A hollow tube surrounds a solid bar and both are supported by a rigid floor. Ref
ID: 3279468 • Letter: A
Question
A hollow tube surrounds a solid bar and both are supported by a rigid floor. Refer to the cut-away illustration below, which shows the bar inside the tube. A rigid plate rests on top of the bar. The tube is slightly shorter, with a small GAP existing between the plate and tube BEFORE any load (P) is applied.
(a) Determine the lateral strain in the bar if the load P is just sufficient to close the gap.
(b) Determine the change in length of the tube when the load P has the value shown in the table.
(c) Determine the Factor of Safety (FoS) in the bar for the same overall load P (from the table) with respect to the allowable stress also listed in the table.
(d) Determine the maximum load P which does not cause yielding in either component.
Bar material: Bronze
-Yield Strength= 50 ksi
-Modulus of Elasticity= 15,000 ksi
-Poisson's Ratio= 0.33
-Allowable Normal Stress= 35 ksi
Tube Material: Aluminum
-Yield Strength= 60 ksi
-Modulus of Elasticity= 10,000 ksi
-Poisson's Ratio= 0.35
-Allowable Normal Stress= 40 ksi
Gap: 0.008 inches
Force P: 70 kip
Tube Dimensions:
Length= 1.50 feet
ID= 3.1 inches
Thickness= 0.40 inches
Bar dimensions:
OD= 1.50 inches
A hollow tube surrounds a solid bar and both are supported by a rigid floor. Refer to the cut-away illustration below, which shows the bar inside the tube. A rigid plate rests on top of the bar. The tube is slightly shorter, with a small GAP existing between the plate and tube BEFORE any load (P) is applied. (a) Determine the lateral strain in the bar if the load P is just Include appropriate units: sufficient to close the gap. has the value shown in the table. overall load P (from the table) with respect to the allowable a) (b) Determine the change in length of the tube when the load P c) Determine the Factor of Safety (FoS) in the bar for the same stress also listed in the table. (d) Determine the maximum load P which does not cause yielding in either component. c) Fos = Problem Score d) P = GAP LENGTH OD ID THICKNESSExplanation / Answer
a. given, gap = 0.008 inches
length of rod, l = 1.5 ft = 1.5*12 = 18 in
so when the gap is closed, change in length of rod = dl = 0.008 in
longitudinal strain, sl = dl/l = 0.008/18 = 4.444*10^-4
now, for this material poissions ratio , p = 0.33
sol lateral strain = s2
s2/s1 = p
s2 = p*s1 = 0.33*4.444*10^-4 = 1.466*10^-4 = lateral strain in bar when the gap is just closed
b. Force given = 70 kip = P
let stress in bar = sigma1
stress in tube = sigma2
then sigma1*A1 + sigma2*A2 = P
A1 = pi*(OD)^2/4
OD = 1.5 in
so A1 = 1.7671 in^2
let thickness of tube be t
A2 = pi(ID + t)^2/4 - pi(ID)^2/4
ID = 3.1 in
t = 0.4 in
A2 = 2.0734 in^2
also, strain in bar = s1
strain in tube = s2
change in length of bar, dl1 = s1*l
change in length of tube, dl2 = s2*(l - d)
and dl2 + d= dl1
where d = gap = 0.008 in
l = 1.5*12 = 18 in
so if youngs modulus for bar = Y1 = 15,000 ksi
youings modulus for tube, Y2 = 10,000 ksi
then from hookes law
Y1 = sigma1/s1
Y2 = sigma2/s2
=> Y1*s1*A1 + Y2*s2*A2 = P
=> Y1*dl1*A1/l + Y2*dl2*A2/(l - d) = P
=> Y1*(dl2 + d)*A1/l + Y2*dl2*A2/(l - d) = P
=>15,000*(dl2 + 0.008)*1.7671/18 + 10,000*dl2*2.0734/(18 - 0.008) = 70
=>1472.588*dl2 + 11.780704 + 1152.4*dl2 = 70
=> dl2 = change in length of tube = 0.02217 IN
C. Stress in bar for the same load as in previous problem
sigma1 = Y1*s1 = Y1*dl1/l = 15,000*(0.02217+0.008)/18 = 25.141 ksi
allowed stress = 40 ksi
factor of safety = 40/25.141 = 1.59
D. let maximum load be P
Y1*s1*A1 + Y2*s2*A2 = P
as the rod has lower yield stress and has more strain ( always ), so it is the weaker link
hence sigma1 = 35 ksi
so, s1 = sigma1/Y1 = 35/15000 = 0.00233
so dl1 = s1*l = 0.042 in
so dl2 = dl1 - 0.008 = 0.034 in
15,000*(0.034 + 0.008)*1.7671/18 + 10,000*0.034*2.0734/(18 - 0.008) = P
=> 1472.588*0.034 + 11.780704 + 1152.4*0.034 = P
P = 101.030 kips
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