Page 1 Support C , .0100m vertically downwards. Assume ; . GPa E 200 = 4 3 10 35 . 1 m I - × = Solution: Assume that supports and are locked and calculate fixed end moments due to externally applied load and support settlements. The fixed end beam moments due to externally applied loads are, D C B A , , 5 100 41.67 kN.m; 12 F AB M × == 41.67 kN.m F BA M =- 41.67 kN.m; F BC M =+ 41.67 kN.m F BC M =- 41.67 kN.m; F CD M =+ (1) 41.67 kN.m F DC M =- , the chord joining joints and In the spanAB A B rotates in the clockwise direction as moves vertical downwards with respect to (see Fig. 19.3b). B A Page 2 Support C , .0100m vertically downwards. Assume ; . GPa E 200 = 4 3 10 35 . 1 m I - × = Solution: Assume that supports and are locked and calculate fixed end moments due to externally applied load and support settlements. The fixed end beam moments due to externally applied loads are, D C B A , , 5 100 41.67 kN.m; 12 F AB M × == 41.67 kN.m F BA M =- 41.67 kN.m; F BC M =+ 41.67 kN.m F BC M =- 41.67 kN.m; F CD M =+ (1) 41.67 kN.m F DC M =- , the chord joining joints and In the spanAB A B rotates in the clockwise direction as moves vertical downwards with respect to (see Fig. 19.3b). B A 0.0005 radians AB ? =- (negative as chord ' AB rotates in the clockwise direction from its original position) 0.0005 radians BC ? =- 0.001 radians CD ? = (positive as chord rotates in the counterclockwise direction). D C' Now the fixed end beam moments due to support settlements are, 93 6 6 200 10 1.35 10 ( 0.0005) 10 81000 N.m 81.00 kN.m S AB AB AB AB EI M L ? - ×× × × =- =- - == 81.00 kN.m S BA M = 81.00 kN.m SS BC CB MM == 162.00 kN.m SS CD DC MM ==- (3) In the next step, calculate stiffness and distribution factors. For span AB and modified stiffness factors are used as supports CD A and are hinged. Stiffness factors are, D Page 3 Support C , .0100m vertically downwards. Assume ; . GPa E 200 = 4 3 10 35 . 1 m I - × = Solution: Assume that supports and are locked and calculate fixed end moments due to externally applied load and support settlements. The fixed end beam moments due to externally applied loads are, D C B A , , 5 100 41.67 kN.m; 12 F AB M × == 41.67 kN.m F BA M =- 41.67 kN.m; F BC M =+ 41.67 kN.m F BC M =- 41.67 kN.m; F CD M =+ (1) 41.67 kN.m F DC M =- , the chord joining joints and In the spanAB A B rotates in the clockwise direction as moves vertical downwards with respect to (see Fig. 19.3b). B A 0.0005 radians AB ? =- (negative as chord ' AB rotates in the clockwise direction from its original position) 0.0005 radians BC ? =- 0.001 radians CD ? = (positive as chord rotates in the counterclockwise direction). D C' Now the fixed end beam moments due to support settlements are, 93 6 6 200 10 1.35 10 ( 0.0005) 10 81000 N.m 81.00 kN.m S AB AB AB AB EI M L ? - ×× × × =- =- - == 81.00 kN.m S BA M = 81.00 kN.m SS BC CB MM == 162.00 kN.m SS CD DC MM ==- (3) In the next step, calculate stiffness and distribution factors. For span AB and modified stiffness factors are used as supports CD A and are hinged. Stiffness factors are, D EI EI K EI EI K EI EI K EI EI K CD CB BC BA 075 . 0 10 4 3 ; 10 . 0 10 10 . 0 10 ; 075 . 0 10 4 3 = = = = = = = = (4) At joint : 0 . 1 ; 075 . 0 = = ? AB DF EI K A At joint : 571 . 0 ; 429 . 0 ; 175 . 0 = = = ? BC BA DF DF EI K B At joint C : 429 . 0 ; 571 . 0 ; 175 . 0 = = = ? CD CB DF DF EI K At joint : 0 . 1 ; 075 . 0 = = ? DC DF EI K D The complete procedure of successively unlocking the joints, balancing them and locking them is shown in a working diagram in Fig.19.3c. In the first row, the distribution factors are entered. Then fixed end moments due to applied loads and support settlements are entered. In the first step, release joints A and . The unbalanced moments at D A and are 122.67 kN.m, -203.67 kN.m respectively. Hence balancing moments at D A and are -122.67 kN.m, 203.67 kN.m respectively. (Note that we are dealing with beam end moments and not joint moments). The joint moments are negative of the beam end moments. Further leave D and unlocked as they are hinged joints. Now carry over moments and to joint A D -61.34 kN.m kN.m 101.84 B and respectively. In the next cycle, balance joints C and C. The unbalanced moment at joint B B is . Hence balancing moment for beam 100.66 kN.m BA is 43.19 ( 100.66 0.429) - -× and for is . The balancing moment on gives a carry over moment of to joint C . The whole procedure is shown in Fig. 19.3c and in Table 19.2. It must be noted that there is no carryover to joints BC BC 57.48 kN.m (-100.66 x 0.571) - 26.74 kN.m - A and as they were left unlocked. D Page 4 Support C , .0100m vertically downwards. Assume ; . GPa E 200 = 4 3 10 35 . 1 m I - × = Solution: Assume that supports and are locked and calculate fixed end moments due to externally applied load and support settlements. The fixed end beam moments due to externally applied loads are, D C B A , , 5 100 41.67 kN.m; 12 F AB M × == 41.67 kN.m F BA M =- 41.67 kN.m; F BC M =+ 41.67 kN.m F BC M =- 41.67 kN.m; F CD M =+ (1) 41.67 kN.m F DC M =- , the chord joining joints and In the spanAB A B rotates in the clockwise direction as moves vertical downwards with respect to (see Fig. 19.3b). B A 0.0005 radians AB ? =- (negative as chord ' AB rotates in the clockwise direction from its original position) 0.0005 radians BC ? =- 0.001 radians CD ? = (positive as chord rotates in the counterclockwise direction). D C' Now the fixed end beam moments due to support settlements are, 93 6 6 200 10 1.35 10 ( 0.0005) 10 81000 N.m 81.00 kN.m S AB AB AB AB EI M L ? - ×× × × =- =- - == 81.00 kN.m S BA M = 81.00 kN.m SS BC CB MM == 162.00 kN.m SS CD DC MM ==- (3) In the next step, calculate stiffness and distribution factors. For span AB and modified stiffness factors are used as supports CD A and are hinged. Stiffness factors are, D EI EI K EI EI K EI EI K EI EI K CD CB BC BA 075 . 0 10 4 3 ; 10 . 0 10 10 . 0 10 ; 075 . 0 10 4 3 = = = = = = = = (4) At joint : 0 . 1 ; 075 . 0 = = ? AB DF EI K A At joint : 571 . 0 ; 429 . 0 ; 175 . 0 = = = ? BC BA DF DF EI K B At joint C : 429 . 0 ; 571 . 0 ; 175 . 0 = = = ? CD CB DF DF EI K At joint : 0 . 1 ; 075 . 0 = = ? DC DF EI K D The complete procedure of successively unlocking the joints, balancing them and locking them is shown in a working diagram in Fig.19.3c. In the first row, the distribution factors are entered. Then fixed end moments due to applied loads and support settlements are entered. In the first step, release joints A and . The unbalanced moments at D A and are 122.67 kN.m, -203.67 kN.m respectively. Hence balancing moments at D A and are -122.67 kN.m, 203.67 kN.m respectively. (Note that we are dealing with beam end moments and not joint moments). The joint moments are negative of the beam end moments. Further leave D and unlocked as they are hinged joints. Now carry over moments and to joint A D -61.34 kN.m kN.m 101.84 B and respectively. In the next cycle, balance joints C and C. The unbalanced moment at joint B B is . Hence balancing moment for beam 100.66 kN.m BA is 43.19 ( 100.66 0.429) - -× and for is . The balancing moment on gives a carry over moment of to joint C . The whole procedure is shown in Fig. 19.3c and in Table 19.2. It must be noted that there is no carryover to joints BC BC 57.48 kN.m (-100.66 x 0.571) - 26.74 kN.m - A and as they were left unlocked. D Page 5 Support C , .0100m vertically downwards. Assume ; . GPa E 200 = 4 3 10 35 . 