Continuous Beam 11Objectives: Derive the Clapeyron’s theorem of three moments Analyze continuous beam with different moment of in- ertia with unyielding supports Analyze the continuous beam with different moment of inertia in different spans along with support settlements using three moment equation. 11.0 INTRODUCTION A beam is generally supported on a hinge at one end and a roller bearing at the other end. The reactions are determined by using static equilibrium equations. Such as beam is a statically determinate structure. If the ends of the beam are restrained/clamped/encastre/fixed then the moments are included at the ends by these restraints and this moments make the structural element to be a statically indeterminate structure or a redundant structure. These restraints make the slopes at the ends zero and hence in a fixed beam, the deflection and slopes are zero at the supports. A continuous beam is one having more than one span and it is carried by several supports (minimum of three supports). Continuous beams are widely used in bridge construction. Consider a three bay of a building which carries the loads W1 , W2 and W3 in two ways. W1 W2 W3 FIG. 11a Simply supported beam FIG. 11b Bending moment diagrams 546 • Basic Structural Analysis Continuous Beam • 545 W1 W2 W3 A B C D FIG. 11c Continuous beam + + − − + FIG. 11d Bending moment diagram If the load is carried by the first case then the reactions of individual beams can be obtained by equilibrium equations alone. The beam deflects in the respective span and does not depend on the influence of adjacent spans. In the second case, the equilibrium equations alone would not be sufficient to determine the end moments. The slope at an interior support B must be same on either side of the support. The magnitude of the slope can be influenced by not only the load on the spans either side of it but the entire loads on the span of the continuous beam. The redundants could be the reactions or the bending moments over the support. Clapeyron (1857) obtained the compatibility equation in term of the end slopes of the adjacent spans. This equation is called theorem of three moments which contain three of the unknowns. It gives the relationship between the loading and the moments over three adjacent supports at the same level. 11.1 DERIVATION OF CLAPEYRON’S THEOREM (THEOREM OF THREE MOMENTS) Figure 11e shows two adjacent spans AB and BC of a continuous beam with two spans. The settlement of the supports are ∆A , ∆B and ∆C and the deflected shape of the beam is shown in A B C (Fig. 11f). l1 l2 A B C FIG. 11e A¢G = dAB 548 • Basic Structural Analysis C¢H = dCB Continuous Beam • • 549 547 GD = DB – DA + d B A The primary structure is consisting ofHFsimply = DCsupported – DB – dbeams B with imaginery hinges over each C support (Fig 11g). Fig 11h shows the simply beam bending moment diagrams and Fig 11i shows the A¢G = d B support moment diagram for the Asupports. A compatibility Page 549C ¢H =isdderived equation C B based on the fact that the end slopes of adjacent spans are equal in magnitude but opposite GDin =sign. D Using – D +Figd B11f and the property similar triangles B A A HF = DC – DB –GD dCB == CF HF DB DB¢ B F B¢ F ∆B − ∆A + δBA ∆C − ∆B + δCB Page 549 = l1 1 l2Ï 1 l1 1 ¸ dAB = Ì A1 x1 + M A l1 ◊ + M B ◊ l1 ◊ 2 l1 /3 ˝ B δA δC B E I ∆A 1− 1∆ÓB ∆C − ∆B 2 3 2 ˛ GD i.e. HF l1 + l2 = + (i) l1 l2 = DB¢ B¢ F The displacements are obtained as follows. Page 556 BB 11 Ï 11 l1l1 11 ¸ dδAA = AA11 x¯x11 + + M MAAl1l1· ◊ ++ MMB lB2 ◊·l12l◊ 22 /3 l1 /3 ˝ (ii) EE11 II11 Ì = Ó 22 33 22 ˛ 1 1 l 1 2 δCB = A2 x¯2 + MC l2 · + MB l2 · 2l2 /3 E2 I2 2 3 2 Combining the equations (i) and (ii) Page 556 MA l1 l1 l2 l2 A1 x¯1 A2 x¯2 + 2MB + + MC +6 + E1 I1 E1 I1 E2 I2 E2 I2 E1 I1 l1 E2 I2 l1 ∆A − ∆B ∆C − ∆B =6 + (iii) l1 l2 The above equation is called as Clapeyron’s equation of three moments. In a simplified form of an uniform beam section (EI = constant); when there are no settlement of supports A1 x¯1 A2 x¯2 MA l1 + 2MB (l1 + l2 ) + MC l2 = −6 + (iv) l1 l2 It is to be mentioned here that x¯1 and x¯2 are measured outwards in each span from the loads to the ends. 11.1.1 Procedure for Analysing the Continuous Beams using Theorem of Three Moments (1) Draw simple beam moment diagram for each span of the beam. Compute the area of the above diagrams viz, A1 , A2 . . . An and locate the centroid of such diagrams x¯1 , x¯2 . . . x¯n . It must be re- membered that the distances x¯1 , x¯2 . . . x¯n are the centroidal distances measured towards the ends of each span as shown in Fig. 11j. 548 • Basic Structural Analysis Continuous Beams • 549 A1 A2 − x1 − x2 FIG. 11j Simple beam moment diagrams (2) Identify the support moments which are to be determined viz, MA , MB and MC (3) Apply three moment equation for each pair of spans which results in an equation or equations which are to be solved simultaneously. If the beam is of uniform section (EI = constant) and no support settlements apply equation (iv) and in case the beam is non-uniform and the support settles/raises apply equation (iii). (4) The solution of the equations gives the values of the support moments and the bending moment diagram can be drawn. (5) The reactions at the supports and the shear force diagram can be obtained by using equilibrium equations. 