The paper deals with problematic of gear units and machine foundating 

and consequences that occur by foundation screws releasing. Namely, 

reliable operating of gear units, as well as other stationary machines, 

requires properly foundating. However, after certain period of time, 

relaxation of screw material, self-untightening of nuts and looseness of 

foundation screws influence on reducing the force of preload screw 

tightening and unequal load distribution. This unequal load distribution 

causes unequal deformations, which generate occuring vibrations. In 

extreme cases, breakages of screws and heavier accidents of systems 

driven by that gear units or machines can happen. With this paper 

authors want to indicate influence of deterioration degree on its load 

capacity

Milan Rackov

Ružica Trbojević

Siniša Kuzmanović

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 211    Siniša Kuzmanović, Ružica Trbojević, Milan Rackov  Causes Analyse of Breakages of Foundation Screws The paper deals with problematic of gear units and machine foundating and consequences that occur by foundation screws releasing. Namely, reliable operating of gear units, as well as other stationary machines, requires properly foundating. However, after certain period of time, relaxation of screw material, self-untightening of nuts and looseness of foundation screws influence on reducing the force of preload screw tightening and unequal load distribution. This unequal load distribution causes unequal deformations, which generate occuring vibrations. In extreme cases, breakages of screws and heavier accidents of systems driven by that gear units or machines can happen. With this paper authors want to indicate influence of deterioration degree on its load capacity. 1. Introduction Properly foundating of gear units, as well as other stationary machines, is very important for their reliable operating requires. However, after certain period of time, relaxation of screw material, self-untightening of nuts and looseness of foundation screws influence on reducing the force of preload screw tightening and unequal load distribution. This unequal load distribution causes unequal deformations, which generate occuring vibrations. In extreme cases, breakages of screws and heavier accidents of systems driven by that gear units or machines can happen. With this paper authors want to indicate influence of deterioration degree on its load capacity. 2. Problem Description It is well known that foundation of almost all stationary machines is required for their reliable operating. However, during certain period of time, relaxation of screw material and self-untightening of nuts are occuring which influences that force of preload screw tightening is significantly reduced. Beside this, due to ANALELE UNIVERSIT ĂłII   “EFTIMIE MURGU” RE ŞIłA  ANUL XIII, NR. 1, 2006, ISSN 1453 - 7397   212 operating forces, foundation screws become more and more loose, which also influences on reducing of preload screw tightening force (Fig. 1).                   3  Figure 1. Shematic review of reducing  of preload screw tightening force (Fp1) due to: (1) material relaxation and self-untightening of nuts, (2) elasticity modifying of screw, (3) elasticity modifying of screw and plates  The example with dimension of foundation screws M30 is discussed. Here, the preload tightening force is calculated:   Fp0 = As  σp = As  0,6  ReH = 561  0,6  240 = 80784 N (1)  ReH = 240 N/ mm2 – yield strength for the material of strength class 4.6  (Č 0370) As = 561 mm2 – area of cross section of screw dimension M30  When screw becomes loose, its force of preload tightening (Fp1) is reduced and can be determined from the condition of equal deformations of screw and plates:   λz0 + λb0 = λz1 + λb1 (2)   11110000bpzpbpzpcFcFcFcF+=+  (3)   Fp0 F λ Fp1 Fp0 F λ Fp1 Fp0 F λ Fp1  