Mechanics of Solids — Complete Equation Reference

01 Stress

Average Normal Stress

Definition — uniform axial load over cross-section \[ \sigma = \frac{P}{A} \]

P = internal axial force (N or lb), A = cross-sectional area (m² or in²). Positive P = tension → positive (tensile) stress.

Average Shear Stress

Direct / transverse shear \[ \tau_{avg} = \frac{V}{A} \]

Bearing stress (pin/bolt) \[ \sigma_b = \frac{P}{A_b} = \frac{P}{t \cdot d} \]

Stress on an Oblique / Inclined Plane

For a uniaxially loaded bar with cross-section area \(A_0\), the plane whose normal makes angle \(\theta\) with the bar axis carries:

Normal stress on inclined plane \[ \sigma_\theta = \frac{P}{A_0}\cos^2\theta = \sigma_x \cos^2\theta \]

Shear stress on inclined plane \[ \tau_\theta = -\frac{P}{A_0}\sin\theta\cos\theta = -\frac{\sigma_x}{2}\sin 2\theta \]

Maximum shear stress occurs at \(\theta = 45°\): \(\tau_{max} = \sigma_x/2\).
Maximum normal stress occurs at \(\theta = 0°\): \(\sigma_{max} = \sigma_x\).

Stress components on inclined plane at angle θ from the axis

Force & Moment Resultants from Stress

Resultant force from normal stress distribution \[ P = \int_A \sigma \, dA \]

Resultant moments \[ M_y = -\int_A z\,\sigma\,dA, \qquad M_z = \int_A y\,\sigma\,dA \]

Example — Shear Pin

A 20 mm diameter steel pin is in double shear, carrying P = 60 kN. Find the average shear stress.

Area of one shear plane: \(A = \pi(0.020)^2/4 = 3.14\times10^{-4}\ \text{m}^2\)

Double shear → \(V = P/2 = 30\ \text{kN}\)

\(\tau_{avg} = V/A = 30{,}000\,/\,3.14\times10^{-4} = 95.5\ \text{MPa}\)

02 Strain

Normal Strain

Average axial strain \[ \varepsilon = \frac{\delta}{L_0} = \frac{\Delta L}{L_0} \]

True (infinitesimal) strain at a point \[ \varepsilon_x = \frac{du}{dx}, \quad \varepsilon_y = \frac{dv}{dy}, \quad \varepsilon_z = \frac{dw}{dz} \]

Shear Strain

Engineering shear strain (change in right angle) \[ \gamma_{xy} = \frac{du}{dy} + \frac{dv}{dx}, \quad \gamma_{yz} = \frac{dv}{dz}+\frac{dw}{dy}, \quad \gamma_{xz} = \frac{du}{dz}+\frac{dw}{dx} \]

Lateral Strain & Poisson's Ratio

Poisson's ratio (definition) \[ \nu = -\frac{\varepsilon_{lat}}{\varepsilon_{axial}} \]

Lateral strain in a bar loaded axially along x: \(\varepsilon_y = \varepsilon_z = -\nu\,\varepsilon_x\)

Typical values: steel ≈ 0.28–0.30; aluminum ≈ 0.33; rubber ≈ 0.5 (incompressible).

Volumetric / Dilatational Strain

\[ e = \varepsilon_x + \varepsilon_y + \varepsilon_z = \frac{1-2\nu}{E}\left(\sigma_x+\sigma_y+\sigma_z\right) \]

Example — Bar Elongation

A 2 m steel bar (E = 200 GPa, ν = 0.29, A = 500 mm²) is pulled with P = 80 kN. Find axial and lateral strain.

\(\sigma = P/A = 80{,}000/(500\times10^{-6}) = 160\ \text{MPa}\)

\(\varepsilon_x = \sigma/E = 160/200{,}000 = 8\times10^{-4}\)

\(\varepsilon_{lat} = -\nu\varepsilon_x = -0.29\times8\times10^{-4} = -2.32\times10^{-4}\)

\(\delta = \varepsilon_x L = 8\times10^{-4}\times2 = 1.6\ \text{mm}\)

03 Constitutive Laws — Hooke's Law

1-D Hooke's Law

\[ \sigma = E\,\varepsilon, \qquad \tau = G\,\gamma \]

Constant	Symbol	Typical Steel	Typical Aluminum
Young's (Elastic) Modulus	\(E\)	200 GPa	70 GPa
Shear Modulus	\(G\)	75 GPa	26 GPa
Poisson's Ratio	\(\nu\)	0.29	0.33
Bulk Modulus	\(K\)	—	—

Relationship between elastic constants \[ G = \frac{E}{2(1+\nu)}, \qquad K = \frac{E}{3(1-2\nu)} \]

2-D Hooke's Law (Plane Stress: \(\sigma_z = 0\))

\[ \varepsilon_x = \frac{1}{E}\left(\sigma_x - \nu\sigma_y\right) \] \[ \varepsilon_y = \frac{1}{E}\left(\sigma_y - \nu\sigma_x\right) \] \[ \varepsilon_z = -\frac{\nu}{E}\left(\sigma_x + \sigma_y\right) \] \[ \gamma_{xy} = \frac{\tau_{xy}}{G} \]

