L) 56_5 U. S. Department of Agriculture, Forest Servic e FOREST PRODUCTS LABORATOR Y In cooperation with the University of Wisconsi n MADISON, WISCONSI N WOOD-BEAM DESIGN METHOD PROMISES ECONOMIE S By J . A . NEWLI N Principal Enginee r G. E . HEC K Engineer and H. W . MARC H Special Consulting Mathematician Published i n ENGINEERING NEWS-RECORD May 11, 1933 By J . A. NE?LIN, Principal Enginee r G. E . HECK, Engil .ee r and H. W . MARCH, Special Consulting Mathematicia n A new design method for calculating the horiz-ontal smear im wooden beams, developed by the Forest Products Laboratory th--reu .mathematical analysis and tests, assumes that, because of the s+hear distortion in the vicinity of the ' base of checks or fissures0uza :t ar e present in all beams, the upper and ldwer halves of the beam act t o some extent as independent beams . The result is to relieve the .mean shearing stress in the neutral plane, and since this reduced sheatin g stress is not comerok in present design methods, they are often direr conservative and therefore uneconomical to use . In certain cases, suc h as floor beams of highway bridges and railway ties, usual design method s predict stresses that are two or three times the ultimate sheaxing :stres s of the material . still these members are carrying their loacl .s withou t failure . In the discussion presented here an attempt is male to emp3 :4im i the elastic behavior of a checked beam under load and this to 3 :Aai4 ;ie pr*d:io = the discrepancy existing between the facts of experience and t . The background fo t tions of the usual methods of calculating shear this explanation is furnished by an approximate mathematical andlys s o f the problem combined with the results of a series of about 200 tests, i m which the loads causing shearing failure were observed on built-u p artificially checked beams varying from 3/4 x 1-1/2 in . to $ x 16 in. in cross section and with varying amounts of checking . On the basis of the theory and the results of tests, practical directions are given for th e more realistic calculation pf loads that will cause shearing failure* In the tests, built-up artificially checked beams of cares Iy matched material were used . A typical section is shown in Figure 1 . The checks are placed in the middle of the lateral faces, as this is th e position in which they are most likely to occur and to cause failur• .e by shear . A characteristic failure, by horizontal shear, of a naturall y checked structural timber loaded at two points is shown in Fire 2 . IA all other tests referred to in this paper the beams were loaded at one point only . Two-Beam Actio n An explanation of the behavior of a checked beam is feuxid iYn. . the fact that, owing to the shear distortion in the material in t Rl005 I 'I i Figure 1 .---(left)--Cross-section of a built-up artificiall y checked beam such as was used in the test, The four parts a were , glued to the central portion or web b . The joints c were' paraffine d to prevent sticking of any glue that might protrude from adjacen t joints and to minimize friction . Figure 2 .---(right)---Characteristic failure of a timbe r beam, failure occurring in horizontal shear in the neighborhood o f the neutral axis . 81005 f vicinity of the base of the checks, the upper and lower halves of th e -beam act to some extent as independent beads . An approximate mathemati= cal analysis indicates a vertical distribution of shearing stress as w shown in Figure 3 . The . shearing stress _plotted . is the mean sl a The exstress at failure averaged over the full width of the beam . istence of the "two beam action" is evident from the curves . The 43 action mean shearing stress in the neutral plane is relieved . of the upper and lower halves of ihd beam . as partly i epO .ent beams . The significance of this action is emphasized by the f, t:11er- result of the mathematical analysis that the reaction R at the supp-ett . nearer the load may be expressed as the slam of two terms, of which only th e first is associated with shearing stress in the neutral plane . Thus it is found that - k fE r I , ~M _11 :. (1) R = B + A/a2 where (e, B - 2/3 Jbh, 1 1 J being the moan of the shearing , stress in the neutral plane over t-V 1 ~■ full width of the beano ; and b and h being the width and depth . 