AXLE LOAD IN TONNES AXLE LOAD IN TONNES AXLE LOAD IN TONNES 4 m M.L K.I K.213.213.213.2 13.6 12.4 12.4 12.112.1 8.1 13.213.213.213.2 10.5 12.1121 121 12.1 + 1776 2197 134713471347 21.33 260 812 2206 1312 148 1778 2197 1391 1346134 2131 3680 19380 OVER BUFFERS 30740 LAME 1312 221312, 1673 19380 OVER BUFFERS C B.L 6.6 10.7 10.7 10.7 10.7 9.3 989898 ME 10,7 10.7 1.7 10.7 98 23 93 93 0000 18288 OVER BUFFERS 18288 OVER BUFFERS C 61 7.6 2 6.1 6.1 6.1 6.1 7.6 4m 17278 OVER BUFFERS AXLE LOAD IN TONNES 17276 OVER BUFFERS METRE GAUGE STANDARD LOADING OF 1929 forces DAG TRIART САВИЙОТ 6.1 6.1 6.2 6.1 B-BOR Bo-Bo TYPE 6.1 6.1 bax moveo inuol a borers TRAIN LOAD OF 3 87 TONNES PER LIN METER TRAIN LOAD OF MOT 3.87 TONNES PER LIN METER TRAIN LOAD OF 3.87 TONNES PER LIN METERS MOT tipoq ganibel bisboste s voda 21. 1636 2433 1676 1676 2438 1676 8261 Over buffers 8261 Over buffers STEAM ENGINE (TANDER LO CO) AXLE LOAD IN TONNES 5.1 6,1 6.16.1 6.1 6.1 6.5.1 6.16.1 6.1 6.1 6.1 6.L AXLE LOAD IN TONNES AXLE LOAD IN TONNES 0000 TRAIN LOAD OF 2.83 TONNES PER LIN METER TRAIN LOAD OF 2.83 TONNES PER LIN METER 1753 9659 4.6 6.1 6.1 6.1 5.6 1753 965 965 965 2083 2083 1270 9976 Over buffers 9976 Over buffers STEAM ENGINE (TANK LO CO) 4.6 6.1 6.1 6.1 6.1 6.1 6.1 1270 5.6 6.1 6.16.1 2210 1753 1010 2337 067106 12837 Over buffers - 1753 2210 2337 12837 Over buffers 1061067 6.16.16.1 DIESEL ELECTRIC 6.16.16.1 6.16.16.1 6.16.16.1 914914 2438 914914 914914 2438 88.39 Over buffers 88.39 Over buffers NARROW GAUGE -8 CLASS LOADING TRAIN LOAD OF 2.83 TONNES PER LIN METER TRAIN LOAD OF 2.83 TONNES PER LIN METER P₁-P+2P+2P+2P+P-8P; P= P 8 Similarly consider the vertical members. A2A3, B2B3, C2C3, D₂D, and E₂E, mined by the condition, :22 Let Obe the horizontal shear in each of the exterior members. The horizontal shear in each of the interior members will be 20. Qcan 5-951 P₁+ P₂ =Q+20+20+20+0=80;Q= P₁+P₂ 8 Similarly, for the vertical members AA BB, CC, D,D, and E,E A1 Pi Choper 42 F2- A3 Py To string sibbin P B₁ 2P C₁ 2P 2P 2P B2 C₂ Q 20 01 By 205 D₁ E₁ 2P 2P E2 D2 2020 h 20 -20 Q C₁ D E3 R 2R 2R 2R R 2R 2.R 2R Ba C4 DA EA P4- 2S -2S 25 25 -2S 2.S As Bs Cs Ds דללון R₁₂+B+P P+ P+P-8R; R= 8 Fig. 5.789 Es חלה 01 Similarly, for the vertical members 444,, BB, CC,. D&D, and EE, P+P₂+ P3+ P=8S;S= R+R+R+P 8 Now the moments for the columns at top and bottom of each storey can be easily determined. = Pet Ph Mol Mazal a2 Ma2 a3 = Ma3a2 (numerically) = 2 MADE- Oh (numerically) 2 h Mon 2Pph (numerically) Mbl b2 M62b1 = = 2 M62 63 M6362=20=Qh(numerically) Moments in beams. Consider the beam end at A Qh B.M. in column just above A3" 2 MARI 42 10 P A h 2 R B.M. in column just below A3= Rh 2 44 Fig. 5.791 Ph Az Fig. 5.790 Fig. 5.792 B2 20 B3 2R Ba UGE 向 ROW 群 40 ROW 30 ROW -agos COMM 3 ROW 廿 CHESHIRE 45 ROW ROW SHERMORAINE ROW H - UGE COMM ds = ds EI M, -xds= xds EI and M Syds Sayas = Now let us imagine a short column (Fig. 9.83) of thickness ΕΙ plan follows the profile of the portal frame. and whose Let the upper face of this column be subjected to a vertical loading given by M. Due to this loading the pressure distribution at the bottom of the column will follow a linear law. This linear law for the pressure variation for the column base is analogous to the corresponding linear diagram for M For the equilibrium of the analogous column, the following conditions must be satisfied: (i) EV=0; (ii) EM about any line parallel to AB-0; (iii) EM about any line parallel to BC-0. If at any point (x, y) of the column base the pressure intensity is n the above three conditions lead to the following conditions: Fig. 9.83 ds = Snds ΕΙ ΕΙ xds=xds ΕΙ Syds=yds ...