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FEBRUARY 2021

www.fluidpowerjournal.com

ON for I T A L L A T S N I EASY

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IN THIS ISSUE

FEBRUARY 2021

VOLUME 28 • ISSUE 2

Features 12 Stressed Out: Extending

the Life of Hydraulic Cylinder

Position Sensors Mitigating conditions that lead to

12

36

premature failure. 16 Cover Story

Making Installation Easier with

Hydraulic Plug-In Connections Improve production-line efficiency in hose and tube assembly components.

34 Keeping Cool with Hydraulic

Case Drains High fluid temperatures diminish a system’s performance.

36 Test Your Skills

Calculate the Kinetic Energy

16 Departments Publisher’s Note: The information provided in this publication is for informational purposes only. While all efforts have been taken to ensure the technical accuracy of the material enclosed, Fluid Power Journal is not responsible for the availability, accuracy, currency, or reliability of any information, statement, opinion, or advice contained in a third party’s material. Fluid Power Journal will not be liable for any loss or damage caused by reliance on information obtained in this publication.

CELEBRATING 60 YEARS

4

Notable Words

6

IFPS Update

19

Web Marketplace

20

Systems Integrator

Directory Listing

33

Air Teaser

39

Classifieds


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N OTA B L E WO R D S

NFPA Guides Students to Fluid Power Careers By Lynn Beyer, Vice President of Workforce Development Programs, National Fluid Power Association

»

IN BIENNIAL SURVEYS, National Fluid Power Association members consistently rank workforce development as the most challenging issue their companies face. One reason is that not enough technical colleges and universities are teaching fluid power. So how can we get more students interested in fluid power careers? It is something we continually work on at NFPA. NFPA identified growth of the workforce as one of its primary strategic priorities. It is important to our mission of strengthening the industry. NFPA needs to work to increase the number of educated technicians and engineers embarking on careers in fluid power and connect them to industry members. We have made our way down many different avenues by supporting scholarships, grants, and competitions, among other efforts. And while we’ve often raised student interest, we were not always able to bring these programs together and work alongside each other to lead students into fluid power careers. We realized we needed to form complete pathways, not just a resource here or there along the way. That approach left us no place to lead students if they were interested in the next step. Here are some of the programs we’ve created to help accomplish these goals and begin to answer this burning question of how to raise interest in fluid power among students at all educational levels. Fast Track to Fluid Power is a workforce development pathway that brings companies and technical colleges together with middle and high school teachers and students. This network creates awareness and interest in fluid power and leads students along a path to careers in the industry. There are four main participants in this program: • A technical or community college with a validated fluid power and mechatronics degree program. NFPA provides financial support to

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the school to help add fluid power learning outcomes. • A ring of local high schools that NFPA equips with hands-on fluid power training equipment as well as train-the-trainer classes for high school teachers. • A community-wide Fluid Power Action Challenge event for middle school students in which they learn about and have fun with fluid power. • A network of industry partners with facilities near the community. Fast Track is an opportunity for NFPA members to make a direct connection to their future workforce. The University Power Partner Program is another pathway we created to connect our members with universities that teach fluid power. To be considered as a Power Partner, a university must: • participate in the Speaker’s Bureau program, • participate in the Fluid Power Vehicle Challenge, • start a fluid power club, • host a student-industry connection event, and • teach the nine core fluid power competencies. Here is a description of these programs and resources NFPA has created to provide help. • Speaker’s Bureau: NFPA members speak about fluid power careers at local universities. NFPA provides a draft presentation, information on types of jobs, salaries, job outlooks, and the degrees needed for each job. NFPA also works with the university and the NFPA member to find a time that works for both. • Vehicle Challenge: A unique engineering design competition embedded in the capstone design course at participating universities. It strives to promote original thinking in a competitive setting by combining human-powered vehicles and fluid

power technology. The program also connects university students to NFPA industry members through mentorship and judging. Fluid power clubs: University students in NFPA-sponsored clubs work on fluid power projects and connect with NFPA industry members for technical support and career mentorship. Nine core competencies: NFPA's university education committee identified nine core competencies with which mechanical engineering students should graduate and that embody the skills industry members look for in job candidates. Fluid power curricula: Academic colleagues use NFPA resources to teach the nine core competencies in university engineering programs. University fluid power grants: NFPA awards grants to support fluid power curricula.

While these programs are new and may take a few years to gain traction, we have already seen growth in hiring as a result of connections between students and our members in the fluid power industry. With the goal of providing future fluid power professionals for the 21st century, NFPA continues working with students at all levels. 

WWW.FLUIDPOWERJOURNAL.COM • WWW.IFPS.ORG


We Make Valve Automation Easy! PUBLISHER Innovative Designs & Publishing, Inc. 3245 Freemansburg Avenue, Palmer, PA 18045-7118 Tel: 800-730-5904 or 610-923-0380 Fax: 610-923-0390 • Email: Art@FluidPowerJournal.com www.FluidPowerJournal.com Founders: Paul and Lisa Prass Associate Publisher: Bob McKinney Editor: Michael Degan Technical Editor: Dan Helgerson, CFPAI/AJPP, CFPS, CFPECS, CFPSD, CFPMT, CFPCC - CFPSOS LLC Art Director: Quynh Fisher Eastern Region Acct Executive: Norma Abrunzo Director of Creative Services: Erica Montes Accounting: Donna Bachman, Sarah Varano Circulation Manager: Andrea Karges

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INTERNATIONAL FLUID POWER SOCIETY 1930 East Marlton Pike, Suite A-2, Cherry Hill, NJ 08003-2141 Tel: 856-489-8983 • Fax: 856-424-9248 Email: AskUs@ifps.org • Web: www.ifps.org

EXECUTIVE DIRECTOR (EX-OFFICIO) Donna Pollander, ACA HONORARY DIRECTORS (EX-OFFICIO) Paul Prass, Fluid Power Journal Liz Rehfus, CFPE, CFPS Robert Sheaf, CFPAI/AJPP, CFC Industrial Training

IFPS STAFF Executive Director: Donna Pollander, ACA Communications Director: Adele Kayser Technical Director: Thomas Blansett, CFPS, CFPAI Assistant Director: Stephanie Coleman Certification Coordinator: Kyle Pollander Bookkeeper: Diane McMahon Administrative Assistant: Beth Borodziuk

Fluid Power Journal (ISSN# 1073-7898) is the official publication of the International Fluid Power Society published monthly with four supplemental issues, including a Systems Integrator Directory, Off-Highway Suppliers Directory, Tech Directory, and Manufacturers Directory, by Innovative Designs & Publishing, Inc., 3245 Freemansburg Avenue, Palmer, PA 18045-7118. All Rights Reserved. Reproduction in whole or in part of any material in this publication is acceptable with credit. Publishers assume no liability for any information published. We reserve the right to accept or reject all advertising material and will not guarantee the return or safety of unsolicited art, photographs, or manuscripts.

WWW.IFPS.ORG • WWW.FLUIDPOWERJOURNAL.COM

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DIRECTORS-AT-LARGE Chauntelle Baughman, CFPHS - OneHydraulics, Inc. Stephen Blazer, CFPE, CFPS, CFPMHM, CFPIHT, CFPMHT Altec Industries, Inc. Randy Bobbitt, CFPAI, CFPHS - Danfoss Power Solutions Steve Bogush, CFPAI/AJPP, CFPHS, CFPIHM - Poclain Hydraulics Cary Boozer, PE, CFPE - Motion Industries, Inc. Lisa DeBenedetto, CFPS - GS Global Resources Daniel Fernandes, CFPECS, CFPS - Sun Hydraulics Brandon Gustafson, PE, CFPE, CFPS, CFPIHT, CFPMHM - Graco, Inc. Garrett Hoisington, CFPAI/AJPP, CFPS, CFPMHM Open Loop Energy Brian Kenoyer, CFPHS - Five Landis Corp. Jon Rhodes, CFPAI, CFPS, CFPECS - CFC Industrial Training Mohaned Shahin, CFPS - Parker Hannifin Randy Smith, CFPHS - Northrop Grumman Corp.

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I F P S U P D AT E

A Message from Donna Pollander, ACA, IFPS Executive Director

»

ONE OF THE things I’ve learned in my 20-plus-year tenure as executive director is that people get certified for many reasons. I’ve heard everything from “I want to be as professional as I can be” to “My company made me do it!” From company managers I’ve heard, “I want to benchmark our employees’ knowledge” and “I want a safer and more efficient workforce.” All valid reasons. I want to share an excerpt from a 2013 Fluid Power Journal article written by Scott Gower, CFPS, CFPECS, of Gulf Controls Co. I couldn’t say it better myself! Scott stated: “People in the fluid power field would likely agree with the following: • Earning an IFPS certification isn’t easy. This speaks to the value the certification holds. Sissies need not apply. • Passing the certification test proves a base understanding of the topic. If you’ve earned one and you meet anyone else who also has, you can instantly speak 'fluid-ese' and have meaningful conversations. • Employers who request/require certification can trust they’re meeting people who know what they’re talking about. • Each certification’s benefit is this: It targets a specific and valuable skill set. It’s not a four-year college degree built to create a well-rounded person. This is a meaningful certification you can add to your resume with a short and dutiful study period.” IFPS wants you to succeed and provides a plethora of study aids (more on that later) to help you prepare for a certification test, including the new Fluid Power Reference Handbook. This 360page reference handbook, offered as soft-cover and hardcover, can 6

FEBRUARY 2021

be used during a certification test. The color-coded table of contents and index match 17 color-coded sections, making information easy to find. The handbook contains full-color graphics, is ISO, ANSI, and SAE compliant, and contains troubleshooting algorithms and general safety guidelines. It is truly the ultimate, must-have resource for fluid power professionals. We also provide a Math for Certification booklet. Both references are offered at a discount price for test applicants.

The ULTIMAT E resource for hydraulic, pneu matic, and motion contr ol profession als

Fluid Power

Reference Handbook FIRST EDITI ON

The IFPS Board of Directors is committed to making fluid power certifications more globally aligned. There are testing sites around the world and many throughout Canada. We are in the final stages of updating our mechanic and technician mobile

and industrial certifications to align with CETOP, a European fluid power committee. I want to review some test preparation options, all designed to help you succeed. Certifications do not exist so you can say, “I took a class and got a certificate”; they are objective, third-party assessments of your fluid power knowledge. So I urge you to take the time to prepare using resources available on our website, www.ifps.org. If you are unsure of which certification test to take, start with the self-assessment tool. It can help determine which certification test to choose based on your everyday job functions. Once you’ve determined which certification test to take and submit a test application, you receive an electronic study manual. As you prepare, take time to test yourself. In addition to the pretest included in the study manuals, IFPS offers corresponding online pretests. They give you an indication of whether you’re ready for the test. We also offer web seminars covering more difficult review questions. Suppose you are more of a visual, auditory learner. In that case, IFPS offers hydraulic specialist and pneumatic specialist interactive study manuals, which turn the traditional black/white print into a full color, animated, interactive online learning platform. Animated hydraulic and pneumatic circuits are also available. These are color-coded, narrated, and animated mp4 and wmv files of circuit operations using ANSI-recognized color designations. We’ve also seen certification success with in-house study groups that meet during lunch breaks or after work and have fun with our free online "Jeopardy!"-style games. Also, if you have a group

CELEBRATING 60 YEARS

of ten or more individuals interested in training and testing at your location, IFPS offers customized training, which can range in length from two days to several weeks and can be customized to your machinery and personnel. Fundamental hydraulic and pneumatic training programs are generally one week in duration but can range from three days to two weeks. If you are unsuccessful in passing a certification test, don’t give up! Our core-competency reports provide a breakdown of test results to help you pinpoint your weak spots. Successful candidates also receive the report to let them know their strengths and weaknesses. To boil it all down, it doesn’t matter if you have been in the industry for thirty days or thirty years. To advance your career and become that well-rounded fluid power expert, you must have some skin in the game. IFPS celebrates its 61st anniversary this year – we’ve got skin in the game! Make a commitment to get certified and join nearly 10,000 active certification holders who speak “fluid-ese” with confidence. For more information, visit www.ifps.org.

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I F P S U P D AT E

IFPS 2021 Spring Meeting

»

IFPS WILL HOST a condensed virtual spring meeting March 8-11, with plans to hold an in-person meeting May 4-7 in San Antonio, Texas, at the Embassy Suites, San Antonio Riverwalk Downtown. Members may register for the condensed spring meeting by visiting www.ifps.org.

Schedule of events Monday, March 8 02:00 PM - 03:00 PM

Strategic Planning Committee Meeting

Tuesday, March 9 11:00 AM - 12:00 PM 02:00 PM - 03:00 PM

Education Committee Meeting Membership Committee Meeting

Wednesday, March 10 11:00 AM - 12:00 PM 02:00 PM - 03:00 PM

Certification Committee Meeting Marketing Committee Meeting

Thursday, March 11 11:00 AM - 12:00 PM 02:00 PM - 03:00 PM

Finance Committee Meeting Board of Directors Meeting

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Congratulations to our newly Certified Fluid Power Accredited Instructors (CFPAI) and Authorized Job Performance Proctors (AJPP)

»

APPROVED CANDIDATES hold at least one certification, conducted a preplanned presentation, and were evaluated by a panel of subject-matter experts and peers during the recent instructor-training workshop in Kansas City, Missouri.

Bruce Bowe, CFPAI/AJPP, CFPMHM, Altec Industries, Inc. Jeremey Goodman, CFPAI/AJPP, CFPMHM, Altec Industries, Inc. Jerry McClain, CFPAI/AJPP, CFPMHM, Altec Industries, Inc. Samuel McClellan, CFPAI/AJPP, CFPMHM, Altec Industries, Inc. Don Mitch, CFPAI/AJPP, CFPMHM, Altec Industries, Inc. Travis Sybesma, CFPAI/AJPP, CFPMHM, Altec Industries, Inc.

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Compact, Leak-Free Hydraulic Systems IFPS is accepting nominations for its 2021 Fluid Power Hall of Fame awards. The Hall of Fame acknowledges individuals who have made significant contributions to fluid power technology and dedicated their careers to the industry.

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Anyone can nominate one living and one deceased individual. The nominee must have 25 or more years of service in the fluid power industry. Living nominees will be asked to complete an application. A panel of judges will select the inductees, who will be announced on Fluid Power Professionals Day, June 19. Nominations are being accepted until April 15 and can be made at www.fluidpowerhalloffame.org.

WWW.FLUIDPOWERJOURNAL.COM • WWW.IFPS.ORG


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I F P S U P D AT E

Certification Testing Locations Individuals wishing to take any IFPS written certification tests can select from convenient locations across the United States and Canada. IFPS is able to offer these locations through its affiliation with The Consortium of College Testing Centers provided by National College Testing Association. Contact headquarters if you do not see a location near you. Every effort will be made to accommodate your needs. If your test was postponed due to the pandemic, please contact headquarters so that we may reschedule.

TENTATIVE TESTING DATES FOR ALL LOCATIONS: March 2021 Tuesday 3/2 • Thursday 3/25 April 2021 Tuesday 4/6 • Thursday 4/22 May 2021 Tuesday 5/4 • Thursday 5/20 June 2021 Tuesday 6/1 • Thursday 6/24

ALABAMA Auburn, AL Birmingham, AL Calera, AL Decatur, AL Huntsville, AL Jacksonville, AL Mobile, AL Montgomery, AL Normal, AL Tuscaloosa, AL ALASKA Anchorage, AK Fairbanks, AK ARIZONA Flagstaff, AZ Glendale, AZ Mesa, AZ Phoenix, AZ Prescott, AZ Scottsdale, AZ Sierra Vista, AZ Tempe, AZ Thatcher, AZ Tucson, AZ Yuma, AZ ARKANSAS Bentonville, AR Hot Springs, AR Little Rock, AR CALIFORNIA Aptos, CA Arcata, CA Bakersfield, CA Dixon, CA Encinitas, CA Fresno, CA Irvine, CA Marysville, CA Riverside, CA Salinas, CA San Diego, CA San Jose, CA San Luis Obispo, CA Santa Ana, CA Santa Maria, CA Santa Rosa, CA Tustin, CA Yucaipa, CA COLORADO Aurora, CO Boulder, CO Springs, CO Denver, CO Durango, CO Ft. Collins, CO Greeley, CO Lakewood, CO Littleton, CO Pueblo, CO DELAWARE Dover, DE Georgetown, DE Newark, DE FLORIDA Avon Park, FL Boca Raton, FL Cocoa, FL Davie, FL Daytona Beach, FL Fort Pierce, FL Ft. Myers, FL Gainesville, FL Jacksonville, FL Miami Gardens, FL Milton, FL New Port Richey, FL Ocala, FL Orlando, FL Panama City, FL Pembroke Pines, FL Pensacola, FL Plant City, FL Riviera Beach, FL Sanford, FL

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FEBRUARY 2021

Tallahassee, FL Tampa, FL West Palm Beach, FL Wildwood, FL Winter Haven, FL GEORGIA Albany, GA Athens, GA Atlanta, GA Carrollton, GA Columbus, GA Dahlonega, GA Dublin, GA Dunwoody, GA Forest Park, GA Lawrenceville, GA Morrow, GA Oakwood, GA Savannah, GA Statesboro, GA Tifton, GA Valdosta, GA HAWAII Laie, HI IDAHO Boise, ID Coeur d ‘Alene, ID Idaho Falls, ID Lewiston, ID Moscow, ID Nampa, ID Rexburg, ID Twin Falls, ID ILLINOIS Carbondale, IL Carterville, IL Champaign, IL Decatur, IL Edwardsville, IL Glen Ellyn, IL Joliet, IL Malta, IL Normal, IL Peoria, IL Schaumburg, IL Springfield, IL University Park, IL INDIANA Bloomington, IN Columbus, IN Evansville, IN Fort Wayne, IN Gary, IN Indianapolis, IN Kokomo, IN Lafayette, IN Lawrenceburg, IN Madison, IN Muncie, IN New Albany, IN Richmond, IN Sellersburg, IN South Bend, IN Terre Haute, IN IOWA Ames, IA Cedar Rapids, IA Iowa City, IA Ottumwa, IA Sioux City, IA Waterloo, IA KANSAS Kansas City, KS Lawrence, KS Manhattan, KS Wichita, KS KENTUCKY Ashland, KY Bowling Green, KY Erlanger, KY Highland Heights, KY Louisville, KY Morehead, KY

LOUISIANA Bossier City, LA Lafayette, LA Monroe, LA Natchitoches, LA New Orleans, LA Shreveport, LA Thibodaux, LA MARYLAND Arnold, MD Bel Air, MD College Park, MD Frederick, MD Hagerstown, MD La Plata, MD Westminster, MD Woodlawn, MD Wye Mills, MD MASSACHUSETTS Boston, MA Bridgewater, MA Danvers, MA Haverhill, MA Holyoke, MA Shrewsbury, MA MICHIGAN Ann Arbor, MI Big Rapids, MI Chesterfield, MI Dearborn, MI Dowagiac, MI East Lansing, MI Flint, MI Grand Rapids, MI Kalamazoo, MI Lansing, MI Livonia, MI Mount Pleasant, MI Sault Ste. Marie, M Troy, MI University Center, MI Warren, MI MINNESOTA Alexandria, MN Brooklyn Park, MN Duluth, MN Eden Prairie, MN Granite Falls, MN Mankato, MN MISSISSIPPI Goodman, MS Jackson, MS Mississippi State, MS Raymond, MS University, MS MISSOURI Berkley, MO Cape Girardeau, MO Columbia, MO Cottleville, MO Joplin, MO Kansas City, MO Kirksville, MO Park Hills, MO Poplar Bluff, MO Rolla, MO Sedalia, MO Springfield, MO St. Joseph, MO St. Louis, MO Warrensburg, MO MONTANA Bozeman, MT Missoula, MT NEBRASKA Lincoln, NE North Platte, NE Omaha, NE NEVADA Henderson, NV Las Vegas, NV North Las Vegas, NV Winnemucca, NV

CELEBRATING 60 YEARS

NEW JERSEY Branchburg, NJ Cherry Hill, NJ Lincroft, NJ Sewell, NJ Toms River, NJ West Windsor, NJ NEW MEXICO Albuquerque, NM Clovis, NM Farmington, NM Portales, NM Santa Fe, NM NEW YORK Alfred, NY Brooklyn, NY Buffalo, NY Garden City, NY New York, NY Rochester, NY Syracuse, NY NORTH CAROLINA Apex, NC Asheville, NC Boone, NC Charlotte, NC China Grove, NC Durham, NC Fayetteville, NC Greenville, NC Jamestown, NC Misenheimer, NC Mount Airy, NC Pembroke, NC Raleigh, NC Wilmington, NC NORTH DAKOTA Bismarck, ND OHIO Akron, OH Cincinnati, OH Cleveland, OH Columbus, OH Fairfield, OH Findlay, OH Kirtland, OH Lima, OH Maumee, OH Newark, OH North Royalton, OH Rio Grande, OH Toledo, OH Warren, OH Youngstown, OH OKLAHOMA Altus, OK Bethany, OK Edmond, OK Norman, OK Oklahoma City, OK Tonkawa, OK Tulsa, OK OREGON Bend, OR Coos Bay, OR Eugene, OR Gresham, OR Klamath Falls, OR Medford, OR Oregon City, OR Portland, OR White City, OR PENNSYLVANIA Bloomsburg, PA Blue Bell, PA Gettysburg, PA Harrisburg, PA Lancaster, PA Newtown, PA Philadelphia, PA Pittsburgh, PA Wilkes-Barre, PA York, PA

SOUTH CAROLINA Beaufort, SC Charleston, SC Columbia, SC Conway, SC Graniteville, SC Greenville, SC Greenwood, SC Orangeburg, SC Rock Hill, SC Spartanburg, SC TENNESSEE Blountville, TN Clarksville, TN Collegedale, TN Gallatin, TN Johnson City, TN Knoxville, TN Memphis, TN Morristown, TN Murfreesboro, TN Nashville, TN TEXAS Abilene, TX Arlington, TX Austin, TX Beaumont, TX Brownsville, TX Commerce, TX Corpus Christi, TX Dallas, TX Denison, TX El Paso, TX Houston, TX Huntsville, TX Laredo, TX Lubbock, TX Lufkin, TX Mesquite, TX San Antonio, TX Victoria, TX Waxahachie, TX Weatherford, TX Wichita Falls, TX UTAH Cedar City, UT Kaysville, UT Logan, UT Ogden, UT Orem, UT Salt Lake City, UT VIRGINIA Daleville, VA Fredericksburg, VA Lynchburg, VA Manassas, VA Norfolk, VA Roanoke, VA Salem, VA Staunton, VA Suffolk, VA Virginia Beach, VA Wytheville, VA WASHINGTON Auburn, WA Bellingham, WA Bremerton, WA Ellensburg, WA Ephrata, WA Olympia, WA Pasco, WA Rockingham, WA Seattle, WA Shoreline, WA Spokane, WA WEST VIRGINIA Ona, WV WISCONSIN La Crosse, WI Milwaukee, WI Mukwonago, WI

WYOMING Casper, WY Laramie, WY Torrington, WY CANADA ALBERTA Calgary, AB Edmonton, AB Fort McMurray, AB Lethbridge, AB Lloydminster, AB Olds, AB Red Deer, AB BRITISH COLUMBIA Abbotsford, BC Burnaby, BC Castlegar, BC Delta, BC Kamloops, BC Nanaimo, BC Prince George, BC Richmond, BC Surrey, BC Vancouver, BC Victoria, BC MANITOBA Brandon, MB Winnipeg, MB NEW BRUNSWICK Bathurst, NB Moncton, NB NEWFOUNDLAND AND LABRADOR St. John’s, NL NOVA SCOTIA Halifax, NS ONTARIO Brockville, ON Hamilton, ON London, ON Milton, ON Mississauga, ON Niagara-on-the-Lake, ON North Bay, ON North York, ON Ottawa, ON Toronto, ON Welland, ON Windsor, ON QUEBEC Côte Saint-Luc, QB Montreal, QB SASKATCHEWAN Melfort, SK Moose Jaw, SK Nipawin, SK Prince Albert, SK Saskatoon, SK YUKON TERRITORY Whitehorse, YU UNITED KINGDOM Elgin, UK GHAZNI Kingdom of Bahrain, GHA Thomasville, GHA EGYPT Cairo, EG JORDAN Amman, JOR NEW ZEALAND Taradale, NZ

WWW.FLUIDPOWERJOURNAL.COM • WWW.IFPS.ORG


I F P S U P D AT E

AVAILABLE IFPS CERTIFICATIONS CFPAI Certified Fluid Power Accredited Instructor CFPAJPP Certified Fluid Power Authorized Job Performance Proctor CFPAJPPCC Certified Fluid Power Authorized Job Performance Proctor Connector & Conductor CFPE Certified Fluid Power Engineer CFPS Certified Fluid Power Specialist (Must Obtain CFPHS & CFPPS) CFPHS Certified Fluid Power Hydraulic Specialist CFPPS Certified Fluid Power Pneumatic Specialist CFPECS Certified Fluid Power Electronic Controls Specialist CFPMT Certified Fluid Power Master Technician (Must Obtain CFPIHT, CFPMHT, & CFPPT) CFPIHT Certified Fluid Power Industrial Hydraulic Technician CFPMHT Certified Fluid Power Mobile Hydraulic Technician CFPPT Certified Fluid Power Pneumatic Technician CFPMM Certified Fluid Power Master Mechanic (Must Obtain CFPIHM, CFPMHM, & CFPPM) CFPIHM Certified Fluid Power Industrial Hydraulic Mechanic CFPMHM Certified Fluid Power Mobile Hydraulic Mechanic CFPPM Certified Fluid Power Pneumatic Mechanic CFPMIH Certified Fluid Power Master of Industrial Hydraulics (Must Obtain CFPIHM, CFPIHT, & CFPCC) CFPMMH Certified Fluid Power Master of Mobile Hydraulics (Must Obtain CFPMHM, CFPMHT, & CFPCC) CFPMIP Certified Fluid Power Master of Industrial Pneumatics (Must Obtain CFPPM, CFPPT, & CFPCC) CFPCC Certified Fluid Power Connector & Conductor CFPSD Fluid Power System Designer CFPMEC (In Development) Mobile Electronic Controls CFPIEC (In Development) Industrial Electronic Controls

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Tentative Certification Review Training In-house Review Training – an IFPS Accredited Instructor will come to your company (minimum 10 individuals) HYDRAULIC SPECIALIST CERTIFICATION REVIEW Onsite review training for small groups – contact kpollander@ifps.org for details March 8-11, 2021 - Milwaukee, WI, MSOE | Written test: March 12, 2021 March 23-26, 2021 - Fairfield, OH - CFC Industrial Training | Written test: March 26, 2021 September 13-16, 2021 - Fairfield, OH - CFC Industrial Training | Written test: September 16, 2021 PNEUMATIC SPECIALIST Onsite review training for small groups – contact kpollander@ifps.org for details July 27-29, 2021 - Fairfield, OH - CFC Industrial Training | Written test: July 29, 2021 ELECTRONIC CONTROLS CERTIFICATION REVIEW Onsite review training for small groups – contact kpollander@ifps.org for details August 9-12, 2021 - Fairfield, OH - CFC Industrial Training | Written test: August 12, 2021 CONNECTOR & CONDUCTOR CERTIFICATION REVIEW Onsite review training for small groups – contact kpollander@ifps.org for details May 18-19, 2021 - Fairfield, OH - CFC Industrial Training | Written and JP test: May 20, 2021 November 16-17, 2021 - Fairfield, OH - CFC Industrial Training | Written and JP test: November 18, 2021 MOBILE HYDRAULIC MECHANIC CERTIFICATION REVIEW Onsite review training for small groups – contact kpollander@ifps.org for details Online Mobile Hydraulic Mechanic Certification Review (for written test) offered through info@cfcindustrialtraining.com. This course takes you through all chapters of the MHM Study Manual (6.5 hours) and every outcome to prepare you for the written MHM test. Members receive 20% off. (Test fees are additional - separate registration required.) April 13-15, 2021 - Fairfield, OH - CFC Industrial Training | Written and JP test: April 16, 2021 August 30 - September 1, 2021 - Fairfield, OH - CFC Industrial Training | Written and JP test: September 2, 2021   INDUSTRIAL HYDRAULIC MECHANIC CERTIFICATION Onsite review training for small groups – contact kpollander@ifps.org for details June 14 - 16, 2021 - Fairfield, OH - CFC Industrial Training | Written and JP test: June 17, 2021   INDUSTRIAL HYDRAULIC TECHNICIAN CERTIFICATION REVIEW TRAINING Onsite review training for small groups – contact kpollander@ifps.org for details Call for dates. Phone: 513-874-3225 - CFC Industrial Training, Fairfield, Ohio   MOBILE HYDRAULIC TECHNICIAN CERTIFICATION REVIEW TRAINING Onsite review training for small groups – contact kpollander@ifps.org for details Call for dates. Phone: 513-874-3225 - CFC Industrial Training, Fairfield, Ohio   PNEUMATIC TECHNICIAN and PNEUMATIC MECHANIC CERTIFICATION REVIEW TRAINING Onsite review training for small groups – contact kpollander@ifps.org for details Call for dates. Phone: 513-874-3225 - CFC Industrial Training, Fairfield, Ohio   JOB PERFORMANCE TRAINING Onsite review training for small groups – contact kpollander@ifps.org for details Online Job Performance Review - CFC Industrial Training offers online JP Reviews which includes stations 1-6 of the IFPS mechanic and technician job performance tests. Members may e-mail askus@ifps.org for a 20% coupon code off the list price or get the code in our Members Only area for the entire IFPS Job Performance Review; test not included. LIVE DISTANCE LEARNING JOB PERFORMANCE STATION REVIEW Onsite review training for small groups – contact kpollander@ifps.org for details E-mail info@cfcindustrialtraining.com for information. FEBRUARY 2021

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Stressed

Out Extending the Life of Hydraulic Cylinder Position Sensors By Henry Menke, Operational Sales Support & Digital Information Manager, Balluff Inc.

M

agnetostrictive linear position sensors are the dominant technology for hydraulic cylinder position feedback. They are preferred over alternatives due to their high accuracy, wide variety of electrical interfaces, signal stability across temperature variation, wear-free operation, and tolerance of shock and vibration. This makes them popular in a wide range of industrial hydraulic applications for smart cylinders and servo-hydraulic systems. Despite the inherent ruggedness of the sensors, adverse operating conditions in industrial applications can stress and sometimes exceed the survivability of standard product offerings, resulting in premature sensor failure and shortened preventive-maintenance intervals. Fortunately, specialized product variations can deliver a longer and more reliable service life under adverse operating conditions. And best practices for installation and mounting can also help alleviate exposure to the most damaging effects of application environments.

Protecting against heat High temperatures are a fact of life in industrial processes such as steel making, with ambient environments that can destroy electronics. Solid-state electronic components are stressed by 12

FEBRUARY 2021

elevated temperatures and can fail at much higher rates than they would at room temperature. Ambient heat can affect sensors through three mechanisms: 1. Conduction – direct transfer from the mechanical mounting, 2. Convection – circulating hot air impinging on the sensor housing, and 3. Radiation – line-of-sight infrared heating of the sensor housing at a distance. To protect against heat, first evaluate what can be done to reduce the sensor’s exposure to it. To combat conduction, try to locate the sensor as far as possible from the heat source. If practical, install insulators to isolate the sensor mechanically from direct contact with hot objects. Alleviate convection by installing guards or deflectors to block hot air from the sensor housing. Often these same guards aid in blocking radiating infrared energy, but keep in mind that with prolonged exposure, energy will get through and be reradiated on the back side. If the heat source is intermittent, such as a passing red-hot slab of steel on rollers, the heat only needs to be blocked for a few seconds to protect the sensor. Next, be sure to specify the right kind of cylinder position sensor for hot applications. Standard cylinder sensors typically have a maximum

specified temperature rating of 75°C (167°F), but enhanced variants can withstand up to 100°C (212°F). Don’t forget that many standard cable materials will melt or break down quickly at higher temperatures, so it’s important to use specialized cable jacket materials such as PTFE that can withstand up to 200°C (392°F). Another strategy is to insert a double-ended cordset between the sensor and the home run cable. If this short sacrificial cable becomes damaged, it can be quickly replaced without the time-consuming task of pulling a new home run cable. Finally, it may be that the hot environment is just too severe for any electronics to survive for very long. In these cases, place the sensor in a protective enclosure designed for air purge or water-jacket cooling.

Cold In some industries, particularly oil and gas, operations are conducted in arctic or subarctic regions where temperatures can fall to -40°C (-40°F) or lower. At these temperatures, polymers intended to remain pliable, such as O-rings, gaskets, cable insulation and outer jackets, and sensor element damping materials, can become brittle. Semiconductor materials start to lose conductivity and circuits can behave unpredictably. WWW.FLUIDPOWERJOURNAL.COM • WWW.IFPS.ORG


Standard cylinder position sensors are typically constructed with a stainless steel flange and an extruded-aluminum housing secured by a few screws. In cases where the housing may be exposed to physical impact, such as a falling log in a lumber mill, standard sensors can easily be damaged. There are three basic approaches to improve survivability to a mechanical impact. 1. Guard. Many cylinder manufacturers can provide accessory guards for cylinder sensors. These are essentially lengths of steel pipe attached to the cylinder end cap that covers the sensor. A disadvantage is that they often have to be removed to service the sensor and are mistakenly or intentionally left off after the service procedure, leaving the sensor vulnerable to impact. 2. Embed. Compact cylinder sensors can be embedded inside the cylinder, so the rugged cylinder itself protects the sensor from damage. These sensors are usually called embeddable types. Often installed in welded cylinders for mobile hydraulic applications, they are entirely suitable for stationary industrial applications as well. 3. Upgrade. Cylinder sensors with especially robust housings provide an upgrade path from standard types. Thread-in types are available for retrofit applications, as are even more robust bolt-in types. Note that the cylinder must be prepared accordingly by the cylinder manufacturer.

Shock Many industrial hydraulic cylinder applications involve large forces and heavy loads that can generate substantial amounts of shock under normal or abnormal conditions. Shock is defined as a rapid acceleration or deceleration transient and can occur axially or radially relative to the cylinder position sensor depending on the application. Typically, hydraulic cylinders deliver or absorb axial shock along their linear axis of WWW.IFPS.ORG • WWW.FLUIDPOWERJOURNAL.COM

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Physical impact

operation. Trunnion-mounted cylinders often generate radial shock, such as infeed rollers coming down on a log in a lumber mill as they pivot in their mounting. If an application is suffering from a high sensor failure rate due to shock, the first step is to determine if the hydraulic control system can be optimized to reduce the generation of shock. For example, can the motion control profile be tuned to deliver less violent acceleration or deceleration? Can the motion profile be adapted to enable softer starts and stops? The key to reducing shock is to reduce the peak amplitude of the g forces and stretch out the time of the transient so that the acceleration gradient is less severe. A 100-g peak shock delivered over 6 milliseconds is less severe than a 100-g peak shock delivered within 3 milliseconds. Once the application has been evaluated and optimized to minimize shock generation in the first place, the next thing to do is look for sensors with enhanced shock specifications. The minimum specification should be 100 g, with preference given to products rated up to 150 g. For even more severe applications, consult the sensor manufacturer regarding specially engineered versions. Don’t forget that the extra mass of a sensor-mounted quick-disconnect connector can sometimes be a vulnerability under severe shock. Combat broken connectors with so-called pigtail or inline connectors that have a flexible cable coming out of the sensor that goes to a quick-disconnect a few inches or feet away.

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Special hydraulic cylinder sensors designed to operate in temperatures as low as -50°C (-58°F) are available. These products feature polymer materials designed and specified to operate down to those temperatures. To combat electronic malfunction under these conditions, the manufacturer may specify that power should remain applied to the electronics at temperatures below -40°C (-40°F). The heat dissipation of the electronics themselves keeps them within normal operating limits.

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Vibration Vibration is also ever-present from motors, rollers, and moving materials. When combined with heat or thermal shock due to ambient temperature fluctuations, vibration can rapidly age electronic components and internal connections, leading to early failure. Vibration mitigation can be difficult or infeasible for cylinder position sensors in most cases, so enhanced sensor design is required to improve sensor survivability. An effective design and testing methodology for achieving sensors that can withstand extended exposure to vibration is called highly accelerated lifetime test (HALT). Products undergoing HALT are subjected to accelerated aging during their development to uncover weak points early so they can be eliminated. The samples are subjected to ultralow and ultrahigh temperature soaking as well as rapid, extreme temperature changes. (Continued on page 14)

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(Continued from page 13) The temperature range can extend from -100°C to 200°C (-148°F to 392°F) with a temperature gradient of up to 70 K/min. This is followed by vibration testing up to maximum levels of 50 g. The repeated cycles of temperature and vibration testing are continued until the test unit finally reaches its limit and is destroyed. The goal of the design team is to extend as long as possible the period of time until ultimate failure. Look for sensors with excellent specifications for vibration. EN 60068-2-6 provides a good standard of 20 g with a frequency sweep from 10-2000 Hz.

to liquids can’t be avoided, select sensors with enclosures rated IP68 or IP69K. IP68 means that the manufacturer is offering immersion protection beyond IP67 (consult the manufacturer for details) and IP69K means that the sensor is rated for pressurized washdown. Some sensor housings are hermetically welded to provide a 100% leak-proof assembly. Be aware that the cable can represent a vulnerability to liquid ingress. If the cable jacket is cut and liquid enters, it can travel inside the cable and end up inside the sensor housing. It’s a good idea to install protective tubing over the cable to ensure it remains intact in case of incidental contact or abrasion in the application.

Liquid ingress

Corrosion

Applications in which there is continuous exposure to liquids like water, coolants, and oils eventually overcome the housing gaskets found on many standard sensors rated to IP67 or less. Standard sensors are designed to withstand liquid ingress for a specified period of time but are not rated for constant immersion or exposure to wet conditions. As always, mitigation is an effective approach; try to locate sensors out of wet areas or provide effective splash guards. When exposure

Chemicals present in the environment can lead to problems with sensor-housing corrosion. In particular, aluminum housing components can be compromised by prolonged exposure to corrosive elements. Common examples include road salt, seawater, airborne sea spray, acidic food juices, and other contaminants. The recommended housing material to alleviate most concerns about corrosion is stainless steel. For typical conditions, standard grades 303 and 304 stainless steel are sufficient.

In areas where salt or seawater is present, stainless steel grade 316L is preferred. Standard stainless steel is protected from corrosion by a thin layer of chromium III oxide that forms at the surface. Some ions, such as chloride ions found in salt water, can strip away this protective layer and expose the underlying steel to corrosion. 316L stainless steel is formulated to resist attack by chloride ions.

High cylinder pressure Standard cylinder position sensors are rated for hydraulic pressures up to 600 bar (8,700 psi). However, some compact mobile hydraulic systems run at higher pressures, as do heavy-lift systems used, for example, in civil engineering projects. Special high-pressure cylinder sensors are available rated up to 1,000 bar (14,500 psi) to ensure that the system remains safe and operational at elevated system pressures. Be aware that the outer diameter of the pressure tube surrounding the sensor element is necessarily larger – 12.7 mm (0.5 inches) vs. standard units at 10.2 mm (0.40 inches) – so it requires a larger gun drill in the cylinder rod and a target magnet with a larger inside diameter. Also keep in mind that the hydraulic port size of

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M22x1.5 is larger vs. the standard ¾-inch16 UNF or M18x1.5. These differences in mechanical details need to be communicated to the cylinder manufacturer at the time of specifying and ordering, or a standard sensor-prepped cylinder might be delivered.

Redundancy and rapid repair To combat the consequences of sensor failure in critical applications, system designers have a number of options to help keep operations running despite a failure or to get failed units back online quickly. Here are three common approaches: 1. Redundant sensors. Redundant sensors means having two, three, or even four sensors operating in parallel. When one sensor fails,

the others continue functioning. In schemes involving more than two sensors, voting in the controller can be used to determine which sensors are delivering valid position signals and which one is out of synch with the others. Typically the primary sensor will be installed in the cylinder. The redundant sensor or sensors will be installed externally in mounting brackets designed to hold the sensor in place and connect the target magnet to the moving portion of the equipment. 2. Redundant sensor outputs. Specialized cylinder position sensors are available with two or three entirely separate position sensors contained in a single sensor enclosure. Each sensor has its own separate connector. The advantage is ease of installation in a standard cylinder port without the need for external mounting and brackets. A potential disadvantage is its dependency on a single target magnet inside the cylinder and a single pressure tube. Loss of the magnet or damage to the pressure tube can take down all channels together; in practice, however, such an occurrence is unlikely. 3. Rapid repair. Sensors are available with modular electronics that can be quickly removed in the event of failure while leaving the pressure

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tube in place in the cylinder. This eliminates safety concerns such as hot oil spilling onto maintenance workers, loss of hydraulic fluid and resulting clean-up or containment, and the potential for entry of contaminants and air into the hydraulic system. Another rapid repair strategy: Install a sensor with redundant outputs, but connect only one channel for control. Cover the unused connectors to protect them while not in use. If the control channel fails, simply move the connecting cable to another channel and resume operations. The sensor unit with a failed channel can then be scheduled for removal and repair during a planned preventive maintenance Don’t put up with premature or frequent cylinder sensor failure due to adverse conditions in industrial applications. Careful attention to the application conditions and matching the sensor to application requirements can alleviate or eliminate issues related to sensors failing. When failures do eventually occur, strategies such as redundant sensors, sensors with redundant outputs, or sensors designed for rapid repair can help maintain production uptime. 

