Machine Design Formulario

Machine design constraints

The direction of the moment follows the rule of cross product.

Stress tensor

σ = σ_x τ_xy τ_xz τ_yx σ_y τ_yz τ_zx τ_zy σ_z

Bending

σ_x = My / J = M / J d / 2

• Circular section: J = πd⁴ / 64 => σ_x = 32M / πd³
• Rectangular section: J = 1/12 bd³ => σ_x = 6M / bd²

Shear

τ_max = VQ / tJ = 3V / 2A

• Circular section: τ_max = 4/3 V / A

Torque

τ_max = Mt·r / Jp

• Circular section: Jp = πd⁴ / 32 => τ_max = 16Mt / πd³

Notch effect

Theoretical Kt = σ_max / σ_nom

σ_max = F (axial) / A_net, σ_nom = My / J

Machine Design Constraints

• Roller
• Pinned
• Fixed support
• Prismatic joint

The direction of the moment follows the rule of cross product.

Stress tensor

σ = σ_xτ_xyτ_xzτ_yxσ_yτ_yzτ_zxτ_zyσ_z

Bending

σ_x = My / J = M / J d / 2

J = ∫y² dA

• Circular section: J = πd⁴ / 64 => σ_x = 32M / πd³
• Rectangular section: J = 1/12 bd³ => σ_x = 6M / bd²
• Hollow section: J = π/64 (d⁴ - di⁴)

Shear

τ_max = VQ / 2J => 3V / 2A

• Circular section: τ_max = 4/3 V / A

Torque

τ_max = Mt * r / Jp

• Circular section: Jp = πd⁴ / 32 => τ_max = 16Mt / πd³
• Hollow section: Jp = π/32 (d² - di²) => τ_max = 16Mt / (d²-di²) * d/2

Notch Effect

Theoretical Kt = σ_max / σ_nom

σ_max = F (axial), σ_nom = My / J (bending)

Kt = τ_max / τ_nom

τ_nom = V/A (shear), τ_nom = Mt * r / Jp (torque)

Effect of Kt on the safety factor η (static assessment): the constitutive law of the material should be taken into account.

Lb experimental stress concentration factors Ks = Feu / F/emin

Brittle materials: ks=kt

Ductile materials: (full plasticity) ks=1

Static Assessment

σ* ≤ σ_adm = σ_el,uη

• σ* ≤ σ_adm = σ_el,uη
• σ_t*, σ_max (eventually considering kb) for uniaxial shear state
• σ* equivalent → different criteria (eventually considering ks)
• σ_el,u = σ_el, ductile materials ≈ 0.1σ_y
• Brittle materials ≈ 0.58 σ_y

Principal stresses

|σ₁ - σ_I| = 0

σ₁ ≥ σ_I → σ₁ = σ_III in one case σ → plane stress condition

Mohr’s circle if σ’ ≤ C_11^1/2 ≥ 0

σ’ = (σ') + ½ [ (σ')² + 4τ² ]^1/2

In uniaxial stress, relate the equivalent quantity representative of the damage must be minimum if the reference system ... according to the principal stress.

AC - EIGINGER_D^u

Ductile materials - same formulation of VN

η = safety factor, production factor of the material strength 1.25

GALILEO-RANKINE-NAVIER

Brittle materials σ_t = σ_cn^{|σI| ≤ |σc|}_η

GUEST - TRESCA

Ductile materials σΓ = σΓ1 - σΓIII ≤ σ_yη

Plane stress: bending + torsion only

σ_t* = √σ² + 4τ² / 2

σ_VN* = √σ² + 3τ²

HUBER - HENCKY - VON MISES

Ductile materials σ_VN = (( σ₁² + σ₂² + σ_III² - σ₁σ₂ - σ₂σ_III - σ₁σ_III) ^1/2

ROS - ESINGHER

Ductile materials - same formulation of VN

Assessment of sharp joints

1. Bolt shear, shank contact τ = V_mF_ANW
2. Thread contact = use A1_r,s (given according to the diameter)

τ ≤ σ_{admgiven according to the class}

1. Not plastic deformation (bearing failure of the plate) P_ref = V_do ≤ σ_rif t _{= thickness} d_{₀ = hole diameter (generally d + 1mm)} σ_rif = ασ_adm

Tearing (tensile failure of the plate)

σ = F_anet ≤ σ_adm

Anet = l:B - n':do:t

Tension

F_t,z,i = _zA_iz,i = ^M_t/_{∑f i/∑Ai iz̅}

Bending

F_i = b _i/∑_i̅

Low stiffness: y_i from the centerline

High stiffness: y_i from the edge

F_i can be normal or tangential

Combined stress

(^_,/₁)² + (^_0,/₁)² ≤ 4

Friction Joints

Preload (axial) V₀ = 0.8 f_y,k N

Area given according to the class (0.7⋅A⋅100 or A⋅0.10)

F_f = V₀ ⋅ f ⋅ COS 0.3

Weldings: But Welds

Q_{1 1 c1d} = 0 if _{d,0 d 1}+₁

CF/21 class _id ≤ 0 depends on material and thickness

I class _id ≤ 0.85 _adm

Weldings: Fillet Welds

Welding length with full thickness. L = 2•2 2₂C Q₂

Effective throat height: The conventional calculation is done by rotating the throat section on one of the two sides and neglecting ₁.

If ₁, ₁ are present: (₂² + 0² + _c2²) ^1/2 ≤ 0.25 _adm

Fe 360 0,70 _adm Fe 430, Fe 5100,70 _adm Fe 360

[ |₁|+|₁|] ≤ 0,85_adm Fe 430, Fe 510 if _c1 and ₁: |₁| + |₁| = _adm on 0,85 _adm on before |0| ≤ 0,85 _adm or 0.90_adm on 0,90 _adm Fe 360

If only |0| ≤ 0,85 _adm or 0.85 _adm Fs 360 Fe 360

Cases

• F / FƩ
• F/L2Q _2,j F F→Q²→FƩ_1,max = NA
• 31 QQ_{1 1 c1}J = d₂ L₁/L_C + d₂h d d_N F L ^F_2Q →(ሮ)FƩ^0_1,max = E/q eζ + b b 3FL_b F ₂ = J₂/ A 2(ℓ₂ ℓ₁² F A1f F A2qℓ²)

Fatigue

• Stress amplitude
• Stress mean
• Load ratio
• σ_a = (σ_max - σ_min) / 2
• σ_m = (σ_max + σ_min) / 2
• R = σ_min / σ_max

Typical design fatigue ratio (ductile steels)

• σ_fb = (0.4; 0.6) σ_uts
• σ_fa = (0.3; 0.45) σ_uts
• σ_fel = (0.23; 0.33) σ_uts
• Bending | Axial | Torsional

Fatigue stress concentration factor

K_f = 1 + q(n-1)

Fatigue limits

• Axial: σ_fa = σ_feb₂b₃ / K_fa
• Bending: σ_fb = σ_feb₂b₃ / K_fb
• Torsional: σ_ft = σ_feb₂b₃ / K_ft

σ' = {σ_feu, σ_fa, σ_fb, σ_ft} without σ_m

If σ_m is present -> Haigh diagram (simplified)

σ_aσ_faσ_m applied: Θσ_u = σ'_f (1 + σ'_u/σ_us σ_m/σ_a) η must be greater than 2 in fatigue assessment

MUNIALA Fatigue

• GOUGH-POLLARD
• Bending + torsion, loads: synchronous and in-phase
• σ_gP = (σ_a² + π²σ_a )^1/2Θσ_m / σ_feb₃:if σ_m is present: H -> New
• SINES
• σ_S^* ≤ σ_F,_ak depends on the material -> from Haigh diagram

Gears

v: ω₁d₁ / 2

ω₂ : ω₁

Transmission ratio: i = (ω₂/ω₁)