Page 1 Instructional Objectives After reading this chapter the student will be able to 1. Solve continuous beam with support settlements by the moment- distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the continuous beam. 19.1 Introduction In the previous lesson, moment-distribution method was discussed in the context of statically indeterminate beams with unyielding supports. It is very well known that support may settle by unequal amount during the lifetime of the structure. Such support settlements induce fixed end moments in the beams so as to hold the end slopes of the members as zero (see Fig. 19.1). In lesson 15, an expression (equation 15.5) for beam end moments were derived by superposing the end moments developed due to 1. Externally applied loads on beams 2. Due to displacements B A ? ? , and ? (settlements). The required equations are, ? ? ? ? ? ? ? - + + = AB B A AB AB F AB AB L L EI M M 3 2 2 ? ? (19.1a) Page 2 Instructional Objectives After reading this chapter the student will be able to 1. Solve continuous beam with support settlements by the moment- distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the continuous beam. 19.1 Introduction In the previous lesson, moment-distribution method was discussed in the context of statically indeterminate beams with unyielding supports. It is very well known that support may settle by unequal amount during the lifetime of the structure. Such support settlements induce fixed end moments in the beams so as to hold the end slopes of the members as zero (see Fig. 19.1). In lesson 15, an expression (equation 15.5) for beam end moments were derived by superposing the end moments developed due to 1. Externally applied loads on beams 2. Due to displacements B A ? ? , and ? (settlements). The required equations are, ? ? ? ? ? ? ? - + + = AB B A AB AB F AB AB L L EI M M 3 2 2 ? ? (19.1a) ? ? ? ? ? ? ? - + + = AB A B AB AB F BA BA L L EI M M 3 2 2 ? ? (19.1b) This may be written as, (19.2a) [] S AB B A AB F AB AB M K M M + + + = ? ? 2 2 [ ] 22 F BA BA AB B A BA S M MK M ?? =+ + + (19.2b) where AB AB AB L EI K = is the stiffness factor for the beam AB. The coefficient 4 has been dropped since only relative values are required in calculating distribution factors. Note that 2 6 AB AB S BA S AB L EI M M ? - = = (19.3) S AB M is the beam end moments due to support settlement and is negative (clockwise) for positive support settlements (upwards). In the moment-distribution method, the support moments and due to uneven support settlements are distributed in a similar manner as the fixed end moments, which were described in details in lesson 18. S AB M S BA M It is important to follow consistent sign convention. Here counterclockwise beam end moments are taken as positive and counterclockwise chord rotation ? ? ? ? ? ? ? L is taken as positive. The moment-distribution method as applied to statically indeterminate beams undergoing uneven support settlements is illustrated with a few examples. Page 3 Instructional Objectives After reading this chapter the student will be able to 1. Solve continuous beam with support settlements by the moment- distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the continuous beam. 19.1 Introduction In the previous lesson, moment-distribution method was discussed in the context of statically indeterminate beams with unyielding supports. It is very well known that support may settle by unequal amount during the lifetime of the structure. Such support settlements induce fixed end moments in the beams so as to hold the end slopes of the members as zero (see Fig. 19.1). In lesson 15, an expression (equation 15.5) for beam end moments were derived by superposing the end moments developed due to 1. Externally applied loads on beams 2. Due to displacements B A ? ? , and ? (settlements). The required equations are, ? ? ? ? ? ? ? - + + = AB B A AB AB F AB AB L L EI M M 3 2 2 ? ? (19.1a) ? ? ? ? ? ? ? - + + = AB A B AB AB F BA BA L L EI M M 3 2 2 ? ? (19.1b) This may be written as, (19.2a) [] S AB B A AB F AB AB M K M M + + + = ? ? 2 2 [ ] 22 F BA BA AB B A BA S M MK M ?? =+ + + (19.2b) where AB AB AB L EI K = is the stiffness factor for the beam AB. The coefficient 4 has been dropped since only relative values are required in calculating distribution factors. Note that 2 6 AB AB S BA S AB L EI M M ? - = = (19.3) S AB M is the beam end moments due to support settlement and is negative (clockwise) for positive support settlements (upwards). In the moment-distribution method, the support moments and due to uneven support settlements are distributed in a similar manner as the fixed end moments, which were described in details in lesson 18. S AB M S BA M It is important to follow consistent sign convention. Here counterclockwise beam end moments are taken as positive and counterclockwise chord rotation ? ? ? ? ? ? ? L is taken as positive. The moment-distribution method as applied to statically indeterminate beams undergoing uneven support settlements is illustrated with a few examples. Example 19.1 Calculate the support moments of the continuous beam (Fig. 19.2a) having constant flexural rigidity ABC EI throughout, due to vertical settlement of support B by 5mm. Assume ; and . 