RIGGING ENGINEERING CALCULATIONS J. Keith AndersonCopyright RIGGING ENGINEERING CALCULATIONS A practical guide for rigging engineers and lift planners Copyright ‘Copyright ©2018 Kerth Anderson ( Edition ‘st Eaton, 2018 ‘usnor J. oth Anderson SBN; 978-0-692-00692-4 AILRights Reserved 'No par ofthis work covered by copyright hereon may be reproduced or used in ay form ct ‘by any means—graphi,elecone, or mechanical, including photocopying, recording, taping, "Web cstroution, or informaten storage and retrieval systems without the expressed wetan [permission ofthe copyright hed. Permissions For permission to use materia rom this book, please send requests to the Ditibutr, which wil be forwarded tthe Authoc Parnell Sevices Group, ne. MBookstore PO Box 1650 Woodland, WA 98674 ‘bookstowiticom 1.888.567.8472 Al references made to specific sofware appications and/or companies used ‘tin this book arregistred trademarks oftheir respective compares. Distributor For Adtonal Copies Contact Indust! Taining International Bookstore (T Bookstore) PO Box 1680 Woodlena, WA 98674 Itloombackstoe 1.888.567.8472 CGratofl thanks to those who have alowed the use of material ins book anc to those who have contibut thee line, exes dcAlso Available from the ITI Bookstor Rigging Engineering Basics - Revised & Expanded 2nd Edition By. Kath Arcerson 470 pages, 85x 1", softbound A practical quide for Lift Directors, Lit Planners, Figaing Engineers, Site Superintendents, Field Enginoers, Rigging Foremen, Heavy Lit Managers, Heavy Haul Planners, rane Operator, and Advanced Figger. Thi book i intended to inform people planning rigging operations and point them in the rection of best practice, to alr them to pitfalls fand how fo avoid ther, and to instruct on how to do Certain basi rigging engineering tasks. Item #1175 | $149.00 (T)BOOKSTORE | .S88323%.. Disolai The content of this book is offered in good faith for educational purposes, to demystify the concepts behind common and not-so commonly encountered rigging related calculations: to be a ready.reference for useful calculations and a guide to thair application. Rigging is an unforgiving discipine, in preparing rigging calculations, users must recognize the limits oftheir skills and seek expert assistance when require. Although striving fr total accuracy, erors are sure to have crept in I will happily correct those brought to my atten- tion, The author cannot be held responsible for any negative consequences of following the: Information contained herein, Opinions expressed are mine alone. -Keithpag engineering caleuatons CONTENTS A Basie Trigonometry eee 1.1 Polygons internal anges. 7 — 12 Rightandled tangles ono = 7 : 6 1.3 Woscees trang, a ec 7 14 cauilteraltianges, —— 7 : 8 15 Sealan trangles. 8 Le cices 20 17 Degreesandradians an 18 Vector Analysis a 1.8.1 Analysing system of ores 7 : 2 1.82 Guying~moment example. a5 2 Weight and Cots. 2 = a coun 2A Weight : . a7 2.1 Densities selected materi os 7 2.1.2 Volumes of common shapes = 18 213 Weightealuation.... a9 22° center of gravity. 7 20 2.2.1 Basie principles~ center of gravity : cme) 2.2.2 _orationof the centr of gravity constant section otc 20 2.23 Location ofthe contr of gravity revolved” solids with changing section 2 224 Calculating the location ofthe C.G of assembles, 2 225. Weightcontol & report. : son 226 Determining the C6. usingacranels) 29 227 Determining the C6. by electronic weighing. snd 3 Load sharing~2lines of suppor. os 38 3.1 Vertical suspension ~2 iting device... se a8 32 Load sharing ~2 support sods " : as 4 Forces in bridle stings. - sos 38 41 Effect sling incination es 38 42 Degged brides... a Seed 421 General solution : ee 42.2 Twolndlned slings attached at same height, difeent angle. om. 43 423 Twolnlined sings ~attached at same height symmetrical angles... 4 43. Suspension from thre attachment pln ea Paget43a ging nginceing Calestions alelating vertical suppor reactions. 44 Hleegedbrides, 442 443 45 452 453 5.1 Upending using iting device & 2 it-up “hinge” Attachment points at same elevation. Attachment points at fering elevations ‘Attachment point at fering elevations alternative “vecor” analysis 445 Suspension from more than thee attachment pons, Incine sings ‘+tegged bridle equal characteris “4egged bricle—lugs at ferent heights pending using two iting devices ~ fee suspension Fescesin spreader systems 6.1 Equalleg spreader arrangements 62 Unequal eg spreader arrangements 63. Using2 rolling block 63.1 sng drectiytoload 63.2 Rolling blocks with spreader. Assessed capacities of sings 7a Genera 7:11 asc pinciles-al types of sings. 