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COVER STORY

MAKING INSTALLATION EASIER WITH

HYDRAULIC PLUG-IN CONNECTIONS

j Completed assembly installation using hydraulic plug-in connections. j Hydraulic plug-in connectors.

By CEJN

Construction and agricultural equipment OEMs continually seek manufacturing improvements and efficiencies for their products and assembly lines. Areas of opportunity receiving additional attention include hydraulic systems, especially those with hose and tube assemblies. OEMs have been challenging their hydraulic-component manufacturers to help reduce hydraulic hose and tube assembly installation problems, including production line rework concerns and time loss. With traditional hydraulic assemblies, hydraulic plug-in technology benefits OEM partners with both hydraulic hose and tube assemblies.

Hydraulic hose and tube design Before a production employee installs hydraulic components, the OEM’s engineer team completes quite a bit of design work. During overall machine review, engineers focus on the hydraulic hose and tube assemblies for a particular machine build. The engineers identify system needs and calculate component demands. Once a parameter is set on the hose and tube assemblies, they use the STAMPED formula as a guideline: Size Temperature Application Media/Material

As new equipment and model updates occur, OEM engineers address concerns about hose and tube assembly routing in more limited space areas. It’s these difficult-to-work areas that create installation issues for the production line team. Here are two situational examples.

Scenario 1: Rework obstacles Multiple hydraulic hose and tube assemblies in a series with little distance between the end connection fittings, allowing a connection to be incorrectly torqued (under or over).

Pressure Ends

Inner diameter (ID), outer diameter (OD) length. Media conveyed and the environmental temperatures. Conditions in which the assembly will be used. Fluid being conveyed, compatibility, and the flow conditions (continuous or intermittent). Maximum operating pressures, including pressure spikes. Attached end fittings and types, including quick-connect needs.

Delivery

Timing, assembly, testing, certifications, packaging, and so on.

16

FEBRUARY 2021

j Screw fittings incorrectly assembled resulting in hydraulic oil leakages.

In an ideal installation setting, the production employee first reviews the installation instructions. Each connection point is properly marked, and the assembler has the right tools and the needed installation values. The assembler begins installing in a series (logically from left to right when using a typical right-handed thread). The equipment then continues to the next steps of production and eventually arrives for quality testing and inspections. If one or WWW.FLUIDPOWERJOURNAL.COM • WWW.IFPS.ORG


more of the hydraulic assemblies in the series is leaking hydraulic oil due to incorrect torque values, the equipment is tagged for rework. The problem and major frustrater is when a hose or tube assembly leaks within the series, especially when it is near the middle of the series. This requires removing all installed assemblies within the confined series prior to this leakage point, essentially reversing the work completed in the installation process. Once the corrections are made, the technician must reinstall the remaining hydraulic assemblies. Quality testing and inspections are performed again, and the passed equipment proceeds further in the production process. Additional failure situations would repeat the above process.

Scenario 2: Time-loss obstacles Hydraulic hose and tube assemblies in confined work spaces, causing time loss and productivity issues. As in scenario 1, the production employee reviews instructions and has the proper tools and installation values. In this case, the assembler is installing hose and tube assemblies with little clearance for both the installer and the required tools. In areas of confinement – e.g., a front-loader cab being positioned and installed over the chassis – when the worker is ready to move forward, the installer uses the available space and mates up the appropriate assemblies to begin the installation process. This limited space causes time delays as the installer adjusts to gain better positioning and leverage to achieve a good connection fit. Eventually, the installer completes the work, allowing the equipment to progress further down the production line, including quality testing and reviews.

Hydraulic plug-in connections Hydraulic plug-in connection technology has been in use for more than 25 years with millions of units in the field. There are several versions of the connections produced by a handful of major manufacturers. While each version is slightly different, the core operational concept is the same. Instead of relying on an operator to wrench together two threaded connection ends at final assembly, a hydraulic plug-in connector allows the installation of components at subassembly points, eliminating problems experienced later in the assembly process. An installer plugs in the components at final assembly and then validates a complete connection with a simple pull test. No extra tools are required. To better understand the operation of the connections, let's review an example, the WEO Plug-In system. As with all hydraulic plug-in connections, these consist of a male half (nipple/plug) and a female half (coupling/socket). The male half is predominately crimped onto the hydraulic hose assembly while the female half is installed on the connecting hydraulic components (i.e., valves, manifolds, and so on). Both halves are also available in common thread options to allow integration benefits without the need to modify an existing core system. Additionally, they are designed to offer the greatest flow values for each respective size and are acceptable for most pressure rating demands in OEM equipment, considering necessary safety factors.

a Male half of a WEO nipple.

j Front loader cab being installed on the chassis.

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a Female half of a WEO coupling.

It is simple for a production line operator to install a hydraulic plug-in connection once the engineer implements it within the equipment design. The WEO nipple and WEO coupling (or WEO cartridge) are individually installed to their respective hydraulic subcomponents. Logically, this is completed at a production step where there is more room for an operator to work. The WEO nipple, with the assembly stop clip attached, is inserted into the corresponding WEO coupling or cartridge until it firmly stops. The installer then performs a simple pull test to ensure the assembly has been properly connected. (Continued on page 18)

j The installation steps of hydraulic plug-in connections.

a Female half of a WEO cartridge.

d Completed front loader.

FEBRUARY 2021

17


d Cross-cut image of an assembled WEO hydraulic plug-in connection.

primary with an additional backup ring). If the assembly ever needed to be disassembled – for example, for a hose replacement in the field – all that is needed is a flat-head screwdriver. (Continued from page 17) Simply remove the assembly stop clip, push The cross-cut image above shows a fully in the nipple until it stops, and then pull the assembled hydraulic plug-in connection. hose assembly out. Once installed, the assembly is held in place by a system that fits and locks into the respec- Hydraulic plug-in value creation tive nipple. The image also shows that this The ideal placement for hydraulic plug-in connection is completely sealed (typical NBR connections occurs when trying to eliminate

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assembly and installation concerns. Earlier, we explored two scenarios that resulted in costly reworks and time loss. A hydraulic plug-in connection used in those cases would create appropriate value and justification for integration with an OEM's next equipment build or redesign. One advantage of a plug-in connection is elimination of torque issues during hydraulic hose and tube assembly installation. If the standard operating procedure is to just plug the fitting in and perform a simple pull test, it eliminates the need for calibrated tooling and access issues when dealing with series installation. The result is an overall reduction in rework concerns. Another benefit is the reduction of production time lost during installation of assemblies in difficult-to-reach areas. A production line worker will not have to continually reposition or struggle with tooling in tight spaces. Less struggling equals greater productivity and output, resulting in less time lost. An added benefit is a reduction of embedded tension on hydraulic hoses. With plug-in connectors, a hydraulic hose finds its natural relaxed position. In cases where embedded tension on the hydraulic hose causes premature failure, elimination of the tension problem helps increase the life of the hose assembly, especially if it is the same assembly on multiple production machines causing warranty concerns. Equipment OEMs seeking manufacturing improvements and efficiencies on their products and assembly lines creates opportunities for improvements to hydraulic hoses and tube hybrid assemblies. In the right areas, a hydraulic plug-in connection brings significant time and cost savings by eliminating the installation issues discussed above. 

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Each year, the International Fluid Power Society and the Fluid Power Journal hold a photo contest in honor of Fluid Power Professionals' Day. This year’s contest runs from January 1 through March 31, 2021.

THE 2021 PHOTO CATEGORIES ARE: 1. New Perspective: Photos that show fluid power used in novel ways and from views not everyone gets to see. 2. Teamwork: Photos of people working together using fluid power to get the job done. 3. Fun with Fluid Power: Photos that capture the joy of using, learning, or teaching about hydraulics and pneumatics. 4. The Muscle of Fluid Power: Photos that capture the might of fluid power in action. A panel of judges will announce three winners for each category on June 19, 2021 in celebration of Fluid Power Professional’s Day. A People’s Choice recognition will be awarded based on votes during a one-week voting period.

Submit your fluid power photo by visiting www.fluidpowerjournal.com to take your shot at recognition in the Fluid Power Journal.

CELEBRATING 60 YEARS

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SYSTEMS INTEGRATOR 2021 DIRECTORY Check out the online matrix at www.fluidpowerjournal.com

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Your Needs. Our Expertise. Controlled Fluids, Inc. is your trusted partner for fluid power, motion control, and reliability solutions across all industries.

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AIR TEASER

New Problem

Solution to December 2020 problem: Comparing an Electrical Circuit to a Pneumatic Circuit

Blocked Ports and Failing Piston Seals By Ernie Parker CFPAI, CFPSD, CFPS, CFPMM, CFPMT, CFPMIP, CFPMMH, CFPMIH, CFPE

HITACHI FLANGELOCK AND CAP KITS AVAILABLE

»

YOU MAY HAVE seen the brain teaser in your lifetime in which a hydraulic cylinder is halfway extended with a load on it. Both ports are blocked and the piston seal fails. The question is, What will the rod do with a load on it? The answer is that, other than a slight compressibility of the oil, it will stay in place. Part number But now the load is supported only by the area of the rod, and the pressure SWINGFLGLCK2062 may skyrocket. This problem is going to be the same scenario, but with air. SWINGCAP2062 Given: 2” x 8” x 1” cylinder or 50 mm x 200 mm x 25 mm Load is 314 lbs. or 1374 N Rod is extended upward at 4 inches or 6.35 mm Both ports are blocked with no leakage

314 lbs.

TRAVELFLGLCK2462 TRAVELCAP2462

TM

Part description

Applicable machines TM

Swing hose FlangeLock kit

EX3600, EX5600, EX8000

Swing circuit cap kit

EX3600, EX5600, EX8000

Number of parts

Weight (kg) TM

16 x 2062U - red FlangeLock

6.7

16 x 2062 - cap

4.5

Many times, when we try to come up with a pneumatic EX3600, EX5600, EX8000 16 x 2462U - purple FlangeLock 7.7 Travel hose FlangeLock kit circuit to do a particular job, we may think an electrical circuit would do the same job.- cap Is there any 6.4 Travel circuit cap kit EX3600, EX5600, EX8000 16 x 2462 problem with the pneumatic circuit if so, what is 14 x 3262U -and black FlangeLock Front attachment EX3600, EX5600, EX8000 8.9 4 x 3261U - black & silver FlangeLock FlangeLock kit the problem? 14 x 3262 - cap Front attachment cap kit EX3600, EX5600, EX8000isn't a way to vent the pilot 9.5 The answer is that there 4 x 3261 - cap Boomport arch hose when any of the valves are released. EX3600, EX5600, EX8000 20 x 3262U - black FlangeLock 9.9 TM

TM

TM

FRONTATTFLGLCK326162 FRONTATTCAP326162 BOOMARCHFLGLCK3262

Questions: BOOMARCHCAP3262 What will the cylinder rod do if the piston seal starts to leak internally? What is the resting pressure inside the cylinder when the piston stops moving?

TM

TM

TM

FlangeLockTM kit

Boom arch hose cap kit

CONTAMINATION CONTROL

EX3600, EX5600, EX8000 20 x 3262 - cap Visit www.fluidpowerjournal.com/air-teaser to view previous problems.

11.3

Routine and scheduled maintenance of hydraulic systems are vital to getting the most out of your Hitachi Mining Excavator. While maintenance plays the largest role in the prevention of unnecessary machine downtime, it can also expose the hydraulic system to high levels of contamination rapidly decreasing component longevity. The importance of contamination control is sometimes overlooked when performing maintenance due to incorrect practices being used.

CO U T CO NTA LTIM HE NT M A RO INA TE L T TI OO ON L

Stop the Mess

THE FLANGELOCK™ TOOL AND CIRCUIT BLANKING CAPS

The FlangeLock™ tool and caps are the ultimate contamination control tools for protecting your hydraulic system. The FlangeLock™ allows for the simple sealing of open hydraulic flanges without tools while the caps can be bolted in place of a flange connection. Easy on, easy off, they offer a leak-proof solution to hydraulic systems and environmental cleanliness. FlangeLock™ tools and caps stop the mess.

The FlangeLock™ Tool is the ultimate contamination control tool for protecting HITACHI MAKING systems. CONTAMINATION CONTROL EASY sealing of open SAE code 61, 62 your hydraulic It allows for the simple Hitachi have packaged FlangeLock™ tool and caps specifically for Hitachi mining excavators. The Hitachi customised & make CAT-Style hydraulic without Constructed from lightweight aluminum. kits sure no matter whichflanges component routine tools. maintenance is being performed on, you will always have the exact Easyofon, easy off.™*Offers to hydraulic system and environmental number FlangeLocks and capsatoleakproof help reducesolution contamination. cleanliness. FlangeLock™ Tools stop the mess! ™ *Note: FlangeLocks are not to be used under pressure

SAVE SAVE SAVE SAVE

TIME MONEY LABOR OIL

• No tools required • One hand installation Call you local Hitachi Muswellbrook representative or • No expensive hardware needed • Eliminate hydraulic oil spills & clean up onhoses 02 6541 6300 for installation more information. • No more the rags branch stuffed into • Quick & ease of usage • No more messy plastic caps • Safe for personnel & environment • The ultimate contamination control tool • Industry acclaimed

This product is Patented, other Patents pending.

