ASAMBLARI DEMONTABILE 1 s m = σs2 m

ds
tf
Moment de torsiune reactiune
Arbore

Mt

D
Figura 1.47

d
Moment de torsiune actiune
Butuc

σs1 max σs2 max b)

ds
σS tf a)
Figura 1.48
σS
a)
b
D

l
d

σs 2
Mt F a F

b σs 1
F
Figura 1.49 b) M i max
2 σs 2
db F
2 Figura 1.50

Figura 1.51

F F

Figura 1.52

Freza deget

Figura 1.53.a Figura 1.53.b

h

l
b

Figura 1.54.b Figura 1.55

Figura 1.54.a Freza

disc
h

l
b

Figura 1.57 Figura 1.56

1 d 2
 2 ⋅ σ s1max ⋅ 2 ⋅ d s ⋅ 3 ⋅ d = M tc ⇒ σ s1max ≤ σ sa( stift −arbore)

 D−d D+d
σ s 2m ⋅ ⋅ds ⋅ = M tc ⇒ σ s 2m ≤ σ sa( stift −butuc) (1.39)
 2 2
 π 2
τ f ⋅ ⋅ d s ⋅ d = M tc ; τ f ≤ τ af stift
 4
ds D D
Recomandari constructive: = 0,2...0,3 ; = 2 (otel/otel); = 2,5 (fonta/otel).
d d d
 ds d
σ s ⋅ 2 ⋅ l 2 = M tc ⇒ σ s ≤ σ as min d 1
 , recomandari: s = 0,13K0,16; = 1K1,5. (1.40)
d
τf ⋅ d s ⋅ l ⋅ = M tc ⇒ τ f ≤ τ af d d
 2
 2
2 ⋅ σs2 ⋅ d ⋅ b = F ⇒ σ s2 ≤ σs2a min

σ s1 ⋅ d ⋅ a = F ⇒ σs1 ≤ σs1a min

 (1.41)
F ⋅  a + b  = M Mi max
 2  4 2  i max ; σi = ≤ σai
 π ⋅ d 3b
 32
h d h
σ s ⋅ ⋅ l ⋅  ±  = M tc
2 2 4
h d
σ s ⋅ ⋅ l ⋅ ≅ M tc ⇒ σ s ≤ σsa (1.42)
2 2
d
τf ⋅ b ⋅ l ⋅ = M tc ⇒ τ f ≤ τaf
2

tf
σs2 σs
σs1

c
t2 t1
h

c
l d
Mt

Figura 1.58 D
d
Figura 1.59 M
-

Figura 1.60 Figura 1.61

σs max
σs max

Mt σs max

Mt
Mt = 0 Mt > 0 Mt > 0 Mt > 0 Mt > 0
Figura 1.62 1.63.a 1.63.b 1.63.c 1.63.d 1.62.e
Fs
L p = p max . cos n ϕ
2

µ⋅N
F
N d N

Fs pm pc
pmax
2
µ⋅N

Figura 1.64 1.65.a 1.65.b 1.65.c

D +d
rm =
2
M tc
Snec =
rm ⋅ σ as
(1.43)
D−d 
s′ = 0,75 ⋅ z ⋅  − 2⋅ c 
 2 
S nec
L nec ≥
s′
1 a 2 
2 ⋅  ⋅ σ s max ⋅ ⋅ l ⋅ ⋅ a  = M tc ⇒ σ s max ≤ σsa (1.44)
2 2 3 
µ ⋅ N ⋅ d ≥ M tc
(1.45)
M tc = k1 ⋅ k 2 ⋅ F ⋅ L
M tc
N= = Fs (1.46)
µ⋅d
 3
N (1.47)
σs = ≤ σ as
b⋅d
4 3
Pentru cazul din figura 1.65 c): daca n = 1: p max = ⋅ p c ;daca n = 2: p max = ⋅ p c . 2 2
π 2 d2 + d
σs ⋅
∆b l d 22 − d 2
2 σs
S ∆a
2 2 σr

da d1 d2 d1
db d
Figura 1.67

S ∆a σt
N L 2
Fs 2
µ⋅N ∆b
F 2
d
d2
N d2 2 2 − 2⋅σs ⋅
− 2 ⋅ σs ⋅ d + d1 d − d12
2
Fa d 2 − d12 − σs ⋅
µ⋅N d − d12
2
a
A R
N Ft
Figura 1.66 Figura 1.68 Figura 1.67
 d d d
Fs ⋅  a +  − N ⋅ − µ ⋅ N ⋅ = 0 (1.48)
 2 2 2
µ ⋅ N ⋅ d ≥ M tc (conditia de ne-patinare).