72 Wrerope sings. 72.4 Forcesin choke hitch 72.2 Wrapped cnote, 72.3 Bending rope within a choke hitch 724 — Rendered choke. 725 Indining ings. 7.25 Basket itch 0/4 bend ratio 7.2.7 Gacket hitch calculating capacty 73° Synthetic web sings 731 Wrapped chote 732 Rendered choke 733. Basket hitch apace 734 Basket itch ~ pared web sings 7.4 Synthetic round slings. 7.81 Choke andbske hitches 742 Adjusting ten. 7.43 Stang connecting hardware 7.44 tring theless 75. Chain sings Choke hitches. 45 47 a7 52 bebe ag HREB EER RRR RED RRSRRE BRSSE BeES Paget‘ing EngncringCaleationt 752 Basket ches. 7 Crushing ores in basket Htc os Bi Basic basket hitch - 182 Wrapped basket hich 183 Basicbaskethiteh, non-round object. 84 Basichaskt hitch -Incined sings. 9 Dringa ose 9.1 What “driting*aloa?. 9.1.1 Typical ase Hoists attached at the same height 9.1.2 How do the fores change salad is rife. 9.13 Twolftpoints 30 Beam Theory 10.1 Bending Moment and Shear Force Diagrams 10.2 Section properties overviw. 102.1. Neutral axis & frst momento 68. : 10.2.2 Second Moment of rea (Area Moment of neta) neu 1023 sending stress i a 10.24 Section modulus elastic section modulus), : 30.25 Radius of gyration. 10.26 Beam deflection. : 10:3 Beam shear, BM. and deflection formulae i 10.4 Properties of common sections ae 41 Design oft beams and spreader bars. 11.1 Design standards. os 11.11 Background. = 112 Curent standads con = 1113 Designto Brier a 7 111.4 Example beam design to ASME 830.20 THA-2017 ne 411.15 Example spreader bar design to ASME 630.20 ST-12017 41.16 Spreader bar design adequacy check using ASD 32. Forcesin pulley systems... 12.1 Calaating anchor forces for verter sheaves foiled) 122 Reevng & pars oftine 12.2. Parts ofline, mechanical advantage and velocity ratio 32.22 Multsheave arrangements - 132.23 Lossesin eeving systems. 12.24 Overhaul in reeved systems 12.25 Winching system diagram. 12.26 Calculating length of wire ona drum, 86 a7 93 98 103, 10 104 108 : 06 : 207 207 rnd A a so as : nA : ~ us : ol 1 150 12,2 “ 15 eng ngncering Clete 123 Wott, power and eneray 1234 Definitions Load strbution beneath mobi cranes 13:1 Intodveton 13.2 Pressure profiles. 123 sis of caleaton 13:4 “Duet” methodology -wotked example (customary uns). 135 "Over methodology- worked example (metric. 13.6 Stee! plate versus wood mats. 1.7 Sling "standar mats for use under outrigger. Assessing wind forces. 14.1 Wind speed, wind pressure and forces induced 14:11 Windspeed. : 14.12 Windspeed versus height. 14.13 Windspeed topography considerations. ALA Wind pressure 1415 Wins force. 142 Assessing forces and rotational effets sting on a load, 143 Swinging outof plumb. Frietion 15: Basi principles 1514 Co-ffidentofftion. 151.2 Gaaient 15:13 Coefficents of frition~ selected materials 15.14 Assessing pulng/estrain forces. ares. 16:1 Basle buoyancy principles 261.2 Pressure inlgue. 16.1.2 Archimedes. 162 oispacement 1621 Principles 162.2 Load v displacement barges 16.23 Center of buoyancy 163 Stability oftarges. 163.1 ightng of barges 26.3.2 Metacenter & stabiity 264 Pitch and rl (st) 264 Principles. 165 Mooring oftarges. o ra 155 455 156 156 156 487 158ging Engineering aeltins 265. Pines ae 195 2652 aasiemooing appiatin. : 195 17 Heavy wheeled transport related : SS 198 17. Forces acting in road wanspor. — ao 19d 2711 General coe _— 198 3712 comeing : So = 198 2713 Windfoces nnn 199 27.14 Acceleration deceleration ee 200 75 tncines, dene ad COEF nnn 20. 17.2 ashing loads on wheeled transport. cn 172.1 Genedd == : Ss 17.22 Forces inplay during “abnormal anspor. 203 1723. Force combination. 204 17.24 Purposes oflshing and sec. 204 1725 Restraining methods 205 18 Guyng-deasman design z ee 207 38.1 Guylng preamble. 382 Deadman deseh on ee 208 19, Suspended i nnn é a 19.1 Catenary versus parabola ae an 1921 Catena ee i am 18:12 _Invoducton to fal rotation sp6F02¢hncnnnne nnn 23 1913 Patabobenwsnnnnnnnnnnnnn ae 19.4 Comparisons between Parola and Catenaes. = 5 1915. Creuarare. _ 6 19:16 Modulus last of wire rope. a oe 26 19.17 Fercesinhorizotalfestines : z 27Pein Engineering Cleultions 1. Basic Trigonometry A trigonometric refresher for those who nad it Much of the followings, for instance, relevant to determining rigging arrangements 1.1. Polygons - internal angles 1.2. Right-angled trlangles It you know the lengths ofthe sides of a right-angled ‘tangle, youcan determine the angles; = cos @ (2) le 86 the angle that has sine = the length ofthe {Gis known a5 the inverse sine and so on “The sum ofthe internal anges ofa polygon regular or lreguat is given by the formula: ewer 180(N2), where Is the number of sides Ina three sided polygon (viangle}, Sag * 180(2-2)" = 280 Ina44sided polygon (quadrilateral), Bas = 1806-2 “The sum of angles“ trough “ein the Sieg pentagon shown here = S40 260" and s00n Ina right angled triage, the sides are designated as shown relative toangle 8 The proportions ofthe angle (ratios ofthe sides) sare purely 3 function ofthe internal angles and a= Unchanged regardless ofthese ofthe triangle. ‘The primary ratios ar defined as sine (SN), cosine (Cos) andvangent AN, : Ifyou know an angle and the length of side, the site ofthe wiangl is e-ermined and the lengths of the two other sides canbe eaeulate, O = Hxsind or Axtano A= Hx e080 oF 2 =H xe080 or 25 Fromythagoras, Knowing two ies, the thir canbe cleulteding Engineering Caletions Examale Whatis angle 82 Ifthe Adjacent side (A) to angle 8= andthe Opposite sate 4 whatisthe length ofthe Hypotenuse (HI? ( Note: a triangle the lengths of whose sides are in a OTH=8) ‘the proportions 3.4: wll always be 2 ight-angled ‘rangle. Inoseles tangles have two sides of equal le 1.3. Isosceles triangles {and therefore two equal angles) = 90/45/45 ifthe upper angles rght angle (0") and the two ‘base angis are equal a 45" each, then ase = 1.414 side ses Sa side = 0.707 xhase height =base/2;0r 0.8 x base elt = de cous = 0.707 side side = 1.414 ; base = 1.732 de ¥ ole SF tase ae se a 4 side = 0.577 x base eight side corto 03 de 209 tse anal Sart sider ae atoms Base =1.732x 3000 5136mm Fase?ina Enaineeing Cl 1.4 Equilateral triangles ce ana ine a6 2abeose The sum ofthe thre internal anges must equal 180. So, knowing two angles, the third can be fun, Depending on what is already known, you apply the aw of Sins and/or the Law of Cosines to determine the est. Note: the Law of Cosines reduces to the Pythagorean theorem when angle C= 80° ees In an equilateral langle, al the sides are equa, therefore the internal angles areal equal at 180"/ 3=60" height = side x cos30 = 0.066% side side height/cos30 side eight/(0866 > side) orside = 1.155 height Say the sides ave 3000mm ea:h Height = 2000 0.865 = 25981 Ina salen triangle the sdes are unequal and one of he angles are a eht angle. TosoWve, you need to know ether: "Length of two sides and an. angle Two ngles andleneth of side Length of two sides and their included angle or Length ofa three sides The eight ofthe triangle height = ax sinc height = sin A The ares of ay trian: 1 J base x heightging Engineering Calculations ‘To determin the height “H” of an oblique triangle knowing just the length ofthe sides (thout fst working coutan angle), “The Law ofCosines canbe reoranged: aa 4b Dabcose ee) Kawing “by” on “a, “WH can be determined sing Pythagoras fsamale 31539 mm, be 3215 rm, e=2306 mm 15398 + 3215* 23062 £32154 = 2306" «149mm be ot ee [ ee al fot og = SARaF Tras = 102 po-7) eramgie ‘Consider this suspension arrangement. Te load has to hang a this angle such that the righthand attachment points level with theCofG, Sling 2a ven length, 10 (3048) raeing rgnceringCleutont Wat s the required length of Sing 1? What are the two angles ang and ang2) that the slings make tothe vertical? First, determine the incination ofthe load. The angle of incination i Next determine the angle “ang2". ‘The horizontal stance from the Cof Gt the RH attachment point given bt Hor = /3610" > 014 ang2 = sin"*(°996/4y49) = 18.96° 0.6mm ‘The angle Sing 2 makes to the lod 90.0-22.62- 18.96 = 48.02, We now know the lengths of Sing 2(2048mm) and the distance between the attachment points (2048mm) and ‘thence ange (4.42) the Law of Cosines can be used to determine the lngth of lg SLi = (3040 F S040? 2x FOF X BOAT cos 4BAZ) Using the Law of Sines the incudd ange between the slings can be found. 