For more information, call 203-861-9400 or email sales@flangelock.com. www.flangelock.com WWW.IFPS.ORG • WWW.FLUIDPOWERJOURNAL.COM

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M

KEEPING COOL with Hydraulic Case Drains By Tommy Williams, AKG Thermal Systems

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FEBRUARY 2021

aintaining a proper oil temperature in a hydraulic system is essential for successful operation. High fluid temperatures can damage components and significantly alter the way the system performs, resulting in costly repairs and downtime. Hot oil increases internal leakage in pumps and motors, causing the machine to operate more slowly. O-rings also harden at higher temperatures, leading to more leaks in the system. Further, hydraulic fluid temperatures above 180°F (82°C) may damage seal compounds and accelerate oil degradation. Operators should avoid running a hydraulic system at temperatures above 180°F (82°C). Every 18-degree increase in temperature above 140°F cuts the oil life in half. Systems that operate at high temperatures can produce sludge and varnish, which result in damage to the hydraulic system and reduce efficiencies. Pressure-compensating piston pumps are commonly used in industrial hydraulic systems. Tolerances allow for a small amount of oil to bypass, generating heat that is piped back to the reservoir through the case-drain line. There are many other reasons for heat generation in hydraulic components and systems; the inefficiency of the pump, friction in the pipes, joints, line fittings, and so on, are the major contributors to heat generation. The tolerances inside pumps and valves are generally in the ten-thousandths of an inch. These tolerances permit a small amount of oil to continuously bypass the internal components, causing the fluid temperature to rise. Flow controls, proportional valves, and servo valves control the oil's flow rate by causing a pressure loss as oil flows through the valves. This means that higher pressure exists at the valve's inlet port than at its outlet port. Anytime oil flows from higher pressure to a lower pressure, it generates heat that is absorbed in the oil. Gear pumps and motors generally do not have a case drain, but they do have internal leakage, which, by design, usually accumulates in a small cavity just before the shaft seal. Aircooled compact heat exchangers can be used to address the heat in this area. The case-drain line's average flow rate is 1% to 3% of the maximum pump flow. Anyone can permanently install a flow meter in the case-drain line to monitor flow rates. When a pump or motor is worn or damaged, internal leakage increases, and therefore the flow available to do useful work decreases. This means the hydraulic efficiency decreases and results in additional heat generation. Implementing a case-drain cooler can assist with maintaining a viscosity range at which your machinery operates most efficiently. System checks should be performed regularly to determine the amount of bypassing oil. The pump should be changed when the oil flow reaches 10% of the pump volume. By introducing cooling into the system, efficiencies increase, resulting in less power consumption. In some cases, high oil temperatures that are reflected back to the drive motor lead to wasted electricity by forcing the drive motor to pull more current to operate the system. Case-drain coolers are ideal for applications requiring minimal cooling. These coolers offer minimal mounting to reduce installation time, typically with horizontal and vertical options.

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RULES OF THUMB FOR SIZING A HEAT EXCHANGER

The cooler mounts behind an existing TEFC motor, using the electric-motor fan airflow. Cooling flow is obtained from the case drain of the hydraulic machine and is then guided to the integrated electrical machine. Pressure losses in the electrical machine can be seen as negligible. The outgoing flow moves by gravity as long as the outlet and inlet ports are of adequate dimensions. To select the best air oil cooler, provide as much information about the application as possible, including oil heat load in Btu/h or hp, oil flow rate in gpm, maximum inlet oil, and maximum ambient air temperatures during operation. If the required heat dissipation is unknown, it can be estimated that 20% to 30% of the installed horsepower will be converted into heat load. Heat exchangers can be used to remove the excess heat in a hydraulic system. The implementation of a heat exchanger has many variables that need to be taken into account. The heat exchanger and reservoir should be sized when a system is initially designed to remove the generated heat. The reservoir allows some of the heat to dissipate through the walls to the atmosphere. The heat exchangers should be sized to remove the balance of the heat. The heat exchanger needs to be maintained to ensure excess heat is removed. If an air-type heat exchanger is used, the cooler fins should be cleaned regularly. A degreaser may be necessary to clean the fins. To ensure profitable and efficient success in a manufacturing operation, maintaining a proper oil temperature in all hydraulic systems is essential. High hydraulic-fluid temperatures can damage system components and significantly alter the way a hydraulic system performs, resulting in costly repairs and downtime. 

• SIMPLE CIRCUIT WITH FLUID MOTORS: 31% • HYDROSTATIC TRANSMISSIONS: 35%-40% • SIMPLE CIRCUIT WITH MINIMAL VALVES: 25% • SERVO-BASED SYSTEMS: 60%-75% • SIMPLE CIRCUIT WITH CYLINDERS: 28%

TYPICAL CASE-DRAIN APPLICATIONS • GEAR BOXES • HYDRAULIC PRESSES • HYDRAULIC TOOLS • HYDRAULIC POWER UNITS • INDUSTRIAL COMPACTORS • ASSEMBLY CONVEYORS

AKG Case-Drain Series

CD SERIES FEATURES: • tube and fin aluminum design, • coolers mount onto the rear of TEFC frame motors, • competitive pricing and assembles from stock, • premium-quality lightweight construction.

Heat Rejection, (hp) 50 Deg ETD

The AKG Thermal Systems CD Series is designed to fit in tight spaces and conveniently mounts onto a power unit with limited envelope space. The narrow profile conditions and cools for a wide array of applications. The AKG tube fin design offers more significant cooling and does not require a protective guard. This cooling setup is compact, low cost, and low flow with minimal heat removal.

CD SERIES

Oil Flow (gpm) Performance for AKG CD Series oil coolers. Listed performance curves are based on ISO VG 32 oil @ 50F ETD.

For more information, visit www.akgts.com. 

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TEST YOUR SKILLS

S E L ES C ETLSC EEO CS LT M EP L C CO ET OC NM C TEP O NC O M TO NP SM EOF N PNO O TER S N NET P FN S O NTE RFSUO PM FR NO AT EPRUN IM C PEN AT USEM YIUS AT CMTS AT IEC YMS IS CTYES SM YTSE T MESM

S E L E C T C O M P O N E N T S F O R P N E U M AT I C S Y S T E M S Outcome Outcome Outcome Outcome 3.18:3.18: Calculate 3.18: 3.18: Calculate Calculate Calculate the kinetic thethe kinetic the kinetic energy kinetic energy energy required energy required required to required stop to stop to a to load stop a stop load a load a loa with with a shock with a with shock aabsorber. shock a shock absorber. absorber. absorber.

Outcome 3.18: the kinetic energy required to a load It is common It Calculate is It common is practice Itcommon is common practice to position practice practice to position shock to position to position absorbers shock shock absorbers shock to absorbers cushion absorbers to cushion loads to cushion to attached cushion loadsloads attached loads to attached airstop attached cylinders, to airtocylinders, air torather cylinders, air cylinders, than rather rather than rather than th to subject to subject the to subject air to the subject cylinders air thecylinders the air tocylinders air shock cylinders to loading shock to shock toloading resulting shock loading resulting loading from resulting end resulting from offrom end stoke from of end impact. stoke end of stoke of impact. This stoke impact. practice impact. This This practice allows This practice practice sizing allows allows sizing allows sizing siz with a shock absorber. the cylinder the cylinder the to the move cylinder cylinder to the move toload move tothe move and load the sizing the load andload the sizing andshock and sizing the sizing absorber shock the the shock absorber shock toabsorber stop absorber to moving stop to stop moving tothe stop moving load moving the within load the the load the within load required within the within required the the required requir

CALCULATE THE

distance. distance. distance. distance.

KINETIC ENERGY

It is common practice to position shock absorbers to cushion loads attached to air cylinders, rather than motion In motion control, In In control, motion it iscontrol, desirable control, it is it desirable isthat it desirable is a desirable that load that abeload that stopped a load be aofload stopped be smoothly be stopped stopped smoothly and smoothly without smoothly and practice without and adverse and without without adverse effects adverse adverse effects sucheffects assuch effects such as such as to subject the airIncylinders tomotion shock loading resulting from end stoke impact. This allows sizing rebounding rebounding rebounding or stress rebounding or upon stress or stress or the upon stress machine upon theupon machine the orabsorber the machine partmachine or itself. part or The itself. or part speed itself. The itself. of speed The the The speed object ofspeed the ofand object the ofwithin the mass object and object of mass and the and mass object of mass the ofwill object the of the object will object will w the cylinder to move the load and sizing the shock topart stop moving the load the required determine determine the determine best determine the method best the the best method tobest accomplish method method to accomplish to accomplish this. to accomplish Previously this. this. Previously this. inPreviously other Previously inoutcomes other in other in outcomes other there outcomes outcomes has there been there has there discussion been hashas been discussion been of discussion discussion of of distance.

deceleration deceleration deceleration of deceleration a cylinder of a of cylinder using aofcylinder a internal cylinder usingusing internal cushions using internal internal cushions or cushions reducing cushions or reducing or end reducing orofreducing end stroke of end stroke impact end of stroke ofimpact by stroke aimpact polymer by impact a by polymer bumper a bypolymer a polymer bumper bumper bump within the within cylinder. within the within cylinder. the When the cylinder. cylinder. the When velocity, When the When velocity, the mass, the velocity, velocity, and mass, amass, potential and mass, aand potential and applied a potential a potential applied force applied isforce applied high, force isthen force high, isahigh, then is shock high, then a absorber shock then a shock aabsorber shock absorber absorb In motion control, it is that amethod be stopped without adverse effects such as is the best is desirable the method is best the is the best method to best control method toload control the to control deceleration to the control deceleration the the deceleration ofsmoothly deceleration the object of the ofand object without the of the object without object adverse without without adverse effects. adverse adverse effects. The effects. kinetic effects. The kinetic The energy The kinetic energy kinetic energy ener rebounding or stress the machine itself. The speed of consideration the object and mass of object will that isupon that dissipated that is dissipated that is isdissipated isconverted dissipated is or converted ispart into converted is converted heat intoenergy, heat into into heat energy, soheat consideration energy, energy, so so consideration so must consideration be must given must bemust to given be thethe given becycle togiven the torate cycle the to of the cycle rate the cycle rate of the rate of the of t determine the best method to accomplish Previously other outcomes there has been discussion of shock absorber shock shock absorber shock to absorber ensure absorber to ensure that to this. ensure this to that ensure heat that this isthat heat this taken this heat isin into taken heat isaccount. taken isinto taken account. intointo account. account.

deceleration of a cylinder using internal cushions or reducing end of stroke impact by a polymer bumper It is common practice to When position shock absorbers totwo cushion loads attached to–or air cylinders, rather than to subject the airand cylinders to shock Shock absorbers Shock Shock absorbers Shock fall absorbers within absorbers fallmass, within fall fall categories within two within two two –categories fixed categories fixed adjustable. – fixed or – fixed adjustable. or adjustable. or Fixed and Fixed adjustable Fixed and Fixed and shock adjustable adjustable absorbers shock shock absorbers shock absorbers absorb within the cylinder. the velocity, and acategories potential applied force isadjustable. high, then a adjustable shock absorber operateoperate on operate similar operate on This principles on practice on similar principles similar to principles a principles cylinder to a the cylinder cushion atocylinder a cylinder cushion inmove that cushion cushion inthe that in deceleration that the in sizing that deceleration the the deceleration isshock deceleration accomplished is accomplished is to accomplished isstop by accomplished forcing by forcing by by forcing forc loading from end-of-stroke impact. allows sizing cylinder to the load and the absorber moving is the resulting best method to control thesimilar deceleration of theto object without adverse effects. The kinetic energy internalinternal fluid internal over internal fluidafluid over series fluid over a of series over a restrictions series aofseries restrictions of restrictions to of restrictions provide to provide the to provide to damping provide the damping the force the damping damping required. forceforce required. force The required. required. advantage The advantage TheThe of advantage the advantage of the of the of t the load the required distance. that iswithin dissipated is converted into heat so consideration must given to cycle rate of the adjustable adjustable shock adjustable adjustable absorber shock shock absorber shock isenergy, absorber that absorber is the that unit is that the iscan that unit the bethe can unit fine-tuned unit be canfine-tuned can be over fine-tuned be be fine-tuned a over variety over a variety of over athe mass variety a of variety and mass ofvelocity mass of and mass velocity and profiles and velocity velocity profiles profiles profi In motion control, itto is ensure desirable that atoload be stopped smoothly and without adverse effects such rebounding orastress upon the machine shock absorber that this heat is taken into account. to achieve to achieve precise to achieve achieve precise deceleration. precise deceleration. precise deceleration. Another deceleration. Another possible Another Another possible variant possible possible variant is tovariant use is variant to aasself-adjusting is use to isause toself-adjusting use a self-adjusting shock self-adjusting absorber shock shock absorber shock which absorber absorber which which wh

