Stress and Deformation

Lecture Details

  • Instructor: Hongsoo Choi, Ph. D.
  • Email: hongsoochoi@umass.edu
  • Semester: Spring 2026
  • Course: BMED-ENG 241: Strength of Materials
  • Lecture Date: February 2, 2026
  • Lecture Topic: Stress

Basics of Classical Mechanics

Mechanical Behavior of Materials

  • Material Behavior: Refers to the material’s response to applied forces and residual forces.
  • Deformation Definition:
    • Deformation occurs when external forces act on a material, causing its atoms to move from their equilibrium positions.
    • Energy is required for displacement or separation of atoms from equilibrium, provided by applied forces.
    • Tension Effect: When subjected to tension, atoms are drawn apart, activating interatomic attractive forces.
    • Compression Effect: When subjected to compression, atoms are pushed closer, resulting in interatomic repulsive forces.

Key Concepts in Mechanics

  • Strength: Evaluates if an object can withstand applied loads without breaking, fracturing, or failing under repeated applications.
  • Stiffness: Examines whether an object will deform excessively to the point of being unable to perform its intended function.
  • Stability: Considers whether an object will bend or buckle under heightened loads, thus ceasing to perform properly.

Normal Stress Under Axial Loading

  • Definitions:
    • A: Cross-sectional area of the bar.
  • Sign Convention for Normal Stresses:
    • Positive Sign: Indicates tensile normal stress.
    • Negative Sign: Indicates compressive normal stress.
  • Stress Calculation Framework: Consider a small area $ riangle{A}$ on the exposed cross-section of the bar:
    • Let $ riangle{F}$ be the resultant of the internal forces transmitted in $ riangle{A}$.
    • If internal forces are uniformly distributed, stress at a point is defined as $ ext{Stress} = rac{ riangle{F}}{ riangle{A}}$.

Stress Units

  • Definition: Stress is defined as force per unit area.
  • Units in Different Systems:
    • U.S. customary units: Stress expressed in pounds per square inch (psi) or kips per square inch (ksi), with 1 kip = 1,000 lb.
    • International System of Units (SI): Stress expressed in pascals (Pa), where extPa=extNextm2ext{Pa} = \frac{ ext{N}}{ ext{m}^2} (N = newtons, m² = square meters).

Example 1.1: Calculation of Normal Stress

  • Scenario Setup:
    • A solid steel hanger rod with a diameter of 0.5 in. (disregarding its weight) is used to support a walkway beam.
    • The force carried by the rod is 5,000 lb.
  • Normal Stress Calculation:
    • Given: Diameter d = 0.5 ext{ in.}
      ightarrow d = rac{0.5}{12} ext{ ft}
    • Stress formula:
      extStress=FAext{Stress} = \frac{F}{A}
    • Area A=extπd24A = \frac{ ext{π}d^2}{4}
    • Substitute given values and compute:
    • Importance of Significant Digits: The calculated stress value may have 10 significant digits, but it should be rounded off to three or four significant digits based on standard conventions.

Example 1.2: Analysis of Rigid Bar

  • Overview: Rigorous analysis of a rigid bar ABC supported by a pin at A and an axial member with area 540 mm², neglecting the weight of the bar.

  • Load condition: A load of P=8extkNP = 8 ext{ kN} is applied at C.

  • Components of Analysis:

    • A free-body diagram (FBD) assessment needs to be performed to establish moment equilibrium around pin A:
    • Fxi=0F_{xi} = 0
    • Fyi=0F_{yi} = 0
    • Mzi=0M_{zi} = 0

  • a. Normal Stress Calculation:

    • To find stress in member (1), axial force must first be computed from the moment equation:
      extStress=FAext{Stress} = \frac{F}{A}
    • Considering the area:A = 540extmm2=0.00054extm2540 ext{ mm}^2 = 0.00054 ext{ m}^2

  • b. Maximum Load Limitation:

    • If normal stress in member (1) is limited to 50 MPa, calculate the maximum allowable load magnitude using:
    • P<em>max=extStress</em>maximesAP<em>{max} = ext{Stress}</em>{max} imes A

Example 1.3: Axial Loads on a Steel Bar

  • Scenario Setup:

    • A steel bar with a width of 50 mm experiences axial loads at points B, C, and D.
    • The normal stress magnitude in the bar must not exceed 60 MPa.

  • Objectives:

    • Determine minimum thickness of the bar:
    • Identify the maximum absolute value of internal axial force and allowable normal stress to compute the minimum required cross-sectional area.
    • A=extthicknessimes50extmmA = ext{thickness} imes 50 ext{ mm}

  • FBD Considerations:

    • Internal force assumed as tensile, with arrows drawn away from the cut.
    • Positive internal force: Tension; Negative internal force: Compression.

  • Summary Calculation Steps:

    • Compute the required cross-sectional area based on the maximum internal axial force.
    • Calculate minimum thickness:
    • extArea=extThicknessimes50extmmext{Area} = ext{Thickness} imes 50 ext{ mm}

Review & Next Class Preparation

  • Review Points:
    • Stress Definition: Internal response or resistance of a material to external forces.
    • Elastic Deformation: Atoms return to original position post-force release, showcasing material resilience.
    • Plastic Deformation: Permanent deformation occurs when surpassing elastic capability.
  • Next Topic: Introduction of Shear Stress, defined as intensity of an internal force acting on a surface parallel to the internal force. Also, address normal single-shear pin connections.