Machine Design Notes

Machine Design

Structure = ability to sustain and transfer load

2 approaches to design:

• create a simplified model doing a basic assessment
• define a complete model

We consider beam fully constrained to the ground. Degrees of freedom in a plane:

3 in translation + 3 in rotation = 6

Constraints can constrain all 3 DOF. It reacts with 2 forces and 1 bending moment.

Roller: constraint in the only direction

Pinned: constraint like a roller, but no possibility to rotate

Pin: the only possibility is rotation

We can calculate reaction forces simply by equilibrium equations.

Over-constrained structure:

Generally real life structures are over-constrained because they have a higher stiffness than single constrained.

• DOF = 3n
• DOF_real = 2+n
• DOC = 2(n - 1)

The grounded hinge is applied to 2 degrees of freedom for every beam.

• DOF = 3n
• DOF_real = n
• DOC = 2n

x translations + n rotations

6 DOF

5 DOC

Each pin is 2 DOF (n beam - 1)

• Roller: (2+3)-1 = 5 DOC
• Hinge: 4 DOC

=> Single constrained structure

Approximation:

W₁ = P

- 2P - VO₂ * 4 - 0 → VO2 = P/2

VO₁ = VO₂ = P/2

Section 2

Reaching 3620N → some fibers around the hole start to behave plastically.

Only reaching 8280N the second section starts to behave plastically.

We can increase the load till the complete plastification of the section (plastic collapse).

Very stiff structure: stop at 3620 N (only elastic regime).

But: brittle material → failure (suddenly).

^σaux

3.3

Fe_lim → R_u (1 - ^σaux) = 682 KN

Fe_lim = R_m (e - d) : ₁ = 715 KN k_t = 2.25

l₁, d fixed → k_t =

^σaux = 3.3 - 3.095 (^d⁄_e) + 0.309 (^d⁄_e)² + 0.986 (^d⁄_e)³

d = 24 mm → l = 34 mm

Two effects that act in opposite ways → we have to find a joint, a compromise.

σ_x = -40

σ_y = -70

τ_xy = 35

O_ij = [ -40 -35 0-35 70 00 0 0 ]

O_II = -93

O_I = 0

O₁₂ = 16,9

O_ij = [ -40 -35 0-35 70 00 0 0 ]

A. O^max = 85,9 MPa √2 = 4,02

O_II = 93 MPa √2 = 3,91

η_f = R_t/O_f = ∞

η_c = R_c/O_III

η = 7,41

D = 20 mm

F = 100 mm

F = 550 N

P = 8200 N

T = 30 Nm

Fe 430 R_m = 430 MPaR_y = 295 MPa

η_c = 225/101 = 2,72

A:

σ_a = P/A + FL_D/J2 + 32FL/T_d' = 25,5 + 70 = 95,5 MPa

τ_CA = 16T/Td³ = 190 MPa

O_Vm = √6² + 3c² = 101 MPa

η = 225/101 = 2,72

B:

No bending movement (neutral axis)

σ_B = P/A = -25,5 MPa

σ_B = σ_A

O_Vm = 41,8 MPa η = 6,58

The safety factor to be chosen is η = 2,72

NB: we are not considering the stress concentration on A, B

There is plastic deformation - increase in deformation

to reduce deflection and increase the stiffness we have

considered a strengthening intensification factor (normally

negligible for welds)

σ: C - 2F/3A = 0,04 MPa negligible!

and in the most stressed region

1y

Q_N=32 N ⁞_N=11 MPa

T_B3 I

Q₁=32

3

1

2

3

2

Ductile material: K_S=1

Point 1: Brittle material (cast - ironium) K_h1 96 - K_hN(II) = 118 MPa

γ₁=√³²₁₈=4.89

Point 2: We have compression: K_h2 = 96 - K_hN = 11 - 152 MPa

γ_c=√¹⁷⁰₅₂ = 7.9

Point 2: γ_c = √⁹⁰₁₀₈ N₁ = √³³⁰₂₅₀ √⁹⁶₁₀₈√¹²⁵₁₂₀ 11√¹⁵⁰

Extra: Points 3 and 4 presence of a shear and of a nominal stress % due to axial Considering ductile material:

Plane stress condition 6σ_T = √¹²⁺⁴²_2²+411 = 2u MPa

√³⁸⁰ ₃₇ = 33 √¹²⁵⁰ _2u = 50

Let's consider brittle material: we need principal stresses:

σ_T = √²₇ = √^{(√2₂+411) = 2u MPa}

we can also find σ_T for ductile mat but it is not necessary (consider that their "sum" is 2u again)

γ_c=√⁵²⁰₉₇ = √¹³²₉₈

d: 30 mm

Fr: 1000

Fs: 2000 N

Ft: 2000 N

b: 100 mm

h: 200 mm

38 NiCrMo4 Rw: 1000 MPa Rs: 850 MPa

Multi stress fatigue - calculation of an equivalent stress to be compared with the limit stress in uniaxial situation

Stress felt by an infinitesimal cube

• Mf = 100 Nm
• Mt = 150 Nm
• d = 25 mm
• t = 30 mm
• r = 1 mm

Neutral axis

Max compressive

• σa = 32 MPa - 65 MPa amplitude
• τm = 49 MPa mean

equ due to fatigue

equ due to static load

σeq = (σa² + H²τm²)^1/2

H = Θ_em / Θ_fem C