Stiffness-Factor Modifications
Three cases where moment-distribution process can be simplified
1. Member Pin (or Roller) Supported at Far End
carry-over factor = 0
if the far end was fixed supported, the stiffness factor K = 4EI/L would have to be modified by 3/4 to model the case of having the far end pin supported
2. Symmetric Beam and Loading
for center span:
the center span's stiffness factor will be one half that usually determined using K = 4EI/L
3. Symmetric Beam and Antisymmetric Loading
for center span:
the stiffness factor is one and a half times as large as that determined using K = 4EI/L
Procedure for Analysis
- draw FBDs of spans and joints
- determine distribution factors and fixed-end moments (click)
- identify joints (free ends are not joints)
- determine stiffness factor K for each span (click)
- K = 4EI/L for far-end fixed
- K = 3EI/L for far-end pinned or roller supported
- K = 2EI/L for symmetric span and loading
- K = 6EI/L for antisymmetric loading
- K = ∞ for ground
- K = 0 for air
- determine distribution factor DF=K/ΣK for each span (click)
- DF = 0 for fixed end
- DF = 1 for an end pin or roller supported
- DF = 0 for overhung (free) side of joint; DF = 1 for other side
- ΣDF = 1 at each joint
- determine FEMs from inside back cover (positive = clockwise)
- perform moment distribution process (click)
- lock all joints
- determine moment needed for joint equilibrium (switch sign)
- unlock pin and roller joints (fixed joints stay locked) and distribute the counterbalancing moments into the connecting span at each joint
- use +1/2 carry-over factor to carry half of these moments to span's other end (click)
- don't carry from fixed joints
- don't carry to end pin or roller joints
- don't carry to free ends
- repeat until desired degree of accuracy is reached (
don't carry-over final distribution)
- sum columns of FEMs, distributed moments, and carry-over moments
- place summed moments on FBDs
- use equilibrium to determine shear forces and ground forces
- draw V&M diagrams
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