2500 __3048 aina942 ~ amine =, (2048 x sin 40.42) io (06) angl = 65.78 ~ 18.96 = 46.82° 500mm (8.2) INC 5.78 16 ‘The circumference ofa ccle= nO or 2nR, where Dis the ameter and Rs theradus ofthe cre. ‘The area of cle =n0"/4 orm Depending on which properties you know, Seto elaionsips can be usefully expressed as either: cu + aDR* - aR (ACR DR— DR ca DR=Rs0s/@Re= CH ‘ha included angle ofthe sector «28 Pane1.7 Degrees and radians In ordinary life, we ae used to angles being measured in A degrees, 560" corresponding to one complete rotation; 8 - fightanglels 90 degrees forinstance, Asimple calculator wil work wth Segrees. i Inmathematcs angles are often expressed in radians t ald Calculation (Oegrees ae unis of rotation, whereas radians ‘are simplypropoional and are units). ‘radian ithe angle made by an arc whose length sequal tothe rads ofthe cre As the decamference: anadins tradi ‘Be awareif automating some ofthe formulae in this book, tha spreadsheets wockir radians; acaleulated angle wilforinstance be expressed in radians. you want your results In degrees you wll ave to convert the radians ‘sing the “CEGREES| "expression. The radians goin the brackets) ‘Angles nputed2¢ degrees must be converted into radians forthe spreadsheet to we them to convert we the RADIANS() expression, ‘When working in radians, the Inverse sin, Inverse sine and inverse tangent are rfenedtoas arsine, aecosine and arctangent. Ona spreadsheet look forthe ASIN, ACOS, and ATAN functions for iaverse trig functions nt, 9 complete revoution =P : 1.8. Vector Analysis Forces have magitude and ection; for ana they can conveniently be represented a vectors Vectors are tines whese length is proportional to the magnitude ofthe force and whose anges the angle ofthe force The aiection ofthe force is indicated by an arow head, Vectors an be summed or broken down to equivalent constituent vectors. «eg, Force F canbe replaced by vertical and horzontal components which together are equivalent. Form the enclosing rectangle; the lengths of the ‘eral and horizontal sides reprevent the magitae ofthe component forces, mh % “The comporant vectors originate atthe sme pont asthe force they replace and thar rections ae Indicated byte arrow heads. x cos sino fxample “Consider anincined ing with a tension F= 10t at an angle tothe horizontal @ = 6egng Engncaring Cleon: 1.8.1 Analyzing a system of forces Fora system tobe in static equilibrium, there mustbeno net force in ay ection and none rotational eect Lelthe forces up and down and side tose must balance, Consider the attachment point on this oad, Vector analysis is a good too to use to analyze whats going on here. The weight i acting down, Host As pllng up and ef, Host Bs pling up and ight We can draw a vector diagram to represent the system of forces. 10 For equlibum, the vectors must forma dosed dlagram. inthis ease, we know the magrtude and drcton of the weight but only know the dizections of he other two forces we wate determine their magnitude. So how do we proceed? Star with sy the weight and work say antl counter}-lockwise ang vectors from th end of the preceding one asa 7.s2Tt sot] 100 to Draw 2 line fom an origin cretly downwards whose length represents the 10 weight. rom the end ofthat, ted line heacing off at 21.95" to the vertical - the erection of Host 8; we don't know the length ofthis yet Next, we need to add a vector representing the tension in Host A We know is racton and the fat that ithas to terminate atthe origin, so work backwards and craw line at 42.95 terminating atthe origin, Wher the two raeein Engineering Caleltions ray view) thas a constant triangular cross: section; the center of area of the tangle les 2UySalongaline rom 3 “point” to themidof the ‘opposite sie, tus applies to any trangle not Just isosceles triangles as shown hee. As the objects symmetrieal alongts length. the Cof & Isatthe center ofthe lngth The center of aea ofthe triangular cross sections the pint at which that 20 shape would be in balance, As this trlange symmetrical the center of area obviously lies onthe centerline ofthe triangle, but where? Consider a sve ofthe section of very small wicth The ares of that sliver = wa Feaitr Engineering aeultions kb ® 4 Taking moments about the apex ofthe triangle {Ui point), the area-moment