Gear Calculation

M236 MACHINE DESIGN EXCEL SPREAD SHEETS Rev. 9 Mar 09 Copy write, © Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006 MACHINE DESIGN This 8 PDH machine design course uses Excel's calculating and optimizing capabilities. Machine design includes: 1. A description of the needed machine in a written specification. 2. Feasibility studies comparing alternate designs and focused research. 3. Preliminary; sketches, scale CAD drawings, materials selection, appearance and styling. 4. Functional analysis; strength, stiffness, vibration, shock, fatigue, temperature, wear, lubrication. Customer endurance and maintenance cost estimate. 5. Producibility; machine tools, joining methods, material supply and handling, manual vs automated manufacture. 6. Cost to design and manufacture one or more models in small and large quantities. 7. Market place: present competition and life expectancy of the product. 8. Customer service system and facilities. 9. Outsource part or all; engineering, manufacturing, sales, warehousing, customer service.

Backhoe Above is the image in its original context on the page: www.chesterfieldgroup.co.uk/products/mobile.html

Strength and Stiffness Analysis The strength and stiffness analysis of the backhoe begins with a, "Free Body Diagram" of one of the members, shown above : Force F1 = Hydraulic pressure x piston area. Weight W = arm material volume x density. Force F3 = (Moments due to F1 and W) / (L1 x cos A4) Force F2 = ( (F1 cos A1) - (W sin A3) + (F3 cos A4) ) / cos A2 Moment Mmax = F1 x cos A1 x L1 Arm applied bending stress, S = K x Mmax D2 / (2 I) I = arm area moment of inertial at D2 and K = combined vibration shock factor. Safety factor, SF = Material allowable stress / Applied stress The applied stress and safety factor must be calculated at each high stress point.

Pick and Place Robot A gripper is attached at the bottom end of the vertical X direction actuator. The vertical actuator is supported by a horizontal Y direction actuator. The Y direction actuator is moved in the horizontal Z direction by the bottom actuator. This pick-and-place robot can be programmed to move the gripper rapidly from point to point anywhere in the X, Y, Z three dimensional zone. For more click on the, "Pwr Screw" tab at the bottom of the display. Shredder Above is the image in its original context on the page: www.traderscity.com/.../ Material to be shredded falls by gravity or is conveyed into the top inlet. A rotating disc with replicable cutters in its circumference performs the shredding. The tensile stress in a rotating disc, S = V2 x ρ / 3 lbf/in2. The disc is mounted and keyed to a shaft supported by roller bearings on each side. The shaft is directly coupled to a three phase electric motor. The coupling joining the motor and disc shafts is covered by a safety guard.

The replicable bearings have seals to keep the grease or oil lubricant in and the dust and grit out. Quick release access panels are provided for clearing jams and cutter replacement. A large, steel rod reinforced concrete pad, foundation is usually provided for absorbing dynamic shredding forces and shock loads.

Above is the image in its original context on the page: www.mardenedwards.com/custom-packaging-machin…

www.mardenedwards.com/custom-packaging-machin…

Automated Packaging Machine The relatively high cost of labor in the United States requires automated manufacturing and assembly to be price and quality competitive in the world market. The product packaging machine above is one example.

Automobile Independent Front Suspension Above is the image in its original context on the page: www.hyundai.co.in/tucson/tucson.asp?pageName=... Coil springs absorb shock loads on bumps and rough roads in the front suspension above. Double acting shock absorbers dampen suspension oscillations. Ball joints in the linkage provide swiveling action that allows the wheel and axle assembly to pivot while moving up and down. The lower arm pivots on a bushing and shaft assembly attached to the frame cross member. These components are applied in many other mechanisms.

Spur Gears Below is the image in its original context on the page: www.usedmills.net/machinery-equipment/feed/ Select the, "Gears" tab at the bottom of the Excel Worksheet for more information about spur gears.

Wheel and Worm Gears Typical, "C-face worm gearbox below. C-face refers to the round flange used to attach a mating motor flange. Worm gears offer higher gear ratios in a smaller package than any other mechanism. A 40 to 1 ratio increases torque by a factor of 40 while reducing worm gear output shaft speed to 1/40 x input speed.

The worm may have a single, double, or more thread. The axial pitch of the worm is equal to the circular pitch of the wheel. Select the, "Gears" tab at the bottom of the Excel Worksheet for more information about worm gears.

Worm gear Above is the image in its original context on the page: www.global-b2b-network.com/b2b/17/25/751/gear...

www.global-b2b-network.com/b2b/17/25/751/gear...

Laser Jet Printer Above is the image in its original context on the page: news.thomasnet.com/fullstory/531589 The computerized printer above has many moving parts: linkages, gears, shafts, bushings, bearings, etc, for manipulating sheets of paper. The design and analysis of the light weight plastic components of such a printer requires the same principals as do many heavy duty machines with steel and aluminum parts. Observance of functional quality control in the design stage has improved their reliability in recent years.

This is the end of this worksheet.

MACHINE DESIGN EXCEL SPREAD SHEETS Copy write, © Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006 * Machine components are designed to withstand: applied direct forces, moments and torsion. * These loads may be applied gradually, suddenly, and repeatedly. * The design load is equal to the applied load multiplied by a combined shock and fatigue factor, Ks. * The average applied design stress must be multiplied by a stress concentration factor K. * Calculated deflections are compared with required stiffness. * The material strength is compared with the maximum stress due to combinations of anticipated loads.

Math Symbols Spread Sheet Method: 1. Type in values for the input data. 2. Enter. 3. Answer: X = will be calculated. 4. Automatic calculations are bold type.

