Problem 3.3. Suspended bar- additional end weight

An additional mass, M is hanged at the end of the bar given in the previous problem. Determine the strains of the bar at the end and at the midheight.

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Determine parametrically the strain at the end.

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$ε (L) = \frac{G g}{E A}$

Determine parametrically the strain at midheigth.

Check solution

$ε (\frac{L}{2}) = \frac{G g + ρ g A L / 2}{E A}$

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Steps

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Step 1. Give the differential equation and its general solution for a tensile rod.

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The differential equation and the general solution is equivalent to that of the previous problem (suspended rod subjected to its self-weight).

See Eq.(3-26)

$- E A \frac{d^{2} u^{}}{d x^{2}} = ρ g A$

$u = u_{hom} + u_{part} = C_{1} + C_{2} x - \frac{ρ g}{E} \frac{x^{2}}{2}$

Step 2. Determine the constants from the boundary conditions.

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At the support the displacement is zero, at the lower end the normal force is equal to the concentrated load.

$at x = 0 u = 0 \to C_{1} = 0 at x = L N = M g N (L) = E A (C_{2} - \frac{ρ g}{E} L) = M g \to C_{2} = \frac{M g + ρ g A L}{E A}$

Step 3. Give the displacement and strain functions. Calculate the values of the strain function at the end and at the midheight.

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$u = C_{2} x - \frac{ρ g}{E} \frac{x^{2}}{2} = \frac{M g + ρ g A L}{E A} x - \frac{ρ g}{E} \frac{x^{2}}{2} ε = C_{2} - \frac{ρ g}{E} x = \frac{M g + ρ g A L}{E A} - \frac{ρ g}{E} x ε (L) = \frac{M g + ρ g A L}{E A} - \frac{ρ g}{E} L = \frac{M g}{E A} ε (\frac{L}{2}) = C_{2} - \frac{ρ g}{E} x = \frac{M g + ρ g A L / 2}{E A}$

Step 4. Draw the strains and the displacement diagrams.

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Results

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The differential equation and the general solution is equivalent to that of the previous problem (suspended rod subjected to its self-weight).

See Eq.(3-26)

$- E A \frac{d^{2} u^{}}{d x^{2}} = ρ g A$

$u = u_{hom} + u_{part} = C_{1} + C_{2} x - \frac{ρ g}{E} \frac{x^{2}}{2}$

However, one of the the boundary conditions is different. At the support the displacement is zero, at the lower end the normal force is equal to the concentrated load. The constants are

$at x = 0 u = 0 \to C_{1} = 0 at x = L N = M g N (L) = E A (C_{2} - \frac{ρ g}{E} L) = M g \to C_{2} = \frac{M g + ρ g A L}{E A}$

The displacement and strain functions can be determined:

$u = C_{2} x - \frac{ρ g}{E} \frac{x^{2}}{2} = \frac{M g + ρ g A L}{E A} x - \frac{ρ g}{E} \frac{x^{2}}{2} ε = C_{2} - \frac{ρ g}{E} x = \frac{M g + ρ g A L}{E A} - \frac{ρ g}{E} x$

The values of the strain function at the end and at the midheight are given below.

$ε (L) = \frac{M g + ρ g A L}{E A} - \frac{ρ g}{E} L = \frac{M g}{E A} ε (\frac{L}{2}) = C_{2} - \frac{ρ g}{E} x = \frac{M g + ρ g A L / 2}{E A}$

the value of the strain functio

The strains and the displacement diagrams are given in the Figure.