Machine Design Excel

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 gr avity or is conveyed into the top inlet. A rotating disc with replicable cutters cut ters in its circumf erence 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 f or absorbing dynamic shredding forces and shock loads.

Above is the image in its original context on the page: -

.

.

-

-

…

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...

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 f atigue 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 2 x 3 = 2* 3 =6

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

A+ B= A+ B 2 +3 = 2 +3 =5

n

X = X^n 2 = 2^3 = 8

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 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 tim es the average value. This is true for tension +P and compression -P.

Machine Component Maximum Stress Calculation Refer to the diagram above:

Input

External force, ± P = Section height, H = Section width, B = Original length, L = Stress concentration factor, K = Combined shock and fatigue factor, Ks =

2000 3.5 0.5 5 3.0 3.0

Section area, A = = Maximum direct stress, Smax = = Safety factor, SF = =

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

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

Input

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 -) =

= =

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

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.

Input

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 =

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 Input

Horizontal force, Vertical force, Torsion, Cantilever length, Diameter,

H= V= T= L= D=

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 Input

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 = =

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 Input

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 = =

6.000 4.000 175

in in 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 Input

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

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

Hn

9 1.5 6

2 7 3

Calculation A 18 10.5 18 46.5

ΣA =

1 2 3

Yn 11.000 6.500 1.500

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

Input

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

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 M MAX from above: Bending shock & fatigue factor, Kb = Bending stress will be calculated.

3

Data

Input

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 = =

13200 12.00 4.66 1.50

in-lbf in in^4 -

Calculation Kb*M*C / I 50987

lb/in^2

Input 1 2 3

Bn

Hn

2 7 3

9 1.5 6

Calculation A Yn 18.00 1.00 10.50 3.50 18.00 1.50 46.5

ΣA =

Yn 1.000 3.500 1.500

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 3

2 7 3

Calculation A Icg 18.00 6 10.50 43 18.00 14 46.5 62

9 1.5 6 ΣA =

Center of gravity, Ycg = = Section modulus, Ixx = = Center of area, C1 = C2 = =

B1 / 2 1.000 ΣIcg 62 B1 / 2 1.000

in in^4

Enter value of applied moment M MAX from above: Input

P= L = a= b=

= Cantilever, MMAX at B = = Fixed ends, MMAX, at C ( a < b ) = = Pinned ends, M MAX, at C =

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

Input

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 =

58000 36000 Calculate 18% * Su 10440 30% * Sy 10800 75% * Sau 7830

lbf/in^2 lbf/in^2

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

Input

7830 10 300 0 10 1.5

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 Input

Su = Sy = ASME Code Shaft Allowable Stress

Allowable stress based on Su, Sau =

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 Input

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 =

9450 20 600 3 60 10 30 10 8 18 200 20 1.5 1

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

Calculate

Shaft torque, T = = T2 / T1 = B = T1 - T2 = T2 = = T1 = =

HP * 63000 / N 2100 3 T / (D2 / 2) -( T / (D2 / 2) ) / (1 - B) 117 B * T2 350

in-lbf

lbf

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 =

V1 * L1 1954 V4 * L3 2041

lbf-in lbf-in

Horizontal Forces

H2 =Ft =

T / (D1 / 2) 525

lbf

H4 =

(T1 + T2) * Cos( A ) 233 lbf H3 = ( (H4*(L2 + L3)) + (H2*L1) ) / L2 486 H1 = H2 - H3 + H4 272 Horizontal Moments

Mh2 = Mh3 =

H1 * L1 2722 H4 * L3 2334

lbf-in lbf-in

Resultant Moments

Mr2 = Mr3 =

(Mv2^2 + Mh2^2)^0.5 3351 (Mv3^2 + Mh3^2)^0.5 3100

lbf-in 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=

1.431

in

Shaft Material Ultimate & Yield Stresses

Input

Su = Sy = ASME Code Shaft Allowable Stress

Allowable stress based on Su, Sau = Allowable stress based on Sy, Say =

70000 46000

lbf/in^2 lbf/in^2

Calculate

18% * Su 12600 30% * Sy

lbf/in^2

Allowable shear stress based on Su, Ss =

Shaft Power & Geometry

13800 75% * Sau 9450

lbf/in^2 lbf/in^2

Input

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 =

9450 20 600 3 60 10 30 10 8 18 200 20 1.5 1 1.000 3.000 2.000

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

Calculate

Shaft torque, T = = T2 / T1 = B = T1 - T2 = T2 = = T1 = =

HP * 63000 / N 2100 3 T / (D2 / 2) -( T / (D2 / 2) ) / (1 - B) 117 B * T2 350

in-lbf

lbf

lbf

Vertical Forces

H2 =Ft = V2 = Fs = = V4 = = V3 =

V1 =

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

lbf

Vertical Moments

Mv2 = Mv3 =

V1 * L1 6742 V4 * L3 2041 Input

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

Input

7.5 1750 3 1.000 5 11500000

hp rpm in in psi

Calculation Shaft Design Torque, Td = Kt*12*33000*HP / (2*π*N) = 810 in-lbf

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

in-lbf in in psi psi

< GOAL SEEK

in^4 lbf/in^2 radians degrees

< GOAL SEEK

POLAR MOMENT OF AREA AND SHEAR STRESS Input

Torsion, T = Round solid shaft diameter, D = Section polar moment of inertia, J = = Torsion stress, Ft = =

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

in-lbf in

in^4 lb/in^2

Input

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

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

in-lbf in in

in^4 lb/in^2

Input

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

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

in-lbf in

in^4 lb/in^2

Input

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

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

in-lbf in in

in^4 lb/in^2

Cantilever shaft bending moment Input

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

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

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

< GOAL SEEK

Input

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 =

Section moment of inertia, I =

= Moment at, x = Moment at x, M = = Bending stress at x: Sb = Cantilever bend'g deflection at x, Yx = = Bending deflection at x = 0, Y =

740 1.000 10 5 3 29000000

lbf in in in psi

Calculation π*D^4 / 64 0.049 in^4 5 in Km*W*x 11100 in-lbf M*(D/2) / I 113063 lbf/in^2 (-W*x^2/(6*E*I))*((3*L) - x) -0.0541 in -W*L^3 / (3*E*I)

< GOAL SEEK

Y =

Section Moment of Inertia

-0.1733

Input

Round solid shaft diameter, D =

1.000

Section moment of inertia, Izz =

Calculations π*D^4 / 64 0.049

Answer: Izz = Section moment of Inertia

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

Section moment of Inertia

in

in

in^4

Input

1.750 1.5 Calculation π*(Do^4 - Di^4) / 64 0.212

in in

in^4

Input

Square shaft breadth = height, B =

1.750

Section moment of inertia, Izz = Answer: Izz =

Calculation B^4 / 12 0.782

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 = c= Section moment of inertia, Izz = Bending shock & fatigue factor, Kb = Max bending stress, Fb = Answer: Fb =

1000 1.000 2.5 3 Calculation Kb*M*c / I 1200

in-lbf in in^4 -

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

t e

ec an ca

ower ransm ss on ssoc at on tan ar

o.

-

an

. . .

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 2.00 ) 1.90 k K ( 1.80 r o t 1.70 c a F 1.60 s s e 1.50 r t S t 1.40 o l S 1.30 y e 1.20 K

A B C D

1.10 1.00 0.2

0.4

0.6

0.8

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

1.0

Legend

h/R

A B C D

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 = Shaft material yield stress, Fy =

85000 45000

Calculation Ultimate tensile stress design factor, ku = 0.18 Design ultimate shear stress, Ssu = ku* Ft = 15300 Yield stress factor, ky = 0.3 Design yield shear design stress factor, Ssy = ky* Ft = 13500 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 =

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

Input

2.000 0.375 0.25

in in in

Calculation

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 = =

0.188 0.75 0.25 Input

Apply to graph above.

60 300 13500 3.00 1.38

hp rpm lbf/in^2 -

Calculation 12*33000*HP / (2*π*N) 12603 π*D^4 / 32 1.5710 Ss*J / (Kt*Kk*Ds/2) 5123

in-lbf

in^4 in-lbf

2. Square Key Torsion Shear Strength Key Width = Height, H = Key Length, L = Shaft diameter, Ds = Allowable shaft stress from above, Ssu or Ssy = Allowable key bearing stress, Sb = Key shear area, A = = Key stress factor, K = Key shear strength, Pk = = Key torsion shear strength, T k = = Key bearing strength, Tk = =

3. Coupling Friction Torsion Strength Outer contact diameter, Do = Inner contact diameter, Di = Pre-load in each bolt, P = Number of bolts, Nb = Coefficient of friction, f = Number of pairs of friction surfaces, n = Coupling friction radius, Rf = Answer: Rf = Axial force, Fa = Fa = Coupling friction torque capacity, Tf = Answer: Tf =

Input

0.375 3.00 2.000 13500 80000 Calculation H*L 1.125 0.75 K*Fs*A 11390.625 Pk*Ds/2 11391 Sb*L*(D/2 - H/4)*(H/2) 40781

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

in^2

lbf/in^2 in-lbf

in-lbf

Input

10.00 9.00 500 6 0.2 1

in in lbf -

Calculation (2/3)*(Ro^3-Ri^3)/(Ro^2-Ri^2) 4.75 in P*Nb 3000 lbf Fa*f*Rf*n 2853 in-lbf

4. Coupling Bolts Torsion Strength Assume half of bolts are effective due differences in bolt holes and bolt diameters. Input

Torque shock load factor, Kt = Bolt allowable shear stress, Fs = Number of bolts, Nb = Bolt circle diameter, Dc = Bolt diameter, D =

3 6000 4 6.5 0.500

One bolt section area, A = A= Shear stress concentration factor, Ks = Shear strength per bolt, Pb = Answer: Pb = Total coupling bolts torque capacity, Tb = Answer: Tb =

lbf/in^2 in in

Calculation π*D^2/4 0.196 1.33 Fs*A / (Kt*Ks) 295

lbf

Pb*(Dc/2)*(Nb / 2) 1919

in-lbf

in

-

Hub - Shaft Interference Fits These 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.

Shaft in Hub The hub is the outer ring, Do to Dc. The shaft is the inner ring, Dc to Di

.

Input

Hub outside diameter, Do = Shaft outside diameter, Dc = Shaft inside diameter, Di = Hub length, L = Max tangential stress, Ft = Hub modulus, Eh = Shaft modulus, Es = Coefficient of friction, f = Hub Poisson's ratio, μh = Shaft Poisson's ratio, μs =

14.000 4.000 0.000 8 5000 1.50E+07 3.00E+07 0.12 0.3 0.3

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

See input above: Pressure at contact surface, Pc = Pc = C1 = C1 = C2 = C2 = C3 = C3 = C4 = C4 = Maximum diameter interference, δ = δ=

Calculation Ft*((Do^2-Dc^2) / (Do^2+Dc^2)) 4245 (Dc^2+Di^2)/(Es*(Dc^2-Di^2)) 3.33333E-08 (Do^2+Dc^2)/(Eh*(Do^2-Dc^2)) 7.85185E-08 μs / Es 1.00E-08 μh / Eh 2.00E-08 Pc*Dc*(C1 + C2 - C3 + C4) 0.00207 in

Maximum axial load, Fa = f*π*Dc*L*Pc Fa = 51221 Maximum torque, T = f*Pc*π*Dc^2*L / 2 T= 102441

This is the end of this spread sheet.

lbf

in-lbf

Y/H

0.40 A

0.60 B

0.80 C

1.00 D

0.2 0.4 0.6 0.8 1.0

2.01 1.59 1.41 1.37 1.35

1.91 1.50 1.32 1.28 1.25

1.77 1.40 1.25 1.19 1.17

1.62 1.30 1.18 1.10 1.07

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

Motor driven: screw jacks, linear actuators, and clamps are examples of power screws. The essential components are a nut engaging the helical screw threads of a shaft. A nut will advance one screw thread pitch per one 360 degree rotation on a single pitch screw. A nut will advance two screw thread pitches per one 360 degree rotation on a double pitch screw, etc. The actuator nut below advances or retreats as the motor shaft turns clockwise or antclockwise. The nut is prevented from rotating by the upper and lower guide slots. The control system of a stepper motor rotates the shaft through a series of small angles very accurately repeatedly. The linear travel of the lug & nut is precise and lockable.

