# Question 2.6: A tensile test is performed on a brass specimen 10 mm in dia......

A tensile test is performed on a brass specimen 10 mm in diameter using a gauge length of 50 mm. When applying axial tensile load of 25 kN, it was observed that the distance between the gauge marks increase by 0.152 mm, calculate modulus of elasticity of brass.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

The problem is sketched in Figure 2.19.

Modulus of elasticity is given by, $E=\frac{\sigma}{\varepsilon}=\frac{\text { stress }}{\text { strain }}$

\begin{aligned} & \text { Stress }=\frac{P}{A}=\frac{25 \times 10^3}{\pi\left(10^2 / 4\right)}=318.3 N / mm ^2 \\ & \text { Strain }=\varepsilon=\frac{d l}{l}=\frac{0.152}{50}=3.04 \times 10^{-3} \\ & E=\frac{318.30}{3.04 \times 10^{-3}}=104703.95 N / mm ^2 \end{aligned}

Question: 2.22

## Verified Answer:

For the external equilibrium, algebraic sum of hor...
Question: 2.1

## Verified Answer:

As it is clear from Figure 2.15 that the portions ...
Question: 2.2

## Verified Answer:

This problem is also straightforward as the previo...
Question: 2.3

## Verified Answer:

The system shown in Figure 2.17 is in equilibrium....
Question: 2.4

## Verified Answer:

We know that the change in length of a prismatic c...
Question: 2.5

## Verified Answer:

The assembly is in equilibrium. The free-body diag...
Question: 2.7

## Verified Answer:

First, we shall calculate longitudinal strain [lat...
Question: 2.9

## Verified Answer:

The steel pipe is shown in Figure 2.20. (a) To cal...
Question: 2.10

## Verified Answer:

To prove this, we shall consider a cube of a mater...
Question: 2.11

## Verified Answer:

The top covering plate will always have the tenden...