Question 2.4: A steel bar 10 m long used in a control mechanism must trans...
A steel bar 10 m long used in a control mechanism must transmit a tensile force of 5 kN without stretching more than 3 mm or exceeding an allowable stress of 150 MN/m² . What must the diameter of the bar be? State your answer to the nearest millimeter, and use E = 200 GPa.
Given: Dimensions and loading on steel bar.
Find: Required bar diameter to nearest mm.
Assume: Hooke’s law applies.
We will impose both strength and stiffness constraints on the bar and will see which is the limiting case.
Using the definition of normal stress, we must have
\sigma =\frac{P}{A}\leq 150\frac{\textrm{MN}}{\textrm{m}^2} ,
or
A≥ 150\frac{P}{150 \textrm{MN/m}^2}=\frac{5000 \textrm{N}}{150\times 10^6 \textrm{N/m}^2}=33.33 \times 10^{-6} \textrm{m}^2 ,
that is,
A ≥ 33.33 mm² .
If Hooke’s law applies, as we have assumed it does, then
\delta =\frac{PL}{AE} ,
and we must have
\frac{PL}{AE} \leq 3 mm
or
A≥ \frac{PL}{ \delta E} = \frac{5000 \textrm{N}\cdot 10 \textrm{m}}{(0.003 \textrm{m})(200\times 10^9 \textrm{N/m}^2)}= 83.33\times 10^{-6} \textrm{m}^2that is,
A ≥ 83.33 mm² .
We see that stiffness is the limiting case and that we must have a crosssectional area greater than or equal to 83.33 mm² to safely meet our constraint. This is all we need to find the required diameter of the steel bar:
\frac{\pi }{4}d^2≥83.33 \textrm{mm}^2so
d≥10.3 mm.
So to the nearest millimeter, we must use an 11-mm-diameter bar.