Free Conveyor Design

Belt conveyor design

Tutorial

This example shows the steps necessary to design an industrial belt conveyor. Conveyor requirements: Length, Lb: 20 ft (6.1 m)

Load: cardboard boxes

Width, wb: 16 in (0.0406 m)

Total load weight, W: 400 lbs (181.4 kg)

Slope, α: +15°(rising)

Speed, s: 60 ft/min (18.3 m/min)

Step 1: Belt selection Step 2: Pulley and drive design Step 3: Frame design Step 4: Supports design www.packexone.com

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Step 1: Belt selection 1.A. Belt applicable operational constraints: •Overall length. •Width. •Temperature range. •Food grade. •Chemical resistance. •Antistatic. •Flammability rating. •Conveying side adhesiveness. •Suitable for a slider bed or carrying rollers. Our application requires an adhesive belt in order to convey the boxes on an upward slope. It will be supported by a slider bed. The adhesiveness between the load and a belt sample is to be tested; this by placing the load on top of the belt, both on an angled surface and increasing the angle. A rubber/polyester belt is a good candidate.

1.B. Belt required data: Thickness, t: 0.06 in. (1.5 mm) Belt mass, m: 0.39 lbs/ft² (2 kg/m²)

Friction coefficient on driving pulley, fp: 0.15

Min. pulley diameter: 1 in (25 mm)

Friction coefficient on slider bed, fb: 0.2

Tensile force, k1%: 40 lbf/in (7 N/mm) Max. tensile force, Tm: 63 lbf/in (11 N/mm)

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Step 1: Belt selection 1.C. Geometry of the driving pulley: Pulley diameter dp: 4 in (101.6 mm) {higher than min. pulley diameter, final value to be validated in Step 2 } Pulley width, wp: 18 in (457.2 mm) Belt wrapping angle, θ: 180°(smaller value than actual angle) Driven (tensioning) pulley has the same geometry for economic reason.

1.D.1 Operation forces and torque: The tension in the belt is at its maximum level when the fully loaded conveyor is accelerating to its rated speed after a stop. The time to accelerate, ta, is to be: Short enough: •To minimize overload current in the electric motor and its amplifier (consult ratings). •To maximize conveyor output.

Long enough: •To prevent conveyed load from slipping or tilting. •To minimize motor horsepower.

For our application, ta = 3 sec.

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Step 1: Belt selection 1.D.2 Operation forces and torque: Pulley speed, ω = 12 s / (30 dp) = 6 rad/sec (π = 3.1416) ω = 1000 s / (30 dp) = 6 rad/sec Pulley acceleration, a = ω / ta = 2 rad/s2 Pulley weight, Wp = (0.283 lbs/in3) π dp2 wp / 4 = 64 lbs per pulley Wp = (7833 Kg/m3) π dp2 wp / 4 = 29 kg per pulley Pulleys mass inertia, Jp = 2 (Wp dp2 / 8) = 256 lbs-in2 (0.075 kg-m2) Belt weight, Wb = m wb (24 Lb + π dp) / 144 = 21.3 lbs Wb = m wb ( 2 Lb + π dp) / 1000 = 9.68 kg Belt mass inertia, Jb = Wb dp2 / 4 = 85.4 lbs-in2 (0.025 kg-m2) Load mass inertia, J = W dp2 / 4 = 1600 lbs-in2 (0.468 kg-m2) Total mass inertia, Jt = Jp + Jb + J = 1941 lbs-in2 (0.568 kg-m2)

Acceleration torque, Ta = Jt a / 386 = 10.06 lbf-in Ta = Jt a = 1.136 N-m

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Step 1: Belt selection 1.D.3 Operation forces and torque: Friction force between belt and slider bed, Ff = (W + 12 m Lb wb / 144) fb cos(α) = 79.3 lbf Ff = 9.81 (W + m Lb wb) fb cos(α) = 353 N

Friction torque, Tf = Ff dp / 2 = 158.6 lbf-in (17.9 N-m) Lifting force, Fl = W sin(α) = 103.5 lbf (460.7 N)

Lifting torque, Tl = Fl dp / 2 = 207 lbf-in (23.4 N-m) Total torque, Tt = Ta + Tf + Tl = 375.7 lbf-in (42.5 N-m) Power, P = Tt ω / 6600 = 0.342 Hp P = Tt ω = 255 Watts

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Step 1: Belt selection 1.E Belt tension and stress

Tension (carrier side) F1 = {Tt / (dp / 2)} / { 1 – 1 / (2.718 (fp θ) ) } = 499 lbf (2221 N) Tension (return side) F2 = F1 / (2.718 (fp θ) ) = 311.5 lbf (1386 N)

Belt stress, σb = F1 / Lb = 31.9 lbf/in (5.46 N/mm) Belt safety factor, SFb = Tm / σb = 2.02, selected belt is adequate. Safety factor for electric motor shall be lower than belt safety factor to prevent breakage. Safety factors for all other conveyor parts shall be higher than belt safety factor.

