1 MachinesM236 MACHINE DESIGN EXCEL SPREAD SHEETSRev. 9 Mar 09Copy write, Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006This is the end of this worksheet.

BackhoeAbove is the image in its original context on the page: www.chesterfieldgroup.co.uk/products/mobile.htmlThe 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

Automated Packaging MachineThe 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.Wheel and Worm GearsTypical, "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.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.Strength and Stiffness AnalysisThe 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.Automobile Independent Front SuspensionAbove 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 GearsBelow 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.Worm gearAbove is the image in its original context on the page: www.global-b2b-network.com/b2b/17/25/751/gear...Laser Jet PrinterAbove 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.Pick and Place RobotA 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.

ShredderAbove 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 / 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.

2 StressMACHINE DESIGN EXCEL SPREAD SHEETSCopy 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 SymbolsA x B =A*BA / B =A / BSpread Sheet Method:2 x 3 =2 * 33 / 2 =3 / 21. Type in values for the input data.=6=1.52. Enter.3. Answer: X = will be calculated.A + B =A + BXn =X^n4. Automatic calculations are bold type.2 + 3 =2 + 323 =2^3=5=8When using Excel's Goal Seek, unprotect the spread sheet by selecting:Drop down menu: Tools > Protection > Unprotect Sheet > OKWhen Excel's Goal Seek is not needed, restore protection with:Drop down menu: Tools > Protection > Protect Sheet > OKTENSION AND COMPRESSIONAs shown below, + P =Tension- P =CompressionTwo 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 CalculationUse if: D/H > 0.5 or R/H > 0.5Refer to the diagram above:InputExternal force, P =2000lbfSection height, H =3.5inSection width, B =0.5inOriginal length, L =5inStress concentration factor, K =3.0-Combined shock and fatigue factor, Ks =3.0-CalculationsSection area, A =H*B=1.75in^2Maximum direct stress, Smax =K*Ks*P / A=10286lbf/in^2Safety factor, SF =Sa / Smax=2.14-MaterialE x 10^6 lbf/in^2G x 10^6Brass15.05.80Bronze16.06.50ASTM A47-52 Malleable Cast Iron25.010.70Duralumin10.54.00Monel Metal26.010.00ASTM A-36 (Mild Steel)29.011.50Nickel-Chrome Steel28.011.80InputTension ( + ) Compression ( - ), P =22000lbf/in^2Section Area, A =2.00in^2Original length, L =10inOriginal height, H =3inMaterial modulus of elasticity, E =29000000lbf/in^2See table above.CalculationStress (tension +) (compression -), S =P / A=11000lbf/in^2Strain, e =S / E=0.00038-Extension (+), Compression ( - ), X =L*e=0.0038inPoisson's Ratio, Rp = 0.3 =((H - Ho) / H) / eFor most metalsTransverse (contraction +) (expansion -) =(H - Ho)=0.3*e*H=0.00034inInputExternal shear force, P =2200lbfSection height, H =3.500inSection width, B =1.250inShear modulus, G =1150000lbf/in^2Length, L =12inCalculationSection area, A =H*BA =4.375in^2Shear stress concentration factor, k =1.5-Maximum shear stress, Sxy =k*P / A=754lbf/in^2Shear strain, e =Fs / G-=0.00066-Shear deflection, v =e*L=0.0079inSHEAR STRESS IN ROUND SECTION BEAMRefer to the diagram above:Solid shafts: K = 1.5 & d = 0.Thin wall tubes: K = 2.0 & d is not zero.InputExternal shear force, P =4000lbfSection outside diameter, D =1.500inSection inside diameter, d =0.000inShear stress concentration factor, k =1.33-Shear modulus, G =1.15E+06lbf/in^2Length, L =5inCalculationSection area, A =*( D^2 - d^2 )/ 4A =1.7674in^2Maximum shear stress, Fs =k*P / AFs =3010lbf/in^2Shear strain, e =Fs / G-e =0.00262-Shear deflection, v =e*Lv =0.0131inCOMPOUND STRESSPRINCIPAL STRESSESPrincipal 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)] / 2Principal plane angle, A =( ATAN(2*Sxy / (Fy - Fx) ) / 2Power Shaft with: Torque T, Vertical Load V, & Horizontal Load HInputHorizontal force, H =3000lbfVertical force, V =600lbfTorsion, T =2000in-lbfCantilever length, L =10inDiameter, D =2inProperties at section A-BCalculation =3.1416-Area, A =*D^2 / 4A =3.142in^2Section moment of inertia, I =*D^4 / 64I =0.7854in^4Polar moment of inertia, J =*D^4 / 32J =1.5708in^4AT POINT "A"Horizontal direct stress, Fd =H / AFd =955lbf/in^2Bending stress, Fb =M*c / IFb =7639lbf/in^2Combined direct and bending, Fx =H/A + M*c / IFx =8594lbf/in^2Direct stress due to, "V", Fy =0lbf/in^2Torsional shear stress, Sxy =T*(D / 2) / JSxy =1273lbf/in^2Max normal stress at point A, F1 =(Fx+Fy)/2 + [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ]F1 =8779lbf/in^2Min normal stress at point A, F2 =(Fx+Fy)/2 - [ ((Fx-Fy)/2)/2)^2 + Sxy^2 )^0.5 ]F2 =-185lbf/in^2Max shear stress at point A, Sxy =[Fn(max) - Fn(min)] / 2=4482lbf/in^2AT POINT "B"Horizontal direct stress, Fd =H/AFd =955lbf/in^2Bending stress, Fb =-M*c / IFb =-7639lbf/in^2Combined direct and bending, Fx =H/A + M*c / IFx =-6684lbf/in^2Direct stress due to, "V", Fy =0lbf/in^2Torsional shear stress, Sxy =T*D / (2*J)Sxy =1273lbf/in^2Max normal stress at B, F1 =(Fx+Fy)/2 + [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ]F1 =234lbf/in^2Min normal stress at B, F2 =(Fx+Fy)/2 - [ ((Fx-Fy)/2)^2 + Sxy^2 )^0.5 ]F2 =-6919lbf/in^2Max shear stress at B, Sxy(max) =[Fn(max) - Fn(min)] / 23577lbf/in^2Curved Beam-Rectangular SectionInputOutside radius, Ro =8.500inInside radius, Ri =7.000inSection width, B =1.500inApplied moment, M =500in-lbfCalculationSection height, H =Ro - Riin=1.500inSection area, A =2.250in^2Section neutral axis radius =RnaRadius of neutral axis, Rna =H / Ln(Ro / Ri)=7.726ine =Ri + H/2 - Rna=0.024inInside fiber bending stress, Si =M*(Rna-Ri) / (A*e*Ri)=950lbf/in^2Outside fiber bending stress, So =M*(Ro-Rna) / (A*e*Ri)=1013lbf/in^2Curved Beams-Circular SectionCurved Beam-Section diameter, D =Ro - Ri=1.500inSection radius of neutral axis, Rna =0.25*(Ro^0.5 + Ri^0.5)^2=7.732ine =Ri + D/2 - Rna=0.018inInside fiber bending stress, Si =M*(Rna-Ri) / (A*e*Ri)=1626lbf/in^2Outside fiber bending stress, So =M*(Ro-Rna) / (A*e*Ro)=1406lbf/in^2Curved Beam-2 Circular SectionInputOutside radius, Ro =6.000inInside radius, Ri =4.000inApplied moment, M =175in-lbfCalculationCurved Beam-Section diameter, D =Ro - RiD =2inSection radius of neutral axis, Rna =0.25*(Ro^0.5 + Ri^0.5)^2Rna =4.949ine =Ri + D/2 - Rnae =0.051inInside fiber bending stress, Si =(P*(Rna+e))*(Rna-Ri) / (A*e*Ri)=1309lbf/in^2Outside fiber bending stress, Fo =M*(Ro-Rna) / (A*e*Ro)=193lbf/in^2Rectangular Section PropertiesInputBreadth, B =1.500inHeight, H =3.000inCalculationSection moment of inertia, Ixx =B*H^3 / 12=3.375in^4Center of area, C1 = C2 =H / 2=1.5inI and C SectionsInputCalculationBnHnAYn192181121.5710.56.5363181.5A =46.5CalculationYnA*YnA*Yn^2Icg111.000198.002178.006.0026.50068.25443.6342.8831.50027.0040.5013.50 =293.252662.1362.38CalculationSection modulus, Ixx =A*Yn^2 + Icg=2724.50in^4Center of area, C1 =A*Yn/A=6.306inC2 =Y1 + H1/2=12.000inInputP =2200lbfL =6ina =2inCalculationb =L - a4Cantilever, MMAX at B =P * L13200in-lbsFixed ends, MMAX, at C ( a < b ) =P * a * b^2 / L^21956in-lbsPinned ends, MMAX, at C =P * a * b / L2933in-lbsEnter value of applied moment MMAX from above:Bending shock & fatigue factor, Kb =3DataBending stress will be calculated.InputApplied moment from above, MMAX =13200in-lbfLarger of: C1 and C2 = C =12.00inSection moment of inertia, Ixx =4.66in^4Bending shock & fatigue factor, Kb =1.50-CalculationMax moment stress, Sm =Kb*M*C / I=50987lb/in^2InputCalculationBnHnAYn12918.001.00271.510.503.5033618.001.50A =46.5CalculationsYnA*YnA*Yn^2Icg1.0009.004.50121.503.50018.3832.161.971.50013.5010.1354.00 =40.8846.78177.47Section modulus, Ixx =A*h^2 + Icg=224.25in^4Center of area, C1 =A*Yn/A=0.879inC2 =B1 - C1=1.121inSymmetrical H Section PropertiesInputCalculationBnHnAIcg12918.006271.510.504333618.0014A =46.562Center of gravity, Ycg =B1 / 2=1.000inSection modulus, Ixx =Icg=62in^4Center of area, C1 = C2 =B1 / 2=1.000Enter value of applied moment MMAX from above:InputP =1800lbfL =12ina =3inCalculationb =L - a=9Cantilever, MMAX at B =P * L=21600in-lbsFixed ends, MMAX, at C ( a < b ) =P * a * b^2 / L^2=3038in-lbsPinned ends, MMAX, at C =P * a * b / L4050in-lbsEnter values for applied moment at a beam section given: C, Ixx and Ycg.Bending stress will be calculated.InputApplied moment from above, MMAX =13200in-lbfLarger of: C1 and C2 = C =1.750inSection moment of inertia, Ixx =4.466in^4Bending shock & fatigue factor, Kb =1.5-Shaft material elastic modulus, E =29000000lb/in^2CalculationBeam length from above, L =12inBeam load from above, P =1800lbfMax moment stress, Sm =Kb*M*C / I=7759lb/in^2Cantilever deflection at A, Y =P*L^3 / (3*E*I)0.0080inFixed ends deflection at C, Y =P*a^3 * b^3 / (3*E*I*L^3)0.000053inPinned ends deflection at C, Y =P*a^2 * b^2 / (3*E*I*L)0.000281inThis is the end of this worksheet

