Mechanics of materials

This chapter covers the following topics:

• Stress
• Strain
• Elongation
• Hooke’s law
• Shear stress and strain
• Bulk modulus
• Uniaxial loading and deformation
• Thermal deformations
• Thin-walled cylindrical pressure vessels
• Mohr’s circle
• Torsion
• Thin-walled hollow shaft
• Shear force and bending moment conventions
• Beam differential equations
• Bending stress in beams
• Shear stress in beams
• Beam deflection
• Critical buckling load (Euler’s formula)
• Critical buckling stress
• Elastic strain energy

Stress ($\sigma$)

Stress describes how intensely internal forces act within a material. It’s defined as internal force per unit area.

Definitions

Stress is the internal force per unit area acting within a material.

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

Here:

• $F$ is the applied (internal) force
• $A$ is the cross-sectional area carrying the force

Example If a force of 1000 N acts on a cross-sectional area of 0.01 m²:

$$\sigma =\frac{1000}{0.01}=100,000\text{Pa}$$

Strain ($\epsilon$)

Strain measures deformation relative to the original size. For axial loading, it’s the change in length divided by the original length.

Definitions

Strain is the deformation per unit original length.

$$\epsilon =\frac{\Delta L}{L_0}$$

Here:

• $\Delta L$ is the change in length
• $L_0$ is the original length

Example If a rod of initial length $2m$ elongates by $0.5mm$:

$$\epsilon =\frac{0.0005}{2}=2.5\times 10^{-4}$$

Elongation

For a prismatic bar under axial load (linear elastic behavior), elongation can be computed directly from force, geometry, and material stiffness:

$$\Delta L=\frac{F{L}_0}{AE}$$

Where:

• $F$ = Axial force
• $L_0$ = Original length
• $A$ = Cross-sectional area
• $E$ = Young’s modulus

Example For $F=10kN$, $L_0=1m$, $A=100m{m}^2=1\times 10^{-4}{m}^2$, $E=200GPa$:

$$\Delta L=\frac{10000\times 1}{1\times 10^{-4}\times 200\times 10^9}=5\times 10^{-4}m=0.5mm$$

Hooke’s law

Hooke’s law links stress and strain in the linear elastic range.

$$\sigma =E\epsilon$$

Here:

• $\sigma$ is normal stress
• $\epsilon$ is normal strain
• $E$ is Young’s modulus

Example For $E=210GPa$ and $\epsilon =1\times 10^{-3}$:

$$\sigma =210\times 10^9\times 10^{-3}=210MPa$$

Shear stress and strain

Shear quantities describe deformation and internal forces that act tangentially to a surface.

Shear stress:

Definitions

Shear stress is the tangential force per unit area.

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

Shear strain:

Definitions

Shear strain is angular deformation.

$$\gamma =\frac{\delta }{L}$$

Hooke’s Law for Shear:

$$\tau =G\gamma$$

Example If $V=2000N$ acts on area $0.01m^2$, shear stress is:

$$\tau =\frac{2000}{0.01}=200,000Pa=200kPa$$

Bulk modulus ($K$)

The bulk modulus relates pressure to volumetric strain (how much the volume changes under pressure).

$$K=-\frac{p}{\Delta V/V}$$

Where:

• $p$ = Applied pressure
• $\Delta V$ = Volume change
• $V$ = Original volume

Example If $p=50MPa$ causes $\Delta V/V=0.002$:

$$K=-\frac{50\times 10^6}{0.002}=25\times 10^{9}Pa=25GPa$$

Uniaxial loading and deformation

For a member loaded axially, stress and strain are commonly computed using:

$$\sigma =\frac{F}{A},\epsilon =\frac{\Delta L}{L}$$

Example $F=15kN$, $A=300m{m}^2=3\times 10^{-4}{m}^2$:

$$\sigma =\frac{15000}{3\times 10^{-4}}=50MPa$$

Thermal deformations

A temperature change causes a free (unrestrained) change in length given by:

$$\Delta L=\alpha L_0\Delta T$$

Where:

• $\alpha$ = Thermal expansion coefficient
• $\Delta T$ = Temperature change

Example For $L_0=2m$, $\alpha =12\times 10^{-6}{/}^{\circ }C$, $\Delta T=50^{\circ }C$:

$$\Delta L=12\times 10^{-6}\times 2\times 50=0.0012m=1.2mm$$

Thin-walled cylindrical pressure vessels

For a thin-walled cylinder under internal pressure, the two common normal stresses are hoop (circumferential) and longitudinal (axial).

Hoop Stress:

$${\sigma }_h=\frac{pr}{t}$$

Longitudinal Stress:

$${\sigma }_l=\frac{pr}{2t}$$

Example $p=2MPa$, $r=0.5m$, $t=10mm=0.01m$:

$${\sigma }_h=\frac{2\times 10^6\times 0.5}{0.01}=100MPa$$

$${\sigma }_l=\frac{2\times 10^6\times 0.5}{0.02}=50MPa$$

Mohr’s circle

Mohr’s circle provides formulas for principal stresses and maximum shear stress for a 2D stress state.

