PLAIN AND CIVIL: EXAMPLE 2.1.2. DEFORMATION OF JOINED ELEMENTS

Saturday 16 March 2019

A rod consisting of two cylindrical portions AB and BC is restrained at both ends. AB is steel [E=200GPa, $\alpha =11.7\mu m/m^{o}C$ ] and BC is made f bronze [E=105GPa, $\alpha =21.6\mu m/m^{o}C$ ]. Knowing that the rod is initially unstressed, determine the following:

The compressive force induced in ABC with a temperature rise of $50^{o}C$
Change in length (free expansion) of steel
Change in length (free expansion of bronze
If a gap of 0.10mm is placed at point A, what compressive force is induced?

The formula for deformation (strain) will always be applied

$\delta _{total}=\delta _{load}+\delta _{temp}$

This total deformation is true as a free expansion. For every material, the total deformation incurred will be the summation of mechanical deformation, that is, the deformation due to the applied load, and the thermal deformation, that is deformation due to the change in temperature.

With two elements in the rod, the basic formula which is derived is:

$\delta _{total}=\delta _{total}_{steel}+\delta _{total}_{bronze}$

But the condition states that the rod is restrained. This would mean that $\delta _{total}=0$ . Thus the working equation becomes:

$0=\delta _{total}_{steel}+\delta _{total}_{bronze}$

So expanding the terms in order to come up with the basic parameters, the formula turns to be:

$0=\left [ \delta _{load} +\delta _{temp}\right ]_{steel}+\left [ \delta _{load} +\delta _{temp}\right ]_{bronze}$

where: $\delta _{load}=\frac{PL}{AE}$

$\delta _{temp}=\alpha \Delta TL$

So if we substitute these formulas into the working equation:

$0=\left [ \frac{PL}{AE} +\alpha \Delta TL\right ]_{steel}+\left [ \frac{PL}{AE} +\alpha \Delta TL\right ]_{bronze}$

By this equation, we can readily plug in the given parameters, leaving P, so the equation will be a function of the load P.

$0=\frac{P\left ( 0.25 \right )}{\frac{\pi }{4}0.03^{2}\left ( 200x10^{9} \right )}+11.7x10^{-6}\left ( 50 \right )\left ( 0.25 \right )+\frac{P\left ( 0.30 \right )}{\frac{\pi }{4}0.05^{2}\left ( 105x10^{9} \right )}+21.6x10^{-6}\left ( 50 \right )\left ( 0.30 \right )$

$0=3.2235x10^{-9}P+4.7025x10^{-4}$

${\color{Red} \mathbf{P=-145881.8055N\simeq -145.882kN}}$

The negative sign denotes compression.

By using the P to the individual material's mechanical deformation, you will arrive to the free expansion of each material.

For steel:

$\delta _{totalsteel}=\frac{PL}{AE}+\alpha \Delta TL$

$\delta _{totalsteel}=\frac{145881.8055\left ( 0.25 \right )}{\frac{\pi }{4}0.03^{2}\left ( 200x10^{9} \right )}+11.7x10^{-6}\left ( 50 \right )\left ( 0.25 \right )$

${\color{Red} \mathbf{\delta _{totalsteel}=4.0423x10^{-4} m}}$

Similar for bronze:

   $\delta _{totalbronze}=\frac{PL}{AE}+\alpha \Delta TL$

$\delta _{totalbronze}=\frac{145881.8055\left ( 0.30 \right )}{\frac{\pi }{4}0.05^{2}\left ( 105x10^{9} \right )}+21.6x10^{-6}\left ( 50 \right )\left ( 0.30 \right )$

${\color{Red} \mathbf{\delta _{totalbronze}=5.3627x10^{-4} m}}$

The last requirement requires a different condition. Instead of restraining both ends (making the total deformation=0, it states that a gap of 0.10mm is allowed. This makes our working equation:
   $0.0001=\left [ \delta _{load} +\delta _{temp}\right ]_{steel}+\left [ \delta _{load} +\delta _{temp}\right ]_{bronze}$

Nothing from the right side of the equation will change, everything stays as it is, just change 0 into 0.0001m.

   $0.0001=3.2235x10^{-9}P+4.7025x10^{-4}$

${\color{Red} \mathbf{P=-114859.6246N\simeq -114.860kN}}$

RELATED TOPIC:

2.0. CONCEPT OF STRAIN

PLAIN AND CIVIL

Pages

Saturday 16 March 2019

EXAMPLE 2.1.2. DEFORMATION OF JOINED ELEMENTS

No comments:

Post a Comment