A x B = A*B 2x3= 2*3 =6

A/B= A/B 3/2= 3/2 = 1.5

A+B= A+B 2+3= 2+3 =5 When using Excel's Goal Seek, unprotect the spread sheet by selecting: Drop down menu: Tools > Protection > Unprotect Sheet > OK When Excel's Goal Seek is not needed, restore protection with: Drop down menu: Tools > Protection > Protect Sheet > OK

Xn = X^n 23 = 2^3 =8

TENSION AND COMPRESSION As shown below, + P = Tension - P = Compression

Reference: Design of Machine Elements, by V.M. Faires, published by: The Macmillan Company, New York/Collier-Macmillan Limited, London, England.

Two machine components, shown above, are subjected to loads P at each end. The force P is resisted by internal stress S which is not uniform. At the hole diameter D and the fillet radius R stress is 3 times the average value. This is true for tension +P and compression -P.

Machine Component Maximum Stress Calculation Refer to the diagram above: External force, ± P = Section height, H = Section width, B = Original length, L = Stress concentration factor, K = Combined shock and fatigue factor, Ks = Section area, A = = Maximum direct stress, Smax = = Safety factor, SF = =

Input 2000 3.5 0.5 5 3.0 3.0 Calculations H*B 1.75 K*Ks*P / A 10286 Sa / Smax 2.14

Material Brass Bronze ASTM A47-52 Malleable Cast Iron Duralumin Monel Metal ASTM A-36 (Mild Steel) Nickel-Chrome Steel

E x 10^6 lbf/in^2 15.0 16.0 25.0 10.5 26.0 29.0 28.0

Tension ( + ) Compression ( - ), P = Section Area, A = Original length, L = Original height, H = Material modulus of elasticity, E = Stress (tension +) (compression -), S = = Strain, e = = Extension (+), Compression ( - ), X = = Poisson's Ratio, Rp = 0.3 = Transverse (contraction +) (expansion -) =

= =

Input 22000 2.00 10 3 29000000 Calculation P/A 11000 S/E 0.00038 L*e 0.0038 ((H - Ho) / H) / e (H - Ho) 0.3*e*H 0.00034

Use if: D/H > 0.5 or R/H > 0.5 lbf in in in -

in^2 lbf/in^2 G x 10^6 5.80 6.50 10.70 4.00 10.00 11.50 11.80

lbf/in^2 in^2 in in lbf/in^2

See table above.

lbf/in^2 in For most metals

in

Shear Stress Distribution A stress element at the center of the beam reacts to the vertical load P with a vertical up shear stress vector at the right end and down at the other. This is balanced by horizontal right acting top and left acting bottom shear stress vectors. A stress element at the top or bottom surface of the beam cannot have a vertical stress vector. The shear stress distribution is parabolic. Reference: Mechanical Engineering Reference Manual (for the PE exam), by M.R. Lindeburg, Published by, Professional Publications, Inc. Belmont, CA. External shear force, Section height, Section width, Shear modulus, Length,

P= H= B= G= L=

Section area, A = A= Shear stress concentration factor, k = Maximum shear stress, Sxy = = Shear strain, e = = Shear deflection, v = =

Input 2200 3.500 1.250 1150000 12 Calculation H*B 4.375 1.5 k*P / A 754 Fs / G 0.00066 e*L 0.0079

lbf in in lbf/in^2 in

in^2 lbf/in^2 in

SHEAR STRESS IN ROUND SECTION BEAM Refer to the diagram above: Solid shafts: K = 1.5 & d = 0. Thin wall tubes: K = 2.0 & d is not zero. External shear force, P = Section outside diameter, D = Section inside diameter, d = Shear stress concentration factor, k = Shear modulus, G = Length, L = Section area, A = A= Maximum shear stress, Fs = Fs = Shear strain, e = e= Shear deflection, v = v=

Input 4000 1.500 0.000 1.33 1.15E+06 5 Calculation π*( D^2 - d^2 )/ 4 1.7674 k*P / A 3010 Fs / G 0.00262 e*L 0.0131

COMPOUND STRESS Stress Element The stress element right is at the point of interest in the machine part subjected to operating: forces, moments, and torques. Direct Stresses: Horizontal, +Fx = tension, -Fx = compression. Vertical, +Fy = tension, -Fy = compression. Shear stress: Shear stress, Sxy = normal to x and y planes.

Principal Stress Plane: The vector sum of the direct and shear stresses, called the principal stress F1, acts on the principal plane angle A degrees, see right. There is zero shear force on a principal plane. Angle A may be calculated from the equation: Tan 2A = 2 x Sxy / ( Fy - Fx)

lbf in in lbf/in^2 in

in^2 lbf/in^2 in

Principal Stresses: Two principal stresses, F1 and F2 are required to balance the horizontal and vertical applied stresses, Fx, Fy, and Sxy. The maximum shear stress acts at 45 degrees to the principal stresses, shown right. The maximum shear stress is given by: Smax = ( F2 - F1 ) / 2 The principal stress equations are given below.

PRINCIPAL STRESSES Principal stress, F1 = (Fx+Fy)/2 + [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ] Principal stress, F2 = (Fx+Fy)/2 - [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ] Max shear stress, Sxy = [Fn(max) - Fn(min)] / 2 Principal plane angle, A = ( ATAN(2*Sxy / (Fy - Fx) ) / 2

See Math Tab below for Excel's Goal Seek. Use Excel's, "Goal Seek" to optimize shaft diameter.

