338 Momentdistribution Method

The moment-distribution method is one of several displacement methods for analyzing continuous beams and rigid frames. Moment distribution, however, provides an alternative to solving the system of simultaneous equations that result with other methods, such as slope deflection. (See Arts. 3.36, 3.37, and 3.39.)

Moment distribution is based on the fact that the bending moment at each end of a member of a continuous frame equals the sum of the fixed-end moments due to the applied loads on the span and the moments produced by rotation of the member ends and of the chord between these ends. Given fixed-end moments, the moment-distribution method determines moments generated when the structure deforms.

Figure 3.79 shows a structure consisting of three members rigidly connected at joint O (ends of the members at O must rotate the same amount). Supports at A, B, and C are fixed (rotation not permitted). If joint O is locked temporarily to prevent rotation, applying a load on member OA induces fixed-end moments at A and O. Suppose fixed-end moment MOAF induces a counterclockwise moment on locked joint O. Now, if the joint is released, MOAF rotates it counterclockwise. Bending moments are developed in each member joined at O to balance MOAF. Bending moments are also developed at the fixed supports A, B, and C. These moments are said to be carried over from the moments in the ends of the members at O when the joint is released.

The total end moment in each member at O is the algebraic sum of the fixed-end moment before release and the moment in the member at O caused by rotation of the joint, which depends on the relative stiffnesses of the members. Stiffness of a prismatic fixed-end beam is proportional to EI/L, where E is the modulus of elasticity, I the moment of inertia, and L the span.

When a fixed joint is unlocked, it rotates if the algebraic sum of the bending moments at the joint does not equal zero. The moment that causes the joint to rotate is the unbalanced moment. The moments developed at the far ends of each member of the released joint when the joint rotates are carry-over moments.

FIGURE 3.79 Straight members rigidly connected at joint O. Dash lines show deformed shape after loading.

In general, if all joints are locked and then one is released, the amount of unbalanced moment distributed to member i connected to the unlocked joint is determined by the distribution factor Di the ratio of the moment distributed to i to the unbalanced moment. For a prismatic member,

2 Eh/Li

where S"=1 E^Ij/Lj is the sum of the stiffness of all n members, including member i, joined at the unlocked joint. Equation (3.144) indicates that the sum of all distribution factors at a joint should equal 1.0. Members cantilevered from a joint contribute no stiffness and therefore have a distribution factor of zero.

The amount of moment distributed from an unlocked end of a prismatic member to a locked end is z/2. This carry-over factor can be derived from Eqs. (3.135a and b) with 0a = 0.

Moments distributed to fixed supports remain at the support; i.e., fixed supports are never unlocked. At a pinned joint (non-moment-resisting support), all the unbalanced moment should be distributed to the pinned end on unlocking the joint. In this case, the distribution factor is 1.0.

To illustrate the method, member end moments will be calculated for the continuous beam shown in Fig. 3.75a. All joints are initially locked. The concentrated load on span AB induces fixed-end moments of 9.60 and -14.40 ft-kips at A and B, respectively (see Art. 3.37). The uniform load on BC induces fixed-end moments of 18.75 and -18.75 ft-kips at B and C, respectively. The moment at C from the cantilever CD is 12.50 ft-kips. These values are shown in Fig. 3.80a.

The distribution factors at joints where two or more members are connected are then calculated from Eq. (3.144). With EIAB/LAB = 200E/120 = 1.67E and EIBC/LBC = 600E/ 180 = 3.33E, the distribution factors are DBA = 1.67E/(1.67E + 3.33E) = 0.33 and DBC = 3.33/5.00 = 0.67. With EICD/LCD = 0 for a cantilevered member, DCB = 10E/(0 + 10E) = 1.00 and Dcd = 0.00.

Joints not at fixed supports are then unlocked one by one. In each case, the unbalanced moments are calculated and distributed to the ends of the members at the unlocked joint according to their distribution factors. The distributed end moments, in turn, are "carried over'' to the other end of each member by multiplication of the distributed moment by a carry-over factor of z/2. For example, initially unlocking joint B results in an unbalanced moment of -14.40 + 18.75 = 4.35 ft-kips. To balance this moment, -4.35 ft-kips is distributed to members BA and BC according to their distribution factors: MBA = -4.35DBA = -4.35 X 0.33 = -1.44 ft-kips and MBC = -4.35DBC = -2.91 ft-kips. The carry-over moments are MAB = MBA/2 = -0.72 and MCB = MBC/2 = -1.46. Joint B is then locked, and the resulting moments at each member end are summed: MAB = 9.60 - 0.72 = 8.88, Mba = -14.40 - 1.44 = -15.84, MBC = 18.75 - 2.91 = 15.84, and MCB = -18.75 -1.46 = -20.21 ft-kips. When the step is complete, the moments at the unlocked joint balance, that is, -Mba = Mbc.

The procedure is then continued by unlocking joint C. After distribution of the unbalanced moments at C and calculation of the carry-over moment to B, the joint is locked, and the process is repeated for joint B. As indicated in Fig. 3.80b, iterations continue until the final end moments at each joint are calculated to within the designer's required tolerance.

There are several variations of the moment-distribution method. This method may be extended to determine moments in rigid frames that are subject to drift, or sidesway.

(C. H. Norris et al., Elementary Structural Analysis, 4th ed., McGraw-Hill, Inc., New York; J. McCormac and R. E. Elling, Structural Analysis—A Classical and Matrix Approach, Harper and Row Publishers, New York.)

Fixed End Moment
FIGURE 3.80 (a) Fixed-end moments for beam in Fig. 3.75a. (b) Steps in moment distribution. Fixed-end moments are given in the top line, final moments in the bottom line, in ft-kips.
Renewable Energy 101

Renewable Energy 101

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.

Get My Free Ebook

Post a comment