200 GPa E = 44 410 m I - =× Solution There is no load on the beam and hence fixed end moments are zero. However, fixed end moments are developed due to support settlement of B by 5mm. In the spanAB , the chord rotates by AB ? in clockwise direction. Thus, 5 10 5 3 - × - = AB ? ? ? ? ? ? ? ? ? × - × × × × - = - = = - - 5 10 5 5 10 4 10 200 6 6 3 4 9 AB AB AB S BA S AB L EI M M ? 96000 Nm 96 kNm. == (1) In the span , the chord rotates by BC BC ? in the counterclockwise direction and hence taken as positive. 5 10 5 3 - × = BC ? Page 4 Instructional Objectives After reading this chapter the student will be able to 1. Solve continuous beam with support settlements by the moment- distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the continuous beam. 19.1 Introduction In the previous lesson, moment-distribution method was discussed in the context of statically indeterminate beams with unyielding supports. It is very well known that support may settle by unequal amount during the lifetime of the structure. Such support settlements induce fixed end moments in the beams so as to hold the end slopes of the members as zero (see Fig. 19.1). In lesson 15, an expression (equation 15.5) for beam end moments were derived by superposing the end moments developed due to 1. Externally applied loads on beams 2. Due to displacements B A ? ? , and ? (settlements). The required equations are, ? ? ? ? ? ? ? - + + = AB B A AB AB F AB AB L L EI M M 3 2 2 ? ? (19.1a) ? ? ? ? ? ? ? - + + = AB A B AB AB F BA BA L L EI M M 3 2 2 ? ? (19.1b) This may be written as, (19.2a) [] S AB B A AB F AB AB M K M M + + + = ? ? 2 2 [ ] 22 F BA BA AB B A BA S M MK M ?? =+ + + (19.2b) where AB AB AB L EI K = is the stiffness factor for the beam AB. The coefficient 4 has been dropped since only relative values are required in calculating distribution factors. Note that 2 6 AB AB S BA S AB L EI M M ? - = = (19.3) S AB M is the beam end moments due to support settlement and is negative (clockwise) for positive support settlements (upwards). In the moment-distribution method, the support moments and due to uneven support settlements are distributed in a similar manner as the fixed end moments, which were described in details in lesson 18. S AB M S BA M It is important to follow consistent sign convention. Here counterclockwise beam end moments are taken as positive and counterclockwise chord rotation ? ? ? ? ? ? ? L is taken as positive. The moment-distribution method as applied to statically indeterminate beams undergoing uneven support settlements is illustrated with a few examples. Example 19.1 Calculate the support moments of the continuous beam (Fig. 19.2a) having constant flexural rigidity ABC EI throughout, due to vertical settlement of support B by 5mm. Assume ; and . 200 GPa E = 44 410 m I - =× Solution There is no load on the beam and hence fixed end moments are zero. However, fixed end moments are developed due to support settlement of B by 5mm. In the spanAB , the chord rotates by AB ? in clockwise direction. Thus, 5 10 5 3 - × - = AB ? ? ? ? ? ? ? ? ? × - × × × × - = - = = - - 5 10 5 5 10 4 10 200 6 6 3 4 9 AB AB AB S BA S AB L EI M M ? 96000 Nm 96 kNm. == (1) In the span , the chord rotates by BC BC ? in the counterclockwise direction and hence taken as positive. 5 10 5 3 - × = BC ? ? ? ? ? ? ? ? ? × × × × × - = - = = - - 5 10 5 5 10 4 10 200 6 6 3 4 9 BC BC BC S CB S BC L EI M M ? . 96 96000 kNm Nm - = - = (2) Now calculate stiffness and distribution factors. EI L EI K AB AB BA 2 . 0 = = and EI L EI K BC BC BC 15 . 0 4 3 = = (3) Note that, while calculating stiffness factor, the coefficient 4 has been dropped since only relative values are required in calculating the distribution factors. For span , reduced stiffness factor has been taken as support C is hinged. BC AtB : EI K 35 . 0 = ? 571 . 0 35 . 0 2 . 0 = = EI EI DF BA 429 . 0 35 . 0 15 . 0 = = EI EI DF BC (4) At support C : EI K 15 . 0 = ? ; . 0 . 