11.2 APPLICATION OF THREE MOMENT EQUATION IN CASE OF BEAMS WHEN ONE OR BOTH OF THE ENDS ARE FIXED 11.2.1 Propped Cantilever Beam Consider the propped cantilever beam of span AB, which is fixed at A and supported on a prop at B. It is subjected to uniformly distributed load over the entire span. The fixed end moment at the support A can be determined by using theorem of three moments. w/m l A′ zero span A B FIG. 11k Propped cantilever beam As the A is fixed support, extend the beam form A to A of span ‘zero length’ and A is simply supported. (1) The simple beam moment diagram is a parabola with a central ordinate of (wl 2 /8). The centroid of this bending moment diagram (symmetrical parabola) is at a distance ‘l/2’ from the supports A and B. wl2/8 A B FIG. 11l Simple beam moment diagram 550 • Basic Structural Analysis Continuous Beam • 549 2 2 wl wl 3 It’s area is A = (l) = . 3 8 12 (2) The support moment diagram is drawn as MA l FIG. 11m Pure moment diagram (3) Apply three moment theorem for the span AB. wl 3 l MA (0) + 2MA (0 + l) + 0 = −6 12 2 ∴ MA = −wl 2 /8 (4) The support reactions are computed by drawing the free body diagram as wl 2/8 w/m A B l VA VB FIG. 11n Free body diagram ∑ V = 0; VA +VB = wl −wl 2 wl 2 ∑ MA = 0; 8 + 2 −VB l = 0 and hence 3wl VB = 8 5wl VA = 8 (5) Using the reactions, the shear force diagram and bending moment diagrams are obtained as ( ( 5wl 8 B A ( ( 3wl 8 FIG. 11o Shear force diagram 25l The location of maximum positive bending moment from support A is obtained by equating the shear force to zero.25 l B − C 3l/8 ( ) wl 2 8 FIG. . 11p Bending moment diagram 11.2. 5wl − wx = 0 8 5l x= 8 At this location. the maximum positive bending moment is obtained from −wl 2 5wl 5l w(5l/8)2 Max + ve BM = + − 8 8 8 2 wl 2 25wl 2 25wl 2 9wl 2 MC = − + − = = 0.2 Beams with Both the Ends Fixed Consider a beam AB of span l is fixed at both the ends. extend this A to A of span (l ) of zero length and is also simply supported at A . Likewise the end B is extended to B .07 wl 2 +ve A 0. As the end A is a fixed support. The beam is carrying a concentrated load of W at a distance of ‘l/3’ from the fixed end A. l The centroid of the unsymmetrical triangle is shown in Fig. The simply supported bending moment diagram is drawn with the maximum ordinate as W × (l/3) × (2l/3) = 2W l/9.3j.07wl 2 8 64 128 128 0.550 • Basic Structural Analysis Continuous Beams • 551 The point of contraflexure is determined by equating the bending moment expression to zero and hence 5wl wx2 wl 2 x− − =0 8 2 8 l 2 + 4x2 − 5lx = 0 Solving the above equation we get x = l and x = 0. 11. 552 • Basic Structural Analysis Continuous Beam • 551 W l/3 2l/3 l′ = O l l′ = O A′ A B B1 FIG. 11s Centroid of an unsymmetrical triangle 4l 5l The centroid of the simply supported BMD is obtained using the above as from A and 9 9 from B.37 W l . Thus MA MB l FIG. 11q Fixed beam 2Wl/9 FIG. 2 9 9 The support moment diagram can be drawn by identifying the support moments as MA and MB . 1 2W l W l2 The area of the bending moment diagram is (l) = . 11r Simple beam moment diagram a b CG ( ( l+a 3 ( ( l+b 3 FIG. 11t Pure moment diagram Applying three moment theorem for a pair of spans of A AB (Ref Eq (iv)) 2 Wl 5l MA (0) + 2MA (0 + l) + MB (l) = 0 − 6 × 1/l > 9 9 2MA + MB = −0. − 0. 11w Bending moment diagram .26 W FIG.296 W l Thus the support moments are obtained by solving the above equations MA = −0.074 Wl FIG.148 Wl − 0.148 W l +W −VB l + 0.74W 0.26W VA = 0.074 Wl l/3 2l/3 C VA VB FIG. VA +VB = W l ∑ MA = O.74 W + − 0.074 W l = O 3 VB = 0. 11v Shearforce diagram 0.074 W l Free body diagram to determine the reactions 0.148 Wl W 0. ∑ V = O.148 W l MB = −0. 11u Using the static equilibrium.0986 Wl + − − − 0.552 • Basic Structural Analysis Continuous Beams • 553 Considering the next pair of spans ABB W l2 4l MA l + 2MB (l + 0) + MB (0) = > −6 9 9 MA + 2MB = −0. 33 kNm2 A2 = × 5 × 31.1) MB = −26.33 × 2 104.1a 20 31. The moment of inertia is I throughout the span.0m Applying three moment equation for the span ABC A1 x¯1 A2 x¯2 MA l1 + 2MB (l1 + l2 ) + MC l2 > > = −6 + l1 l2 53.26 kNm.5m l1 = 4m l2 = 5.25 kNm A1 A2 A C − x1 − x2 FIG.3 NUMERICAL EXAMPLES ON CONTINUOUS BEAMS EXAMPLE 11.67 + 52.554 • Basic Structural Analysis Continuous Beam • 553 11. 11. Analyse the continuous beam and draw SFD and BMD.25 = 104. 10 kN/m A 4m B 5m C FIG.5 2MB (4 + 5) = −6 + 4 5 18MB = −6(26. 11.1b Simple beam moment diagram MB FIG.17 kNm2 3 3 x¯1 = 2m x¯2 = 2.1: A continuous beam ABC is simply supported at A and C and continuous over support B with AB = 4m and BC = 5m. A uniformly distributed load of 10 kN/m is acting over the beam. 11. .1c Pure moment diagram Properties of the simple beam BMD 2 2 A1 = × 4 × 20 = 53.17 × 2. Draw SFD and BMD. 11.0 A2 = × 6 × 15 = 45 2 2 6+2 x¯1 = = 2.33 = 40. 11.33 15 kNm A D B E C FIG.33 kNm 6 6 ME = W l/4 = 10 × = 15 kNm 4 13.2: Analyse the continuous beam by three moment theorem.2a S OLUTION The simple beam moment diagram is drawn as 10 × 2 × 4 MD = Wab/l = = 13. 10 kN 10 kN 2 4m 3 3m D 6m E 6m A B C FIG. 11.556 • Basic Structural Analysis Continuous Beam • 555 EXAMPLE 11.67 m x¯2 = 3 m 3 l1 = 6 m l2 = 6 m .2c Pure moment diagram Properties of the simple beam BMD 1 1 A1 = (6)13.2b Simple beam moment diagram MB A B C FIG. 4c Pure moment diagram Properties of the simple beam BMD 2 1 A1 = × 5 × 50 = 167.8 kN VC = 5. 11.33 m 3 l1 = 5.0m l2 = 10 m 167.0 10 30 MB = −6 (83. VA +VB1 = 80 (i) VB2 +VC = 25 (iii) 16(5)2 ∑ M = 0. 11.5 300 × 5.75 + 159.5 kNm2 A2 = × 10 × 60 = 300 kNm2 3 2 10 + 6 x1 = 2.5 × 2.73 kNm Properties of the simple beam BMD ∑ V = 0.560 • Basic Structural Analysis Continuous Beams • 561 50 60 kNm + 2 + A B C FIG.2 kN VB2 = 19.1 kN .9) MB = −48.5m x2 = = 5. 5VA + 49 − = 0 (ii) 10VB2 − 25(6) − 49 = 0 (iv) 2 VA = 30.9 kN VB1 = 49.33 5MA + 2MB (5 + 10) + 10MC = −6 > > + 5.4b Simple beam moment diagram MB A C FIG. 8 kNm . Apply three moment theorem for the spans A AB.875) 10MA + 36MB = −5749.5b Simple beam moment diagram Properties of simple beam BMD 2 1 A1 = × 10 × 250 = 1666.562 • Basic Structural Analysis Continuous Beams • 563 The simple beam moments are MD = 20 × 102 /8 = 250 kNm 50 × 6 × 2 ME = = 75 kNm 8 250 kNm 75 A D B E C FIG. 11.7 × 5 MA (0) + 2MA (0 + 10) + MB (10) = −6 +0 10 20MA + 10MB = −5000 2MA + MB = −500 (i) Apply three moment theorem for the spans ABC. 1666.7 × 5 300 × 3.35 (ii) Solving equations (i) and (ii) MA = −197.0 m Since A is fixed imagine a span A A of zero length and A as simply supported.33 m 3 l1 = 10 m l2 = 8. 1666.6 kNm MB = −104.7 kNm2 A2 = × 8 × 75 = 300 kNm2 3 2 8+2 x1 = 5 m x2 = = 3.35 + 124.33 MA (10) + 2MB (10 + 8) + 8MC = −6 > + 10 8 10MA + 36MB = −6(833. 11.7 kN 4m 13.5 kN.7 kN. VA +VB1 = 24 (i) VB2 +VC = 24 (iii) ∑ M = 0.67 m FIG.3 kN.5 + 9.3 + x A B D C − − 14.7e FIG. 11. 10.5 6m FIG. VC = 9. ∴ Maximum +ve BM = 14.7 − 4x = 0 x = 3. 11. 4VA + 16 − 10 − 48 = 0 (ii) ∑ MB = 0. 62 VA = 10. Page 570 VB2 = 14.67)2 /2 − 16 = 11 kNm 24 16 kNm 11 + 10 + − − C A B 3.7f Shear force diagram The zero shear location in span BC is 14.67) − 4(3.67 m. 16 + 6VC − 4 × =0 2 VB1 = 13.7g Bending moment diagram Page 572 . 568 • 570 • Basic Structural Analysis Continuous Beams • 569 Free body diagram of spans AB and BC 24 kN 16 kNm 4 kN/m 10 kNm 16 kNm B C Page 568A 2 2 6m B VB2 VC VA VB1 FIG.7d Static equilibrium of AB Static equilibrium of BC ∑ V = 0. 11.5.7(3. 45 + 10.9 kNm 30 kN/m A B B C 2m 2m A VB1 VB2 VC FIG 11.5 FIG. 30 × 22 5.45 kNm.2 kN VB2 = 35.9 kNm 10.45 kNm 10.2 kN VC = 24.8e Static equilibrium of span AB Static equilibrium of span BC ∑ V = 0.2 kN 24. MB = −2MA and putting in eq (ii) MA − 12MA = −60 ∴ MA = 5.8f Shear force diagram Page 572 10.45 kNm FIG.9 + 2VB2 − =0 2 VA = −8.8d FIG 11. ∑V = 0 VA +VB1 = 0 (i) VB2 +VC = 60 (iii) ∑ MB = 0. 11. 11. MB = −10. From eq (i). Free body diagram of span AB and BC 5.5 kN 35.9 + 2VA = 0 (ii) −10.9 9.5 kN/m A + C B − − 8.8g Bending moment diagram Page 575 .55 − A + B D C 5. ∑ MB = 0. 570 • 572 • Basic Structural Analysis Continuous Beams • 571 Solving (i) and (ii).5 kN VB1 = +8.9 kNm. 67 kNm 30 kNm 6m VBC VCB FIG.5I 1.76 kNm.76 + 25.5 kNm. 5VAB + 66.67 kNm. Member AB 80 kN 3m 2m 66.67 kNm D 33.9e . Shear forces and moments in members AB and BC.67 − 33.67 360 × 3 = −6 + 5I 6 × 1.76 kNm VAB VBA FIG.42 ∴ VBA = 54.58 MD = − 33.5I 240 × 2.76 − 80(2) = 0 (ii) VAB = 25. VAB +VBA = 80 (i) ∑ MB = 0. 11.16 + 120) 5MA + 18MB = −1488. 11.96 (ii) Solving equations (i) and (ii) MA = −33.96 + 120 5MA + 18MB = −1368.9d ∑ V = 0. Member BC 20 kN/m 66.42(3) = 42.572 • Basic Structural Analysis Continuous Beams • 573 Applying three moment theorem for the span ABC 5 5 6 6 MA + 2MB + − 30 × I I 1. MB = −66.5I 5MA + 18MB − 120 = −6 (128. 20 kN 10 kN/m 10 kN 4m 2m E F 1m D A 6m B 6m C I 2I 2I FIG.0 kNm 8 MC = −10 × 1 = −10 kNm 26. 11. A concentrated load of 10 kN is acting at D. An overhang CD is of 1 metre length.10: A continuous beam ABCD is simply supported at A and continuous over spans B and C.10a The simple beam moments are 20 × 4 × 2 ME = = 26.9i Elastic curve EXAMPLE 11.574 • Basic Structural Analysis Continuous Beams • 575 A D B E C FIG.7 kNm 6 62 MF = 10 × = 45. 11. 11. The span AB is 6 m and BC are of length 6 m respectively. An uniformly distributed load of 10 kN/m is acting on the span BC.7 45 kNm A E B F C FIG. A concentrated load of 20 kN is acting at 4 m from support A.10b Simply suppored BMD . 98 VBA = 18.15 kNm. 11.10d Using equilibrium conditions. x2 = 3 m.15 − 20(2) = 0 (ii) VAB = 1.02 ∴ ME = VAB (4) = 7. VAB +VBA = 20 (i) ∑ M = 0.576 • Basic Structural Analysis Continuous Beam • 575 MB MC A B C D FIG.45 + 45) MB = 28.1 × 3. 6VAB + 28.7 = 80. l2 = 6 m. ∑ V = 0. (−10) 6 6 6 6MC 80.10c Pure moment diagram Considering spans ABC Properties the simple beam BMD 1 2 A1 = × 6 × 26. Shear force and bending moment values for the spans AB and BC 20 kN 28. 3 l1 = 6 m.15 kNm 4 2 A E B VAB VBA FIG.33 180 × 3 > > MA + 2MB + + = −6 + I I 2I 2I 6 6×2 18MB − 30 = −6 (44.33 m.1 kNm2 A2 = × 6 × 45 = 180 kNm2 2 3 6+4 x1 = = 3.9 kNm . 11. Page 575 33 1.10f Shear force diagram Page 578 26 7. 