"Consider a cylindrical specimen of a steel alloy..." solved and explained

High School
Other

Pamela Meyer

Answered question

2022-01-10

Consider a cylindrical specimen of a steel alloy 8.5 mm (0.33 in.) in diameter and 80 mm (3,15 in) long that is pulled in tension. Determine its elongation when a load of 65,250 N

(14, 500 l b_{f})

is applied.

Answer & Explanation

Donald Cheek

Beginner2022-01-11Added 41 answers

The equation for stress is:

σ = \frac{F}{A}

The area if a cylinder is given by:

A = π {(\frac{d}{2})}^{2}

Which results in:

σ = \frac{F}{π {(\frac{d}{2})}^{2}} = \frac{65250}{π {(\frac{3.5 \times 10^{- 3}}{2})}^{2}} = 678 M P A

Refer to Fig 6.22 on page 208 we can see the associated stress is 0.00125. We can find the amount elongated from:

△ L = ϵ L_{0} = 0.00125 (80) = 0.1 m m

sonSnubsreose6v

Beginner2022-01-12Added 21 answers

This problemasks that we calculate the elongation $△ l$ of a specimen of steel the stress-strain behavior of which is shown in Figure 6.21. First it becomes necessary to compute the stress when a load of 65,250 N is applied using Equation 6.1 as
$σ = \frac{F}{A_{0}} = \frac{F}{π {(\frac{d_{0}}{2})}^{2}} = \frac{65.250 N}{π {(\frac{8.5 \times 10^{- 3} m}{2})}^{2}} = 1150 M P a (170.000 ψ)$
Referring to Figure 6.21, at this stress level we are in the elastic region on the stress-strain curve, which corresponds to a strain of 0.0054. Now, utilization of Equation 6.2 to compute the value of $△ l$
$△ l = ϵ l_{0} = (0.0054) (80 m m) = 0.43 m m (0.017 i n .)$

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