1 = CB DF Now joint moments are balanced as discussed previously by unlocking and locking each joint in succession and distributing the unbalanced moments till the joints have rotated to their final positions. The complete procedure is shown in Fig. 19.2b and also in Table 19.1. Page 5 Instructional Objectives After reading this chapter the student will be able to 1. Solve continuous beam with support settlements by the moment- distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the continuous beam. 19.1 Introduction In the previous lesson, moment-distribution method was discussed in the context of statically indeterminate beams with unyielding supports. It is very well known that support may settle by unequal amount during the lifetime of the structure. Such support settlements induce fixed end moments in the beams so as to hold the end slopes of the members as zero (see Fig. 19.1). In lesson 15, an expression (equation 15.5) for beam end moments were derived by superposing the end moments developed due to 1. Externally applied loads on beams 2. Due to displacements B A ? ? , and ? (settlements). The required equations are, ? ? ? ? ? ? ? - + + = AB B A AB AB F AB AB L L EI M M 3 2 2 ? ? (19.1a) ? ? ? ? ? ? ? - + + = AB A B AB AB F BA BA L L EI M M 3 2 2 ? ? (19.1b) This may be written as, (19.2a) [] S AB B A AB F AB AB M K M M + + + = ? ? 2 2 [ ] 22 F BA BA AB B A BA S M MK M ?? =+ + + (19.2b) where AB AB AB L EI K = is the stiffness factor for the beam AB. The coefficient 4 has been dropped since only relative values are required in calculating distribution factors. Note that 2 6 AB AB S BA S AB L EI M M ? - = = (19.3) S AB M is the beam end moments due to support settlement and is negative (clockwise) for positive support settlements (upwards). In the moment-distribution method, the support moments and due to uneven support settlements are distributed in a similar manner as the fixed end moments, which were described in details in lesson 18. S AB M S BA M It is important to follow consistent sign convention. Here counterclockwise beam end moments are taken as positive and counterclockwise chord rotation ? ? ? ? ? ? ? L is taken as positive. The moment-distribution method as applied to statically indeterminate beams undergoing uneven support settlements is illustrated with a few examples. Example 19.1 Calculate the support moments of the continuous beam (Fig. 19.2a) having constant flexural rigidity ABC EI throughout, due to vertical settlement of support B by 5mm. Assume ; and . 200 GPa E = 44 410 m I - =× Solution There is no load on the beam and hence fixed end moments are zero. However, fixed end moments are developed due to support settlement of B by 5mm. In the spanAB , the chord rotates by AB ? in clockwise direction. Thus, 5 10 5 3 - × - = AB ? ? ? ? ? ? ? ? ? × - × × × × - = - = = - - 5 10 5 5 10 4 10 200 6 6 3 4 9 AB AB AB S BA S AB L EI M M ? 96000 Nm 96 kNm. == (1) In the span , the chord rotates by BC BC ? in the counterclockwise direction and hence taken as positive. 5 10 5 3 - × = BC ? ? ? ? ? ? ? ? ? × × × × × - = - = = - - 5 10 5 5 10 4 10 200 6 6 3 4 9 BC BC BC S CB S BC L EI M M ? . 96 96000 kNm Nm - = - = (2) Now calculate stiffness and distribution factors. EI L EI K AB AB BA 2 . 0 = = and EI L EI K BC BC BC 15 . 0 4 3 = = (3) Note that, while calculating stiffness factor, the coefficient 4 has been dropped since only relative values are required in calculating the distribution factors. For span , reduced stiffness factor has been taken as support C is hinged. BC AtB : EI K 35 . 0 = ? 571 . 0 35 . 0 2 . 0 = = EI EI DF BA 429 . 0 35 . 0 15 . 0 = = EI EI DF BC (4) At support C : EI K 15 . 0 = ? ; . 0 . 1 = CB DF Now joint moments are balanced as discussed previously by unlocking and locking each joint in succession and distributing the unbalanced moments till the joints have rotated to their final positions. The complete procedure is shown in Fig. 19.2b and also in Table 19.1. Table 19.1 Moment-distribution for continuous beam ABC Joint A B C Member BA BC CB Stiffness factor 0.2EI 0.15EI 0.15EI Distribution Factor 0.571 0.429 1.000 Fixd End Moments (kN.m) 96.000 96.000 -96.000 -96.000 Balance joint C and C.O. to B 48.00 96.000 Balance joint B and C.O. to A -13,704 -27.408 -20.592 Final Moments (kN.m) 82.296 68.592 -68.592 0.000 Note that there is no carry over to joint as it was left unlocked. C Example 19.2 A continuous beam is carrying uniformly distributed load as shown in Fig. 19.3a. Compute reactions and draw shear force and bending moment diagram due to following support settlements. ABCD m kN / 5 , 0.005m vertically downwards. Support BRead More

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30 videos|122 docs|28 tests

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- The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2
- Slope Deflection Method
- The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 3
- The Moment Distribution Method: Frames without Sidesway
- The Moment Distribution Method: Frames with Sidesway - 1