Pitch (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 Torque

Input

Motor Power, HP = Shaft speed, N =

30 1750

hp rpm

Calculation Motor shaft torque, Tm = 12*33000*HP / (2*π*N) Answer: Tm = 1080 in-lbf

Power Screw Torque Screw outside diameter, D = Screw thread turns per inch, TPI = Thread angle, At = Thread multiple pitch lead number, n = Thread friction coefficient, Ft = Bearing friction coefficient, Fb = Bearing mean radius, Rb = Load to be raised by power screw, W = Acme thread depth, H = Answer: H = Thread mean radius, Rm = Rm = Thread helix angle, Tan (Ah) = Answer: Tan (Ah) = Answer: Ah = Thread normal force angle, Tan (An) = Answer: Tan (An) = Answer: An =

Input

3.000 3 5.86 2 0.15 0 2 500 Calculation 0.5*(1/ TPI )+0.01 0.177 (D - H) / 2 1.412

in threads/in degrees

in lbf

in in

n*(1/ TPI ) / (2*π*Rm) 0.0752 4.31 degrees Tan (At)*Cos (Ah) 0.0749 4.29

degrees

X = (Tan (Ah) + Ft/ Cos (An)) 0.2257 Y = 1- Ft*Tan (Ah)/ Cos (An)) 0.9887 Power screw torque, T = W*(Rm*( X / Y) + Fb*Rb) Answer: T = 161 in-lbf Force W will cause the screw to rotate (overhaul) if, (-Tan (Ah) + Ft/ Cos (An)) is negative. (-Tan (Ah) + Ft/ Cos (An)) = 0.0751

SCREW THREAD AVERAGE PRESSURE

Load to be raised by power screw, W = Nut length, L = Screw thread turns per inch, TPI = Thread height, H = Thread mean radius, Rm =

Input

2000 4 3 0.18 0.9

lbf in threads/in in

Calculation Screw thread average pressure, P = W / (2*π*L*Rm*H*TPI) Answer: P = 164 lbf/in^2


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

DISC BRAKE A sectional view of a generic disc brake with calipers is illustrated right. Equal and opposite clamping forces, F lbf acting at mean radius Rm inches provide rotation stopping torque T in-lbf.

Calculate Brake Torque Capacity Clamping force, F = Coefficient of friction, μ = Caliper mean radius, Rd = Number of calipers, N =

Braking torque, T =

SHOE BRAKE

stopping capacity is proportional to the normal force of brake shoe against the drum and coefficient of friction.

Input

50 0.2 7.00 1 Calculation 2*μ*F*N*Rm 140

lbf in -

in-lbf

Calculate Brake Torque Capacity

Coefficient of friction, f = Brake shoe face width, w = Drum internal radius, Rd = Shoe mean radius, Rs = Shoe heel angle, A1 = Shoe angle, A2 = Shoe mean angle, Am = Right shoe maximum shoe pressure, Pmr = Left shoe maximum shoe pressure, Pml = C=

X= X= Right shoe friction moment, Mr = Mr =

Input

0.2 2 6 5 0 130 90 150 150 9

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

Calculation (Rd - Rd*Cos(A2)) - (Rs/2)*Sin^2(A2)) 8.3892 ((f*Pm*w*Rd)/(Sin(Am))*(X) 3020 in-lbf

Y = (0.5*A2) - (0.25*Sin(2*A2)) Y = 1.3806 Right normal forces moment, Mn = ((Pm*w*Rd*Rs)/(Sin(Am))*(Y) Mn = 12426 in-lbf Brake cylinder force, P = Answer: P =

(M n - M r) / C 1045

lbf

Z = ((Cos(A1)-Cos(A2)) / Sin(Am) Z= 1.6427 Right shoe brake torque capacity, Tr = f*Pm*w*Rd^2*(Z) Tr = 3548 in-lbf

This is the end of this work sheet.


V-BELT DRIVES V-belts are used to transmit power from motors to machinery. Sheaves have a V-groove. Pulleys have a flat circumference. A V-belt may be used in combination with a drive sheave on a motor shaft and a pulley on the driven shaft.

Angle B

Input

Small sheave pitch circle radius, R1 = Large sheave pitch circle radius, R2 = Center distance, C = Sin (B) Sin (B) B B

= = = =

4 6 14 Calculation (R2-R1) / C 0.1429 0.1433 8.21

in in in

radn. degrees

V-Belt Drive

Input

Drive power, HP = Motor speed, N = Drive sheave pitch diameter, D1 = Driven sheave pitch diameter, D2 = Center distance, C = Sheave groove angle, A = Sheave to V-belt coefficient of friction, f1 = Pulley to V-belt coefficient of friction, f2 = B1 = B2 = D = V-belt weight per cubic inch, w = Tight side V-belt allowable tension, T1 = V-belt C.G. distance, x = = Driven sheave pitch diameter, D2 = =

30 1800 10 36 40 40 0.2 0.2 0.75 1.5 1 0.04 200

hp rpm in in in deg in in in lbm/in^3 lbf

Calculation D*(B1+ 2*B2)/ 3(B1+B2) 0.556 D2 + 2*x 37.11

in in

Angle of Wrap An Small sheave pitch radius, R1 = Large pulley pitch radius, R2 = Sin (B) = Sin (B) = B = B = Small sheave angle of wrap, A1 = A1 = Large pulley angle of wrap, A2 = A2 = e =

5.00 18.56 (R2-R1) / C 0.3389 0.3457 19.81 180 - 2*B 140.38 180 + 2*B 219.62 2.7183

in in

radn. degrees degrees degrees

Sheave capacity Cs = e^(f1*A1/ Sin(A/2)) = 4.77 Pulley capacity, Cp = e^(f2*A2/ Sin(90/2)) = 2.15

The smaller of Cs and Cp governs design. Belt section area, Ab = = V-belt weight per ft, W = = V-belt velocity, V = V = g =

(B1 + B2)/ (2*D) 1.125 Ab*w*12 0.54 π*(D1/12)*(N/60) 78.55 32.2

in^2 lbm/ft ft/sec ft^2/sec

Slack side belt tension, T2 T 2 = (T1-W*V^2/g)/(Csp)+ (W*V^2/g) = 148 lbf Horsepower wer per belt, HPb = (T2-T T2-T1 1)*V / 550 = 7.4 hp Number of belts, Nb Nb = HP / HPb = 4.1 belts Input

Use

This is the end of this work sheet.