is the area ‘ltd by the moment arm" ‘area moment = wy.2-dx mr total area momen eax whe ‘otal area moment = ‘Moment arm =Total area moment / total area _wht 22 moment arm =" 2 2 te the Balance pont or center of area lesa 2/3 from the apex (or 1/3" up fom the base) onthe center tine. famale Where does the centr of area ofthis box section le? ‘inspection i les on the vertical centerline at is symmetrical about that. ‘Take arasmements about any convenient axis, say the bottom edge (the moment ams ofeach plat are frm ther ing anters of area tothe bottom edge) Meuse arenes ® 406% 32 «12982? ge ht a0 2netomtarge= 305 x 25 = 7625 Br areasenwno= 19x 857 = 16283? 2 thus) ees ded mone bas Nement rm ttl rea memento area 263th? 290 ‘ee. the center of area ofthe section ies 499mm above the bottom edge, ‘onthe centerline.Rein rgnering Cleats 2.23 Location ofthe center of gravity “revolved slide with changing section fan object canbe created by revolving profile around a central als as you might do on a athe) andthe x- section changes along the length, how do you calulte the location ofthe C of G? Consider the cone, weknow how to calculate its volume, but where does ts C of le? Again, we ae trying te find the point of balance; e's caleulat the volume-moment ofthe objec about say the apex then dvde by te volume to determine the moment arm ofthe volume from the apex. eI the mass “where al concentrate at that point, t would act the same a the actual object with ts distributed mass doe ‘that wil give us the point of Balance. Again, considera very thin ic sation of the cone atx" from the apex and of width a, ‘Conner: We can jus work with volume 0 fnd the balance pone rather than weght ana ignore density os it vould jut bean equal multiplier on both sides of the clevlation and would vance vs. arax() 20 ‘The moment of that litle volume about the apex= 1(Q)P-nsran(S) re Integrating ti rm x20 to xh wil gve us the total sum of ‘the moment ofl the sce" forming the total vlume “Glee ta atone momen = (2) 28 Divide this bythe volume derived previous, ‘moment arm = — Qe a le. the Cof 6 ies of the height down fom the apex (or of the way up from the base) along the avs of ‘This methodology canbe used anywhere you have a revolved solid and have formula describing the profile relative tos" (slong as your math sls ae up to Integrating complex formulae, 2.24 Gaeulating the location of the CG. of assemblies “Many objects we itor move are assemblies of many object, structural, mechanical, electrical, piping and so fon, We want to reat the object as single ety, so how do we clelate the location ofits of G? Remembering that the Cof Gis a location at which the total weight ofthe object wil act the same athe sum of ‘the constituent parts, we an select convenient reference pont and sum the moments ofthe constituent pats about that point. y definition, the total moment must be the same a the moment created by the total weight, about the same pont. So, diving the total moment by the weight yields the moment arm ar distance the Cof| {Gis rom the reference in that plane. De the same thing inal tee planes and you have the 3 location ofthe cols, raeASSEMBLED SKID samy ‘Conse thie skid mounted water pump. te comprises 2 motor, a pump and some associated ‘leacal equipment mounted on 2 frame. We wart to identify the C of location o postion iting lugs onthe base of the frame and analyze the stability of the singing arrangement. For the exercise, we will assume that “in and out” ofthe page the © of is centrally located. MOTOR choose a convenient datum point -t doesnt matter where but preferably one where the moment ams wll allbe postive (it makes it easier to ep track that way) In this cse the bottom let comer i selected as 2 reference Locate the pump and motor on the fame, note the C of 6's of the constituent pars and add their and y stances from the reference; note ther weights Mathematical method Inthe" direction, the moment are the weightsxthe horontal distances from the reference. ‘Bu " 0590 828+ 1.290.571 + 24416 Ban065+1.2+24=605¢ Faeating Ergreeirg Cleans sah Les ee Inthe" iection, the moment are the weighs xthe verti distances fromthe reference. Bu 204590.215 + 1.290.843 6 2.440.665 «2.708 Sn2045-+1.2+24=805¢ "¥ moment arm = Fu / Ew =2.704/8.05 = 668m Grophical method [pps It