Exhaustive Study Notes on Machine Design: Curved Beams, Thick Cylinders, and Helical Springs

Axis Relationship

The Centroidal Axis ($CA$) and Neutral Axis ($NA$) coincide.

The Neutral Axis ($NA$) does not coincide with the Centroidal Axis ($CA$); $NA$ shifts toward the center of curvature.

Stress Distribution

Linear distribution across the section (Tensile and Compressive).

Hyperbolic and non-linear distribution across the section.

Eccentricity ($e$)

No eccentricity ($e = 0$) because $CA$ and $NA$ coincide.

The distance between $CA$ and $NA$ is called eccentricity ($e$).

Bending Stress Formula

Calculated using $\sigma_b = \frac{M \times y}{I}$

Calculated using $\sigma_b = \frac{M \times y}{A \times e \times (R_n + y)}$

Symmetry Effects

For symmetrical sections (circle, $, rectangle, I-beam), max tensile stress equals max compressive stress. | Max tensile and compressive stresses are different even for symmetrical sections because$NA$and$CA differ. |\n| **Stress at Axes** | Bending stress is 0$at both$CA$and$NA$. | Bending stress is$0$at$NA$, but not at$CA. |\n\n# Derivation of Bending Stresses: Winkler-Bach Formula\n\nTo find the bending stresses at the inner and outer fibers of a curved beam, consider an element AB$of area$dA$at a distance$y$from the neutral axis. This element subtends an angle$d\theta$due to the application of a bending moment$M$. The rotation shifts section$BB$to$B'B'$at angle$d\phi.\n\n## Assumptions\n1. The material is homogeneous and isotropic.\n2. Sections that are plane before bending remain plane after bending.\n3. The modulus of elasticity (E) is the same in tension and compression.\n4. The material obeys Hooke's Law.\n\n## Geometric Parameters\n- M: Applied bending moment.\n- R_i: Inner radius of the curved beam.\n- R_o: Outer radius of the curved beam.\n- R_n$: Radius of the Neutral Axis ($NA).\n- R$: Radius of the Centroidal Axis ($CA).\n- e$: Eccentricity ($R - R_n).\n- c_i$: Distance from$NA$to inner fiber ($R_n - R_i).\n- c_o$: Distance from$NA$to outer fiber ($R_o - R_n).\n\n## Stress and Strain Relationships\n- Original length of element AB = (R_n + y) \, d\theta.\n- Change in length due to bending B'B' = y \, d\phi.\n- Strain (\epsilon):\n\epsilon = \frac{\text{Change in length}}{\text{Original length}} = \frac{y \, d\phi}{(R_n + y) \, d\theta}\n- Bending Stress (\sigma_b):\n\sigma_b = \epsilon \times E = \frac{y \, d\phi \, E}{(R_n + y) \, d\theta}\n\n## Equilibrium Conditions\nAt equilibrium, the sum of forces (\sum F$) and the total moment ($\sum M) across the section must equal zero.\n\n### Case a) Force Equilibrium (\sum F = 0)\nIntegrating the elemental force df = \sigma_b \, dA$over the area$A:\n\int_A \frac{y \, d\phi \, E \, dA}{(R_n + y) \, d\theta} = 0\nSince E$,$d\phi$, and$d\theta are constants:\n\frac{E \, d\phi}{d\theta} \int_A \frac{y \, dA}{(R_n + y)} = 0\n\int_A \frac{y}{(R_n + y)} \, dA = 0\n\n### Case b) Moment Equilibrium (\sum M = M)\nThe total moment is the integral of the elemental moment dM = df \times y:\nM = \int_A \frac{y^2 \, d\phi \, E \, dA}{(R_n + y) \, d\theta}\nM = \frac{E \, d\phi}{d\theta} \int_A \frac{y^2}{(R_n + y)} \, dA\nTo simplify, rewrite y^2$as$y^2 + y R_n - y R_n:\nM = \frac{E \, d\phi}{d\theta} \int_A (y - \frac{R_n y}{R_n + y}) \, dA\nFrom equation 4 (\int_A \frac{y}{R_n + y} \, dA = 0), this reduces to:\nM = \frac{E \, d\phi}{d\theta} \int_A y \, dA\nSince \int_A y \, dA$is the total moment of area on the neutral axis, if the entire area is concentrated at the geometrical axis ($Ae):\nM = \frac{E \, d\phi}{d\theta} A e\nTherefore, \frac{E \, d\phi}{d\theta} = \frac{M}{A e}.