Question 9.1.5: The nonplanar frame shown in Figure 1 is anchored by a ball-...
NONPLANAR FRAME ANALYSIS
The nonplanar frame shown in Figure 1 is anchored by a ball-and-socket joint at E. It is prevented from rotating by the links attached to the ground at joints A, B, and C. Assume the weight of the frame members can be ignored.
(a) Find an expression for the forces acting on members AB, AD, and AE.
(b) For a = 1.0 m, b = 1.0 m, P_{1} = 15.0 kN, P_{2} = 15.0 kN, find the forces in members AB, AD, and AE.
(c) Define the safety factor as the ratio: SF = \frac{force capacity of a member}{force in a member}
The force capacity of a member is the force at which the member fails, and is a function of the member’s cross-sectional area and the material from which it is constructed. If the force capacity of members AB, AD, and AE is an axial force of 20 kN, calculate safety factors for these members for the conditions defined in (b). Confirm that this ratio is greater than unity for each of the three members (this means that failure is not predicted in the member).
Goal Find expressions for the forces acting on members AB, AD, and AE (b). Also calculate the safety factors for these members (c).
Given Information about the supports and geometry of the nonplanar frame and a definition of safety factor. An axial force of 20 kN causes failure in members AB, AD, and AE.
Assume members AB, BC, AD, AE, BE, DE, and CE act as two-force members because they are pinned at both ends and no forces act on them between the ends. Note that member CD is not a two-force member because a force P_{2} acts along its length.
Draw We draw a free-body diagram of the entire frame (Figure 2).
Formulate Equations and Solve To find the forces in members AB, AD, and AE, we need to first calculate the six unknown forces at the supports, as drawn in Figure 2. using the six equilibrium equations we would find that:
F_{A}=- \frac{P_{2}}{2\cos 45^{\circ }} , F_{Bx}=-P_{1} + \frac{a}{b}\frac{P_{2}}{2} , F_{Cy}= – \frac{P_{2}}{2}
F_{Ex}=- \frac{a}{b}\frac{P_{2}}{2} , F_{Ey}=0 , F_{Ez}=\frac{P_{2}}{2}
(a) We analyze joint A to find the forces acting on members AB, AD, and AE (Figure 3). We have assumed that all of the members are in tension and thus pull away from the joint.
Equilibrium at joint A requires that:
\sum{F} =P_{1} + F_{A}+ F_{AB}+ F_{AD}+ F_{AE}=0
where
P_{1}=P_{1}i F_{A}=- \frac{P_{2}}{2j}- \frac{P_{2}}{2 k}
F_{AB} = – F_{AB} i F_{AD}=F_{AD}j
F_{AE}= – F_{AE}\frac{a}{\sqrt{a^{2}+ b^{2}} }i+ F_{AE}\frac{b}{\sqrt{a^{2}+ b^{2}}} k (Figure 4)
Three equations result from applying equilibrium in the i, j, and k directions:
In the i direction: P_{1} – F_{AB} – F_{AE}\frac{a}{\sqrt{a^{2}+ b^{2}} } =0
In the j direction: – \frac{P_{2}}{2} + F_{AD}=0
In the k direction: – \frac{P_{2}}{2} + F_{AE}\frac{b}{\sqrt{a^{2}+ b^{2}}} =0
We solve these for F_{AB} , F_{AD} , andF_{AE} :
F_{AB}=P_{1}- \frac{a}{b} \frac{P_{2}}{2} F_{AD}=\frac{P_{2}}{2} F_{AE}=\frac{P_{2}}{2}\frac{\sqrt{a^{2}+ b^{2}}}{b}
(b) We now consider a specific geometry and loading of the frame; namely, a = 1.0 m, b = 1.0 m, and P_{1} = P_{2} = 15.0 kN. The forces in members AB, AD, and AE are found from substituting into the results of (a) giving
F_{AB}=7.5 kN F_{AD}= 7.5 kN F_{AE}=\frac{15}{2} kN
(c) We have been told that the force capacity of each member AB, AD, and AE is 20 kN (we will assume that this is its capacity in both tension and compression). Therefore the safety factor in the given loading is:
\left(Safety factor\right) _{AB} =SF_{AB} =\frac{20 kN}{7.5 kN}\Rightarrow SF_{AB}=2.7
SF_{AD}=\frac{20 kN}{7.5 kN}\Rightarrow SF_{AD}=2.7
SF_{AE}=\frac{20 kN}{\frac{15}{\sqrt{2} }}\Rightarrow SF_{AE}=1.9
This ratio is greater than unity for each member, which means failure is not predicted in any of these members.