N ⋅ ⋅ (1 + µ )
d
M tc 2
⇒N= ; Fs = (1.49)
µ⋅ d d
a+
2
M tc ⋅ (1 + µ)
⇒ Fs = → Cu Fs se dimensioneaza suruburile. (1.50)
µ ⋅ (2 ⋅ a + d )
N
σs = ≤ σ as (1.51)
d⋅b
Fa
p min = σs min = (1.52)
µ ⋅ π⋅ d ⋅ l
2 ⋅ Mt
p min = σ s min = (1.53)
µ⋅ π ⋅ d2 ⋅ l
R N
R = Fa2 + Ft2 ; N = ; p min = (1.54)
µ π⋅ d ⋅ l
2
 2 ⋅ Mt 
Fa2 +   (1.55)
 d 
p min = σs min =
µ ⋅π ⋅d ⋅ l
Smin = ∆d + ∆b = d a − d b (1.56)
Ka d + d 12
2
∆a = σ s min ⋅ ⋅ d; K a = − µa
Ea d 2 − d 12
(1.57)
Kb d 2 + d12
∆b = σs min ⋅ ⋅ d; K b = − µb
Eb d 2 − d12
K K 
Smin = σ s min ⋅  a + b  ⋅ d (1.58)
 Ea E b 
Smin tot = Smin + Sr + S t (1.59)
Sr = 1,2 ⋅ (R maxa + R max b ) (1.60)
St = [α b ⋅ (t b − t 0 ) − αa ⋅ (t a − t 0 )] ⋅ d (1.61)
σ t max − σ r max
= 0,5 ⋅ σ c → dupa teoria tensiunilor tangentiale (1.62)
2
 4
σ t max − σ r max
= 0,75 ⋅ σ c → dupa teoria energetica (1.63)
2
d 22 + d 2
Pentru butuc: σ t max b = σ s ⋅ ; σ r max b = − σ s
d 22 − d 2
d2
Pentru arbore: σ t max a = −2 ⋅ σ s ⋅ ; σ r max a = σ s
d − d12
2

 d2 + d 2 
σ s max ⋅  2 − (− 1)
 d 2 − d
2 2
 σ (1.64)
b

= 0,5 ⋅ cb
2 cc
 d2 
σ s max a ⋅ − 2 ⋅ 2 − (1)
 d − d1
2
 σ
= 0,5 ⋅ ca (1.65)
2 cc

(
σ s max = min σ s max b , σs max a ) (1.66)
K K 
Smax = σs max ⋅  a + b  ⋅ d (1.67)
 a Eb 
E
Smax tot = Smax + Sr + St (1.68)
SSTAS
max ≤ Smax tot si SSTAS
min ≥ Smin tot (1.69)

SSTAS
max + J = α b ⋅ (t − t 0 )⋅ d ⇒ t (1.70)

SSTAS
max + J = α a ⋅ (t 0 − t ) ⋅ d ⇒ t (1.71)

µN Q
α
V
N α
ϕ
dm

σs max
σs m = σs α
Q σs max

1.71.a 1.71.b 1.71.c 1.72

Figura 1.70
l

Q = N ⋅ (µ ⋅ cos α + sin α )
2 ⋅ M tc (1.72)
N=
µ ⋅ dm
Mtc = k1 ⋅ k 2 ⋅ M t nomin al (1.73)
N
σs = ≤ σ sa (1.74)
π ⋅d m ⋅ l
M tc = K1 ⋅ K 2 ⋅ M tn
2 ⋅ M tc N
N= ; σs = ≤ σ sa (1.75)
µ⋅d π ⋅ d ⋅l
Fa1 = N ⋅ [tg(α + ϕ1 ) + tgϕ] (1.76)
E ⋅S
Fa 0 = ⋅ ∆l r (1.77)
l0
∆l r
tg(α + ϕ1 ) = (1.78)
∆l a

2⋅E⋅
(
π⋅ D2 − d 2 )
Fa 0 = 4 ⋅ ∆l a ⋅ tg(α + ϕ1 ) (1.79)
D+d
4
d
σ sm ⋅ π ⋅ d ⋅ l ⋅ µ ⋅ = M tc (1.80)
2
 5
µ