4

belts


SPUR GEARS Circular pitch (CP) is the pitch circle arc length

between a point on one tooth and the corresponding point on the adjacent tooth. Diametral pitch (P) is the number of teeth per inch of pitch circle diameter.

Spur Gear Dimensions Input

Pressure angle, Pa = Diametral pitch, Pd = Number of gear teeth, N = Gear hub diameter = Gear hub width = Bore diameter =

14.5 or 20 N/D -

14.5 6 12 3.00 1.50 1.875

deg. in in in

Calculation

Pitch circle diameter, D = N / Pd Addendum, A = 1 / Pd Dedendum, B = 1.157 / Pd Whole depth= Addendum+Dedendum, d = 2.157 / Pd Clearance, C = .157 / Pd Outside diameter, OD = D + (2*A) or OD = (N + 2) / Pd Root circle diameter, RD = D - (2*B) RD = or (N - 2.314) / Pd Base circle, BC = D*Cos(Pa*.01745) Circular pitch, CP = π*D / N or CP = π / Pd Chordal thickness, TC = D*Sin(90*.01745/N) Chordal addendum, AC = A + N^2 / (4*D) Working depth, WD = 2*A

2.000 in 0.167 in 0.193 in 0.360 in 0.026 in 2.333 in 2.333 in 1.614 in 1.614 in 1.936 in 0.524 in 0.524 in 0.167 in 18.167 in 0.333 in Note: Excel requires degrees to be converted to radians. Degrees x .01745 = Radians π= 3.1416 Use the above spread sheet to calculate the dimensions of gears.

Gear Tooth Interference

Base circle radius, Rbc = CP/2 = Outside radius, Ros = OD/2 = Pressure angle, Pa =

Pinion base circle radius = Gear addendum radius = There will be no interference if, Rbc < Rbc < Rbc < Addendum radius, Ra =

Input

4.65 9.3 20

in in deg.

Calculation Rbc Ra Ra (Rbc^2 + Rc^2*(Sin(Pa))^0.5 5.63 6.00

GEAR TEETH STRENGTH

Gear Tooth Bending Stress Tooth base thickness, t = Moment arm length, h = Tooth load, W = Tooth face width (into paper), b = Base half thickness, c = c=

Input

1.50 0.70 1000 1.00 Calculation t/2 0.75

in in lbf in

in

Section modulus, I = b*t^3 / 12 I= 0.28125 in^3 Tooth bending stress, Sb = M*c / I Sb = 1867 lbf/in^2 The stress calculated above does not include stress concentration or dynamic loading.

Gear Tooth Dynamic Load Pitch line velocity, Vp = Tooth face width, b = Gear torque, T = Circular pitch radius, R = CP / 2 = Deformation factor (steel gears), C =

Input

100 3.13 1836 3.00 2950

ft/min in in-lbf in -

4980

Calculation Static load, F = 2*T / R F= 1224 lbf Dynamic load, Pd = ((0.05*V*(b*C + F)) / (0.05*V + (b*C + F)^.5)) + F Pd = 1711

Use the Lewis form factor, Y below:

Lewis Equation Form Factor Y Pressure Pressure Number of Teeth Angle 14 Angle 20 12 0.067 0.078 14 0.075 0.088 16 0.081 0.094 18 0.086 0.098 20 0.090 0.102 25 0.097 0.108 30 0.101 0.114 50 0.110 0.130 60 0.113 0.134 75 0.115 0.138 100 0.117 0.142 150 0.119 0.146 300 0.122 0.150 Rack 0.124 0.154

Strength of Gear Teeth Strength of Gear Teeth- Lewis Equation - if pitch circle diameter is known Input

Allowable gear tooth tensile stress, S = Tooth width, b = Circular pitch, Pc = Lewis form factor, Y = Allow able gear tooth load, F = F=

5000 3.5 1.0473 0.094

lbf/in^2 in in -

Calculation S*b*Pc*Y 1723

lbf

Strength of Gear Teeth- Lewis Equation - if pitch circle diameter is not known Input

Gear shaft torque, T = Diametral pitch, Pd = Constant, k = Lewis form factor, Y = Number of gear teeth, N =

15300 5.00 4 0.161 100

in-lbf in max -

Calculation Gear tooth tensile stress, S = 2*T*Pd^3 / (k*π^2*Y*N) S= 6016 lbf/in^2 Gear Pitch Line Velocity

Input

Pitch circle diameter, Dp = Rotational speed, n = Gear Pitch Line Velocity, V = V= Allowable gear tooth load, F = Gear Pitch Line Velocity, V =

5.33 800 π*Dp*n / 12 1116 1722 840

Gear horsepower transmitted, HP = HP =

Calculation F*V / 33000 44

Worm & Wheel Gearing

in rpm ft/min lbf ft/min Note: 1.0 HP = hp

33000

ft/min

Lead Angle, A

Input

Lead = Dw = Tan(A/57.2975) = A= Lead angle, A = Answer: A =

2.25 4 Calculation Lead / (π*Dw) 0.1790 Tan-1(a) 10.15

radians degrees

Worm Circular Pitch, Pc AGMA Standard Circular Pitches: 1/8, 5/16, 3/8, 1/2, 5/8, 3/4, 1, 1.25, 1.75, and 2. Input

Worm and wheel center distance, Cd = Wheel diameter, Dw = Dw = Worm circular pitch, Pc = Pc = Use standard, Pc =

16 Calculation Cd^0.875 / 2.2 5.143 Dw / 3 1.71 1.75

in

in in in

Strength of Worm & Wheel Gears - Lewis Equation Input

Pitch circle diameter, Dp = Rotational speed, n = Ultimate stress, Su =

Gear Pitch Line Velocity, Vg = Vg = Worm / Wheel allowable stress, So = So = Worm/gear design stress, Sd = Sd =

5.33 600 20000

in rpm lbf/in^2

Calculation π*Dp*n / 12 837 ft/min Su / 3 6667 lbf/in^2 o*1200 / (1200 + Vg) 3927 lbf/in^2 Input

Sd = Tooth width, b = Circular pitch, Pnc = Lewis form factor, Y = Allowable gear tooth load, F = F=