\n\n## Final Stress Formulas\nSubstituting the constant back into the stress equation:\n\sigma_b = \frac{M \, y}{A \, e \, (R_n + y)}\n\n- **Inner Fiber Stress (\sigma_{bi}):**\n\sigma_{bi} = \frac{M \, c_i}{A \, e \, R_i}\n- **Outer Fiber Stress (\sigma_{bo}):**\n\sigma_{bo} = \frac{M \, c_o}{A \, e \, R_o}\n*(Note: Tension is positive and compression is negative, depending on the bending direction)*.\n\n# Design Problem: Crane Hook Analysis\n\n**Problem:** Find the maximum stress for a crane hook carrying a load of 9600\,\text{N}$with the following cross-sectional dimensions:$b_i = 100\,\text{mm}$,$b_o = 50\,\text{mm}$,$h = 120\,\text{mm}$, inner radius$R_i = 70\,\text{mm}.\n\n### Step 1: Find the Center of Gravity (CG)\nUsing the formula for a trapezoid:\nx = \frac{h}{3} \times (\frac{b_i + 2b_o}{b_i + b_o}) = \frac{120}{3} \times (\frac{100 + 100}{100 + 50}) = 53.33\,\text{mm}\nDistance from COC to the centroidal axis (R):\nR = x + R_i = 53.33 + 70 = 123.33\,\text{mm}\n\n### Step 2: Find the Neutral Axis Radius (R_n)\nUsing the formula for a trapezoidal section:\nR_n = \frac{A}{[\frac{(b_i R_o - b_o R_i)}{h} \ln(\frac{R_o}{R_i})] - (b_i - b_o)}\nArea A = \frac{1}{2} \times h \times (b_i + b_o) = \frac{1}{2} \times 120 \times (100 + 50) = 9000\,\text{mm}^2\nR_n = 113.95\,\text{mm}\n\n### Step 3: Eccentricity and Distances\n- Eccentricity e = R - R_n = 123.33 - 113.95 = 9.38\,\text{mm}.\n- c_i = R_n - R_i = 113.95 - 70 = 43.95\,\text{mm}.\n- c_o = R_o - R_n = 190 - 113.95 = 76.05\,\text{mm}.\n\n### Step 4: Bending Stress Calculation\nBending Moment M = F \times R = 9600 \times 123.33 = 1.183 \times 10^6\,\text{N}\,\text{mm}.\n- **Bending stress at inner fiber (\sigma_{bi}):**\n\sigma_{bi} = \frac{1.183 \times 10^6 \times 43.95}{9000 \times 9.38 \times 70} = 8.79\,\text{N/mm}^2\n- **Bending stress at outer fiber (\sigma_{bo}):**\n\sigma_{bo} = \frac{1.183 \times 10^6 \times 76.05}{9000 \times 9.38 \times 190} = 5.60\,\text{N/mm}^2 \, \text{(Compressive)}\n\n### Step 5: Direct Stress and Maximum Stress\nDirect stress (\sigma_d):\n\sigma_d = \frac{F}{A} = \frac{9600}{9000} = 1.067\,\text{N/mm}^2\nMaximum stress at inner fiber (Tensile):\n\sigma_{\max} = \sigma_d + \sigma_{bi} = 1.067 + 8.79 = 9.857\,\text{N/mm}^2\nMinimum stress at outer fiber:\n\sigma_{\min} = \sigma_d - \sigma_{bo} = 1.067 - 5.60 = -4.533\,\text{N/mm}^2 \, \text{(Compressive)}\n\n# Thick and Thin Cylinders\n\nCylinders and tanks store fluids under pressure. Their design is critical as failure/rupture can lead to significant loss of life and property.\n\n## Classification\n- **Thin Cylinders:** Used in boilers, tanks, and pipes. The wall thickness is small relative to the diameter.