Question 1.5: An elastic spring rests on a base plate that is on top of ri...
An elastic spring rests on a base plate that is on top of rigid tube B (see Fig.1-23). The spring is enclosed by rigid tube A but is longer than tube A by an amount s. Force P is then applied to a cap plate to compress the spring. Both tubes have outer diameter d_{O}, but the inner diameters are d_{A} and d_{B} for tubes A and B, respectively. Assume that spring stiffness k = 24 kips / in., d_{O} = 3 in. , d_{A} = 2.5 in. , d_{B} = 2.25 in., and s = 0.125 in.
(a) If applied load P = 2500 lb, what are the axial normal stresses in tubes A and B?
(b) Repeat part (a) if P = 5000 lb.
(c) What is P if the normal stress in tube A is 800 psi? What is the associated stress in tube B?
Use the following four-step problem-solving approach.
1. Conceptualize : The two possible states of the assemblage are shown in the freebody diagrams in Fig.1-24. In Fig.1-24a, an upper-section cut through both the spring and tube A creates an upper free-body diagram that reveals a spring force (k)(x) for the case of downward cap displacement x that is less than gap width s. In Fig.1-24b, cap displacement x is equal to gap width s, so the spring force now equals (k)(s). Figure1-24 also shows lower free-body diagrams for both cases in which a section cut through tube B shows that the internal compressive force in tube B is equal to applied load P. (Internal forces in tubes A and B are shown as two arrows, one at each tube wall, indicating that N_{A} and N_{B} are actually uniformly distributed forces acting on the circular cross section of each tube.)
2. Categorize : The force P required to close gap s is (k)(s). This is also the maximum force that can be developed in the spring. If applied force P is too small to close the gap s, force P will be transferred to the base plate and into rigid tube B; tube A will be unaffected by the load. However, if force P is large enough to compress the spring to close the gap s, the spring and tube A will share the load P applied to the cap plate and together will transfer it to tube B through the base plate. In summary, the free-body diagrams in Fig.1-24 show that, if the spring is compressed by load P an amount x, the compressive internal forces in the spring and the two tubes are
N_{s}=P=k x , N_{A}=0, N_{B}=P for x < s
N_{s}=k s, N_{A}=P-k s, N_{B}=N_{s}+N_{A}=P for x = s
3. Analyze :
Force P required to close gap s: The gap closes when force P is equal to ks
P=k s=\left(24 \frac{\text { kips }}{\text { in. }}\right)(0.125 \text { in. })=3000 lbTube stresses for applied load P = 2500 lb: The cap will displace downward a distance x = P/k = 0.104 in. (< s), so tube internal forces are N_{A} = 0 and N_{B} = P. Tube cross-sectional areas are
A_{A}=\frac{\pi}{4}\left(d_{O}^{2}-d_{A}^{2}\right)=\frac{\pi}{4}\left(3^{2}-2.5^{2}\right) in ^{2}=2.16 in ^{2}A_{B}=\frac{\pi}{4}\left(d_{O}^{2}-d_{B}^{2}\right)=\frac{\pi}{4}\left(3^{2}-2.25^{2}\right) in ^{2}=3.093 in ^{2}
The resulting axial normal compressive stresses in tubes A and B are
\sigma_{A}=\frac{N_{A}}{A_{A}}=0 \quad \sigma_{B}=\frac{N_{B}}{A_{B}}=808 psiTube stresses for applied load P = 5000 lb: Cap downward displacement is now x = P/k = 0.208 in. (> s), so tube internal forces are N_{A} = P – (k)(s) = (5000 – 3000) lb = 2000 lb and N_{B} = P. The normal stresses in tubes A and B are now:
\sigma_{A}=\frac{2000 lb }{2.16 in ^{2}}=926 psi \quad \sigma_{B}=\frac{5000 lb }{3.093 in ^{2}}=1617 psiApplied load P if stress in tube A is 800 psi: Force P must exceed (k)(s) = 3000 lb for the gap to close, leading to a force in tube A and a normal stress of \sigma_{A}=800 psi. The normal compressive force in tube A is N_{A}=\left(\sigma_{A}\right)\left(A_{A}\right)=1728 lb. It follows that applied force P is now P=N_{A}+k s=1728 lb +3000 lb =4728 lb. Internal force N_{B} = P, so the normal compressive stress in tube B is now \sigma_{B}=\frac{4728 lb }{3.093 in ^{2}}=1529 psi
4. Finalize : If tube A is elastic instead of rigid as assumed here, tube A can be modeled as another spring that is parallel to the spring it encloses. Now a more advanced analysis procedure will be needed to find tube force N_{A} for the case of P > (k)(s). Force N_{A} is no longer equal to P – (k)(s), and downward displacement x can be larger than s.
