1. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
CALCULATION OF THE STRUCTURE
Two hypothesis.
1. We suppose the load will be leant just on one beams.
2. We suppose the load distributed equally on each .
First :
The heaviest mould weighs 3200kg , We suppose a distributed load
3200Kg/540mm =6Kg/mm
6Kg/mm
R1 R2
We use a square hollow beam:
R1 + R 2 = 6 • 540
6 • 540 • 270 = 540 R1
R1 = 1620 Kg
R 2 = 1620 Kg
Maximum moment
4200kg
1176000

2. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
Q = −6 x + 1620
Mf = −3 x 2 + 1620 x
x = 270mm
Mf max = −3 • 2702 + 1670 • 270
Mf max = 218700 Kg • mm
Square hollow beam
Steel A-42
218000 Kg • m  L 
σ adm = 260 Kg 2 260 = • 
mm L
4
2
M f max  
σ adm = •x  12 
Ix L = 18mm
.
Square de 40.2 L=40mm (length) y e=2mm (thickness)
M f max M f max (Kg • mm )
W = 3.40cm3 = 3400mm 3 σ adm = 260 Kg 2 =
W mm 3400mm 3
Mf max = 884000 Kg • mm Mf max > 218000 (Kg • mm ) Cumple

3. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
Second hypothesis :
2Kg/mm
R1 R2
R1 + R 2 = 2 • 540
2 • 540 • 270 = 540 R1
R1 = 1080 Kg
R 2 = 1080 Kg
Maximum moment
2Kg/mm
R1
Q = −2 x + 540
Mf = − x 2 + 540 x
x = 270 mm
Mf max = (− 1) • 270 2 + 540 • 270
Mf max = 72900 Kg • mm

4. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
72900 Kg • m  L 
σ adm = 260 Kg 2 260 = • 
mm L
4
2
M f max  
σ adm = •x  12 
Ix L = 12mm
12mm de lado, the standard which fits best according to the regulation NBE
EA-95
L= 40mm; e = 2mm
M f max M f max (Kg • mm )
W = 3.40cm3 = 3400mm 3 σ adm = 260 Kg 2 =
W mm 3400mm3
Mf max = 884000 Kg • mm Mf max > 72900 (Kg • mm ) Cumple
CALCULATION OF THE LIFT
Calculation of the total load lean on the lift
Ct = Cd + Ces
Cd = depots loaded; Ces = weight metallic structure.

5. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
ρac= 7850 Kg / m 3 (Steel density).
ρpp = 1kg / m 3 (Polypropylene density)
Measurement of the depots:
• ∅235mm x 835
• ∅235mm x 605
Cálculation of the depot´s load
M π • 0.2352
ρ= M = ρ • V : M = 7851• 0.850 • = 284Kg Deposito1
V 4
M π • 0.2352
ρ= M = 7851• 0.605 • = 206Kg Deposito2
V 4
Calculation of the metallic structure load
M = ρ • V Ves = L • as = 3.555 • 0.060 2 = 0.012798 mm 3 Volumen de la estructura metálica
M = 7850 • 0.012798 = 100 Kg
Ct = Cd + Ces Ct = 490 + 100 = 590
Suponiendo que los deposito esta n llenos
590 kg
Seccion A-A
E:1/5

6. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
section A-A
Q=590Kg
Mf=590*50Kgmm
:
• Sheer stress.
• Bending moment.
Sheer stress
a a
590 • •a•
Q • Me
τ≥ ; 2 2 ≤ 26• 0.5 a = 12mm (sec ción maciza)
b•I a 4
a•
12
Momento flector
Mf 590 • 50 • 3 a
σ≥ y : 26 = • ; a = 30mm Con un coeficiente de seguridad N=3
I a4 2
12
Estándar profile according to the regulations is 40.2
Mf 590 • 50 • 3
σ≥ y : 26 ≥ • 20 do not meet the minimum security conditions
I 66000

7. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
profile 60.2
Mf 590 • 50 • 3
σ≥ y : 26 ≥ • 30
I 248000 meet all the specifications
(*)Meeting all the specifications according to the regulations NBE-EA95.
Calculation metal sheet nº7
550 Kg.
59mm
Sección B-B
Section B-B
550 Kg
550*59
b
a
Sheer stress
b b
550 • • 20 •
Q • Me
τ≥ ; 2 2 ≤ 26• 0.5 b = 5mm
b•I 20 • b 4
b•
12

8. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
Bending moment
Mf 550 • 59 • 3 b
σ≥ y : 26 ≥ • b = 10.5mm
I 20 • b 4 2
12
The metal sheet has 11mm of thickness
Calculation metal sheet nº8
550 Kg
30
a
Sección C-C
60
section C-C
550 Kg
550*54
b
a
Bending moment
Mf 550 • 54 • 3 b
σ≥ y : 26 ≥ • b = 8mm N=3
I 40 • b 4 2
12
The thickness is of 8mm.
Screw calculation for piece 9

9. S.M.E.D (Single minute Exchange of day)
Metal structure calculation
N 225(*) • 3 De diámetro
σ= 26 ≥ D = 2 mm
A D •π •8
Sheer screw calculation
Q 225(*) • 3
τ = 13 ≥ D = 8mm Cogemos el mayor, necesitamos un tornillo de
A π •D2
4
métrica 8