Question

A tie-bar has a cross-section area of 125mm² and is subjected to a pull of 10kN. Calculate the stress in Megapascals​

Answer 1

The stress in Magapascals is 80 Magapascals

This question can be solved by applying Young's modulus.

Young's modulus : Young's modulus states that, stress  is directly proportional to strain.

Where Stress is pressure acting the body. The s.i unit of stress is N/m² or Pascal. And it can be expressed mathematically as,

• $P = F/A$................. Equation 1

• Let: P = Stress, F = Force on the tie-bar, A = cross section area of the tie-bar

From the question,

• Given$F = 10 kN$ $= 10000 N$, $A = 125 mm^{2}$ $= (125/10^6)$ $= 1.25*10^{-6} m^{2}$

• Substitute these values into equation 1

$P = 10000/(125*10^{-6} )$

$P =$ $8*10^{7}$ Pascals

$P =$ 80 Magapascals

Hence the stress in Magapascals is 80 Magapascals

Learn more about stress here : https://brainly.com/question/18836872

Answer 2

The stress experimented by the tie-bar is 80 megapascals.

The tie-bar is under axial load. Under the assumption that force is distributed uniformly in the cross-section area, we can use the following definition of normal stress ($\sigma$), in megapascals:

$\sigma = \frac{F}{A}$ (1)

Where:

$F$ - Axial force, in meganewtons.

$A$ - Cross-section area, in square meters.

If we know that $F = 10\times 10^{-3}\,MN$ and $A = 125\times 10^{-6}\,m^{2}$, then the stress experimented by the tie-bar is:

$\sigma = \frac{10\times 10^{-3}\,MN}{125\times 10^{-6}\,m^{2}}$

$\sigma = 80\,MPa$

The stress experimented by the tie-bar is 80 megapascals.