Question

If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the cylinder. Neglect the mass of the cylinder. Express your answer to three significant figures and include appropriate units.

Answer 1

This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

 

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = $\frac{1}{2}$mv²

we substitute

⇒ T = $\frac{1}{2}$ × (550/32.2) × (2)²

T = 34.16149 lb.ft

T =  ( 34.16149 × 12 ) lb.in  

T = 409.93788 lb.in

Now, the volume will be;

V = $\frac{\pi }{4}$d²L

from the diagram; d = 0.5 ft and L = 1.5 ft

so we substitute

V =  $\frac{\pi }{4}$ × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² =  5,312,794,924.8

σ² = 10,438,982.5967  

σ = √10,438,982.5967

σ = 3230.9414  

σ = 3.2309 ksi  ≈ 3.23 ksi    { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi