104 710 ANALYSIS OF CONCRETE SKEW RIGID FRAMES Thesis for flu Dogma (if M. 5. MICHIGAN STATE COLLEGE Afr‘u M. Chowdiah 1949 , __-___ "9.9—... 999,9 .x, ‘1 , THESIS This is to certifg that the thesis entitled ANALYSIS OF CONCRETE SKEW RIGID FRAMES presented hg Atru M. Chowdiah has been accepted towards fulfillment of the requirements for 21.5. degree in C .E. €21é/é4a/ . u u--;: Major professor Date August 30 1949 07169 .A time-“.1 . 'g. .9 ' .u.‘ :41 .9"I.I\L'E n. .' ANALYSIS OF CONCRETE SKEW RIGID FRAMES By Atru M. Chowdiah ~— A THESIS Submitted to the School of Graduate Studies of Michigan State College of Agriculture and Applied Science in partial fulfillment of the requirements for the degree of MASTER.OF SCIENCE Department of Civil Engineering 1949 ({H E533 TABLE OF CONTENTS Introduction Figure l & 2 Figure 3 & 4 Figure 5 Figure 6, 7, & 8 Outline for Determination of Reactions Details of Analysis/ Example Solution Dead Loads for the Frame Table l " 2 " 3 Live Load Table 4 " 5 " 6 7 " 8 9 N 10 " ll " 12 n 13 n 14 n 15 Conclusion Bibliography 218340 , \IO\\JI¥\ 8-19 20 21, 22 23 24 25 26 27 28 29 30 31 32, 33 34, 35 36 37 38, 39 40 41, 42 43-45 46 47 INTRODUCTION The first paper on the subject entitled "Analysis of the Stresses in the Ring of a Concrete Skew Arch" by J. C. Rathbun was published in A.S.C.E. Transactions, 1924, p. 611. At that time Professor Rathbun's theory was questioned and the A.S.C.E. appointed a a special Committee on Concrete and Reinforced Concrete Arches to give a practical test of the theory.Professor G. E. Beggs of Princeton, N. J., was engaged in making deformeter analysis on models of skew arches. At the same time, Professor Rathbun was making mathematical analysis of the structures of the same proportions. After the analysis was completed individually with- out consultation, results were compared and were found to be in close agree- ment, and so Professor Rathbun's theory was accepted. In 1925 Professor Rathbun extended his theory of skew rigid arches to skew rigid frames and an adaptation of his theory was made by Mr. A.G. Hayden and is presented in a book "The Rigid-Frame Bridge" by A. G. Hayden, John Wiley & Sons, Inc., 1940, 2nd Edition. In the presentation made by Mr. Hayden, it is necessary to derive and solve 4 simultaneous equations. ‘ Later Mr. Hodges presented a paper "Simplified Analysis of Skewed Reinforced Concrete Frames and Arches" and it is published in A.S.C.E. Transactions (1944), p. 913. Mr. Hodges has simplified the work by re- ducing the solution of four simultaneous equations to solution of two simultaneous equations. For this thesis, Mr. Hodges' paper is the principal reference al- though references are made to other papers and one example has been worked out to illustrate the application of the theory. The analysis has been made for assumptions that rotation may occur about any axis. There is no justification to assume complete fixity in 2. any direction since the degree of rotation required to change the effect of complete fixity is small. On the other hand, any assumption of com- plete unrestricted rotation about any axis is not correct. Therefore, the frame is analyzed for: Assumption 1 — rotation about Z axis. Assumption 2 - rotation about Z and X axes. Assumption 3 - rotation about Z, X and Y axes. i t .. z e A?\ u . . . 24 . ,_ t i. I . a , I i, A n .I . l 4.x r. . r 1 iv A I , I. I. s w t, .r\ . , II . . r I .383 I i I L. JLI I Ii i V A / a . ItvtIr I. . I .v Hymn, , fl. 7 \1 i. ,I i I.y . i , v, ,a W : ., . _ I 2. . .. _ l e . I . e a; I z I II ._ I I I .I . . _. \, . . .14 \_ , I /. A I .\T r L _ / I I I I ,I I I I ,1 I I Ian. . _\ , .I III . itIi II . \Ii IA? (V II _ I“ I: I I ,. a y I l . )ZIK4II IJIIIIII-nrilliilxsfitt . .Itéy I I . vi . 