PNGE Mechanics Calculator

Engineering mechanics skill for petroleum engineering applications. Covers statics (MAE 201), mechanics of materials (MAE 243), and fluid statics — with worked examples for wellbore design, casing loading, derrick structures, and tubular analysis.

Important: Results are based on classical analytical solutions and standard engineering correlations. For final design decisions, verify against API standards (API 5C3, API RP 7G) and perform full finite-element analysis where required. Always apply appropriate safety factors.

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Module 1 — Statics and Force Equilibrium

Capabilities

• Sum of forces and moments for 2D and 3D systems
• Free body diagram construction (described with ASCII art where useful)
• Support reactions for beams (pin, roller, fixed)
• Truss analysis: method of joints and method of sections
• PNGE applications: crown block loads, derrick leg loading, hook load, traveling block

Equilibrium Equations

For a body in static equilibrium, all forces and moments sum to zero:

ΣFx = 0    ΣFy = 0    ΣFz = 0
ΣMx = 0    ΣMy = 0    ΣMz = 0

For 2D problems only ΣFx, ΣFy, and ΣM_point = 0 are needed.

Beam Support Reactions

Support Type	Reactions Provided
Roller	1 force (normal to surface)
Pin (hinge)	2 forces (Rx, Ry)
Fixed (cantilever)	2 forces + 1 moment (Rx, Ry, M)

Workflow — Support Reactions

1. Draw the free body diagram; replace each support with its reaction forces/moments
2. Label all known loads (magnitudes, directions, positions)
3. Write the three equilibrium equations
4. Solve the system of equations for unknowns
5. Verify: sum all forces and moments independently

Derrick Hook Load Example

Problem: A crown block is mounted at the top of a derrick mast. The hoisting cable makes a 5-degree angle from vertical due to offset. Hook load = 400,000 lb (including traveling block weight).

Free body diagram (ASCII):

          [Crown block]
         /|
      T / | Vertical component
       /  |
      /5° |
     /    v
  [Cable] ----> Horizontal component

Solution:

T_vertical = 400,000 * cos(5°) = 400,000 * 0.9962 = 398,480 lb
T_horizontal = 400,000 * sin(5°) = 400,000 * 0.0872 = 34,880 lb

The crown block must resist both components. Horizontal component creates a bending moment on the mast structure.

Truss Analysis — Method of Joints

1. Find all external reactions using global equilibrium
2. Isolate each joint; draw FBD with all member forces as unknowns
3. At each joint: ΣFx = 0 and ΣFy = 0
4. Solve for two unknowns per joint; proceed joint by joint
5. Positive result = tension; negative = compression

Truss Analysis — Method of Sections

Use when only a few member forces are needed:

1. Pass an imaginary cut through no more than 3 members
2. Isolate one side; draw FBD with cut member forces as unknowns
3. Apply three equilibrium equations to solve for the three unknowns
4. Choose moment centers to reduce simultaneous equations

Example Python — Truss Joint

import math

def solve_joint_2member(F_ext_x, F_ext_y, theta1_deg, theta2_deg):
    """
    Solve for forces in two members meeting at a joint.
    theta = angle from positive x-axis (degrees).
    Returns (F1, F2); positive = tension.
    """
    t1 = math.radians(theta1_deg)
    t2 = math.radians(theta2_deg)
    # [cos(t1) cos(t2)] [F1]   [-F_ext_x]
    # [sin(t1) sin(t2)] [F2] = [-F_ext_y]
    det = math.cos(t1)*math.sin(t2) - math.cos(t2)*math.sin(t1)
    if abs(det) < 1e-10:
        raise ValueError("Members are parallel — joint is unstable")
    F1 = (-F_ext_x * math.sin(t2) + F_ext_y * math.cos(t2)) / det
    F2 = ( F_ext_x * math.sin(t1) - F_ext_y * math.cos(t1)) / det
    return F1, F2

# Example: vertical load 50,000 lb downward at joint
# Member 1 at 135°, Member 2 at 45°
F1, F2 = solve_joint_2member(0, -50000, 135, 45)
print(f"F1 = {F1:,.0f} lb, F2 = {F2:,.0f} lb")
# Positive = tension, negative = compression

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Module 2 — Axial Stress, Strain, and Deformation

Capabilities

• Normal stress and strain in rods, casing, tubing, and drill pipe
• Axial deformation under mechanical and thermal loads
• Statically indeterminate axial systems
• PNGE applications: casing string tension analysis, tubing elongation, packerless completions

Core Equations

See references/equations.md for full derivations and variable definitions.

Normal stress:

sigma = P / A   (psi or MPa)

Strain and deformation:

epsilon = delta / L = sigma / E
delta = P * L / (A * E)

Thermal expansion:

delta_T = alpha * delta_T * L

For combined mechanical and thermal loading:

delta_total = P*L/(A*E) + alpha * delta_T * L

For statically indeterminate problems, compatibility (geometry of deformation) adds the equation needed beyond equilibrium alone.

Typical Material Properties

Material	E (psi)	E (GPa)	alpha (/°F)	Yield (ksi)
Steel (API N-80)	30,000,000	207	6.5e-6	80
Steel (API P-110)	30,000,000	207	6.5e-6	110
Steel (API Q-125)	30,000,000	207	6.5e-6	125
Aluminum drill pipe	10,000,000	69	13.1e-6	60

Workflow — Casing Axial Load Analysis

1. Identify all axial loads: buoyancy-corrected string weight, hook load, packer loads, pressure-induced forces
2. Calculate cross-sectional area A = pi*(OD^2 - ID^2)/4
3. Compute stress: sigma = P/A; compare to yield strength with safety factor
4. Compute elongation (or contraction) for each load case
5. Check against API rating and required safety factor (typically 1.6 for tension)

Example — Casing String Elongation Under Tension

Problem: 7-inch OD, 26 lb/ft casing (ID = 6.276 in), 8,000 ft in air. Calculate axial stress and elongation at surface due to string weight.

import math

# Casing geometry
OD_in = 7.0        # inches
ID_in = 6.276      # inches
L_ft = 8000.0      # feet
weight_lbft = 26.0 # lb/ft (in air)
E_psi = 30e6       # psi

# Area
A = math.pi / 4 * (OD_in**2 - ID_in**2)  # in^2
print(f"Cross-sectional area: {A:.3f} in^2")

# Total weight (load at top of string)
P_total = weight_lbft * L_ft  # lb
print(f"Total string weight: {P_total:,.0f} lb")

# Axial stress at top (maximum)
sigma = P_total / A  # psi
print(f"Axial stress at surface: {sigma:,.0f} psi = {sigma/1000:.1f} ksi")

# Elongation
L_in = L_ft * 12  # convert to inches
delta = P_total * L_in / (2 * A * E_psi)  # average load is half
# More precisely: integrate along length
# delta = integral(W(z)/A/E dz) from 0 to L
# where W(z) = weight_lbft * z * 12 (lb) at depth z
# delta = weight_lbft * 12 * L^2 / (2*A*E) — units: lb/in * in^2 / (in^2 * psi) = in
delta_in = weight_lbft * 12 * (L_ft**2) / (2 * A * E_psi)
print(f"Elongation: {delta_in:.2f} in = {delta_in/12:.3f} ft")

# Safety factor check vs N-80 yield
yield_psi = 80000  # psi
SF = yield_psi / sigma
print(f"Safety factor vs yield: {SF:.2f}x")

Thermal Expansion of Tubing

Problem: 9,500 ft of 2-7/8 inch tubing (alpha = 6.5e-6 /°F). Temperature increases from 70°F to 220°F after production begins.

alpha = 6.5e-6   # /°F, steel
dT = 220 - 70    # 150°F temperature increase
L = 9500 * 12    # inches

delta_thermal = alpha * dT * L
print(f"Free thermal expansion: {delta_thermal:.2f} in = {delta_thermal/12:.2f} ft")
# If tubing is constrained by packer, this thermal expansion induces
# compressive stress: sigma = -E * alpha * dT
sigma_thermal = -30e6 * alpha * dT
print(f"Induced compressive stress (if constrained): {sigma_thermal:,.0f} psi")

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Module 3 — Beam Bending

Capabilities

• Shear force and bending moment diagrams for standard loading cases
• Flexure formula for bending stress in pipe and hollow sections
• Second moment of area for circular and hollow circular sections
• Beam deflection formulas (cantilever, simply supported, overhanging)
• PNGE applications: drill collar bending analysis, derrick beam sizing, casing in curved wellbore

Core Equations

Flexure formula:

sigma_max = M * c / I

Second moment of area — hollow circular (pipe):

I = pi * (d_o^4 - d_i^4) / 64

Section modulus:

S = I / c = I / (d_o/2)

Standard Beam Formulas

Loading	Max Moment	Max Deflection
Simply supported, center point load P	M = P*L/4 (at center)	delta = PL^3/(48E*I) (at center)
Simply supported, UDL w (lb/in)	M = w*L^2/8 (at center)	delta = 5wL^4/(384EI) (at center)
Cantilever, end point load P	M = P*L (at wall)	delta = PL^3/(3E*I) (at free end)
Cantilever, UDL w (lb/in)	M = w*L^2/2 (at wall)	delta = wL^4/(8E*I) (at free end)

Workflow — Shear and Moment Diagrams

1. Find reactions at supports using equilibrium
2. Starting from the left: shear V(x) changes by +R at reactions (upward) and -P at point loads
3. Shear is constant between loads; moment M(x) = integral of V(x)
4. Bending moment is maximum where shear = 0 (changes sign)
5. Compute sigma_max from M_max, c, and I

Example — Drill Collar Bending

Problem: A 6.75-inch OD, 2.25-inch ID drill collar (9.625 in OD housing excluded) spans 30 ft between stabilizers in a horizontal section. Formation side-load = 500 lb/ft distributed. Calculate maximum bending stress.

import math

# Drill collar geometry
OD_in = 6.75    # inches
ID_in = 2.25    # inches
L_ft = 30.0     # feet between stabilizers
w_lbft = 500.0  # distributed side load, lb/ft

# Convert to inch units
L = L_ft * 12   # inches
w = w_lbft / 12 # lb/in

# Second moment of area
I = math.pi * (OD_in**4 - ID_in**4) / 64
print(f"I = {I:.2f} in^4")

# Outer fiber distance
c = OD_in / 2
print(f"c = {c:.3f} in")

# Simply-supported assumption (stabilizers act as pins)
M_max = w * L**2 / 8  # lb-in
print(f"M_max = {M_max:,.0f} lb-in = {M_max/12000:.1f} kip-ft")

# Maximum bending stress
sigma_b = M_max * c / I
print(f"Bending stress: {sigma_b:,.0f} psi = {sigma_b/1000:.1f} ksi")

# Deflection at center
E = 30e6  # psi, steel
delta_max = 5 * w * L**4 / (384 * E * I)
print(f"Max deflection: {delta_max:.3f} in")

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Module 4 — Thick-Walled Cylinder (Lame Equations)

Capabilities

• Radial and hoop stress distribution through cylinder wall (Lame)
• Maximum stress at inner radius for internal pressure loading
• Maximum stress at outer radius for external pressure loading
• Von Mises yield criterion for biaxial stress state
• API casing burst and collapse pressure estimation
• PNGE applications: casing design, tubing burst/collapse, wellhead pressure ratings

Lame Equations

For a thick-walled cylinder under internal pressure p_i and external pressure p_e:

Constants A and B:

A = (p_i * r_i^2 - p_e * r_e^2) / (r_e^2 - r_i^2)
B = (p_i - p_e) * r_i^2 * r_e^2 / (r_e^2 - r_i^2)

Stress at any radius r:

sigma_r = A - B/r^2      (radial stress, compressive at inner wall)
sigma_theta = A + B/r^2  (hoop/circumferential stress)

At inner radius (critical for burst):

sigma_theta_max = (p_i * (r_i^2 + r_e^2) - 2*p_e*r_e^2) / (r_e^2 - r_i^2)

At outer radius (critical for collapse):

sigma_theta_outer = (2*p_i*r_i^2 - p_e*(r_i^2 + r_e^2)) / (r_e^2 - r_i^2)

Von Mises Yield Criterion (Plane Stress)

For biaxial stress state (sigma_r, sigma_theta) at any point:

sigma_vm = sqrt(sigma_r^2 - sigma_r*sigma_theta + sigma_theta^2) <= sigma_yield

With axial stress sigma_z included (triaxial):

sigma_vm = (1/sqrt(2)) * sqrt((s1-s2)^2 + (s2-s3)^2 + (s1-s3)^2)

Workflow — Casing Burst Check

1. Identify: OD, ID (or wall thickness t), internal pressure p_i, external (mud or cement) pressure p_e
2. Compute r_i = ID/2, r_e = OD/2
3. Compute A and B
4. Find sigma_theta at r_i (maximum hoop stress, controlling for burst)
5. Compute Von Mises effective stress
6. Compare to yield strength with safety factor (API minimum = 1.0; design often 1.25)

Example — API 7-inch Casing Burst Check

Problem: 7-inch OD, 23 lb/ft casing (ID = 6.366 inch, t = 0.317 inch), N-80 steel (80,000 psi yield). Internal pressure = 5,000 psi, external (mud) pressure = 3,000 psi at the point of interest.

import math

# Casing geometry
OD_in = 7.000   # inches
ID_in = 6.366   # inches
r_i = ID_in / 2
r_e = OD_in / 2

# Pressures
p_i = 5000.0    # psi, internal
p_e = 3000.0    # psi, external (mud column)

# Lame constants
A = (p_i * r_i**2 - p_e * r_e**2) / (r_e**2 - r_i**2)
B = (p_i - p_e) * r_i**2 * r_e**2 / (r_e**2 - r_i**2)
print(f"Lame constants: A = {A:.1f} psi, B = {B:.1f} psi-in^2")

# Stresses at inner radius (burst-critical)
sigma_r_inner = A - B / r_i**2   # should equal -p_i (BC check)
sigma_t_inner = A + B / r_i**2   # max hoop stress
print(f"\nAt inner wall (r = {r_i:.3f} in):")
print(f"  sigma_r = {sigma_r_inner:.0f} psi (check: should be -{p_i})")
print(f"  sigma_theta = {sigma_t_inner:.0f} psi")

# Stresses at outer radius
sigma_r_outer = A - B / r_e**2   # should equal -p_e
sigma_t_outer = A + B / r_e**2
print(f"\nAt outer wall (r = {r_e:.3f} in):")
print(f"  sigma_r = {sigma_r_outer:.0f} psi (check: should be -{p_e})")
print(f"  sigma_theta = {sigma_t_outer:.0f} psi")

# Von Mises at inner wall (biaxial: assume sigma_z = axial/2 for capped ends)
# For open-ended or free tube, sigma_z ~ A (Lame axial)
sigma_z = A  # Lame plane strain approximation
sigma_vm = math.sqrt(
    sigma_t_inner**2 - sigma_t_inner * sigma_z + sigma_z**2
    - sigma_r_inner * (sigma_t_inner + sigma_z - sigma_r_inner)
)
# Simplified Von Mises for sigma_r, sigma_theta only:
sigma_vm_2d = math.sqrt(
    sigma_r_inner**2 - sigma_r_inner * sigma_t_inner + sigma_t_inner**2
)
print(f"\nVon Mises (biaxial): {sigma_vm_2d:.0f} psi")

yield_psi = 80000
SF = yield_psi / sigma_vm_2d
print(f"Safety factor vs N-80 yield: {SF:.2f}x")

# Net internal pressure approach (API simplified)
p_net = p_i - p_e
burst_approx = 0.875 * 2 * yield_psi * (r_e - r_i) / OD_in
print(f"\nAPI simplified burst rating: {burst_approx:.0f} psi")
print(f"Net pressure: {p_net} psi — {'PASS' if p_net < burst_approx else 'FAIL'}")

API Casing Rating Quick Reference

Grade	Min Yield (psi)	Use Case
J-55	55,000	Surface casing, low pressure
K-55	55,000	Similar to J-55, better weldability
N-80	80,000	Intermediate casing, common
L-80	80,000	Sour service (H2S)
P-110	110,000	Production casing, high pressure
Q-125	125,000	HPHT applications

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Module 5 — Mohr's Circle and Stress Transformation

Capabilities

• Transform stress components from one coordinate system to another
• Find principal stresses (maximum and minimum normal stress)
• Find maximum shear stress
• Mohr's circle construction and reading
• PNGE applications: in-situ stress analysis, failure analysis of tubulars, borehole wall stress state

Stress Transformation Equations

For a 2D stress state (sigma_x, sigma_y, tau_xy), the stress on a plane rotated by angle theta from the x-axis:

sigma_n = (sigma_x + sigma_y)/2 + (sigma_x - sigma_y)/2 * cos(2*theta) + tau_xy * sin(2*theta)
tau = -(sigma_x - sigma_y)/2 * sin(2*theta) + tau_xy * cos(2*theta)

Principal Stresses

sigma_1,2 = (sigma_x + sigma_y)/2  ±  sqrt[((sigma_x - sigma_y)/2)^2 + tau_xy^2]

The angle to the principal plane:

tan(2*theta_p) = 2*tau_xy / (sigma_x - sigma_y)

Maximum Shear Stress

tau_max = sqrt[((sigma_x - sigma_y)/2)^2 + tau_xy^2]
        = (sigma_1 - sigma_2) / 2

Occurs at 45° from the principal planes.

