8. Mechanics of materials

Achievable FE Civil

Mechanics of materials

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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

Exam tip: Every formula in this chapter is listed in the NCEES FE Reference Handbook, which you can search electronically during the exam. The tested skill is finding the right equation and applying it correctly under time pressure, not memorizing it - practice locating each one (stress and strain, Mohr’s circle, torsion, Euler buckling, and so on) in the handbook as you study. All worked examples below use consistent SI units; on the exam, watch for problems that mix SI and USCS values and convert everything to one unit system before substituting into a formula. Carry full precision through intermediate steps rather than rounding early, since rounding errors compound across multi-step calculations.

Stress ( $σ$ )

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

Definitions

Stress: The internal force per unit area acting within a material.

$σ = \frac{F}{A}$

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:

$σ = \frac{1000}{0.01} = 100, 000 Pa$

Strain ( $ε$ )

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

Definitions

Strain: The deformation per unit original length.

$ε = \frac{Δ L}{L _{0}}$

Here:

$Δ L$ is the change in length
$L_{0}$ is the original length

Example

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

$ε = \frac{0.0005}{2} = 2.5 \times 1 0^{- 4}$

Elongation

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

$Δ L = \frac{F L _{0}}{A E}$

Where:

$F$ = Axial force
$L_{0}$ = Original length
$A$ = Cross-sectional area
$E$ = Young’s modulus

Example

For $F = 10 k N$ , $L_{0} = 1 m$ , $A = 100 m m^{2} = 1 \times 1 0^{- 4} m^{2}$ , $E = 200 GP a$ :

$Δ L = \frac{10000 \times 1}{1 \times 1 0 ^{- 4} \times 200 \times 1 0 ^{9}} = 5 \times 1 0^{- 4} m = 0.5 mm$

Hooke’s law

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

$σ = Eε$

Here:

$σ$ is normal stress
$ε$ is normal strain
$E$ is Young’s modulus

Example

For $E = 210 GP a$ and $ε = 1 \times 1 0^{- 3}$ :

$σ = 210 \times 1 0^{9} \times 1 0^{- 3} = 210 MP a$

Shear stress and strain

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

Shear stress:

Definitions

Shear stress: The tangential force per unit area.

$τ = \frac{V}{A}$

Shear strain:

Definitions

Shear strain: Angular deformation.

$γ = \frac{δ}{L}$

Hooke’s law for shear:

$τ = G γ$

Example

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

$τ = \frac{2000}{0.01} = 200, 000 P a = 200 k P a$

Bulk modulus ( $K$ )

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

$K = - \frac{p}{Δ V / V}$

Where:

$p$ = Applied pressure
$Δ V$ = Volume change
$V$ = Original volume

Example

If $p = 50 MP a$ causes $Δ V / V = - 0.002$ (a 0.2% decrease in volume):

$K = - \frac{50 \times 1 0 ^{6}}{- 0.002} = 25 \times 1 0^{9} P a = 25 GP a$

Uniaxial loading and deformation

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

$σ = \frac{F}{A}, ε = \frac{Δ L}{L}$

Example

$F = 15 k N$ , $A = 300 m m^{2} = 3 \times 1 0^{- 4} m^{2}$ :

$σ = \frac{15000}{3 \times 1 0 ^{- 4}} = 50 MP a$

Thermal deformations

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

$Δ L = α L_{0} Δ T$

Where:

$α$ = Thermal expansion coefficient
$Δ T$ = Temperature change

Example

For $L_{0} = 2 m$ , $α = 12 \times 1 0^{- 6} /^{\circ} C$ , $Δ T = 5 0^{\circ} C$ :

$Δ L = 12 \times 1 0^{- 6} \times 2 \times 50 = 0.0012 m = 1.2 mm$

Thin-walled cylindrical pressure vessels

For a thin-walled cylinder under internal pressure, the two common normal stresses are hoop (circumferential) and longitudinal (axial). These formulas only apply when the wall is thin: the FE Reference Handbook treats a cylinder as thin-walled when the wall thickness is about one-tenth or less of the inside radius. Thicker-walled vessels need thick-wall equations instead.

Hoop stress:

$σ_{h} = \frac{p r}{t}$

Longitudinal stress:

$σ_{l} = \frac{p r}{2 t}$

Here $p$ is the internal gauge pressure, $t$ is the wall thickness, and $r$ is the mean radius, measured to the middle of the wall: $r = (r_{i} + r_{o}) /2$ .