11.10g Bending moment diagram .9 + 28.15 10 kNm + − − A B E C D FIG. ∑ V = 0. VCB = 27 kN. VBC +VCB = 10(6) = 60 (iii) 62 ∑ MC = 0 10 − 28.98 10 kN – + + A 6m B 6m – C D – 27 18.02 FIG.15 kNm 10 kN/m 10 kNm B C 6m VBC VCB FIG.576 • 578 • Basic Structural Analysis Continuous Beams • 577 28.10e Page 572 Using equilibrium conditions. 11.15 + 6 VBC − 10 × 2 =0 (iv) VBC = 33 kN. 11. 578 • Basic Structural Analysis Continuous Beam • 577 EXAMPLE 11. 30 kN 40 kN 20 kN 4 2 2 2 E F 2m D A B C FIG. .0 m. 3 l1 = 6.11c Pure moment diagram Properties of simply supported beam BMD 1 1 A1 = × 6 × 40 = 120 kNm2 A2 = × 4 × 40 = 80 kNm2 2 2 6+4 x1 = = 3.11: Analyse the continuous beam shown in figure by three moment theorem. 11.33 x2 = 2 m. 11.11b Simply supported beam BMD MB MC − A B C D FIG. Draw SFD & BMD. 11.11a S OLUTION The simple beam moments at E and F are Wab 30 × 4 × 2 ME = = = 40 kNm l 6 Wl 40 × 4 MF = = = 40 kNm 4 4 40 40 kNm + + A E B F C D FIG.00 l2 = 4. 6VA + 24 − 30(2) = 0 (ii) ∑ MB = 0.11f Shear force diagram .6 MB = −23. VA +VB1 = 30 (i) ∑ V = 0. 40 + 40(2) − 24 − 4VC = 0 (iv) VA = 6 kN VC = 24 kN ∴ VB1 = 24 kN VB2 = 16 kN 20 16 + + 6 + 2m 2m A 4m E B 2m F C 2m D – – 24 24 kN FIG. VB2 +VC = 40 (iii) ∑ MB = 0.00 4 20MB − 160 = −6 (66.11e ∑ V = 0.11d FIG.6 + 40) 20MB = −479. 11. 11.98 kNm Free Body diagrams 30 kN 40 kN 4m 2m 24 kNm 2m 2m 40 kNm E F VA VB1 VB2 VC FIG.578 • Basic Structural Analysis Continuous Beams • 579 Applying three moment theorem for spans AB & BC 1 3. 11.33 80(2) 6MA+ 2MB (6 + 4) + 4MC = −6 × 6 × 40 × > > + 2 6. imagine an imaginery span A A of zero length with no load and A is simply supported. Page 580 • 581 Basic Structural Analysis Continuous Beam • • 581 579 40 40 40 kNm −24 − + A E B F C D FIG.5 180 × 3 3MA + 2MB (3 + 6) + 6MC = > −6 + 3 6 3MA + 18MB = −6 (11.12: Draw the shear force diagram and bending moment diagram for the beam shown in figure.5 (ii) .25 + Page 583 A B C FIG.25 + 90) 3MA + 18MB = −607.5 Considering the span ABC 22. 10 kN/m A' A 3m B 6m C FIG. 11. Considering the span A AB 2 0 + × 3 × 11.5 × 1.5 3 > MA (0) + 2MA (0 + 3) + 3MB = −6 (i) 3 6MA + 3MB = −67.11g Bending moment diagram Page 582 EXAMPLE 11.25 1. 11.12b Simply supported beam BMD S OLUTION As the end A is fixed. 11.12a 45 kNm + 11. 77 + 6.77 + + − A B C + 6.77 kNm A 3m B VAB VBA FIG. VAB +VBA = 30 (i) 2 3 ∑ MB = 0.12e .77 kNm.14 kNm FIG. VBA = 28. 11.580 • 582 • Basic Structural Analysis Continuous Beams • 581 Solving (i) & (ii) MB = −34.36 kN.12d ∑ V = 0.12c Bending moment diagram Shear force and BM values for spans AB and BC Page 583 Static equilibrium of AB 6.25 34. The BMD Page 582 is drawn using the above end moments as 45 kNm 11.14 kNm.14 + 3VAB − 10 × 2 =0 (ii) VAB = 1.14 kNm 10 kN/m 34.77 kNm 10 kN/m B C 6m VBC x2 VCB FIG. MA = 6. Static equilibrium of BC 34. 34. 11.64 kN. 11. VCB = 24.3 − + + A + B E C 2.12f Shear force diagram Page 583 34.8 1. VBC +VCB = 10(6) = 60 (iii) Page 581 62 ∑ MC = 0.8 kN.77 kNm 6. x1 = 0.19 + 1.77 + 6VBC − 10 × 2 =0 VBC = 35.2 kN FIG.36 x2 C A D B E x1 28.36 − 10x1 = 0.12g Bending moment diagram MX1X1 = 6.28 kNm.64 24.135 m Page 582 35. − 34.35 (0.135)2 /2 = 6.28 29.14 FIG.42 m 6. . 11. Maximum positive BM is span AB The location of zero shear force in AB zone is 1.135) − 10 (0. 582 • Basic Structural Analysis Continuous Beam • • 583 581 ∑ V = 0. 11.2 kN. 3 kNm. B and C. 11.582 • Basic Structural Analysis Continuous Beams • 583 Maximum positive BM in span BC The location of zero shear force in BC zone is 24. Apply three moment theorem for the spans A AB > MA(0) + 2MA (0 + 5) + 5MB = −6(0 + 0) 10MA + 5MB = 0 2MA + MB = 0 (i) Apply three moment theorem for the spans ABC 5MA + 2MB (5 + 5) − 60(5) = −6(0 + 0) 5MA + 20MB = 300 (ii) Solving (i) and (ii) MA = −8. MB = +17.57 kNm. Sketch the shear force and bending moment diagram and mark in the salient values. determine the moments and reactions at A. AB = BC = 5 m and CD = 2 m.42 m M×2×2 = 24. 30 kN 5m 5m 2m A' A B C D FIG.14 kNm.2(2.13: A continuous beam ABCD is of uniform section.13a S OLUTION As the end A is fixed imagine a imaginery span A A of zero length and A is simply supported.42) − 10(2. simply supported at B and C and CD is an overhang. It is fixed at A.42)2 /2 = 29.2 − 10 x2 = 0 x2 = 2. EXAMPLE 11. If a concentrated load of 30 kN acts at D. . 60.0 + 17.14 + 5VBC = 0 VBC = −15.584 • Basic Structural Analysis Continuous Beam • 583 Shear force and BM values for spans AB and BC Span AB 8.57 kNm 17. 11.14 = 0 VAB = 5.13d Shear force diagram .14 kN. VCB = +15. Span BC 17. VBA = −5.43 kN. VBC +VCB = 0 ∑ M = 0. 30 kN 5.14 + + A B C D − 15. 11.14 kN.13c ∑ V = 0.43 kN. 11.57 − 17. 5VAB − 8. VAB +VBA = 0 (i) ∑ MB = 0.13b ∑ V = 0.14 kNm 60 kNm B 5m C VBC VCB FIG.43 FIG.14 kNm A 5m B VAB VBA FIG. 11.57 A B C D 17.14: Analyse the continuous beam by the theorem of three moments. 20 KN/m 16 kN 40 kN 2m 2m 2m 4m 4m A' A C D E FIG. Clearly indicate all the salient values.14b Simple beam moment diagram . 11.10 kNm FIG.14a S OLUTION The simple beam moments are wl 2 wl 42 4 MB = + = 20 × + 16 × = 56 kNm 8 4 8 4 wl 4 MD = = 40 × = 40 kNm 4 4 40 40 A B C D E FIG.584 • Basic Structural Analysis Continuous Beams • 585 60 8. 