3-D Generalized Hooke's Law

\[ \varepsilon_x = \frac{1}{E}\bigl[\sigma_x - \nu(\sigma_y + \sigma_z)\bigr] \] \[ \varepsilon_y = \frac{1}{E}\bigl[\sigma_y - \nu(\sigma_x + \sigma_z)\bigr] \] \[ \varepsilon_z = \frac{1}{E}\bigl[\sigma_z - \nu(\sigma_x + \sigma_y)\bigr] \] \[ \gamma_{xy} = \frac{\tau_{xy}}{G},\quad \gamma_{yz} = \frac{\tau_{yz}}{G},\quad \gamma_{xz} = \frac{\tau_{xz}}{G} \]

Thermal Effects (Temperature Change \(\Delta T\))

Free thermal strain (no restraint) \[ \varepsilon_T = \alpha\,\Delta T \]

Total axial strain (stress + thermal) \[ \varepsilon_{total} = \frac{\sigma}{E} + \alpha\,\Delta T \]

\(\alpha\) = coefficient of thermal expansion (steel ≈ 12×10⁻⁶ /°C; aluminum ≈ 23×10⁻⁶ /°C)

Restrained bar thermal stress: if \(\delta = 0\) then \(\sigma = -E\alpha\Delta T\) (compressive for +ΔT).

04 Mechanics of Axially Loaded Bars

Deformation — Constant P, A, E

\[ \delta = \frac{PL}{AE} \]

Deformation — Varying P, A, or E

Segmented bar (N segments) \[ \delta_{total} = \sum_{i=1}^{N} \frac{P_i L_i}{A_i E_i} \]

Continuously varying (integrate) \[ \delta = \int_0^L \frac{P(x)}{A(x)\,E(x)}\,dx \]

Thermal Elongation

\[ \delta_T = \alpha\,\Delta T\,L \]

Total deflection (mechanical + thermal) \[ \delta_{total} = \frac{PL}{AE} + \alpha\,\Delta T\,L \]

Statically Indeterminate Axial Problems

Requires compatibility (geometry of deformation) + equilibrium:

Compatibility condition (fixed-fixed bar) \[ \delta_{AB} + \delta_{BC} = 0 \quad \text{(or prescribed value)} \]

3 Types of indeterminate axial problems:

Fixed–fixed bar with intermediate load
Bar in rigid frame / statically indeterminate truss
Bar with temperature change and restraint

Procedure: ① Free body diagram → equilibrium. ② Draw deformed shape → write compatibility (e.g., \(\delta_1 + \delta_2 = 0\)). ③ Express each \(\delta\) via \(PL/AE\). ④ Solve simultaneously.

Stress on Inclined Plane (Axial Member)

\[ \sigma_\theta = \sigma_x\cos^2\theta = \frac{P}{A}\cos^2\theta \] \[ \tau_\theta = -\frac{\sigma_x}{2}\sin 2\theta \]

Example — Statically Indeterminate Bar (Thermal)

A steel bar (E = 200 GPa, α = 12×10⁻⁶ /°C, A = 600 mm², L = 1.5 m) is fixed at both ends. Temperature rises 80°C. Find the thermal stress.

Equilibrium: \(R_A + R_B = 0 \Rightarrow R_A = -R_B = R\) (compressive)

Compatibility: \(\delta_T - \delta_{mech} = 0 \Rightarrow \alpha\Delta T L = RL/AE\)

\(R = AE\alpha\Delta T = 600\times10^{-6}\times200\times10^9\times12\times10^{-6}\times80 = 115.2\ \text{kN}\)

\(\sigma = R/A = 115{,}200\,/\,600\times10^{-6} = 192\ \text{MPa}\ (\text{compressive})\)

05 Torsion of Circular Sections

Shear Stress Distribution

Torsion formula (shear stress at radius ρ) \[ \tau = \frac{T\,\rho}{J} \]

Maximum shear stress (at outer radius c) \[ \tau_{max} = \frac{T\,c}{J} \]

Polar Moment of Inertia J

Solid circular cross-section (radius c or diameter d) \[ J = \frac{\pi c^4}{2} = \frac{\pi d^4}{32} \]

Hollow circular cross-section (inner c₁, outer c₂) \[ J = \frac{\pi}{2}(c_2^4 - c_1^4) \]

Angle of Twist

Constant T, J, G \[ \phi = \frac{TL}{GJ} \]

Segmented shaft \[ \phi_{total} = \sum_i \frac{T_i L_i}{G_i J_i} \]

Varying T, J, G \[ \phi = \int_0^L \frac{T(x)}{G\,J(x)}\,dx \]

Composite (Multi-Material) Circular Shafts

For coaxial shafts sharing the same angle of twist:

\[ T_{total} = T_1 + T_2 + \cdots, \qquad \phi_1 = \phi_2 \] \[ \frac{T_1}{G_1 J_1} = \frac{T_2}{G_2 J_2} \]

Statically Indeterminate Torsion

Type 1: Fixed-fixed shaft with intermediate torque

Compatibility: total twist = 0 (or given) \[ \phi_{A/C} = \sum_i \frac{T_i L_i}{G J_i} = 0 \]

Type 2: Coaxial hollow + solid shaft with bonded end plates

\[ T_{inner} + T_{outer} = T_{applied}, \qquad \phi_{inner} = \phi_{outer} \]

Power-Torque relation: \(P = T\omega = 2\pi n T / 60\) where n = rpm, P in watts, T in N·m.

