Page 1 Example 19.3 Analyse the continuous beam shown in Fig. 19.4a by moment-distribution method. The support ABC B settles by below mm 5 and C. Assume A EI to be constant for all members ; and . GPa E 200 = 4 6 10 8 mm I × = Solution: Calculate fixed end beam moments due to externally applied loads assuming that support and C are locked. B m kN M m kN M m kN M m kN M F CB F BC F BA F AB . 67 . 2 ; . 67 . 2 . 2 ; . 2 - = + = - = + = (1) In the next step calculate fixed end moments due to support settlements. In the span AB , the chord ' AB rotates in the clockwise direction and in span , the chord rotates in the counterclockwise direction (Fig. 19.4b). BC C B' Page 2 Example 19.3 Analyse the continuous beam shown in Fig. 19.4a by moment-distribution method. The support ABC B settles by below mm 5 and C. Assume A EI to be constant for all members ; and . GPa E 200 = 4 6 10 8 mm I × = Solution: Calculate fixed end beam moments due to externally applied loads assuming that support and C are locked. B m kN M m kN M m kN M m kN M F CB F BC F BA F AB . 67 . 2 ; . 67 . 2 . 2 ; . 2 - = + = - = + = (1) In the next step calculate fixed end moments due to support settlements. In the span AB , the chord ' AB rotates in the clockwise direction and in span , the chord rotates in the counterclockwise direction (Fig. 19.4b). BC C B' radians AB 3 3 10 25 . 1 4 10 5 - - × - = × - = ? radians BC 3 3 10 25 . 1 4 10 5 - - × = × = ? (2) ? ? ? ? ? ? ? ? × - × × × × - = - = = - - 4 10 5 4 10 8 10 200 6 6 3 6 9 AB AB AB S BA S AB L EI M M ? . (3) 3 3000 kNm Nm = = m kN M M S CB S BC . 0 . 3 - = = In the next step, calculate stiffness and distribution factors. EI EI K EI K K BC BA AB 1875 . 0 25 . 0 4 3 25 . 0 = = = = (4) At joint : 429 . 0 ; 571 . 0 ; 4375 . 0 = = = ? BC BA DF DF EI K B At joint C : 0 . 1 ; 1875 . 0 = = ? CB DF EI K At fixed joint, the joint does not rotate and hence no distribution moments are developed and consequently distribution factor is equal to zero. The complete moment-distribution procedure is shown in Fig. 19.4c and Table 19.3. The diagram is self explanatory. In this particular case results are obtained in two cycles. In the first cycle joint is balanced and carry over moment is taken to joint C . In the next cycle , joint B B is balanced and carry over moment is taken to joint . The bending moment diagram is shown in fig. 19.4d. A Page 3 Example 19.3 Analyse the continuous beam shown in Fig. 19.4a by moment-distribution method. The support ABC B settles by below mm 5 and C. Assume A EI to be constant for all members ; and . GPa E 200 = 4 6 10 8 mm I × = Solution: Calculate fixed end beam moments due to externally applied loads assuming that support and C are locked. B m kN M m kN M m kN M m kN M F CB F BC F BA F AB . 67 . 2 ; . 67 . 2 . 2 ; . 2 - = + = - = + = (1) In the next step calculate fixed end moments due to support settlements. In the span AB , the chord ' AB rotates in the clockwise direction and in span , the chord rotates in the counterclockwise direction (Fig. 19.4b). BC C B' radians AB 3 3 10 25 . 1 4 10 5 - - × - = × - = ? radians BC 3 3 10 25 . 1 4 10 5 - - × = × = ? (2) ? ? ? ? ? ? ? ? × - × × × × - = - = = - - 4 10 5 4 10 8 10 200 6 6 3 6 9 AB AB AB S BA S AB L EI M M ? . (3) 3 3000 kNm Nm = = m kN M M S CB S BC . 0 . 3 - = = In the next step, calculate stiffness and distribution factors. EI EI K EI K K BC BA AB 1875 . 0 25 . 0 4 3 25 . 0 = = = = (4) At joint : 429 . 0 ; 571 . 0 ; 4375 . 0 = = = ? BC BA DF DF EI K B At joint C : 0 . 1 ; 1875 . 