Example

$p = 2 MP a$ , $r = 0.5 m$ , $t = 10 mm = 0.01 m$ :

$σ_{h} = \frac{2 \times 1 0 ^{6} \times 0.5}{0.01} = 100 MP a$

$σ_{l} = \frac{2 \times 1 0 ^{6} \times 0.5}{0.02} = 50 MP a$

Mohr’s circle

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

Principal stresses:

$σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}$

Maximum in-plane shear stress (the radius $R$ of Mohr’s circle, which the FE Reference Handbook writes as $τ_{in} = R$ ):

$τ_{in} = R = (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}$

This is the largest shear stress in the x-y plane, not necessarily the largest at the point. The absolute maximum shear stress is $τ_{m a x} = \frac{σ _{1} - σ _{3}}{2}$ , where $σ_{1}$ and $σ_{3}$ are the algebraically largest and smallest of the three principal stresses. For plane stress those three are the two in-plane values from the formula above and zero (the Handbook labels the in-plane pair $σ_{a}$ and $σ_{b}$ before sorting them). When the in-plane principal stresses have opposite signs, $τ_{m a x} = R$ ; when they share a sign, zero becomes one of the extremes and $τ_{m a x}$ exceeds $R$ .

Example

If $σ_{x} = 60 MP a$ , $σ_{y} = 20 MP a$ , $τ_{x y} = 15 MP a$ :

$σ_{1, 2} = 40 \pm (20)^{2} + 1 5^{2} = 40 \pm 25 = 65, 15 MP a$

Here $R = 25 MP a$ is the maximum in-plane shear stress. Both principal stresses are tensile, so the third principal stress, $0$ , is the smallest: $τ_{m a x} = \frac{65 - 0}{2} = 32.5 MP a$ , larger than $R$ .

Torsion

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

Torsional shear stress:

$τ = \frac{T r}{J}$

Angle of twist:

$θ = \frac{T L}{G J}$

Example

For $T = 500 N m$ , $r = 0.05 m$ , $J = 2 \times 1 0^{- 6} m^{4}$ :

$τ = \frac{500 \times 0.05}{2 \times 1 0 ^{- 6}} = 12.5 \times 1 0^{6} P a = 12.5 MP a$

Thin-walled hollow shaft

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

$τ = \frac{T}{2 A _{m} t}$

Where:

$A_{m}$ = Area enclosed by midline
$t$ = Wall thickness

Example

$T = 2000 N m$ , $A_{m} = 0.01 m^{2}$ , $t = 5 mm = 0.005 m$ :

$τ = \frac{2000}{2 \times 0.01 \times 0.005} = 20 \times 1 0^{6} P a = 20 MP a$

Shear force and bending moment conventions

Positive shear force: the right portion of the beam tends to shear downward with respect to the left (the shear forces on a small element form a clockwise couple)
Positive bending moment: causes sagging (concave up)

Watch out: Some textbooks and exam problems define positive shear and moment the opposite way. Whatever convention a problem states, apply it consistently across the whole diagram - switching partway through flips the sign of every value that follows, even though the magnitudes stay correct.

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.

$\frac{d V}{d x} = - w (x), \frac{d M}{d x} = V$

Example

If $w (x) = 5 k N / m$ , then:

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

Bending stress in beams

Bending stress varies linearly with distance from the neutral axis.

$σ = \frac{M y}{I}$

Example

$M = 10 k N m$ , $y = 0.05 m$ , $I = 8 \times 1 0^{- 6} m^{4}$ :

$σ = \frac{10000 \times 0.05}{8 \times 1 0 ^{- 6}} = 62.5 MP a$

Shear stress in beams

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

$τ = \frac{V Q}{I b}$

Where:

$Q$ = First moment of area about the neutral axis, taken for the portion of the cross-section between the point of interest and the outer edge
$b$ = Width at point of interest

Transverse shear stress is maximum at the neutral axis and drops to zero at the outer fibers of the cross-section - the opposite of how bending stress behaves.

