{I}$

  $\begin{array}{l}{\sigma }_x=-\frac{450}{1000}-\frac{900\times 25\times 1000}{208333.33}\\ {\sigma }_x=-0.45-108\\ {\sigma }_x=-108.45{\rm M}{\rm P}a\end{array}$

  ${\sigma }_y=0$

And at principle plane where there is maximum normal stress the shear stress is zero. So,

  ${\tau }_{xy}=0$

Principle stresses:

  $\begin{array}{l}{\sigma }_1=\frac{{\sigma }_x+{\sigma }_y}{2}+\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_1=\frac{-108.45}{2}+\sqrt{{\left(\frac{\left(-108.45\right)-0}{2}\right)}^2+{\left(0\right)}^2}\\ {\sigma }_1=-54.225+\sqrt{{\left(-54.225\right)}^2+0}\\ {\sigma }_1=-54.225+54.225\\ {\sigma }_1=0\end{array}$

  $\begin{array}{l}{\sigma }_2=\frac{{\sigma }_x+{\sigma }_y}{2}-\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_2=\frac{-108.45}{2}-\sqrt{{\left(\frac{\left(-108.45\right)-0}{2}\right)}^2+{\left(0\right)}^2}\\ {\sigma }_2=-54.225-\sqrt{{\left(-54.225\right)}^2+0}\\ {\sigma }_2=-54.225-54.225\\ {\sigma }_2=-108.45{\rm M}{\rm P}a\end{array}$

Maximum in-plane shear stress

  $\begin{array}{l}{\tau }_{max}=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{max}=\frac{{\sigma }_x}{2}\\ {\tau }_{max}=-54.225{\rm M}{\rm P}a\end{array}$

To determine

Maximum stresses located at the mid-height of the beam at A.

Answer to Problem 8.5.34P

Maximum normal stresses, ${\sigma }_1=0.703{\rm M}{\rm P}a$ , ${\sigma }_2=-1.153{\rm M}{\rm P}a$

Maximum in-plane shear stress ${\tau }_{\max }=0.928{\rm M}{\rm P}a$

Explanation of Solution

Given information: A compound beam ABCD has a cable having force P anchored at C as shown in the figure below:

Cable force $P=450N$

Dimension variable $L=0.25m$

The cross section of the beam $b=20mm$ and $h=50mm$

Area of the rectangular beam $A=bh$

  $\begin{array}{l}A=20\times 50\\ A=1000m{m}^2\end{array}$

Moment of inertia of the rectangular beam is $I=\frac{b{h}^3}{12}$

  $\begin{array}{l}I=\frac{20\times {(50)}^3}{12}20\times 50\\ I=208333.333m{m}^4\end{array}$

Now reactions at fixed support A

Horizontal force $H_A=P$

  $H_A=450{\rm N}$

Shear force $V_A=\frac{4L}{3L}P$

  $\begin{array}{l}{V}_A=\frac{4}{3}\times 450\\ V_A=600{\rm N}\end{array}$

The maximum normal stresses at the mid-point of the beam at A is,

  ${\sigma }_x=\frac{-H_A}{A}$

  $\begin{array}{l}{\sigma }_x=-\frac{450}{1000}\\ {\sigma }_x=-0.45{\rm M}{\rm P}a\end{array}$

  ${\sigma }_y=0$

And, ${\tau }_{xy}=-\frac{3}{2}\frac{V_A}{A}$

  $\begin{array}{l}{\tau }_{xy}=-\frac{3}{2}\times \frac{600}{1000}\\ {\tau }_{xy}=-0.9{\rm M}{\rm P}a\end{array}$

Principle stresses:

  $\begin{array}{l}{\sigma }_1=\frac{{\sigma }_x+{\sigma }_y}{2}+\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_1=\frac{-0.45}{2}+\sqrt{{\left(\frac{\left(-0.45\right)-0}{2}\right)}^2+{\left(-0.9\right)}^2}\\ {\sigma }_1=-0.225+\sqrt{{\left(-0.225\right)}^2+{\left(-0.9\right)}^2}\\ {\sigma }_1=-0.225+0.927698766\\ {\sigma }_1=0.7026\approx 0.703{\rm M}{\rm P}a\end{array}$

