Notes on Simple Stress – Mech 313 (Introduction to Strength of Materials)
Introduction
Strength of materials (SOM) definition:
When an external force acts on a body, the body tends to deform.
Internal resisting forces develop due to cohesion between molecules, opposing deformation. This phenomenon is the essence of strength of materials.
Elastic limit concept:
Within the elastic limit, resistance offered by the material is proportional to the deformation caused by the external force.
Within the elastic limit, the resistance is equal to the external force (i.e., load).
Beyond the elastic stage:
The resistance offered by the material is less than the applied load.
Deformation continues until failure occurs.
Stress definition (within elastic stage):
The resisting force per unit area is called stress.
In symbols: stress is the ratio of the applied force to the resisting area.
Historical Development
Strength of materials dates back to ancient times (design of bridges, buildings, tools) but formal science emerged in the 17th century.
1638: Galileo Galilei studied breaking strength of materials.
18th century: Robert Hooke formulated Hooke's Law (linear stress–strain relation within elastic limits).
Leonhard Euler contributed to theories on column buckling.
19th century: Augustin-Louis Cauchy advanced the mathematical foundations of stress and strain analysis.
Industrial Revolution spurred safer, stronger structures and major field advancements.
Modern context: Strength of Materials is essential in the design of buildings, machines, vehicles, and other structures.
Purpose and Key Concepts
One of the basic problems in engineering: select proper material and proportion it to enable a structure or machine to function efficiently.
This requires determining material properties: strength, stiffness, etc.
Stress vs. Strain:
Stress is associated with the strength of the material.
Strain is a measure of the deformation of the body.
Simple Stresses (Definition and Units)
Simple stress is expressed as the ratio of the applied force to the resisting area:
S = rac{P}{A}
Where:
P = external force or load
A = cross-sectional area
Units: stress has the unit of force per area.
ext{Unit of stress} = rac{N}{m^{2}} = ext{Pa}
Common engineering units: rac{N}{mm^{2}} = ext{MPa}
Classification of Simple Stress
1) Normal stress: develops when a force is applied perpendicular to the cross-sectional area of the material.
a) Tensile stress – force that pulls the material (pulling apart).
b) Compressive stress – material is compressed by two opposing forces.
2) Shear stress – develops when the applied force is parallel to the resisting area.
3) Bearing stress – contact pressure between two bodies.
Illustrative Example (Visual Reference)
Good example often cited: Suspension Bridge (illustrates load transfer and stresses in structural elements).
Contextual note: The interaction of forces in real structures demonstrates normal, shear, and bearing stresses in practice.
Normal Stress: Details and Variables
Where:
P = the applied normal load in Newtons (N)
A = the cross-sectional area in square millimeters (mm^2)
Relationship basics:
S_{normal} = rac{P}{A}
Unit considerations: load in N, area in mm^2, yielding stress in MPa when using N/mm^2.
Shear Stress: Details
Definition:
Shear stress arises when shear force acts parallel to the surface area being sheared.
Where:
V = the resultant shearing force that passes through the centroid (N)
A = area being sheared (mm^2)
Relationship: S_{shear} = rac{V}{A} (not explicitly shown in the transcript, but implied by the standard definition of shear stress)
Bearing Stress: Details
Definition:
Bearing stress is the contact pressure between two bodies in contact.
Common example: pressure between a rivet/bolt and the plate surface that it presses against.
Relevance: governs design of joints and contact interfaces.
Problem Solving (Normal Stress) – Practice Problems
Problem 1:
A hollow steel tube with an inside diameter of 100~ ext{mm} must carry a tensile load of 400~ ext{kN}.
Determine the outside diameter of the tube if the allowable stress is limited to 120~ ext{MN/m}^{2}.
Given data to use:
Inner diameter: d_i = 100~ ext{mm}
Load: P = 400~ ext{kN} = 4.0 imes 10^{5}~ ext{N}
Allowable stress: au_{allow} = 120~ ext{MPa} = 120~ ext{N/mm}^{2} (since 1~ ext{MPa} = 1~ ext{N/mm}^2)
Approach (outline): compute required outer cross-sectional area from $$S = rac{P}{A} \