Question

Consider a bar having the following properties, a = 6.6x10-06/°F, E = 29x100 psi, having a diameter of 0.25 in and length 2 ft is fixed from one end and other end is attached to a spring as shown below: mm The spring constant is k=1000 lb/in, the temperature is raised by 40°F. Find the force developed in the bar
a) 6.23 lb
b) 10.29 lb
c) 0 lb
d) 2.33 lb

Answer 1

Answer:

Consider a bar having the following properties, ... a diameter of 0.25 in and length 2 ft is fixed from one end and - 26922…

D)2.33ib

Answer 2

Answer:

(a)Force developed in bar is 6.23lb

Explanation:

Given data is

Diameter of bar(D)=0.25in

Length of bar(L)=2ft

Spring constant(k)=1000lb/in

Temperature raised(∆T)=40°F

When temperature of bar increases, spring is compressed and force developed in rod will be compressive in nature

let ∆ be the expansion in rod and F be the force developed

$∆ = Lα∆T - \frac{FL}{AE} \\ since \: F = k∆ = > ∆ = \frac{F}{k} \\ = > \frac{F}{k} = Lα(∆T) - \frac{FL}{AE} \\ = > F( \frac{1}{k} + \frac{L}{AE} ) = Lα(∆T) \\ = > F = \frac{Lα(∆T)}{\frac{1}{k} + \frac{L}{AE} }$

By substituting the known values we get

$F = \frac{24 \times 6.6 \times 10 {}^{ - 6} \times 40 }{ \frac{1}{1000} + \frac{24}{ \frac{\pi}{4} \times 0.25 {}^{2} \times 29 \times10 {}^{6} } } = 6.23lb$

Therefore the force developed in the bar is 6.23lb(a)

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