(i) ...(ii) Comparing equations (i) and (ii) we conclude that the pressure or stress intensity at any point of the column section is equal to the statically indeterminate moment M, at the corresponding section of the portal frame. The imaginary column introduced in this discussion is called the analogous column. The stress n at any point of the analogous column section is given by, + y+ N M M, M = where уу A 1xx Ta x. N=Total vertical load on the analogous column ds-Volume of the M, diagram A=Area of the analogous column section M Moment of the load N about the centroidal axis XX of the analogous column section = Ne, = M,y ds ΕΙ My Moment of the load N about the centroidal axis YY of the analogous column section. M,xds == Corresponding to the points, A, B, C and D, we can determine the stresses MM, Mc and M For example Stress at A, n-Ma PM 41+ Ya+ M A 1xx After determining MM, Mic and Mid the final moments are determined by the general equation M-M,-M For example, M.- M-MM-Msh-M and so on. It should be noted that the stresses given by the above formulae are valid only when this analogous column is symmetrical about at least one axis. It the analogous column section is not symmetrical about any axis, then the stress at any point of the analogous column section follows the law N n+ax+by where A N A represents the axial stress component and (ax + by) represents the bending stress component. Infamterer Internal Redundancy Minimum number of members needed for stability n-21-3-h=2(8)-3-0-13 number of members provided-15 Degree of internal redundancy n-n, -15-13-2 Total degree of redundancy -2+2-4. Second Method (Joint Equilibrium Method) Number of equations of equilibrium of the joints-2/ Number of unknown forces in members Number of unknown reactions Total number of unknown forces Total degree of redundancy -2x8-16 -15 =r+2H=1+2(2)-5 =15+5=20 -20-16-4. 3.7 STRESSES DUE TO ERRORS IN LENGTHS Suppose ABCD is a redundant square frame. Let AB BC-CD-DA=1. Length of each diagonal-1√2 C 8 Let the member AC be the last member to be fitted. Suppose the length of this member is a little shorter (or a little longer) than When this member is forced into position, a force X is induced in this member. This will also produce forces in the members of the fra We say in such a case there is a lack of fit in the member AC. Hence a lack of fit will produce forces in the members of a redundant frame. Let S, S₂, Sy.. yonubau be the forces in the members of the frame due to lack of fit. All these forces can be expressed in terms of X S² 24,E Total strain energy stored-W, with the usual notations. .. Displacement of one end of the diagonal AC with respect to the other = aw ax =8 where 8 is the lack of fit. In the lack of fit & is known the force X in the member AC can be determined and thus, the forces in the other members also c determined. Problem 35. While fabricating the pin-joined frame shown in Fig. 3.114, the member AC was the last member to be fitted, and was found to be 1 mm short of the required length. Find the forces in all the members of the frame when the member AC is forced into position. The diagonal members are each 1000 mm² in area, while the remaining members are 2000 mm² in area. Take E-200 kN/mm² Solution. Let the inclination of each diagonal with the horizontal be 0 B C 2 m tan 0 1.5 3 sin 0 and cos 0= 3 to shoot telforts 5 1.5 m H Let the tension in the member AC after forcing it into position beX. Resolving vertically at C, Sed = X sin 0 X(compressive) 5 Resolving horizontally at C, Sch X cos0X(compressive) Resolving vertically at D, Sasin 0 X S-X (tensile) 3 Resolving horizontally at D, SX cos 0=X(compressive) Resolving vertically at A, Sab X sin0X(compressive) X E 0 D C واحد ple the J (1) Trusses stable without the assistance of supports. Let Then no least number of members needed for stability j-total number of joints (inclusive of joints at supports) no-21-3 Two types of supports namely hinged supports and roller supports are provided for a truss. At a roller support the direction of reaction is always normal to the roller base irrespective of the loading on the truss. Hence at a roller support, the number of reactions is unity. To a hinged support if only one member is connected, the reaction at that support is always in line with the member. Hence, the number of reactions at such a hinged support is unity. To a hinged support if two or more members are connected, then the number of reactions at the support is equal to 2. (1) Trusses stable without the assistance of supports Let Then no least number of members needed for stability j- Total number of joints (inclusive of joints at supports) no-2j-3. (ii) Trusses stable only with the assistance of supports Let Then антриз и ановка от виз агазита та BUG 3 no least number of members needed for stability j-total number of joints (inclusive of joints at supports) h'- Number of single reaction hinged supports que nod of tim -21-3-h Degree of External Redundancy Let r=number of roller supports h-number of double reaction hinged supports h' number of single reaction hinged supports The degree of external redundancy-r+2h+h'-3. Degree of Internal Redundancy Let n number of members provided no least number of members needed for stability The degree of internal redundancy-m-n, n-12/-3-h (Neglect 'where not applicable) Total degree of redundancy where, +2h+h'-3+n-[2-3-h] +2h+2h+n-2/=r+2(h+h)+n-2/-r+2H+n-2) H=h+h' total number of hinged supports. This method of determining the degree of redundancy is called the stability method. To determine the total degree of redundancy we may also proceed as follows. To Ifj Total number of joints, then at each joint the number of available equations of equilibrium - 2 .. Total number of available equations = 2j Total number of unknown forces to be determined number of members + number of supports=n+r+2H .. Total degree of redundancy n+r+2H-2) This is the same result as obtained before. This method of determining the degree of redundancy is called the joint equilibrium method. Problem 34. Find the degree of redundancy for the truss shown in Fig. 3.112. Solution. number of roller supports number of single reaction hinged supports number of double reaction hinged supports -1-0 h-2 --8 -n=15. number of joints number of members First Method (Stability Method) External Redundancy Number of unknown reactions r+N+2h-1+0+2(2)-5 Number of equations of equilibrium -3 Degree of external redundancy -5-3-2. 120 THEORY OF STRUCTURE REDUNDANT the val Let R The Figs. 3.2 (), (i) and (ii) show different types of statically indeterminate structures. of X. For instance assuming any value of X, the reactions and V, can be determined. Hence, infinite values of X with corresponding values of and along with the given loads, satisfy the conditions of equilibrium. Hence, our problem is to find out the actual vala of X. This can be determined from the condition that the deflection at A is zero. But, by the first theorem of Castigliano the deflection a where, W, is the strain energy stored by the structure. Hence, the value of X is given by the condition, A is given by Similarly in Fig. 3.1 (i), the horizontal thrust. X for the two-hinged arch cannot be determined by using the conditions of equilibriu But X can be determined from the condition that the horizontal movement of the end B with respect to A equals zero. aw ax aw -0. i.e., by the condition, ax aw -0. ax LOITATE Similarly the tension X in the member AC of the redundant frame of Fig. 3.2 (ii) cannot be determined by using the conditions of equilibrium. But X can be determined from the condition that the horizontal movement of C with respect to A equals zero. i.e., by the condition,- aw ax -0. Hence, if W, be the strain energy stored by a redundant structure which is a function of a redundant quantity X, then the redunda quantity X is given by the condition, Ow ax -0. The strain energy stored W, being a function of redundant quantity X, the condition minimum value of the function Wi aw ax is a condition for maximum or B. I In general, the strain energy stored by a structure subjected to bending or axial loading is given by rea By i W - Σ M²ds 2E1 Σ SL 24E where, M-B.M. at any section and S-Axial force in any member. The first part of the above expression will hold good for the case in which energy is stored due to bending. The second part of the above expression will hold good for the case in which energy is stored due to axial loading. For the condition of maximum or minimum value of W, we have ow ax =0. aw ax M. дм ax L AE ax as =0 To test for the condition for maximum or minimum value of W, differentiating again, - ar² ΣΙ • M 0M дм ax² ax L Σ a²s AE ax? ax The bending moment M at any section, or the force S in any member of the redundant truss, is a linear function of X ам as and are constants. ax ax a2M a's 4 =0 ar ax OM as And, further, the terms and are positive. ax ax Hence, a²w ax² is positive. aw Le, the condition =0 is a condition for minimum value of W. ax Hence of all the infinite values of X satisfying the conditions of equilibrium the actual value of X is given by that value for whi the strain