Stress ElementThe 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)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.See Math Tab below for Excel's Goal Seek.

Use Excel's, "Goal Seek" to optimize shaft diameter.Shear Stress DistributionA 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.Ref: AISC Manual of Steel Construction.Ref: AISC Manual of Steel Construction.Reference: Design of Machine Elements, by V.M. Faires, published by: The Macmillan Company, New York/Collier-Macmillan Limited, London, England.

3 ShaftMACHINE DESIGN EXCEL SPREAD SHEETSCopy write, Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006Rev: 26Sep09Spread Sheet Method:1. Type in values for the input data.2. Enter.3. Answer: X = will be calculated.4. Automatic calculations are bold type.1. ASME Code Shaft Allowable StressInputSu =58000lbf/in^2Sy =36000lbf/in^2CalculateAllowable stress based on Su, Sau =18% * Su10440lbf/in^2Allowable stress based on Sy, Say =30% * Sy10800lbf/in^2Allowable shear stress based on Su, Ss =75% * Sau7830lbf/in^22. ASME Code Shaft DiameterInputLowest of Sau, Say, & Ss: Sa =7830lbf/in^2Power transmitted by shaft, HP =10hpShaft speed, N =300rpmShaft vertical load, V =0lbfShaft length, L =10inKb =1.5Kt =1CalculateShaft torque, T =HP * 63000 / N=2100in-lbfVertical Moment, M =V * L0lbf-inASME Code for shaft with keyway, D^3 =(16 / (*Sa) ) * ( (Kb*Mb)^2 + ( Kt*T)^2 )^0.5=1.366in^3Minimum shaft diameter, D =1.109inShaft Material Ultimate & Yield StressesInputSu =70000lbf/in^2Sy =46000lbf/in^2ASME Code Shaft Allowable StressCalculateAllowable stress based on Su, Sau =18% * Su12600lbf/in^2Allowable stress based on Sy, Say =30% * Sy13800lbf/in^2Allowable shear stress based on Su, Ss =75% * Sau9450lbf/in^2Shaft Power & GeometryInputLowest of Sau, Say, & Ss: Sa =9450lbf/in^2Power transmitted by V-Belt, HP =20hpShaft speed, N =600rpmT1 / T2 =3A =60degL1 =10inL2 =30inL3 =10inD1 =8inD2 =18inV-Pulley weight, Wp =200lbsSpur gear pressure angle, (14 or 20 deg) B =20degKb =1.5-Kt =1-CalculateShaft torque, T =HP * 63000 / N=2100in-lbfT2 / T1 = B =3T1 - T2 =T / (D2 / 2)T2 =-( T / (D2 / 2) ) / (1 - B)=117lbfT1 =B * T2=350lbfVertical ForcesV2 = Fs =Ft * Tan( A )=191lbfV4 =( (T1 + T2) * Sin( A ) )-Wp=204lbfV3 =( (V4*(L2 + L3)) - (V2*L1) ) / L2208lbfV1 =V2 + V3 - V4195lbfVertical MomentsMv2 =V1 * L11954lbf-inMv3 =V4 * L32041lbf-inHorizontal ForcesH2 =Ft =T / (D1 / 2)525lbfH4 =(T1 + T2) * Cos( A )233lbfH3 =( (H4*(L2 + L3)) + (H2*L1) ) / L2486H1 =H2 - H3 + H4272Horizontal MomentsMh2 =H1 * L12722lbf-inMh3 =H4 * L32334lbf-inResultant MomentsMr2 =(Mv2^2 + Mh2^2)^0.53351lbf-inMr3 =(Mv3^2 + Mh3^2)^0.53100lbf-inInputLarger of: Mr2 & Mr3 = Mb =3351lbf-inCalculate Shaft DiameterCalculateASME Code for shaft with keyway, D^3 =(16 / (*Sa) ) * ( (Kb*Mb)^2 + ( Kt*T)^2 )^0.5=2.936in^3D =1.431inShaft Material Ultimate & Yield StressesInputSu =70000lbf/in^2Sy =46000lbf/in^2ASME Code Shaft Allowable StressCalculateAllowable stress based on Su, Sau =18% * Su12600lbf/in^2Allowable stress based on Sy, Say =30% * Sy13800lbf/in^2Allowable shear stress based on Su, Ss =75% * Sau9450lbf/in^2Shaft Power & GeometryInputLowest of Sau, Say, & Ss: Sa =9450lbf/in^2Power transmitted by V-Belt, HP =20hpShaft speed, N =600rpmT1 / T2 =3A =60degL1 =10inL2 =30inL3 =10inD1 =8inD2 =18inV-Pulley weight, Wp =200lbsSpur gear pressure angle, (14 or 20 deg) B =20degKb =1.5-Kt =1-Left side shaft diameter, SD1 =1.000inCenter shaft diameter, SD2 =3.000inRight side shaft diameter, SD3 =2.000inCalculateShaft torque, T =HP * 63000 / N=2100in-lbfT2 / T1 = B =3T1 - T2 =T / (D2 / 2)T2 =-( T / (D2 / 2) ) / (1 - B)=117lbfT1 =B * T2=350lbfVertical ForcesH2 =Ft =T / (D1 / 2)525lbfV2 = Fs =Ft * Tan( A )=909lbfV4 =( (T1 + T2) * Sin( A ) )-Wp=204lbfV3 =( (V4*(L2 + L3)) - (V2*L1) ) / L2-31lbfV1 =V2 + V3 - V4674lbfVertical MomentsMv2 =V1 * L16742lbf-inMv3 =V4 * L32041lbf-inInputLarger of: Mr2 & Mr3 = Mb =6742lbf-inCalculate Shaft DiameterCalculateASME Code for shaft with keyway, D^3 =(16 / (*Sa) ) * ( (Kb*Mb)^2 + ( Kt*T)^2 )^0.5=5.567in^3D =1.771inPower Shaft TorqueInputMotor Power, HP =7.5hpShaft speed, N =1750rpmTorque shock & fatigue factor, Kt =3Shaft diameter, D =1.000inShaft length, L =5inShaft material shear modulus, G =11500000psiCalculationShaft Design Torque, Td =Kt*12*33000*HP / (2**N)=810in-lbfDrive Shaft Torque Twist AngleInputShaft Design Torque from above, Td =1080in-lbfShaft diameter, D =0.883in< GOAL SEEKShaft length, L =10inShaft material tension modulus, E =29000000psiShaft material shear modulus, G =11500000psiCalculationSection polar moment of area, J =*D^4 / 32=0.060in^4Shear stress due to Td, ST =Td*D / (2*J)=8000lbf/in^2< GOAL SEEKShaft torsion deflection angle, a =Td*L / (J*G)=0.0158radians=0.90degreesPOLAR MOMENT OF AREA AND SHEAR STRESSInputTorsion, T =360in-lbfRound solid shaft diameter, D =2.000inCalculationSection polar moment of inertia, J =*D^4 / 32=1.571in^4Torsion stress, Ft =T*(D/2) / J=229lb/in^2InputTorsion, T =1000in-lbfRound tube shaft outside dia, Do =2.250inRound tube shaft inside dia, Di =1.125inCalculationSection polar moment of inertia, J =*(Do^4 - Di^4) / 32J =2.359in^4Torsion stress, Ft =T*(Do/2) / J=477lb/in^2InputTorsion, T =1000in-lbfSquare shaft breadth = height, B =1.750inCalculationSection polar moment of inertia, J =B^4 / 6=1.563in^4Torsion stress, Ft =T*(B/2) / J=560lb/in^2InputTorsion, T =1000in-lbfRectangular shaft breadth, B =1.000inHeight, H =2.000inCalculationSection polar moment of inertia, J =B*H*(B^2 + H^2)/ 12=0.833in^4Torsion stress, Ft =T*(B/2) / J=600lb/in^2Cantilever shaft bending momentInputShaft transverse load, W =740lbfPosition in shaft, x =5inBending shock & fatigue factor, Km =3Shaft diameter, D =1.000inCalculationMoment at x, Mx =W*xin-lbsDesign moment at x, Md =Km*Mx=11100in-lbsSection moment of inertia, I =*D^4 / 64=0.049in^4Bending stress for shaft, Fb =M*D / (2*I)=113049lbs/in^2< GOAL SEEKCantilever shaft bending deflectionInputShaft transverse load at free end, W =740lbfShaft diameter, D =1.000inShaft length, L =10inDeflection location, x =5inBending moment shock load factor, Km =3Modulus of elasticity, E =29000000psiCalculationSection moment of inertia, I =*D^4 / 64=0.049in^4Moment at, x =5inMoment at x, M =Km*W*x=11100in-lbfBending stress at x: Sb =M*(D/2) / I113063lbf/in^2< GOAL SEEKCantilever bend'g deflection at x, Yx =(-W*x^2/(6*E*I))*((3*L) - x)=-0.0541inBending deflection at x = 0, Y =-W*L^3 / (3*E*I)Y =-0.1733inSection Moment of InertiaInputRound solid shaft diameter, D =1.000inCalculationsSection moment of inertia, Izz =*D^4 / 64Answer: Izz =0.049in^4Section moment of InertiaInputRound tube shaft diameter, Do =1.750inDi =1.5inCalculationSection polar moment of inertia, Izz =*(Do^4 - Di^4) / 64Answer: Izz =0.212in^4Section moment of InertiaInputSquare shaft breadth = height, B =1.750CalculationSection moment of inertia, Izz =B^4 / 12Answer: Izz =0.782in^4BENDING STRESSEnter values for applied moment at a beam section, c, Izz and Kb. Bending stress will be calculated.InputApplied moment at x, M =1000in-lbfc =1.000inSection moment of inertia, Izz =2.5in^4Bending shock & fatigue factor, Kb =3-CalculationMax bending stress, Fb =Kb*M*c / IAnswer: Fb =1200lb/in^2This is the end of this worksheet