1 N a 3A .7, K. .7 ., . .3 N IIIIII IIIIIII 1C aw .N I L , a» a , ., .14 W m i. a I . _. W . ,r w _ ,L , w,/ .a. 5 m i a a . . I» . :\ 9 r I a. I, l. ,N e 9 .. _ l. e a, , _, I. ,_. I) I. .V m a 4 It III I «III I III . .. i. .w, . I u. r r _ / .3 Ifri ,, H I. 0 ,I . Ii. ”i / P M i a A ATIIII. r. V I Al .I N. i a, I H I .a _ l . . a w m a h, _ I . III .I H In .I I /. , I , .V I I .I\ . r I ”I a .. I, , : I I” , I _, , n ,Ii . , , . _ I_ N- w\\ . , / , fi \ , . x x , .I. I /. , I, , . .u . / , . _ I a , n » . .II I I II. .a I I _ I , n I. r: J I I l r a I . r I a I .,. S. . :7 U ,. I IN , I. , I I. w , I a. . IIIw a , ., I” _ I I I , f .I . _ . x, I“ I . A I . .I C? .1 , Is a ,I, II , I a . I ,I (\I‘ rIII IIIII _ l 9 ---_- 9—9.... D ., , - .9——_’ OUTLINE FOR DETEBMINATION OF REACTIONS The skew frame,if restrained against translation and rotation in all three principal directions, would have six reaction components. Let the principal axes be designated by X, Y, and Z. The X axis is horizontal and is perpendicular to the axis of the abutments, Y axis is vertical, and Z axis is parallel to the axis of the abutment. The reaction components at the right footing are Rx, Mx’ Ryr, ”yr, Rz, Hz. The usual assumption for footings on earth foundations is that there can be free rotation about the Z axis. So Mz = O. Ryr is obtained by statics and will be the same as the reaction for a simply supported beam over a span = the span of the corresponding rectangular frame. The reactions to be determined are Rx’ “x! Myr’ Rz. It has been proven before that the component H of a two hinged rect- angular rigid frame is so nearly equal to the Ex of a two hinged skewed rigid frame of equal right span, that the difference is negligible for purposes of design. This assumption has greatly reduced the amount of labor in the solution of the problem. The reaction components H and its equilibriant GH are the most important in the design. The equilibrantwéH is the principal part of the R2 component. The remaining part of RZ which will be designated RE: the reactions Mx and My constitute a separate group which depend on the torsional elastic de- formation,so we have to solve three simultaneous equations for the three 1 2 so we need to unknowns. However, My turns out to be a function of R solve only two simultaneous equations for the remaining redundants. The solution of two separate elastic systems is certainly less likely to be of error than a combined elastic system, and so it is definitely an advantage. DETAILS OF ANALXSIS The forces acting on the structure are shown in Figure 1 for vertical loading and in Figure 2 for balanced earth pressure. Unbalanced earth pressure is omitted due to the fact that it is not usually considered in practical design. Let us determine the reaction components at the point of origin at the right support. The solution of the reaction components is done in two stages. Stage 1: This consists of computing the horizontal component H of the reaction at the footing of a two-hinged rectangular rigid frame (designated as the "corresponding rectangular frame".) which is the projection of the skew frame on a plane parallel to the XY plane for loads on one foot width of the bridge. The H is equal to the R1 of the skew for vertical loading and earth pressure and H(l +(s2) = Rx in case of temperature change. Stage 2: This comprises of setting up equations and solving for the redundants R%, Mx, “yr for the structure in equilibrium under the forces H, CH, Ry, Rz, Mx’ and ”yr acting at the origin. The equations are set up to satisfy the following conditions: There can be no displacement along any of the axes. There can be no rotation about any axis except for Assumption 1 ~ about the Z axis. 'Assumption 2 - about Z and X axes. Assumption 3 - about X and'Y axes,and Z axis. The internal stresses acting at any point P are represented as the re- actions exerted by the portion to the left of P on the portion to the right of P. The reactions whose directions are unknown are shown in Figure 1 in their assumed positive directions. The subscript on each moment term indi- cates the axis about which the moment occurs. 9 Geometric Relations and Definitions. The coordinate axes, the applied loads, the reaction components, the components of the internal stresses acting at any point P lie in a plane parallel to or perpendicular to the XY plane. The profiles shown in Figures 1 and 2 are projections of the skew frame on a plane parallel to the XY plane. The point P is on the neutral axis and at the center of any subdivision S. The internal stress components at P are along the axes u, v, and z axes with origin at P. The angle 9 is the slope of the neutral axis (v axis) at P. The quantity u is the component parallel to the u.axis of the distance between P and the origin 0. Also: c = coefficient of linear expansion for concrete and reinforcing steel. to: temperature rise or fall, in degrees Fahrenheit. w = equivalent fluid pressure of earth, in pounds per square foot per foot of depth. M = simple span moment on the corresponding rectangular frame, for vertical loading. Modulus of elasticity for concrete under axial stress Modulus of elasticity for concrete in shear = E G = 2.67 (assumed) F = factor of torsion (corresponding to the moment of inertia I in case of flexure) :33 3.58 H = horizontal thrust in the corresponding rectangular frame for the particular loading. h = height of the structure from bottom of footing to the highest point of the structure (crown). (6 = the summation sign Z (- = 9-: tan—C- . . _ . ‘ l Derivation of Equations for the Redundants R 2’ Myr’ Myl! Mxr’ Three sets of equations are derived to satisfy the three different 10. assumptions. Each set of equations is derived for vertical loading, bal— anced earth pressure and temperature change. The assumption is made that the effects of all thrusts and shears are negligible in considering deflections. In the following equations, the effect of only the torsional moment Mv at all points of the structure is con- sidered. Let Mv -moment at P about the v—axis due to all forces acting. mv(z) = moment at P about the v—axis, due to a unit force acting 1 along and in the direction of R s' mv(0x)= moment at P about the v-axis due to a unit moment acting along and in the direction of Mx' mv(oy)= moment at P about the v-axis due to a unit moment acting along and in the direction of My. The equation for no deflection along the Z axis is: '3 (S Mug" =0 The equation for no rotation about X-axis is: E. (S Mvmgfox)= 0 The equation for no rotation about Y—axis is: g (s Mvm;(o.l)=o Assumption 1: The expression for ”V for vertical loading from Figure l is: Mvzpx Q.x sin 5 - Rgu + ”x cos B - Myr sin b + Ryr (-x cos E - W (-(x-xl) cos 5 = Rx (-x sin 5 - Rgu + “X cos 5 But for vertical loading Rx = H (la) (lb) (lo) (2) Also mv(s) mv(ox) mV(oy) Due to symmetry From Figure l: 2 “u (38) = cos 5 (3b) = -sin 5 (30) s sin 5 cos 6 _ ( F - 0 (Be) and (s y Si; 6 cos 5 = 0 (3f) R3 = t H + R1z V (38.) = x sin 6 + y cos 6 (3h) 2 u - y 0013; (3k) sin 5 Substituting Eqs. 2 and 3 in Eq. la, we have for vertical loading, -H 4. ‘5 mm?“ + R1. (5 Biwen <5 i3 (s ggg§§_§ + Myr (s u sin 5 (s u 0:8 5 0 2 (s g3+ H(- (51%:‘13ifi... (-H (S BF (8 -% cosi + Myr ('8 Leigh (sugosib + R15 (3 if”: 0 (SI-1:3 ‘Mx (8 %cosb + Myr (8 LEI—£33 (8. Mo ufcos 6 _ H (s uchos 5 i (43) By similar substitution of Eqs. 2 and 3, in Eq. (lb) we have for vertical loading 3 6 x sin b cosgg (s 4£_H u cos 6 ( H F - F l _(s Ra ucosh + s M c0525 8 Moi-008235 F ( F + ( T‘ = 0 (s H G— x sin Qgcos L _ (s (:H x sing§:cos E . F F 2 1 s _ (s eHyCOS as _ (s 12311005 F F + (S ”X C082 5 + (8 MO 4 0032 E F F = 0 l s u cos 5 8 cos2 B . -——'*'—‘ - M -————- Pie ( F x ( F 2 ( s M cos 5 s 2 = - _ ycos E < ( ( J? H ( -—-—-I-,I————- I (a) By similar substitution of Eqs. 2 and 3 in Eq. (lc), we have for vertical loading - (3 “X81134 I. (s W F F _(s {lisinBcosh +(s Myrsinzb F F _ (s Moésinicosb F II C) In the expression on the L.H.S., the 3rd term becomes 0 and substituting u = x sin 5 + y cos 6 in the 2nd and 5th terms, we have .. (3 E E2; ginz E _ (8 EH X 111295. F F A 2 + (S EHYFsingcosg +(SRlsxliin 15 R13 y sin 6 cos 5 F 13. . 2 2 l s x Sin 6 s sin 5 s M sin 6 cos b R ---———— -‘ _ O 3 ( F + arm. ( F — (_ ( F __ (4c) Equations 4 are modified for application to earth pressure and temperature change as follows: 1. Balanced earth pressure in Eq. 2 Rx = Egg - H Ryr = O W = O .3 Mo = 0 In equations 4, Mo term is omitted and H is replaced by E§3-- H. 2. Temperature rise: The increase in span = C to L Deflection of right end in direction of R5 = - (—-ctO L (See Figure 6.) Equation la represents the deflection of the right footing in the direction of Rs .. (la) becomes (S EXEXLfil - é-cto L = 0 film multi plying by the constant Q s 1 1 (8 31:12:19-3 (-ct°L II C M m n I O Also W: 0 : Ryr = O 3. M0 = O 2 ande=H(l+(-) .1 In Eq. 4a Mo term is omitted, H (l + 9f) is substituted for H and the term LIE-E- (- ct°L is added to the L.H.S. (or 1%,, (- ct°L can be added. to the R.H.S.) In Eqs. (4b) and (4c) omit the Mo term and replace H by H (l + éfi3. (See Figure 7.) For a symmetrical frame as shown in the Figure, there are pairs of points like P and P' in such a way that . 2 . 2 Sin 5 for P = Sln E for P' and P is at a distance x from origin and P' is at a distance (L-x) from origin 0 2 '25 25 .3 For the 2 points, (S §_§in E ) filQ——— = L §i%——» .3 For the whole frame, o 2 ' o 2 where (S 51; 5 is the sum of 'ELEFHQ for half the number of sin2 5 . . 3 points on the frame. For sake of convenience, 1f ( F has to be s- 2 the sum of -=$%~—§ for all the points on the frame, then s x sin2.§, L s sin2 E ( ._._._..... __ .... ( ____.__F .._.. For symmetrical loading in Eq. (4c) Mo term becomes zero. - 2 . l L s filfl__§ s sin 6 yr 15. Also R1; and Mx at right support = R1; and MX at left support Taking moments about left support (GH+R%)L+My1+Myr—H(—L=o; My1+Myr=-R12L L 0.0 Myr=uyl=-R]é 2 For unsymmetrical loading . . 2 s M 8111 Bcos§ 3 sm 35 _e < °F - Bl; a < F r 4 My 51112 E (s ---- F .. 8 MO sin 5 Cos? - G ( ~41; - R1 L (S STHFE a E F _. l L (- 3 Mo sin 6 €03? where C = ( F (S §ll12 5 F but My. + M: -n1.L 0.. W1 = - C '- R13 ~123- Now supposing the unsymmetrical loading is made symmetrical by adding equal loading placed symmetrically with the unsymmetrical loading, then Myr due to additional loading = -C - £23}. Myl" " " " =C-R15L 2 ... The total Myr due to the combined balanced loading = - ng L and M71 = 41]; L. Designating the Myr and Myl for the combined balanced loading by My, we have My = _, R13 L o L = .. R33 2 __ 52L 16. For unsymmetrical loading “yr: 0+? Myl = -C + %1 where my = Myr or Myl for balanced loading. Mx and R13 for unsymmetrical loading are half of the respective quan- tities for symmetrical loading. So in case of unsymmetrical loading also it is only necessary to solve only 2 equations in which quantities of the balanced loading are used, and to apply the correction discussed. flaking the aforesaid substitutions, L.H.S. of Eq. (4a)for symmetrical loading 2 2 r l s .£_§ifl.ifi =R%(51_1F_ Mx suco;5-%Rs( F l E s u2 L2 s sin2 5 g M (s u cos 5 9.. Eq. (43) becomes u cos 6 1 8 112 L2 3 . 2 5 - S __..._.._........ R,“ E( -F"‘Z—‘ ( F 3 Mx ( F for symmetrical vertical loading, M u cos 5 s 21;£2§J§ _1 .. S .. - G E ( Q“’F H ( 'F 3 (5a ) for balanced earth pressure, 2 .. wh 15 ..-_ - H 3 8 ma. _2 ( T ( F (5a ) and for temperature rise, 0 L 2 s u cos 5 =rfflg—- - H(1+H ( 'Lf"; (551-3) Eq. (45) becomes 315 (s u__c_______os 5 _ M (s 0032 5 F X F 17. for symmetrical vertical loading, 2 .2 =_(_£ (s an cos 5 _H (s ycgsbg (Sb-l) for balanced eartg pressure, 2 b =-(.E 2%-11} (Si-figL— (Sb—2) and for temperature rise, = - {-H (1+92 ) (S L—-——’ cgsz '5 (Sb—3) For symmetrical loading, Myr = Myl = - ng L21 (50) For unsymmetrical vertical loading, Myr = g1 + C and I"lyl ‘3 4’55» ‘ C (5d) where My is Myl or “yr for balanced vertical loading. J 9- (8 Mo sin i cos B and C is = F (s sin2 g F in which Mo are values for actual vertical unsymmetrical loading. Also, R%§and ”x for unsymmetrical loading are equal to one half of R13 and MK for balanced vertical loading. Eqs. 5 constitute a set from which Biz, Myr, Myl, and Mx are deter- mined for Assumption 1. Assumption‘g. Since “x: 0, it is only necessary to solve one equation, say Eq. 5a. For vertical loads and temperature rise in the L.H.S. of Eq. 5a, omit the ”x term. For balanced earth pressure loading, we assume that the frictional resistance of the concrete surface also acts. Assuming the coefficient of friction of earth on concrete = .5, the total frictional force =.E%E. acting at a distance of.%.above the bottom 18. of the footing on either faces. 3 .3 The moment at the ri ht end = Wh g 17‘ , wh3 . wh3 .. The contribution towards Mv due to IZ- 13 'I2"' cos 6 .2 In the expression for Mv in Eq. 2, omit Mx—term and add the term 52: cos B 12 . . wh3 s u cos b :. The R.H.S. of Eq.(5a)will have an additional term = '12. F .2 Eq. (5a) becomes: 2 L2 2 R???“ 2‘ <3 T51“ 1") 5 wh3 =-(—H (SW4. 12.. (s ucosm A sum ion 3. Since “X = 0 and My = O, in the expression for Mv, the Mx-term and My-term 3 are omitted and add E2—-cos E. .2 Eq.(5a) becomes: u2 E18 (8 1r“ ' 2 _ ‘12..- s u cos wh3 s u cos B "‘{2 H}(‘Lf~"§+'1‘2”( "‘ Internal fiiresgeg a; g Egint. The stresses at Point P are represented bwav, Tv’ Mg, and T3“ Mg and Tv are determined from the analysis of the corresponding rectangular frame. Tg = Re, and MB is computed after solving for the redundant re- actions and substituting the values in Eq. 2. But for design purposes, the.stresses have to be transformed in di- rections parallel and perpendicular to the skew. Let ML direct moment axial thrust "T = torsional moment shear 19. The transformation is shown in Figure 3 and ML = ”3 Sec. V “ Tv Sec. V *3 I’.“ ! MT=MaTanV-Mv TT=TvTanV—Tg where V = projection of skew angle-G-on the vs plane The “L and TL control the cross sectizlof the structure and the longi— tudinal reinforcement and they are entirely independent of M5 and Tv3 MT and TT determine the transverse reinforcement necessary. This concludes the analysis of the skew rigid frame. 20. EXAMPLE Analyze a rigid frame 16' being span of corresponding rectangular frame, skew angle = 45° for w = 30 lb. per square foot per foot of depth, and live loading H-20-44—no impact, and temperature change t =‘t 45° F. for the frame shown in the Figure 5. 21. Misa- Part 1. Analysis of the corresponding rectangular frame.~ The frame is divided into equal parts with s = 4' The dead loads assumed acting at the center of the subdivisions are shown in the Figure 5. The earth pressure acting is also shown in Figure 5- The origin of coordinate system is chosen at the bottom of right footing. (3M1. By the formula H = I (s .i I where H = horizontal thrust E II simple span bending moment Y = Y coordinate of any point H II moment of inertia of the section at the point whose coordinate is Y. Tables are self-explanatory and Tables 1 to 5 show the complete analysis for obtaining the moments and normal thrusts at the required points of the frame for various loads. Some definitions. In Table l: M = simple span moment at a point in KIP-FT. = horizontal component of reaction at the origin in KIPS H t = depth at any section in ft. I moment of inertia. moment factor which is the influence line ordinate for the moment at the particular point. M. F. Tables 2 and 3: M.F. = moment factor obtained from Table l. m = (M.F.) x load in KIPS H Kip ft. 22. N.F. = normal thrust factor N = normal thrust in KIPS Table 4: V = vertical reaction at the hinges due to the earth pressure. Part 2. This consists of tabulating the maximum moments and thrusts which control the cross-sectional areas and longitudinal reinforcement after applying the necessary skew corrections. This is indicated in Tables 6 and 70 Part 3. This comprises of analysis of the frame for torsional moments and shearing forces acting in the transverse direction. This is carried through in Tables 8 to 15. DEAD LOADS FOR THE FRAME 23. POINT WT. compummows wr. m KIPS 4 = I. x 1 = z. iégl%l.x 1 = 2 3/4 ill—2+1; x 2 = 2.1 1.65 2 11 3 =(a-_3.2+_1_-12)x z. = 8.2 1.23 2 2.5 x 2 = 2.5 2 z 4.92 .737) 3 2 =2ng =a5 ) 2 g21054 = 1.7 x _1_.5_ = 1.28 g 2 = 3.78x1.414=5.35 .802) 1 = 1.7+ 1.3 x 2 __, 3 2 = 122+ 1.1 x 2 = 2.4 = _5_._z_._ x 1.4115165 1.15 o = 1.05 x 4 = 11.12. x 1.4115535 .892 24. SN. u WWW n m 3. n mmmm n m CNN. 2 m3 wmbm I I Imv Nm. I mm. I Nm. I Nm. I m.< m<<. mm.m mb.a mq bm.HI bm.HI $0. I 00. I 0.0 how. o.m o.m an ©.NI o.NI 0.HI o.HI b.b m.H o.mH m.N mN m. I w.NI «1V0 N mN.H+ Nb.HI mb.§ m mNfiuw wNN. «S.N Q...” ma MH.H+ pm.NI N.Hmm ¢ «N. + ob.HI o.moN N m.~ma o.ooI 6H om.b+ me mm.mI mm.~I o 0.6 0.6 0+ 6.6I 6H oo.e+ gm mm.mI mw.mI o e.» o.oa o+ 6.6I ca oo.e+ gm bo.HI ea.eI o.eoa mm.eq me.oa o~.~+ 6.6I NH oo.e+ 4H ea. I e~.eI o.eem m.mma o.HH oa.q+ m.mI m oo.b+ o.aa o 0 mm. mm.a+ 6H..oa oo.o+ o.HI q om.e+ m me.oa am 0 o m. om.e+ mo. m.oa Hm Ho. mo. m. Ho.e+ mm.mI p.06 >.e 0H om.e+ o om.>+ OH mm co. no. N mm.e+ «2. m «a 00. m4. m oc.q+ mm.mI «.mo 0.0 0 m0.o+ o mo.0+ 0 mm 00. me. m 8.6+ 8. 2 mm ou.~ mm.H m mm.~+ we. I o.mH «.4 m mm.m+ o mm.m+ m me mm.H mm.H H mm.H+ mm. H em ma.a mm.a a mo.a 0 mm .302 mm H . .H m: +.<2 n .aoz mvaoq . H I Hmpoa .Nm M .5 Hapog.msm Nb u “owmo 4:. Imo.osH Imm.asm Ammsww soospmm .0 Boga p:H0m_ memos .00.» a .nom .pewam .na spasm m .eoz .wmwm mecca .aom .pgmnm .mnm Apnea ”WWW .wmwm A.w ohamHh momv mmnmmmmm mamdm Q mum¢a IIHApn .em u.a..na om «6 new. on» p. wemeonoeA weapooa one «6 369969 an» as .xua a 6» mop cap as 0_aonm mnahnmp ma vopanOHdo mH onsmmoam 29980 0:90 TABLE 5 SUMMARY OF MOMENTS (III) AND THBUSTS (N) RESULTING FROM VARIOUS LOADS IN THE CORRESPONDING RECTANGULAR FRAME 29. Point 4R Point 3R Point 2R Point 1R Point 0 “wing II N N N N N N N N Dead -l.20 6.02 -3.62 4.37 —5.98 2.65 — .07 .6 +1.56 .6 Earth P. Right +2.55 -.48 +4.60 -.48 +4.01 -.06 +1.83 .39 - .17 .39 Earth P. Left - .78 +.48 -2.33 +.48 -3.88 +.62 -1.97 .39 - .17 .39 Sub-Total + .57 6.02 -1.35 4.37 ~5.85 3.21 - .21 1.38 +1.22 1.38 Live Load 0 —1.67 1.6 -5.03 1.6 -8.32 1.73 -2.56 .83 +3.62 .83 1R -1.02 2.4 -3.07 2.4 ~5.13 2.08 +4.1 .51 +.77 .51 1L -l.02 .8 -3.07 .8 -5.13 .93 —2.3 .51 + .77 .51 Temperature $.24 0 $.73 o +1.2 33.09 31311.12 11.34512 TABLE 6 VALUES OF FUNCTIONS OF THE ANGLE V cos 0-= .707; -G= 45°; (- == 1; (-2= 1; sec-9-= 1.1.14; secz-efi 2; Function Point 0 Point 111 Point 2R Point 3R Point 411 Cos E 1.00 0.99 0.97 0 0 TanV=é- cos 0 1.00 0.99 0.97 0 0 v 45° 4498 44918 0 0 sec V 1.414 1.412 1.395 1.0 1.0 Sec 9- Sec V 2.00 2.00 1.972 1.414 1.414 TABLE 7 QUANTITIES IN TABLE 5 CORRECTED FOR SKEW 31. L d E” Point 4R Point 3R Point 2R Point 1R Point 0 0a 8 .9 :0 Nfl Tv Mg, TY III,a TV Mg Tv N5 Tv_ Dead load plus earth pressure + .57 6.02 —l.35 4.37 ~5.85 3.21 - .21 1.38 +1.22 1.38 Live Load 0 1 -1.18 1.13 —3.56 1.13 -5.88 1.22 -1.81 .59 +2.56 .59 1R 1 - .72 1.70 —2.17 1.7 -3.63 1.47 +2.9 .36 + .54 .36 1L 1 " .72 057 L‘20.].7 057 -3063 066 ‘1063 036 +054 036 Temp. Rise) + Fall) 2 +.48 0 +1.46 0 +2.44 ”1.18 +2.62 -.24 +2.68 $.24 Totals MEX. 3 +095 6002 ’4064 202.1 + . . -1.09 7.15 -6.3 5.5 —14.17 4.61 +5.3 1.5 6 46 l 73 Corrected 3 +1.34 8.51 -6.55 3.12 +9.14 2.44 h'90()O 7078 -2000 6052 -1.54 10.10 ‘ +7.50 2.12 *Note: Step 1: Live load quantities multiplied by cos-6-(.71) Step 2: Temperature load quantities multiplied by secz-G-(2) Step 3: Max. moments & thrusts summed up and multiplied by sec-G-sec V (Table 6) TABLE 8 SOME CONSTANTS “g y u sin 6 sin2 35 cosi cos2 b ucosé y 025" a. 48 2.0 0 -1 1 0 0 0 0 BR 6.0 0 -1 1 O O 0 0 2R 10.0 7.0 49.0 -.71 0.501 0.71 0.501 4.97 5.01 1R 10.75 10.2 104.0 -.122 0.0149 0.99 0.98 10.1 10.55 0 11.0 11.0 121.0 0 0 1.00 1.00 11.0 11.0 1L 10.75 12.2 148 +.122 0.0149 0.99 0.98 12.1 10.55 2L 10.0 18.4 338 +.7l 0.501 0.71 0.501 13.1 5.01 3L 6.0 16.0 256 +1 1.0 0 0 0 0 4L 2.0 16.0 256 +1 1.0 0 0 0 0 :g cos2 6 u cos 0 sin 5 sinb cos 0 (E F F cos E F 1R 1.285 13.2 -.121 -.1585 0 3.58 39.4 0 0 1L 1.285 15.85 ' +.12l +.1585 33. Table 8 (Continued) :3 _1_1_‘_a_ 2.1.93.2 0052 g 11 cos 6 LcosiLchosfi guy cos as F F F F F F 4R 0 O 0.67 0 0 0 0 3R 0 0 0.45 0 0 0 0 2R 49.7 11.30 0.115 0.115 1.145 1.15 11.45 1R 108.6 136.20 0.0195 1.285 13.22 13.82 142.20 0 121-0 433.00 0 3.58 39.40 39.40 433.00 IL 130.0 194.0 0.0195 1.285 15.85 13.82 170.40 2L 131.0 77.7 0.115 0.115 3.015 1.15 30.15 3L 0 115.1 0.45 0 0 0 0 4L 0 171.5 0.67 0 0 0 0 (5 1138.8 2.509 6.38 72.630 69.34 787.20 TABLE 9 LOADING TERM SUMMATIONS FOR EQUATIONS Dead.Load Live Load Live Load Live Load at Point 0 at 1B at 1B and 1L Ia I9 16 IeI I9 4, I3 <0 I9. 00 la 0') m o 0') a ° 8 N I o 8 N01 0 8 I; o (O) NO] 8 a .s gs a 6 8s a «as .s 8 .sss o o o .5 o o E 0 5 :55 L0 :02 2 :5 1R 6.38 84.2 8.2 6.4 84.5 8.22 9.6 -.159 —.153 12.8 169 16.4 0 8.16 322.0 29.2 12.8 504.0 45.8 6.4 0 0 12.8 504 45.8 IL 6.38 101.0 8.2 6.4 101.5 8.22 3.2 +.159 +.51 12.8 203 16.4 Quantities for direct substitution in Equations. Assumption 1. The following values of H are obtained for the rectangular frame. Dead Balanced Temp. Live Live Live Load Earth Rise Load Load Load Pressure Doint 0 Point R Points IR}? 1L .6 .78 .12 .83 .51 1.02 34. 35. Substitutions for the left hand sides of the equations are: Eq. Expression Solution Substitution 5a (S u? L2 (8 sin2 a 1139—64x2.5 981 ‘__ -'._ F F 1* =(1139-158) 5a ggd (5 2.9%21 72.6 5b (S ”60:2 5 .38 Quantities for the right hand sides of the equations are listed in Table 6. For assumptions 2 and 3, quantities for direct substitution: Balanced earth pressure substitution, in the right hand sides of Equations 58 and 5b for assumptions 2 and 3 are: wh2 su cos 0 wh3 s 2.92.91 -tts-m ( 1T1..fl_ ( F e -1(1.98-1.2) 787.2 + 30x11.53 x 72.6 12 = -102 X 787.2 + 3077 X 7206 = -945 + 274 = -671 f The substitution for left hand side (Assumption 3) is (S 171" = 1139. The My correction for Assumption 2 is same as shown in Table 6 for Assumption 1. With the above mentioned exceptions, all substitutions for the R.H.S. of Eqs. 5a for Assumptions 2 and 3 are the same as the corresponding substitutions in Eqs. 5a for Assumption 1. TABLE 10 SUBSTITUTIONS FOR THE R.H.S. OF E0. 5a,ASSUMPTION 1 36. Substi- Loading Expression Computation tution Dead Load: (. f (S W 1(507.2-.6x787.2) __ H (8 314.19.291.12 3 =(507.2—472) 35 2 Balanced i - p (3% - H) x -1(1.98-.78)787.2 -945 earth pressure ( (S 214%9§~§ ) Temperature (- [EfiLEL rise. s _(1+ 92) H (S flggfll] I [.288x106x6.5x106x45X16 2.67x4 - (1+12)(.122 x 787.2) = -(126.4-192) -66 Live Load: - Mou cos B 9 (" ( (S “W 1(690-083X787o2.) 0 - H (3 11.31.932.151) 0059. L707 F =(690-653).707 = 37x.707 26.2 Live load IR " 1(876-1.02x’787.2) (balanced) x.707 =(876—803)x.707 = 73x.707 51.6 I TABLE 11 SUBSTITUTIONS FOR THE R.H.S. 0F EQUATIONS 5b: ASSUMPTION 1. 37. Load at 1R S b ti- Loading Expression Computation tgtion Dead Load: é-E (5 EELJfi§£iJ§ l(45.6-.6x69.34) . 2 ’ Balanced ) _ F $112.1 _ g (S ECOSZ__._6 -l(1098-o78)69034 earth F pressure 9 =-(1.2)69.34 ~83.0 Temperature 2 s COS2 6 rise. - 6 (1+4.) H ( 1L—§¢~—- -1(2).122(69.34) -16.9 . .. Mo cos2 0 lee Load- 6 ”a 1(62.2—.83x69.34) 2 XO707 s 21£Q§__§) 4} 0 - H ( F ) °°S = (62.2-57.5).707 Live load R1 n 1(78.6—l.02x69.34) Lbalanced) x.707 =(78.6-70.7)x.707 5-6 My correction G (8 Mo Si? 5 cos 6 ( 1 02) . 2 1 x - . -. 06 C (Eqée) % (W 51;: 5 2.509 A for Live TABLE 12 SOLUTION OF EQUATIONS FOB VARIOUS ASSUMED FOOTING CONDITIONS: ASSUMPTION 1 38. Eq. Left Hand Dead 231:2:8 Temp. Live Load 1R Sides Load Press Rise 0 Balanced 5a 981311; - -72.63 113:: +35 -945 -66 +26.2 +51.6 5b 72.63 81:. — 6.38 ax: +4.0 -83.0 —16.9 +3.32 + 5.6 Dividing throughout by the coefficients of IX. 58. 13.52 111.. “ "x = +0482 -1300]. -091. +036]. +0711 5b 11.4 R13 -1\Ix = +.62'7 -13.01 ~2.65 +.52 +.878 5b 2.12 R ‘3: “-0145 + 0 +1074 -016 -0167 111.: -.0684 + .0 + .82 -.0755 -.O788 —.0394 So M’-_L R15 2 -—8 81% == +.547J o -6.56 +.604 +.63 +.315(Av.) For live load at IE. 5“ "yL = .315 + .406 =+.721 -11,4315=|.+.78 l 0 l —9.34 +.86 l +.896 Adding -11.4 R]; to 5b .11): = +1.40? -13.01 -11.99 +1.38 +1.774 +.88’7 Mx = -1.407 +13.01 +11.99 71.38 -1.774 -.887 TABLE 12 (Continued) SOLUTION 0F EQUATIONS, FOR ASSUMPTION 2. (b). ROTATION ABOUT x AXIS AND a AXIS 39. 8 ce 0 Eq' L'H'S° Load Press. Rise 0 Balanced 5a 9811113 = + 35 ~67l - 66 +26.2 +51.6 a}, = +.0357 -.684 -.0673 +.0268 +.0526 +.026 So my: —8R13 =J —.286 +5.47 +.538 —.214 -.42 -.21(Av.) For live load at 1B 5d MyL = -.21 + .406 = +.196 (c). SOLUTION OF EQUATIONS FOB ASSUMPTION 3 ROTATION ABOUT x-AXES, y-AXES, and s—AXES 5a 1138.8 11].: +35 -671 -66 +26.2 +51.6 83a =I +.03O8 -.591 -.058 +.O23 +.0454 +.O227(Av.) 40. m Amw + Hv I omHh endpmuoaaov Mom NI m Immm.n madmmmum muses you I0Imoo m n csoH o>4H now m n 684 seen no.4 am so 6348, 43oz 84+ 84+ 34+ 3. + «.0 I 442+ 34+ 4482 o m<.H+ «#.H+ Nb.m+ 00.N+ o.MN+ H0.©I <0.N+ omfim w onspdnomaoa + unannoum spasm + 0800 0800 umdoapmsssm em .0. m8.» R4.“ 8mm.“ goo.» 844 8444 snow 8.“ 3m.“ .36..“ 4468162 2. waspmnogsms mm.+ mmo.+ mm.+ 001+ 0.8.1. mm. + Sm? 423+ .38? on. on. .44 0 mm.+ mm0.+ 0m.+ 0H0.I 0~0.+ mm. + bww.I H00.I 40m0.1 on. 0m. mH m ooo.+ MNO.+ ma0.+ 44 4.“ om.» mm.444 m.O4H 044.” AmmHm mndpwhoqaoa mw.4I 44 m<.mI Nn.©+ m>.m+ oo.~4I ww.I 40. I mm.o+ booo.+ m.44+ mo.b4I mm.I mm.o caoq vwoa .ww<.I n n :4m x_¢ MN.O4 u fi «mm. 