7e, I 6 r of the beam. The portion B of the reaction is precisely the reatt that would be associated with the mean shearing stress J in the neit'ka,l plane in the usual theory of beams and may be referred .+to as the. gars t P - '. beam portion" of the reaction . x the second term of equation 1, a is the distance ftem Ake ?i,, %s the nearer support, and A is a quantity determined c .4 e.y by the dimensions ' of the beam, the longitudinal Youn gs s modulus ,' -aid. the mean longitudinal displaceient oh the lower face of the upper half of the beam at points imt ediatelf o+rer the supports The portAan of th e reaction expressed: by this term is attributed to the inAepend!enn as ±b n of the upper and lover halves of the beam and is not assoc sated wit h shear in the neutral planes It is the "two-beam porttom" of the reaction . The values of the two'beem and single -,beam portions, Of the-reaction at ,a . the ihsta it of failure were determined by testing a :large h-u er of 1eawsi ar4l$er combining the results of tests of is ,of carefully matdhe d beams loaded to failure) the concentrated. loads be '.hg ap aliec ai different points . On entering the results of each pair of tests it egtt ;on 1 twd equations were obi-aimed. that douL be solvdd. &'t r B axle.. s ~f '1 TestResults Suggest Design Procedure 1 - In this procedure it was assumed that tie values o$ A and 66 the mean shearing stress J in the neutral plane at faiiuee we-e ,indepe n of the position of tha load,. The justification of bota ass ns i s found in the fact tkgl 'the results. of the tests were representei . appDxiiec . Th e matteln by eft-ion Z' "th A and B as constants fox the. ash©le - .. . r., i O '61 O Q+ 0 O H p O .~ ~Ocd a) r° 4--I OO 0 O 0 •-1 H H O 4-I 0 zd a) .~ O 0 E-f Fi -p ed -P . H F-t cd cd F+ N Q) -p 0 0 H cd 0 {.' O cd P+ 0 H O faD cd 0 .r4 ,O a) O 4-f cd 0 4+ O ▪ 0 If\ •ra cn H ~ cc 0 r4 •r•I 5 0 •r- 4 F-i . +~ N cd •r+~,-t cn H 0 I rq a) H >4z cd k cd q O 0 R1005 W Cl) -P O O +' U] bS 0 N tests showed that as a concentrated load approaches a support, th e two-beam portion et' the reaction increases rapidly, while the single beam portion associated with shear in the neutral plane, remains prac tically constant and becomes a correspondingly smaller fraction of th e reaction . As a consequence, the point of application of the minimu m concentrated load to produce failure by shear is not just inside th e support, as is commonly assumed because of the usual simple beam theory , but is at some distance from the support . The results of the tests o f three,series of beams are given in the accompanying table and are repre sented graphically in Figure )4e The single-beam and two beam portion s of t64 reaction shown in the table were computed by combining in pair s the mean reaction corresponding to a = 7 with that corresponding t o each of the other points of loading . If these reactions are combine d in pairs in other ways, results varying somewhat from those shown ma y be obtained . The results of the numerous tests may be summarized in th e following statements : For checked beams with a span-depth ratio of 9 to 1 the point of application of the minimum concentrated load causin g failure by shear is at a distance from the support approximately thre e times the height of the beam . The distance of the critical point flor a the support is somewhat greater for longer spans and somewhat less fo r shorter spans ; but in any case, for loads applied at three times th e height of the beam from the support, the two-beam portion of ta e reaction at the nearer support is approximately one-sixth . Design Recommendations 1. If there are moving loads, place them so that the heavies t concentrated moving load is at a distance from the support of thre e times the height of the beam . After this has been done the followin g recommendations apply both to static loads and to the moving loads . 