\n- **Thick Cylinders:** Used for high-pressure storage, gun barrels, etc.\n- **Ends:** Can be "Open" (e.g., cylinder with a piston) or "Closed" (e.g., a tank).\n- **Stresses:** In open-cycle cylinders, only hoop stresses are induced. In closed-end cylinders, both hoop and longitudinal stresses are present.\n\n## Lame's Equations for Thick Cylinders\nStresses are calculated based on radii:\n- **Hoop Stress (\sigma_t):**\n\sigma_t = \frac{P_i d_i^2 - P_o d_o^2 + \frac{d_i^2 d_o^2}{4 r^2} (P_i - P_o)}{d_o^2 - d_i^2}\n- **Radial Stress (\sigma_r):**\n\sigma_r = \frac{P_i d_i^2 - P_o d_o^2 - \frac{d_i^2 d_o^2}{4 r^2} (P_i - P_o)}{d_o^2 - d_i^2}\nWhere P_i$is internal pressure and$P_o is external pressure.\n\n## Clavarino's Equation\nUsed specifically for closed cylinders to account for lateral deformation (\mu = Poisson's ratio):\nt = \frac{d_i}{2} \left[ \sqrt{\frac{\sigma_t + (1 - 2\mu)P_i}{\sigma_t - (1 + \mu)P_i}} - 1 \right]\nFor cast iron, Poisson's ratio \mu \approx 0.27.\n\n# Helical Springs\n\nSprings are mechanical elements that deflect under load, absorbing and storing energy within elastic limits. This energy is released when the load is removed.\n\n## Applications\n- Shock absorbers in automobiles.\n- Mechanical damping in machine tools.\n- Inlet/outlet valves in IC engines.\n- Torsion springs in watches and toys.\n- Impact cushions or spring balances.\n\n## Nomenclature\n- **Wire Diameter (d):** The diameter of the wire used to form the spring.\n- **Mean Diameter (D$):** Measured from the center of the wire ($D = D_o - d).\n- **Outer Diameter (D_o$):** Maximum outside dimension ($D + d).\n- **Inner Diameter (D_i$):** Inside dimension ($D - d).\n- **Spring Index (C$):** Ratio of mean diameter to wire diameter ($C = D/d$). Typically ranges from$6 - 12.\n- **Active Coils (n):** Coils that are free to deflect under load.\n- **Free Length (FL$):** Total unloaded length ($FL = (n + n')d + y + a$), where$n'$are inactive coils and$a is clearance.\n\n## Stress and Deflection Formulas\n\n### Wahl Stress Factor (k)\nAccounts for curvature and transverse shear:\nk = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}\n\n### Induced Shear Stress (\tau)\n\tau = \frac{8 F D k}{\pi d^3} = \frac{8 F C k}{\pi d^2}\n\n### Axial Deflection (y)\ny = \frac{8 F D^3 n}{G d^4}\nWhere G is the Modulus of Rigidity.\n\n### Variable Load Wire Diameter\nFor dynamic loads, a reliability factor (R$) or factor of safety ($FOS) is applied:\nd = \left[ \frac{1.35 R (3 F_{\max} - F_{\min})}{\tau_{\text{allowable}}} \right] \times C^{0.43}\n\n# Questions & Discussion\n\n**Q: How does the maximum stress in a thick cylinder compare to a thin cylinder?**\n**A:** In a thick cylinder, the stress varies across the wall thickness (highest at the inner surface), whereas in a thin cylinder, it is assumed to be uniform across the wall.\n\n**Q: What is the benefit of nested (parallel) springs?**\n**A:** Nested springs can withstand higher loads within a restricted space. The total load F$is the sum of the loads shared by the individual springs ($F = F_{\text{inner}} + F_{\text{outer}}$), while the deflection$y$$ remains the same if the free lengths are adjusted properly.