Power Shaft with: Torque T, Vertical Load V, & Horizontal Load H Horizontal force, Vertical force, Torsion, Cantilever length, Diameter,

H= V= T= L= D=

Input 3000 600 2000 10 2

lbf lbf in-lbf in in

Properties at section A-B π= Area, A = A= Section moment of inertia, I = I= Polar moment of inertia, J = J= AT POINT "A" Horizontal direct stress, Fd = Fd = Bending stress, Fb = Fb = Combined direct and bending, Fx = Fx = Direct stress due to, "V", Fy = Torsional shear stress, Sxy = Sxy = Max normal stress at point A, F1 = F1 = Min normal stress at point A, F2 = F2 = Max shear stress at point A, Sxy = = AT POINT "B" Horizontal direct stress, Fd = Fd = Bending stress, Fb = Fb = Combined direct and bending, Fx = Fx = Direct stress due to, "V", Fy = Torsional shear stress, Sxy = Sxy =

Calculation 3.1416 π*D^2 / 4 3.142 π*D^4 / 64 0.7854 π*D^4 / 32 1.5708 H/A 955 M*c / I 7639 H/A + M*c / I 8594 0 T*(D / 2) / J 1273

in^2 in^4 in^4

lbf/in^2 lbf/in^2 lbf/in^2 lbf/in^2 lbf/in^2

(Fx+Fy)/2 + [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ] 8779 lbf/in^2 (Fx+Fy)/2 - [ ((Fx-Fy)/2)/2)^2 + Sxy^2 )^0.5 ] -185 lbf/in^2 [Fn(max) - Fn(min)] / 2 4482 lbf/in^2

H/A 955 -M*c / I -7639 H/A + M*c / I -6684 0 T*D / (2*J) 1273

lbf/in^2 lbf/in^2 lbf/in^2 lbf/in^2 lbf/in^2

Max normal stress at B, F1 = (Fx+Fy)/2 + [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ] F1 = 234 lbf/in^2 Min normal stress at B, F2 = (Fx+Fy)/2 - [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ] F2 = -6919 lbf/in^2 Max shear stress at B, Sxy(max) = [Fn(max) - Fn(min)] / 2 3577 lbf/in^2

Curved Beam-Rectangular Section Outside radius, Ro = Inside radius, Ri = Section width, B = Applied moment, M = Section height, H = = Section area, A = Section neutral axis radius = Radius of neutral axis, Rna = = e= = Inside fiber bending stress, Si = = Outside fiber bending stress, So = =

Input 8.500 7.000 1.500 500 Calculation Ro - Ri 1.500 2.250 Rna H / Ln(Ro / Ri) 7.726 Ri + H/2 - Rna 0.024 M*(Rna-Ri) / (A*e*Ri) 950 M*(Ro-Rna) / (A*e*Ri) 1013

in in in in-lbf in in in^2

in in lbf/in^2 lbf/in^2

Curved Beams-Circular Section Curved Beam-Section diameter, D = Ro - Ri = 1.500 Section radius of neutral axis, Rna = 0.25*(Ro^0.5 + Ri^0.5)^2 = 7.732 e= Ri + D/2 - Rna = 0.018 Inside fiber bending stress, Si = M*(Rna-Ri) / (A*e*Ri) = 1626 Outside fiber bending stress, So = M*(Ro-Rna) / (A*e*Ro) = 1406

in in in lbf/in^2 lbf/in^2

Curved Beam-2 Circular Section Outside radius, Ro = Inside radius, Ri = Applied moment, M = Curved Beam-Section diameter, D = D= Section radius of neutral axis, Rna = Rna = e= e= Inside fiber bending stress, Si = = Outside fiber bending stress, Fo = =

Input 6.000 in 4.000 in 175 in-lbf Calculation Ro - Ri 2 in 0.25*(Ro^0.5 + Ri^0.5)^2 4.949 in Ri + D/2 - Rna 0.051 in (P*(Rna+e))*(Rna-Ri) / (A*e*Ri) 1309 lbf/in^2 M*(Ro-Rna) / (A*e*Ro) 193 lbf/in^2

Rectangular Section Properties Breadth, B = Height, H = Section moment of inertia, Ixx = = Center of area, C1 = C2 = =

Input 1.500 3.000 Calculation B*H^3 / 12 3.375 H/2 1.5

in in

in^4 in

I and C Sections Input 1 2 3

Bn 9 1.5 6

1 2 3

Yn 11.000 6.500 1.500

Hn 2 7 3 ΣA =

Calculation A 18 10.5 18 46.5

Calculation A*Yn A*Yn^2 198.00 2178.00 68.25 443.63 27.00 40.50 Σ = 293.25 2662.13

Yn 11 6.5 1.5

Icg 6.00 42.88 13.50 62.38

Calculation Section modulus, Ixx = ΣA*Yn^2 + ΣIcg = 2724.50 in^4 Center of area, C1 = ΣA*Yn/ΣA = 6.306 in C2 = Y1 + H1/2 = 12.000 in

P= L= a= b= Cantilever, MMAX at B = Fixed ends, MMAX, at C ( a < b ) = Pinned ends, MMAX, at C =

Input 2200 6 2 Calculation L-a 4 P*L 13200 P * a * b^2 / L^2 1956 P*a*b/L 2933

lbf in in

in-lbs in-lbs in-lbs

Ref: AISC Manual of Steel Construction.

Enter value of applied moment MMAX from above: Bending shock & fatigue factor, Kb = Bending stress will be calculated. Applied moment from above, MMAX = Larger of: C1 and C2 = C = Section moment of inertia, Ixx = Bending shock & fatigue factor, Kb = Max moment stress, Sm = =

3 Input 13200 12.00 4.66 1.50 Calculation Kb*M*C / I 50987

Data in-lbf in in^4 -

lb/in^2

Input 1 2 3

Bn 2 7 3

Yn 1.000 3.500 1.500

Calculation A Yn 18.00 1.00 10.50 3.50 18.00 1.50 46.5

Hn 9 1.5 6 ΣA =

Calculations A*Yn A*Yn^2 9.00 4.50 18.38 32.16 13.50 10.13 Σ= 40.88 46.78

Icg 121.50 1.97 54.00 177.47

Section modulus, Ixx = ΣA*h^2 + ΣIcg = 224.25 in^4 Center of area, C1 = ΣA*Yn/ΣA = 0.879 in C2 = B1 - C1 = 1.121 in Symmetrical H Section Properties Input Bn Hn 1 2 9 2 7 1.5 3 3 6 ΣA = Center of gravity, Ycg = = Section modulus, Ixx = = Center of area, C1 = C2 = =

B1 / 2 1.000 ΣIcg 62 B1 / 2 1.000

Calculation A Icg 18.00 6 10.50 43 18.00 14 46.5 62

in in^4

Enter value of applied moment MMAX from above: P= L= a= b= = Cantilever, MMAX at B = = Fixed ends, MMAX, at C ( a < b ) = = Pinned ends, MMAX, at C =

Input 1800 12 3 Calculation L-a 9 P*L 21600 P * a * b^2 / L^2 3038 P*a*b/L 4050

lbf in in

in-lbs in-lbs

Ref: AISC Manual of Steel Construction.

in-lbs

Enter values for applied moment at a beam section given: C, Ixx and Ycg. Bending stress will be calculated. Applied moment from above, MMAX = Larger of: C1 and C2 = C = Section moment of inertia, Ixx = Bending shock & fatigue factor, Kb = Shaft material elastic modulus, E = Beam length from above, L = Beam load from above, P = Max moment stress, Sm = = Cantilever deflection at A, Y = Fixed ends deflection at C, Y = Pinned ends deflection at C, Y =

This is the end of this worksheet

Input 13200 1.750 4.466 1.5 29000000 Calculation 12 1800 Kb*M*C / I 7759 P*L^3 / (3*E*I) 0.0080 P*a^3 * b^3 / (3*E*I*L^3) 0.000053 P*a^2 * b^2 / (3*E*I*L) 0.000281

in-lbf in in^4 lb/in^2 in lbf lb/in^2 in in in

MACHINE DESIGN EXCEL SPREAD SHEETS Copy write, © Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006 Rev: 26Sep09 Spread Sheet Method: 1. Type in values for the input data. 2. Enter. 3. Answer: X = will be calculated. 4. Automatic calculations are bold type.

DESIGN OF POWER TRANSMISSION SHAFTING The objective is to calculate the shaft size having the strength and rigidity required to transmit an applied torque. The strength in torsion, of shafts made of ductile materials are usually calculated on the basis of the maximum shear theory. ASME Code states that for shaft made of a specified ASTM steel: Ss(allowable) = 30% of Sy but not over 18% of Sult for shafts without keyways. These values are to be reduced by 25% if the shafts have keyways. Shaft design includes the determination of shaft diameter having the strength and rigidity to transmit motor or engine power under various operating conditions. Shafts are usually round and may be solid or hollow. Shaft torsional shear stress: Ss = T*R / J Polar moment of area:

J = π*D^4 / 32 J = π*(D^4 - d^4) / 32

Shaft bending stress: Moment of area:

for solid shafts for hollow shafts

Sb = M*R / I I = π*D^4 / 64 I = π*(D^4 - d^4) / 64

for solid shafts for hollow shafts

The ASME Code equation for shafts subjected to: torsion, bending, axial load, shock, and fatigue is: Shaft diameter cubed, D^3 = (16/π*Ss(1-K^4))*[ ( (KbMb + (α*Fα*D*(1+K^2)/8 ]^2 + (Kt*T)^2 ]^0.5 Shaft diameter cubed with no axial load, D^3 = (16/π*Ss)*[ (KbMb)^2 + (Kt*T)^2 ]^0.5 K = D/d

D = Shaft outside diameter,

Kb = combined shock & fatigue bending factor Kt = combined shock & fatigue torsion factor

d = inside diameter

α = column factor = 1 / (1 - 0.0044*(L/k)^2 for L/k < 115 L = Shaft length

k = (I/A)^0.5 = Shaft radius of gyration

A = Shaft section area For rotating shafts: Kb = 1.5, Kt = 1.0 for gradually applied load Kb = 2.0, Kt = 1.5 for suddenly applied load & minor shock Kb = 3.0, Kt = 3.0 for suddenly applied load & heavy shock

Power Transmission Shaft Design Calculations Input shaft data for your problem below and Excel will calculate the answers, Excel' "Goal Seek" may be used to optimize the design of shafts, see the Math Tools tab below.

1. ASME Code Shaft Allowable Stress Su = Sy = Allowable stress based on Su, Sau = Allowable stress based on Sy, Say = Allowable shear stress based on Su, Ss =

2. ASME Code Shaft Diameter Lowest of Sau, Say, & Ss: Sa = Power transmitted by shaft, HP = Shaft speed, N = Shaft vertical load, V = Shaft length, L = Kb =

Input 58000 36000 Calculate 18% * Su 10440 30% * Sy 10800 75% * Sau 7830 Input 7830 10 300 0 10 1.5

lbf/in^2 lbf/in^2

lbf/in^2 lbf/in^2 lbf/in^2

lbf/in^2 hp rpm lbf in

Kt = Shaft torque, T = = Vertical Moment, M = ASME Code for shaft with keyway, D^3 = =

1 Calculate HP * 63000 / N 2100 in-lbf V*L 0 lbf-in (16 / (π*Sa) ) * ( (Kb*Mb)^2 + ( Kt*T)^2 )^0.5 1.366 in^3

Minimum shaft diameter, D = 1.109

in

Shaft Material Ultimate & Yield Stresses Su = Sy = ASME Code Shaft Allowable Stress Allowable stress based on Su, Sau =

Input 70000 46000 Calculate 18% * Su

lbf/in^2 lbf/in^2

Allowable stress based on Sy, Say = Allowable shear stress based on Su, Ss =

12600 30% * Sy 13800 75% * Sau 9450

lbf/in^2 lbf/in^2 lbf/in^2

Shaft Power & Geometry Lowest of Sau, Say, & Ss: Sa = Power transmitted by V-Belt, HP = Shaft speed, N = T1 / T2 = A= L1 = L2 = L3 = D1 = D2 = V-Pulley weight, Wp = Spur gear pressure angle, (14 or 20 deg) B = Kb = Kt =

Input 9450 20 600 3 60 10 30 10 8 18 200 20 1.5 1 Calculate HP * 63000 / N 2100 3 T / (D2 / 2) -( T / (D2 / 2) ) / (1 - B) 117 B * T2 350

lbf/in^2 hp rpm deg in in in in in lbs deg -

Shaft torque, T = = in-lbf T2 / T1 = B = T1 - T2 = T2 = = lbf T1 = = lbf Vertical Forces V2 = Fs = Ft * Tan( A ) = 191 lbf V4 = ( (T1 + T2) * Sin( A ) )-Wp = 204 lbf V3 = ( (V4*(L2 + L3)) - (V2*L1) ) / L2 208 lbf V1 = V2 + V3 - V4 195 lbf

Vertical Moments Mv2 = Mv3 = Horizontal Forces H2 =Ft =

V1 * L1 1954 V4 * L3 2041 T / (D1 / 2) 525

lbf-in lbf-in

lbf

H4 =

(T1 + T2) * Cos( A ) 233 lbf H3 = ( (H4*(L2 + L3)) + (H2*L1) ) / L2 486 H1 = H2 - H3 + H4 272 Horizontal Moments Mh2 = H1 * L1 2722 lbf-in Mh3 = H4 * L3 2334 lbf-in Resultant Moments Mr2 = (Mv2^2 + Mh2^2)^0.5 3351 lbf-in Mr3 = (Mv3^2 + Mh3^2)^0.5 3100 lbf-in Input Larger of: Mr2 & Mr3 = Mb = 3351 lbf-in

Calculate Shaft Diameter

Calculate ASME Code for shaft with keyway, D^3 = (16 / (π*Sa) ) * ( (Kb*Mb)^2 + ( Kt*T)^2 )^0.5 = 2.936 in^3

D=

Shaft Material Ultimate & Yield Stresses Su = Sy = ASME Code Shaft Allowable Stress Allowable stress based on Su, Sau = Allowable stress based on Sy, Say =

1.431

Input 70000 46000 Calculate 18% * Su 12600 30% * Sy

in

lbf/in^2 lbf/in^2

lbf/in^2

Allowable shear stress based on Su, Ss =

Shaft Power & Geometry Lowest of Sau, Say, & Ss: Sa = Power transmitted by V-Belt, HP = Shaft speed, N = T1 / T2 = A= L1 = L2 = L3 = D1 = D2 = V-Pulley weight, Wp = Spur gear pressure angle, (14 or 20 deg) B = Kb = Kt = Left side shaft diameter, SD1 = Center shaft diameter, SD2 = Right side shaft diameter, SD3 = Shaft torque, T = = T2 / T1 = B = T1 - T2 = T2 = = T1 = = Vertical Forces H2 =Ft = V2 = Fs = = V4 = = V3 =

V1 = Vertical Moments Mv2 = Mv3 =

13800 75% * Sau 9450 Input 9450 20 600 3 60 10 30 10 8 18 200 20 1.5 1 1.000 3.000 2.000 Calculate HP * 63000 / N 2100 3 T / (D2 / 2) -( T / (D2 / 2) ) / (1 - B) 117 B * T2 350

lbf/in^2 lbf/in^2

lbf/in^2 hp rpm deg in in in in in lbs deg in in in

in-lbf

lbf lbf

T / (D1 / 2) 525 lbf Ft * Tan( A ) 909 lbf ( (T1 + T2) * Sin( A ) )-Wp 204 lbf ( (V4*(L2 + L3)) - (V2*L1) ) / L2 -31 lbf

V2 + V3 - V4 674 V1 * L1 6742 V4 * L3 2041 Input

lbf

lbf-in lbf-in

Larger of: Mr2 & Mr3 = Mb =

6742

lbf-in

Calculate Shaft Diameter

Calculate ASME Code for shaft with keyway, D^3 = (16 / (π*Sa) ) * ( (Kb*Mb)^2 + ( Kt*T)^2 )^0.5 = 5.567 in^3

D=

1.771

in

Power Shaft Torque

Input 7.5 1750 3 1.000 5 11500000 Calculation Shaft Design Torque, Td = Kt*12*33000*HP / (2*π*N) = 810

Motor Power, HP = Shaft speed, N = Torque shock & fatigue factor, Kt = Shaft diameter, D = Shaft length, L = Shaft material shear modulus, G =

Drive Shaft Torque Twist Angle Shaft Design Torque from above, Td = Shaft diameter, D = Shaft length, L = Shaft material tension modulus, E = Shaft material shear modulus, G =

Section polar moment of area, J = = Shear stress due to Td, ST = = Shaft torsion deflection angle, a = = =