DESIGN OF POWER TRANSMISSION SHAFTINGThe 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 = *D^4 / 32 for solid shafts

J = *(D^4 - d^4) / 32 for hollow shafts

Shaft bending stress: Sb = M*R / I

Moment of area: I = *D^4 / 64 for solid shafts

I = *(D^4 - d^4) / 64 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/*Ss(1-K^4))*[ ( (KbMb + (*F*D*(1+K^2)/8 ]^2 + (Kt*T)^2 ]^0.5

Shaft diameter cubed with no axial load, D^3 = (16/*Ss)*[ (KbMb)^2 + (Kt*T)^2 ]^0.5

K = D/d D = Shaft outside diameter, d = inside diameter

Kb = combined shock & fatigue bending factor

Kt = combined shock & fatigue torsion factor = 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 CalculationsInput 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.TYPICAL BULK MATERIAL BELT CONVEYOR SHAFTING SPECIFICATIONSee 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.

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.

4 CouplingMACHINE DESIGN EXCEL SPREAD SHEETSCopy write, Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006COUPLINGSLegendh / RY/H0.400.600.801.00A0.2ABCDB0.30.22.011.911.771.62C0.40.41.591.501.401.30D0.50.61.411.321.251.180.81.371.281.191.101.01.351.251.171.07Design StressCoupling Design Shear Stress =Design allowable average shear stress.InputMaterial ultimate tensile stress, Ft =85000lbf/in^2Shaft material yield stress, Fy =45000lbf/in^2CalculationUltimate tensile stress design factor, ku =0.18-Design ultimate shear stress, Ssu =ku* Ft-=15300lbf/in^2Yield stress factor, ky =0.3-Design yield shear design stress factor, Ssy =ky* Ft-=13500lbf/in^2Use the smaller design shear stress of Fsu and Fsy above.1. Shaft Torsion Shear StrengthInputShaft diameter, D =2.000inKey slot total width = H =0.375inKey slot depth, h =0.25inCalculationKey slot half width, y =0.188Key slot half width / Slot depth, y / h =0.75Slot depth / Shaft radius, h / R =0.25InputMotor Power, HP =60hpShaft speed, N =300rpmAllowable shaft stress from above, Ssu or Ssy =13500lbf/in^2Torque shock load factor, Kt =3.00-Key slot stress factor from graph above, Kk =1.38CalculationMotor shaft torque, Tm =12*33000*HP / (2**N)=12603in-lbfSection polar moment of inertia, J =*D^4 / 32=1.5710in^4Allowable shaft torque, Ts =Ss*J / (Kt*Kk*Ds/2)=5123in-lbf2. Square Key Torsion Shear StrengthInputKey Width = Height, H =0.375inKey Length, L =3.00inShaft diameter, Ds =2.000inAllowable shaft stress from above, Ssu or Ssy =13500lbf/in^2Allowable key bearing stress, Sb =80000lbf/in^2CalculationKey shear area, A =H*L=1.125in^2Key stress factor, K =0.75Key shear strength, Pk =K*Fs*A=11390.625lbf/in^2Key torsion shear strength, Tk =Pk*Ds/2=11391in-lbfKey bearing strength, Tk =Sb*L*(D/2 - H/4)*(H/2)=40781in-lbf3. Coupling Friction Torsion StrengthInputOuter contact diameter, Do =10.00inInner contact diameter, Di =9.00inPre-load in each bolt, P =500lbfNumber of bolts, Nb =6-Coefficient of friction, f =0.2-Number of pairs of friction surfaces, n =1-CalculationCoupling friction radius, Rf =(2/3)*(Ro^3-Ri^3)/(Ro^2-Ri^2)Answer: Rf =4.75inAxial force, Fa =P*NbFa =3000lbfCoupling friction torque capacity, Tf =Fa*f*Rf*nAnswer: Tf =2853in-lbf4. Coupling Bolts Torsion StrengthAssume half of bolts are effective due differences in bolt holes and bolt diameters.InputTorque shock load factor, Kt =3-Bolt allowable shear stress, Fs =6000lbf/in^2Number of bolts, Nb =4-Bolt circle diameter, Dc =6.5inBolt diameter, D =0.500inCalculationOne bolt section area, A =*D^2/4A =0.196inShear stress concentration factor, Ks =1.33-Shear strength per bolt, Pb =Fs*A / (Kt*Ks)Answer: Pb =295lbfTotal coupling bolts torque capacity, Tb =Pb*(Dc/2)*(Nb / 2)Answer: Tb =1919in-lbf.InputHub outside diameter, Do =14.000inShaft outside diameter, Dc =4.000inShaft inside diameter, Di =0.000inHub length, L =8inMax tangential stress, Ft =5000lbf/in^2Hub modulus, Eh =1.50E+07lbf/in^2Shaft modulus, Es =3.00E+07lbf/in^2Coefficient of friction, f =0.12-Hub Poisson's ratio, h =0.3-Shaft Poisson's ratio, s =0.3-See input above:CalculationPressure at contact surface, Pc =Ft*((Do^2-Dc^2) / (Do^2+Dc^2))Pc =4245C1 =(Dc^2+Di^2)/(Es*(Dc^2-Di^2))C1 =0.0000000333C2 =(Do^2+Dc^2)/(Eh*(Do^2-Dc^2))C2 =0.0000000785C3 =s / EsC3 =1.00E-08C4 =h / EhC4 =2.00E-08Maximum diameter interference, =Pc*Dc*(C1 + C2 - C3 + C4) =0.00207inMaximum axial load, Fa =f**Dc*L*PcFa =51221lbfMaximum torque, T =f*Pc**Dc^2*L / 2T =102441in-lbfThis is the end of this spread sheet.

