In Example 12.2, for the beam with an overhang, the balancing at joint B requires an internal moment of -4000 lb-ft to the right of B. What is the source of this balancing moment?
Explanation
When a beam has an overhang, the moment created by the overhang at the first interior support is calculated by statics. The moment distribution process at that support must then ensure that the final moments in the continuous spans balance this known moment from the overhang.
Other questions
Who developed the moment-distribution method of analysis for beams and frames, and in what year?
What is the sign convention for moments used in the moment-distribution method?
What is the term for the moments at the fixed joints of a loaded member before any distribution occurs?
For a prismatic beam member that is fixed at its far end, what is the formula for the member stiffness factor, K?
A prismatic beam of length 10 m is fixed at both ends and subjected to a concentrated load of 800 N at its center. What is the value of the Fixed-End Moment (FEM) at the near end, A?
How is the Distribution Factor (DF) for a member at a joint calculated?
A frame joint A has three members connected: AB, AC, and AD. Their stiffness factors are K_AB = 4000, K_AC = 5000, and K_AD = 1000 units, respectively. What is the distribution factor (DF) for member AB?
When analyzing a continuous beam or frame made from the same material, why is it often easier to use the member's relative-stiffness factor, KR = I/L, for computations?
For a prismatic beam member with the far end fixed, what is the value of the carry-over factor?
What is the theoretical distribution factor for a member framing into a fixed wall support?
In a continuous beam, span AB has a moment of inertia I = 300 in^4 and length L = 15 ft. Span BC has I = 600 in^4 and L = 20 ft. Calculate the relative stiffness factor, KR, for span BA.
In the first step of the moment distribution procedure for the beam in Example 12.1, what is the total unbalanced moment at joint C before any distribution takes place?
In Example 12.1, a balancing moment of +240 kN-m is applied to joint B. Given that the distribution factors DF_BA and DF_BC are both 0.5, what is the moment distributed to member BA?
What is the primary assumption made about all joints at the beginning of the moment distribution process?
For a beam member where the far end is a pin or roller support, what is the modified member stiffness factor?
What is the carry-over factor for a beam member with a pin or roller support at its far end?
For a symmetric beam with symmetric loading, the stiffness factor for the center span can be modified to simplify the analysis. What is this modified stiffness factor?
For a symmetric beam subjected to antisymmetric loading, what is the modified stiffness factor for the center span?
In Example 12.4, a beam is analyzed where the far end C is a roller support. For span BC, a single fixed-end moment is used with the formula -wL^2/8. Why is this formula used instead of the standard -wL^2/12?
When does a frame have a tendency to sidesway?
What is the general principle used to analyze frames with sidesway using the moment-distribution method?
In the superposition analysis of a frame with sidesway, what is the purpose of the 'artificial joint support'?
In the analysis of a frame with sidesway, after the no-sidesway case is solved, a restraining force R is found. In the second step, an equal and opposite force is applied. How are the moments from this second step typically determined?
In Example 12.6, after the no-sidesway moment distribution, the horizontal reactions at the column bases A and D are found to be 1.73 kN and 0.81 kN, respectively. The external load is 16 kN. What is the magnitude of the restraining force R?
In the sidesway analysis part of Example 12.6, an arbitrary Fixed-End Moment of -100 kN-m is assumed for the columns. Why is a negative sign necessary for this assumed moment?
In Example 12.6, the no-sidesway analysis yields MBA = 5.78 kN-m. The sidesway analysis yields MBA' = -60 kN-m for a force R' = 56.0 kN. The actual restraining force was R = 0.92 kN. What is the final moment MBA?
What is the key difference in the analysis of multistory frames with sidesway compared to single-story frames?
In Example 12.3, a symmetric beam with symmetric loading is analyzed using stiffness modifications. The left half of the beam consists of span AB (pinned at A) of length 15 ft and half of span BC of length 20 ft. What relative stiffness factor is used for span AB?
In Example 12.3, analyzing a symmetric beam, what relative stiffness factor is used for the center span BC?
Based on the modified relative stiffness factors for Example 12.3, where KR_BA is based on 3I/15 and KR_BC is based on 2I/20, what is the distribution factor DF_BA?
For the beam in Example 12.2, with an overhang AB, the distribution factor (DF)_BA is set to 0. Why is this?
What is the primary goal of the successive locking and unlocking of joints in the moment distribution process?
According to the procedure for analysis, which of the following is NOT a step in the 'Moment Distribution Process'?
For the beam in Fig. 12-6, which has a roller support at C, what is the distribution factor DFCB at joint C?
In the analysis of the symmetric frame in Example 12.5 (Fig. 12-15), a 20-k load is applied at joint B. Why does this load not contribute a FEM to the analysis?
In Example 12.5, the stiffness of member CD is computed using K = 3EI/L. What is the reason for using this modified stiffness factor?
The moment distribution method is described as a displacement method. What does this mean?
Consider the beam from Example 12.1. It has spans AB (L=12m), BC (L=12m), and CD (L=8m). Supports A and D are fixed. Using relative stiffness factors (I/L), what is the distribution factor DF_BC at joint B? Assume I is constant for all spans.
A prismatic beam with far end fixed has a stiffness factor K = 4EI/L. A similar beam with far end pinned has a modified stiffness factor K_mod = 3EI/L. By what fraction must the standard stiffness factor be multiplied to obtain the modified factor for the pinned-end case?
In the tabular method for moment distribution shown in Figure 12-6c, the process involves simultaneous distribution at multiple joints in each step. How does the convergence of this simultaneous method compare to the successive, one-joint-at-a-time method?
What is the physical meaning of the 'total stiffness factor' at a joint, KT = sum(K)?
If a moment M is applied to a fixed connected joint, and a specific member 'i' has a stiffness factor Ki and a distribution factor DFi, what is the resisting moment Mi supplied by that member?
In the first distribution step of the beam in Example 12.1, a moment of +120 kN-m is distributed to member BA at joint B. What is the corresponding carry-over moment at the fixed support A?
For the symmetric beam in Example 12.3, the Fixed-End Moment for span BA, (FEM)BA, is calculated using the formula wL^2/15. Why is this formula used?
If a multistory frame has two independent joint displacements (sidesway at each floor), how many simultaneous equations must be solved to find the correction factors C' and C'' for the sidesway moments?
In the context of the moment distribution method, what is meant by an 'unbalanced moment' at a joint?
What does a negative sign on a final, calculated end moment indicate?
A standard prismatic beam member with a fixed far end has a stiffness factor K=4EI/L and carry-over factor COF=0.5. A symmetric beam's center span with symmetric loading has a modified stiffness factor K_mod=2EI/L. What is the carry-over factor associated with this modified stiffness?
In the final step of the sidesway analysis for Example 12.7, the final moment M_AB is calculated by adding the no-sidesway moment (9.58) to a scaled version of the sidesway moment (-69.91). The scaling factor is R/R' = 1.89/12.55. What is the calculated final moment M_AB?