0 = = ? CB DF EI K At fixed joint, the joint does not rotate and hence no distribution moments are developed and consequently distribution factor is equal to zero. The complete moment-distribution procedure is shown in Fig. 19.4c and Table 19.3. The diagram is self explanatory. In this particular case results are obtained in two cycles. In the first cycle joint is balanced and carry over moment is taken to joint C . In the next cycle , joint B B is balanced and carry over moment is taken to joint . The bending moment diagram is shown in fig. 19.4d. A Table 19.3 Moment-distribution for continuous beam ABC Joints A B C Member AB BA BC CB Stiffness factor 0.25 EI 0.25 EI 0.1875 EI 0.1875 EI Distribution Factor 0.571 0.429 1.000 Fixed End Moments due to applied loads (kN.m) 2.000 -2.000 2.667 -2.667 Fixed End Moments due to support settlements (kN.m) 3.000 3.000 -3.000 -3.000 Total 5.000 1.000 -0.333 -5.667 Balance joint C and C.O. 2.835 5.667 Total 5.000 1.000 2.502 0.000 Balance joint B and C.O. to A -1.00 -2.000 -1.502 Final Moments (kN.m) 4.000 -1.000 1.000 0.000 Page 4 Example 19.3 Analyse the continuous beam shown in Fig. 19.4a by moment-distribution method. The support ABC B settles by below mm 5 and C. Assume A EI to be constant for all members ; and . GPa E 200 = 4 6 10 8 mm I × = Solution: Calculate fixed end beam moments due to externally applied loads assuming that support and C are locked. B m kN M m kN M m kN M m kN M F CB F BC F BA F AB . 67 . 2 ; . 67 . 2 . 2 ; . 2 - = + = - = + = (1) In the next step calculate fixed end moments due to support settlements. In the span AB , the chord ' AB rotates in the clockwise direction and in span , the chord rotates in the counterclockwise direction (Fig. 19.4b). BC C B' radians AB 3 3 10 25 . 1 4 10 5 - - × - = × - = ? radians BC 3 3 10 25 . 1 4 10 5 - - × = × = ? (2) ? ? ? ? ? ? ? ? × - × × × × - = - = = - - 4 10 5 4 10 8 10 200 6 6 3 6 9 AB AB AB S BA S AB L EI M M ? . (3) 3 3000 kNm Nm = = m kN M M S CB S BC . 0 . 3 - = = In the next step, calculate stiffness and distribution factors. EI EI K EI K K BC BA AB 1875 . 0 25 . 0 4 3 25 . 0 = = = = (4) At joint : 429 . 0 ; 571 . 0 ; 4375 . 0 = = = ? BC BA DF DF EI K B At joint C : 0 . 1 ; 1875 . 0 = = ? CB DF EI K At fixed joint, the joint does not rotate and hence no distribution moments are developed and consequently distribution factor is equal to zero. The complete moment-distribution procedure is shown in Fig. 19.4c and Table 19.3. The diagram is self explanatory. In this particular case results are obtained in two cycles. In the first cycle joint is balanced and carry over moment is taken to joint C . In the next cycle , joint B B is balanced and carry over moment is taken to joint . The bending moment diagram is shown in fig. 19.4d. A Table 19.3 Moment-distribution for continuous beam ABC Joints A B C Member AB BA BC CB Stiffness factor 0.25 EI 0.25 EI 0.1875 EI 0.1875 EI Distribution Factor 0.571 0.429 1.000 Fixed End Moments due to applied loads (kN.m) 2.000 -2.000 2.667 -2.667 Fixed End Moments due to support settlements (kN.m) 3.000 3.000 -3.000 -3.000 Total 5.000 1.000 -0.333 -5.667 Balance joint C and C.O. 2.835 5.667 Total 5.000 1.000 2.502 0.000 Balance joint B and C.O. to A -1.00 -2.000 -1.502 Final Moments (kN.m) 4.000 -1.000 1.000 0.000 Summary The moment-distribution method is applied to analyse continuous beam having support settlements. Each step in the numerical example is explained in detail. All calculations are shown at appropriate locations. The deflected shape of the continuous beam is sketched. Also, wherever required, the bending moment diagram is drawn. The numerical examples are explained with the help of free- body diagrams.Read More

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