Chapter 5, Problem 5.5.28P

A foot bridge on a hiking trail is constructed using two timber logs each having a diameter d = 0.5 m (see figure a). The bridge is simply supported and has a length L = 4 m. The top of each log is trimmed to form the walking surface (see Fig, b)_LA simplified model of the bridge is shown in Fig. g. Each log must carry its own weight w = 1.2 kN/m and the weight (P = 850 N) of a person at mid-span, (see Fig. b).

1. Determine the maximum tensile and compressive stresses in the beam (Fig, b) due to bending.

If load h is unchanged, find the maximum permissible value of load ... if the allowable normal stress in tension and compression is 2.5 M Pa.

  

  

To determine

The maximum tensile stress in beam.

The maximum compressive stress in beam.

Answer to Problem 5.5.28P

The maximum tensile stresses in beam are $0.335\text{MPa}$

The maximum compressive stresses in beam are $0.288\text{MPa}$

Explanation of Solution

Given information:

The diameter of the timber log is $0.5\text{m}$.

The length of the bridge is $4\text{m}$.

The weight of the log is $1.2\text{kN}/\text{m}$.

The following figure shows the force distribution on the wooden log.

Figure 1

Write the expression for the area of the circle.

$A_{\text{cir}}=\frac{\pi d^2}{4}$...... (I)

Here, the diameter of the circle is $d$, the area of the circle is $A_{cir}$.

Write the expressions for the polar moment of inertia of the circle.

$I_{z_{\text{cir}}}=\frac{\pi d^4}{64}$...... (II)

Here, the polar moment of inertia is $I_{z_{cir}}$.

Write the expression for centroidal distance.

$y_{\text{cir}}=\frac{d}{2}$...... (III)

Here, the centroidal distance of the circle is $y_{cir}$.

Write the expressions for radius of circle.

$r=\frac{d}{2}$...... (IV)

Here, the radius of circle is $r$.

Write the expressions for the area of the circular segment.

$A_{\text{cut}}=r^2\left(\alpha -\sin \alpha \cos \alpha \right)$...... (V)

Here, the radius of the circle is $r$and the subtended angle is $\alpha$, and the area of the segment is $A_{cut}$.

Write the expression for the centroidal distance of the segment.

$y_{\text{cut}}=\frac{2r}{3}\left(\frac{{\sin }^3\alpha }{\alpha -\sin \alpha \cos \alpha }\right)$...... (VI)

Here, the centroidal distance of the segment is $y_{cut}$.

Write the expression for moment of inertia of the segment about x axis.

$I_{x_{\text{cut}}}=\frac{r^4}{4}\left(\alpha -\sin \alpha \cos \alpha +2{\sin }^3\alpha \cos \alpha \right)$...... (VII)

Here, the moment of inertia about x axis is $I_{x_{\text{cut}}}$.

Write the expression for moment of inertia about z axis.

$I_{z_{\text{cut}}}=I_{x_{\text{cut}}}-A_{\text{cut}}{y}_{\text{cut}}^2$...... (VIII)

Here, the moment of inertia about z axis is $I_{z_{\text{cut}}}$.

Write the expression for the distance from the centroid to the bottom.

$y_{\text{bot}}=\frac{A_{\text{cir}}{y}_{\text{cir}}-A_{\text{cut}}\left(\frac{d}{2}+y_{\text{cut}}\right)}{A_{\text{cir}}-A_{\text{cut}}}$....... (IX)

Here, the distance of centroid from the bottom is $y_{bot}$.

Write the expression for the distance from the centroid to the top.

$y_{\text{top}}=d-\left(\frac{d}{2}-\frac{d}{2}\cos \alpha \right)-y_{\text{bot}}$

...... (X)

Here, the distance from centroid to the top is $y_{\text{top}}$.

Write the expression for moment of inertia of the wooden beam.

$I_{z_{\text{net}}}=I_{z_{\text{cir}}}-I_{z_{\text{cir}}}+A_{\text{cir}}{\left(y_{\text{cir}}-y_{\text{bot}}\right)}^2-A_{\text{cut}}\left(\frac{d}{2}+y_{\text{cut}}-y_{\text{bot}}\right)$..... (XI)

Here, the net moment of inertia of the wooden beam is $I_{z_{net}}$.

Write the expression for maximum moment.

$M_{\text{max}}=\frac{w{L}^2}{8}+\frac{PL}{4}$...... (XII)

Here, the distributed load is $w$and the applied load is $p$, the maximum moment is $M_{\max }$

Write the expression for maximum tensile stress in mid span.