11.13e Bending moment diagram EXAMPLE 11. Draw neat sketches of SFD and BMD. 11.67 × 4 8MA + 4MC = −416.01 (i) Applying theorem of three moments for the spans ACE 138.67 A2 = × 4 × 40 = 80 kNm2 3 2 2 l1 = 4 m x2 = 2 m x1 = 2 m l2 = 4 m Applying three moment theorem for span A AC A x A x 1 1 2 2 MA l1 + 2MA l1 + l2 + MC l2 = −6 + l1 l2 2 2MA (4) + 4MC = −6 × 138. 11.14e FIG.14c Simple beam moment diagram Properties of simple beam BMD 2 1 1 A1 = × 4 × 40 + × 4 × 16 = 138. 11.67 × 2 80 × 2 > MA (4) + 2MC (4 + 4) + ME (4) = −6 + 4 4 4MA + 16MC = −6(109.01 (ii) MA = −36 kNm Solving Equations (i) and (ii) MC = −32 kNm Free body diagram 2m 16 kN 20 kN/m 32 kNm 40 kN 36 kNm 32 kNm 2m 4m 4m VC2 VE VA B VC1 C D E A C FIG.14d .335) = −656.586 • Basic Structural Analysis Continuous Beam • 585 16 A B C E FIG. 15a . 11.15: Sketch the BMD for the continuous beam shown in figure. 11. VA +VC1 = 16 + 4(20) = 96 (i) VC2 +VE = 40 (iii) ∑ MA = 0 ∑ ME = 0. 42 −36 + 32 + 4VA − 16(2) − 20 × = 0 (ii) −32 + 4VC2 − 40(2) = 0 (iv) 2 VA = 49 kN ∴ VC2 = 28 kN VC1 = 47 kN VE = 12 kN 49 28 + 9 + B D E A − C − 7 47 kN 12 FIG.14g Bending moment diagram Page 590 EXAMPLE 11.586 • 588 • Basic Structural Analysis Continuous Beams • 587 Static equilibrium of AC Static equilibrium of CE ∑ V = 0.14f Shear force Diagrams Page 588 56 40 kNm + + 36 32 − − A B C D E FIG. 11. ∑ V = 0. 60 kN 20 kN/m 30 kN 1m 3m O 4m 1m A' A D 3I B E 4I C F FIG. Apply three moment equation for the span A AB.3 kNm. .18 (ii) Solving (i) and (ii).338) 1. 11.33 + 1.7 × 2 MA + 2MB + + MC = −6 + .33 MA (0) + 2MA + MB = −6 3I 3I 4 × 3I 8MA + 4MB = −315 (i) Applying three moment theorem for the spans AB and BC.588 • Basic Structural Analysis Continuous Beam • 587 45 40 A D B E C FIG.863)6 1.0) − 30 = −6 (12. 0+4 4 0 + 90 × 2.15c Pure moment diagram S OLUTION Properties of the simple beam BMD 1 2 A1 = × 4 × 45 = 90 kNm2 A2 = × 4 × 40 = 106. 3I 3I 4I 4I 4 × 3I 4 × 4I 1. MA = −30. 11.15b Simple beam moment diagram MB MC MA A B C FIG. Assume A as simply supported.525 + 13.33MA + 4.66MB = 30 − (25.66MB = −125.1 kNm.67 m x2 = 2 m 3 l1 = 4 m l2 = 4 m Since A is fixed assume an imaginery span of A A of zero length with no loading. 4 4 4 4 90 × 1.67 106.7 kNm2 2 3 4+1 x1 = = 1. MB = −18.33MA + 2MB (1.33MA + 4. ∑ MB = 0 42 −30. 11.02 kN.1 kNm 18. 11.02 + + + A D − B E C F 1.98 kN FIG.15g Bending moment diagram .3 + 18.15f Shear force diagram + 15.1 kNm 30 kNm 1m 3m B C A B 4m VA VB1 VB2 VC Page 588 FIG 11.1 30.95 kN VC = 42.1 + 4VA − 60(3) = 0 −18.98 kN.15e Static equilibrium of AB Static equilibrium of BC ∑ V = 0.95 + 18.0 − − − A D B E C F FIG.95 42.0 kNm 30.15d FIG 11.85 − 11. Page 590 48. 588 • 590 • Basic Structural Analysis Continuous Beams • 589 Free body diagrams of span AB and BC 60 kN 20 kN/m 30. VA +VB = 60 (i) VB2 +VC = 80 (iii) ∑ MB = 0.05 30 37.10 + 30 + 4VB2 − 20 × =0 2 VA = 48. ∑ V = 0.3 kNm 18.3 45.05 kN (ii) VB2 = 37. VB2 = 11. 80 kN 20 kN/m 2m 1m A' 3m E 4m F 1m D 3I A 3I B 2I C 2I FIG. 11.16a 53.06 (ii) .590 • Basic Structural Analysis Continuous Beam • 589 EXAMPLE 11. 11.67 106.33 40 kNm A E B F C D FIG.84 + 26. Apply three moment theorem for spans A AB 0 0 3 3 80 × 1.16c Pure moment diagram S OLUTION As the end A is fixed assume an imaginery span A of zero length with no load and A is simply supported. E is constant.67 × 2 MA + 2MB + − 10 = −6 + 3I 3I 2I 2I 3 × 3I 2I × 4 MA + 6MB − 20 = −6(14. 11.16: Analyse the continuous beam by three moment theorem.16b Simple beam BMD MB 10 kNm MA A B C D FIG.33 > > MA + 2MA + + MB = −6 0 + 3I 3I 3I 3I 3 × 3I 2MA + MB = −70.93 (i) Applying three moment theorem for spans ABC 3 3 4 4 80 × 1. Draw the bending moment diagram.67) MA + 6MB = −249. 79 kN ME = −17.05 kNm 80 kN 38.88 kNm At the midspan of BC.16e ∑ V = 0.16d FIG.07 kN VB2 = 47.21 − 20x = 0 x = 2. MB = −38.590 • Basic Structural Analysis Continuous Beams • 591 Solving (i) and (ii) MA = −16.84 + 47.84 kNm E VB2 4m VC VA VB1 FIG.05 kNm.362 ∴ Max +ve BM = −38. 11.36 − 20 × 2 = 16. VA +VB1 = 80 (i) VB2 +Vc = 80 (iii) ∑ MB = 0 ∑ MC = 0 20 × 42 −16.84 − 80(1) + 3VA = 0 (ii) 10 − 38.58 kNm 2 .84 + 47. ∑ V = 0.36 m 2.84 kNm 20 kN/m 10 kNm 2m 1m 38. 11.86 + 20.21 kN VB1 = +60. 22 MF = −38.21 × 2 − 20 × = 15.84 + 4VB2 − =0 (iv) 2 VA = +19.05 + 38.88(2) = 23.93 kN ∴ VC = 32.9 kNm The location of zero shear in zone BC is obtained from 47.21 × 2. Free Body diagrams of span AB and BC 16.84 kNm. Span AB is 5 m while BC span is 6 m. 60 kN 10 kN/m 3m 2m 5m 6m A' A B C C EI 1.5EI Page 599 FIG. The ratio of flexural rigidity of span BC to BA is 1.33 15. Sketch the shear force and bending moment diagram. 11.05 10 kNm − − − A 2m E 1m B 2.16f Shear force diagram 53.07 47. It is subjected to a concentrated load of 60 kN at 3 m from A and the span BC is subjected to uniformly distributed load of 10 kN/m.17: EXAMPLE A continuous beam ABC is fixed at A and C.5.21 20 + + + A E B C D − − 60.17a S OLUTION The simple beam moments are Wab 60 × 3 × 2 MD = = = 72 kNm l 5 wl 2 10 × 62 ME = = = 45 kNm 8 8 . Use Clapeyron’s theorem of three moments. 11.93 33.592 Basic Structural Analysis Page•593 Continuous Beam • • 593 591 19. 11.32 m C 1m D FIG. It is continuous over a simple support B.84 + 16.79 FIG.58 + 38.16g Bending moment diagram Page 59611. 