Circular shaft under torque T — shear stress linear in ρ

Example — Solid vs. Hollow Shaft

Compare the max shear stress in a solid shaft (d = 60 mm) vs a hollow shaft (outer 60 mm, inner 40 mm), both carrying T = 1200 N·m.

Solid: \(J_s = \pi(0.03)^4/2 = 1.272\times10^{-6}\ \text{m}^4\), \(\tau_s = 1200\times0.03/1.272\times10^{-6} = 28.3\ \text{MPa}\)

Hollow: \(J_h = \pi/2\,(0.03^4 - 0.02^4) = 1.021\times10^{-6}\ \text{m}^4\), \(\tau_h = 1200\times0.03/1.021\times10^{-6} = 35.3\ \text{MPa}\)

The hollow shaft is 27% lighter but only 25% higher stress — excellent efficiency.

06 Torsion of Thin-Walled Closed Sections

Shear Flow

Shear flow q (constant around closed section by Bredt's theorem) \[ q = \tau \cdot t = \frac{T}{2\,A_m} \]

\(A_m\) = area enclosed by the median line of the wall (centroidal area of the thin wall), NOT the cross-sectional material area.

Shear Stress at Any Point

\[ \tau = \frac{q}{t} = \frac{T}{2\,A_m\,t} \]

For varying thickness, maximum shear stress occurs at minimum thickness \(t_{min}\).

Angle of Twist (Bredt Formula)

\[ \phi = \frac{TL}{4\,G\,A_m^2}\oint \frac{ds}{t} \]

For constant wall thickness t and perimeter p: \(\phi = \frac{T L\, p}{4\,G\,A_m^2\,t}\)

Key: Median-Area \(A_m\)

Section Shape	\(A_m\) formula
Rectangle (b × h median dims)	\(A_m = b \cdot h\)
Circle (median radius r)	\(A_m = \pi r^2\)
Triangle (base b, height h)	\(A_m = \tfrac{1}{2}bh\)

Thin-walled closed section: median area A_m, shear flow q constant

Example — Rectangular Tube

A closed rectangular tube: outer 80 mm × 120 mm, wall t = 5 mm, T = 4 kN·m. Find τ_max.

Median dimensions: \(b_m = 75\ \text{mm},\ h_m = 115\ \text{mm}\)

\(A_m = 0.075\times0.115 = 8.625\times10^{-3}\ \text{m}^2\)

\(\tau_{max} = T/(2 A_m t) = 4000/(2\times8.625\times10^{-3}\times0.005) = 46.4\ \text{MPa}\)

07 The Beam Flexure Formula

Flexure Formula

Normal stress at distance y from neutral axis \[ \sigma_x = -\frac{M\,y}{I} \]

Maximum bending stress (at extreme fiber |y| = c) \[ \sigma_{max} = \frac{M\,c}{I} = \frac{M}{S}, \quad S = \frac{I}{c} \]

\(S\) = section modulus (m³ or in³). Sign convention: positive M curves beam concave up; y positive upward → tensile stress below NA, compressive above.

Centroid — Neutral Axis Location

Centroid of composite area \[ \bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i} \]

Moment of Inertia (Second Moment of Area)

Definition \[ I = \int_A y^2 \,dA \]

Shape	I about centroid
Rectangle (b wide, h tall)	\(I = \tfrac{1}{12}bh^3\)
Circle (radius r)	\(I = \tfrac{\pi r^4}{4} = \tfrac{\pi d^4}{64}\)
Hollow circle	\(I = \tfrac{\pi}{4}(r_o^4 - r_i^4)\)
Triangle (base b, height h, from base)	\(I_{base} = \tfrac{1}{12}bh^3;\ I_c = \tfrac{1}{36}bh^3\)

Parallel-axis theorem (transfer to new axis) \[ I = \bar{I} + A\,d^2 \]

\(d\) = distance between centroidal axis and new axis. Always adds — \(I\) increases when shifted from centroid.

Example — T-Section Beam

T-section: flange 120×20 mm (top), web 20×80 mm. M = 15 kN·m. Find σ_max.

