To determine the diameter of a cable that can carry a tensile force, we can use the following relationship between the tensile force, the shearing stress, and the cross-sectional area of the cable.

Step 1: Formula for tensile stress

The tensile force $F$ and the cross-sectional area $A$ of the cable are related by the following equation:

$\tau = \frac{F}{A}$

Where:

• $F$ is the tensile force (in Newtons),
• $A$ is the cross-sectional area (in square meters),
• $\tau$ is the shearing stress (in Pascals, $Pa$).

The cross-sectional area $A$ for a circular cable is given by the area of a circle:

$A = \frac{\pi d^2}{4}$

Where $d$ is the diameter of the cable.

Step 2: Solving for the diameter

From the equation for the shearing stress:

$\tau = \frac{F}{A}$

We can substitute $A = \frac{\pi d^2}{4}$ into this equation:

$\tau = \frac{4F}{\pi d^2}$

Rearranging the equation to solve for the diameter $d$:

$d = \sqrt{\frac{4F}{\pi \tau}}$

Step 3: Substituting the given values

We are given:

• $F = 1,468 \, \text{kN} = 1,468 \times 10^3 \, \text{N}$,
• $\tau = 115,000 \, \text{kPa} = 115,000 \times 10^3 \, \text{Pa}$.

Now substitute these values into the equation for the diameter:

$d = \sqrt{\frac{4 \times 1,468 \times 10^3}{\pi \times 115,000 \times 10^3}}$

I'll calculate that now.The diameter of the cable that can carry a tensile force of 1,468 kN with a unit shearing stress of 115,000 kPa is approximately $0.1275$ meters, or $127.5$ millimeters.

Would you like further details or have any questions? Here are 5 related questions to explore:

1. How would the cable's diameter change if the tensile force increased to 2,000 kN?
2. What is the relationship between tensile force and shearing stress for different materials?
3. How can we calculate the maximum allowable stress for a given cable diameter?
4. What is the importance of material properties (like elasticity) in determining tensile stress?
5. How would thermal expansion affect the tensile force in cables?

Tip: Always ensure units are consistent when solving engineering problems to avoid calculation errors.