For P = 5.2 kN, determine the factor of safety for the structure shown in Figure 8.6. Assume E = 200 GPa and consider buckling only in the plane of the structure.

Question

Length (m)	Diameter (mm)	Member
1.2	18.0	AB
1.2	18.0	AC
1.2 √2	22.0	BC

Accepted Answer

Let us draw the overall free-body diagram first as shown in Figure 8.7.

From the consideration of equilibrium,

[latex] \sum F_x=0=A_x+5200 \cos 70^{\circ} [/latex]

or [latex] A_x=-1778.5 N [/latex]

[latex] \sum F_y=0=A_y+C_y-5200 \sin 70^{\circ} [/latex]

or [latex] A_y+C_y=4886.4 N [/latex]

[latex] \sum M_{ A }=0=1.2 C_y-1.2\left(5200 \cos 70^{\circ} ight) [/latex]

or [latex] C_y=1778.5 N [/latex]

Substituting value of [latex] C_y ext {, we get } A_y [/latex] = 3107.9 N. Now we consider joint A, and its free-body diagram is shown in Figure 8.7(b). From the figure, we see that

[latex] F_{ AC }=1778.5 N ext { and } F_{ AB }=3107.9 N [/latex]

Thus, AC is in tension while AB is in compression. Similarly, consider the free-body diagram of joint B as shown in Figure 8.8.

From the figure, we can write

[latex] F_{ BC } \cos \frac{\pi}{4}=1778.5 [/latex]

or [latex] F_{ BC }=2515.18 N [/latex]

So, member BC is in compression.
In summary, AB and BC are in compression and the loads are 3107.9 N and 2515.18 N, respectively.
We need to check these loads for possible buckling of the respective members. Now, for AB, critical load for buckling is given by

[latex] P_{ Cr }=\left\lgroup \frac{\pi^2 E I}{L^2} ight group_{ AB } [/latex] [as both ends are pinned]

Substituting the values, we get

[latex] P_{ Cr }=\frac{\pi^2\left(200 imes 10^9 ight) imes \frac{\pi}{64} imes\left(18 imes 10^{-3} ight)^4}{(1.2)^2}=7063.62 N [/latex]

For BC, critical load for buckling can be similarly found out,

[latex] \begin{aligned} P_{ Cr } & =\left\lgroup \frac{\pi^2 E I}{L^2} ight group_{ BC } \ & =\frac{\pi^2\left(200 imes 10^9 ight) imes \frac{\pi}{64} imes\left(22 imes 10^{-3} ight)^4}{(1.2 \sqrt{2})^2} \ & =7881.3 N \end{aligned} [/latex]

Both these values are well above the compressive loads acting. So, the factor of safety for AB as far as buckling of the member is concerned, is

[latex] ext { Factor of safety for } AB =\frac{7063.62}{3107.9}=2.27 [/latex]

Similarly for BC,

[latex] ext { Factor of safety for } BC =\frac{7881.3}{2515.18}=3.13 [/latex]

The minimum of these two, that is, 2.27 is the desired factor of safety of the entire structure against buckling.

Question 8.2: For P = 5.2 kN, determine the factor of safety for the struc...

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