## Related questions with answers

A simple log bridge in a remote area consists of two parallel logs with planks across them. The logs are Douglas fir with average diameter 300 mm. A truck moves slowly across the bridge, which spans 2.5 m. Assume that the weight of the truck is equally distributed between the two logs. Because the wheelbase of the truck is greater than 2.5 m, only one set of wheels is on the bridge at a time. Thus, the wheel load on one log is equivalent to a concentrated load W acting at any position along the span. In addition, the weight of one log and the planks it supports is equivalent to a uniform load of $850\ \mathrm{N} / \mathrm{m}$ acting on the log. Determine the maximum permissible wheel load W based upon (a) an allowable bending stress of 7.0 MPa, and (b) an allowable shear stress of 0.75 MPa.

Solution

VerifiedTo determine the maximum load $W$, let's firstly determine the maximum shear force and maximum bending moment in the beam. Note that in the support $A$ one vertical reaction $R_A$ and one horizontal reaction $H_A$. In the support $B$ there will be one vertical reaction $R_B$. Since the position of the load $W$ can change, let's call the distance from the point $A$ to the load as distance $x_1$. Let's set an equation for sum of moments around point $B$:

$\begin{aligned} &-R_A\cdot L+W\cdot (L-x_1)+(q\cdot L)\Big(\frac{1}{2}L\Big)=0\\\\ &R_A=\frac{W\cdot (L-x_1)+(q\cdot L)\Big(\frac{1}{2}L\Big)}{L}\\\\ &R_A=W\Big(1-\frac{x_1}{L}\Big)+\frac{qL}{2} \end{aligned}$

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