A_flange = 120×20 = 2400 mm², ȳ_f = 90 mm from bottom. A_web = 20×80 = 1600 mm², ȳ_w = 40 mm

\(\bar{y} = (2400\times90 + 1600\times40)/(2400+1600) = 70\ \text{mm from bottom}\)

\(I = [\tfrac{1}{12}(120)(20)^3 + 2400(20)^2] + [\tfrac{1}{12}(20)(80)^3 + 1600(30)^2]\)

\(I = [80{,}000 + 960{,}000] + [853{,}333 + 1{,}440{,}000] = 3.333\times10^6\ \text{mm}^4\)

\(c_{top} = 30\ \text{mm},\ c_{bot} = 70\ \text{mm}\) → \(\sigma_{max} = 15\times10^6\times70/3.333\times10^6 = 315\ \text{MPa (bottom)}\)

08 Shear and Moment Diagrams

Differential Relationships

Load ↔ Shear \[ \frac{dV}{dx} = -w(x) \]

Shear ↔ Moment \[ \frac{dM}{dx} = V(x) \]

Combined (Euler-Bernoulli beam equation) \[ \frac{d^2M}{dx^2} = -w(x) \]

Integral (Area) Method

\[ V_B - V_A = -\int_A^B w\,dx = -(\text{area of load diagram}) \]

\[ M_B - M_A = \int_A^B V\,dx = (\text{area of shear diagram}) \]

Discontinuities

Concentrated force P (downward): V jumps down by P.
Concentrated moment M₀ (counterclockwise when viewed from right): M jumps up by M₀.
M is maximum where V = 0 (or changes sign).

Professor's note: Reactions will NOT be given in general — you must solve for them first using \(\sum F = 0\) and \(\sum M = 0\).

Simply supported beam — central point load P: SFD (blue) and BMD (gold)

Common Beam Cases

Loading	Max V	Max M	Location of M_max
SS beam, central load P	\(P/2\)	\(PL/4\)	Center
SS beam, UDL w (total W = wL)	\(wL/2\)	\(wL^2/8\)	Center
Cantilever, end load P	\(P\)	\(PL\)	Fixed end
Cantilever, UDL w	\(wL\)	\(wL^2/2\)	Fixed end
SS beam, end moment M₀	\(M_0/L\)	\(M_0\)	At applied moment

09 Normal Stress in Beams (My/I) & Stress Resultants

Normal Stress Distribution

\[ \sigma(y) = -\frac{M\,y}{I} \]

Linear distribution across the cross-section; zero at the neutral axis (NA), maximum at extreme fibers.

Resultant Force over Part of Cross-Section

Resultant normal force over partial area A' \[ F = \int_{A'} \sigma\,dA = -\frac{M}{I}\int_{A'} y\,dA = -\frac{M\,\bar{y}'\,A'}{I} \]

\(\bar{y}'\) = centroid of the partial area A' measured from the neutral axis.

First Moment of Area Q

\[ Q = \int_{A'} y\,dA = \bar{y}'\,A' \]

Q is used both for resultant computation and for shear stress (§10). Q = 0 at extreme fibers and maximum at the neutral axis.

Unsymmetric (Biaxial) Bending

Bending about both axes simultaneously (principal axes) \[ \sigma = -\frac{M_z\,y}{I_z} + \frac{M_y\,z}{I_y} \]

Neutral axis angle φ from z-axis \[ \tan\phi = \frac{I_z}{I_y}\tan\theta, \qquad \theta = \text{angle of } M \text{ from } z \]

Example — Resultant Force on Partial Area

Rectangular beam 60 mm wide × 120 mm tall, M = 8 kN·m. Find the resultant compressive force in the top half.

\(I = \tfrac{1}{12}(60)(120)^3 = 8.64\times10^6\ \text{mm}^4\)

Top half: A' = 60×60 = 3600 mm², \(\bar{y}' = 30\ \text{mm above NA}\)

\(Q = \bar{y}'A' = 30\times3600 = 108{,}000\ \text{mm}^3\)

\(F = MQ/I = 8\times10^6\times108{,}000/8.64\times10^6 = 100\ \text{kN (compression)}\)

10 Shear Stresses in Beams (VQ/It)

Shear Formula

Horizontal (and vertical) shear stress at level y \[ \tau = \frac{V\,Q}{I\,t} \]

\(V\) = transverse shear force at the cross-section
\(Q = \bar{y}'A'\) = first moment of area above (or below) the cut line, about the NA
\(I\) = second moment of entire cross-section about NA
\(t\) = width of section at the cut line

Q for Common Shapes

Shape	Q at neutral axis (max Q)
Rectangle (b × h)	\(Q_{max} = \tfrac{b h^2}{8}\) → \(\tau_{max} = \tfrac{3V}{2A}\)
Circle (radius r)	\(Q_{max} = \tfrac{2r^3}{3}\) → \(\tau_{max} = \tfrac{4V}{3A}\)
Wide-flange (I-beam)	Computed separately for flange and web

Shear Flow in Built-Up Sections

Shear flow q (force per unit length along longitudinal axis) \[ q = \frac{V\,Q}{I} \]

Force in fasteners (bolts/nails/glue) spaced at distance s \[ F_{fastener} = q \cdot s = \frac{V\,Q\,s}{I} \]

For multiple fasteners in a row: divide force by number of fasteners.

Parabolic shear stress distribution in rectangular cross-section

Example — Bolt Spacing in Built-Up Beam

A built-up T-beam carries V = 25 kN. Top flange (100×20 mm) is attached to web with bolts (shear capacity 8 kN each). Find max allowable bolt spacing s.