Hub - Shaft Interference FitsThese ridged or, "shrink fits" are used for connecting hubs to shafts, sometimes in addition to keys. Often the computed stress is allowed to approach the yield stress because the stress decreases away from the bore.

4 Coupling00000000000000000000

ABCDKey half slot width / Slot depth (y / h)Key Slot Stress Factor (Kk)KEY SLOT STRESS FACTOR

5 Pwr ScrewMACHINE DESIGN EXCEL SPREAD SHEETSCopy write, Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006Pitch (P) is the distance from a point on one thread to the corresponding point on the next thread.Lead (n*P) is the distance a nut advances each complete revolution.Multiple pitch number (n) refers to single (n=1), double (n=2), triple (n=3) pitch screw.Motor Shaft TorqueInputMotor Power, HP =30hpShaft speed, N =1750rpmCalculationMotor shaft torque, Tm =12*33000*HP / (2**N)Answer: Tm =1080in-lbfPower Screw TorqueInputScrew outside diameter, D =3.000inScrew thread turns per inch, TPI =3threads/inThread angle, At =5.86degreesThread multiple pitch lead number, n =2Thread friction coefficient, Ft =0.15Bearing friction coefficient, Fb =0Bearing mean radius, Rb =2inLoad to be raised by power screw, W =500lbfCalculationAcme thread depth, H =0.5*(1/ TPI )+0.01Answer: H =0.177inThread mean radius, Rm =(D - H) / 2Rm =1.412inThread helix angle, Tan (Ah) =n*(1/ TPI ) / (2**Rm)Answer: Tan (Ah) =0.0752Answer: Ah =4.31degreesThread normal force angle, Tan (An) =Tan (At)*Cos (Ah)Answer: Tan (An) =0.0749Answer: An =4.29degreesX =(Tan (Ah) + Ft/ Cos (An))0.2257Y =(1- Ft*Tan (Ah)/ Cos (An))0.9887Power screw torque, T =W*(Rm*( X / Y) + Fb*Rb)Answer: T =161in-lbfForce W will cause the screw to rotate (overhaul) if, (-Tan (Ah) + Ft/ Cos (An)) is negative.(-Tan (Ah) + Ft/ Cos (An)) =0.0751SCREW THREAD AVERAGE PRESSUREInputLoad to be raised by power screw, W =2000lbfNut length, L =4inScrew thread turns per inch, TPI =3threads/inThread height, H =0.18inThread mean radius, Rm =0.9CalculationScrew thread average pressure, P =W / (2**L*Rm*H*TPI)Answer: P =164lbf/in^2This is the end of this spread sheet.