Worm Gear Dynamic Load Static load, F = Gear Pitch Line Velocity, Vg =

3927 1.5 1.0473 0.094

lbf/in^2 in in -

Calculation Sd*b*Pnc*Y 580

lbf

Input

1723 800

Calculation Worm Gear Dynamic Load, Fd = F*(1200+Vg) / (1200) Fd = 2872

Worm Gear Endurance Load Worm/gear design stress, Sd = Tooth width, b = Lewis form factor, Y = Worm wheel pitch circle diameter, Dp = Worm Gear Endurance Load, Fe = Fe =

Worm Gear Wear Load Gear pitch diameter, Dg = Tooth width, b = Material wear constant, B = Worm Gear Wear Load, Fw = Fw =

lbf ft/min

lbf

Input

4000 1.5 0.094 5.3 Calculation Sd*b*Y*π / Pnd 334

lbf/in^2 in in

lbf

Input

5.3 1.5 60

in in -

Calculation Dg*b* 477

lbf

Worm Gear Efficiency Material Wear Constant Worm Gear

Hardened steel 250 BHN steel Hardened steel Hardened steel Cast iron

Cast iron Phosphor bronze Phosphor bronze Antimony bronze Phosphor bronze

B

50 60 80 120 150

Input Data

Coefficient of friction, f = Lead angle, A =

0.1 12

degrees

Calculation Worm gear efficiency, e = (1 - f*Tan(A/57.2975) / (1 + f/Tan(A/57.2975) e= 0.986 AGMA Worm Gear Heat Dissipation Limit Input

Worm to wheel center distance, C = Transmission ratio, R = M aximum horse power limit, HPm = HPm =


3 25

in -

Calculation 9.5*C^1.7 / (R + 5) 2.05

hp

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

One gallon = 231 cu in W

Input Pressure, P = Weight, W =

1000 3000

psi lbs

Output Cylinder area, A = W / P = Cylinder diameter, D = (4*A / 3.142 )^0.5 =

3.00 1.95

sq in in

Input Weight, W = Cylinder diameter, D =

300 2

P

Output Cylinder area, A = 3.142 x D^2 / 4 = Pressure, P = W / A =

3.142 95

psi

Input Piston extends, x = Time to extend, t = Cylinder diameter, d = Hydaulic pipe internal diameter, pd =

10 2 4 0.5

in sec in in

Output Piston speed, S = 60*x / t = Cylinder area, A = 3.142 x D^2 / 4 = Piston extention volume, v = A * x = Volume in gallons, V = v / 231 = Time in minutes to extend, T = t / 60 = Flow rate, GPM = V / T = Pipe internal area, pa = 3.142 x pd^2 / 4 = Fluid speed in pipe, fs = v / (12*t*A) =

300 12.568 125.68 0.544 0.033 16.32

0.196 0.42

in/min sq-in cu-in gal min gpm sq-in ft/sec

Input Pump flow, GPM = Pump displacement, d = Pump speed, RPM = GPM x 231 / d =

20 4.20 Output 1100

gpm cu in / rev rpm

Input Hydraulic motor flow, GPM = Hydraulic motor displacement, d = Hydraulic motor speed, RPM = GPM x 231 / d =

20 2 Output 2310

gpm cu in / rev rpm

Input Pump flow, GPM = Pump pressure, P =

20 1000

gpm psi

Pump efficiency pecent, e = Pump power, HP = 100*GPM x P / (1741 x e%) = This is the end of this spread sheet.

70.00

%

Output 16.4

hp


Damped Vibrations With Forcing Function The inertia forces of rotating and oscillating machinery cause elastic supports to vibrate. Vibration amplitudes can be reduced by installing vibration damping m ounting pads or springs.

Simple Vibrating Systems External forcing function F(t) varies with time and is externally applied to the mass M. We will assume, F(t) = Fm*Sin(ωt) Fm is the maximum applied force. M is the mass of the vibration object that is equal to W/g. Omega, ω is the angular frequency as defined below. g is the gravitational constant, 32.2 ft/sec^2. X is the displacement from the equilibrium position. C is the damping constant force per second velocity and is proportional to velocity. K is the spring stiffness force per inch. See, "Math Tools" for Vibration Forcing Function Calculations.

Undamped Vibrations If the mass M shown above is displaced through distance x and r eleased it will vibrate freely. Undamped vibrations are called free vibrations. Both x and g are measured in inch units. Input

Weight, W = Spring stiffness, k = Gravitational Content, g = π= Static Deflection, x = = Mass, M = = Natural Frequency, fn = = Angular frequency, ω = =

2 10 Calculation 32.2 3.142 W/k 0.20 W / (g*12) 0.005 (1/2*π)*(k*/M)^.5 69.05 2*π*fn 434

lb lb/in ft/sec^2

in lbm-sec^2/in Hz Hz radn/sec

Displacement vs Time Graph

Forced Undamped Vibrations Motor weight, W = Motor speed, N = Gravitational content (ft), g = Gravitational content (in), g = Periodic disturbing force, Fd = Motor mount stiffness, k = Angular lar natural fre frequency, fn = = Disturbing force frequency, f = = Disturbing force angular frequency, frequency, fd = = Pseudo-static deflection, x = = Amplitu mplitude de magn magnifi ifica catio tion n fact factor or,, B = = Vibration amplitude = Pick cell B84, Tools, Goal Seek,

Input

50 1150 32.2 386.4 840 500 Calculation (k* (k*g / W)^. )^.5 62.2 N 1150 f*2*π / 60 120.4 Fd / k 1.68000 1 / ( (1 - (fa / fn)^ fn)^2) 2) 0.363 B*(Fd / k) 0.610

lb rpm ft/sec^2 in/sec^2 lb lb/in

rad/sec cycles/min rad/sec rad/sec in in

in in

Damped, (Viscous) Forced Vibrations Input

Motor W eight, W = Motor Speed, N = Gravitational Content (ft), g = Gravitational Constant (in), g = Isolation mount combined stiffness, k = Rotating imbalance mass, W i = Rotating imbalance eccentricity, e = Viscous damping ratio, C = Static deflection of the mounts, d = = Undamped natural natural frequency, frequency, fn = =

500 1750 32.2 386.4 20000 40 1.5 0.2 Calculation W/k 0.0250 (1 / 2*π)*(g / d)^.5 19.784

lbm rpm ft/sec^2 in/sec^2 lb/in lbm in in in Hz

"Math Tools" tab.