Compute I for entire section (use parallel axis theorem — assume centroid 30 mm from bottom of web for this example).

\(Q_{flange} = \bar{y}'_{flange}\times A_{flange} = 55\times(100\times20) = 110{,}000\ \text{mm}^3\) (distance from NA to flange centroid ≈ 55 mm)

Shear flow: \(q = VQ/I\). With \(I = 4.5\times10^6\ \text{mm}^4\): \(q = 25{,}000\times110{,}000/4.5\times10^6 = 611\ \text{N/mm}\)

For two bolts per row: \(s = 2\times8000/611 = 26.2\ \text{mm}\) maximum spacing

11 Stresses in Composite Beams

Transformed Section Method

Transform all materials into one equivalent material (usually reference material 1 with modulus E₁).

Modular ratio \[ n = \frac{E_2}{E_1} \]

Width transformation: material 2 becomes equivalent width \[ b_{transformed} = n \cdot b_{actual} \]

Finding NA of Transformed Section

\[ \bar{y} = \frac{\sum (A_i)_T \bar{y}_i}{\sum (A_i)_T} \]

where \((A_i)_T\) uses the transformed widths.

Stresses in Each Material

Stress in the reference material (1) \[ \sigma_1 = -\frac{M y}{I_T} \]

Stress in transformed material (2) \[ \sigma_2 = -n\,\frac{M y}{I_T} \]

\(I_T\) = moment of inertia of the transformed section about its neutral axis.

Key principle: The strain \(\varepsilon\) is the same for both materials at a given height (compatibility); the stress is scaled by the ratio of moduli.

Compatibility Equation (strain continuity)

\[ \varepsilon_1 = \varepsilon_2 \quad \Rightarrow \quad \frac{\sigma_1}{E_1} = \frac{\sigma_2}{E_2} \quad \Rightarrow \quad \sigma_2 = n\,\sigma_1 \]

Example — Steel-Reinforced Concrete Beam

Rectangular concrete beam 200 mm wide × 300 mm deep, with 2 steel bars (As = 1000 mm²) at 250 mm from top. E_c = 25 GPa, E_s = 200 GPa, n = 8. M = 40 kN·m.

Transform steel: equivalent concrete width = n·As/d_bar = place as area at same height

Transformed steel area: \(nA_s = 8\times1000 = 8000\ \text{mm}^2\) at y = 250 mm from top

Find NA (kd from top): \(200(kd)(kd/2) = 8000(250-kd)\) → solve quadratic for kd ≈ 106 mm

\(I_T = \tfrac{1}{3}(200)(106)^3 + 8000(250-106)^2 = 7.52\times10^8 + 1.66\times10^8 = 9.18\times10^8\ \text{mm}^4\)

Concrete: \(\sigma_c = 40\times10^6\times106/9.18\times10^8 = 4.62\ \text{MPa}\); Steel: \(\sigma_s = 8\times40\times10^6\times144/9.18\times10^8 = 50.1\ \text{MPa}\)

12 Deflections by Direct Integration of the 4th-Order ODE

Governing Equations (Euler-Bernoulli Beam)

Moment-curvature relation \[ EI\,y'' = M(x) \]

Shear relation \[ EI\,y''' = V(x) \]

Load relation (4th-order ODE) \[ EI\,y'''' = -w(x) \]

\(y'' = d^2y/dx^2 \approx\) curvature \(\kappa\) for small deflections. Sign: y positive upward, w positive downward.

Integration Procedure

Step 1: Start from w(x) or M(x) \[ EI\,y'''' = -w(x) \xrightarrow{\int} EI\,y''' = V(x) + C_1 \xrightarrow{\int} EI\,y'' = M(x) + C_1 x + C_2 \] \[ \xrightarrow{\int} EI\,y' = \theta(x) + \frac{C_1 x^2}{2} + C_2 x + C_3 \xrightarrow{\int} EI\,y = \delta(x) + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4 \]

ALWAYS FOUR boundary conditions — combinations of: y = 0 (no deflection), y' = 0 (no slope/rotation), M = EIy'' = 0 (no moment), V = EIy''' = 0 (no shear).

Common Boundary Conditions

Support Type	Conditions
Pinned / Roller (simple support)	\(y = 0\), \(M = EIy'' = 0\)
Fixed (clamped) end	\(y = 0\), \(y' = 0\)
Free end	\(M = EIy'' = 0\), \(V = EIy''' = 0\)
Guided / internal roller	\(y = 0\), \(y' = 0\) or \(y = \delta_{given}\)

Macaulay Brackets (Singularity Functions)

\[ \langle x - a \rangle^n = \begin{cases} 0 & x < a \\ (x-a)^n & x \geq a \end{cases} \]

Integrate as: \(\int \langle x-a\rangle^n dx = \frac{\langle x-a\rangle^{n+1}}{n+1}\). Use separate constant of integration only once per integration.

