Tiêu đề 25 Moment Distribution Method.pdf ✅

PSAD 25: Moment Distribution Method This module discusses one method of analyzing frames and beams. This method deals with the conversion of fixed end moments from the external loads and then distributing these moments along each span. 1. Fixed End Moment For sign convention, positive FEM values are in the clockwise direction while for counterclockwise FEM values are negative. FEM values are determined by assuming that the ends of each unsupported lengths are fixed ends. Hence, these unsupported lengths are analyzed as fixed-ended beams. Shown below are the FEM values of typical loadings. Loading FEMA FEMB Pab 2 L 2 Pa 2b L 2 wL 2 12 wL 2 12 wL 2 20 wL 2 30 ∫ w(x) ⋅ x(L − x) 2 L 2 dx L 0 ∫ w(x) ⋅ x 2 (L − x) L 2 dx L 0 2. Member Stiffness Factor The member stiffness factor is the amount of moment to rotate an end of the beam by 1 radian. 2.1. General Formula K = 4EI L Since in most cases, 4E is a common term, then for the MDM, the relative stiffness factor is utilized. K ′ = I L
2.2. Modified Stiffness For cases wherein the end support is a hinge or roller, a modification factor is introduced to the relative stiffness of that span. This simplifies the MDM table. K ′′ = 3 4 ⋅ I L 3. Distribution Factor For external pin/roller support, DF = 1. For fixed support, DF = 0. For other ends, considering at each joint. The distribution factor of each span at each joint is based on the formula DF = K ∑ K 4. Balancing of Moments To balance end moments in a joint, Step 1: Take the resultant of all moments at each joint Step 2: Take the additive inverse of that resultant Step 3: Multiply the result to the corresponding distribution factor 5. Carry Over Moments For members with DF ≠ 0, balanced moment values of one end of each span are transferred to the other end. MCO = 1 2 Mbal Note that ends with DF = 1 do not receive carry over moment (apply modified K values).
6. Example Member stiffness kAB = kBA = I 4 kBC = kCB = 3 4 ⋅ I 4 = 3I 16 Distribution factor DFA = 0 DFBA = I 4 I 4 + 3I 16 = 0.5714 DFBC = 3I 16 I 4 + 3I 16 = 0.4286 DFCB = 1 Fixed End Moments FEMAB = − (30 kN)(3 m)(1 m) 2 (4 m) 2 = −5.625 kNm FEMBA = (30 kN)(3 m) 2 (1 m) (4 m) 2 = 16.75 kNm FEMBC = − (5 kN m ) (4 m) 2 12 = −6.6667 kNm FEMCB = (5 kN m ) (4 m) 2 12 = 6.6667 kNm
Construct the table Joint A B C Member AB BA BC CB k 0.25 0.25 0.1875 0.1875 DF 0 0.5714 0.4286 1 FEM -5.625 16.875 -6.6667 6.6667 Bal -5.8333 -4.3750 -6.6667 CO -2.9167 -3.3333 Bal 1.9048 1.4286 CO 0.9524 BAL TOTAL -7.5893 12.9464 -12.9464 0 This means that MA = 7.5893 kNm, MB = 12.9464 kNm.