# 29- Moment redistribution for Continuous steel beams.

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## Moment Redistribution For Continuous Steel Beams.

### Elastic redistribution for Moment values.

A new subject is the design of continuous beams. The first method is the elastic method of moment redistribution.

Let us have a look at the shapes W-shapes, M shapes, S-shapes, Hp-shapes and c-shapes, L-SHAPES, wt-shapes, HSS-shapes, and pipe-shapes, and the ASTM designation, the values of both Fy and Fult for each of these shapes. For instance, for wide flanged beam Fy=50-65 ksi based on the steel of A992. For hollow structural sections, rectangular shapes Fy=46 ksi.

This is a brief revision of the values of Fy and Fult for the different types of ASTM.

The next image shows the values of shear and moment for continuous beams with multiple spans.

We are aware of this method; the code permits utilizing this moment by making redistribution for the moment.
The AISC has issued Table 3-23 in which, let us review the fixed end,  the beam of span= L under uniformly distributed load, the fixed end moments will be w*L^2/12 from both ends, the reaction will be WL/2 at both ends, the shear magnitude for will be w*L/2 the values for the shear, positive, and negative moment values, for the different loading cases Condition.

For instance, simply supported beams, fixed at both ends, continuous beams supported on multiple spans and loading conditions. For example, concentrated and uniformly distributed, the fixed end moments will be w*L^2/12 from both ends, as shown in the sketches.

The reaction will be W*L/2 at both ends; the shear force diagram is also drawn.
The free-body diagram is created after removing the supports from the ends and replacing them with reaction forces and moments.

The bending moment diagram has a negative value at the ends, wL2/12, as drawn at the tail of the moment arrow.
The positive value of the moment will be w*L2/24 since we subtract w*L2/12 from the W*L2/8; this moment will be in the middle.

The main point will be how to redistribute the bending moments of continuous steel and fixed-end beams.

I quote from both Profs. Bowles and Alan Williams’s books. The principle of moment redistribution can be performed if the section of the beam is compact and not a cantilever.

The principle of moment redistribution can be applied if the beam’s section is compact and not a cantilever.

How do we perform redistribution?

A section may be designed using either 0.90* the largest negative moment or a positive moment based on a maximum positive moment from the moment diagram + 10%* average negative span moments, whichever is larger.

When the beam or girder is rigidly framed into a column, the design moment value. The column at that point can also be reduced by 10%.

### Elastic moment distribution for a continuous beam with two equal spans.

If we want to apply the 0.90 rule for the fixed-end moment beam. Apply the moment redistribution, then the net value of the fixed end moment will be Wl^2*0.1125.

The positive bending moment will be increased by 0.5* 0.10*(w*L^2)*0.125=0.00625 WL^2.

### Elastic moment distribution for Fixed end beam.

This sketch shows the shear and bending moment values for the fixed end beam before the moment is redistributed.

If we want to apply the 0.90 rule for moment redistribution. For the fixed-end moment beam, the net value of the fixed-end moment will be Wl^2/13.33.

The positive bending moment will be increased by 0.10*(w*L2)/12=0.0083333 WL2. The value of the moment after using the 0.90 rule is shown in the slide image.

The next post is a  Solved Problem 4-15. The post includes how to design a continuous beam of two spans.

For a useful link about the distribution of continuous beams.

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