4.2 Stress and Strain Analysis
4.2 Stress and Strain Analysis
Introduction to Stress and Strain
When external forces act on a deformable body, they induce internal forces within the material. Stress is the measure of these internal forces per unit area, while Strain quantifies the resulting deformation relative to the original dimensions. The analysis of stress and strain is fundamental to predicting how engineering materials and structures will behave under load, ensuring they remain within safe limits of strength and deformation. This section explores various states of stress, the critical concepts of principal stresses and maximum shear stress, material behavior through stress-strain curves, and the specific case of torsional loading.
1. Normal and Shear Stresses
1.1 Definitions and Concepts
Stress (σ or τ):
Definition: The intensity of internal force distributed over a given area. It is force per unit area.
Units: Pascal (Pa = N/m²), Megapascal (MPa = N/mm²), Gigapascal (GPa).
Stress=AreaInternal Force
Normal Stress (σ):
Definition: The stress component acting perpendicular (normal) to a given plane or cross-section.
Sign Convention:
Tensile Stress (+σ): Tends to elongate the material. Force is directed away from the section.
Compressive Stress (-σ): Tends to shorten the material. Force is directed into the section.
Formula for Prismatic Bar under Axial Load (P): σ=AP where A is the original cross-sectional area.
Shear Stress (τ):
Definition: The stress component acting parallel (tangential) to a given plane. It tends to cause one layer of material to slide over an adjacent layer.
Causes: Direct shear forces, torsion, bending.
Sign Convention: Positive shear stress tends to rotate the element clockwise. Often, direction is more important than sign convention for shear.
Formula for Direct Shear (e.g., bolt in single shear): τavg=AsV where V is the shear force and As is the shear area.
General State of Stress at a Point (2D):
Represented by a stress element.
Requires two normal stresses (σ_x, σ_y) and one shear stress (τ_xy) to fully define. The complementary nature of shear requires τxy=τyx.
2. Principal Stresses and Principal Planes
2.1 Concept and Importance
Principal Stresses (σ₁, σ₂):
The maximum and minimum normal stresses that exist at a point when the stress element is rotated to a specific orientation.
On these special planes, the shear stress is zero. τ=0
σ₁ is the maximum principal stress (algebraically largest).
σ₂ is the minimum principal stress (algebraically smallest). In 2D, it can be compressive.
These are extremely important as failure theories (like Maximum Normal Stress Theory) often use these values to predict yielding or fracture.
Principal Planes:
The specific planes (orientations) on which the principal stresses act.
Perpendicular to each other.
Defined by the angle θp from a reference axis (e.g., x-axis).
2.2 Formulas for Principal Stresses and their Orientation
For a 2D state of stress (σ_x, σ_y, τ_xy):
Magnitude of Principal Stresses: σ1,2=2σx+σy±(2σx−σy)2+τxy2
σ1 uses the '+' sign.
σ2 uses the '-' sign.
Orientation of Principal Planes (θ_p): tan2θp=σx−σy2τxy
This equation gives two angles, 90° apart, corresponding to the planes for σ₁ and σ₂.
2.3 Mohr's Circle for Stress (Graphical Method)
A powerful graphical tool to determine principal stresses, maximum shear stress, and stresses on any inclined plane.
Center (C): C=(2σx+σy,0)
Radius (R): R=(2σx−σy)2+τxy2
Principal Stresses from Mohr's Circle:
σ1=C+R
σ2=C−R
3. Maximum Shear Stress and Corresponding Plane
3.1 Concept and Formula
Maximum In-Plane Shear Stress (τ_max):
The maximum shear stress that exists at the point for any orientation of the stress element.
Formula: τmax=(2σx−σy)2+τxy2=R (Radius of Mohr’s Circle)
Note that τmax=2σ1−σ2
Planes of Maximum Shear Stress:
Oriented at 45° to the principal planes.
On these planes, normal stresses are equal to the average stress: σavg=2σx+σy=C (Center of Mohr’s Circle)
Orientation (θ_s) relative to principal axes: θs=θp±45∘
3.2 Relationship Summary
The planes of maximum shear stress and principal planes are always 45 degrees apart.
On principal planes (θ_p): τ=0, σ=σ1,2 (extreme normal stresses).
On max shear planes (θ_s): τ=τmax, σ=σavg.
4. Stress-Strain Curves
Stress-strain curves graphically represent a material's mechanical behavior under load. They are obtained from a standard tensile test.