Example

$V = 20 k N$ , $Q = 5 \times 1 0^{- 4} m^{3}$ , $I = 8 \times 1 0^{- 6} m^{4}$ , $b = 0.1 m$ :

$τ = \frac{20000 \times 5 \times 1 0 ^{- 4}}{8 \times 1 0 ^{- 6} \times 0.1} = 12.5 MP a$

Beam deflection

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

$δ = \frac{P L ^{3}}{48 E I}$

Example

$P = 10 k N$ , $L = 2 m$ , $E = 200 GP a$ , $I = 5 \times 1 0^{- 6} m^{4}$ :

$δ = \frac{10000 \times 8}{48 \times 200 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 6}} = 1.67 \times 1 0^{- 3} m = 1.67 mm$

Critical buckling load (Euler’s formula)

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

$P_{cr} = \frac{π ^{2} E I}{( K L ) ^{2}}$

Example

$E = 210 GP a$ , $I = 5 \times 1 0^{- 6} m^{4}$ , $L = 3 m$ , $K = 1$ :

$P_{cr} = \frac{π ^{2} \times 210 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 6}}{9} \approx 1.15 \times 1 0^{6} N$

Critical buckling stress

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

$σ_{cr} = \frac{P _{cr}}{A} = \frac{π ^{2} E}{( \frac{K L}{r} ) ^{2}}$

Example

For $E = 200 GP a$ , $K L / r = 100$ :

$σ_{cr} = \frac{π ^{2} \times 200 \times 1 0 ^{9}}{10 0 ^{2}} = 197 MP a$

Elastic strain energy

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

Axial:

$U = \frac{F ^{2} L}{2 A E}$

Bending:

$U = \int \frac{M ^{2}}{2 E I} d x$

Example (Axial)

$F = 20 k N$ , $L = 2 m$ , $A = 1000 m m^{2} = 1 \times 1 0^{- 3} m^{2}$ , $E = 200 GP a$ :

$U = \frac{2000 0 ^{2} \times 2}{2 \times 1 \times 1 0 ^{- 3} \times 200 \times 1 0 ^{9}} = 2 J$

Stress ( $σ$ )

Internal force per unit area: $σ = \frac{F}{A}$
Units: Pascals (Pa)
Used to quantify intensity of internal forces

Strain ( $ε$ )

Deformation per unit original length: $ε = \frac{Δ L}{L _{0}}$
Dimensionless quantity
Measures relative change in length

Elongation

Axial elongation: $Δ L = \frac{F L _{0}}{A E}$
Depends on force, length, area, and Young’s modulus

Hooke’s law

Linear relation: $σ = Eε$
$E$ = Young’s modulus (material stiffness)

Shear stress and strain

Shear stress: $τ = \frac{V}{A}$
Shear strain: $γ = \frac{δ}{L}$
Shear Hooke’s law: $τ = G γ$ ( $G$ = shear modulus)

Bulk modulus ( $K$ )

Resistance to uniform compression: $K = - \frac{p}{Δ V / V}$
Relates pressure to volumetric strain

Uniaxial loading and deformation

Axial stress: $σ = \frac{F}{A}$
Axial strain: $ε = \frac{Δ L}{L}$

Thermal deformations

Free expansion: $Δ L = α L_{0} Δ T$
- $α$ = thermal expansion coefficient
- $Δ T$ = temperature change

Thin-walled cylindrical pressure vessels

Hoop (circumferential) stress: $σ_{h} = \frac{p r}{t}$
Longitudinal (axial) stress: $σ_{l} = \frac{p r}{2 t}$
$p$ = internal pressure, $r$ = radius, $t$ = wall thickness

Mohr’s circle

Principal stresses: $σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}$
Maximum shear stress: $τ_{m a x} = (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}$

Torsion

Shear stress: $τ = \frac{T r}{J}$
Angle of twist: $θ = \frac{T L}{G J}$
- $T$ = torque, $r$ = radius, $J$ = polar moment of inertia

Thin-walled hollow shaft

Shear stress: $τ = \frac{T}{2 A _{m} t}$
- $A_{m}$ = area enclosed by midline, $t$ = wall thickness

Shear force and bending moment conventions

Positive shear: clockwise on left section
Positive bending moment: causes sagging (concave up)

Beam differential equations

$\frac{d V}{d x} = - w (x)$ (shear from distributed load)
$\frac{d M}{d x} = V$ (moment from shear)

Bending stress in beams

$σ = \frac{M y}{I}$
- $M$ = moment, $y$ = distance from neutral axis, $I$ = moment of inertia

Shear stress in beams

$τ = \frac{V Q}{I b}$
- $Q$ = first moment of area, $b$ = width at point

Beam deflection

Max deflection (center load, simply supported): $δ = \frac{P L ^{3}}{48 E I}$

Critical buckling load (Euler’s formula)

$P_{cr} = \frac{π ^{2} E I}{( K L ) ^{2}}$
- $K$ = effective length factor

Critical buckling stress

$σ_{cr} = \frac{P _{cr}}{A} = \frac{π ^{2} E}{( K L / r ) ^{2}}$
- $r$ = radius of gyration

Elastic strain energy

Axial: $U = \frac{F ^{2} L}{2 A E}$
Bending: $U = \int \frac{M ^{2}}{2 E I} d x$

Next | 9.1 Fluid statics

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 ( $σ$ )

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

Definitions

Stress: The internal force per unit area acting within a material.

