Question

Part A Review The rod is made of A-36 steel and has a diameter of 0.29 in. If the rod is 4 ft long when the springs are compressed 0.7 in. and the temperature of the rod is T 60 °F, deterihine the force in the rod when its temperature is T 120 F. (Figure 1) nν ΑΣφ vec F Figure of 1 Submit Request Answer Provide Feedback Next> k=1000 Ib/in. k 1000 lb/in. 4 ft

Answer 1

Ans) Given,

Diameter of rod = 0.29 in

Length of rod (L) = 4 ft

Compression in spring = 0.70 in

Spring constant of both spings (K) = 1000 lb/in

Due to increase in temperature the rod will expand but at a same time due to spring , it experience compression .So compatibility equation for the system is :

x = $\delta$ _T - $\delta$ _F ( expansion as positive)

$\delta$_T = L $\alpha\Delta$ T ,

where , $\alpha$ = coefficient of thermal expansion of steel = 6.60 x 10^-6 / F

$\Delta$ T = change in temperature = 120 - 60 = 60 F

=> $\delta$ _T = (2 x 12) x 6.60 x 10^-6 x 60

= 0.0095 in  

Now, compression due to force in spring ,

$\delta$_F = K a L / A E

where, a = compression in spring

L = length of rod

A = area of rod = ($\pi$/4)(0.29)² = 0.066 in²

E = Modulus of elasticity of steel

NOTE : Since, spring is on both side consider length of rod is half = L/2 = 2 ft in both above cases

Putting values,

$\delta$_F = (1000 x 0.5) (2 x 12) / (0.066 x 29 x 10⁶)

= 0.00627 in

=> x = 0.0095 - 0.00627

= 0.00323 in

Force in rod = K (x + a)

= 1000 ( 0.00323 + 0.5)

= 503.23 lb

or 0.503 kip