Load type	w(x) in Macaulay notation
Point load P at x = a (downward)	\(w = P\langle x-a\rangle^{-1}\)
UDL from x = a onward	\(w = w_0\langle x-a\rangle^0\)
Point moment M₀ at x = a (CCW)	Add \(M_0\langle x-a\rangle^0\) to M(x)

Standard Deflection Results

Beam / Loading	Max Deflection \(\delta_{max}\)	Location
SS beam, central load P	\(\dfrac{PL^3}{48EI}\)	Center
SS beam, UDL w	\(\dfrac{5wL^4}{384EI}\)	Center
Cantilever, end load P	\(\dfrac{PL^3}{3EI}\)	Free end
Cantilever, UDL w	\(\dfrac{wL^4}{8EI}\)	Free end
SS beam, off-center load P at a (a>b)	\(\dfrac{Pb(L^2-b^2)^{3/2}}{9\sqrt{3}\,EIL}\)	\(x=\sqrt{(L^2-b^2)/3}\)

Example — Cantilever Beam, UDL (Full Integration)

Cantilever of length L, fixed at x=0, UDL w (downward). Find y(x). Four B.C.s: y(0)=0, y'(0)=0, M(L)=EIy''(L)=0, V(L)=EIy'''(L)=0.

\(EIy'''' = -w \xrightarrow{\int} EIy''' = -wx + C_1\)

B.C. V(L)=0: \(-wL + C_1 = 0 \Rightarrow C_1 = wL\)

\(EIy'' = -wx^2/2 + wLx + C_2\); B.C. M(L)=0: \(-wL^2/2 + wL^2 + C_2 = 0 \Rightarrow C_2 = -wL^2/2\)

\(EIy' = -wx^3/6 + wLx^2/2 - wL^2 x/2 + C_3\); B.C. y'(0)=0 → C₃=0

\(EIy = -wx^4/24 + wLx^3/6 - wL^2 x^2/4 + C_4\); B.C. y(0)=0 → C₄=0

At free end: \(y(L) = \frac{w}{EI}\left(-\frac{L^4}{24}+\frac{L^4}{6}-\frac{L^4}{4}\right) = -\frac{wL^4}{8EI}\) ✓

13 Mohr's Circle for Plane Stress

Stress Transformation Equations

\[ \sigma_{x'} = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \]

\[ \sigma_{y'} = \frac{\sigma_x+\sigma_y}{2} - \frac{\sigma_x-\sigma_y}{2}\cos 2\theta - \tau_{xy}\sin 2\theta \]

\[ \tau_{x'y'} = -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \]

Mohr's Circle Parameters

Center of circle (on σ-axis) \[ C = \sigma_{avg} = \frac{\sigma_x + \sigma_y}{2} \]

Radius of circle \[ R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} \]

Principal Stresses

\[ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = C \pm R \]

Angle of principal planes from x-axis (2θ on Mohr's circle → θ physically) \[ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} \]

At principal planes: \(\tau = 0\). Principal directions are mutually perpendicular.

Maximum Shear Stress

\[ \tau_{max,\,in\text{-}plane} = R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} \]

Absolute maximum shear (including out-of-plane) \[ \tau_{abs,max} = \frac{\sigma_1 - \sigma_3}{2} \]

where \(\sigma_3 = 0\) for plane stress unless \(\sigma_1\) and \(\sigma_2\) have the same sign.

Angle of max shear plane from principal axes \[ \theta_s = \theta_p \pm 45° \]

At max shear planes: \(\sigma = C = (\sigma_1+\sigma_2)/2\) (not zero unless \(\sigma_1 = -\sigma_2\)).

Mohr's Circle: center C = (σ_avg, 0), radius R. Point A = original x-face. Rotate CCW on circle by 2θ → rotate stress element CCW by θ.

Mohr's Circle Sign Convention

Horizontal axis: normal stress σ (tension positive)
Vertical axis: shear stress τ (positive downward by convention OR upward — be consistent with textbook)
Rotation on circle is twice the physical angle
CCW rotation on circle = CCW rotation of element

Strain Transformation & Mohr's Circle for Strain

\[ \varepsilon_{x'} = \frac{\varepsilon_x+\varepsilon_y}{2} + \frac{\varepsilon_x-\varepsilon_y}{2}\cos 2\theta + \frac{\gamma_{xy}}{2}\sin 2\theta \]

\[ \frac{\gamma_{x'y'}}{2} = -\frac{\varepsilon_x-\varepsilon_y}{2}\sin 2\theta + \frac{\gamma_{xy}}{2}\cos 2\theta \]

Same circle formula but replace \(\sigma \to \varepsilon\) and \(\tau \to \gamma/2\).

Principal strains \[ \varepsilon_{1,2} = \frac{\varepsilon_x+\varepsilon_y}{2} \pm \sqrt{\left(\frac{\varepsilon_x-\varepsilon_y}{2}\right)^2 + \left(\frac{\gamma_{xy}}{2}\right)^2} \]

Example — Mohr's Circle

State of stress: σx = 80 MPa, σy = -40 MPa, τxy = 30 MPa. Find σ₁, σ₂, τ_max, θp.