  $\begin{array}{l}{\sigma }_2=\frac{{\sigma }_x+{\sigma }_y}{2}-\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_2=\frac{-0.45}{2}-\sqrt{{\left(\frac{\left(-0.45\right)-0}{2}\right)}^2+{\left(-0.9\right)}^2}\\ {\sigma }_2=-0.225-\sqrt{{\left(-0.225\right)}^2+{\left(-0.9\right)}^2}\\ {\sigma }_2=-0.225-0.927698766\\ {\sigma }_2=1.1526\approx 1.153{\rm M}{\rm P}a\end{array}$

Maximum in-plane shear stress

  $\begin{array}{l}{\tau }_{max}=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{max}=\sqrt{{\left(\frac{\left(-0.45\right)-0}{2}\right)}^2+{\left(-0.9\right)}^2}\\ {\tau }_{max}=\sqrt{{\left(-0.225\right)}^2+{\left(-0.9\right)}^2}\\ {\tau }_{max}=0.9276\approx 0.928{\rm M}{\rm P}a\end{array}$

To determine

The maximum permissible value of the cable force P.

Answer to Problem 8.5.34P

Maximum permissible value of P $P_{\max }=348.5{\rm N}$

Explanation of Solution

Given information: A compound beam ABCD has a cable having force P anchored at C as shown in the figure below:

Dimension variable $L=0.25m$

Maximum tensile stress at point A, ${\sigma }_a=90{\rm M}{\rm P}a$

Maximum shear stress at point A, ${\tau }_a=42{\rm M}{\rm P}a$

The cross section of the beam $b=20mm$ and $h=50mm$

Area of the rectangular beam $A=bh$

Moment of inertia of the rectangular beam is $I=\frac{b{h}^3}{12}$

And moment at point A, $M_A=8PL$

In the above figure the maximum cable force P is controlled by the tensile and maximum shear force at the bottom of the beam at point A.

Hence, tensile stress ${\sigma }_x={\sigma }_a$ (given)

Then, ${\sigma }_x=\frac{P}{A}-\frac{-M_A\left(\frac{h}{2}\right)}{I}$

  $\begin{array}{l}{\sigma }_a=\frac{P}{bh}+\frac{8PL\left(\frac{h}{2}\right)}{\left(\frac{b{h}^3}{12}\right)}\\ {\sigma }_a=\frac{P}{bh}+\frac{48PL}{b{h}^2}or,\\ P=\frac{{\sigma }_a}{\left(\frac{1}{bh}+\frac{48L}{b{h}^2}\right)}\end{array}$

By substituting the values we get,

  $\begin{array}{l}P=\frac{90}{\left(\frac{1}{1000}+\frac{48\times 0.25\times 1000}{20{\left(50\right)}^2}\right)}\\ P=\frac{90}{\left(\frac{1}{1000}+\frac{48\times 250}{20{\left(50\right)}^2}\right)}\\ P=\frac{90}{\left(\frac{1}{1000}+\frac{12000}{50000}\right)}\\ P=\frac{90}{\left(0.001+0.24\right)}\\ P=\frac{90}{0.241}\\ P_{\max }=373.44{\rm N}\end{array}$

And,

  $\begin{array}{l}{\tau }_{max}=\frac{{\sigma }_x}{2}\\ \end{array}$

Then, $\begin{array}{l}{\sigma }_x={\sigma }_a=\frac{P}{bh}+\frac{48PL}{b{h}^2}\\ \end{array}$

  $\begin{array}{l}{\tau }_{max}=\frac{{\sigma }_x}{2}\\ \end{array}$

  $\begin{array}{l}{\tau }_a=\frac{P}{2bh}+\frac{24PL}{b{h}^2}or,\\ P_{\max }=\frac{{\tau }_a}{\left(\frac{1}{2bh}+\frac{24L}{b{h}^2}\right)}\\ P_{\max }=\frac{42}{\left(\frac{1}{2\times 20\times 50}+\frac{24\left(0.25\times 1000\right)}{20{\left(50\right)}^2}\right)}\\ P_{\max }=\frac{42}{\left(\frac{1}{2000}+\frac{6000}{50000}\right)}\\ P_{\max }=\frac{42}{\left(0.0005+0.12\right)}\\ P_{\max }=\frac{42}{0.1205}\\ P_{\max }=348.5{\rm N}\end{array}$

Conclusion:

Hence from the obtained values the maximum permissible value of the force P is

  $P_{\max }=348.5{\rm N}$