Principal Stresses:

$${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$$

Max shear stress:

$${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$$

Example If ${\sigma }_x=60MPa$, ${\sigma }_y=20MPa$, ${\tau }_{xy}=15MPa$:

$${\sigma }_{1,2}=40\pm \sqrt{(20{)}^2+15^2}=40\pm 25=65,15MPa$$

Torsion

Torsion formulas relate applied torque to shear stress and twist in a circular shaft.

Torsional shear stress:

$$\tau =\frac{Tr}{J}$$

Angle of Twist:

$$\theta =\frac{TL}{GJ}$$

Example For $T=500Nm$, $r=0.05m$, $J=2\times 10^{-6}{m}^4$:

$$\tau =\frac{500\times 0.05}{2\times 10^{-6}}=12.5\times 10^{6}Pa=12.5MPa$$

Thin-walled hollow shaft

For a thin-walled closed section, shear stress under torque can be approximated by:

$$\tau =\frac{T}{2A_{m}t}$$

Where:

• $A_m$ = Area enclosed by midline
• $t$ = Wall thickness

Example $T=2000Nm$, $A_m=0.01m^2$, $t=5mm=0.005m$:

$$\tau =\frac{2000}{2\times 0.01\times 0.005}=20\times 10^{6}Pa=20MPa$$

Shear force and bending moment conventions

• Positive shear force: clockwise rotation on the left section
• Positive bending moment: causes sagging (concave up)

Example A simply supported beam with a downward load at midspan will have a positive bending moment (sagging) at the center and a shear force changing sign at midspan.

Beam differential equations

These relationships connect distributed load $w(x)$, shear force $V$, and bending moment $M$ along a beam.

$$\frac{dV}{dx}=-w(x),\frac{dM}{dx}=V$$

Example If $w(x)=5kN/m$, then:

$$\frac{dV}{dx}=-5\Rightarrow V=-5x+C$$

Bending stress in beams

Bending stress varies linearly with distance from the neutral axis.

$$\sigma =\frac{My}{I}$$

Example $M=10kNm$, $y=0.05m$, $I=8\times 10^{-6}{m}^4$:

$$\sigma =\frac{10000\times 0.05}{8\times 10^{-6}}=62.5MPa$$

Shear stress in beams

Transverse shear stress in a beam cross-section can be found using:

$$\tau =\frac{VQ}{Ib}$$

Where:

• $Q$ = First moment of area
• $b$ = Width at point of interest

Example $V=20kN$, $Q=5\times 10^{-4}{m}^3$, $I=8\times 10^{-6}{m}^4$, $b=0.1m$:

$$\tau =\frac{20000\times 5\times 10^{-4}}{8\times 10^{-6}\times 0.1}=12.5MPa$$

Beam deflection

For a simply supported beam with a center load, the maximum deflection at midspan is:

$$\delta =\frac{P{L}^3}{48EI}$$

Example $P=10kN$, $L=2m$, $E=200GPa$, $I=5\times 10^{-6}{m}^4$:

$$\delta =\frac{10000\times 8}{48\times 200\times 10^9\times 5\times 10^{-6}}=1.67\times 10^{-3}m=1.67mm$$

Critical buckling load (Euler’s formula)

Euler’s formula gives the elastic critical load for a slender column.

$$P_{cr}=\frac{{\pi }^{2}EI}{(KL{)}^2}$$

Example $E=210GPa$, $I=5\times 10^{-6}{m}^4$, $L=3m$, $K=1$:

$$P_{cr}=\frac{{\pi }^2\times 210\times 10^9\times 5\times 10^{-6}}{9}\approx 1.15\times 10^{6}N$$

Critical buckling stress

Critical buckling stress can be written in terms of critical load and area, or in terms of slenderness ratio.

$${\sigma }_{cr}=\frac{P_{cr}}{A}=\frac{{\pi }^{2}E}{{\left(\frac{KL}{r}\right)}^2}$$

Example For $E=200GPa$, $KL/r=100$:

$${\sigma }_{cr}=\frac{{\pi }^2\times 200\times 10^9}{100^2}=197MPa$$

Elastic strain energy

Strain energy is the energy stored in a body due to elastic deformation.

Axial:

$$U=\frac{F^{2}L}{2AE}$$

Bending:

$$U=\int \frac{M^2}{2EI}dx$$

Example (Axial) $F=20kN$, $L=2m$, $A=1000m{m}^2=1\times 10^{-3}{m}^2$, $E=200GPa$:

$$U=\frac{20000^2\times 2}{2\times 1\times 10^{-3}\times 200\times 10^9}=2J$$

Example problem

Given

• $E=200\text{GPa}$
• $L=2m$
• $A=500m{m}^2=5\times 10^{-4}{m}^2$
• $F=20kN=20,000N$

Stress:

$$\sigma =\frac{F}{A}=\frac{20000}{5\times 10^{-4}}=40\times 10^6\text{Pa}=40\text{MPa}$$

Strain:

$$\epsilon =\frac{\sigma }{E}=\frac{40\times 10^6}{200\times 10^9}=2\times 10^{-4}$$

Elongation:

$$\Delta L=\epsilon L=2\times 10^{-4}\times 2=0.0004m=0.4mm$$