Document Sample

School of the Built Environment Structural Analysis Notes Moment Distribution These notes are designed to complement the lecture material, not replace it. They should serve as a reminder of what is already in your mind and are not intended as a self-teaching aid. Moment distribution is an iterative (i.e. numerical) method for obtaining the bending moments at appropriate points in continuous beams and rigidly-jointed frames. Although largely superseded by computer methods, it is still used to obtain rapid solutions so that decisions can be made before a full computer analysis is attempted. It is also useful for the design of support beams for sheet pile walls and formwork. Starting Point: Fixed End Moments The object of the method is to split a problem up into simple standard parts for which a solution is known and then to fit them all together in such a way that the real structure is modelled correctly. Look at the continuous beam below: A B C Figure 1: Continuous two-span beam. Moment distribution starts by fixing all joints against rotation, so the beam in figure 1 would be simplified as follows: A B C Figure 2: Beam split into two fully-fixed single span beams. 1 These are now similar (they are both fully fixed beams), and we can therefore use the same formulae to calculate the moments at each end of the span. These are called the Fixed End Moments or FEMs for short. The only two standard cases with which you need concern yourself are shown below: MAB MBA P A B a b L MAB MBA MAB = -Pab2/L2 MBA = Pa2b/L2 MAB MBA udl p (kN/m) A B L MAB MBA MAB = -pL2/12 MBA = pL2/12 Figure 3: Standard single-span fully-fixed beams for calculation of FEMs. You may notice the similarities between the first standard case in figure 3 and the problem with which we finished in the notes on Moment-Are Theorems. Notice also two important differences. Firstly, we have adopted a sign convention. This is absolutely essential with a numerical method: 2 Sign convention: Clockwise is positive There are no exceptions to this rule. The second difference is in the way the fixing moments are named – we use a two- letter suffix. The first letter refers to the end of the beam and the two letters together refer to the beam span. So for span B-C in figure 2, we would be calculating MBC and MCB. Equilibrium at joints Unless there is a moment applied directly at a joint, then the bending moments immediately either side of the joint must be in equilibrium, i.e. when added up they must equal zero. This will not be the case when we have split the beam or frame artificially into single span fully-fixed beams, so the purpose of the distribution process is to restore this equilibrium by allowing the joints to rotate. This is done in stages, one joint at a time, until all joints are in equilibrium. This lack of equilibrium at each joint is called the out-of-balance moment. Before we can remove this, we need to know something about the effect of the rotation. Member Stiffness and Carry-over The moment required to rotate one end of a beam can be calculated as follows: CBASBA SBA A B rotation 1 radian L CBASBA M/EI diagram SBA Figure 4: Member Stiffness and carry-over. In order to rotate the beam at B by 1 radian, we must apply a moment, which we will call SBA. Because end A is still fixed, there is a reaction which is proportional to the 3 applied moment. The constant of proportionality is denoted CBA. The M/EI diagram is re-drawn in figure 5 to make it easier to apply the moment-area theorems. SBA + CBASBA (-) CBASBA Figure 5: M/EI diagram drawn in parts. We can see that the deflection of B from the tangent at A is zero, hence from the second moment-area theorem, [((SBA + CBASBA) × L/2) × L/3 - (CBASBA × L) × L/2 ] /EI = 0 Hence CBA = ½. This is called the carry-over factor. The area under the M/EI diagram is equal to the rotation at B, i.e. 1, so from the first moment-area theorem: ((SBA + CBASBA) × L/2 - CBASBA × L)/EI = 1 and since CBA = ½, then SBA = 4EI/L. This is called the member stiffness. Before we go on, note that the value of 1 radian for the rotation at B is purely notional – its actual value is unimportant. 4 Distribution Factors When we allow a joint to rotate, then all of the members meeting at that joint rotate by the same amount. The amount of moment which must be applied to each member at a particular joint is proportional to the member stiffness, so we can now calculate the amount of moment which is distributed to each member as a joint is rotated. M A rotation θ C B Figure 6: Distribution Factors. The actual rotation applied to the joint is θ, so the moment applied to member BC must be SBC/ θ since θ is a fraction of the 1 radian angle used to derive the member stiffnesses. Similarly, the moment applied to member BA is SBA/ θ. For equilibrium, M = SBC/ θ + SBA/ θ Because we are not concerned with the actual size of θ, we calculate the proportions of M which go to each beam: moment distributed to beam BC = SBC/ θ = SBC SBC/ θ + SBA/ θ SBC + SBA moment distributed to beam BC = SBA/ θ = SBA SBC/ θ + SBA/ θ SBC + SBA This is known as the distribution factor for each member meeting at this joint. It should be clear that the sum of the distribution factors at a joint must equal unity. The Moment Distribution Process We now have all of the tools necessary for the analysis. The order of calculation is as follows: 1. Split the structure up into full-fixed single-span beam elements by fixing all of the joints. 2. Calculate fixed-end moments for all beam elements which support a transverse load. 3. For each beam element, calculate its member stiffness. 4. For each joint, calculate the distribution factors using the member stiffnesses. 5. Transfer all distribution factors and fixed-end moments to a diagram of the structure or to a calculation table. 5 6. Starting at one joint, sum all of the FEMs at that joint to obtain the out-of- balance moment at the joint. 7. Remove the out-of-balance moment by applying an equal and opposite moment to the joint – distribute this to the members in proportion to their distribution factor at the joint. 8. Carry over half of the distributed moment to the other end of each of the beams affected. 9. Move to another joint and repeat the process from step 6 until the out-of- balance moment is small (e.g 0.1 or 0.2 kNm). The process is illustrated in the following example. 50kN 30kN A B C 3m 2m 8m 6m Figure 7: Two-span continuous beam (EI = constant). FEMs: MAB 50kN MBA MAB = -50 5 32 / 82 A B = -35.2 kNm 5m 3m MBA = 50 52 3 / 82 8m = 58.6 kNm Similarly, MBC = -30 4 22 / 62 = -13.3 kNm MCB = 30 42 2 / 62 = 26.7 kNm Member stiffnesses: sAB = sBA = 4EI/L 4/8 = ½ and sBC = sCB = 4EI/L 4/6 = 2/3 6 Distribution factors: We have only artificially fixed one joint, B, as A and C are fixed anyway in the real beam, so we only need distribution factors at this joint: DFBA = 1/2 = 3/7 DFBA = 1/2 = 4/7 1/2 + 2/3 1/2 + 2/3 (check: 3/7 + 4/7 = 1 ) Distribution factors at A and C are set to zero, because they represent support reaction moments. This gives us all we need for the distribution. Distribution A B C 0 3/7 4/7 0 FEMs -35.2 58.6 -13.3 26.7 Dist. -19.4 -25.9 C/O -9.7 -12.9 Total -44.9 39.2 -39.2 13.9 Note: only one distribution has been needed. The out-of-balance moment at joint A is 58.6 – 13.3 = 45.3. This is distributed 3/7 x 45.3 = 19.4 to span BA and 4/7 x 45.3 = 25.9 to span BC. Note the change of sign during distribution. Carry-over factor is always ½ and the sign remains the same during carry-over to the opposite end of each beam. Each column is totalled to obtain the bending moment at that point. Note that the sum of the moments at a joint (e.g. B) must be zero to ensure equilibrium. 7 Modified Stiffness If the end of a beam is pinned, then it will be free to rotate without inducing a bending moment: Applied Moment No carried- = 3EI/L over Moment 1 radian Figure 8: No carry-over towards a pinned end. The stiffness of the member containing the pinned end is calculated as 3EI/L instead of the usual value, and no carry-over then occurs towards the pin. Consider the problem in figure 7, but with C as a pinned (simple) support: 50kN 30kN A B C 3m 2m 8m 6m Figure 9:Beam with pinned end. Remember, we must still calculate the FEMs as before. However, the distribution factors at B will be different because we will use a modified stiffness for BA. sAB = sBA = 4EI/L 4/8 = ½ ( as before) but sBC = 3EI/L 3/6 = ½ and our distribution factors at B will therefore be ½ and ½ since the stiffness either side is now the same (NOTE: do not confuse the stiffness values, which have the dimensions of kNm with the distribution factors which are dimensionless). 8 The distribution proceeds as before but with the major difference that there is no carry-over from B to C: A B C 0 1/2 1/2 0 FEMs -35.2 58.6 -13.3 26.7 Dist 1: -26.7 C/O -13.3 Dist 2: -16.0 -16.0 C/O -8.0 Total -43.2 42.6 -42.6 0 Notice that we have needed to perform two distributions, the first to remove the moment at the pinned end and the second one to produce equilibrium at joint B. Note that there is no carry-over from B to C after the second distribution, though we still can carry over away from the pinned end from C to B. Exercise: Use the distribution factors in the first example (page 7) to deal with the pinned end without the modified stiffness. You will need to make the DF of C equal to 1 so that each time you carry over from B to C you must immediately distribute at C to bring the moment there back to zero. You will obtain the same moments as the above example, but it will take much longer. 9

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 55 |

posted: | 8/9/2012 |

language: | English |

pages: | 9 |

OTHER DOCS BY Jeqtax

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.