energy stored is a minimum. A beam of span I is fixed at one end and simply supported at the other end. It carries a uniformly distributed load w per unit run over the whole span. Find the reaction at the simply supported end, by the principle of least work. Fig. 3.3 shows the beam AB, fixed at A and simply supported at B and carrying the uniformly distributed load. 178 We will regard the portal frame as a closed loop or circuit. Thus a closed circuit or loop like a portal frame with ends fixed is redundant to the third degree. Now for any multistoreyed frame the total degree of redundancy-3x number of loops-Number of releases at the supports. Consider the multistoreyed frame shown in Fig. 3.108. Number of loops Number of releases at the supports Total degree of redundancy Now consider the multistoreyed frame Number of loops Number of releases at the supports .. Total degree of redundancy <-3 0 -X-3(3)-0-9 shown in Fig. 3.109. -7 1+1-2 -x-3x7-2-19. Fig. 3.107 Loop 3 Loop 5 Loop 6 Loop 7 X-9 Loop 2 8-9 Loop 1 Loop 2 Loop 3 Loop 4 Fig. 3.108 Loop I Fig. 3.109 Degree of Redundancy for Pin-jointed Plane Trusses The stability of a pin-jointed truss depends on the number and arrangement of its members. In order a truss may be stable, it shou least be provided with a certain minimum number of members and these members must be so provided so that no part of the truss she behave as a mechanism ie., the number and arrangement of the members should be such that the geometry of the truss should be sta Pin-jointed trusses may be classified into the following two types, viz. (1) Trusses having stability of geometry without the need for supports. (i) Trusses whose geometry is stable only with the assistance of supports. The trusses shown in Fig. 3.110 are those which maintain their geometry without the assistance of supports. (a) (b) (c) (s) Fig. 3.110. Trusses whose geometry is stable without the assistance of supports. Fig. 3.111 shows trusses whose geometry will remain stable with the assistance of supports. If the supports are remov geometry of the trusses will become unstable. (a) (b) (c) Fig. 3.111. Trusses whose geometry is stable only with the assistance of supports. Infamterer Internal Redundancy Minimum number of members needed for stability n-21-3-h=2(8)-3-0-13 number of members provided-15 Degree of internal redundancy n-n, -15-13-2 Total degree of redundancy -2+2-4. Second Method (Joint Equilibrium Method) Number of equations of equilibrium of the joints-2/ Number of unknown forces in members Number of unknown reactions Total number of unknown forces Total degree of redundancy -2x8-16 -15 =r+2H=1+2(2)-5 =15+5=20 -20-16-4. 3.7 STRESSES DUE TO ERRORS IN LENGTHS Suppose ABCD is a redundant square frame. Let AB BC-CD-DA=1. Length of each diagonal-1√2 C 8 Let the member AC be the last member to be fitted. Suppose the length of this member is a little shorter (or a little longer) than When this member is forced into position, a force X is induced in this member. This will also produce forces in the members of the fra We say in such a case there is a lack of fit in the member AC. Hence a lack of fit will produce forces in the members of a redundant frame. Let