Question

1. a tie rod of cross - sectional area 0.002 m² carries a load of 200 kn. calculate the stress in mpa.
2. two rods a and b carry the same load of 40 kn. the area of rod a is 200 mm² and the area of rod b is 500 mm². calculate the stress in each rod and state

Explanation:

Problem 5:

Step 1: Recall the stress formula

Stress ($\sigma$) is calculated by the formula $\sigma = \frac{F}{A}$, where $F$ is the force (load) and $A$ is the cross - sectional area.
Given that $F = 200\space kN=200\times10^{3}\space N$ and $A = 0.002\space m^{2}$.

Step 2: Substitute the values into the formula

$\sigma=\frac{200\times 10^{3}\space N}{0.002\space m^{2}}$
First, calculate the numerator and denominator: $200\times10^{3}=200000$ and $0.002$
Then, $\frac{200000}{0.002}=100000000\space Pa$
Since $1\space MPa = 10^{6}\space Pa$, convert Pascals to Megapascals: $\frac{100000000}{10^{6}} = 100\space MPa$

Step 1: Recall the stress formula

Stress ($\sigma$) is given by $\sigma=\frac{F}{A}$, where $F$ is the load and $A$ is the cross - sectional area. The load $F = 40\space kN=40\times10^{3}\space N$ for both rods.

Step 2: Calculate stress in rod A

For rod A, $A_{A}=200\space mm^{2}=200\times 10^{-6}\space m^{2}$ (since $1\space mm^{2}=10^{-6}\space m^{2}$)
$\sigma_{A}=\frac{F}{A_{A}}=\frac{40\times 10^{3}\space N}{200\times 10^{-6}\space m^{2}}$
First, simplify the fraction: $\frac{40\times 10^{3}}{200\times 10^{-6}}=\frac{40}{200}\times10^{3 + 6}=\frac{1}{5}\times10^{9}=2\times 10^{8}\space Pa = 200\space MPa$ (since $1\space MPa = 10^{6}\space Pa$)

Step 3: Calculate stress in rod B

For rod B, $A_{B}=500\space mm^{2}=500\times 10^{-6}\space m^{2}$
$\sigma_{B}=\frac{F}{A_{B}}=\frac{40\times 10^{3}\space N}{500\times 10^{-6}\space m^{2}}$
Simplify the fraction: $\frac{40\times 10^{3}}{500\times 10^{-6}}=\frac{40}{500}\times10^{3+6}=\frac{4}{50}\times10^{9}=\frac{2}{25}\times10^{9}=8\times 10^{7}\space Pa = 80\space MPa$

Answer:

The stress in the tie rod is $100\space MPa$

Problem 6: