Question 3.7: A hoist is constructed using two wooden bars (E = 12.4 GPa) ...

By inspection, we see that member BC will be in compression. Thus, to determine the maximum value of W, we need to consider buckling failure of member BC and strength failure of both members due to the axial stress exceeding the given allowable stress. The internal forces in members BC and CD can be found in terms of W. The axial stresses in the members are then compared with the given allowable values to determine one set of limits on W. To determine the critical buckling load, the smaller second moment of area about the perpendicular axes through the section centroid should be used and the upper limit on W to prevent buckling failure can be found. The maximum value of W that satisfies the strength and buckling criteria can now be determined.
The free-body diagram for the pulley is shown in Figure 3.72, with F_{BC} drawn compressive and F_{CD} as tensile. The internal axial forces are found from

\sum{F_{\gamma }} = 0       \Rightarrow    F_{CD} \sin  30 = 2  W           \Rightarrow    F_{CD} = 4  W

\sum{F_{x}} = 0           \Rightarrow     F_{BC} = F_{CD} \cos  30                 \Rightarrow      F_{BC} = 3.46  W

The cross-sectional areas for the two members are A_{BC} = 5000  mm^{2} and A_{CD} = 6000  mm^{2}. Thus the axial stresses in terms of W can be found and these must be less than 13.8 MPa, giving two limits on W:

\sigma _{CD} = \frac{F_{CD}}{A_{BC}} = \frac{4  W}{6000} \leq 13.8  MPa          or           W \leq 20.7  kN

\sigma _{BC} = \frac{F_{BC}}{A_{BC}} = \frac{3.464  W}{5000} \leq 13.8  MPa          or       W \leq 17.9  kN

For cross section A-A, we note that

I_{min} = \frac{1}{12} \times 100 \times 50^{3} = 1041  666.7  mm^{4}

Thus, for length (a),

P_{crit} = \frac{\pi ^{2}EI}{L^{2}} = \frac{\pi ^{2}  \times  12.4  \times  10^{3}  \times  1041  667}{1200^{2}} = 56658.8  N

Thus, a second limit on F_{BC} is F_{BC} < 88 529 N, thus 3.464 W
< 88 529 N or W < 25.6 kN. Since this is greater than the value that causes yielding, the yielding value will occur first, so that the maximum permissible value of W for L = 1.2 m is 17.9 kN.

Then, for length (b),

P_{crit} = \frac{\pi ^{2}EI}{L^{2}} = \frac{\pi ^{2}   \times   12.4   \times  10^{3}   \times   1041  667}{1500^{2}} = 88  529.4  N

Thus, a second limit on F_{BC} is F_{BC} < 56 658.8 N, thus 3.464 W
< 56 658.8 N or W < 16.36 kN. Since this is less than the value that causes yielding, the buckling value becomes limiting, so that the maximum permissible value of W for L = 1.5 m is 16.4 kN.

Note: In case (a) the design was governed by material strength whereas in case (b) buckling governed the design. If there were several bars of different lengths and cross sectional dimensions, it would save significant time to calculate the slenderness ratio that separates long columns from short columns, i.e. the buckling failure regime from the yielding failure regime. Using \sigma _{cr} = 13.8 MPa in \sigma = \frac{\pi ^{2}E}{(L/k)^{2}} gives L/k = 94.2 as the threshold ratio separating

long columns from short columns. For case (a) the slenderness ratio is 83.1, so that it is classified as a short column and material strength governs W_{max·} For case (b) the slenderness ratio is 103.9 so that buckling governs W_{max·}.