or part itself. The speed the is object mass of object determine the best method totypically accomplish the mass, and ait so isofnow offered now isand now offered isby now several offered offered bythe several manufacturers. by several by will several manufacturers. manufacturers. manufacturers. These typically These These typically These have a typically have range have a this. of range have self-adjustment, aWhen range aofrange self-adjustment, ofvelocity, self-adjustment, of self-adjustment, so it is so still is it still soisit still is s important and important necessary important and necessary andand to necessary calculate necessary calculate to the calculate to amount calculate the amount the of kinetic the amount of kinetic energy of kinetic ofand energy to kinetic energy absorbed. to energy betoabsorbed. be toshock absorbed. be adverse absorbed. Shock absorbers fall within categories –to fixed or adjustable. Fixed adjustable absorbers potential applied forceimportant is high, then atwo shock absorber is the best method to control theamount deceleration ofbe the object without effects. The operate onthat similar principles to ainto cylinder cushion in that must the bedeceleration is rate accomplished by forcing kinetic energy is dissipated is converted heat energy, so consideration given to the cycle of the shock absorber to ensure Safety Safety tip: Safety Energy tip: Safety Energy tip: that tip: Energy isto that Energy input that istoinput that the is input is shock to input the toabsorber shock the to the shock absorber shock should absorber absorber should be accurately should be should accurately be calculated accurately be accurately calculated ifcalculated possible calculated if possible or if possible if possible or or internal fluid over a series of restrictions provide the damping force required. The advantage of the that this heat is taken into account. conservatively conservatively conservatively conservatively estimated estimated for estimated worst estimated for case. worst for for worst Acase. shock worst case. A case. absorber shock A shock Aabsorber shock that absorber “bottoms-out” absorber that that “bottoms-out” that “bottoms-out” “bottoms-out” in service in service because in service in because service of because because of of adjustable shock absorber is insufficient that the unit can be fine-tuned over amounting variety of mass and velocity profiles Shock absorbers fall within two categories – fixed or adjustable. Fixed and adjustable shock absorbers operate on similar principles a cylinder insufficient insufficient energy insufficient energy capacity energy capacity energy willcapacity cause capacity will the cause will mounting will cause the cause mounting the structure the mounting structure ofstructure the structure of shock the of to shock the ofabsorb the shock toshock absorb the to to absorb remaining to the absorb remaining the the remaining remain to achieve deceleration. Another possible variant tomounting use amounting self-adjusting shock absorber which energy. Thisenergy. may energy. This result This may This in may result damage may result inresult damage to in the damage inover damage mounting to ais the to the to structure, the mounting structure, the structure, shock, structure, the the shock, or the both. the shock, or shock, both. or both. orrequired. both. cushion in thatprecise the deceleration is energy. accomplished by forcing internal fluid series of restrictions to provide damping force The is now offered by several manufacturers. These typically have a range of self-adjustment, so it is still advantage of the adjustable shock absorber is that the unit can be fine-tuned over a variety of mass and velocity profiles to achieve precise decelimportant and necessary to calculate the amount of kinetic energy to be absorbed. eration. Another possible variant is to use a self-adjusting shock absorber, now offered by several manufacturers. These typically have a range of self-adjustment, so it is still important and necessary to calculate the amount of kinetic energy to be absorbed. Safety tip: Energy that is input to the shock absorber should be accurately calculated if possible or SAFETY TIP: Energy that is inputfor to the shockcase. absorber be absorber accurately calculated if possible or conservatively estimated for worst conservatively estimated worst A should shock that “bottoms-out” in service because of case. A shock energy absorber that “bottoms-out” in servicethe because of insufficient energy capacity cause the of the shock insufficient capacity will cause mounting structure of thewill shock to mounting absorb structure the remaining to absorbThis the remaining energy.in This may result damage to the mounting structure, theshock, shock, oror both. energy. may result damage tointhe mounting structure, the both.

Fig. 3-30 Fig. Shock 3-30 Fig.Absorber Fig. 3-30 Shock 3-30 Shock Absorber Applications Shock Absorber Absorber Applications Applications Applications

The typical The typical characteristics TheThe typical characteristics typical characteristics of characteristics stopping of stopping of a load stopping of stopping fit a load into a load fit one a into load fit of one the into fit into six of one the scenarios one ofsix the of scenarios the sixshown six scenarios scenarios shown in Fig. shown 3-30: inshown Fig. in 3-30: Fig. in Fig. 3-30: 3-30: There are There four There are There (4)four are parameters are four (4) four parameters (4) (4) parameters that parameters must thatbe that must known that must bemust known tobeaccurately known beto known accurately to size accurately to accurately a shock size a size absorber. shock size a shock aabsorber. shock absorber. absorber. • Mass• toMass be • decelerated Mass •to Mass betodecelerated be to decelerated be m decelerated (kg) m or (kg) (slugs) m (kg) or m (slugs) (kg) or (slugs) or (slugs) • Impact • Impact velocity • Impact • velocity Impact v (m/s) velocity velocity vor (m/s) (ft./sec.) v (m/s) or v (m/s) (ft./sec.) or (ft./sec.) or (ft./sec.) • Propelling • Propelling • orPropelling •driving Propelling or driving force or driving orF force driving (N) or force F (lb.) (N) force For (N) F(lb.) (N) or (lb.) or (lb.) • Number • Number of • impact Number • Number of impact cycles of impact ofper cycles impact hour cycles per cycles C hour (#/hr) per per hour C (#/hr) hour C (#/hr) C (#/hr)

Figure 1: applications. The total The energy total TheThe total energy that total energy willthat energy need that will to that need will beShock will absorbed need to absorber need betoabsorbed be to by absorbed be the absorbed by shock the byabsorber shock the by the shock absorber shock is absorber theabsorber sum is the of issum the the is the kinetic sum of the sum ofkinetic energy the of the kinetic energy and kinetic energy and energy anda Fig. Shock Absorber any applied any applied any force any applied orforce applied thrust force or 3-30 force energy thrust or thrust orenergy imparted thrust energy imparted energy toimparted theApplications imparted object. to the toobject. the The to the object. first object. The step first The inThe first step these first step inscenarios these step in these in scenarios these isscenarios to scenarios calculate is to is calculate toiscalculate to calcul the kinetic kinetic energy the the kinetic energy of kinetic the moving of energy the ofof moving the load of moving which load moving is load which given load which isby which given isfigure given is by formula: given the by formula: the byare the formula: formula: The typical characteristics ofthe stopping a load fitenergy into one thethe six scenarios shown inthe 1. There four parameters that must be known

The typical characteristics of stopping a load fit into one of the six scenarios shown in Fig. 3-30:

to accurately a shock absorber: There are size four (4) parameters that must be known to accurately size a shock absorber. Kinetic Kinetic Kinetic energy Kinetic energy energy E = Kinetic E = energy Kinetic E = EKinetic = energy Kinetic energy energy Eft.-lb. = ft.-lb. Kinetic E = energy Kinetic E = EKinetic =Joules energy Kinetic energy energy ft.-lb. ft.-lb. Joules Joules Joules energy • Mass to be to decelerated m (kg) or (slugs) • Mass be decelerated m or (slugs) of an object of an of object an ofobject an object 1 (kg) 1 21 2 2 Eq. Eq. Eq. Eq. 1 2 to be dissito be to dissibe todissibe dissimass m= mass m =mmass = mass m= m =mmass = mass E =3.22 •(ft./sec.) Emv = E •=E mv =•m mv •= mv • Impact velocity v (m/s) orv(ft./sec.) slugs slugs slugsslugsm = mass kgmass kg kg kg • Impact velocity (m/s) or 3.22 3.22 3.22 pated bypated shock pated by pated shock by shock by shock 2 2 2 2 • Propelling or driving F (N)force or (lb.) F (N) or (lb.) absorber. absorber. absorber. v = velocity v = velocity v = vvelocity =ft./sec. velocity v = velocity v = velocity v = vvelocity =m/sec. velocity • Propelling or force driving ft./sec. ft./sec. ft./sec. m/sec. m/sec. m/sec. absorber. • Number of impact cycles percycles hour C (#/hr) • Number of impact per hour C (#/hr) The total energy that will need to be absorbed by the shock absorber is the sum of the kinetic energy and any applied force or thrust energy imparted to the object. that The first step in scenarios is calculate theshock kinetic energy of theis moving load asofshown in the formula above. The total energy will need toManual be absorbed by the absorber the sum the kinetic energy and Study Manual Study Study Manual •these 06/08/17 Study •Manual 06/08/17 • 06/08/17 • to 06/08/17 Pneumatic Pneumatic Specialist Pneumatic Pneumatic Specialist Certification Specialist Specialist Certification Certification • 3 Certification - 53• 3 - •533 •- 53 3

any applied force or thrust energy imparted to the object. The first step in these scenarios is to calculate WWW.FLUIDPOWERJOURNAL.COM • WWW.IFPS.ORG of the moving load which is given by the formula:

36 the FEBRUARY kinetic2021 energy


As shown in Fig. 3-30, the secnario is moving a horizontal load (by pushing with a cylinder), there is the As shown in 3-30, the secnario is a horizontal load (by pushing with a cylinder), there is the As shown in Fig. Fig. the secnario is moving moving a force horizontal load (by pushing with cylinder),there there isthe the kinetic energy of 3-30, thethe load itself plus the applied of the cylinder that must be by theisshock As shown Fig. 3-30, secnario is moving a horizontal load (by pushing with a adissipated cylinder), kinetic in energy of the the load load itself plus plus the applied applied force of of the the cylinder that must must be dissipated by the the shock shock kinetic energy of itself the force cylinder that be dissipated by absorber. kinetic energy of the load itself plus the applied force of the cylinder that must be dissipated by the shock absorber. absorber. absorber. As shown inEq. figure 1, provides the scenario is moving a the horizontal (byof pushing with cylinder). kinetic the energy of the load itself plus the 3.22 for calculating kinetic load energy the load anda Eq. 3.23 isThere used is tothe calculate applied Eq. 3.22 3.22 provides provides for for calculating calculating the the kinetic kinetic energy energy of of the the load load and and Eq. Eq. 3.23 3.23 is is used used to to calculate calculate the the applied applied Eq. force. Each ofthat these is be additive and the resulting sum is theand total energy toused be dissipated. applied force the cylinder must dissipated by the shock absorber. Eq.of 3.22 provides for calculating the kinetic energy of the load Eq. 3.23 is to calculate the applied force. force. Each Each of of these these is is additive additive and and the the resulting resulting sum sum is is the the total total energy energy to to be be dissipated. dissipated. force.onEach of thesepage is additive the resulting is the total energy bethe dissipated. The equation the previous providesand for calculating thesum kinetic energy of the load,to and equation below is used to calculate the applied Note: The shorter the shock absorber stroke, the more abrupt the deceleration force that must be taken Note: isThe The shorter the shock absorber stroke, the moretoabrupt abrupt the deceleration deceleration force force that that must must be be taken taken force. Each of these additive and the resulting sum is the total energy be dissipated. Note: shorter the shock absorber stroke, the more the up by shockthe absorber. Note: Thethe shorter shock absorber stroke, the more abrupt the deceleration force that must be taken up the absorber. up by by the shock shock absorber. NOTE: The shorter the shock absorber stroke, the more abrupt the deceleration force that must be taken up by the shock absorber. up by the shock absorber. Eq. Eq. Eq. 3.23 Eq. 3.23 3.23 3.23

AF = Thrust force ft.-lb. A AFF = = Thrust Thrust force force ft.-lb. ft.-lb. F= Applied force AF = Thrust force lb. ft.-lb. F = Applied lb. F = Applied force force lb. = Stroke length of shock F =S Applied force lb. of shock S S= = Stroke Stroke length length of shock S =absorber Stroke length of shock ft. absorber absorber ft. ft. absorber ft.

A AFF = =F F ••• S S A = F S AF = FF • S

AF = Thrust force Joules A AFF = = Thrust Thrust force force Joules Joules Applied force AF F == Thrust force N Joules F = Applied N F = Applied force force N S Applied = Strokeforce length of shock F= N of shock S S= = Stroke Stroke length length of shock absorber S= Stroke length of shock m absorber m absorber m absorber m

Applied force Applied force Applied to force energy be Applied force energy to be energy to dissipated be by energy to be dissipated by by dissipated shock dissipated by shock shock absorber. absorber. shock absorber. absorber.

The next two scenarios reflect applications where the load is being moved vertically – either up or down. The next two scenarios reflect applications where the load being moved vertically – up or The next twonot scenarios reflect applications where the load is ismass beingand moved vertically – either either or down. down. In this case, only does the kinetic energy of the moving the applied force if any,up to be The next scenarios reflect applications where load is being moved – either up need or this down. In thistwo case, not only only does the the kinetic energy of the the moving mass and the thevertically applied force ifdown. any, need tocase, be not only does the In this case, not does kinetic energy of the moving mass and applied force if any, need be The next two scenarios reflect applications where the load is being moved vertically – either up or In accounted for, but gravity is also at work. In lifting the load, gravity is acting to resist the movementtoand accounted for, but gravity is also at work. In lifting the load, gravity is acting to resist the movement and In this case, not only does the kinetic energy of the moving mass and the applied force if any, need to be accounted for, somewhat but gravity also at work. Inany, lifting the load, gravityWhen is for, acting to resist the movement and provides ofis aapplied damping effect to the applied force. the load is is hanging downward, kinetic energythus of the moving mass and the force, if need to be accounted but gravity also at work. In lifting the load, gravity thus provides a damping to force. isWhen hanging downward, accounted for, butsomewhat gravity is of also at work. effect In lifting the applied load, gravity actingthe to load resistis movement and thus provides somewhat of aaccelerate damping effect to the the applied When the load isthe hanging downward, however, gravity isand acting toprovides the load and adding force. to the applied force. is acting tothus resist the movement thus somewhat of a damping effect to the applied force. When the load however, gravity is acting acting to accelerate theto load and adding to the theWhen applied force. providesgravity somewhat of a to damping effect theand applied force. theforce. load is hanging downward,is hanging downward, however, is accelerate the load adding to applied however, gravity isaccelerate acting tothe accelerate the load to and adding toforce. the applied force. however, gravity is acting to load and adding the applied The two equations Eq.’s 3.24 and 3.25 provide for calculations to address the vertical load orientations. below provide for calculations to address Eq.’s 3.24 Eq.’s 3.24 and and 3.25 3.25 provide provide for for calculations calculations to to address address the the vertical vertical load load orientations. orientations. the vertical load orientations. Eq.’s 3.24 and 3.25 provide for calculations to address the vertical load orientations. Eforce = Applied force energy ft.-lb. Eforce = Applied force energy Joules Eforce = = Applied Applied force force energy energy ft.-lb. E Eforce = = Applied Applied force force energy energy Joules Applied force E Applied force force ft.-lb. Joules Applied F force ==Applied F force ==Applied force energy (liftN energy Eforce Applied force force lb. energy ft.-lb. Eforce Applied force F= = Applied Applied force F= = Applied Applied force energy (liftJoules Applied lb. N F force F force energy (liftforce lb. N ing a load S = Stroke length of shock absorberft. S = Stroke length of shock absorberm ing a load Eq. E F =SApplied Applied = Stroke Strokeforce lengthlb. of shock shock absorber absorberft.F = S= Stroke force length of shock absorber energy ing a load (lift-to = F •S − m• g •S Eq. vertically) = length of Eq. E ft. S = Stroke lengthNof shock absorberm m vertically) to Eforce = F •S − m• g •S S 3.24 force = F • S − m • g • S m = mass slugs m = mass kg vertically) adissipated load to force 3.24 be = mass m = mass S =m Stroke length of shock absorberft. S = Stroke length of shock absorberm ing Eq. 3.24 slugs kg be dissipated m = mass m = mass slugs kg be shock dissipated E = F •S − m• g •S to by g = gravitational acceleration 32.2ft./ g = gravitational acceleration 9.8m/ vertically) 3.24 force by shock g gravitational acceleration gravitational acceleration m= mass m g==mass bydissipated shock g= = gravitational acceleration 32.2ft./ acceleration 9.8m/ slugs kg absorber. 32.2ft./ g = gravitational 9.8m/ be absorber. 2 2 sec. sec. shock g =sec. gravitational acceleration 32.2ft./ g = gravitational acceleration 9.8m/ byabsorber. 2 2 sec. 2 sec.2 sec. absorber. 2 sec.2 E sec.E = Applied force energy ft.-lb. = Applied force energy Joules Applied Eforce = Applied Applied force force energy ft.-lb. Eforce = Applied Applied force force energy Joules Applied force = force = E force energy F force = Applied force lb. energy ft.-lb. E F force = Applied force N energy Joules Applied force F = Applied force F = Applied force force energy energy lb. N F = Applied force F = Applied force (with load Eforce = Applied force energy E = Applied force lb. S = Stroke length of shock S = Stroke length Nofenergy shock Joules Applied ft.-lb. force (with load (withenergy load S= = Stroke Stroke length length of of shock shock S= = Stroke Stroke length length of of shock shock hanging Eq. Eforce   F  S    m  g  SF =S force S Applied force lb. F= Applied force N Eq. hanging absorber absorber E  F  S  m  g  S     ft. m Eq. Eforce hanging absorber absorber  F  S  m  g  S down) to be 3.25     (with load force ft. m absorber absorber S = Stroke length of shock S = Stroke length of shock 3.25 down) ft. m m = mass slugs m = mass kg down) to to be be dissipated m= = mass mass slugs m= = mass mass kg hanging Eq. 3.25 Eforce   F  S    m  g  S  absorber m m dissipated absorber slugs acceleration kg dissipated g = gravitational g = gravitational acceleration ft. m by shock 32.2ft./ g = gravitational acceleration9.8m/ down) to be 3.25 g = gravitational acceleration by shock 32.2ft./m g 9.8m/ by shock g= gravitational acceleration32.2ft./ gravitational acceleration9.8m/ m= mass ==mass absorber. slugs kg dissipated sec.2 sec.2 absorber. 2 2 sec. sec. 2 2 g =sec. gravitational acceleration32.2ft./ g = gravitational acceleration9.8m/ byabsorber. sec. shock absorber. The fourth scenario is onesec. where there is not a constantsec. applied force to the load (a possible transfer 2 2