working CAD, you might indi easier ho = a ‘to derive the location this way. mel hh | fist apa pace the pump and mA Tina ‘motor and locate ther centers of gravity. : raw tne joining two of them (say the pump and motor) and measure it lenth Divide i in Inverse proportion of the 08 ‘weights at either end, That locates the C 1 fof tnetwo tems. Le tne is 1093mm long, the C of 6 locaton is 1083 31200/3600 = 364mm {rom the Rend Page 2ein raneering Caletions Draw ane joining the pont you have ust derived othe thir point (the frame). Measure its length (87). Divide that ine in inverse proportion ofthe welghts at either end (3600:450). ie. the CofG locations 6873450/L050'= 76mm rom the RH end “That locates the of ofthe assemble three items You can them dimension tin CAD from a reference point “This method yields the same answer In what might be fewer steps. “You could repeat the ercite forthe electrical boxes et, but thelr total weight I usiely tobe mare than 1004 or so andthe stibuion unlikely to shift the C of G very muh. It would be reasonable to leave the C af 'G where tis and simply athe known extra weight and a small contingency fr msealaneous items Knowing where the Col Gi, to keep the ling lengths and forces the same you might want to lcate iting ‘runnions on the base ecultant either side generallyasshown. Use an overhead single spreader with inclined slings above of adequate length to maintain stably considering the height ofthe Cof 6 above the trunnions ~ ‘ee stably section hein PaneFein Engineering Clestions This Ash Hopper Panel was tobe transported on house moving doles inside a coal fired Boller, Physi constraints entering the boller determined the only possible arangement. Unfortunately this placed the C of Gof the boiler ands transport fame very eccentric relative to the wheels It was decided to ballast the transport frame using cane counterweight to move the C of G left (as shown) to center it up beter. Two beams were located onthe frame (you see the ands of them on the frame et). The ballast eto pan the two beams event, leealy, how much ballast in adtion tothe weight of ‘the two beams) should be used to center the CofG of the arangement between the doles? ‘The ponel weighs 212,084bs (96,1834, the frame weighs 1900s (6,8434g) & the two beams carrying the ballast weigh total of 22,50 (10,200) Ting moments abou the centerline ofthe transport In tipe (0203) ana Comment: it doesn't hove tobe the center, onpwhere Clockwise =232.084hpex5.215K = 110i Counter (an) clockwise 335k 1 = 19.5kips x 1.719% © Nett ckwse (w/o ballast beams ete! = 2106- To counter this at 6.281 tothe lft ofthe zeterstne requires 1072.5/6.281 = 1702p. The beams supply 22Stps, $0 1707-225 = 148.2kips ballast required (approx. 670), 6410¢ plus bse crane counterweight blocks center the arrangement toa close approximation Comment you could e xorlen metic units and see how it works out for you. The metric equivelent masse ofthe vrous components ing) ere provided above.Ring Fgineering Cations Example _A4Qot module it be transported ona double-with SPMT ar shown in end view below. r veh) 1 | = | I weep a af ee eo page| fa : °f ‘The OG ofthe module is offset by 500mm laterly; we want to center itup on the taller. There we some 10 counterweight slabs available and they can be stacked and lashed) two-igh on the deck of the module along the length on the axis shown, |L How many blocks should be used? |i How much weight woul this add to the raller? i Where would the combined Cf G now be located lteraly? 'v. How much would the combined of G be lowered? ‘Taking moments about the center ofthe traller (where we want the combined load tobe balanced), 400560 Counterweight x2240 Counterweight=228,000/2,240 = 100t Le, 10 No, 10 blocks ave requied, tacked 2-igh slong the module ‘his ads 100 of weight to the tral. ‘he Cof G would be exacty centered horizontally “Taking moments about the base of the module, the height ofthe Cof is (400 x 2700) + (100 810) " 300 322mm the Cof Gis lowered by 378m. Page 27ging Ergnerrg Cleutstone 2.25 Weight control & report To caeuate the theoretical weght and Cf G location of say @ process sid comprising 2 struct feu ts of plant, pipe andso on, a move sophisticates approach needed, Fist it is necessary to establish a datum and co-ordinate system and identify