Disturbing force frequency, f = = Disturbing force angular frequency, frequency, fa = = Out of balance force F due to rotating mass F= =

N / 60 29.17 2*π*f 183.3

Hz Hz rad/sec rad/sec

Wi*fa^2*e / g 5216

lbf

Forcing frequency / Natural frequency = r = f / fn = 1.474 Amplitude Amplitude magnification magnification factor, factor, MF = 1/( (1 -r^2)+ (2*Cr)^2) (2*Cr)^2) = 0.761 Vibration amplitude, x = (MF)*(F / k) = 0.1986 Transmissibi Transmissibility, lity, TR = (MF)*(1 + (2*r*C)^2) (2*r*C)^2)^.5 ^.5 = 0.884 Transmissibility Force, Ft Ft r = (TR)*F = 4611

Critical Damping Critical damping occurs when the vibration amplitude is stable: C = Damp Dampin ing g Coe Coeff ffic icie ient nt Ccrit = Critical Critical Damping Damping Coeff. Coeff. Ccrit = 2*(K*M)^.5 K= System stiffness M= Vibrating Mass

in in

lbf

Transmissibility (TR) Transmissibility is the ratio of the force transmitted to a machine's supports due to a periodic imbalance in an; engine, pump, compressor, pulverizer, motor, etc. The amplitude of vibrations in machinery mountings can be reduced with resilient pads or springs called isolators. The isolated system must have a natural frequency less than 0.707 x the disturbing periodic imbalance force. The vibration amplitude will increase if the isolated system has a natural frequency fr equency higher than 0.707 x the disturbing frequency. f requency. Transmissibility ratio is equal to the, mass displacement amplitude / base displacement amplitude. TR =

X2 / X1

The transmissibility ratio TR, is the vibration amplitude reduction. Input

Disturbing force frequency, fd = Undamped natural frequency, fn =

16.0 12.0

Calculation Tra Transmis missibility lity,, TR = 1/(1 /(1-(fd -(fd//fn)^ fn)^2 2) TR = -1.286 If mounting damper pad natural frequency is known: Input

Transmissibility, TR = Disturbing force frequency, fd = Syste System m natu natura rall freq freque uenc ncy, y, fn = Answer: fn =

Hz Hz

-

0.5 14

Hz

Calculations fd / (1+ (1+(1 (1/TR /TR))^ ))^0.5 0.5 8.1

Hz

Springs are employed as vibration isolators.

Series Springs Combined Stiffness k1 = k2 = 1/k= k = Answer: k =

Input

10 15

lbf/in lbf/in

Calculation 1 / k1 + 1 / k2 (k1*k2) / (k1 + k2) 6

lbf/in

Parallel Springs Combined Stiffness Input

k1 = k2 = Answer: k = k =

12 24

lbf/ in lbf/ in

Calculation k1 + k2 36

lbf/ in

Critical Speed of Rotating Shaft The critical speed of a shaft is its natural frequency. The amplitude of any vibrating system will increase if an applied periodic force has the same or nearly same frequency. Resonance occurs at the critical speed.

Input

Flywheel mass, W = Shaft diameter, D = Steel Shaft, E = Bearing center distance, L2 = Flywheel overhang, L1 = Gravitational constant (ft), g = Gravitational constant (in), g = Shaft radius, r = = Shaft section moment of inertia, I = =

50 1.000 29000000 20 8 32.2 386.4

lbm in lb/sq in in in ft/sec^2 in/sec^2

Calculation D/2 0.500

in in

π*r^4 / 4 0.0491

in^4 in^4

The ball bearings act as pivoting supports Flywheel static deflection is; x = W*L1^2*(L1+L2) /3*E*I = 0.021 Natural frequency, f = =

(1 / 2*π)*(g / x)^.5 21.6

in in Hz Hz

Beam Stiffness (k), Deflection (x), and Natural Frequency ( f ) Cantilever, load W at Free End

Load at Free End, W = Length, L = Young's Modulus, E = Moment of Inertia, I = Deflection, x = Answer: x = Stiffness, k = Answer: k = Natural frequency, f = f= Cantilever, Uniform Load w

Input

600 30 29000000 4.000 Calculation W*L^3 / (3*E*I) 0.047 3*E*I/L^3 12889 (1/2π)*(g / x)^0.5 1321 Input

Uniform Load, w = Length, L = Young's Modulus, E = Moment of Inertia, I =

450 4 29000000 2.000

Deflection, x = Answer: x = Stiffness, k = Natural frequency, f = f=

Calculation w*L^4 / (8*E*I) 0.001 8*E*I/L^3 (1/2π)*(g / x)^0.5 92887

Beam, Pinned ends, W at Mid Span

Load at Mid Span, W = Length, L = Young's Modulus, E = Moment of Inertia, I = Deflection, x = Answer: x = Stiffness, k = Answer: k = Natural frequency, f = f= Beam, Pinned ends, Uniform Load w

lbf in lb/sq in in^4 in in lbf/in lbf/in Hz

lbf/in in lb/sq in in^4 in in lbf/in Hz

Input

400 60 29000000 3.000 Calculation W*L^3 / (48*E*I) 0.021 48*E*I/L^3 19333.33333 (1/2π)*(g / x)^0.5 2972 Input


500 40 29000000 2.000

Deflection, x = Answer: x = Stiffness, k = Answer: k = Natural frequency, f = f=

Calculation 5*w*L^4 / (384*E*I) 0.287 384*E*I/(5*L^3) 69600 (1/2π)*(g / x)^0.5 214


lbf/in in lb/sq in in^4 in in lbf/in lbf/in Hz

Beam, Fixed Ends, Load W at Mid Span

Load at Mid Span, W = Length, L = Young's Modulus, E = Moment of Inertia, I = Deflection, x = Answer: x = Stiffness, k = Answer: k = Natural frequency, f = f= Beam, Fixed ends, Uniform Load w

Input

700 80 29000000 2.000 Calculation W*L^3 / (192*E*I) 0.032 192*E*I/L^3 21750 (1/2π)*(g / x)^0.5 1911 Input