17c Pure moment diagram Since A is fixed imagine a span of zero length A A with no load and A is simply supported. Apply three moment theorem for the spans A AB Properties of the simple beam BMD 1 A1 = 0 × 5 × 72 = 180 A2 = 2 5+2 x1 = 0 x2 = = 2. 11.67 m x2 = 3 m 3 l1 = 5 m l2 = 6 m . 11.33 3 l1 = 0 l2 = 5.33 2MA + MB = −6 I I 5×I 10MA + 5MB = −503.17b Simple beam BMD MB MA MC A B C FIG.28 (i) Apply three moment theorem for the spans ABC Properties of the simple beam BMD 2 A1 = 180 kNm2 A2 = × 6 × 45 = 180 kNm2 3 5+3 x1 = = 2.592 • Basic Structural Analysis Continuous Beams • 593 72 45 kNm A D B E C FIG.0 l1 l1 l2 l2 A1 x1 A2 x2 MA + 2MA + + MB = −6 + I1 I1 I2 I2 l1 l2 5 5 180 × 2. 62 kNm 60 kN 37. MB = 37.62 kNm.67 180 × 3 MA + 2MB + + MC = −6 + I I 1. Shear force and bending moment values for the spans AB and BC respectively. (ii) and (iii) MA = 31.5I 6 × 1.5I 1.4 kNm 3m 2m A B D VAB VBA FIG.62 = 36. Span AB 31.5I 1. VAB +VBA = 60 (i) ∑ MB = 0.5I 1.594 • Basic Structural Analysis Continuous Beam • 593 5 5 6 6 180 × 2.16 kN Span BC MD = 22. 5VAB − 60(2) − 31. 11.4 = 0 (ii) VAB = 22.5I 5 6 × 1.4 kNm.5 5MA + 18MB + 4MC = −6 (96.62 + 37. 11.72 (ii) Applying three moment theorem BCC As the end C is fixed imagine a span CC of zero length and C is simply supported 6 6 0 180 × 3 > MB + 2MC + 0 + MC = −6 +0 1.29 kNm B C 6m E VBC VCB FIG.12 + 60) 5MA + 18MB + 4MC = −936.4 kNm 10 kN/m 26.17d ∑ V = 0.84 kN VBA = 37.17e . MC = 26.29 kNm.5 4MB + 8MC = −360 (iii) Solving equations (i).9 kNm 37.84(3) − 31. 62 26.4 + 13.594 • 596 • Basic Structural Analysis Continuous Beams • 595 ∑ V = 0.4 + 26.15 − 10x = 0 x = 2.85 kN VCB = 28.15(2. VBC +VCB = 60 (iii) 62 ∑ MC = 0.17f Shear force diagram The location of zero shear in span CB is obtained by equating the shear force equation to zero as (SF)xx = 28.85(3) − 37.29 kNm − − A D B F C FIG.29 = 13.15 kN Page 593 32 ME = 31.2 kNm Page 596 + 37.15 kN 2.815 m MF = 28.16 28. 11.815) − 10(2.15 31.815)2 /2 − 26.29 + 6VBC − 10 × 2 =0 VBC = 31.17g Bending moment diagram Page 599 . 11.85 22. − 37.815 FIG.49 37.4 − 10 × = 13.15 kNm 2 31.84 + + F A D B C − − 37. 18a S OLUTION The simple beam moments are 10 × 62 ME = MF = MG = = 45 kNm 8 45 45 45 kNm A B C D FIG. 11.18b Simple beam BMD MB MC A D B C FIG. EI is constant. Draw the SFD and BMD 10 kN/m 6m 6m 6m A E B F C G D FIG. 11. 11.18: A continuous beam ABCD is of uniform section as shown in figure.18c Pure moment diagram Considering spans ABC 2 A1 = × 6 × 45 = 180 KNm2 3 x1 = 3 m 180 × 3 180 × 3 > 6MA + 2MB (6 + 6) + 6MC = −6 + 6 6 24MB + 6MC = −6(90 + 90) = −1080 (1) .596 • Basic Structural Analysis Continuous Beam • 595 EXAMPLE 11. VBC +VCB = 60 (iii) 62 ∑ MC = 0.18d ∑ V = 0. 6VBC − 36 + 36 − 10 × 2 =0 (iv) VBC = 30 kN. 11. 11. VAB +VBA = 6(10) = 60 (i) 62 ∑ MB = 0. BCD Consider span AB 10 kN/m 36 kNm A B 6m VAB VBA FIG. 6VAB + 36 − 10 × 2 =0 (ii) VAB = 24 kN ∴ VBA = 36 kN Consider span BC 36 kNm 10 kN/m 36 kNm B C VBC 6m VCB FIG.596 • Basic Structural Analysis Continuous Beams • 597 Considering span BCD 180 × 3 180 × 3 > 6MB + 2MC (6 + 6) + 6MD = −6 + 6 6 6MB + 24MC = −1080 (2) Solving Equations (1) and (2) MB = −36 kNm MC = −36 kNm Shear force and bending moment values in the spans ABC.18e ∑ V = 0. VCB = 30 kN . 18h Bending moment diagram .4 m ME = 24(2. 11.4)2 /2 = 28. 11.18g Shear force diagram The location of zero shear is calculated as 24 − 10x1 = 0 x1 = 2.8 36 36 28.0 kNm MG = 24 (2.8 kNm MF = 30(3) − 36 − 10 × 32 /2 = 9.18f ∑ V = 0.4) − 10(2. VCD +VDC = 10 × 6 = 60 (v) 62 ∑ MD = 0. 6VCD − 36 − 10 × 2 =0 VCD = 36 kN Page 599 VDC = 24 kN 24 30 36 + + + E F G A B C D − − − x1 3m 36 30 24 kN 6m 6m 6m 2.4) − 10(2.598 • Basic Structural Analysis Continuous Beam • • 599 597 Span CD Page 596 36 kNm 10 kN/m D C VCD 6m VDC FIG.4 FIG.8 kNm + 9 28.4)2 /2 = 28.8 kNm + − − + A E B F C G D FIG. 11. Also draw SFD and BMD.1 A E B F C G D FIG.8 11.8MC = −6(28.2 × 1.4 > MA (3) + 2MB (3 + 2. 11.47 × 1.47 kNm2 A3 = 80 kNm2 3 3 x1 = 1.8 × 49 = 91.1 + 45.4 m x3 = 2 m l1 = 3 m l2 = 2.598 • Basic Structural Analysis Continuous Beams • 599 EXAMPLE 11.19: Analyse the continuous beam by three moment theorem.8 m 4m A E B F C G D FIG.8MC = −443 (i) .19c Pure moment diagram S OLUTION Properties of the simple beam BMD 2 2 A1 = × 3 × 28. 11. 11. 25 kN/m 50 kN/m 15 kN/m 3m 2.8MC = −6 + 3 2.1 = 56.5 91.8 m l3 = 4 m Applying three moment theorem for spans ABC 56.6MB + 2.5 m x2 = 1.19b Simply supported BMD MB MC A B C D FIG.6MB + 2.2 kNm2 A2 = × 2.74) 11.8) + 2.19a 49 30 28. 19f FIG.19g Shear force diagram .0 m 3m VB2 VC1 VC1 VD VA VB1 FIG.88 kN VB1 = 47.53 kNm Free body diagrams of AB. ∑ MD = 0.44 (ii) Solving (i) and (ii) MB = −30.600 • Basic Structural Analysis Continuous Beam • • 599 601 Applying three moment theorem for spans BCD 91.19e FIG.6MC = −6 (45. BC and CD ∑ V = 0.7 70.53 kNm 15 kN/m 25 kN/m 30.34 kN VD = 22.12 kN FIG.82 −50 × = 0 (iv) 2 VA = 27.7 kN Page 601 VC = 70. 11.3 kN VB2 = 69.6MC = −514.58 kNm.58 kNm 50 kN/m 31.8 + 4) + 4MD = −6 + 2.58 + 2.47 × 1.53 kNm 31.53 + 4VC2 − 15 × =0 (vi) 2 2 2. ∑ V = 0.66 37. MC = −31.8MB + 2MC (2. 11.8 4 2.4 80 × 2 > 2.53 − 30.74 + 40) 2.34 22.12 kN 27.53/4 = 37.19d Static equilibrium of spans AB.8MB + 13.3 69.66 kN VC2 = 151.58 kNm 2.88 + + + x1 x2 x3 − − − 47. VA +VB1 = 75 (i) VB2 +VC1 = 140 (iii) VC2 +VD = 60 (v) ∑ MB = 0. ∑ MC = 0.58 − × 32 = 0 (ii) 31. ∑ V = 0.8 m 4.8VB2 −31. BC and CD 30. 11. 11.8MB + 13. 25 42 3VA + 30. 88 − 15x3 = 0 x1 = 1.9 30.1 + 30 kNm − − 1.94 31.66(1.092 /2 = 14.09 m x2 = 1.66 − 50x2 = 0 37.600 • 602 • Basic Structural Analysis Continuous Beams • 601 The locations of shear forces in zones AB.52) − 15 × 2.09) − 25 × 1.53 + 16.3 28.20b Simple beam BMD for span ABC .39) − 50 × 1. 11.3(1.09 1.19h Bending moment diagram EXAMPLE 11.39 2.58 + 17.3 − 25x1 = 0 69.53 + 37.3 kNm Page 602 14.9 kNm M2 = −30. 11.5 kNm A B C FIG.88(2. 11.392 /2 = 17.52 m FIG. BC and CD are 27. EI is constant.39 m x3 = 2.522 /2 = 16.58 + 69.94 kNm M3 = −31.52 m M1 = 27.20: Analyse the continuous beam by theorem of three moments and draw SFD and BMD.20a 10 22. 10 kN 5 kN/m 15 kN 2m 3m 2m 4m 6m 5m A E B F C G D FIG. 5 18 kNm B C D FIG.20 b Properties of simple beam BMD 1 2 A1 = × 10 × 4 = 20 kNm2 A2 = × 6 × 22. 11.5 = 90 kNm2 2 3 x1 = 2 m x2 = 3 m l1 = 4 m l2 = 6 m Applying three moment theorem for spans ABC. 11.20c Simple beam BMD for span BCD MC MB FIG.602 • Basic Structural Analysis Continuous Beam • 601 22.20 c Properties of simple beam BMD 2 1 A1 = × 6 × 22.33 m 3 l1 = 6 m l2 = 5 m . 20 × 2 90 × 3 4MA+ 2MB (4 + 6) + 6MC = −6 > + 4 6 20MB + 6MC = −6(10 + 45) 20MB + 6MC = −330 (i) Referring to Fig.20d Pure moment diagram S OLUTION Referring to Fig.5 = 90 kNm2 A2 = × 5 × 18 = 45 kNm2 3 2 5+2 x1 = 3 m x2 = = 2. 11. 11. 10 kN 2m 2m 12.20e ∑ V = 0. VBC +VCB = 6(5) = 30 kN (iii) 2 5×6 ∑ MB = 0. 4VAB + 12.82 (ii) Solving (i) and (ii) MB = 12.69 kNm B C VBC 6m VCB FIG.20f ∑ V = 0.56 kN VCB = 15.09 kNm A B E VAB VBA FIG.09 − 2 =0 VBC = 14.98(2) = 3. MC = 14.02 kN ME = 1. BC and CD.09 kNm.69 − 12.96 kNm Span BC 12.97) 6MB + 22MC = −395.69 kNm.98 kN VBA = 8.33 6MB + 2MC (6 + 5) + 5MD= −6 > + 6 5 6MB + 22MC = −6(45 + 20. 6VBC + 14. 11. Shear force and bending moment values for spans AB. 11.09 − 10(2) = 0 (ii) Solving (i) and (ii) VAB = 1.602 • Basic Structural Analysis Continuous Beams • 603 Applying three moment theorem for spans BCD Considering span BCD 90 × 3 45 × 2.44 kN . VAB +VBA = 10 (i) ∑ MB = 0.09 kNm 5 kN/m 14. 06 kN 8. 11.02 14.91) − 12.09 − 5 × = 9.69 kNm 3m 2m C D VCD VCD FIG.11 15.44 kNm 12.56 8.91 m 2.69 FIG.94 + + + D A E B C − − − 6.20h Shear force diagram + 9.09 + + 12.604 • Basic Structural Analysis Continuous Beam • • 605 603 The location of shear force is zero is found out as 14.56 − 5x = 0 x = 2.06 kN 1.94 kN Page 605 VDC = 6. 11.20g ∑ V = 0.56(2.12 3. 11.912 Hence Max +ve BM = 14.11 kNm 2 Span CD 15 kN 14. 5VCD − 14.96 − − A E B C D FIG. VCD +VDC = 15 (iv) ∑ MD = 0.69 − 15(2) = 0 VCD = 8.98 14.20i Bending moment diagram Page 608 . 21a S OLUTION The simple beam moment 8 × 42 ME = = 16 kNm 8 16 kNm A B E C D FIG.604 • Basic Structural Analysis Continuous Beams • 605 EXAMPLE 11. It is subjected to uniformly distributed load of 8 kN/m over the span BC. 11. AB = BC = CD = 4 m. EI is constant.21: A continuous beam ABCD is simply supported at A and D.21b Simple beam bending moment diagram MB MC A B C D FIG.21c Pure moment diagram Consider span ABC Applying three moment theorem. It is continuous over supports B and C. 11. Draw the shear force diagram and bending moment diagram. 11. 2 4 × 16 × 2 4MA + 2MB (4 + 4) + 4MC = −6 0 + × 3 4 16MB + 4MC = −128 (i) Consider span BCD 2 4 × 16 × 2 4MB + 2MC (4 + 4) + 4MD= +0 > −6 × 3 4 4MB + 16MC = −128 (ii) . 8 kN/m 4m 4m 4m A B E C D FIG. 21f . 11.6 kN span BC 6.6 kN VBA = +1.4 + 4VBC − 2 =0 (iv) VBC = 16 kN VCB = 16 kN span CD 6. VAB +VBA = 0 (i) ∑ MB = 0. − 6.4 kNm A B 4m VAB VBA FIG.21e ∑ V = 0.21d ∑ V = 0. VBC +VCB = 8(4) = 32 (iii) 2 8×4 ∑ MC = 0.4 kNm C D 4m VCD VCD FIG. 11.4 kNm B C 4m VBC VCB FIG. BC and CD span AB 6.4 + 6.4 kNm MC = −6.606 • Basic Structural Analysis Continuous Beam • 605 Solving (i) and (ii) MB = −6. 4VAB + 6.4 kNm 8 kN/m 6.4 = 0 (ii) VAB = −1. 11.4 kNm Shear force and bending moment values of spans AB. 4 − − A B C D FIG.6 kN + + B A C D − − −1. 11.606 • 608 • Basic Structural Analysis Continuous Beams • 607 ∑ V = 0.22: Analyse the beam shown in figure by SFD and BMD. 10 kN/m 6m 6m 6m 6m A B C D E FIG.21g Shear force diagram 16 kNm + 6. EI is constant.22a 45 45 A B C D E FIG. 11.4 6.6 kN Page 608 VDC = −1.4 + 4VCD = 0 (vi) VCD = 1.6 16 FIG. VCD +VDC = 0 (v) ∑ MD = 0.22b Simple beam BMD . 11. − 6.6 kN 16 1.21h Bending moment diagram EXAMPLE 11. 11. 86 kNm. Free body diagram of AB.86 kNm A B 6m VAB VBA FIG.57 kNm. BC.22c Pure moment diagram Properties of simple beam BMD 2 2 A1 = × 6 × 45 = 180 A2 = × 6 × 45 = 180 KNm2 3 3 x1 = 3 m x2 = 3 m l1 = 6 m l2 = 6 m Applying 3 moment theorem for the spans ABC 180 × 3 6MA+ 2MB (6 + 6) + 6MC = −6 0 + > 6 24MB + 6MC = −540 (i) Applying 3 moment theorem for the spans BCD 180 × 3 180 × 3 6MB + 2MC (6 + 6) + 6MD = −6 + 6 6 6MB + 24MC + 6MD = −6(180) = −1080 MB + 4MC + MD = −180 (ii) Applying 3 moment theorem for spans CDE 180 × 3 6MC + 2MD (6 + 6) + 6ME = −6 +0 6 6MC + 24MD = −540 (iii) solving equations (i). CD and DE span AB 12. 11. MC = −38.608 • Basic Structural Analysis Continuous Beam • 607 MB MC MD A B C D E FIG. (ii) and (iii) MB = −12.22d . MD = −12.86 kNm. 11. 12.72 kN span DE 12. 6VAB + 12.57 + 6VCD − 10 × 2 =0 (vi) VCD = 34. VCD +VDC = 6(10) = 60 kN (v) 2 6 ∑ MD = 0. VBC +VCB = 60 (iii) 2 10 × 6 ∑ MC = 0.86 + 38.86 kNm C D 6m VCD VDC FIG.57 kNm 10 kN/m 12.28 kN span CD 38.28 kN VDC = 25.14 kN. 11.57 kNm B C 6m VBC VCB FIG.22g . 11.72 kN VCB = 34.86 kNm 10 kN/m 38.608 • Basic Structural Analysis Continuous Beams • 609 ∑ V = 0.86 kNm 6m VDE VED FIG.22f ∑ V = 0.86 = 0 (ii) VAB = −2. 11.57 + 6VBC − 2 =0 (iv) VBC = 25. VAB +VBA = 0 (i) ∑ MB = 0. − 12.86 − 38. span BC 12.22e ∑ V = 0.14 kN VBA = +2. 33 + 60) = −680 (ii) Solving (i) and (ii) MC = −6.23b (b) Simple beam BMD MD MC 20 B C D E FIG.610 • Basic Structural Analysis Continuous Beams • 611 S OLUTION MB = −20(1) = −20 kNm The simple beam moments are 20 × 42 MF = = 40 kNm 8 60 × 4 MG = = 60 kNm 4 40 kNm 60 kNm B C F D G E FIG. 11.92 kNm MD = −40.77 kNm .23c (c) Pure moment diagram Apply 3 moment theorem for the spans BCD 2 2 −20(3) + 2MC (3 + 4) + MD (4) = −6 0 + × 4 × 40 × 3 4 −60 + 14MC + 4MD = −320 14MC + 4MD = −260 (i) Apply 3 moment theorem for the spans CDE 2 2 1 2 MC (4) + 2MD (4 + 4) + 4ME = × 4 × 40 × + × 4 × 60 × > −6 3 4 2 4 4MC + 16MD = −6(53. 11. hogging BM.56 A B C 94.33 kNm Solving (1) and (2).44 kN FIG. 11. MB = −123.25d Free body diagram of spans AB and BC 130 kN 140.33 kNm 2m 2m A B B O 6m C 130 kN 110 kN 140.33 − + 180 − FIG.25f Bending moment diagram . 11.614 • Basic Structural Analysis Continuous Beam • 613 δA δC + 6EI + L1 L2 A1 x1 = 960 2 A2 x2 = × 6 × 180 × 3 = 2160 3 960 2160 240 120 Substituting.33 kNm + 123. MA × 4 + 2MB (4 + 6) = −6 + + 6EI + 4 6 EI × 4 EI × 6 4MA + 20MB = −3120 → MA + 5MB = −780 (2) MA = −163.33 kNm 240 kN 40 kN/m 163.44 110 FIG.56 kN 94. 11.33 kNm 123.25e Shear force diagram 240 163. . . kN.. Draw 240 120 SFD and BMD. 627 (11.. . . . I = 6(10)7 mm4 PREFACE TO THE FIRST EDITION 40 kN Preface to the Second Edition 10 kN/m v 20 kNm Preface to the First Edition 3m vi A Acknowledgements 4m B 6m C 1m D vii Preamble EI = Constant xiii Concept Map xiv 12 Ans: Inuence Lines for Indeterminate Beams 622 12. .. . at the supports. . . determine determine the resultants the resultants and moments and questions at the supports.0 . . VAB = +16. .25. . 6 EI d If the support B is higher by d. . During loading.5kN (11. . . . . as shown If in gure.4 INFLUENCE Determine LINESatUSING the reactions A. ..1 INTRODUCTION . MB = −3.31. BA . . . .75 kN. . Assume E = 2(10)5 N/mm2 . . . . .44 . RB = 165. . . . . . .. . . . . . . . . . . .5. VBA = 1.4)12.BC. . B andMOMENT DISTRIBUTION C of the continuous beam shown METHODin figure.3 NUMERICAL EXAMPLES.. CB . . . . V.+21. 664 References 8 kN 674 3 kN/m Index 1m 3m 675 A 4m 5m I B C 1. RC = 69.. If the the support support B sinks B sinks bymm. . . the100 endkN moments will be marked by + 30 kN/m because each end l2 moment is clockwise direction.. .=. . . .+23.. . EI EI EI is uniform. VBC = 6.. the end moments must be marked by − since each end moment 616 • 618 • Basic Structural Analysis l2 Continuous Beams • 617 acts in anticlockwise direction.V. +19. . . . . . .25I Ans: VAB = 6. Ans: Page 618 RA = 45kN..31 kNm. . 626 12.5kN.3) A continuous continuous beam beamofofuniform uniformsection sectionABCD ABCD is supported andand is supported loaded as shown loaded in Figure. . . . .V. . =. .. .5EI B continuous beam shown by EI theorem of three moments. VCB = 8.. . . . l 6 EI d If support B is lower by d. . . .69 MA = −3. 622 12.= .2 DEVELOPMENTMB = −14OFkNm INFLUENCE LINE DIAGRAM .24: Analyse the2. . the supports at B and C settle by and respectively. . 4m A 8m 6m C E XAMPLE 11.87. . . . by 10 10 mm. .5 . . . MCB = +7. . MD = 0 .5 m 2. RB = 127.618 • Basic Structural Analysis Continuous Beam • 617 (11.68 kN. MB = −75.5) Analyse the continuous beam shown in Figure and determine the reactions 80 kN 50 kN/m 40 kN 1m 2m 2m 2m 4m 4m 4m A 1.03. VDC = 11.85 kN.42. MC = −41. MC = −55. MA = −17.93.44 kNm.32.98 kNm.2 kNm. VCB = 97.17 kN MA = −75.78 kNm. Determine the end moments and reactions EI is constant.29 kNm (11. MB = −73. Modulus of elasticity is constant. 50 kN 50 75 kN/m 25 kN/m 2m 2m A 6m B 6m C 4m D EI = Constant Ans: (i) RA = 75.25 m 2.97. MB = −52.39 kN.47 kNm. 100 kN 30 kN/m 80 kN 40 kN 1.25 m 2I 3I 4I A B 6m D C Ans: MA = −56.5I B 2I C I D Ans: VAB = 41.5 m 1.88.59 kN.6) Analyse three span continuous beam by three moment theorems. VBC = 102.12 VCD = 28. The values of second moment area of each span are indicated along the members. RD = 99. MC = −94. RC = 97.55 kNm. MD = −9. Md = −55.7) Analyse and draw BMD and SFD for the beam shown in Figure.02 kNm.3 kNm. VBA = 38. (11. Draw the BMD and shear force diagram.5 m 2. VB = 0.188 kNm. MB = −0. MB = −2.620 • Basic Structural Analysis Continuous Beam • 619 (11. determine the support moments.12) Analyse the continuous beam by three moment theorem.5 kN 1m 3m 8m 6 mD 3m A I B I C I Ans: MA = −1.09 kNm.5 kNm . No loads on span AB. MC = −7. 0. Using Clapeyron’s three moment theo- rem.81 kN. 10 kN 2m 2m 30 kN/m 3m 3m A B C EI = Constant Ans: VA = 4.72 kNm (11.88 kN.31. VC = 4.11) Determine the reactions and the support moment at B.5 kN 2.