How to Determine the Minimum Diameter of an Alloy Cable for a Specific Load?

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SUMMARY

The minimum diameter of an alloy cable required to support a load of 15 kN, given a tensile strength of 75 MPa, is calculated to be 1.6 cm. The tensile strength is defined as 75 MPa, which equates to 75,000,000 Newtons per square meter. To determine the diameter, one must understand that tensile strength represents stress, or force per unit area, and apply the formula for stress to find the necessary cross-sectional area to support the specified load.

PREREQUISITES
  • Understanding of tensile strength and stress calculations
  • Familiarity with basic physics concepts related to force and area
  • Knowledge of units of measurement, particularly Newtons and square meters
  • Ability to perform algebraic manipulations to solve for diameter
NEXT STEPS
  • Review calculations for tensile strength and stress in materials
  • Learn about the relationship between force, area, and diameter in structural engineering
  • Investigate different materials and their tensile strengths for cable applications
  • Explore engineering design principles for selecting cable sizes based on load requirements
USEFUL FOR

Engineers, material scientists, and students studying structural mechanics or materials engineering who need to determine appropriate cable sizes for load-bearing applications.

benhorris
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Having a bit of a problem with this question,

Find the minimum diameter of an alloy cable, tensile strength 75 MPa, needed to support a load of 15 kN.

I know the answer is 1.6cm but i need someone to help explain how to get it.

Ive got this far...

Tensile strength of cable = 75 MPa = 75 000 000 Newtons per square meter
= 7.5 x 10^7 m^-2
The weight material required to hold = 15kN = 15 000 N
 
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Look at the units.

Tensile Strength is a measure of stress, or force per unit area.
 

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