stress & strain

[logistic_guy]

The bar \(\displaystyle DA\) is rigid and originally held in the horizontal position when the weight \(\displaystyle W\) is supported from \(\displaystyle C\). If the weight causes \(\displaystyle B\) to be displaced downward \(\displaystyle 0.025 \ \text{in}\), determine the strain in wires \(\displaystyle DE\) and \(\displaystyle BC\). Also, if the wires are made of \(\displaystyle \text{A-36}\) steel and have a cross-sectional area of \(\displaystyle 0.002 \ \text{in}^2\), determine the weight \(\displaystyle W\).

 

[logistic_guy]

When point \(\displaystyle B\) was displayed \(\displaystyle 0.025 \ \text{in}\) downward, it gave us an idea. Thanks to Geometry, we can find the change in length in the wire \(\displaystyle DE\).

\(\displaystyle \frac{\Delta L_{DE}}{0.025} = \frac{5 \times 12}{3 \times 12}\)

This gives:

\(\displaystyle \Delta L_{DE} = 0.04167 \ \text{in}\)

 

[logistic_guy]

\(\displaystyle \Delta L_{DE} = 0.04167 \ \text{in}\)

From previous exercises, we know that strain is given by:

\(\displaystyle \epsilon = \frac{\Delta L}{L}\)

Then, the strain in the wire \(\displaystyle DE\) is:

\(\displaystyle \epsilon_{DE} = \frac{\Delta L_{DE}}{L_{DE}} = \frac{0.04167}{3 \times 12} = \textcolor{blue}{0.0011575 \ \text{in/in}}\)

 

[logistic_guy]

There is a beautiful formula that relates stress and strain (Hooke's Law).

\(\displaystyle \sigma_{DE} = E \ \epsilon_{DE}\)

where \(\displaystyle E\) is Young's modulus.

For steel, \(\displaystyle E = 29000 \ \text{ksi}\).

Then,

\(\displaystyle \sigma_{DE} = E \ \epsilon_{DE} = 29000 \times 0.0011575 = 33.5675 \ \text{ksi}\)

 

[logistic_guy]

Stress and pressure have the same formula, then

\(\displaystyle \sigma_{DE} = \frac{F_{DE}}{A_{DE}}\)

Or

\(\displaystyle F_{DE} = \sigma_{DE}A_{DE} = 33.5675(0.002) = 0.067135 \ \text{kip}\)

 

[logistic_guy]

determine the weight \(\displaystyle W\).

Playing with the torque around the bar gives us:

\(\displaystyle -5F_{DE} + 3W = 0\)

Then, the weight is:

\(\displaystyle W = \frac{5F_{DE}}{3} = \frac{5(0.067135)}{3} = 0.11189 \ \text{kip} = \textcolor{blue}{111.89 \ \text{lb}}\)

 

[logistic_guy]

Stress and pressure have the same formula, then

\(\displaystyle \sigma_{BC} = \frac{F_{BC}}{A_{BC}} = \frac{W}{A_{BC}}\)

Plug in numbers.

\(\displaystyle \sigma_{BC} = \frac{W}{A_{BC}} = \frac{0.11189}{0.002} = 55.945 \ \text{ksi}\)

 

[logistic_guy]

Thanks to the Young's moduluswe finally can solve the last part.

\(\displaystyle \sigma_{BC} = \epsilon_{BC}E\)

determine the strain in wire \(\displaystyle BC\).

\(\displaystyle \epsilon_{BC} = \frac{\sigma_{BC}}{E} = \frac{55.945}{29000} = \textcolor{blue}{0.001929 \ \text{in/in}}\)