Snub rollers Snub rollers are used to: •Increase the angle of contact between the belt and the driving pulley and improve the torque transmitted. •Decrease the distance between the belt carrying side and return side. •Improve belt tracking with adjustable snub rollers.

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Step 2: Pulley and drive design 2.A Pulley design Pulleys with cylindrical-conical shape are to be used in order to help belt tracking and prevent runoff. The geometry is derived from experience and is sometimes supplied by belt manufacturers. For our application, the following pulley is used for the drive end and the idler end.

Pulley material: UNS G43400

Surface finish: Ra 60 µin (1.6 µm)

Pulley deflection under maximum load Section inertia, Ip = π (dp4 – di4) / 64 = 12.52 in4 (5.211 x 10-6 m4) Deflection, y = {5 (F1 + F2) wp3} / {384 x 29000000 Ip) = 0.0002 in y = {5 (F1 + F2) wp3} / {384 x 2.07 x 1011 Ip) x 1000 = 0.004 mm The maximum deflection is smaller than the slope height in the conical portion of the pulley. The pulley will be able to track the belt.

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Step 2: Pulley and drive design 2.B.1 Shaft design

Drive shaft design is subjected to torsion stress from the motor torque and bending stress from the pulley tension, the drive sprocket, and the bearing supports

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Step 2: Pulley and drive design 2.B.2 Drive shaft loads

Force from gearmotor torque, Fm = 2 Tt / Ds = 188 lbf (838 N) (sprocket pitch diameter, Ds= 3.989 in (101 mm) Along the conveyor axis: Fx = Fm cos(17°) = 180 lbf (801 N) Q = (F1 + F2) / C = 47.7 lbf/in (8.35 N/mm) (distributed belt force) R1 = ( Q C (C/2 + D) + F (A + B + C + D) ) / (B + C + D) = 606 lbf (2696 N) (left bearing reaction) R2 = Fx + Q C – R1 = 386 lbf (1718 N) (right bearing reaction) M1x = -Fx A = -391 lbf-in (-44.2 N-m) Perpendicular to conveyor axis: Fy = Fm sin(17°) = 41 lbf (183 N) M1y = -Fy A = -119 lbf-in (-13.5 N-m) Bending torque at left bearing : M1 = (M1x2 + M1y 2)0.5 = 409 lbf-in (46.2 N-m)

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Step 2: Pulley and drive design 2.B.3 Drive shaft design Drive shaft geometry

Shaft material: UNS G10450 Tensile yield strength, Sy = 58700 psi (405 MPa) Ultimate strength, Su = 97900 psi (675 MPa) Endurance strength Se = 0.55 x 1 x 0.814 x 1 x 1 x (0.5 Su) = 21914 psi (151 MPa) Shaft safety factor, SFd = 2.1 (greater than belt safety factor) Minimum shaft diameter dm = { (32 SFd / π) x ( (Tt / Sy)2 + (M1 / Se)2 )0.5 }0.33 = 0.750 in (19.05 mm)

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Step 2: Pulley and drive design 2.C Commercial parts

Gearmotor: integrated worm gear reducer and AC motor with electrical brake, ratio 22.5:1 , power 0.5 Hp (0.37 kW), maximum output torque 340 lbf-in (38.5 N-m ), service factor 1.8 . Overload protection from motor amplifier. Electrical installation must be conformed to local government regulations. Bearings: ball bearing oval flanged unit, bore 25 mm, dynamic load rating 14 kN, static load rating 7 kN, limiting speed 7000 rpm, grub screw locking, with bearing shields. Sprockets: roller chain steel sprocket, no.40 (1/2”pitch), 25 teeth, keyway + 2 setscrews. Roller chain: no.40 (1/2”pitch) steel roller chain. To be lubricated with oil (refer to manufacturer for oil viscosity)

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Steps 3 and 4 explain the design of the conveyor frame, supports and components with: •Loads and reactions. •Stresses and deflections. •Detailed drawings.

To access Steps 3 and 4 (14 pages), please subscribe to www.packexone.com

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