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I m H m I H m H IQESmmd pom .m and 4 m:04umasmm< Mom cmpw4 o I d o n moo 4 a :4m 0 M IdUHdo mud .mN .m4 .0 mpc4om How >2 ”mmaoz m4 cad mm mpd40m N aOHQQESmm< Adv o ' I. 0 II 0 II 0 O I. 0 II II “jam on 4+. 8 D. on m... 3 .4» mm M... «4 R 392 ohflpmnmaame oo.NI I I oo.NI I om.NI I N4m.+ me. I 4N.NI I I 44 pc4om wo.NI I I ow.NI I M®.NI I mqoo.I me. I 44 No.0I A I mo.N+ 0b.wI I bq.MI I mmm.+ «N.w+ 4.N4I I I waoq came .0 n 9 :4m N.¢ .b H d 44>. n a moo 44b.I n a n4m no N x umN pq4om 40V a mm “MW. La N" .mfl n" mm o X Al N I.) H HI 4) H n... A I o + are + an I AN I o + s I m + an I x 41 m I o o S I o m. S S W D. S S m. S m. 0 — O. U. H G. G. U. D. - ImFI m :04pmasmm4 4 :04pqasmm4 Acuan4paoov 44 049.9 >: how mammwwadaaoo 43- mmo.+ oo.q+ 4mm.+ 4>.m+ :04pwessm I I + + + I + I I I + + 444mm mmo. + om4. + pm. I mo.mI mm. I $0.4 + ob.mI >4. + mmm. + 4m. + mom.mI mo.mI Aom4x oadpwnoaaoe «mo. I mm. + 4m. + mm.4I one. + mm. + um. + mo.4I omm.+ mm. + 440.4I mo.4I 44 4:4om «mo. I mm. + mm. + mo.m+ omo. + mm. + 04. + mm.~+ 0mm. + om. + obw.m+ m.m+ m4 4n4om mmo. I moo. + 4m. + mo.mI «me. + 4n. + Na. + 4m.~I 4mm. + mm. + 4ms.4I 4m.4I o 4s4om 430.4 344 m4. + 4m.4+ m~.m+ mq.mI cm. I mu.4+ m4. + 4c. I bom.4+ mm.4+ mom.I 4m. I .44 agnam ms4m umoq naan mo.o n > qua «m4 pn4om 44v be. + mb.m+ mm. + om.m+ qo4pmaaam I I + + + I + I I I + + 4444.44 no. + om4.+ «o. I «o.mI mm. I oo.4+ 4o.mI mm. + 4m. + 4m. + wo.mI mo.mI 4mm4m mhspmnogaoa mo. I mm. + mo. + 04. + «o. + mm. + M4. + 44. + om. + om. + 4n. + 4n. + 4 pa4om m4o.I moo.+ om. + om.m+ mo. + 4m. + .m4. + no.m+ mm. + mm. + om.~+ om.m+ o pa4om uwoq m>44 44. + 44.4+ 4o.m+ 00.4I mm. I m>.4+ 04. + on. + mm.4+ mm.4+ mm.4+ -.4+ .4; £44.. ms4a v.04 ammo 4 n > say no pn4om Adv 44 we a: p: as we as >2 >adp >9 >9 > 4.4 as a: ma4cwo4 m :04pmasmm< 4 :04pmanmm4 ea mm4mmm 424 a: mazmzo: qquHmmoa mmmm>mz4ma m4 Hands mw.4I 40.mI 00. I 00.? 84.90530 + I + I + I I + A44wm 004. I 004. + 000. I 000. + 00.4I 004+ 000+ 00.0I 434m 09590900809 00. I 0m. + 004. I 004. + Nm. I NM. + 4N0.I 4N0.+ 04 pn4om 00. I 00. + 040. + 040. I mm. I mm. + 40o.+ 400.I m4 0:400 040. I 040. + 4N. + .440... I 40. I 40. + .4 .I 000% 0 .9594 6.6qu OPHQ 0m.4I 0m.4+ 04.9I m4.m+ m9.4I m9.4+ 949.I 949.+ .99 309.. mn49 0094 0000 N no4pmasmm< 4 :04pgazmmd 0 n > nae .m0 .mm 040400 40V 03.? 04.~I .40. N+ 00:7 84.93.4490 + I + + + I I + I I + + 4440.9 440. I 004. + 000.4I 004I 000. I 004+ 0044+ 00.0I 004. + m4.+ m0m.NI 3.? 484m manpwnoaama 0N. + mm. + 00. I 00.NI mm. + mm. + 04.4I 0m.NI 0.40. + 00. + 0.0.? 00.0I .44 .9500 040.4+ mm. +, 00. I 00.NI 004.4+ Nm. + 00. I m0.NI mN0.4+ 00.4+ N0.MI m0.MI m4 0:400 3.0. + 000. + 3.17 0N..4I 000. + 4n. + 00.4I N417 0044+ .Nm4+ 00.? 00.? o 43494 0004 m>44 mbw.4+ 0N.4+ mm. + No.0I mwm.4+ m0.4+ 0N.NI 00.0I m44.m+ 4N.m+ 00.0I mm.mI .90 £9900 0040 0004 0000 4.0. n > :09. “mm 9:43 3 99 u9 9: >2 99 m9 9: >2 > cm> >9 >9 9 gap as as ma4c.Q4 m a04pmesmm4 4 n04pmazmm« Acosc4pcoov 04 «40.9 45- Notes for Table 15: MT: MatanV-Mv Values of M3 and Tv are given in Table 7, Ta = R5 in Table 13, Mv in Table 14. 46. QQNCLUSION A look at Table 15 in our example indicates that ”T and TT for Assumption 1 are greater than the corresponding quantities for any other assumption. So it appears that values of MT and TT have to be computed for all the three assumptions for any skew rigid frame, to determine the worst condition. Also, the MT due to temperature change, is not less than .74 of the total MT for Assumption 1 and is not more than .17 of the MT for Assump— tion 2 or 3 for all points except for Point 2B. Thus torsional moment due to temperature change constitutes the principal part of the final torsional moment in Assumption 1, and constitutes a minor part of the final torsional moment for Assumptions 2 and 3. 47. BIBLIOGRAPHY Hodges, R. M., "Simplified Analysis of Skewed Reinforced Concrete Frames and Arches", A.S.C.E. Transactions, 1944, p. 913. Hayden, A. G., The Rigid Frame Bridge, 1940 Edition. Gifford, E. 0., "Approximate Design Method for Concrete Skew Rigid Frames", E.N.R., May 3, 1934. mom wise ONLY. MITIWHWHWIWIIl”111|1|111|11Y|111|i|111|1llES 3129313046 3511