µ1 N

µN
Fa1
σs

D
d
α
l N ϕ1
α N′
Figura1.76

µN′
H′ = N⋅(µ1⋅cos α +sin α)

Figura 1.73

1 1 1
M t1 ⋅ M t1 ⋅ M t1 ⋅ M t1 K
4 4 8
Figura 1.74 Figura 1.75

d D
l
d

σs max
Figura 1.77 Figura 1.78 Figura 1.79

Tabelul 1.2
σc σR Alungirea la
Marca STAS Tipuri de arcuri
[MPa] [MPa] rupere, A%
OLC 55A 880 1080 6 lamelare, cu foi, spirale
795-80
OLC 65A 780 980 10
OLC 75A 880 1080 9 lamelare, cu foi, spirale, elicoidale
OLC 85A 980 1130 8
51 Si 18A 1080 1180 cu foi, elicoidale
795-80
56 Si 17A 1270 1480 6 cu foi, elicoidale, disc, inelare
60 Si 15A 1080 1270
51 VCr 11A 1180 1320 cu foi, elicoidale, disc, inelare, bara de torsiune
61 Si 2 WA 11514-80 1668 1864 5 elicoidale, bara de torsiune
40 Cr130 3583-80 1250 1650 − elicoidale
12 TiNiCr 180 11523-80 1450 1650 − elicoidale
F F F F F
- l M
(M) (M) (M) (M) (M)

f f f f f Mt
arcuri metalice arcuri metalice arc disc arcuri metalice arcuri din
θ de înaltimiθ arc θ cu foi θ cauciuc θ
d θ
(Hooke) θ
diferite spiral (histerezis) (histerezis)
1.80.a 1.80.b 1.80.c 1.80.d 1.80.e 1.81.a 1.81.b
dF dM
c= sau c = (1.81)
df dθ

∫
L = F ⋅ df sau L = M ⋅ dθ ∫ (1.82)
L
kv = (1.83)
V
 6
L
ku = V , pentru solicitarea de torsiune, sau
τ 2max
G (1.84)
L
ku = 2 V , pentru solicitarea de încovoiere sau de tractiune-compresiune.
σ max
E
G
ku = kv ⋅ , pentru solicitarea de torsiune, sau
τ 2max
(1.85)
E
ku =kv ⋅ , pentru solicitarea de încovoiere sau de tractiune-compresiune.
σ 2max
Lh = L − L′ (1.86)
L′ L
η= =1 − h (1.87)
L L
L − L′ 1 − η
δ= = (1.88)
L + L′ 1 + η
Mt
τt = ≤ τ at (1.89)
Wp
M t ⋅ l 32 ⋅ M t ⋅ l
θ= = (1.90)
G ⋅ Ip π ⋅ G ⋅d 4
π ⋅d 4 ⋅ G
c= (1.91)
32 ⋅ l
2
 π 3 

L=
Mt ⋅ θ
=
M 2t ⋅l
=
(Wp ⋅ τ at )
2
⋅l
=
 ⋅ d ⋅ τ at  ⋅ l
 16  (1.92)
2 2 ⋅ G ⋅ Ip 2 ⋅ G ⋅ Ip π ⋅d
4
2⋅G⋅
32
2
τ t max
L = 0,25 ⋅ ⋅V (1.93)
G
16 ⋅ M t π ⋅ d4 ⋅ θ ⋅G
d nec = 3 sau l nec = (1.94)
π ⋅ τa t 32 ⋅ M t
F
caneluri F1 F2 Fb
F

f1
t; t 0 d
f2
Dm fb
H0 H1 H2 Hb

f
Figura 1.82
Figura 1.83
F

F·cos
a
a
tt
F·sina
Dm
k·t t
Figura 1.84 Figura 1.85
F ⋅ cos α ⋅ D m
Mt = (1.95)
2
F ⋅ sin α ⋅ D m
Mi = (1.96)
2
 7
F⊥ = F ⋅ cos α (1.97)
FN = F ⋅ sin α (1.98)
M t 8 ⋅ F ⋅ Dm
τt = = (1.99)
Wp π ⋅d3
F 4⋅ F
τf = = , (1.100)
A π ⋅ d2
4 ⋅ F  Dm  4⋅ F
⋅ (2 ⋅ i + 1) , unde: i = m este indicele arcului (raportul de înfasurare)
D
τ= ⋅ 2 ⋅ + 1 = (1.101)
2  2
π ⋅d  d  π⋅ d d
8 ⋅ F⋅ i
τ = τt = (1.102)
π ⋅ d2
τ max = k ⋅ τ (1.103)
i + 0,5
k= (1.104)
i − 0,5
8⋅ k ⋅D m ⋅ F 8⋅ k ⋅i ⋅ F
τ max = = ≤ τa (1.105)
π ⋅ d3 π ⋅ d2
8 ⋅ k ⋅ F⋅ i
d nec ≥ ; cu D m = i ⋅ d (1.106)
π ⋅ τ at
Dm M ⋅l
f = ⋅ θ; θ = t (1.107)
2 G ⋅ Ip