Mohr's Circle Construction

1. Plot point X = (sigma_x, -tau_xy) and point Y = (sigma_y, +tau_xy)
2. The center C is at ((sigma_x + sigma_y)/2, 0)
3. The radius R = tau_max = sqrt[((sigma_x-sigma_y)/2)^2 + tau_xy^2]
4. Principal stresses: sigma_1 = C + R (rightmost), sigma_2 = C - R (leftmost)
5. The angle XC to the sigma axis = 2*theta_p (counterclockwise on Mohr's circle = counterclockwise on element)

Example — In-Situ Stress Analysis

Problem: At a depth of 7,000 ft in the Marcellus, the in-situ stress state in the horizontal plane is: S_Hmax = 6,500 psi, S_hmin = 4,800 psi. A borehole is drilled at 30° to the S_Hmax direction. Find the normal and shear stress on the borehole axis.

import math

# In-situ stresses (horizontal plane, compressive positive convention)
S_Hmax = 6500  # psi
S_hmin = 4800  # psi
tau_xy = 0     # principal stresses, so no shear in original frame

theta_deg = 30   # borehole axis angle from S_Hmax
theta = math.radians(theta_deg)

# Stress transformation
sigma_n = (S_Hmax + S_hmin)/2 + (S_Hmax - S_hmin)/2 * math.cos(2*theta) + tau_xy * math.sin(2*theta)
tau_nt  = -(S_Hmax - S_hmin)/2 * math.sin(2*theta) + tau_xy * math.cos(2*theta)

print(f"Normal stress on plane perpendicular to borehole axis: {sigma_n:.0f} psi")
print(f"Shear stress on that plane: {tau_nt:.0f} psi")

# Principal stresses (these are already principal since tau_xy=0)
sigma_1 = S_Hmax
sigma_2 = S_hmin
tau_max = (sigma_1 - sigma_2) / 2

print(f"\nPrincipal stresses: sigma_1 = {sigma_1} psi, sigma_2 = {sigma_2} psi")
print(f"Maximum shear stress: {tau_max:.0f} psi")
print(f"Center of Mohr's circle: {(sigma_1+sigma_2)/2:.0f} psi")
print(f"Radius of Mohr's circle: {tau_max:.0f} psi")

Torsional Shear Stress

For circular shaft or drill string under torque T:

tau = T * r / J

Where J = polar moment of inertia:

J = pi * (d_o^4 - d_i^4) / 32 = 2 * I

Maximum shear stress at outer radius r = d_o/2:

tau_max = T * (d_o/2) / J = 2*T / (pi * (d_o^4 - d_i^4) / 16)

Angle of twist:

phi = T * L / (G * J)

Where G = shear modulus = E / (2*(1+nu)) ≈ 11.5e6 psi for steel.

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Module 6 — Fluid Statics and Wellbore Pressure

Capabilities

• Hydrostatic pressure at depth for single and multi-fluid columns
• Wellbore pressure profiles with changing mud weights
• Buoyancy force on casing string in fluid
• ECD (equivalent circulating density) estimates
• PNGE applications: mud weight selection, casing design pressure loads, kick detection

Hydrostatic Pressure

Single fluid:

P = rho * g * h   (SI: Pa)
P = gradient * depth   (field: psi, where gradient in psi/ft)

Pressure gradient conversion:

gradient (psi/ft) = mud_weight (ppg) * 0.052
gradient (psi/ft) = density (g/cc) * 0.4335

Multi-fluid column:

P_bottom = P_surface + sum[gradient_i * depth_i]

Buoyancy Force on Casing String

The buoyant weight of a casing string in fluid (Archimedes principle):

W_buoyant = W_air - W_fluid_displaced
W_buoyant = W_air * (1 - rho_fluid / rho_steel)

For a steel casing string (rho_steel = 65.4 lb/gal = 7,850 kg/m^3) in 10.5 ppg mud:

buoyancy_factor = 1 - 10.5/65.4 = 0.8394

Workflow — Wellbore Pressure Profile

1. Identify all fluid columns in the annulus (water, mud, cement, spacer)
2. Starting from surface (known pressure or atmospheric), add gradient * depth for each section
3. At each casing shoe, record the pressure
4. Compare to formation pore pressure (must exceed) and fracture gradient (must be below)
5. Plot: depth vs pressure with pore pressure and fracture gradient lines

Example — Wellbore Pressure Profile

Problem: 10,000 ft well. Mud weight = 9.8 ppg from surface to 10,000 ft. Surface pressure = 0 psi. Calculate pressure profile and buoyancy factor.