4.1 Key Regions and Points on a Ductile Metal Curve (e.g., Mild Steel)
Elastic Region (O to A):
Behavior: Stress is proportional to strain. Material returns to its original shape upon unloading.
Hooke's Law: σ=Eϵ
Proportional Limit (A): Point up to which Hooke's Law holds exactly.
Yield Point:
Upper Yield Point (B): Stress peaks briefly.
Lower Yield Point (C): Stress drops and remains nearly constant while strain increases significantly without an increase in load. This is the yield strength (σ_y). Material yields (deforms plastically).
Plastic Region (C to D):
Behavior: Permanent, irreversible deformation occurs. Strain hardening (increase in stress required to continue deformation) takes place.
Ultimate Tensile Strength (UTS) (Point D):
The maximum stress on the curve. Necking (localized reduction in cross-section) begins at or just after this point.
Fracture Point (F):
The stress at which the specimen finally breaks. The breaking strength is less than UTS due to necking.
4.2 Important Material Parameters from the Curve
Modulus of Elasticity (Young's Modulus, E):
Slope of the linear elastic region. E=ϵσ (in elastic zone). Measures stiffness.
High E = Stiff material (Steel ≈ 200 GPa).
Low E = Flexible material (Aluminum ≈ 70 GPa).
Yield Strength (σ_y):
The stress at which a material begins to deform plastically. This is the practical limit for elastic design.
Ultimate Tensile Strength (σ_u):
The maximum stress the material can withstand.
Percent Elongation & Percent Reduction in Area:
Measures of ductility.
% Elongation: L0Lf−L0×100
% Reduction in Area: A0A0−Af×100
4.3 Comparison of Material Behaviors
Ductile Materials (Mild steel, Aluminum): Exhibit significant plastic deformation and yield plateau before fracture. Stress-strain curve has a distinct yield point or a smooth transition (0.2% offset yield for many metals).
Brittle Materials (Cast iron, Glass, High-strength concrete): Fracture with little or no plastic deformation. No distinct yield point. The curve is nearly linear up to fracture.
5. Torsion
Torsion refers to the twisting of a structural member when it is loaded by couples (moments) that produce rotation about its longitudinal axis.
5.1 Basic Concepts and Assumptions (for Circular Shafts)
Torque (T): The twisting moment applied about the longitudinal axis. Units: N.m.
Angle of Twist (φ): The angle through which one end of the shaft rotates relative to the other. Units: Radians.
Pure Torsion: A state where the shaft is subjected to torque only (no axial load, bending, etc.).
Assumptions for Circular Shafts (from Strength of Materials):
Plane sections remain plane and circular.
Radial lines remain radial.
The shaft material is homogeneous, isotropic, and obeys Hooke's Law.
The shear stress is proportional to the shear strain.
5.2 Torsional Shear Stress Formula
For a solid circular shaft of radius R subjected to a torque T:
Shear Stress Variation:
Shear stress varies linearly with radial distance from the center.
Maximum shear stress (τ_max) occurs at the outer surface (r = R).
At the center (r = 0), shear stress is zero.
Torsion Formula: JT=rτ=LGϕ
For Maximum Shear Stress: τmax=JTR=ZpT
For Shear Stress at any radius (r): τ=JTr
Where:
T = Applied torque.
J = Polar Moment of Inertia of the cross-section (resistance to torsion).
For a solid circular shaft: J=32πd4
For a hollow circular shaft (outer dia. D, inner dia. d): J=32π(D4−d4)
r = Radial distance from center.
R = Outer radius of shaft.
Zp=J/R = Polar Section Modulus.
5.3 Angle of Twist Formula
The angle of twist (in radians) for a shaft of length L and constant cross-section under constant torque T is: ϕ=JGTL Where:
G = Shear Modulus or Modulus of Rigidity of the material. It relates shear stress to shear strain: τ=Gγ.
5.4 Power Transmission by Shafts
A common application of torsion is in power transmission shafts (e.g., in motors, turbines). P=Tω Where:
P = Power (Watts).
T = Torque (N.m).
ω = Angular speed (radians/sec).
If speed is in RPM (N): ω=602πN Therefore: P=602πNT or T=2πN60P
Conclusion: Stress and strain analysis provides the theoretical framework for understanding material response. From identifying critical principal stresses that may initiate failure to interpreting the ductile or brittle nature of a material from its stress-strain curve, this knowledge is directly applied in design. Torsion analysis completes the picture for members subjected to twisting, ensuring shafts and other components are sized correctly to transmit required torques without excessive stress or deformation.
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