Input 1080 0.883 10 29000000 11500000

Calculation π*D^4 / 32 0.060 Td*D / (2*J) 8000 Td*L / (J*G) 0.0158 0.90

hp rpm in in psi

in-lbf

in-lbf in in psi psi

< GOAL SEEK

in^4 lbf/in^2 radians degrees

< GOAL SEEK

POLAR MOMENT OF AREA AND SHEAR STRESS Torsion, T = Round solid shaft diameter, D = Section polar moment of inertia, J = = Torsion stress, Ft = =

Torsion, T = Round tube shaft outside dia, Do = Round tube shaft inside dia, Di = Section polar moment of inertia, J = J= Torsion stress, Ft = =

Torsion, T = Square shaft breadth = height, B = Section polar moment of inertia, J = = Torsion stress, Ft = =

Torsion, T = Rectangular shaft breadth, B = Height, H = Section polar moment of inertia, J = = Torsion stress, Ft = =

Input 360 2.000 Calculation π*D^4 / 32 1.571 T*(D/2) / J 229

Input 1000 2.250 1.125 Calculation π*(Do^4 - Di^4) / 32 2.359 T*(Do/2) / J 477

Input 1000 1.750 Calculation B^4 / 6 1.563 T*(B/2) / J 560

Input 1000 1.000 2.000 Calculation B*H*(B^2 + H^2)/ 12 0.833 T*(B/2) / J 600

in-lbf in

in^4 lb/in^2

in-lbf in in

in^4 lb/in^2

in-lbf in

in^4 lb/in^2

in-lbf in in

in^4 lb/in^2

Cantilever shaft bending moment Shaft transverse load, W = Position in shaft, x = Bending shock & fatigue factor, Km = Shaft diameter, D = Moment at x, Mx = Design moment at x, Md = = Section moment of inertia, I = = Bending stress for shaft, Fb = =

Cantilever shaft bending deflection Shaft transverse load at free end, W = Shaft diameter, D = Shaft length, L = Deflection location, x = Bending moment shock load factor, Km = Modulus of elasticity, E =

Input 740 5 3 1.000 Calculation W*x Km*Mx 11100 π*D^4 / 64 0.049 M*D / (2*I) 113049

Input 740 1.000 10 5 3 29000000

Calculation π*D^4 / 64 0.049 5 Km*W*x 11100 M*(D/2) / I 113063 Cantilever bend'g deflection at x, Yx = (-W*x^2/(6*E*I))*((3*L) - x) = -0.0541 Bending deflection at x = 0, Y = -W*L^3 / (3*E*I) Section moment of inertia, I = = Moment at, x = Moment at x, M = = Bending stress at x: Sb =

lbf in in in-lbs in-lbs in^4 lbs/in^2

< GOAL SEEK

lbf in in in psi

in^4 in in-lbf lbf/in^2 in

< GOAL SEEK

Y=

Section Moment of Inertia Round solid shaft diameter, D = Section moment of inertia, Izz = Answer: Izz = Section moment of Inertia Round tube shaft diameter, Do = Di = Section polar moment of inertia, Izz = Answer: Izz =

Section moment of Inertia Square shaft breadth = height, B = Section moment of inertia, Izz = Answer: Izz =

-0.1733

Input 1.000 Calculations π*D^4 / 64 0.049 Input 1.750 1.5 Calculation π*(Do^4 - Di^4) / 64 0.212

Input 1.750 Calculation B^4 / 12 0.782

in

in

in^4

in in

in^4

in^4

BENDING STRESS Enter values for applied moment at a beam section, c, Izz and Kb. Bending stress will be calculated. Input Applied moment at x, M = 1000 in-lbf c= 1.000 in Section moment of inertia, Izz = 2.5 in^4 Bending shock & fatigue factor, Kb = 3 Calculation Max bending stress, Fb = Kb*M*c / I Answer: Fb = 1200 lb/in^2

TYPICAL BULK MATERIAL BELT CONVEYOR SHAFTING SPECIFICATION See PDHonline courses: M262 an M263 by the author of this course for more information. 1.1 Pulley Shafts: 1.2 All shafts shall have one fixed type bearing; the balance on the shaft shall be expansion type. 1.3 Pulleys and pulley shafts shall be sized for combined torsional and bending static and fatigue stresses.

1.4 Shaft keys shall be the square parallel type and keyways adjacent to bearings shall be round end, all other keyways may be the run-out type. 2.1 Pulleys: 2.2 The head pulley on the Reclaim Conveyor shall be welded 304-SS so as not to interfere with tramp metal removal by the magnet. 2.3 All pulleys shall be welded steel crown faced, selected in accordance with ratings established by the Mechanical Power Transmission Association Standard No.301-1965 and U.S.A.

2.3 All pulleys shall be welded steel crown faced, selected in accordance with ratings established by the Mechanical Power Transmission Association Standard No.301-1965 and U.S.A. Standard No.B105.1-1966. In no case shall the pulley shaft loads as listed in the rating tables of these standards be exceeded. 2.4 All pulleys shall be crowned. 2.5 All drive pulleys shall be furnished with 1/2 inch thick vulcanized herringbone grooved lagging. 2.6 Snub pulleys adjacent to drive pulleys shall have a minimum diameter of 16 inches.

This is the end of this worksheet

MACHINE DESIGN EXCEL SPREAD SHEETS Copy write, © Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006

COUPLINGS

RIGID COUPLING DESIGN Couplings are used to connect rotating shafts continuously. Clutches are used to connect rotating shafts temporarily. Rigid couplings are used for accurately aligned shafts in slow speed applications. Refer to ASME code and coupling vendor design values.