# Wellbore pressure profile
mw_ppg = 9.8          # mud weight, ppg
depth_ft = 10000      # total depth, ft
surf_P = 0            # surface pressure, psi

# Pressure gradient
grad_psi_ft = mw_ppg * 0.052
print(f"Mud gradient: {grad_psi_ft:.4f} psi/ft")

# Pressure at various depths
depths = [0, 2000, 4000, 6000, 8000, 10000]
print("\nDepth (ft) | Pressure (psi) | Pressure (ppg equiv)")
print("-" * 55)
for d in depths:
    P = surf_P + grad_psi_ft * d
    P_ppg = P / (0.052 * d) if d > 0 else mw_ppg
    print(f"{d:10,} | {P:14,.0f} | {P_ppg:.2f}")

# Buoyancy factor for steel casing
rho_steel_ppg = 65.4
BF = 1 - mw_ppg / rho_steel_ppg
print(f"\nBuoyancy factor: {BF:.4f}")
print(f"7-inch 26 lb/ft casing string: air weight = {26*10000:,} lb")
print(f"Buoyant weight = {26*10000*BF:,.0f} lb")

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Workflow Summary

Step 1 — Identify Module

User Request	Module
Force equilibrium, reactions, truss, derrick loads	Module 1 — Statics
Axial stress, strain, elongation, thermal, casing tension	Module 2 — Axial
Bending moment, shear diagram, deflection	Module 3 — Beams
Casing burst/collapse, thick-walled cylinder, hoop stress	Module 4 — Thick-Walled Cylinder
Mohr's circle, principal stresses, transformation, torsion	Module 5 — Mohr's Circle
Mud pressure, hydrostatic, buoyancy, ECD	Module 6 — Fluid Statics

Step 2 — Gather Inputs

Collect required parameters. Always confirm units. Provide sensible defaults from the typical Appalachian/Marcellus parameter set if user omits values.

Step 3 — Calculate

Perform step-by-step arithmetic, showing each equation and substitution. Include Python code using only math and statistics (stdlib). Label all intermediate results with units.

Step 4 — Output

Format as:

1. Input summary table — echo back all parameters and assumptions
2. Equations used — state the governing equations before substituting
3. Step-by-step solution — each calculation step with units
4. Results table — final computed values in both field and SI units
5. Interpretation — is the design adequate? What is the safety factor?
6. Caveats — assumptions, limits of validity, what to check next

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Output Format

## [Calculation Title]

### Problem Statement
[Restate what is being solved and why it matters]

### Input Parameters
| Parameter | Symbol | Value | Unit |
|-----------|--------|-------|------|
| ... | ... | ... | ... |

### Governing Equations
[State equations explicitly before substituting numbers]

### Solution
Step 1: [calculation with units]
Step 2: [calculation with units]
...

### Results
| Result | Value | Unit |
|--------|-------|------|
| ... | ... | ... |

**Summary:** [1-3 sentences on adequacy, safety factor, recommendation]

**Caveats:**
- [Key assumptions, e.g., "assumes purely axial loading — bending neglected"]
- [Validity range of any correlations used]
- [Recommend API standard or full analysis for design decisions]

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Error Handling

Condition	Action
Missing geometry (OD, ID, length)	Request from user; provide common API casing tables as reference
Pressure exceeds yield with SF less than 1	Flag FAIL clearly; suggest next heavier weight or higher grade
Division by zero (zero area, zero moment arm)	Report degenerate geometry; ask user to verify
Value outside typical range	Warn with typical range; calculate anyway with disclaimer
Statically indeterminate without deformation data	Request E, A, or geometry for compatibility equation

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Units Reference

All calculations default to field units. Present final answers in both field and SI.

Quantity	Field Unit	SI Unit	Conversion
Force	lb	N	1 lb = 4.448 N
Pressure/Stress	psi	MPa	1 psi = 0.006895 MPa
Length	ft or in	m	1 ft = 0.3048 m; 1 in = 0.0254 m
Moment	lb-ft or lb-in	N-m	1 lb-ft = 1.356 N-m
Young's modulus	psi	GPa	30e6 psi = 207 GPa (steel)
Density	ppg	kg/m3	1 ppg = 119.8 kg/m3
Mud gradient	psi/ft	kPa/m	1 psi/ft = 22.62 kPa/m
Temperature	°F	°C	°C = (°F - 32) * 5/9

See references/equations.md for the complete equation library with derivations.