600 50 29000000 2.000

Deflection, x = Answer: x = Stiffness, k = Answer: k = Natural frequency, f = f=

Calculation w*L^4 / (384*E*I) 0.168 384*E*I/(L^3) 178176 (1/2π)*(g / x)^0.5 365


lbf/in in lb/sq in in^4 in in lbf/in lbf/in Hz

Plate Natural Frequency (f) Rectangular plate natural frequency, f = (K / 2*π)*((D*g)/(w*a^4)) Rectangular Plate, simply supported edges = K, ss Rectangular Plate, fixed edges = K, fixed Vibration Coefficients a/b Circular Stiffness Factors 1.0 Circular Plate, simply supported 0.8 edges, K = 4.99. 0.6 0.4 Circular Plate, fixed supported edges, 0.2 K = 10.2. 0.0

K, ss 19.7 16.2 13.4 11.5 10.3 9.87

Rectangular Plate Natural Frequency (f) Input

Modulus of elasticity, E = Plate thickness, t = Poisson's ratio, v = Plate short side, a = Plate long side, b = From the table above, K,ss or Kfixed = Load per unit area, w =

2.90E+07 0.5 0.3 36 45.0 16.2 50

Calculation

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

K, fixed 36.0 29.9 25.9 23.6 22.6 22.4

Answer: a / b = 0.80 D = E*t^3 / (12*(1 - ν^2)) Answer: D = 331960 π= 3.142 Gravitational acceleration, g = 386.4 in/sec^2 Rectangular Plates, f = (K / 2*π)*((D*g)/(w*a^4)) Answer: f = 3.938 Hz

Circular Plate Natural Frequency (f) Load per unit area, w = Modulus of elasticity, E = Plate thickness, t = Poisson's ratio, v = Plate radius, r = From the table above, K,ss = Kfixed = π= g= D= Answer: D =

Input

50 2.90E+07 0.5 0.3 36 4.99 10.2 Calculation 3.142 386.4 E*t^3 / (12*(1 - ν^2)) 331960

lb/in^2 lb/in^2

in

in/sec^2

Simply supported edges, f = (K / 2*π)*((D*g)/(w*r^4)) Answer: f = 1.213

Hz

Fixed edges, f = (K / 2*π)*((D*g)/(w*r^4)) Answer: f = 2.479

Hz

Balancing Rotating Shafts Masses in the Same Plane For static balance: Two masses, M1 and M2 must be in the same plane and 180 degrees out of phase and moments must balance: Σmi*Ri = 0 M1*R1+ M2*R2 = 0

Masses in Different Planes For static and dynamic balance there must be no unbalanced moments and couples. When the masses are in the same plane static and dynamic balance occurs when: Σmi*Ri*Xi = 0 M2*R2*X2+ M3*R3*X3 + M4*R4*X4 = 0

The crank (Mc) is statically and dynamically balanced by two counter weights, M1 & M2, all three masses are in the same plane. Find the masses of the two counterweights. Input

Mass 1 C.G. radius, R1 = X1 = Mass 2 C.G. radius, R2 = X2 = Crank Mass, Mc = Crank Mass Eccentricity, E =

10 16 14 30 450 2.5

Dynamic balance about mass M1: Calculation Mc*E*X1 = M2*R2*(X1+X2) M2 = Mc*E*X1 / R2*(X1+X2) Answer: M2 = 27.95031056 Condition for static balance: Σmi*Ri = 0 0 = M1*R1+M2*R2-Mc*E Mass required to balance Mc, M1 = (-M2*R2+Mc*E) / R1 Answer: M1 = 73.36956522

Example only

in in in in lbm in

lbm

lbm

12 18 12 36 57 0 3.96

Forced, Steady State Vibration Example

Calculate the two spring support stiffness (k) if the horizontal vibration amplitude is to be no more than 0.25 inches. Estimated friction is 5% of the critical damping factor (Cc).

Input

Motor speed, N = Motor+Compressor+Table Mass, W = Critical damping coefficient = Friction damping coefficient = (Friction/ Critical) damping factor ratio, DR =

Allowable vibration amplitude, Y = Motor speed, ω = Answer: ω = g= M= Answer: M = Total spring support stiffness, Kt = Kt = Answer: Kt = K= Answer: K = Critical value of damping factor, Cc = Answer: Cc = Friction damping factor, Cf = Answer: Cf = The motor periodic imbalance force, F = The motor peak imbalance force, Fo = At resonance, Y = Fo = Answer: Fo =

Vertical Vibration Damper Selection A metal tumbling drum driven by an electric motor-gear, right, rotates at 1080 rpm causing a disturbing vibration to the floor on which it is

360 80 Cc Cf Cf / Cc 0.05 0.25 Calculation 2*π*N / 60 37.704 386.4 W/g 0.2070 2*K M*ω^2 294.3 Kt / 2 147.2 2*(Kt*M)^.5 15.61 Cc*DR 0.781 Fo*Sin(ω*t) Cf*ω*Y Fo / Cc*ω Cf*ω*Y 7.36

rpm lbm

in

rad / sec in/sec^2 m-sec^2/in

lbf / in lbf / in

lbf lbf in lbf

mounted. The loaded drum, motor, and support base . weigh 400 lbm.

Vibration Isolator Selection Select 4 vibration isolators that will provide 80% vibration reduction applied to the floor.

Input

System weight, W = Number of isolators, N = Vibration reduction, VR = Disturbing frequency, Fd =

200 4 0.80 1080

Weight per isolator, w = Answer: w =

Calculation W/N 50

Transmissibility, T = Answer: T = Answer: Fd = Transmissibility, T = System natural frequency, Fn = Answer: Fn = g= Stiffness, K = Deflection, x = Undamped natural frequency, Fn = Fn = Fn = Solving for deflection in the above, x = Answer: x =

lbm

rpm

lbm

1 - VR 0.20 18 (1 / (1-(Fd / Fn)^.5) Fd / (1 +(1/T))^.5 7.35

rps

386.4 W/x W/K (1 / 2π)*(K*g / W)^.5 (1 / 2π)*(g / x)^.5 3.128*(1 / x)^.5

ft / sec^2

(3.128)^2 / (Fn)^2 0.181

Hz

Hz

in

Suggested max transmissibility, Tmax = 10 Ref. "Engineered Solutions" a Barry Controls publication.