S, S₂, Sy.. yonubau be the forces in the members of the frame due to lack of fit. All these forces can be expressed in terms of X S² 24,E Total strain energy stored-W, with the usual notations. .. Displacement of one end of the diagonal AC with respect to the other = aw ax =8 where 8 is the lack of fit. In the lack of fit & is known the force X in the member AC can be determined and thus, the forces in the other members also c determined. Problem 35. While fabricating the pin-joined frame shown in Fig. 3.114, the member AC was the last member to be fitted, and was found to be 1 mm short of the required length. Find the forces in all the members of the frame when the member AC is forced into position. The diagonal members are each 1000 mm² in area, while the remaining members are 2000 mm² in area. Take E-200 kN/mm² Solution. Let the inclination of each diagonal with the horizontal be 0 B C 2 m tan 0 1.5 3 sin 0 and cos 0= 3 to shoot telforts 5 1.5 m H Let the tension in the member AC after forcing it into position beX. Resolving vertically at C, Sed = X sin 0 X(compressive) 5 Resolving horizontally at C, Sch X cos0X(compressive) Resolving vertically at D, Sasin 0 X S-X (tensile) 3 Resolving horizontally at D, SX cos 0=X(compressive) Resolving vertically at A, Sab X sin0X(compressive) X E 0 D C K=3.3h n = 4.5 (n) 11.632 Area of the catchment = 300 km² 6 7 8 9 2 a(t) lagged 1-h S₁- Time tin (f) u(t) by 1 UH Curve of S₁- hours (t/K) (cm/h) (m/s) hour (m³/s) addition Curve Curve 3-h DRH of Ord. Ordinate lagged 3 cm in of 3-h S₁-3 hours 10 11 UH 0 0.000 0.0000 0.000 0.000 0.000 0.000 0.000 0.00 1 0.303 0.0003 0.246 0.000 0.123 0.000 0.123 0.123 2 0.606 0.0025 2.054 0.246 1.150 0.40 0.123 1.273 1.273 0.42 3 0.909 0.0075 6.271 2.054 4.162 1.273 5.435 0.000 5.435 1.81 4 1.212 0.0152 12.676 6.271 9.473 5.435 14.909 0.123 14.786 4.93 1,515 0.0245 20.444 12.676 16.560 14.909 31.469 1.273 30.196 10.07 6 1.818 0.0343 28.583 20.444 24.513 31.469 55.982 5.435 50.547 16.85 7 2.121 0.0434 36.209 28.583 32.396 55.982 88.378 14.909 73.469 24.49 8 2.424 0.0512 42.676 36.209 39.442 88.378 127.820 31.469 96.351 32.12 9 2.727 0.0571 47.600 42.676 45.138 127.820 172.958 10 11 12 3.636 0.0629 52.490 52.411 52.451 273.798 3.030 0.0610 50.834 47.600 49.217 3.333 0.0628 52.411 50.834 51.623 222.175 273.798 127.820 145.978 48.66 55.982 116.975 38.99 172.958 222.175 88.378 133.797 44.60 13 3.939 0.0615 51.303 52.490 51.897 326.248 172.958 153.291 51.10 326.248 14 4.242 0.0589 49.112 51.303 50.207 378.145 15 4.545 0.0554 46.180 49.112 47.646 428.353 16 17 4.848 0.0513 42.751 46.180 44.466 5.152 0.0468 39.039 42.751 40.895 18 5.455 0.0422 35.219 39.039 37.129 475.998 520.464 561.359 19 5.758 0.0377 31.431 35.219 33.325 598.488 20 6.061 0.0333 27.779 31.431 29.605 631.812 21 24 25 35 36 37 38 39 40 222222222333334 27 8.182 0.0114 9.520 11.295 10.408 7.273 0.0188 15.647 18.253 16.950 729.924 7.576 0.0160 13.332 15.647 14.489 26 7.879 0.0135 11.295 13.332 12.313 6.364 0.0292 24.337 27.779 26.058 22 6.667 0.0254 21.153 24.337 22.745 687.475 23 6.970 0.0219 18.253 21.153 19.703 710.221 378.145 222.175 155.970 51.99 428.353 273.798 154.555 51.52 475.998 326.248 149.750 49.92 520.464 378.145 142.319 47.44 561.359 428.353 133.006 44.34 598.488 475.998 122.489 40.83 631.812 520.464 111.348 37.12 661.417 561.359 100.058 33.35 661.417 687.475 598.488 88.988 29.66 710.221 631.812 78.408 26.14 746.875 729.924 661.417 68.507 22.84 746.875 687.475 59.399 19.80 761.364 710.221 51.143 17.05 761.364 773.677 729.924 43.753 14.58 29 30 31 28 8.485 0.0096 7.986 9.520 8.753 8.788 0.0080 6.669 7.986 7.328 792.838 9.091 0.0067 5.546 6.669 6.108 800.166 9.394 0.0055 4.594 5.546 5.070 806.273 9.697 0.0045 3.792 4.594 4.193 811.344 10.000 0.0037 3.119 3.792 3.456 815.537 3.119 2.838 818.992 10.303 0.0031 2.558 10.606 0.0025 10.909 0.0020 11.212 0.0017 11.515 0.0013 11.818 0.0011 12.121 0.0009 2.091 2.558 2.324 821.831 1.704 2.091 1.897 824.155 1.385 1.704 1.545 826.052 1.123 1.385 1.254 827.597 0.909 1.123 1.016 828.851 773.677 784.085 746.875 37.210 12.40 784.085 792.838 761.364 31.474 10.49 800.166 773.677 26.488 8.83 806.273 784.085 811.344 792.838 22.188 7.40 18.505 6.17 815.537 800.166 15.371 5.12 818.992 806.273 12.719 4.24 821.831 811.344 10.487 3.50 824.155 815.537 8.618 2.87 2.35 826.052 818.992 7.060 1.92 827.597 821.831 5.766 1.57 828.851 824.155 4.696 1.27 829.867 826.052 3.815 0.733 0.909 0.821 829.867 1.03 830.688 827.597 3.091 Table A.3(a) (Contd) Ads (LSM) sida M 20 M/bd (MPa) M 25 Fe 250 Fe 415 Fe 500 Fe 250 Fe 415 Fe 500 1.162 0.700 2.16 0.581 1.118 0.674 0.559 1.175 0.708 2.18 0.587 1.130 0.681 0.565 1.188 0.716 2.20 0.594 1.142 0.688 0.571 1.201 0.723 2.22 0.600 1.154 0.695 0.577 1.214 0.731 2.24 0.607 1.166 0.702 0.583 2.26 1.227 0.739 0.614 1.178 0.710 0.589 1.241 228 0.747 0.620 1.190 0.717 0.595 2.30 1.254 0.755 0.627 1.202 0.724 0.601 2.32 1.267 0.764 0.634 1.214 0.731 0.607 2.34 1.281 0.772 0.640 1.226 0.739 0.613 2.36 1.295 0.780 0.647 1.238 0.746 0.619 2.38 1.308 0.788 0.654 1.251 0.753 0.625 2.40 1.322 0.796 0.661 1.263 0.761 0.631 2.42 1.336 0.805 0.668 1.275 0.768 0.638 2.44 1.349 0.813 0.675 1.288 0.776 0.644 2.46 1.363 0.821 0.682 1.300 0.783 0.650 2.48 1.377 0.830 0.689 1.312 0.791 0.656 2.50 1.391 0.838 0.696 1.325 0.798 0.662 2.52 1.406 0.847 0.703 1.338 0.806 0.669 2.54 1.420 0.855 0.710 1.350 0.813 0.675 2.56 1.434 0.864 0.717 1.363 0.821 0.681 2.58 1.448 0.873 0.724 1.375 0.829 0.688 2.60 1.463 0.881 0.731 1.388 0.836 0.694 2.62 1.477 0.890 0.739 1.401 0.844 0.700 2.64 1.492 0.899 0.746 1.414 0.852 0.707 2.66 1.507 0.908 0.753 1.426 0.859 0.713 2.68 1.522 0.917 1.439 0.867 0.720 2.70 1.536 0.926 1.452 0.875 0.726 2.72 1.551 0.935 1.465 0.883 0.733 2.74 1.567 0.944 1.478 0.891 0.739 2.76 1.582 0.953 1.491 0.898 0.746 2.78 1.597 1.505 0.906 0.752 2.80 1.612 1.518 0.914 0.759 2.82 1.628 1.531 0.922 0.765 284 1.643 1.544 0.930 0.772 2.86 1.659 1.558 0.938 0.779 2.88 1.675 1.571 0.946 0.785 2.90 1.691 1.584 0.954 0.792 2.92 1.707 1.598 0.963 0.799 2.94 no 1.723 1.611 0.971 0.806 2.96 0.812 1.739 1.625 0.979 2.98 1.755 1.639 0.987 0.819 3.00 0.826 3.02 1.652 0.995 0.833 3.04 1.666 1.004 0.840 3.02 1.680 1.012 (Contd) Composed of Weight Sectional One steel Joist Each Flange to form Designation W Plates per Moduli of Section Radii of Gyration Area metre Width Thick- ness ZXX Zyy tyy N mm mm N mm² 10³ x mm³ 103 x mm³ mm mm (1) (2) (3) (4) (5) ISHB 225 431-0 350 25.0 1687.0 21494-0 2208-2 937-9 118-9 83-6 32.0 2039.0 25974-0 2717-7 1176-9 123-0 85-1 40-0 2441-0 31094-0 3315-7 1449-9 127-5 86-4 ISHB 225 468-0 320 12.0 1071-0 13646-0 1307-0 496-9 109-2 76-3 16.0 1272-0 16206-0 1585-2 633-4 112-1 79-1 20.0 1073.0 18766-0 1866-4 770-0 114-8 81-0 25.0 1724-0 21966-0 2222-7 940-6 118-0 82-8 32.0 2076-0 26466.0 2731-5 1179-1 122-2 24-5 40-0 2478-0 31566-0 3328-8 1452-6 126-8 85-6 ISHB 250 510-0 320 12.0 1113-0 14176-0 1527-4 532-2 121-5 77-5 16-0 1314.0 16736-0 1834-9 668-7 124-3 80-0 20-0 1515-0 19296-0 2145-3 806-2 127-0 81-7 25-0 1766-0 72496-0 2538-0 975-9 130-1 83.9 32.0 2118-0 26976-0 3097-3 1214-8 134-3 84-9 40-0 2528-0 32096-0 3751-6 1487-9 138-9 86-1 ISHB 250 547-0. 320 12.0 1150-0 14851-0 1545-5 535-3 120-2 76-5 16-0 1351-0 17211-0 1852-4 671-9 123-2 79-0 20-0 1552-0 19771-0 2162-4 808-4 125-9 80-9 25-0 1803-0 22971-0 2554-5 779-1 129-2 82-6 32-0 2155-0 27451.0 3113-1 1218-0 133-4 84-3 40-0 2557-0 32571-0 3766-6 1491-1 738-1 85-6 ISHB 250 510-0 400 12-0 1264-0 16096-0 1768-1 738-1 122-7 96.3 16-0 1515-0 19256-0 2156-4 951-4 125-5 99-3 20-0 1766-0 22496-0 2548-3 1164-7 128-2 101-8 25.0 2080-0 26496-0 3043-5 1431-4 131-3 103-9 32-0 2520-0 32096-0 3748-9 1804-7 135-4 106-0 40.0 3022-0 38496-0 4572 3 2231-4 140-0 107.7 ISHB 250 547-0 400 12-0 1301-0 16571-0 1786-1 740-6 121-5 94.5 16-0 1552-0 19771-0 2174-0 953-9 124:5 93-2 20.0 1803-0 22971-0 2565-3 1167-3 127-2 100-8 25-0 2117-0 26971-0 3060-0 1433-9 130-5 103-1 32-0 2557-0 32571-0 3764-2 1807-3 134-7 105-1 40.0 3059-0 38971-0 4587-3 2233-9 139-4 107-1 Contd.. AXLE LOAD IN TONNES AXLE LOAD IN TONNES AXLE LOAD IN TONNES 4 m M.L K.I K.213.213.213.2 13.6 12.4 12.4 12.112.1 8.1 13.213.213.213.2 10.5 12.1121 121 12.1 + 1776 2197 134713471347 21.33 260 812 2206 1312 148 1778 2197 1391 1346134 2131 3680 19380 OVER BUFFERS 30740 LAME 1312 221312, 1673 19380 OVER BUFFERS C B.L 6.6 10.7 10.7 10.7 10.7 9.3 989898 ME 10,7 10.7 1.7 10.7 98 23 93 93 0000 18288 OVER BUFFERS 18288 OVER BUFFERS C 61 7.6 2 6.1 6.1 6.1 6.1 7.6 4m 17278 OVER BUFFERS AXLE LOAD IN TONNES 17276 OVER BUFFERS METRE GAUGE STANDARD LOADING OF 1929 forces DAG TRIART САВИЙОТ 6.1 6.1 6.2 6.1 B-BOR Bo-Bo TYPE 6.1 6.1 bax moveo inuol a borers TRAIN LOAD OF 3 87 TONNES PER LIN METER TRAIN LOAD OF MOT 3.87 TONNES PER LIN METER TRAIN LOAD OF 3.87 TONNES PER LIN METERS MOT tipoq ganibel bisboste s voda 21. 1636 2433 1676 1676 2438 1676 8261 Over buffers 8261 Over buffers STEAM ENGINE (TANDER LO CO) AXLE LOAD IN TONNES 5.1 6,1 6.16.1 6.1 6.1 6.5.1 6.16.1 6.1 6.1 6.1 6.L AXLE LOAD IN TONNES AXLE LOAD IN TONNES 0000 TRAIN LOAD OF 2.83 TONNES PER LIN METER TRAIN LOAD OF 2.83 TONNES PER LIN METER 1753 9659 4.6 6.1 6.1 6.1 5.6 1753 965 965 965 2083 2083 1270 9976 Over buffers 9976 Over buffers STEAM ENGINE (TANK LO CO) 4.6 6.1 6.1 6.1 6.1 6.1 6.1 1270 5.6 6.1 6.16.1 2210 1753 1010 2337 067106 12837 Over buffers - 1753 2210 2337 12837 Over buffers 1061067 6.16.16.1 DIESEL ELECTRIC 6.16.16.1 6.16.16.1 6.16.16.1 914914 2438 914914 914914 2438 88.39 Over buffers 88.39 Over buffers NARROW GAUGE -8 CLASS LOADING TRAIN LOAD OF 2.83 TONNES PER LIN METER TRAIN LOAD OF 2.83 TONNES PER LIN METER