1                                                          2   213  1110001111pbzpbzFccFcc +=+  (4)   286008078410755,0110167,111044,4110828,1111116666011001 =⋅⋅+⋅⋅+⋅=++= pbzbzp FccccF N (5)  Overall stiffness of tensile screw (its lenght is l = 60 mm) is calculated from the expression:   6660 10828,11107110474,211111⋅=⋅+⋅=++=nsgz cccc (6)   60 10828,1 ⋅=zc  (7)  In this case, stiffness of screw head can be neglected, so that overall screw stiffness is calculated from different parts, as folows:  1. screw shank stiffness:   ∑= ⋅=⋅==ni iiss AlEc 1625 10474,2143060101,2111π (8) 2. nut stiffness:  ( )Edc n18,095,01 −=   for   d / P = 6 – 10  (which is the case in this example). (9)   65 107130101,219,01⋅=⋅=nc (10)  Stiffness of tightened plates can be approximately calculated using expression:   651044,46012,1268101,2 ⋅=⋅⋅==bbbbo lAEc  (11)  where, for the approximate calculation (if lb < D0), although this isn’t such case here:   ( ) ( )42,12684 33524 22202=−=−= ππDDAb  mm2 (12)  214 D = s + lb / a = s + 0,1 lb = 46 + 0,1  60 = 52 mm D0 = 33 mm – hole diameter s = 46 mm – width across flats of nut a = 10 – for steel  Stiffness of the relaxed screw (l = 106 mm in the analysed example):   6661 10167,111071104,111111⋅=⋅+⋅=++=nsgz cccc (13)   61 10167,1 ⋅=zc  (14)  Stiffness of the screw head can be also neglected, so that overall screw stiffness is calculated from the expressions:  1. screw shank stiffness:   ∑= ⋅=⋅==ni iiss AlEc 1625 104,11430106101,2111π (15) 2. nut stiffness:  ( )Edc n18,095,01 −=   for d / P = 6 – 10  (which is such a case here). (16)   65 107130101,219,01⋅=⋅⋅⋅=nc (17)  Stiffness of the most relaxated „plate“ (in analysed example lb = 106 mm) can be approximately calculated from two parts (one is screw shank – 60 mm, and the other is in concrete – 46 mm), as folows:   66612111 10755,011091,011044,41111⋅=⋅+⋅=+=bbb ccc (18)   cb11 = cb0 = 4,44 · 106 (19)   6522212 1091,04649,1396103,0 ⋅=⋅⋅==bbbb lAEc  (20)   ( ) ( )49,13964 3075,514 222022 =−=−= ππDDAb  mm2 (21)  Eb2 = 0,3  105 N/mm2 – elasticity module for concrete lb2 = 106 mm – 60 mm = 46 mm  215 D = s + lb2 / a = s + 0,125 lb2 = 46 + 0,125  46 = 51,75 mm D0 = 30 mm – diameter of hole in concrete s = 46 mm – width across flats of nut a = 8 – for concrete  This looseness of foundation screws significantly influences on increasing of their elasticity and on modifying of their loading (Fig. 2) and thus also on their reliability.          Figure 2. Shematic review of deformations of screws with different elasticity  Theoretically, the bigest alowable operating force of untightened screw is:   2692838078400 === γprFF  N (22)  where: γ = 2 ÷ 4 – coefficient of tightening for variable loading,    it is adopted γ  = 3. and the bigest alowable operating force of loose screw is:   953332860011 === γprFF  N (23)  Since machine or gear units housings are compact, approximately equal additional deformations occur to all screws (Fig. 3). Thus, it follows that screws with biger elasticity are less loaded and receive smaller operating force (Fr1).  Because screws have to transmit operating load, and stiffer screws receive biger loading, than it is calculated, which means they are exposed to greater possibility of occuring cracks than untightened elastic screws are. Since both screws, tightened andf loose, have equal deformations:   ∆λ = ∆λz0 = ∆λz1 (24)  it follows:   11zzzozocFcF ∆=∆ (25) Fz0  Fr1 Fz1  λ F Fp1 Fp0 Fr 0  216 which can be written and calculated as:   5013785310828,110167,1660011 =⋅⋅⋅=∆=∆ zzzz FccF N (26)  where:   7853269281044,410828,110828,166600000 =⋅⋅+⋅⋅=+=∆ rbzzz FcccF N (27)  and the real operating load of loose screw is:   8256501310167,110755,010167,166611111 =⋅⋅⋅+⋅=∆+= zzbzr FcccF N (28)  which is a little smaller than it is foreseen for loose and relaxed screw (9533 N), but it is much smaller than real operating load of tightened tensile screw.                    