\(C = (80 + (-40))/2 = 20\ \text{MPa}\)

\(R = \sqrt{((80-(-40))/2)^2 + 30^2} = \sqrt{60^2+30^2} = \sqrt{4500} = 67.1\ \text{MPa}\)

\(\sigma_1 = 20 + 67.1 = 87.1\ \text{MPa},\quad \sigma_2 = 20 - 67.1 = -47.1\ \text{MPa}\)

\(\tau_{max} = R = 67.1\ \text{MPa}\)

\(\tan 2\theta_p = 2(30)/(80-(-40)) = 60/120 = 0.5 \Rightarrow 2\theta_p = 26.6° \Rightarrow \theta_p = 13.3°\)

14 Thin-Walled Pressure Vessels

Thin-wall assumption: valid when \(r/t \geq 10\). Wall thickness t is negligible compared to radius r.

Spherical Pressure Vessel

Hoop (circumferential) = meridional (longitudinal) — equal biaxial \[ \sigma_1 = \sigma_2 = \frac{p\,r}{2t} \]

Radial stress (through-thickness) — neglected for thin wall \[ \sigma_3 \approx 0 \text{ (inner face: } {-p}\text{, outer face: }0\text{)} \]

Maximum shear stress in wall \[ \tau_{max} = \frac{\sigma_1}{2} = \frac{pr}{4t} \quad \text{(out of plane)} \]

Cylindrical Pressure Vessel

Hoop (circumferential) stress σ₁ \[ \sigma_1 = \frac{p\,r}{t} \]

Axial (longitudinal) stress σ₂ — closed ends \[ \sigma_2 = \frac{p\,r}{2t} \]

Open ends (no axial load) \[ \sigma_2 = 0 \]

Hoop stress is twice the axial stress in a closed cylinder. This is why pipes tend to burst with longitudinal splits (hoop failure).

Strains in Pressure Vessel Walls

Cylinder (biaxial stress state) \[ \varepsilon_1 = \frac{1}{E}(\sigma_1 - \nu\sigma_2) = \frac{pr}{Et}\left(1 - \frac{\nu}{2}\right) \] \[ \varepsilon_2 = \frac{1}{E}(\sigma_2 - \nu\sigma_1) = \frac{pr}{2Et}(1 - 2\nu) \]

Volumetric strain of cylinder \[ \frac{\Delta V}{V} = \varepsilon_{vol} = 2\varepsilon_1 + \varepsilon_2 = \frac{pr}{Et}\left(\frac{5}{2} - 2\nu\right) \]

Cylindrical pressure vessel: hoop σ₁ = pr/t (gold), axial σ₂ = pr/2t (blue)

Mohr's Circle for Pressure Vessels

Cylinder: \(\sigma_1 = pr/t\), \(\sigma_2 = pr/2t\), \(\tau_{max,in-plane} = (\sigma_1-\sigma_2)/2 = pr/4t\)

Sphere: \(\sigma_1 = \sigma_2 = pr/2t\) → in-plane τ = 0 (Mohr's circle is a point). Out-of-plane: \(\tau_{abs,max} = \sigma_1/2 = pr/4t\).

Example — Cylindrical Tank Design

Steel tank: internal pressure p = 2.5 MPa, r = 0.6 m, allowable σ = 120 MPa. Find minimum wall thickness t.

Hoop stress governs: \(\sigma_1 = pr/t \leq 120\ \text{MPa}\)

\(t \geq pr/\sigma_{allow} = 2.5\times10^6\times0.6/120\times10^6 = 12.5\ \text{mm}\)

Check axial: \(\sigma_2 = pr/2t = 60\ \text{MPa} < 120\ \text{MPa}\) ✓

15 Buckling — Euler's Formula

Critical (Euler) Buckling Load

\[ P_{cr} = \frac{\pi^2 E I}{(K L)^2} = \frac{\pi^2 E I}{L_e^2} \]

\(L_e = KL\) = effective length; \(I\) = minimum moment of inertia of the cross-section (buckling occurs about the weak axis).

Effective Length Factor K

End Conditions	K (theoretical)	K (recommended)
Pinned – Pinned (both ends simply supported)	1.0	1.0
Fixed – Fixed (both ends clamped)	0.5	0.65
Fixed – Pinned (one clamped, one pinned)	0.7	0.8
Fixed – Free (cantilever, flagpole)	2.0	2.1

Always use the smaller I (weak-axis bending). Check both axes if they have different effective lengths.

Critical Buckling Stress & Slenderness Ratio

\[ \sigma_{cr} = \frac{P_{cr}}{A} = \frac{\pi^2 E}{(L_e/r)^2} \]

Slenderness ratio (governs buckling mode) \[ \lambda = \frac{L_e}{r}, \quad r = \sqrt{\frac{I}{A}} \text{ (radius of gyration)} \]

Euler formula valid only when \(\sigma_{cr} < \sigma_{yield}\) (i.e., column is long / slender). For short columns, use material yield limit.

Four column end conditions and their effective length factors K (blue = buckled shape, green = support)

Example — Critical Load for W-Section Column

Steel W-section column (E = 200 GPa): I_min = 8.6×10⁶ mm⁴, L = 4 m, fixed-pinned. Find P_cr.

