The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter
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Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
<h3>The static equilibrium is given as:</h3>
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area =
d = 1.5
= 1.767 inches²
<h3>Solve for inertia</h3>
= 0.2485inches⁴
Solve for the tensile force from here
30x10³ =
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
Read more on tensile force here: brainly.com/question/25748369