everything inthe skid or module, Its weight and where itis (x2) in reference to the datum. Using a spreadsheet (ora piece of paper and & talelator) the moment each component induces about the reference point canbe calculated by multilvngits weight by sx. coordinates. running total canbe kept of the weight and the tree moments box with a To determine the location ofthe C of G, the thee cumlative moments are eath divide by the cumulative weight to give the co-ordinates ofthe Cof referenced from the datum point. Thespreadehest can be designed toad sto members by length and wi/length, late by sorve much volume at sucha density, Items by lndvgval weights and soon Examale-consder a simple structural boxer the following schematic fron sa 1 pee eae Desa a era TEFeng Engnearing scltions Iminstances suchas iting an cilrig mate offshore with ting rane, knowing the weigh and Cot Glocation 's paricolry vital and 2 sophisticated welght contol program will be instigated durng the design and fabrication of the module. Everything that goes into the module wil be racked and they wil weigh the module several tes including immediately before ladout. Normally weight conto wllincudealowances fr roling tolerances and the like and realty may ifr sight from the prediction, ‘Note: modeling a structure ina structural design program such as TEKLA or STAAD PRO, twill ikely be abet ive youtheC of location of your design dct. 2.2.6 Determining the C6. using a crane(s) you need to know the approximate location f the Cof Gof an object for say shipping purposes, ican be determined (ina single una! une |! direction) 3 reasonable degree of acuracy ty suspension rom Following principles described erie, a aya we( ond : 2 a= We (Bs bvdsn, | ° sot Re L Le. the horizontal distance between the suspersion point is split ie by the vertal ine tough the Cin te nese rato of he reactions (uspesion fre, What des ths mean? Follow the ete ‘exam beow ‘The reactions Ra & a in ree suspension (at thelift lugs) are show here. (Remember to deduct hoakbock, gen ee. from the nl) aol sosndinarns eal ume asl aah Br —z" py oe Distance between lugs = 12'=1'~2 “Therelore,x= 602-8 -As the plein fee suspension from two lines of support the weight isthe sum of the resctions: \We= 20,0008 + 10,0008 = 3.0008 Paes‘tample= Metis eng Engineering Clestions Distance between lugs = 3658 ~ 610-208 = 2743%mm Therefore, = 18284610 « 2.428 We= 4535490070 13,605ke This work within the accuracy ofthe weghlon inlstos of he cranes and the accuracy wth wich you know the weigh ofthe hokblock and other suspended items you want greater accuracy incorporate stain gauge shackle or link at each it pont How could the location ofthe C of G be determined sing a single crane? Principles of leverage canbe used. Ling one end at a time and pivoting about the opposite Rottom edge works Ike 2 Class 2 lever (hin wheelbarrow), Doing this for ether end and knowing the hor zontal distances from lugs to “hinges” allows determination ofthe weight and locaton ofthe CoG. ‘Consider this instance; note thatthe lugs are not lcated atthe ‘extremes ofthe lead whereas the pivot axes obviously are Fiat if cay] the lefthand end ofthe pce just clear using A ‘only, pluating onthe bottom RH ede ofthe piece Hote the it reaction tthe Ly Repeat the proces iting from the lugs wth the piece pivoting bout the btton left edge ofthe pace. Note the it reaction Can we simply ue the reactions obtained in ths two-stage process and proceed as outined above for atwo-rane lift? Nol ‘Atleast not unless the lugs were to be located onthe extremes ofthe load directly above the pivt axes, Fist iting at A, taking moments about thebottom RH comer; Ra(L 14) = We(L 2) Next, lifting at 8, taking moments about thebottom LH corner: Rg(L~ bp) = 20) ivding resort, Rall Lae = Rall byNL—2) y(t ty) = Ryle by) Rall Uy x a( RCL = 1) Rg(L~ Uy) = Ryll = LDL pe Pall to WyUE= L)Ry(L= Ea) Ute A showed £2,000 5.442) and 8 showed 21,8184 (88084), whats"? us (620mm), 2 (3656) & Le = 1 (305mm). Pagesigre Eee Cleans Working in US Customary Units. 