RIGID COUPLING DESIGNCouplings 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.Apply to graph above. Protection > Unprotect Sheet > OKWhen Excel's Goal Seek is not needed, restore protection with:Drop down menu: Tools > Protection > Protect Sheet > OKAB5Input6ADJ =4.007OPP =3.008Calculations9HYP =(ADJ^2 + OPP^2)^(1/2)10=5.00AB5Input6ADJ =4.007OPP =3.008Calculations9HYP =(ADJ^2 + OPP^2)^(1/2)10=5.00Excel's Goal Seek ExampleDrive Shaft DesignInputMotor Power, HP =5.0hpShaft speed, N =1750rpmTorque shock & fatigue factor, Kt =3Shaft diameter, D =0.500inShaft length, L =10inMaterial shear modulus, G =11500000psiCalculationApplied motor shaft torque, Ta =12*33000*HP / (2**N)=180.05in-lbfSection polar moment of inertia, J =*D^4 / 32J =0.006in^4Answer: Design Torque, Td =Kt*Ta=540in-lbfShear stress for shafts, St =Td*D / (2*J)=22005lbf/in^2Shaft torsion deflection angle, a =Td*L / (J*G)a =0.0765radiansa =4.39degreesExcel's Goal Seek ProblemUse Excel's, "Goal Seek" in the duplicate example below to calculate a new shaft diameter D that willreduce 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, 20005Step 2.Select drop-down menu, Tools > Goal SeekStep 3.Pick the "To value" box and type, 12000Step 4.Pick the, "By changing cell" box and pick the shaftdiameter D cell B78 initially containing, 0.500Step 5.Click, OKStep 6.Use the same spread sheet below:The shaft torsion stress St will is set at 12000 lbf/in^2the shaft diameter D has changed from 0.500 to 0.612inches and the shaft twist will change from 4.39 to1.95 degrees.Drive Shaft DesignInputMotor Power, HP =5hpShaft speed, N =1750rpmTorque shock & fatigue factor, Kt =3Shaft diameter, D =0.612inShaft length, L =10inMaterial shear modulus, G =11500000psiCalculationApplied motor shaft torque, Ta =12*33000*HP / (2**N)=180.05in*lbfSection polar moment of inertia, J =*D^4 / 32J =0.014in^4Answer: Design Torque, Td =Kt*Ta=540in-lbfShear stress for shafts, St =Td*D / (2*J)=12000lbs/in^2Shaft torsion deflection angle, a =Td*L / (J*G)a =0.0341radiansa =1.95degreesMax kinetic energy, K.E. =(1/2)*M1^2* ^2 + (1/2)*M2^2* ^2Max potential energy, P.E. =(1/2)*K1*X1^2 + (1/2)*K2*(X2 - X1)^2Neglecting friction, Max K.E. =Max P.E.-^2 =[K1+K2*((X2/X1) - 1)^2]/ [(M1+M2*(X2/X1)^2]1. This equation will give the first and lowest natural frequency ().2. The solution for is by trial and error for various values of X2/X1.InputMass, M1 =0.1Mass, M2 =0.1K1 =20k2 =20X2 / X1 =1.6180Calculation-^2 =[K1+K2*((X2/X1) - 1)^2]/ [(M1+M2*(X2/X1)^2]-^2 =76.3932 =8.740radn/sec3. Use Excel's Solver for a trial and error solution to the above forcing function example.4. Start above solution by typing, X2 / X1 = 05. Use drop down menu, Tools > Solver > Set Target Cell: > B144 > Equal to Min6. By Changing Cell > B140 > Solve > Keep Solver SolutionCD5Problem6Guess X =1.478Y =2*X^5 - 3*X^2 - 59=-0.1235CD5Solution6Solved X =1.404178Y =2*X^5 - 3*X^2 - 59=0.0004Simultaneous Equations Using Excel's, "Solver"Reference: www.dslimited.biz/excel_totorialsEquations to be solved:u + v + w + x + y = 5.5u + 2v + w - 0.5x + 2y = 22.52v + 2w - x - y = 302u - w + 0.75x + 0.5y = -11u + 0.25v + w - x = 17.51. Insert the equations below into column B cells:EquationsConstantsSolution=E146+E147+E148+E149+E1500.05.5u ==E146+2*E147+E148-0.5*E149+2*E1500.022.5v ==2*E147+2*E148-E149-E1500.030w ==2E146-2E148-E149-E1500.0-11x ==E146+0.25E147+E148-E1490.017.5y =2. Select cells, E146 to 1503. Click on drop down menu: Tools > Solver >4. Delete contents of; Set Target Cell5. Pick: By Changing Cells: > Select cells E146 to E150EquationsConstantsSolutionRow 1465.55.5u =1.00Row 14722.522.5v =4.00Row 14830.030w =7.50Row 149-11.0-11x =-8.00Row 15017.517.5y =1.00You may use the table below to solve the 5 simultaneous equations.EquationsConstantsSolutionRow 1460.05.5u =0.00Row 1470.022.5v =0.00Row 1480.030w =0.00Row 1490.0-11x =0.00Row 1500.017.5y =0.00This is the end of this spread sheet.

Insert the Microsoft Office CD for Add-InsIf Excel's, "Goal Seek" or "Solver" are not installed you will need to select drop-down menu: Tools > Add-Ins > Goal Seek Tools > Add-Ins > Solver To open select Tools.The Vibration Forcing FunctionOne 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.Reference: Machine Design by A.S. Hall, A.R. Holowenko, H.G. Laughlin, Published byMcGraw-Hill.What if CalculationsExcel will make a, what if calculation using, "Goal Seek" when the calculated formula value needs to be changed.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.

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.Solver Example1. 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.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.To Create the Above TableType, 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.Goal Seek ExampleThe 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 7.00.5. Cell B10 below contains the calculated value 5.00.What if CalculationsExcel 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.