\(K = 0.7 \Rightarrow L_e = 0.7\times4000 = 2800\ \text{mm}\)

\(P_{cr} = \pi^2 E I_{min}/L_e^2 = \pi^2\times200{,}000\times8.6\times10^6/(2800)^2\)

\(P_{cr} = 9.87\times200{,}000\times8.6\times10^6/7.84\times10^6 = 2{,}165\ \text{kN}\)

16 Combined Loadings — Superposition

Prof's note: Problems will be basic and tied directly to class material. Simple, obvious, and low-value — but still important to know the method.

Principle of Superposition

For linear elastic materials under small deformations, stresses and displacements from multiple loads can be added algebraically at any point.

\[ \sigma_{total} = \sigma_{axial} + \sigma_{bending} + \sigma_{pressure} + \cdots \] \[ \tau_{total} = \tau_{torsion} + \tau_{shear\,force} + \cdots \]

Axial + Bending (Eccentric Loading)

\[ \sigma = \frac{P}{A} \pm \frac{M\,c}{I} = \frac{P}{A} \pm \frac{P\,e\,c}{I} \]

\(e\) = eccentricity (distance from centroid to line of action of P). Maximum compressive or tensile stress occurs at extreme fibers.

Kern (no tension zone for eccentricity) \[ e \leq \frac{I}{A\,c} = \frac{r^2}{c} \]

Bending + Torsion in Shafts

Normal stress from bending \[ \sigma = \frac{Mc}{I} \]

Shear stress from torsion \[ \tau = \frac{Tc}{J} \]

Maximum principal stress (using stress transformation) \[ \sigma_{1,2} = \frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2} \]

Maximum shear stress in shaft \[ \tau_{max} = \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau_{torsion}^2} \]

Combined: Axial + Bending + Torsion + Shear Force

At the critical point (usually the outer surface of a shaft or extreme fiber of a beam), identify:

\(\sigma_x = N/A \pm Mc/I\) (axial + bending)
\(\tau_{xy} = Tc/J + VQ/It\) (torsion + transverse shear — note VQ/It is often negligible vs torsion)

Full stress state → use Mohr's circle or transformation equations \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \]

Shaft under bending (P) + torsion (T): critical point at top/bottom of cross-section has both σ and τ

Example — Eccentric Axial Load on Column

Rectangular column 100×150 mm carries P = 80 kN at eccentricity e = 20 mm from the centroid (about the 100 mm axis). Find max/min stress.

\(\sigma_{axial} = P/A = 80{,}000/(0.1\times0.15) = 5.33\ \text{MPa (compression)}\)

\(M = P\times e = 80{,}000\times0.020 = 1600\ \text{N·m}\)

\(I = \tfrac{1}{12}(0.1)(0.15)^3 = 28.1\times10^{-6}\ \text{m}^4,\quad c = 0.075\ \text{m}\)

\(\sigma_{bending} = Mc/I = 1600\times0.075/28.1\times10^{-6} = 4.27\ \text{MPa}\)

\(\sigma_{max} = -5.33 + 4.27 = -1.06\ \text{MPa (comp)}\), \(\sigma_{min} = -5.33 - 4.27 = -9.60\ \text{MPa (comp)}\)

★ Quick-Reference Formula Card

Topic	Key Equation	Notes
Normal stress	\(\sigma = P/A\)	Positive = tension
Shear stress	\(\tau = V/A\)	Average
Hooke's Law	\(\sigma = E\varepsilon,\;\tau = G\gamma\)	\(G = E/2(1+\nu)\)
Thermal strain	\(\varepsilon_T = \alpha\Delta T\)	Steel: α≈12e-6/°C
Axial deformation	\(\delta = PL/AE\)	Sum for segments
Torsion stress	\(\tau = Tc/J\)	J solid: πc⁴/2
Angle of twist	\(\phi = TL/GJ\)
Thin-wall torsion	\(\tau = T/(2A_m t)\)	A_m = median enclosed area
Bending stress	\(\sigma = -My/I\)	S = I/c section modulus
Beam shear	\(\tau = VQ/It\)	Q = ȳ'A' first moment
Shear flow	\(q = VQ/I\)	Bolt force: F = qs
Composite beam	\(\sigma_2 = n\,My/I_T\)	n = E₂/E₁
Beam deflection	\(EIy'' = M(x)\)	4th order: EIy'''' = -w
Principal stresses	\(\sigma_{1,2} = C \pm R\)	\(C = (\sigma_x+\sigma_y)/2\)
Max in-plane τ	\(\tau_{max} = R\)	\(R = \sqrt{((\sigma_x-\sigma_y)/2)^2+\tau_{xy}^2}\)
Cylinder hoop	\(\sigma_1 = pr/t\)	Axial: pr/2t
Sphere	\(\sigma = pr/2t\)	Equal biaxial
Euler buckling	\(P_{cr} = \pi^2EI/L_e^2\)	Use min I; \(L_e = KL\)
Combined loading	\(\sigma_{tot} = N/A \pm Mc/I\)	Then use Mohr's circle