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The fourth scenario is one where there is not a constant applied force to the load (a possible transfer

The application) fourth scenario is one therehas is applied not a constant applied to thetransfer load (aline possible transfer line but the moving kinetic energy the force load moving across a conveyor must The fourth scenario is one where there iswhere not aobject constant force to and the load (a possible application) but the moving object line application) but the moving object kinetic energy and load across a must line application) but the moving objectishas has kinetic energy and the the loadtomoving moving across a conveyor conveyor must overcome a frictional resistance. The fourth scenario is one where there not a constant applied force the load (a possible transfer has kinetic energy and the load moving across a conveyor must overcome a frictional resistance. After calculating the kinetic energy, then the overcome a frictional frictional resistance. overcome a resistance. line application) but the moving object has kinetic energy and the load moving across a conveyor must following equation be used the to calculate the additional applied energy. After can calculating kinetic energy, then Eq. 3.26 can be used to calculate the additional applied energy.

overcome a frictional resistance. After After calculating calculating the the kinetic kinetic energy, energy, then then Eq. Eq. 3.26 3.26 can can be be used used to to calculate calculate the the additional additional applied applied energy. energy. After calculating the kinetic energy, then Eq. 3.26 can ft.-lb. be usedEfriction to calculate the additional applied energy. Efriction = Applied energy = Applied energy Joules

Efriction = = Applied Applied energy energy ft.-lb. Efriction = = Applied Applied energy energy Joules Energy E E Energy Cfriction Coefficient of frictionft.-lb. Cfriction Coefficient of frictionJoules Energy fricagainst f = f = C Coefficient of friction C Coefficient of friction against fricff = ff = C = Coefficient of friction C = Coefficient of friction against fricS == Stroke length of shock S = Stroke length of shock Efriction Applied energy E = Applied energy tion resisft.-lb. friction Joules S= = Stroke Stroke length length of of shock shock S= = Stroke Stroke length length of of shock shock Eq. tion Energy S S tion resisresisEq. Efriction = m • g • C • S absorber ft. tance, to be absorber Eq. m f •S C = Coefficient of friction Cf absorber = Coefficient of friction E = m • g • C tance, to 3.26 E against fricfriction = m • g • C f • S f absorber tance, to be be ft. m absorber absorber dissipated by 3.26 friction ft. f = mass m = massm 3.26 dissipated by slugs of shock kg S =m Stroke length S = Stroke length of shock tion resism = mass m = mass dissipated S Emass L E Cslugs T Cacceleration OMPONEN T Smass F Okg R Pacceleration N E U M AT I Cshock S Y S Tby EMS m m Eq. slugs kg g == gravitational g == gravitational shockto be E = m • g • C • S tance, shock absorber absorber g= = gravitational gravitational acceleration g= = gravitational gravitational acceleration absorber. friction ft. m f 3.26 g acceleration g acceleration absorber. 2 2 dissipated by absorber. S E L E C T C O M P O N E N T S F O R P N E U M AT I C S Y S T EMS 32.2ft./sec. m= mass slugs m 9.8m/sec. = mass kg 2 32.2ft./sec. 9.8m/sec.2 2 2 shock 32.2ft./sec. 9.8m/sec. g = gravitational acceleration g = gravitational acceleration The fifth scenario is the impact from a free falling object. Eq. 3.27 is used to calculateabsorber. this and for this

The fifth scenario is the impact from a free falling2object. The equation below is used to calculate this and for this scenario the total energy forces 32.2ft./sec. scenario the total energy forces are due simply to gravity9.8m/sec. acting2on the mass. The fifth scenario is on the impact are due simply to gravity acting the mass. from a free falling object. Eq. 3.27 is used to calculate this and for this Etotal = Total to free acting Etotal = the Totalmass. energy due to free Total energy scenario the total energy forces are due energy simply due to gravity on 3 - 54 • Pneumatic Specialist Certification falling objectft-lb. 3 Certification 3 -- 54 54 • • Pneumatic Pneumatic Specialist SpecialistE Certification total= Total energy due to free

Study Manual • 06/08/17

m = massslugs Eq. E = m • g • h + m • g • S falling object ft-lb. 3 - 54 total • Pneumatic Specialist Certification g = gravitational acceleration32.2ft./sec2 m = massslugs 3.27 Eq. E = m • g • h + m • g • S h = Height of object above shock g = gravitational acceleration32.2ft./sec2 total 3.27 absorberft. h = Height of object above shock S = Stroke length of shock absorber ft. absorberft. S = Stroke length of shock absorber ft.

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falling objectJoules of free fallStudy Manual Study Manual • • 06/08/17 06/08/17 Etotal= Total energy due to free Total energy ing object m = mass kg falling objectJoules of free fallStudy Manual •206/08/17 to be disg = gravitational acceleration 9.8m/sec ing object m = mass kg sipated h = Height of object above shock g = gravitational acceleration9.8m/sec2 to be disby shock absorberm sipated h = Height of object above shock S = Stroke length of shock absorber m absorber. by shock absorberm S = Stroke length of shock absorber m absorber.

The sixth and last scenario is the total energy due to rotational forces. The total energy to dissipate is the kinetic energy due to the moment of inertia and rotational speed. (This is the same formula that was The and last scenario is the energy forces. due to rotational forces. The is total energy to dissipate The sixth andsixth last scenario is the total energy duetotal to rotational The total energy to dissipate the kinetic energy due to theismoment of inertia in Outcome 3.17 Eq. 3.21). Additive to this is the propelling force (torque) imparted to the object while the kinetic energy due to the moment of inertia and rotational speed. (This is the same formula that was and rotational speed. Additive to this is theusing propelling force (torque) imparted to the object while rotating, which is calculated using this equation. rotating, which is calculated Eq. 3.28. in Outcome 3.17 Eq. 3.21). Additive to this is the propelling force (torque) imparted to the object while rotating, which is calculated using Eq. 3.28. Eq. 3.28 Eq. 3.28

Thrustenergy = T • Thrustenergy = T •

S R S R

Thrustenergy = Applied torquelb.-ft. S = Stroke length of shock Thrust energy = Applied torquelb.-ft. absorber ft. length of shock S = Stroke T = Rotational torque appliedlb-ft absorber ft. R = Rotational Radius from rotating center to T= torque applied lb-ft impact pointfrom of shock absorber ft. to R = Radius rotating center

Thrustenergy = Applied torqueNm S = Stroke length of shock Thrust energy = Applied torqueNm absorber m length of shock S = Stroke T = Rotational torque appliedNm absorber m R = Rotational Radius from rotating center to T= torque applied Nm impact pointfrom of shock absorber m to R = Radius rotating center

impact point of shock absorber

impact point of shock absorber

Thrust energy of rotatThrust enering object to gy of rotatbe dissipating object to ed by shock be dissipatabsorber. ed by shock absorber.

m onunknown page 38) but can be calculated In all the scenarios, it is likely that the actual(Continued velocityft. is based on the distance covered in a set amount of time. In all the scenarios, it is likely that the actual velocity is unknown but can be calculated based on theFEBRUARY 2021 WWW.IFPS.ORG • WWW.FLUIDPOWERJOURNAL.COM distance covered in a set amount of time. Consider a crate, like that shown in Fig. 3-31, rolling into a shock absorber mounted on a conveyor. Since

37


Consider a crate, like that shown in Fig.of3-31, rolling into of a shock theallstarting velocity is andthat assuming that the crate is accelerates average velocity the distance covered a set amount time. absorb In the scenarios, it zero is likely the actual velocity unknown uniformly, but can bethe calculated based on in the the starting velocity is zero and assuming that the crate accelerates unifo crate may be calculated by the following equations. distance covered in a set amount of time. crate may be calculatedConsider by the following equations. a crate, like that shown in Fig. 3-31, roll (Continued from page 37) The average velocity is the distance traveled divided by the time it takes to travel the distance: the starting velocity is zero and assuming that the Consider a crate, like that shown in Fig. 3-31, rolling into a shock absorber mounted on a conveyor. Since In all the scenarios, it is likely that the actual velocity is unknown but can beThe calculated based on theisdistance covered in a set amount of time. average velocity the distance divided the timeequation it takes crate may be traveled calculated by thebyfollowing the starting velocity is zero and assuming that the crate accelerates uniformly, the average velocity of the Considercrate a crate, like that shown in figure 2, rolling into a shock absorber mounted on a conveyor. Since the starting velocity is zero and assuming may be calculated by the Vfollowing equations. average ft./sec. Vavg = Velocity average m/sec. Calculate teravg = Velocity that the crate accelerates uniformly, equations. average velocity is the Vavg = Velocity average Vavg = Velocity The average velocity is the distance traveled divid D the average velocity of the crate may be calculated by the following minal orThe final ft./sec. Eq. Dthe distance: velocity for vvelocity D to = travel Distance = Distance distance The traveled divided by= the time it takes the ft. distance: average is the distance traveled divided by theDtime it takes m to travel Eq. avg 3.29

t

t = Time sec. Vavg = Velocity average

v

ft./sec.

=

avg 3.29 t = Time sec. Vavg = Velocity average

kinetic energy ft. D = Distance calculations.

t

t = Time Eq.Calculate tersec.

v

m/sec.

=

D = Distance m Vavg = Velocity averag

D

t = Time D = Distance

sec.

ft. D minal or avg final Since is zero, the average velocity may also be expressed by the 3.29 equation: Eq. the starting velocity t velocity for average v = D = Distance ft. D =Since Distance the starting velocity is zero, the velocity may also be expr m avg 3.29

vavg =

(vterm − vinit )t

t = Time

sec.

t = Time

t = Time

kinetic energy calculations.

(vterm − vinit )

sec.

sec.