$σ = \frac{F}{A}$

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:

$σ = \frac{1000}{0.01} = 100, 000 Pa$

Strain ( $ε$ )

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

Definitions

Strain: The deformation per unit original length.

$ε = \frac{Δ L}{L _{0}}$

Here:

$Δ L$ is the change in length
$L_{0}$ is the original length

Example

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

$ε = \frac{0.0005}{2} = 2.5 \times 1 0^{- 4}$

Elongation

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

$Δ L = \frac{F L _{0}}{A E}$

Where:

$F$ = Axial force
$L_{0}$ = Original length
$A$ = Cross-sectional area
$E$ = Young’s modulus

Example

For $F = 10 k N$ , $L_{0} = 1 m$ , $A = 100 m m^{2} = 1 \times 1 0^{- 4} m^{2}$ , $E = 200 GP a$ :

$Δ L = \frac{10000 \times 1}{1 \times 1 0 ^{- 4} \times 200 \times 1 0 ^{9}} = 5 \times 1 0^{- 4} m = 0.5 mm$

Hooke’s law

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

$σ = Eε$

Here:

$σ$ is normal stress
$ε$ is normal strain
$E$ is Young’s modulus

Example

For $E = 210 GP a$ and $ε = 1 \times 1 0^{- 3}$ :

$σ = 210 \times 1 0^{9} \times 1 0^{- 3} = 210 MP a$

Shear stress and strain

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

Shear stress:

Definitions

Shear stress: The tangential force per unit area.

$τ = \frac{V}{A}$

Shear strain:

Definitions

Shear strain: Angular deformation.

$γ = \frac{δ}{L}$

Hooke’s law for shear:

$τ = G γ$

Example

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

$τ = \frac{2000}{0.01} = 200, 000 P a = 200 k P a$

Bulk modulus ( $K$ )

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

$K = - \frac{p}{Δ V / V}$

Where:

$p$ = Applied pressure
$Δ V$ = Volume change
$V$ = Original volume

Example

If $p = 50 MP a$ causes $Δ V / V = - 0.002$ (a 0.2% decrease in volume):

$K = - \frac{50 \times 1 0 ^{6}}{- 0.002} = 25 \times 1 0^{9} P a = 25 GP a$

Uniaxial loading and deformation

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

$σ = \frac{F}{A}, ε = \frac{Δ L}{L}$

Example

$F = 15 k N$ , $A = 300 m m^{2} = 3 \times 1 0^{- 4} m^{2}$ :

$σ = \frac{15000}{3 \times 1 0 ^{- 4}} = 50 MP a$

Thermal deformations

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

$Δ L = α L_{0} Δ T$

Where:

$α$ = Thermal expansion coefficient
$Δ T$ = Temperature change

Example

For $L_{0} = 2 m$ , $α = 12 \times 1 0^{- 6} /^{\circ} C$ , $Δ T = 5 0^{\circ} C$ :

$Δ L = 12 \times 1 0^{- 6} \times 2 \times 50 = 0.0012 m = 1.2 mm$

Thin-walled cylindrical pressure vessels

Hoop stress:

$σ_{h} = \frac{p r}{t}$

Longitudinal stress:

$σ_{l} = \frac{p r}{2 t}$

Here $p$ is the internal gauge pressure, $t$ is the wall thickness, and $r$ is the mean radius, measured to the middle of the wall: $r = (r_{i} + r_{o}) /2$ .

Example

$p = 2 MP a$ , $r = 0.5 m$ , $t = 10 mm = 0.01 m$ :

$σ_{h} = \frac{2 \times 1 0 ^{6} \times 0.5}{0.01} = 100 MP a$

$σ_{l} = \frac{2 \times 1 0 ^{6} \times 0.5}{0.02} = 50 MP a$

Mohr’s circle

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

Principal stresses:

$σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}$

Maximum in-plane shear stress (the radius $R$ of Mohr’s circle, which the FE Reference Handbook writes as $τ_{in} = R$ ):

$τ_{in} = R = (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}$

Example

If $σ_{x} = 60 MP a$ , $σ_{y} = 20 MP a$ , $τ_{x y} = 15 MP a$ :

$σ_{1, 2} = 40 \pm (20)^{2} + 1 5^{2} = 40 \pm 25 = 65, 15 MP a$

Here $R = 25 MP a$ is the maximum in-plane shear stress. Both principal stresses are tensile, so the third principal stress, $0$ , is the smallest: $τ_{m a x} = \frac{65 - 0}{2} = 32.5 MP a$ , larger than $R$ .