Figure 3. Shematic review of deformations of screws with different elasticity values for the same screws deformations  This means that for transmission of overall loading (Fr = z1 Fr1 + z2 Fr2 + ...), tightened screws have to receive biger loading.  Analysing the example with two screws (one is tightened and the other is loose), screw force in tightened tensile screw can be calculated from the expression: Fr 0 Fz0 Fp0 λ F  Fz1 Fp1 F Fr1  λ  217 92801412051044,410828,110828,180784666000000 =⋅⋅+⋅⋅+=′++= rbzzpz FcccFF  N (29) and screw force in loose elastic screw is:  33613825610755,010167,110167,128600666111111 =⋅⋅+⋅⋅+=++= rbzzpz FcccFF  N (30)  So, it follows that loose and relaxed screw is less loaded for 64% than tightened screw, and thus possibility of its breakage is much smaller.  Maintaining the same force of preload screw tightening (by periodical tightening nuts of foundation screws), this problem can be corrected, ie. operating load in each screw can be nearly equated (Fig. 4), but the problem always remains actuel and present.           Figure 4. Shematic review of deformations of screws with different elasticity for the same preload tightening force (Fp0 = Fp1)   For providing the same loading of foundation screws, it should be necessary to determine preload tightening force, ie. tightening torque of each screw particularly. 3. Conclusion On the basis of implemented analyse, it follows that looseness and relaxation of foundation screws have significant influence on reducing their operating load. Because of that, tightened tensile screws receive biger loading than it is foreseen, and thus cracks and breakages of those screws can more easier occur. So, it can be concluded that periodical maintaining and checking the λ Fz0 Fz1 F Fp0 = Fp1 Fr 0 Fr1   218 tighteness of foundation screws is necessary, so that calculated screw tightening torque and system reliability can be preserved. This analyse is done for a small foundation damaging (l = 106 mm), but when there is a biger damaging, the situation is much worse because breakages easily happens. Also, when there is a problem with much more screws with different damaging, ie. relaxation, the calculation is more complicated. References [1] Kuzmanović Siniša, Machine Elements 1, Faculty of Technical Sciences, Novi Sad, 2005 [2] Kuzmanović Siniša, Constructing, Shaping and Design 1, Faculty of Technical Sciences, Novi Sad, 2006 [3] Kuzmanović Siniša, Constructing, Shaping and Design 2, Faculty of Technical Sciences, Novi Sad, 2005 [4] Kuzmanović Siniša, Milan Rackov, Ružica Trbojević, Analysis Of Influ-ence of Relaxation of Foundation Screws on Their Load Capacity, Annals of the Faculty of Engineering Hunedoara, 2006, Tome IV, Fascicole 2, (ISSN 1584 – 2665), Faculty of Engineering Hunedoara, 5, Revolutiei, 331128, Hunedoara, Romania, p. 167-172 [5] Miltenović Vojislav, Machine Elements, Faculty of Mechanical Engineer-ing, Niš, 2006 [6] Robert L. Norton, Machine Design – An Integrated Approach, Worces-ter Polytechnic Institute, Worcester, Massachusetts, Prentice-Hall, 2000 [7] Milosav Ognjanović, Machine Elements, Faculty of Mechanical Engi-neering, Belgrade, 2006 Addresses: • Prof. Dr. Eng. Siniša Kuzmanović, University of Novi Sad, Faculty of Technical Sciences, Trg Dositeja Obradovića 6, 21000 Novi Sad, Serbia, kuzman@uns.ns.ac.yu • Ass. Mr. Eng. Ružica Trbojević, University of Novi Sad, Faculty of Technical Sciences, Trg Dositeja Obradovića 6, 21000 Novi Sad, Serbia, ruzica@uns.ns.ac.yu • Ass. Eng. Milan Rackov, University of Novi Sad, Faculty of Technical Sciences, Trg Dositeja Obradovića 6, 21000 Novi Sad, Serbia, racmil@uns.ns.ac.yu 

Causes Analyse of Breakages of Foundation Screws

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Causes Analyse of Breakages of Foundation Screws

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