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# Metal Structure Calculations

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### Metal Structure Calculations

1. 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. 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. 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. 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. 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. 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. 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. 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. 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

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• #### JackWilliam24

Oct. 23, 2020

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