Question

1. a material can withstand a maximum stress of 100 n/mm². if its cross - sectional area is 8 cm², calculate the maximum load it can carry in kn.
2. a compressive load of 1.5 mn acts on a concrete column of cross - section 0.25 m². find the compressive stress developed in mpa.
3. a rod is required to carry a load of 90 kn without exceeding a stress of 120 mpa. calculate the minimum cross - sectional area required in mm².
4. a wire of diameter 10 mm carries a tensile load of 25 kn. calculate the stress in the wire.

Explanation:

Problem 7:

Step 1: Recall the stress formula

Stress ($\sigma$) is given by the formula $\sigma = \frac{F}{A}$, where $F$ is the load and $A$ is the cross - sectional area. We need to find $F$, so we can rearrange the formula to $F=\sigma\times A$.

Step 2: Convert units

The stress $\sigma = 100\ N/mm^{2}$ and the area $A = 8\ cm^{2}$. First, convert the area from $cm^{2}$ to $mm^{2}$. Since $1\ cm = 10\ mm$, then $1\ cm^{2}=(10\ mm)^{2}=100\ mm^{2}$. So, $A = 8\ cm^{2}\times100\ mm^{2}/cm^{2}=800\ mm^{2}$.

Step 3: Calculate the load

Using the formula $F=\sigma\times A$, substitute $\sigma = 100\ N/mm^{2}$ and $A = 800\ mm^{2}$. Then $F=100\ N/mm^{2}\times800\ mm^{2}=80000\ N$. Convert this to $kN$ (since $1\ kN = 1000\ N$), so $F=\frac{80000\ N}{1000\ N/kN}=80\ kN$.

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.

Step 2: Convert units

The load $F = 1.5\ MN=1.5\times10^{6}\ N$ and the area $A = 0.25\ m^{2}$. First, convert the area to $mm^{2}$. Since $1\ m = 1000\ mm$, then $1\ m^{2}=(1000\ mm)^{2}=10^{6}\ mm^{2}$. So, $A = 0.25\ m^{2}\times10^{6}\ mm^{2}/m^{2}=250000\ mm^{2}$. And $F = 1.5\times10^{6}\ N$.

Step 3: Calculate the stress

Using the formula $\sigma=\frac{F}{A}$, substitute $F = 1.5\times10^{6}\ N$ and $A = 250000\ mm^{2}$. Then $\sigma=\frac{1.5\times10^{6}\ N}{250000\ mm^{2}} = 6\ N/mm^{2}$. Since $1\ N/mm^{2}=1\ MPa$, the stress is $6\ MPa$.

Step 1: Recall the stress formula

Stress ($\sigma$) is given by $\sigma=\frac{F}{A}$, and we need to find the area $A$, so we can rearrange the formula to $A=\frac{F}{\sigma}$.

Step 2: Convert units

The load $F = 90\ kN = 90\times10^{3}\ N$ and the stress $\sigma = 120\ MPa=120\ N/mm^{2}$.

Step 3: Calculate the area

Using the formula $A=\frac{F}{\sigma}$, substitute $F = 90\times10^{3}\ N$ and $\sigma = 120\ N/mm^{2}$. Then $A=\frac{90\times10^{3}\ N}{120\ N/mm^{2}}=750\ mm^{2}$.

Answer:

$80\ kN$

Problem 8: