Question 8.2: For P = 5.2 kN, determine the factor of safety for the struc...
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.
| Length (m) | Diameter (mm) | Member |
| 1.2 | 18.0 | AB |
| 1.2 | 18.0 | AC |
| 1.2√2 | 22.0 | BC |
Let us draw the overall free-body diagram first as shown in Figure 8.7.
From the consideration of equilibrium,
\sum F_x=0=A_x+5200 \cos 70^{\circ}
or A_x=-1778.5 N
\sum F_y=0=A_y+C_y-5200 \sin 70^{\circ}
or A_y+C_y=4886.4 N
\sum M_{ A }=0=1.2 C_y-1.2\left(5200 \cos 70^{\circ}\right)
or C_y=1778.5 N
Substituting value of C_y \text {, we get } A_y = 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
F_{ AC }=1778.5 N \text { and } F_{ AB }=3107.9 N
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
F_{ BC } \cos \frac{\pi}{4}=1778.5
or F_{ BC }=2515.18 N
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
P_{ Cr }=\left\lgroup \frac{\pi^2 E I}{L^2} \right\rgroup_{ AB } [as both ends are pinned]
Substituting the values, we get
P_{ Cr }=\frac{\pi^2\left(200 \times 10^9\right) \times \frac{\pi}{64} \times\left(18 \times 10^{-3}\right)^4}{(1.2)^2}=7063.62 N
For BC, critical load for buckling can be similarly found out,
P_{ Cr }=\left\lgroup \frac{\pi^2 E I}{L^2} \right\rgroup_{ BC }
=\frac{\pi^2\left(200 \times 10^9\right) \times \frac{\pi}{64} \times\left(22 \times 10^{-3}\right)^4}{(1.2 \sqrt{2})^2}
= 7881.3 N
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
\text { Factor of safety for } A B=\frac{7063.62}{3107.9}=2.27
Similarly for BC,
\text { Factor of safety for BC }=\frac{7881.3}{2515.18}=3.13
The minimum of these two, that is, 2.27 is the desired factor of safety of the entire structure against buckling.