${\sigma }_{\text{ten}}=\frac{M_{\text{max}}{y}_{\text{bot}}}{I_{z_{\text{net}}}}$...... (XIII)

Here, the maximum tensile stress is ${\sigma }_{ten}$.

Write the expression for maximum compressive stress in mid span.

${\sigma }_{\text{comp}}=\frac{M_{\text{max}}{y}_{\text{top}}}{I_{z_{\text{net}}}}$...... (XIV)

Here, the maximum compressive stress is ${\sigma }_{comp}$.

Calculation:

Substitute $0.5\text{m}$for $d$in Equation (I).

$\begin{array}{c}{A}_{\text{cir}}=\frac{\pi {\left(0.5\text{m}\right)}^2}{4}\\ =0.1963{\text{m}}^2\times \left(\frac{{1000}^2{\text{mm}}^2}{1{\text{m}}^2}\right)\\ =1.963\times {10}^5{\text{mm}}^2\end{array}$

Substitute $0.5\text{m}$for $d$in Equation (II).

$\begin{array}{c}{I}_{z_{\text{cir}}}=\frac{\pi {\left(0.5\right)}^4}{64}\\ =3.068\times {10}^{-3}{\text{m}}^4\times \left(\frac{{1000}^4{\text{mm}}^4}{1{\text{m}}^4}\right)\\ =3.068\times {10}^9{\text{mm}}^2\end{array}$

Substitute $0.5\text{m}$for $d$in Equation (III).

$\begin{array}{c}{y}_{\text{cir}}=\frac{0.5\text{m}}{2}\\ =0.25\text{m}\times \left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =250\text{mm}\end{array}$

Substitute $0.5\text{m}$for $d$in Equation (IV).

$\begin{array}{c}r=\frac{0.5\text{m}}{2}\\ =0.25\text{m}\times \left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =250\text{mm}\end{array}$

Substitute $\text{250}\text{mm}$for $r$and $45°$for $\alpha$in Equation (V).

$\begin{array}{c}{A}_{\text{cut}}={\left(250\text{mm}\right)}^2\left(\frac{\pi }{4}-\sin 45°\cos 45°\right)\\ =62500{\text{mm}}^2\left(\frac{\pi }{4}-\sin 45°\cos 45°\right)\\ =1.784\times {10}^4{\text{mm}}^2\end{array}$

Substitute $\text{250}\text{mm}$for $r$and $45°$for $\alpha$in Equation (VI).

$\begin{array}{c}{y}_{\text{cut}}=\frac{2\times 250\text{mm}}{3}\left(\frac{{\sin }^{3}45°}{\frac{\pi }{4}-\sin 45°\cos °}\right)\\ =\frac{500\text{mm}}{3}\left(\frac{{\sin }^{3}45°}{\frac{\pi }{4}-\sin 45°\cos °}\right)\\ =206.468\text{mm}\end{array}$

Substitute $\frac{r^4}{4}\left(\alpha -\sin \alpha \cos \alpha +2{\sin }^3\alpha \cos \alpha \right)$for $I_{x_{\text{cut}}}$

$\text{250}\text{mm}$for $r$and $45°$for $\alpha$, $1.784\times {10}^4{\text{mm}}^2$for $A_{\text{cut}}$and $206.468\text{mm}$for $y_{\text{cut}}$in Equation (VII).

$\begin{array}{c}{I}_{z_{\text{cut}}}=\left[\begin{array}{c}\frac{{\left(250\text{mm}\right)}^4}{4}\left(\frac{\pi }{4}-\sin 45°\cos 45°+2{\sin }^{3}45°\cos 45°\right)\\ -\left(1.784\times {10}^4{\text{mm}}^2\times {206.468}^2\right)\end{array}\right]\\ =6.6\times {10}^4{\text{mm}}^4\end{array}$

Substitute $1.784\times {10}^4{\text{mm}}^2$for $A_{\text{cut}}$, $206.468\text{mm}$for $y_{\text{cut}}$

$0.5\text{m}$for $d$

$250\text{mm}$for $y_{\text{cir}}$and $1.963\times {10}^5{\text{mm}}^2$for $A_{\text{cir}}$in Equation (VIII).