Since the starting velocity is zero, the average ve vavg = 2 2 Since the starting velocity is zero, the average velocity may also be expressed by the equation: − vinit ) (v Since the starting is zero, the average may alsomay be expressed by theas: equation: Since the velocity initial velocity is zero, thevelocity last equation be rewritten vavg = term (vterm − vinit ) Since the initial velocity is zero, the 2 last equation may be rewritten as: vavg = vterm = 2 • 2 vavg vterm = 2 • vavg Since the initial velocity is zero, the last equation may be rewritten as: Therefore: Since the initial velocity is zero, the last equation Therefore: equation may beSrewritten S E L ESince C T the C Oinitial M P Ovelocity N E N T is S zero, F O RtheP last NEU M AT I C SY T ETherefore: M S as: Calculate tervterm = 2 • vavg vterm= Velocity terminal ft./sec. vterm= Velocity terminal m/sec. minal final vtermor = Velocity terminal ft./sec. vterm= Velocity 2•D Eq.= 2S•EvLavg vterm E C=T C O M P ODN=E Distance N T S Fft.O R P N E U M D AT C S Y SmT E M S2 • DTherefore: velocity for = IDistance vterm Eq. 3.30 kinetic energy ft. D = Distance D = Distance m vterm = t me = mass Calculatecalculations. mass me= mass 3.30 Therefore: t =equivalent Time sec. slugs t =equivalent Time vterm= Velocity termin sec. kg t

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equivalentt of = Time sec. 2 • D t = Time sec. Etotal = Total (sum 2 • Etotal Eq. Calculate tervtermenergy = Velocity terminalEft./sec.= Total vterm= Velocity terminal Eq. object using energy (sum ofequivalent KE m/sec. D = Distance ft. total m = Calculate mass m = mass equivalent m = mass of KE and applied force vabsorber. = minal or final e manufacturers Many of shock absorbers use an additional in shock e slugsstep e selecting the proper kg3.30 2 • D term 2 and applied force energy) total energy to 3.31 Eq. Many manufacturers use an additional in selecting the one. They calculatevelocity the mass equivalent of the entire Joules S E L Ecalculate C T vvCofOshock Oabsorbers N Eequivalent Nft.-lb. TS OEof R the P=ft. NTotal E Ustep M AT IC S=Y S Tproper E MS equivalent energy =F Distance D Distance Many manufacturers use an additional step in selec m =M Pmass They the entire kinetic energy imparted to an of object at absorbers the for speed oft of energy (sum t = Time be shock absorbed 3.30 2at•the EtotalDspeed total sec. kinetic Eq. term object using Ecalculate = Totalthe energy (sum of KE energy kinetic energy imparted to anm object of impact. That is calculated using: total They mass equivalent of the entire kinetic energy impart = t of KE and applied force andthe velocity. v = velocity impact. 3.31 That is calculated using Eq. An important fact to use when calculating mass equivalent e v = velocity Time3.31. t = Time calculations. m/sec. 2 t = ft./sec. and total energy to sec. sec. applied force energy)Joules v impact.acceleration That is calculated using Eq.Eq. 3.31. Anofimportant fact to use energy for a free falling object is that the velocity isft.-lb. due to the gravitational and Many height using 3.32. manufacturers shock absorbers usewhe an be absorbed v = velocity v = velocity a vfree falling objectCalculate is thatCalculate the velocity ismass due toequivalent the gravitational ft./sec. m/sec. mass mabsorbers mestep =for mass equivalent v =equivalent velocity = velocity and velocity. They calculate the of the accele entire e = mass slugs kg Many manufacturers ofg shock use an additional in selecting the proper shock absorber. ft./sec. m/sec. g =gravitational acceleration equivalent 9.8m/ velocity S E L calculate E C T C Othe M Pmass O N=gravitational Eequivalent NT S F OofR the P N E U(sum Mkinetic AT I C energy S Y S Timparted EMS Eq. They impact. That isofcalculated using Eq. 3.31. An imp 2 Etotal = Total energy to an object atofthe speed of 2 • Etotalacceleration Study 06/08/17 Specialist Certification v = Manual 2m • g =• h 2 entire Eq. object using• 3 - 55 E = Total energy offree KE falling 32.2ft./sec. sec v = velocity v = Pneumatic velocity(sum 3.32 impact. total ft./sec. m/sec. for a free falling object is that the velocity is due to of KE and applied force Calculate That using Eq. 3.31. An important fact to use when calculating the mass equivalent e is calculated Study Manual • 06/08/17 Pneu h = Height of object above h = Height object above shockacceleration and of applied force energy) energy to 3.31 g =gravitational g =gravitational Joules object. total 9.8m/ velocity of v 2 shock energy Eq. ft.-lb. for a free falling object is that the velocity is due to the gravitational acceleration and height using Eq. 3.32. absorber absorber 2 be absorbed ft.acceleration32.2ft./sec.2 m v = 2• g •h sec free 3.32 Calculate massfalling m mass m mass equivalent v e==velocity v e==velocity velocity. h =equivalent Height ofslugs object above h= Height of shock kgobject above ft./sec. m/sec. Studyand Manual • 06/08/17 equivalentobject. of Consider the following example: shock absorber absorber E = Total energy (sum ft. m 2 • Etotal total Study • 06/08/17 Pneumatic Specialist Certification • 3 - 55 Eq. Manual object using v = velocityft./sec. v total = velocity E = Totalm/sec. energy (sum of KE me = of KE and applied force Calculate 2 100 kg gis=gravitational and applied force energy) total energy to 3.31 g =gravitational acceleration A An crate that has a mass of pushed from Joules 9.8m/ important to use when calculating the mass equivalent for a free falling object is that the velocity is due to theof gravitational acceleration v following Consider the example: energy velocity Eq. fact ft.-lb. 2 be absorbed acceleration v = 2to • ganother •h one horizontal conveyor by a cylinder (not 2 32.2ft./sec. sec free falling and height3.32 using: v = velocity v = velocity and velocity. h = Height of object above h = Height of object above shock m/sec. shown but indicated bythat the arrow). A crate has a mass of 100ft./sec. kg is pushed from object. shock absorberft. absorberm one horizontal conveyor tovelocity another by a cylinder (not v= v = velocity ft./sec. m/sec. Calculate The coefficientshown of friction of the conveyor is 0.2 and but indicated byg the arrow). =gravitational g =gravitational acceleration9.8m/ the following example: Eq. velocity of toPbe is 5 Rm.P N The time it takes S Ethe L Edistance C TConsider C OM OvNtraveled E N T S F O E U M AT I C S Y S T E M S SELE 2 acceleration32.2ft./sec.2 = 2 The • g •crate h sec free falling to travel 3.32 this is 1 sec. was pushed onto The coefficient of friction of the of conveyor is 0.2 and h = Height object above h = Height of object above shock A crate by that cylinder has a mass of 100resulted kg is pushed object. the conveyor that in The an from theadistance toand be traveled is 5 m. time it takes shock by absorber absorberm ft. one horizontal conveyor to another a cylinder (not

(

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initial amount to of travel kineticm energy 100The Joules. The this is 1 of sec. cratem was pushed onto Calculate mass mass equivalent e =by slugs e= mass equivalentkg but stroke indicated the arrow). shock shown absorber length is 50 mm. What equivalent of the conveyor by a energy cylinder and thatis resulted in an Consider the following example: E = Total (sum 2 • Etotalthat the totalshock absorber will need Eq. object using the total energy Etotal = Joules. Total energy (sum of KE initial amount of kinetic energy of 100 The m = of KE and applied force e 2 THE The coefficient of friction ofthese the conveyor is 0.2 and and applied force energy) totalfrom energy to horizontal conveyor to another by a cylinder 3.31 FOLLOWING EXAMPLE: Aiscrate that has a mass of 100 kg is pushed one to be CONSIDER capable of dissipating? Of factors, only Joules vshock absorber stroke length is 50 mm. What is energy A crate that has a mass of 100 kg pushed from ft.-lb. is 5 m. The time it takes be absorbed the distance towill be be traveled the terminal velocity an unknown factor in the (not shown buttotal indicated by to the arrow). coefficient of(not friction of the conveyor and is 0.2 and the distance to be traveled is 5 m. The time it the energy that the The shock absorber will need one horizontal another cylinder velocity v = velocity velocity. ft./sec. to travel this isconveyor 1v = sec. The crate by wasa pushed ontom/sec. problem. to be capable dissipating? Of these factors, only shown but indicated byof the arrow). takes to travel this The crate was pushed onto in the conveyor by a cylinder and that resulted in an initial amount of kinetic energy of the conveyor byisva1=sec. cylinder and that resulted an v = velocity the terminal velocity velocity will be an unknown factor in the ft./sec. m/sec. initial amount of gkinetic energy ofis100 100 Joules. 100 Joules. The shock absorber stroke length 50 mm. isThe the total energy that theCalculate shock absorber will need to be capable of dissipating? Solution: The kinetic energy was given as Joules. =gravitational gisWhat =gravitational acceleration problem. 9.8m/ velocity of The coefficient of friction of the conveyor 0.2 and Eq. 2 shock absorber stroke length is 50 mm. What is acceleration v = 2 • g • h 2 To solve for the applied force energy due to the mass Of thesethe factors, only the terminal velocity will beThe an unknown factor in the problem. sec it takes 32.2ft./sec. distance to be traveled is 5 m. time 3.32 free falling total energywith the ofshock need hthat = Height object absorber above h will = Height of object above shock movingthe over akinetic surface friction, use Eq. 3.31. Solution: The kinetic energy was given 100 Joules. to thisenergy isof1shock sec. The crate was pushed onto SOLUTION: Thecapable was given as 100 Joules. To solve foras the applied force energy due toobject. the mass moving over a surface with friction, use this equation. absorber absorber to travel be dissipating? only ft.Of these factors, m the conveyor by a cylinder and that resulted in an To solve for the applied force energy due to the mass the terminal velocity will be an unknown factor in the kinetic energy of friction, 100 Joules. The3.31. a surface with use Eq. Consider following Efrictionthe =initial m • gamount • moving Cf example: • S ofover problem. shock absorber stroke length is 50 mm. What is Fig. 3-31 Kinetic Energy Of A Rolling Object A crate that hastotal a mass of 100 kg the is pushed from the energy that shock absorber willJoules. need Solution: The kinetic energy was given as 100 E = m • g • C • S E = 100 kg • 9.8 • 0.2 • 0.05 TEST YOUR SKILLS one horizontal conveyor to another by fa cylinder (not friction capable ofm/sec. dissipating? Of these factors, only frictionto be m To solve by for thearrow). applied force energy due to the mass Fig. 3-31 Kinetic Energy Of A Rolling Object shown but indicated the terminalthe velocity will be an unknown factor in the Efrictionmoving = 9.8Joules over a surface with friction, use Eq. 3.31. E problem. friction = 100 kg • 9.8 m/sec. • 0.2 • 0.05m What is the total energy needed to be

The coefficient of friction of the conveyor is 0.2 and Efriction +toEbe=traveled Total Energy the distance is 59.8 m. The time it takes dissipated by a shock absorber that is Efriction Solution: The energy was given Efriction m • gThe •kinetic C=fcrate • S Joules to travel this is 1= sec. was pushed onto as 100 Joules. positioned Fig. 3-31 Kinetic Energy Of A Rolling Object under a vacuum cup that To by solve for the applied force energy due to the mass + 100 =a109.8 the 9.8 conveyor cylinder and that resulted in an Joules Efriction + E = Total Energy initial amount of kinetic energy of 100 The moving over a kg surface withJoules. friction, use Eq. 3.31. releases a free-falling load of 2.5 kg, 0.01 Efriction = 100 • 9.8 • 0.2 • 0.05 m/sec. m as a selection criteria, then the Eq. 3.31 would be used: If using absorbers that use mass equivalent shock stroke is 50 mm. What is Ifabsorber usingshock shock absorbers that mass equivalent 9.8 length + 100 =use 109.8 meter above the shock absorber, and the Joules the total energy that the shock absorber will need E = 9.8 2Efriction •friction Eoftotal=dissipating? as capable criteria, equation would m • gJoules •then Cf Of •this Sthese stroke of the shock to be factors, only mea =selection If using shock absorbers that use mass equivalent as selection criteria, the Eq. 3.31 would beabsorber used: is 0.05 m? Fig.a 3-31 Kinetic Energy then Of A Rolling Object 2 the terminal an unknown be used:E velocity v a. 0.17Joules +will E =beTotal Energy factor in the friction problem. 2 • Etotal

Efriction =m100 • 9.8 • 0.2 calculated • 0.05m from the distance and time stated. The m is 2.2 m/sec. = kgbut b. 0.67 Velocity was not given could be easily kgs. Joules e 2 9.8 + 100 e= 109.8

See the v Joules Solution: The kinetic energy was c. 0.96Joules Efriction = 9.8Joules given as 100 Joules. solution on To solve forIfthe applied force energy due to use the mass was given couldequivalent be easily calculated fromcriteria, the distance andEq. time stated. The m is 2.2 kgs. shock absorbers that mass as a selection then the 3.31 would be used: Velocityusing was Velocity not given butnot could be but easily e d. 1.12 page 39. moving over friction, Eq.energy 3.31. needed to be dissipated by a shock absorber that is positioned Joules E a surface + Ewith = Total Review 3.18.1: What is Energy theuse total friction 2the • Edistance calculated from and time stated. The e. 1.47 total undermae vacuum cup that releases a free falling load of 2.5 kg, 0.01 meter above the shock absorber Joules = 2 9.8 + 100 = 109.8 m is 2.2 kgs.  Efriction =and g •C • S Joules v Figure 2: Kinetic energy of a rolling object. e m •the stroke of the shock absorber is 0.05 m? Review 3.18.1: What is the total energy needed to be dissipated by a shock absorber that is positioned f Fig. 3-31 Kinetic Energy Of A Rolling Object under agiven vacuum cupuse that releases a freeasfalling load of 2.5 and kg,then 0.01 meter above the absorber was not but could be easily calculated from the distance time stated. Thewould m 2.2used: kgs. a.Velocity If 0.17 usingJoules shock absorbers that mass equivalent a selection criteria, the Eq. 3.31 be e isshock Efriction = 100 kg • 9.8and • 0.2 • 0.05 m/sec. m the stroke of the shock absorber is 0.05 m? 38 FEBRUARY WWW.FLUIDPOWERJOURNAL.COM • WWW.IFPS.ORG b. 0.672021 2Joules • Etotal = Efriction = 9.8 a. 0.17Joules e Joules c. m 0.96 2


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PA Series

More NFPA Actuator Options Than Ever Before

TAS Series

Specification

PA Series

TA Series

Bore Range

1.50 – 4.00”

1.50 – 12.00”

Rod Diameters

One diameter per bore size

Multiple diameters per bore size

Mounting Options

12

20*

Flush Mount Option

Standard

Optional

Configurability

Basic

High

Cushioning

Auto-Cushion Standard

Adjustable & Fixed Optional

Seals

Original Ecology Seal

Bumper Piston Seal Optional

Temperature Rating

+25 – 175°F

-20 – 200°F

Steel Body Option

N/A

Available

Market Price

$

$$

FM Series

TD Series

*with TAS steel body option

Bimba’s PA Series is the perfect complement to our NFPA actuator line. It offers set bore sizes and rod diameters for a simple, straightforward counterpoint to the highly configurable TA Series. Standard designs and features provide the versatility your pneumatic applications need, at a price that can’t be beat. Find out more www.bimba.com

TA Series

Profile for Innovative Designs & Publishing, Inc.

Fluid Power Journal February 2021