Question

A cylindrical specimen of aluminum having a diameter of 19 mm and length of 200 mm is deformed elastically in tension with a force of 48,800 N Using the data contained in Table 6.1, determine the following:

(a) The amount by which this specimen will elongate in the direction of the applied stress.
(b) The change in diameter of the specimen. Will the diameter increase or decrease?

Answer 1

To determine the elongation and change in diameter of the cylindrical aluminum specimen, we can use the formulas for elongation in tension and Poisson's ratio.

Step-by-step explanation:

To determine the amount of elongation and change in diameter of the cylindrical aluminum specimen, we can use the formula for elongation in tension and the formula for Poisson's ratio:

(a) Elongation:

The formula for elongation in tension is: AL = (F × L0) / (A × Y)

Where:

• AL is the change in length
• F is the applied force
• L0 is the original length
• A is the cross-sectional area
• Y is the Young's modulus

Plugging in the given values, we have: AL = (48,800 N × 200 mm) / (π * (9.5 mm)^2 * (70 GPa))

(b) Change in diameter:

The Poisson's ratio (v) for aluminum is approximately 0.35. The formula for change in diameter is: Ad = -v(AL / L0)

Plugging in the calculated values, we have: Ad = -0.35 * (AL / L0)

Since the value of Ad is negative, the diameter of the specimen will decrease.