$\begin{array}{c}{y}_{\text{bot}}=\frac{\left[\begin{array}{l}1.963\times {10}^5{\text{mm}}^2\times 250\text{mm}\\ -1.784\times {10}^4{\text{mm}}^2\left(\frac{500\text{mm}}{2}+206.468\text{mm}\right)\end{array}\right]}{1.963\times {10}^5{\text{mm}}^2-1.784\times {10}^4{\text{mm}}^2}\\ \frac{\left[\begin{array}{l}1.963\times {10}^5{\text{mm}}^2\times 250\text{mm}\\ -1.784\times {10}^4{\text{mm}}^2\left(250\text{mm}+206.468\text{mm}\right)\end{array}\right]}{1.963\times {10}^5{\text{mm}}^2-1.784\times {10}^4{\text{mm}}^2}\\ =229.369\text{mm}\end{array}$

Substitute $500\text{mm}$for $d$, $229.369\text{mm}$for $y_{\text{bot}}$and $45°$for $\alpha$in Equation (X)

$\begin{array}{c}{y}_{\text{top}}=500\text{mm}-\left(\frac{500\text{mm}}{2}-\frac{500\text{mm}}{2}\cos 45°\right)-229.369\text{mm}\\ \text{=}\left(500\text{mm}\right)-\left(\left(250\text{mm}\right)-\left(250\text{mm}\cos 45°\right)\right)-\left(229.369\text{mm}\right)\\ =197.407\text{mm}\end{array}$

Substitute $3.068\times {10}^9{\text{mm}}^2$for $I_{z_{\text{cir}}}$, $6.6\times {10}^6{\text{mm}}^4$for $I_{z_{\text{cut}}}$, $1.963\times {10}^5{\text{mm}}^2$for $A_{\text{cir}}$, $250\text{mm}$for $y_{\text{cir}}$, $229.369\text{mm}$for $y_{\text{bot}}$, $1.784\times {10}^4{\text{mm}}^2$for $A_{\text{cut}}$, $500\text{mm}$for $d$, $206.468\text{mm}$for $y_{\text{cut}}$in Equation (XI).

$\begin{array}{c}{I}_{z_{\text{net}}}=\left[\begin{array}{l}3.068\times {10}^9{\text{mm}}^2-6.6\times {10}^6{\text{mm}}^4\\ +1.963\times {10}^5{\text{mm}}^2{\left(250\text{mm-229}\text{.369}\right)}^2\\ -1.784\times {10}^4{\text{mm}}^2\left(\frac{500\text{mm}}{2}+206.468\text{mm-229}\text{.369}\text{mm}\right)\end{array}\right]\\ =2.225\times {10}^9{\text{mm}}^4\end{array}$

Substitute $1.2\text{kN}/\text{m}$for $w$, $4\text{m}$for $L$and $850\text{N}$for $p$in Equation (XII).

$\begin{array}{c}{M}_{\text{max}}=\left(\frac{1.2\text{kN}\times {\left(4\text{m}\right)}^2}{8}\right)+\left(\frac{85\text{N}\times 4\text{m}}{4}\right)\\ =\left(\frac{\left(1.2\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)\times {\left(4\text{m}\right)}^2}{8}\right)+\left(\frac{85\text{N}\times 4\text{m}}{4}\right)\\ =3250\text{N}\cdot \text{m}\times \left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =3.25\text{kN}\cdot \text{m}\end{array}$

Substitute $3.25\text{kN}\cdot \text{m}$for $M_{\text{max}}$

$229.369\text{mm}$for $y_{\text{bot}}$and $2.225\times {10}^9{\text{mm}}^4$for $I_{z_{\text{net}}}$.in Equation (XIII).

$\begin{array}{c}{\sigma }_{\text{ten}}=\frac{\left(3.25\text{MPa}\right)\times \left(229.369\text{mm}\right)}{\left(2.225\times {10}^9{\text{mm}}^4\right)}\\ =\frac{\left(3.25\text{MPa}\right)\left(\frac{1\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\times \left(197.407\text{mm}\right)}{\left(2.225\times {10}^9{\text{mm}}^4\right)}\\ =0.335\frac{\text{N}}{{\text{mm}}^2}\times \left(\frac{1\text{MPa}}{\text{1}\text{N}/{\text{mm}}^2}\right)\\ =0.335MPa\end{array}$

Substitute $3.25\text{kN}\cdot \text{m}$for $M_{\text{max}}$

$197.407\text{mm}$for $y_{\text{top}}$and $2.225\times {10}^9{\text{mm}}^4$for $I_{z_{\text{net}}}$.in Equation (XIV).