KEY SLOT STRESS FACTOR 2.10

Key Slot Stress Factor (Kk)

2.00 1.90 1.80 1.70 1.60

A B C D

1.50 1.40 1.30 1.20 1.10 1.00 0.2

0.4

0.6

0.8

Key half slot width / Slot depth (y / h)

1.0

Legend A B C D

h/R 0.2 0.3 0.4 0.5

Design Stress Coupling Design Shear Stress = Design allowable average shear stress. Input Material ultimate tensile stress, Ft = 85000 lbf/in^2 Shaft material yield stress, Fy = 45000 lbf/in^2 Calculation Ultimate tensile stress design factor, ku = 0.18 Design ultimate shear stress, Ssu = ku* Ft = 15300 lbf/in^2 Yield stress factor, ky = 0.3 Design yield shear design stress factor, Ssy = ky* Ft = 13500 lbf/in^2 Use the smaller design shear stress of Fsu and Fsy above.

1. Shaft Torsion Shear Strength Shaft diameter, D = Key slot total width = H = Key slot depth, h = Key slot half width, y = Key slot half width / Slot depth, y / h = Slot depth / Shaft radius, h / R = Motor Power, HP = Shaft speed, N = Allowable shaft stress from above, Ssu or Ssy = Torque shock load factor, Kt = Key slot stress factor from graph above, Kk = Motor shaft torque, Tm = = Section polar moment of inertia, J = = Allowable shaft torque, Ts = =

Input 2.000 in 0.375 in 0.25 in Calculation 0.188 0.75 Apply to graph 0.25 above. Input 60 hp 300 rpm 13500 lbf/in^2 3.00 1.38 Add-Ins > Goal Seek Tools > Add-Ins > Solver To open select Tools.

Spread Sheet Method: 1. Type in values for the input data. 2. Enter. 3. Answer: X = will be calculated. 4. Automatic calculations are bold type.

When using Excel's Goal Seek, unprotect the spread sheet by selecting: Drop down menu: Tools > Protection > Unprotect Sheet > OK When Excel's Goal Seek is not needed, restore protection with: Drop down menu: Tools > Protection > Protect Sheet > OK

What if Calculations Excel will make a, “what if calculation” using, "Goal Seek" when the calculated formula value needs to be changed.

Goal Seek Example The hypotenuse of the right angle triangle above is calculated in the table below. Columns, A and B are intercescted by rows 5 through 10 forming cells. Cell B6 contains the value 4.00. Cell B10 contains the formula, "= (B6^2 + B7^2) ^ (1/2)". The hypotenuse is found to be 5.00 when the other two sides are: 3.00 and 4.00. However the, "Optimum Value" for hypotenuse is 7.00. Select the formula cell, B10 and Goal Seek will calculate a new value (target value) for cell B7 that will change the hypotenuse to 7.00.

A 5 6 7 8 9 10

To Create the Above Table

ADJ = OPP = HYP = =

B Input 4.00 3.00 Calculations (ADJ^2 + OPP^2)^(1/2) 5.00

To Create the Above Table Type, “Input” in cell B5 as shown below. “ADJ =” in cell A6. “4” in cell B6. Complete the spreadsheet table below in columns A and B down to row 9. 1. Select cell B9 with the mouse pointer. 2. Press keys: ctrl and C together. 3. Pick cell B10, Enter. The formula, ( ADJ^2 + OPP^2 )^(1/2) will be copied into cell B10. 4. Press: f2, home , =. Function key f2 enables editing a cell. Home key moves the mouse pointer to the left side of the cell. Type the, = sign and press, "Enter" to enable cell B10 to do the math calculation. See cell below B10.

5. Cell B10 below contains the calculated value 5.00.

A 5 6 7 8 9 10

What if Calculations

ADJ = OPP = HYP = =

B Input 4.00 3.00 Calculations (ADJ^2 + OPP^2)^(1/2) 5.00

Excel will make a, “what if calculation” when the calculated formula value needs to be changed. 1. While in Excel 2007 pick the, “Data” tab shown below.

2. To the right of the Data tab pick, “What-If Analysis” followed by, “Goal Seek” illustrated below.

3. Goal Seek allows you to pick the formula cell with the 5.00 result followed by entering the desired value, 7.00 in the, “Goal Seek” dialog box below. 4. Next pick an input number, 3.00 in this example then pick, OK.

5. Excel has iteratively changed cell B7 to 5.74 at which point cell B10 is equal to the desired result of 10.00, below.

Excel's Goal Seek Example Drive Shaft Design Motor Power, HP = Shaft speed, N = Torque shock & fatigue factor, Kt = Shaft diameter, D = Shaft length, L = Material shear modulus, G =

Input 5.0 1750 3 0.500 10 11500000

hp rpm in in psi

Calculation Applied motor shaft torque, Ta = 12*33000*HP / (2*π*N) = 180.05 in-lbf Section polar moment of inertia, J = π*D^4 / 32 J= 0.006 in^4 Answer: Design Torque, Td = Kt*Ta = 540 in-lbf Shear stress for shafts, St = Td*D / (2*J) = 22005 lbf/in^2 Shaft torsion deflection angle, a = Td*L / (J*G) a= 0.0765 radians a= 4.39 degrees

Excel's Goal Seek Problem Use Excel's, "Goal Seek" in the duplicate example below to calculate a new shaft diameter D that will reduce the above torsion stress of 22005 lbf/in^2 to 12000 lbf/in^2, keeping the same 5 hp motor. Answer: 0.612 inch diameter. Step 1. Pick the torsion shear stress (St) cell B90, 20005 Step 2. Select drop-down menu, Tools > Goal Seek… Step 3. Pick the "To value" box and type, 12000 Step 4. Pick the, "By changing cell" box and pick the shaft diameter D cell B78 initially containing, 0.500 Step 5. Click, OK Step 6. Use the same spread sheet below: The shaft torsion stress St will is set at 12000 lbf/in^2 the shaft diameter D has changed from 0.500 to 0.612 inches and the shaft twist will change from 4.39 to 1.95 degrees.