8 ⋅ F ⋅ i3 ⋅ n
f = (1.108)
G⋅d
G ⋅ d4
c= (1.109)
8 ⋅ n ⋅ D 3m
f ⋅ G ⋅ d4
n= (1.110)
8 ⋅ D 3m ⋅ F
F⋅ f
W= (1.111)
2
π ⋅ d 3 ⋅ τa
F= (pentru k = 1)
8⋅ D m
(1.112)
π ⋅ d 2 ⋅ τa
F= (pentru k real )
8 ⋅ k ⋅ Dm
Dm
F⋅ ⋅ π ⋅ Dm ⋅ n
M ⋅ π ⋅ Dm ⋅ n 2 π ⋅ n ⋅ D 2m ⋅ F (1.113)
f = t = =
G ⋅ Ip G ⋅ Ip 2 ⋅G ⋅ Ip

 0,25  τ 2
W =  2  ⋅ max ⋅ V (1.114)
 k  G

F F

Figura 1.86 1.87.a 1.87.b 1.87.c 1.87.d

6 ⋅ F⋅ l
σ i max = ≤ σ ai (1.115)
b⋅h2
M 6 ⋅F ⋅l
σ i max = max = ≤ σa i (1.116)
Wz b ⋅ h2
Mx F⋅x F⋅ l
σi x = = = = σ i max
Wz x x Wz (1.117)
Wz ⋅
l
b x ⋅ h 2x x b⋅h2 x
Wz x = = Wz ⋅ = ⋅ (1.118)
l l 6 l

b x = b 0 + (b − b 0 )⋅
x
(1.119)
l
6 ⋅ F⋅ l
S nec =
3  b (1.120)
  ⋅σa i
h
 8
(1.121)
F ⋅ l3 b ⋅ h3
f1 = ; Iz = (figura 1.78.a)
3⋅ E ⋅ Iz 12 l
F ⋅ l3
f2 = ; f 2 > f1 (figura 1.78.b)
2⋅ E ⋅ Iz b F

f3 =
F ⋅l
⋅
(
3 ⋅ 3 ⋅ β − 4 ⋅ β + 2 ⋅ β ⋅ ln β
3 2 2
)
; (figura 1.78.c) a) F
2 ⋅ (1 − β )3
3⋅ E ⋅ Iz h

f 2 > f 3 > f1
2
1 σ i max l
L1 = ⋅ ⋅ V;
18 E
2 b F
1 σ i max
L2 = ⋅ ⋅ V; (1.122)
6 E F
( )
b)
1 3 ⋅ β 2 − 4 ⋅ β + 1 − 2 ⋅ β 2 ⋅ ln β σ 2i max h
L3 = ⋅ ⋅ ⋅V
2 ⋅ (1 − β ) ⋅ (1 + β )
6 3 E
l
Ee = k e ⋅ E (1.123)

F (M) F b F b0
energie disipata prin frecari 1 2 3
încarcare
interne (max 40%)
c) F
h

descarcare
f (?) f
Figura 1.89 Figura 1.90 Figura 1.88
F
f β
a) dupa încarcare, sub sarcina F = ct;
b) dupa îndepartarea sarcinii.
a)
h 10
b)
timp
5
Figura 1.91 a
5 10 a/h
Figura 1.92 F Figura 1.93
A
F= E ⋅ ⋅f (1.124)
h
A a
F = β ⋅ E ⋅ ⋅ f , unde β = f   cu variatia prezentata în figura 1.93 (1.125)
h h
F F f
τf = = γ ⋅ G; γ = tgγ = (1.126)
A A⋅ G s

f
?
F
s

Figura 1.94

Figura 1.95 Figura 1.96

Figura 1.97