$\begin{array}{c}{\sigma }_{\text{com}}=\frac{\left(3.25\text{MPa}\right)\times \left(197.407\text{mm}\right)}{\left(2.225\times {10}^9{\text{mm}}^4\right)}\\ =\frac{\left(3.25\text{MPa}\right)\left(\frac{1\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\times \left(197.407\text{mm}\right)}{\left(2.225\times {10}^9{\text{mm}}^4\right)}\\ =0.288\frac{\text{N}}{{\text{mm}}^2}\times \left(\frac{1\text{MPa}}{1\text{N}/{\text{mm}}^2}\right)\\ =0.288\text{MPa}\end{array}$

Conclusion:

The maximum tensile stress in beam are $0.335\text{MPa}$.

The maximum compressive stress in beam are $0.288\text{MPa}$.

To determine

The maximum permissible value of load $P_{\text{max}}$.

Answer to Problem 5.5.28P

The maximum permissible value of load $P_{\text{max}}$is $21.9\text{kN}$.

Explanation of Solution

Given Information:

The allowable normal tensile stress is $2.5\text{MPa}$.

The allowable normal compressible stress is $2.5\text{MPa}$.

Write the expression for the maximum tensile moment.

$M_{\text{max ten}}=\frac{{\sigma }_a{I}_{z\text{}\text{net}}}{y_{\text{bot}}}$...... (XV)

Here the maximum tensile moment is $M_{\max ten}$.

Write the expression for the maximum compressible moment.

$M_{\text{max com}}=\frac{{\sigma }_a{I}_{z\text{net}}}{y_{\text{top}}}$...... (XVI)

Here the maximum compressive moment is $M_{\max comp}$.

Write the expression for the tensile force corresponding to the permissible stress.

$P_{\text{max ten}}=\frac{4}{L}\left(M_{\text{max ten}}-\frac{1.2\times 4^2}{8}\right)$

...... (XVII)

Calculation:

Substitute $2.5\text{MPa}$for ${\sigma }_a$and $2.225\times {10}^9{\text{mm}}^4$for $I_{z\text{net}}$and $229.369\text{mm}$for $y_{\text{bot}}$in Equation (XV).

$\begin{array}{c}{M}_{\text{max ten}}=\frac{\left(2.5\text{MPa}\right)\times \left(2.225\times {10}^9{\text{mm}}^4\right)}{\left(229.369\text{mm}\right)}\\ =\frac{\left(2.5\text{MPa}\right)\left(\frac{{10}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\times \left(2.225\times {10}^9{\text{mm}}^4\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^4}{\left(229.369\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =24.251\times {10}^3\text{N}\cdot \text{m}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =24.251\text{kN}\cdot \text{m}\end{array}$

Substitute $2.5\text{MPa}$for ${\sigma }_a$and $2.225\times {10}^9{\text{mm}}^4$for $I_{z\text{net}}$and $197.407\text{mm}$for $y_{\text{top}}$in Equation (XVI).

$\begin{array}{c}{M}_{\text{max ten}}=\frac{\left(2.5\text{MPa}\right)\times \left(2.225\times {10}^9{\text{mm}}^4\right)}{197.407\text{mm}}\\ =\frac{\left(2.5\text{MPa}\right)\left(\frac{{10}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\times \left(2.225\times {10}^9{\text{mm}}^4\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^4}{\left(197.407\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =28.178\times {10}^3\text{N}\cdot \text{m}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =28.178\text{kN}\cdot \text{m}\end{array}$

Substitute $1.2\text{kN}/\text{m}$for $w$, $24.251\text{kN}\cdot \text{m}$for $M_{\text{max ten}}$, $4\text{m}\text{}\text{for}\text{L }$, and $850\text{N}$for $P$in Equation (XVII).

$\begin{array}{c}{P}_{\text{max ten}}=\frac{4}{4\text{m}}\left(\left(24.251\text{kN}/\text{m}\right)-\frac{\left(1.2\text{kN}\cdot \text{m}\right)\times {\left(4\text{m}\right)}^2}{8}\right)\\ =\left(24.251\text{kN}/\text{m}\right)-\frac{\left(1.2\text{kN}\cdot \text{m}\right)\times {\left(4\text{m}\right)}^2}{8}\\ =21.9\text{kN}\end{array}$

Conclusion:

The maximum permissible value of load $P_{\text{max}}$is $21.9\text{kN}$.