2ig18(12 932 TapwOTEE = 2) + DBI) Re-arranging fst equation, 120000122) _ wr EE 2 so.008 Working in Metric Units 9995(3658 ~ 308)3650 SAT TESE— 610) FSBRST ISU TS si42¢36s8~ 610) ‘Ges8 — 2438) Note that you were to simply sum Re +R, You would get atta of 33/818 (15,3374) which is greater than ‘the actual weight of 300008 (13,600Kg). This rests fom loss of mechanical advantage in the Class? lever _arrangement used, caused by the lift points being closer to the Cf than the “hinge” points. For this reason, ‘ness te lugs We arecly over the hinge pons, simply summing the It reactions wil not gve an accurate ‘weight they are cosncident, then Ls and be wee 3.600 and the equation simplifies tothe ara. 2.27 Determining the C6. by electronic weighing If youneedto know the weght of an oblect and Cof 6 locaton toa high degree of accuracy, you could consider weighing it electronical. A vesel ike this could, for instance, be supported on four load etl, twounder each saddle. As the vessel is basicaly symmetrical we wil reat this as 2-0 problem to determine the Caf Gin the Xelrection ony ! 1 iia ciara Rt ' Fe am Considering Ri and R2 to eachbe the sum ofthe wo support reactions unde a sade inorder to deer ‘the locaton ofthe CofGalongthe length ofthe vessel x2 Rm We= Ri +R RUXL= R22 so, The BE op xa xt eRe lathe Cof is locate at 9 pint thet spits the tance Between the supports the fnvese puja ‘he euctionsRein riveting Cleltione What does this mean? eg. IF RL = 6 and Ri the proportions X21» 60; 9, the distance between suppet 0 then the distance between supports splitby the C of G in 4, K2= dm and X= 9.6m, Comment Realty check- te Cof Gis closer lroportionlly tthe sod wth the heavier reaction sample =3 suppor points same EY Three sain gauge load coll are used located a5 shown, eT (One ofthe cls is used as the 0,0 datum, By convention I put that one bottom lft and measure x and ¥ from that L point. T simply aleuation,erentatethe plan view such that one ofthe other el ies “due North” of the eign on ae a the pace = T 4 owerninetexandy cordate ofthe weighs fom 1 BZ eon pot tthe sxctonsobtsina ater flowing the weighing | pred ar shown F edn 7 Wheres the CofG located (cand fromthe oii? Mevodt Taking moments in the Xeon, 22,9672 3500nm = We ‘The total welght the sum ofthe reactions = 1.553 + 14S + 23.967 = St 2%» 23.967 3800/58 = 1446mm Taking momentsin the ¥ direction (23.9670 187Snm) + (14.5t33500mm) = WX (23.967x1875 + 14518500) / 58 = 2650mm Method? ‘Alternatively, using CAD, you might use a graphial approach raw a line between two ofthe cals, measure its son length (500mm). The two reactions of 14.5t + 19.5331 can be replaced by an equivalent 34.033t locate somewhere along tis ine ‘vide the line ininvrsepropotionsofthereactions a either end to find ut where. The distance from ‘the bottom eft ell= 145 x 3500/34 033 = 149mm, Craw a line rom that point tothe remaining el: measure its length (352imm), the two reactions of 2.967 + 34.033 can be replaced by an equivalent single (tal) weight of $8 at some point along that lin the Co oetion, ‘The distance from the left end = 23.967 x 3523/58 = 2455mm, Having idenied that pont the x and coordinates rom whatever reference you are using can be measured if needed eeeaging Ernceing Caleltions Example 4 suppor points when weighing say 2 module using strain gauge load cell, you wil be able te determine the weight andthe location ofthe Cf in plan to ahigh degree of accuracy. Typleally, the weighlg points willbe ona rectangular id. You should establish one ofthe comers of the grid as your xy datum (0,0) - conventionally take the lower leftcorer of your plan view. [Note thatthe item being weighed is si and the fur suppor load cells se independent of each ther ‘hydraulically, the reactions can “jump around” f the lad is not raised absolutely level, Ths Is because with stiff loads and sti supports, 2 &poin suppor arrangement statically indeterminate unike 23 point support. This wilnothoweverafect the total recorded weight or the calculation ofthe CofGlecation, bu the rdistrisution of weight may be a structural problem forthe piece being weighed Most modern weighing systems now Incorporate closed-circuit extension feedback o maintain evel iting to close trance. b Lona + ‘The weight = RIMRZORIARA hex. “Taking moments aboutthe RRA ai, Rey RB (aoa xtength= WeihxX 50 sa+na length “Toking moments abou the R-R? aus, wie PS length ‘Example 1624198, 2900 -7200m0 BETS opt 398+ 132. 5900.— s950mm o Comment: These general principles can be odopted to most weighing arrengements Tage