Drive Shaft Design Motor Power, HP = Shaft speed, N = Torque shock & fatigue factor, Kt = Shaft diameter, D = Shaft length, L = Material shear modulus, G =

Input 5 1750 3 0.612 10 11500000

hp rpm in in psi

Calculation Applied motor shaft torque, Ta = 12*33000*HP / (2*π*N) = 180.05 in*lbf Section polar moment of inertia, J = π*D^4 / 32 J= 0.014 in^4 Answer: Design Torque, Td = Kt*Ta = 540 in-lbf Shear stress for shafts, St = Td*D / (2*J) = 12000 lbs/in^2 Shaft torsion deflection angle, a = Td*L / (J*G) a= 0.0341 radians a= 1.95 degrees

The Vibration Forcing Function One end of a spring having stiffness K1 is connected to mass M1 on wheels and the other end is connected to a vertical wall. One end of a second spring having stiffness K2 is connected to mass M2 on wheels and the other end is connected to mass M1. A force applied to mass M1 initiates the vibration. Friction is small enough to be neglected.

Max kinetic energy, K.E. = (1/2)*M1^2* ω^2 + (1/2)*M2^2* ω^2 Max potential energy, P.E. = (1/2)*K1*X1^2 + (1/2)*K2*(X2 - X1)^2 Neglecting friction, Max K.E. =

Max P.E.

-ω^2 = [K1+K2*((X2/X1) - 1)^2]/ [(M1+M2*(X2/X1)^2] 1. This equation will give the first and lowest natural frequency (ω). 2. The solution for ω is by trial and error for various values of X2/X1. Input Reference: Machine Mass, M1 = 0.1 Design by A.S. Hall, Mass, M2 = 0.1 A.R. Holowenko, H.G. K1 = 20 Laughlin, Published k2 = 20 byMcGraw-Hill. X2 / X1 = 1.6180 Calculation -ω^2 = [K1+K2*((X2/X1) - 1)^2]/ [(M1+M2*(X2/X1)^2] -ω^2 = 76.3932 ω= 8.740 radn/sec 3. Use Excel's Solver for a trial and error solution to the above forcing function example. 4. Start above solution by typing, X2 / X1 = 0 5. Use drop down menu, Tools > Solver > Set Target Cell: > B144 > Equal to Min 6. By Changing Cell > B140 > Solve > Keep Solver Solution

Excel's, Equation "Solver" Excel's Solver can solve one equation of the form: y equals a function of x, y = f(x). The function of x can be a polynomial; ( a + bx + cx2 + dx3 +…. zxn ), an exponential: ( aenx ), a logarithmic: a(logx), trigonometric: ( aSin x + bCos x), or any other function of x.

logarithmic: a(logx), trigonometric: ( aSin x + bCos x), or any other function of x. Also Excel's Solver can solve multple simultaneous equations; linear, non-linear, or a mixture of the two. Excel iteratively adjusts one input value of x to cause one calculated formula cell value of y to equal a target value of y.

C 5 6 7 8 9

Guess X =

D Problem 1.4

Y = 2*X^5 - 3*X^2 - 5 = -0.1235

Solver Example 1. The input value of X is 1.4 and this value of X causes Y to equal -0.1235 in the spreadsheet table above. 2. Excel's Solver will adjust the input value of X, in this case1.4 in blue cell D6, by iteration (repeatedly) until the calculated value of Y in the yellow cell D9 approaches the target value of zero, ( 0 ). 3. Select the calculated answer in yellow cell, ( D9 ) below. 4. Select: Tools > Goal Seek > Target Cell [ $D$9 ] > Equal to: > Value of: > 0 > By changing cells: Select [ $D$6 ] > Add (Constraints) > Cell Reference > $D$9 = 0 > OK. C 5 6 7 8 9

Solved X =

D Solution 1.4041

Y = 2*X^5 - 3*X^2 - 5 = 0.0004

5. The completed calculation above shows that if X = 1.4041 then Y = 0.0004 or 4 / 10,000 which is close enough to 0 for engineering purposes.

Simultaneous Equations Using Excel's, "Solver" Reference: www.dslimited.biz/excel_totorials Equations to be solved: u + v + w + x + y = 5.5 u + 2v + w - 0.5x + 2y = 22.5 2v + 2w - x - y = 30 2u - w + 0.75x + 0.5y = -11 u + 0.25v + w - x = 17.5 1. Insert the equations below into column B cells:

=E146+E147+E148+E149+E150 =E146+2*E147+E148-0.5*E149+2*E150 =2*E147+2*E148-E149-E150 =2E146-2E148-E149-E150 =E146+0.25E147+E148-E149

Equations 0.0 0.0 0.0 0.0 0.0

2. Select cells, E146 to 150 3. Click on drop down menu: Tools > Solver > 4. Delete contents of; Set Target Cell 5. Pick: By Changing Cells: > Select cells E146 to E150

Constants 5.5 22.5 30 -11 17.5

Solution u= v= w= x= y=

Row Row Row Row Row

146 147 148 149 150

Equations 5.5 22.5 30.0 -11.0 17.5

Constants 5.5 22.5 30 -11 17.5

Solution u= v= w= x= y=

1.00 4.00 7.50 -8.00 1.00

Solution u= v= w= x= y=

0.00 0.00 0.00 0.00 0.00

You may use the table below to solve the 5 simultaneous equations.

Row Row Row Row Row

146 147 148 149 150

This is the end of this spread sheet.

Equations 0.0 0.0 0.0 0.0 0.0

Constants 5.5 22.5 30 -11 17.5