Torsion

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

Torsional shear stress:

$τ = \frac{T r}{J}$

Angle of twist:

$θ = \frac{T L}{G J}$

Example

For $T = 500 N m$ , $r = 0.05 m$ , $J = 2 \times 1 0^{- 6} m^{4}$ :

$τ = \frac{500 \times 0.05}{2 \times 1 0 ^{- 6}} = 12.5 \times 1 0^{6} P a = 12.5 MP a$

Thin-walled hollow shaft

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

$τ = \frac{T}{2 A _{m} t}$

Where:

$A_{m}$ = Area enclosed by midline
$t$ = Wall thickness

Example

$T = 2000 N m$ , $A_{m} = 0.01 m^{2}$ , $t = 5 mm = 0.005 m$ :

$τ = \frac{2000}{2 \times 0.01 \times 0.005} = 20 \times 1 0^{6} P a = 20 MP a$

Shear force and bending moment conventions

Positive shear force: the right portion of the beam tends to shear downward with respect to the left (the shear forces on a small element form a clockwise couple)
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.

$\frac{d V}{d x} = - w (x), \frac{d M}{d x} = V$

Example

If $w (x) = 5 k N / m$ , then:

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

Bending stress in beams

Bending stress varies linearly with distance from the neutral axis.

$σ = \frac{M y}{I}$

Example

$M = 10 k N m$ , $y = 0.05 m$ , $I = 8 \times 1 0^{- 6} m^{4}$ :

$σ = \frac{10000 \times 0.05}{8 \times 1 0 ^{- 6}} = 62.5 MP a$

Shear stress in beams

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

$τ = \frac{V Q}{I b}$

Where:

$Q$ = First moment of area about the neutral axis, taken for the portion of the cross-section between the point of interest and the outer edge
$b$ = Width at point of interest

Transverse shear stress is maximum at the neutral axis and drops to zero at the outer fibers of the cross-section - the opposite of how bending stress behaves.

Example

$V = 20 k N$ , $Q = 5 \times 1 0^{- 4} m^{3}$ , $I = 8 \times 1 0^{- 6} m^{4}$ , $b = 0.1 m$ :

$τ = \frac{20000 \times 5 \times 1 0 ^{- 4}}{8 \times 1 0 ^{- 6} \times 0.1} = 12.5 MP a$

Beam deflection

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

$δ = \frac{P L ^{3}}{48 E I}$

Example

$P = 10 k N$ , $L = 2 m$ , $E = 200 GP a$ , $I = 5 \times 1 0^{- 6} m^{4}$ :

$δ = \frac{10000 \times 8}{48 \times 200 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 6}} = 1.67 \times 1 0^{- 3} m = 1.67 mm$

Critical buckling load (Euler’s formula)

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

$P_{cr} = \frac{π ^{2} E I}{( K L ) ^{2}}$

Example

$E = 210 GP a$ , $I = 5 \times 1 0^{- 6} m^{4}$ , $L = 3 m$ , $K = 1$ :

$P_{cr} = \frac{π ^{2} \times 210 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 6}}{9} \approx 1.15 \times 1 0^{6} N$

Critical buckling stress

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

$σ_{cr} = \frac{P _{cr}}{A} = \frac{π ^{2} E}{( \frac{K L}{r} ) ^{2}}$

Example

For $E = 200 GP a$ , $K L / r = 100$ :

$σ_{cr} = \frac{π ^{2} \times 200 \times 1 0 ^{9}}{10 0 ^{2}} = 197 MP a$

Elastic strain energy

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

Axial:

$U = \frac{F ^{2} L}{2 A E}$

Bending:

$U = \int \frac{M ^{2}}{2 E I} d x$

Example (Axial)

$F = 20 k N$ , $L = 2 m$ , $A = 1000 m m^{2} = 1 \times 1 0^{- 3} m^{2}$ , $E = 200 GP a$ :

$U = \frac{2000 0 ^{2} \times 2}{2 \times 1 \times 1 0 ^{- 3} \times 200 \times 1 0 ^{9}} = 2 J$

Achievable FE Civil