1 m I - × = Solution: Assume that supports and are locked and calculate fixed end moments due to externally applied load and support settlements. The fixed end beam moments due to externally applied loads are, D C B A , , 5 100 41.67 kN.m; 12 F AB M × == 41.67 kN.m F BA M =- 41.67 kN.m; F BC M =+ 41.67 kN.m F BC M =- 41.67 kN.m; F CD M =+ (1) 41.67 kN.m F DC M =- , the chord joining joints and In the spanAB A B rotates in the clockwise direction as moves vertical downwards with respect to (see Fig. 19.3b). B A 0.0005 radians AB ? =- (negative as chord ' AB rotates in the clockwise direction from its original position) 0.0005 radians BC ? =- 0.001 radians CD ? = (positive as chord rotates in the counterclockwise direction). D C' Now the fixed end beam moments due to support settlements are, 93 6 6 200 10 1.35 10 ( 0.0005) 10 81000 N.m 81.00 kN.m S AB AB AB AB EI M L ? - ×× × × =- =- - == 81.00 kN.m S BA M = 81.00 kN.m SS BC CB MM == 162.00 kN.m SS CD DC MM ==- (3) In the next step, calculate stiffness and distribution factors. For span AB and modified stiffness factors are used as supports CD A and are hinged. Stiffness factors are, D EI EI K EI EI K EI EI K EI EI K CD CB BC BA 075 . 0 10 4 3 ; 10 . 0 10 10 . 0 10 ; 075 . 0 10 4 3 = = = = = = = = (4) At joint : 0 . 1 ; 075 . 0 = = ? AB DF EI K A At joint : 571 . 0 ; 429 . 0 ; 175 . 0 = = = ? BC BA DF DF EI K B At joint C : 429 . 0 ; 571 . 0 ; 175 . 0 = = = ? CD CB DF DF EI K At joint : 0 . 1 ; 075 . 0 = = ? DC DF EI K D The complete procedure of successively unlocking the joints, balancing them and locking them is shown in a working diagram in Fig.19.3c. In the first row, the distribution factors are entered. Then fixed end moments due to applied loads and support settlements are entered. In the first step, release joints A and . The unbalanced moments at D A and are 122.67 kN.m, -203.67 kN.m respectively. Hence balancing moments at D A and are -122.67 kN.m, 203.67 kN.m respectively. (Note that we are dealing with beam end moments and not joint moments). The joint moments are negative of the beam end moments. Further leave D and unlocked as they are hinged joints. Now carry over moments and to joint A D -61.34 kN.m kN.m 101.84 B and respectively. In the next cycle, balance joints C and C. The unbalanced moment at joint B B is . Hence balancing moment for beam 100.66 kN.m BA is 43.19 ( 100.66 0.429) - -× and for is . The balancing moment on gives a carry over moment of to joint C . The whole procedure is shown in Fig. 19.3c and in Table 19.2. It must be noted that there is no carryover to joints BC BC 57.48 kN.m (-100.66 x 0.571) - 26.74 kN.m - A and as they were left unlocked. D Table 19.2 Moment-distribution for continuous beam ABCD Joint A B C D Members AB BA BC CB CD DC Stiffness factors 0.075 EI 0.075 EI 0.1 EI 0.1 EI 0.075 EI 0.075 EI Distribution Factors 1.000 0.429 0.571 0.571 0.429 1.000 FEM due to externally applied loads 41.670 -41.670 41.670 -41.670 41.670 -41.670 FEM due to support settlements 81.000 81.000 81.000 81.000 - 162.000 - 162.000 Total 122.670 39.330 122.670 39.330 - 120.330 - 203.670 Balance A and D released - 122.670 203.670 Carry over -61.335 101.835 Balance B and C -43.185 -57.480 -11.897 -8.94 Carry over -5.95 -26.740 Balance B and C 2.552 3.40 16.410 12.33 Carry over to B and C 8.21 1.70 Balance B and C -3.52 -4.69 -0.97 -0.73 C.O. to B and C -0.49 -2.33 Balance B and C 0.21 0.28 1.34 1.01 Carry over 0.67 0.14 Balance B and C -0.29 -0.38 -0.08 -0.06 Final Moments 0.000 -66.67 66.67 14.88 -14.88 0.000Read More

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