2. Neglect all concentrated loads within a distance from th e support equal to the height of the beam . 3. Consider all concentrated loads that are from one to thre e times the height of-the beam from the support as being at three time s the height of the beam from the support and compute the resultin g reaction . Neglect one-sixth of the reaction . This one-sixth is th e two beam portion of the reaction and is not associated with shear i n the neutral plane . (Note : For very small span-depth ratios -- less than G _to I -place the loads designated in recommendations 1 and 3 above at. the midffle + of the span .) 4. R1005 Consider all other concentrated loads in the usual manner . -3- '11 1 "" >1 H i 0. o H al d 'a3 o S p o Ii l r-I H H 0 +3 "-' a) +3 lk . II) a) +3 0] H 0 11 a R1005 11 .r fait I . Results of tests of built-up, artificially checked spruce beam s (Various lengths ; breadth, 2-1/2 in . ; height 4-1j2 in . ; depth of checks, I in . ) Distance Load at failure (a) froth } .. : load to i 2 g-in. : 42-in. : 63--in . : nearer : span : span : span : support Inches 7 .0 :Pounds, :Pounds : Pounds : : 5,770 Mean reaction at nearer support Pounds : Single- : Two- : Two: beam : beam : beam : action : action : action : x a2 : Pounds : Pounds Pounds 1,872 : 91,70 0 4,226 10 .5 5,460 : 4,240 : 3,560 : • : 3,18 6 2,354 832 : 91 :700 14.0 5,680 : 4,430 . 3,780 • 2,91 1 2,473 43 g 85, 80 0 : 2,64 9 : 2,349 300 : 91,900 4,800 : 3,630 • 2,41 0 2,183 : 221 100,100 • 2,140 : 170 102,000 17 .5 5,050 : 4,660 : 4,800 : 3,460 21 .0 24.5 ; . . : : 3,780 : 2,31 0 28 .0 : ; : . . ; 4,050 • 2,250 31 .5 : : : 3 :970 K O05 1)985 -4- Rai1odAn bensioav • 1,869 : 116 115,100 S 5. Neglect 10 percent of the uniform load . This 10 percent is taken care of by two-beam action . 6. Assume lateral distribution of the various loads to adjacent parallel beams or stringers on the basis of their assumed placement . 7. Calculate the shearing stress in the neutral plane by th e usual formula, using the full width of the beam and all of the reaction , obtained as directed, that is not included in the two-beam portion . S . Use 90 percent of the safe shearing stresses previousl y recommended by the Forest Products Laboratory, since these recommenda tions did not take into account the effect of two-beam action , The .use of these recommendations will result in a considerabl e saving of material . The highway bridge, on account of the effects o f lateral distribution of the loads to adjacent stringers, is probably th e structure that will show the greatest difference in the results of th e application of the new and old methods . An application of the tw o methods of calculation to a highway bridge follows : Consider a single concentrated load P moving along a floo r beam or joist . Assume the span to be 16 ft ., the height of the joist , 11, 16 in ., and its width, by 5 in., and all dimensions full n fina l size . Assume also that when the load is in the center of the span th e floor carries one-fourth of it to each of the two adjacent beams, on e on either side, In such a case it has been calculated that he sid e beams will carry a little more than 40 percent of the load when th e load is at three times the height of the beam from the support . The shearing stress, J, in the neutral plane is given b y J=x 2 6 x bh (4) where R = reaction due to the load P placed at three times the heigh t of the beam from the support . But, according to the assumptions as to lateral distribution , R= 10 . x1~}xP= OP (5) Assuming 100 lb . per sq. in. as the tabular value for the safe shearing stress and reducing it by 10 percent as explained earlier , it follows from (4) and (5) that 90 = 3 x.2x x_ P 2 20 5 x 1 6 (6 ) Hence, P = 12,800 lb ., the safe load in shear . R1005 -5- plane Itlt ' tabular value noting that at the en d off' tl? beh m the safe load P is found frog r s r.• 1 1 I • 1r • I • 11- 1 • ' I II I ~I - a 1 1 I r. III+I LiL I 77 J II This method of computatio n '. one-half that permitted b the new_ m11d,. r 11 90 2 x " Then P = 7,680 i4 . P 5 x 16 lb . This load is more than 40 percent greater than that by the old method of calculation s 'R'1005 -6- permitte d