At resonance transmissibility, T = C / Ccrit = Answer: C / Ccrit =

1/ (2*C / Ccrit) 1/ (2*T) 0.05

Isolator Selected: Go to the Barry Controls home page at: 4 Barry Controls vibration isolators http://www.barrycontrols.com/ Part No. 633A-100 Graphical Values

Deflection due to static load of 100 lb = Isolator frequency =

0.275 7.2

in Hz

The "Barry Controls" information presented here may be found on the web at: www.barrycontrols.com "Barry 633A Series Mounts are medium weight mounts normally used for vertically applied loads to prevent transm ission of noise and vibration caused by rotation of imbalanced equipment (i.e. generators, blowers, pumps, etc...) Low-profile, low frequency elastomeric noise and vibration isolators for medium weight industrial equipment."

The above graph shows a static load of 100 lbs produces a deflection of 0.275 inches.


69.05255


Shock Loads A shock load is caused by a nearly instantaneous rise and fall of acceleration.

Shock input pulse is normally expressed in g's.

Free Fall Impact Shock

A typical free fall shock test is an 11 millisecond second half sine waveform with a peak acceleration of 15 g. The above graph shows a static load of 100 lbs produces a natural frequency of 7.2 Hz. Shock Impulse Deflection

An electronic device is to be subjected to a 15g half sine shock lasting 11 milliseconds. The unit is mounted on a 10 Hz natural frequency isolation system. Determine the maximum shock transmission Input

Half sine shock acceleration, a = Shock pulse time, t = g= Isolator natural frequency, Fn =

12 0.018 386.4 20 Calculation

g sec in/ sec^2 Hz

Half sine pulse max peak velocity, Vmax = 2*g*a*t / π Answer: Vmax = 53.13 in/ sec^2 Max acceleration, G = Vmax*(2*π*Fn)/ g Answer: G = 17.3 g's Dynamic isolator deflection: Dd = Vmax/ (2*π*Fn) Answer: Dd = 0.423 in Transmissibility Ratio, TR = Ftransmitted/ Fapplied TR = d*(1+(2*r*C)^2)^.5 Notes: Magnification factor Bd must be greater than 1.00 or vibrations will be amplified. Magnification factor, Bd = Bd = D= Fo = K=

1/((1-r^2)^2+(2*C*r)^2)^.5 D /(Fo / K) Vibration amplitude Peak disturbing force Support stiffness

Isolator Selection

http://www.baldor.com/support/product_specs/generators/Vibration_Isolators/01_Korfund_Catalog.pdf Input

Equipment Weight, W = Number of Isolators, N = Applied Vertical Shock Acceleration, Gv = Shock Half Sine Pulse time, t = Allowable sway space, Xv = Isolator Roll Stiffness, Kr = Isolator Shear Stiffness, Kh = Isolator Compression Stiffness, Kv = Isolator Combined Total Stiffness, Kt = Equipment Fragility g Limit, Af =

13.3 4 50 0.003 1.4 0 0 133 133 10

Load per Isolator, Wi = Answer: Wi = Required Isolation Factor, If = Answer: If =

Calculation W/N 3.317 Af / Gv 20.00

Required Transmissibility, Tr = Answer: Tr =

1 - (If /1000) 0.8000

The spring type vibration and shock isolator information shown here may be found at: http://www.baldor.com Korfund division of Baldor Motor corp.

lbm g sec in lbf/in lbf/in lbf/in lbf/in g

lbm lbm %

Flexmount CB1260-39

" " "

and at the direct link above. "Effective vibration control for loads up to . Static deflections up to 1.36". Available with, or without adjustable snubbing." "Applications include: Stationary equipment, HVAC, Compressors, Pumps, Motor Generators, Fans, Blowers, etc." Vibration Damper Selection Gravitational constant, g = Isolator Vertical Natural frequency, Fn = Answer: Fn = Half Sine Shock Pulse Frequency, Fp = Answer: Fp =

Calculations continued 386 in/sec^2 3.13*(Kv / Wi)^.5 19.8 Hz 1/ (2 * t) 166.7 Hz

Shock Absorber Selection Max Vertical Shock Transmitted, Gv = Wi *(2*π*Fn)/ g Answer: Gv = 9.0 Required Average Spring Rate, Ks = (2*π*Fn)^2*(W/g) Answer: Ks = 133 Combined Isolator Vertical Frequency, Fc = Answer: Fc =

3.13*(Ks / Wi) 19.8

Maximum Dynamic Travel, Dt = Gv*g / (2*π*Fs)^2 Answer: Dt = 0.22 Max Half Sine Pulse Velocity, Vv = Answer: Vv =

2*g*Gv*t / π 36.9

g

lb/in

Hz

in

in/sec

Above: Korfund division of Baldor Motor corp.


MACHINE DESIGN EXCEL SPREAD SHEETS Copy write, © Machine Design Spreadsheet Calculations by John R Andrew, 6 July 2006 EXCEL MATH TOOLS Useful math tools applicable to this course are given below. Insert the Microsoft Office CD for Add-Ins If Excel's, "Goal Seek" or "Solver" are not

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.

installed you will need to select drop-down menu: Tools > Add-Ins > Goal Seek Tools > Add-Ins > Solver To open select Tools.

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

ADJ = OPP = HYP = =

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

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

ADJ = OPP = HYP = =

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

What if Calculations 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. =

M ax 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

Mass, M1 = Mass, M2 = K1 = k2 = X2 / X1 =

0.1 0.1 20 20 1.6180

Reference: Machine Design by A.S. Hall, A.R. Holowenko, H.G. Laughlin, Published byMcGraw-Hill.

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.

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

D

5 6

Guess X =

Problem 1.4

7

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

8 9

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

D

5 6

Solved X =

Solution 1.4041

7 8 9

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

Constants

Solution

0.0 0.0 0.0 0.0 0.0

5.5 22.5 30 -11 17.5

u= v= w= x= y=

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

Row Row Row Row Row

146 147 148 149 150

Equations

Constants

Solution

5.5 22.5 30.0 -11.0 17.5

5.5 22.5 30 -11 17.5

u= v= w= x= y=

1.00 4.00 7.50 -8.00 1.00

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

Row Row Row Row Row

146 147 148 149 150


Equations

Constants

Solution

0.0 0.0 0.0 0.0 0.0

5.5 22.5 30 -11 17.5

u= v= w= x= y=

0.00 0.00 0.00 0.00 0.00

Machine Design Excel

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